Q
stringlengths 4
3.96k
| A
stringlengths 1
3k
| Result
stringclasses 4
values |
|---|---|---|
Let \( K/F \) be a separable extension, and let \( L/F \) be a purely inseparable extension. Then \( K \) and \( L \) are linearly disjoint over \( F \). | To prove this, note that if \( \operatorname{char}\left( F\right) = 0 \), then \( L = F \), and the result is trivial. So, suppose that \( \operatorname{char}\left( F\right) = p > 0 \) . We first consider the case where \( K/F \) is a finite extension. By the primitive element theorem, we may write \( K = F\left( a\right) \) for some \( a \in K \) . Let \( f\left( x\right) = \min \left( {F, a}\right) \) and \( g\left( x\right) = \min \left( {L, a}\right) \) . Then \( g \) divides \( f \) in \( L\left\lbrack x\right\rbrack \) . If \( g\left( x\right) = {\alpha }_{0} + \cdots + {\alpha }_{n - 1}{x}^{n - 1} + {x}^{n} \), then for each \( i \) there is a positive integer \( {r}_{i} \) with \( {\alpha }_{i}^{{p}^{{r}_{i}}} \in F \) . If \( r \) is the maximum of the \( {r}_{i} \), then \( {\alpha }_{i}^{{p}^{r}} \in F \) for each \( i \), so \( g{\left( x\right) }^{{p}^{r}} \in F\left\lbrack x\right\rbrack \) . Consequently, \( g{\left( x\right) }^{{p}^{r}} \) is a polynomial over \( F \) for which \( a \) is a root. Thus, \( f \) divides \( {g}^{{p}^{r}} \) in \( F\left\lbrack x\right\rbrack \) . Viewing these two divisibilities in \( L\left\lbrack x\right\rbrack \), we see that the only irreducible factor of \( f \) in \( L\left\lbrack x\right\rbrack \) is \( g \), so \( f \) is a power of \( g \) . The field extension \( K/F \) is separable; hence, \( f \) has no irreducible factors in any extension field of \( F \) . This forces \( f = g \), so\n\n\[ \left\lbrack {{KL} : L}\right\rbrack = \left\lbrack {L\left( a\right) : L}\right\rbrack = \deg \left( g\right) \]\n\n\[ = \deg \left( f\right) = \left\lbrack {K : F}\right\rbrack \]\n\nFrom this, we obtain \( \left\lbrack {{KL} : F}\right\rbrack = \left\lbrack {K : F}\right\rbrack \cdot \left\lbrack {L : F}\right\rbrack \), so \( K \) and \( L \) are linearly disjoint over \( F \) by Lemma 20.4.\n\nIf \( K/F \) is not necessarily finite, suppose that \( \varphi : K{ \otimes }_{F}L \rightarrow {KL} \) is not injective. Then there are \( {k}_{1},\ldots ,{k}_{n} \in K \) and \( {l}_{1},\ldots ,{l}_{n} \in L \) with \( \varphi \left( {\sum {k}_{i} \otimes }\right. \) \( \left. {l}_{i}\right) = 0 \) . If \( {K}_{0} \) is the field generated over \( F \) by the \( {k}_{i} \), then the restriction of \( \varphi \) to \( {K}_{0}{ \otimes }_{F}L \) is not injective, which is false by the finite dimensional case. Thus, \( \varphi \) is injective, so \( K \) and \( L \) are linearly disjoint over \( F \). | Yes |
Corollary 20.21 Suppose that \( K = F\left( {{a}_{1},\ldots ,{a}_{n}}\right) \) is finitely generated and separable over \( F \) . Then there is a subset \( Y \) of \( \left\{ {{a}_{1},\ldots ,{a}_{n}}\right\} \) that is a separating transcendence basis of \( K/F \) . | Proof. This corollary is more accurately a consequence of the proof of (3) \( \Rightarrow \) (1) in Theorem 20.18, since the argument of that step is to show that if \( K \) is finitely generated over \( F \), then any finite generating set contains a separating transcendence basis. | Yes |
Corollary 20.22 Let \( F \) be a perfect field. Then any finitely generated extension of \( F \) is separably generated. | Proof. This follows immediately from part 3 of Theorem 20.18, since \( {F}^{1/{p}^{\infty }} = F \) if \( F \) is perfect. | Yes |
Corollary 20.23 Let \( F \subseteq E \subseteq K \) be fields.\n\n1. If \( K/F \) is separable, then \( E/F \) is separable.\n\n2. If \( E/F \) and \( K/E \) are separable, then \( K/F \) is separable.\n\n3. If \( K/F \) is separable and \( E/F \) is algebraic, then \( K/E \) is separable. | Proof. Part 1 is an immediate consequence of condition 2 of Theorem 20.18. For part 2 we use Theorems 20.18 and 20.12. If \( E/F \) and \( K/E \) are separable, then \( E \) and \( {F}^{1/p} \) are linearly disjoint over \( F \), and \( K \) and \( {E}^{1/p} \) are linearly disjoint over \( E \) . However, it follows from the definition that \( {F}^{1/p} \subseteq {E}^{1/p} \), so \( E{F}^{1/p} \subseteq {E}^{1/p} \) . Thus, \( K \) and \( E{F}^{1/p} \) are linearly disjoint over \( E \) . Theorem 20.12 then shows that \( K \) and \( {F}^{1/p} \) are linearly disjoint over \( F \), so \( K \) is separable over \( F \) .\n\nTo prove part 3, suppose that \( K/F \) is separable and \( E/F \) is algebraic. We know that \( E/F \) is separable by part 1 . Let \( L = E\left( {{a}_{1},\ldots ,{a}_{n}}\right) \) be a finitely generated subextension of \( K/E \) . If \( {L}^{\prime } = F\left( {{a}_{1},\ldots ,{a}_{n}}\right) \), then by the separability of \( K/F \) there is a separating transcendence basis \( \left\{ {{t}_{1},\ldots ,{t}_{m}}\right\} \) for \( {L}^{\prime }/F \) . Because \( E/F \) is separable algebraic, \( E{L}^{\prime } = L \) is separable over \( {L}^{\prime } \), so by transitivity, \( L \) is separable over \( F\left( {{t}_{1},\ldots ,{t}_{m}}\right) \) . Thus, \( L \) is separable over \( E\left( {{t}_{1},\ldots ,{t}_{m}}\right) \), so \( \left\{ {{t}_{1},\ldots ,{t}_{m}}\right\} \) is a separating transcendence basis for \( L/E \) . We have shown that \( L/E \) is separably generated for every finitely generated subextension of \( K/E \), which proves that \( K/E \) is separable. | Yes |
Example 21.7 Let \( V = \left\{ {\left( {{t}^{3},{t}^{4},{t}^{5}}\right) : t \in C}\right\} \) . Then \( V \) is a \( k \) -variety, since \( V \) is the zero set of \( \left\{ {{y}^{2} - {xz},{z}^{2} - {x}^{2}y}\right\} \) . | To verify this, note that each point of \( V \) does satisfy these two polynomials. Conversely, suppose that \( \left( {a, b, c}\right) \in {C}^{3} \) is a zero of these three polynomials. If \( a = 0 \), then a quick check of the polynomials shows that \( b = c = 0 \), so \( \left( {a, b, c}\right) \in V \) . If \( a \neq 0 \) , then define \( t = b/a \) . From \( {b}^{2} = {ac} \), we see that \( c = {t}^{2}a \) . Finally, the equation \( {c}^{2} = {a}^{2}b \) yields \( {t}^{4}{a}^{2} = {a}^{3}t \), so \( a = {t}^{3} \) . Thus, \( \left( {a, b, c}\right) = \left( {{t}^{3},{t}^{4},{t}^{5}}\right) \in V \) . | Yes |
Lemma 21.11 The sets \( \left\{ {Z\left( S\right) : S \subseteq k\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack }\right\} \) are the closed sets of a topology on \( {C}^{n} \) ; that is,\n\n1. \( {C}^{n} = Z\left( {\{ 0\} }\right) \) and \( \varnothing = Z\left( {\{ 1\} }\right) \) .\n\n2. If \( S \) and \( T \) are subsets of \( k\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \), then \( Z\left( S\right) \cup Z\left( T\right) = Z\left( {ST}\right) \), where \( {ST} = \{ {fg} : f \in S, t \in T\} \) .\n\n3. If \( \left\{ {S}_{\alpha }\right\} \) is an arbitrary collection of subsets of \( k\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \), then \( \mathop{\bigcap }\limits_{\alpha }Z\left( {S}_{\alpha }\right) = Z\left( {\mathop{\bigcup }\limits_{\alpha }{S}_{\alpha }}\right) \n | Proof. The first two parts are clear from the definitions. For the third, let \( P \in Z\left( S\right) \) . Then \( f\left( P\right) = 0 \) for all \( f \in S \), so \( \left( {fg}\right) \left( P\right) = 0 \) for all \( {fg} \in {ST} \) . Thus, \( Z\left( S\right) \subseteq Z\left( {ST}\right) \) . Similarly, \( Z\left( T\right) \subseteq Z\left( {ST}\right) \), so \( Z\left( S\right) \cup Z\left( T\right) \subseteq Z\left( {ST}\right) \) . For the reverse inclusion, let \( P \in Z\left( {ST}\right) \) . If \( P \notin Z\left( S\right) \), then there is an \( f \in S \) with \( f\left( P\right) \neq 0 \) . If \( g \in T \), then \( 0 = \left( {fg}\right) \left( P\right) = f\left( P\right) g\left( P\right) \), so \( g\left( P\right) = 0 \) , which forces \( P \in Z\left( T\right) \) . Thus, \( Z\left( {ST}\right) \subseteq Z\left( S\right) \cup Z\left( T\right) \) . This proves that \( Z\left( S\right) \cup Z\left( T\right) = Z\left( {ST}\right) \n\nFor the fourth part, the inclusion \( Z\left( {\mathop{\bigcup }\limits_{\alpha }{S}_{\alpha }}\right) \subseteq \mathop{\bigcap }\limits_{\alpha }Z\left( {S}_{\alpha }\right) \) follows from part 1. For the reverse inclusion, take \( P \in \mathop{\bigcap }\limits_{\alpha }Z\left( {S}_{\alpha }\right) \) . Then \( P \in Z\left( {S}_{\alpha }\right) \) for each \( \alpha \), so \( f\left( P\right) = 0 \) for each \( f \in {S}_{\alpha } \) . Thus, \( P \in Z\left( {\mathop{\bigcup }\limits_{\alpha }{S}_{\alpha }}\right) \) . | Yes |
Lemma 21.14 If \( V \) is any subset of \( {C}^{n} \), then \( I\left( V\right) \) is a radical ideal of \( k\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) . | Proof. Let \( f \in k\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) with \( {f}^{r} \in I\left( V\right) \) for some \( r \) . Then \( {f}^{r}\left( P\right) = 0 \) for all \( P \in V \) . But \( {f}^{r}\left( P\right) = {\left( f\left( P\right) \right) }^{r} \), so \( f\left( P\right) = 0 \) . Therefore, \( f \in I\left( V\right) \) ; hence, \( I\left( V\right) \) is equal to its radical, so \( I\left( V\right) \) is a radical ideal. | Yes |
Lemma 21.15 The following statements are some properties of ideals of subsets of \( {C}^{n} \) . | Proof. The first two parts of the lemma are clear from the definition of \( I\left( V\right) \) . For the third, let \( V \) be a subset of \( {C}^{n} \) . If \( f \in I\left( V\right) \), then \( f\left( P\right) = 0 \) for all \( P \in V \), so \( P \in Z\left( {I\left( V\right) }\right) \), which shows that \( V \subseteq Z\left( {I\left( V\right) }\right) \) . Suppose that \( V = Z\left( S\right) \) for some subset \( S \in k\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) . Then \( S \subseteq I\left( V\right) \), so \( Z\left( {I\left( V\right) }\right) \subseteq \) \( Z\left( S\right) = V \) by the previous lemma. Thus, \( V = Z\left( {I\left( V\right) }\right) \) . Conversely, if \( V = Z\left( {I\left( V\right) }\right) \), then \( V \) is a \( k \) -variety by definition. | Yes |
Let \( f \in k\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) be a polynomial, and let \( V = Z\left( f\right) \) . If \( f = {p}_{1}^{{r}_{1}}\cdots {p}_{t}^{{r}_{t}} \) is the irreducible factorization of \( f \), then \( I\left( V\right) = \sqrt{\left( f\right) } \) by the Nullstellensatz. However, we show that \( \sqrt{\left( f\right) } = \left( {{p}_{1}\cdots {p}_{t}}\right) \) | for, if \( g \in \sqrt{\left( f\right) } \), then \( {g}^{m} = {fh} \) for some \( h \in k\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) and some \( m > 0 \) . Each \( {p}_{i} \) then divides \( {g}^{m} \) ; hence, each \( {p}_{i} \) divides \( g \) . Thus, \( g \in \left( {{p}_{1}\cdots {p}_{t}}\right) \) . For the reverse inclusion, \( {p}_{1}\cdots {p}_{t} \in \sqrt{\left( f\right) } \), since if \( r \) is the maximum of the \( {r}_{i} \), then \( {\left( {p}_{1}\cdots {p}_{t}\right) }^{r} \in \left( f\right) \) | Yes |
Let \( V \) be an irreducible \( k \) -variety. By taking complements, we see that the definition of irreducibility is equivalent to the condition that any two nonempty open sets have a nonempty intersection. Therefore, if \( U \) and \( {U}^{\prime } \) are nonempty open subsets of \( V \), then \( U \cap {U}^{\prime } \neq \varnothing \) . One consequence of this fact is that any nonempty open subset of \( V \) is dense in \( V \), as we now prove. | If \( U \) is a nonempty open subset of \( V \), and if \( C \) is the closure of \( U \) in \( V \), then \( U \cap \left( {V - C}\right) = \varnothing \) . The set \( V - C \) is open, so one of \( U \) or \( V - C \) is empty. Since \( U \) is nonempty, this forces \( V - C = \varnothing \) , so \( C = V \) . But then the closure of \( U \) in \( V \) is all of \( V \), so \( U \) is dense in \( V \). | Yes |
Proposition 21.21 Let \( V \) be a \( k \) -variety. Then \( V \) is irreducible if and only if \( I\left( V\right) \) is a prime ideal, if and only if the coordinate ring \( k\left\lbrack V\right\rbrack \) is an integral domain. | Proof. First suppose that \( V \) is irreducible. Let \( f, g \in k\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) with \( {fg} \in I\left( V\right) \) . Then \( I = I\left( V\right) + \left( f\right) \) and \( J = I\left( V\right) + \left( g\right) \) are ideals of \( k\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) containing \( I\left( V\right) \) ; hence, their zero sets \( Y = Z\left( I\right) \) and \( Z = Z\left( J\right) \) are contained in \( Z\left( {I\left( V\right) }\right) = V \) . Moreover, \( {IJ} \subseteq I\left( V\right) \), since \( {fg} \in I\left( V\right) \), so \( Y \cup Z = Z\left( {IJ}\right) \) contains \( V \) . This forces \( V = Y \cup Z \), so either \( Y = V \) or \( Z = V \), since \( V \) is irreducible. If \( Y = V \), then \( I \subseteq I\left( Y\right) = I\left( V\right) \) , and if \( Z = V \), then \( J \subseteq I\left( Z\right) = I\left( V\right) \) . Thus, either \( f \in I\left( V\right) \) or \( g \in I\left( V\right) \) , so \( I\left( V\right) \) is a prime ideal of \( k\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) .\n\nConversely, suppose that \( I\left( V\right) \) is prime. If \( V = Y \cup Z \) for some \( k \) -varieties \( Y \) and \( Z \), let \( I = I\left( Y\right) \) and \( J = I\left( Z\right) \) . Then \( {IJ} \subseteq I\left( {Y \cup Z}\right) = I\left( V\right) \), so either \( I \subseteq I\left( V\right) \) or \( J \subseteq I\left( V\right) \) . This means that \( V \subseteq Z\left( I\right) = Y \) or \( V \subseteq Z\left( J\right) = Z \) . Therefore, \( Y = V \) or \( Z = V \), so \( V \) is irreducible. | Yes |
Let \( V = Z\left( {y - {x}^{2}}\right) \). Then the coordinate ring of \( V \) is \( k\left\lbrack {x, y}\right\rbrack /\left( {y - {x}^{2}}\right) \), which is isomorphic to the polynomial ring \( k\left\lbrack t\right\rbrack \) by sending \( t \) to the coset of \( x \) in \( k\left\lbrack V\right\rbrack \). | Therefore, the function field of \( V \) is the rational function field \( k\left( t\right) \). | No |
Let \( V = Z\left( {{y}^{2} - {x}^{3}}\right) \). Then \( k\left( V\right) \) is the field \( k\left( {s, t}\right) \), where \( s \) and \( t \) are the images of \( x \) and \( y \) in \( k\left\lbrack V\right\rbrack = k\left\lbrack {x, y}\right\rbrack /\left( {{y}^{2} - {x}^{3}}\right) \), respectively. Note that \( {t}^{2} = {s}^{3} \). | Let \( z = t/s \). Substituting this equation into \( {t}^{2} = {s}^{3} \) and simplifying shows that \( s = {z}^{2} \), and so \( t = {z}^{3} \). Thus, \( k\left( V\right) = k\left( z\right) \). The element \( z \) is transcendental over \( k \), since if \( k\left( V\right) /k \) is algebraic, then \( k\left\lbrack V\right\rbrack \) is a field by the argument in Example 19.11, so \( \left( {{y}^{2} - {x}^{3}}\right) \) is a maximal ideal of \( k\left\lbrack {x, y}\right\rbrack \). However, this is not true, since \( \left( {{y}^{2} - {x}^{3}}\right) \) is properly contained in the ideal \( \left( {x, y}\right) \). Thus, \( k\left( V\right) \) is a rational function field in one variable over \( k \). Note that \( k\left\lbrack V\right\rbrack \) is isomorphic to \( k\left\lbrack {{x}^{2},{x}^{3}}\right\rbrack \), a ring that is not isomorphic to a polynomial ring in one variable over \( k \). | Yes |
If \( V \) is an irreducible \( k \) -variety, then \( V \) gives rise to a field extension \( k\left( V\right) \) of \( k \) . We can reverse this construction. Let \( K \) be a finitely generated field extension of \( k \) . Say \( K = k\left( {{a}_{1},\ldots ,{a}_{n}}\right) \) for some \( {a}_{i} \in K \) . Let\n\n\[ P = \left\{ {f \in k\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack : f\left( {{a}_{1},\ldots ,{a}_{n}}\right) = 0}\right\} . | Then \( P \) is the kernel of the ring homomorphism \( \varphi : k\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \rightarrow K \) that sends \( {x}_{i} \) to \( {a}_{i} \), so \( P \) is a prime ideal. If \( V = Z\left( P\right) \), then \( V \) is an irreducible \( k \) -variety with coordinate ring \( k\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack /P \cong k\left\lbrack {{a}_{1},\ldots ,{a}_{n}}\right\rbrack \) | Yes |
5.3 Lemma.\n\n(a) For every fixed nonzero \( u \in {\mathbb{R}}^{n} \), the hyperplane\n\n(*)\n\n\[ {H}_{K}\left( u\right) \mathrel{\text{:=}} \left\{ {x\mid \langle x, u\rangle = {h}_{K}\left( u\right) }\right\} \]\n\nis a supporting hyperplane of \( K \) (Figure 9b).\n\n(b) Every supporting hyperplane of \( K \) has a representation of the form \( \left( *\right) \) . | Proof.\n\n(a) Since \( K \) is compact and \( \langle \cdot, u\rangle \) is continuous, for some \( {x}_{0} \in K \),\n\n\[ \left\langle {{x}_{0}, u}\right\rangle = {h}_{K}\left( u\right) = \mathop{\sup }\limits_{{x \in K}}\langle x, u\rangle .\n\]\nFor an arbitrary \( y \in K \), it follows that \( \langle y, u\rangle \leq \left\langle {{x}_{0}, u}\right\rangle \) ; hence, \( K \subset {H}_{K}^{ - }\left( u\right) \). This proves (a).\n\n(b) Let \( H = \left\{ {x\mid \langle x, u\rangle = \left\langle {{x}_{0}, u}\right\rangle }\right\} \) be a supporting hyperplane of \( K \) at \( {x}_{0} \). We choose \( u \neq 0 \) such that \( K \subset {H}^{ - } \). Then, \( \left\langle {{x}_{0}, u}\right\rangle = \mathop{\sup }\limits_{{x \in K}}\langle x, u\rangle = {h}_{K}\left( u\right) \) which implies (b). | Yes |
6.4 Theorem. Let \( K \) be a convex body with \( 0 \in \operatorname{int}K \) . Then,\n\n(a)\n\[ \n{K}^{* * } = K \n\]\n\n(b) The distance function of \( K \) equals the support function of \( {K}^{ * } \), and, conversely,\n\n\[ \n{d}_{K} = {h}_{{K}^{ * }},\;{d}_{{K}^{ * }} = {h}_{K}. \n\] | Proof.\n\n(a) By definition of \( {H}_{u} \), for every \( u \neq 0 \) of \( K \) ,\n\n\[ \n{H}_{u}^{ - } = \{ x \mid \langle u, x\rangle \leq 1\} \n\]\n\nTherefore,(using the obvious notation \( \langle K, x\rangle \leq 1 \) )\n\n\[ \n{K}^{ * } = \{ x \mid \langle K, x\rangle \leq 1\} \;\text{ and }\;{K}^{* * } = \left\{ {y \mid \left\langle {{K}^{ * }, y}\right\rangle \leq 1}\right\} .\n\]\n\nIf \( y \in K \), then, the definition of \( {K}^{ * } \) yields \( \left\langle {y,{K}^{ * }}\right\rangle \leq 1 \) and, thus, \( K \subset {K}^{* * } \) . Suppose \( K \neq {K}^{* * } \) . Then, let \( x \in {K}^{* * } \smallsetminus K \) . For\n\n\[ \n{x}^{\prime } \mathrel{\text{:=}} {p}_{K}\left( x\right) \;\text{ and }\;u \mathrel{\text{:=}} \frac{x - {x}^{\prime }}{\left\langle {x}^{\prime }, x - {x}^{\prime }\right\rangle },\n\]\n\nLemma 3.5 yields\n\n\[ \nx \in {H}_{u}^{ + } \smallsetminus {H}_{u},\;\text{ but also }K \subset {H}_{u}^{ - },\n\]\n\nwhence \( u \in {K}^{ * } \) . Since \( x \in {K}^{* * } \), it follows that \( \langle u, x\rangle \leq 1 \), i.e., \( x \in {H}_{u}^{ - } \), a contradiction. | Yes |
6.8 Theorem. Every positive homogeneous and convex function \( h : {\mathbb{R}}^{n} \rightarrow \mathbb{R} \) is the support function \( h = {h}_{K} \) of a unique convex body \( K \) (whose dimension is possibly less than \( n \) ). | Proof. Let us write \( {\mathbb{R}}^{n} = U \oplus {U}^{ \bot } \), where \( U \) is the maximal linear subspace of \( {\mathbb{R}}^{n} \) on which \( h \) is linear. Then, there exists \( a \in U \) such that, for \( \left( {x,{x}^{\prime }}\right) \in U \oplus {U}^{ \bot } \) ,\n\n(*)\n\n\[ h\left( {x,{x}^{\prime }}\right) = \langle x, a\rangle + {\left. h\right| }_{{U}^{ \bot }}\left( {x}^{\prime }\right) . \]\n\nMoreover, \( {\Gamma }^{ + }\left( {\left. h\right| }_{{U}^{ \bot }}\right) \) is a cone with apex 0 in \( {U}^{ \bot } \oplus \mathbb{R} \) (see Lemma 5.7). Thus, there exists some \( b \in {U}^{ \bot } \) such that the hyperplane \( H \mathrel{\text{:=}} \left\{ {\left( {y,\langle y, b\rangle }\right) \mid y \in {U}^{ \bot }}\right\} \) in \( {U}^{ \bot } \oplus \mathbb{R} \) intersects \( {\Gamma }^{ + }\left( {\left. h\right| }_{{U}^{ \bot }}\right) \) only in the apex. Now the set\n\n\[ {K}_{0} + \left( {0,1}\right) \mathrel{\text{:=}} \left( {{U}^{ \bot }\times \{ 1\} }\right) \cap {\Gamma }^{ + }\left( {{\left. h\right| }_{{U}^{ \bot }}-\langle \cdot, b\rangle }\right) \]\n\nis a convex body and, by Lemma 5.2, \( {\left. h\right| }_{{U}^{ \bot }} - \langle \cdot, b\rangle \) the support function of \( {K}_{0} - b \) . Finally, \( \left( *\right) \) and Lemma 5.2 yield that \( h \) is the support function of \( K \mathrel{\text{:=}} {K}_{0} - b + \) a. | Yes |
2.8 Theorem. Let \( F \) be a proper face of the polytope \( P \), and let \( {P}^{ * } \) be the polar polytope of \( P \) with respect to a polarity \( \pi \) . For an affine subspace \( U \) of \( {\mathbb{R}}^{n} \), let \( {\pi }_{U} \) denote the restriction of \( \pi \) to \( U \), and set \( {\pi }_{U}\left( {P/F}\right) \mathrel{\text{:=}} {\pi }_{U}\left( {\operatorname{aff}\left( {P/F}\right) }\right) \) .\n\n(a) For any face figure \( P/F \) of \( F \) ,\n\n\[{\pi }_{U}\left( {P/F}\right) \approx {F}^{ * }\]\n\n(b) Any two face figures of \( P \) with respect to \( F \) are combinatorially isomorphic. Therefore, we can consider \( P/F \) to be the equivalence class of these face figures. | Proof. Let \( k \mathrel{\text{:=}} \dim F \), so that \( \dim U = n - k - 1 \) for \( U \) as in 2.6. For any face \( G \) which contains \( F \) properly, we set\n\n\[G\underset{\varphi }{ \mapsto }\;{\pi }_{U}\left( {G \cap U}\right)\]\n\nIf \( g \mathrel{\text{:=}} \dim \left( {G \cap U}\right) \), so that \( g = \dim G - k - 1 \), then, by the construction in Lemma 2.6,\n\n\( \dim \varphi \left( G\right) = \left( {n - k - 1}\right) - g - 1 = n - \left( {g + 1 + k}\right) - 1 = n - \dim G - 1. \)\n\nTherefore \( \dim \varphi \left( G\right) = \dim {G}^{ * } \) . The mapping \( \varphi \) is inclusion-reversing as is the polarity \( \pi \) . So, by\n\n\[ \varphi \left( G\right) \; \mapsto \;{G}^{ * } \in \mathcal{B}\left( {F}^{ * }\right) \]\n\nwe obtain an isomorphism\n\n\[{\mathcal{B}}_{0}\left( {{\pi }_{U}\left( {P/F}\right) }\right) \; \rightarrow {\mathcal{B}}_{0}\left( {F}^{ * }\right)\]\n\nwhence (a) follows. Note that, by Lemma 2.6(c), only faces \( G \) including \( F \) properly contribute to \( \mathcal{B}\left( {{\pi }_{U}\left( {P/F}\right) }\right) \) .\n\nPart (b) of the theorem is a consequence of (a). | Yes |
Let \( U, W \) be subspaces of \( V = {\mathbb{R}}^{4} = U \oplus W \) where \( U = \operatorname{lin}\left\{ {{e}_{1},{e}_{2}}\right\} \) , \( W = \operatorname{lin}\left\{ {{e}_{3},{e}_{4}}\right\} ,{e}_{1},\ldots ,{e}_{4} \) the canonical basis of \( {\mathbb{R}}^{4} \) . We write the coordinates of the elements of \( U, V, W \) as column vectors and those of the elements of \( {V}^{ * } \) , \( {U}^{ * },{W}^{ * } \) as row vectors, respectively. Applying a dual vector of \( {V}^{ * },{U}^{ * } \), or \( {W}^{ * } \) to a vector of \( V, U, W \), respectively, can be carried out as (matrix) multiplication of a row vector and a column vector. | Let\n\n\[ X = \left( {{x}_{1}{x}_{2}{x}_{3}{x}_{4}}\right) = \left( \begin{array}{llll} 1 & 0 & 1 & 2 \\ 0 & 1 & 2 & 1 \end{array}\right) \]\n\n\[ \left( {{b}_{1}{b}_{2}{b}_{3}{b}_{4}}\right) = \left( \begin{array}{llll} 1 & 0 & 1 & 2 \\ 0 & 1 & 2 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right) \]\n\n\( {L}_{1},{L}_{2} \) are given as matrices\n\n\[ {L}_{1} = \left( \begin{array}{llll} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{array}\right) = \left( {{E}_{2} \mid O}\right) \]\n\n\[ {L}_{2} = \left( \begin{array}{ll} 0 & 0 \\ 0 & 0 \\ 0 & 0 \\ 0 & 1 \end{array}\right) = \left( \frac{O}{{E}_{2}}\right) \]\n\nThen,\n\n\[ \left( \begin{array}{l} {b}_{1}^{ * } \\ {b}_{2}^{ * } \\ {b}_{3}^{ * } \\ {b}_{4}^{ * } \end{array}\right) = \left( \begin{matrix} 1 & 0 & - 1 & - 2 \\ 0 & 1 & - 2 & - 1 \\ & & & \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{matrix}\right) \]\n\n\( {L}_{1}\left( {b}_{i}\right) = {x}_{i}, i = 1,\ldots ,4 \) and im \( {L}_{2} = \ker {L}_{1} \) are readily checked. We obtain (Figure 7)\n\n\[ \left( \begin{array}{l} {\bar{x}}_{1} \\ {\bar{x}}_{2} \\ {\bar{x}}_{3} \\ {\bar{x}}_{4} \end{array}\right) = \left( \begin{array}{l} {L}_{2}^{ * }\left( {b}_{1}^{ * }\right) \\ {L}_{2}^{ * }\left( {b}_{2}^{ * }\right) \\ {L}_{2}^{ * }\left( {b}_{3}^{ * }\right) \\ {L}_{2}^{ * }\left( {b}_{4}^{ * }\right) \end{array}\right) = \left( \begin{array}{l} {b}_{1}^{ * } \\ {b}_{2}^{ * } \\ {b}_{3}^{ * } \\ {b}_{4}^{ * } \end{array}\right) {L}_{2} = \left( \begin{array}{rr} - 1 & - 2 \\ - 2 & - 1 \\ 1 & 0 \\ 0 & 1 \end{array}\right) . \] | Yes |
In Example 2, replace the elements \( {x}_{3},{x}_{4} \) by \( \frac{1}{3}{x}_{3},\frac{1}{3}{x}_{4} \), respectively. Then the linear transform obtained by the same calculation as in Example 1 is | \[ \left( \begin{array}{l} {\bar{x}}_{1} \\ {\bar{x}}_{2} \\ {\bar{x}}_{3} \\ {\bar{x}}_{4} \end{array}\right) = \left( \begin{array}{l} {L}_{2}^{ * }\left( {b}_{1}^{ * }\right) \\ {L}_{2}^{ * }\left( {b}_{2}^{ * }\right) \\ {L}_{2}^{ * }\left( {b}_{3}^{ * }\right) \\ {L}_{4}^{ * }\left( {b}_{4}^{ * }\right) \end{array}\right) = \left( \begin{array}{rr} - \frac{1}{3} & - \frac{2}{3} \\ - \frac{2}{3} & - \frac{1}{3} \\ 1 & 0 \\ 0 & 1 \end{array}\right) \] | Yes |
Example 1. We consider a triangular prism \( P \) in \( {\mathbb{R}}^{3} \) and wish to find a Gale transform of \( X \mathrel{\text{:=}} \) vert \( P \) (see Figure 9). Since \( v = 6 \) and \( n = 3 \), the elements of a Gale transform lie in a two-dimensional space. By Theorem 5.3, it is sufficient to find two independent affine dependencies of \( {x}_{1},\ldots ,{x}_{6} \) . Since \( {x}_{4} - {x}_{1},{x}_{5} - {x}_{2} \) , and \( {x}_{6} - {x}_{3} \) coincide, | \[ {\left( {a}_{1}{a}_{2}\right) }^{t} = \left( \begin{array}{rrrrrr} - 1 & 1 & 0 & 1 & - 1 & 0 \\ - 1 & 0 & 1 & 1 & 0 & - 1 \end{array}\right) . \] By Theorem 5.3, the rows of \( \left( {{a}_{1}{a}_{2}}\right) \) provide a Gale transform. | Yes |
5.4 Theorem. A finite sequence \( X \) in \( U = \operatorname{aff}X \) consists of all points of the vertex set of a polytope \( P \) in \( U \) if and only if one (and, thus, every) Gale transform \( {\bar{X}}_{\widehat{U}} \) of \( X \) satisfies the following condition:\n\n(4) For every hyperplane \( H \) in \( \operatorname{lin}{\bar{X}}_{\widehat{U}} \) which contains 0, each of the associated half-spaces \( {H}^{ + } \) and \( {H}^{ - } \) includes at least two points of \( {\bar{X}}_{\widehat{U}} \) in its interior. | Proof. In appropriate coordinates for \( \widehat{U} \), we may identify \( U \) with \( U \times \{ 1\} \subset \widehat{U} \) . First, suppose that \( X \) comes from a polytope. Then, each element \( {\widehat{x}}_{i} \) of \( {X}_{i\prime } \) is a face. By Theorem 4.14, \( {\bar{X}}_{\widehat{U}} \smallsetminus \left\{ {\overline{\widehat{x}}}_{i}\right\} \) satisfies\n\n(5)\n\n\[ 0 \in \text{ relint pos }\left( {{\bar{X}}_{\widehat{U}} \smallsetminus \left\{ {\overline{\widehat{x}}}_{i}\right\} }\right) \]\n\nLet \( H \) be a hyperplane in \( \operatorname{lin}{\bar{X}}_{\widehat{U}} \) with \( 0 \in H \) . Let \( {k}^{ + },{k}^{ - } \) be the number of elements \( {\overline{\widehat{x}}}_{i} \) in int \( {H}^{ + } \), int \( {H}^{ - } \), respectively. If \( {k}^{ + } = {k}^{ - } = 0, H = \operatorname{lin}{\bar{X}}_{\widehat{U}} \) so that \( H \) is not a hyperplane. \( {k}^{ + } \leq 1 \) and \( {k}^{ - } > 0 \) or \( {k}^{ - } \leq 1 \) and \( {k}^{ + } > 0 \) imply a contradiction to (5). So, (4) follows.\n\nConversely, if (4) holds, then, (5) readily follows (since, otherwise, 0 would lie on the boundary of \( \sigma \mathrel{\text{:=}} \operatorname{pos}\left( {{\bar{X}}_{\widehat{U}} \smallsetminus \left\{ {\overline{\widehat{x}}}_{i}\right\} }\right) \), so that, for a supporting hyperplane \( H \) of \( \sigma \) with \( 0 \in H \) and \( \sigma \subset {H}^{ + } \), the half-space \( {H}^{ - } \) would not contain two of the \( {\widehat{x}}_{i} \) ’s in its interior). Therefore, by Theorem 4.14, \( {x}_{i} \in \operatorname{vert}P \) for \( i = 1,\ldots, v \) . | Yes |
The vertex set \( X = \left( {{x}_{1},\ldots ,{x}_{n + 1}}\right) \) of an \( n \) -simplex in \( {\mathbb{R}}^{n} \) is characterized by \( {\overline{\widehat{x}}}_{1} = \cdots = {\overline{\widehat{x}}}_{n} = 0 \) for a Gale transform \( {\bar{X}}_{\widehat{U}} \) of \( X \). | This also follows directly from \( \dim \operatorname{lin}{\bar{X}}_{\widehat{U}} = n + 1 - n - 1 = 0 \) by Lemma 4.5. | Yes |
6.5 Theorem. An \( n \) -dimensional polytope \( P = \operatorname{conv}\left\{ {{x}_{1},\ldots ,{x}_{v}}\right\} \) is simplicial if and only if, for a Gale diagram \( \bar{X} \) of \( X = \left( {\operatorname{vert}P}\right) \), the following condition holds: (1) For any hyperplane \( H \) in \( \operatorname{lin}\bar{X} \), with \( 0 \in H,0 \notin \) relint conv \( \left( {\bar{X} \cap H}\right) \) . | Proof. Suppose \( 0 \in \) relint conv \( \left( {\bar{X} \cap H}\right) \) for some hyperplane \( H \) . Since \( \dim P = n \) and \( P \) lies in a hyperplane of \( U \) which does not contain 0, we have \( \dim U = \operatorname{rank}X = n + 1 \), so, \( \operatorname{rank}\bar{X} = v - \operatorname{rank}X = v - n - 1 \) (Lemma 4.5).\n\n\n\n\n\nThen, by Carathéodory’s theorem (I, Theorem 2.3), for \( k = \operatorname{rank}\left( {H \cap \bar{X}}\right) \leq \) \( v - n - 2 \), there exist \( k + 1 \leq v - n - 1 \) affinely independent elements \( {\bar{x}}_{1},\ldots ,{\bar{x}}_{k + 1} \in \bar{X} \cap H \) such that \( 0 \in \) relint conv \( \left\{ {{\bar{x}}_{1},\ldots ,{\bar{x}}_{k + 1}}\right\} \) . Therefore, \( \left\{ {{\bar{x}}_{1},\ldots ,{\bar{x}}_{k + 1}}\right\} \) is a coface of \( X \), and the corresponding (proper) face has at least \( v - \left( {v - n - 1}\right) = n + 1 \) vertices and cannot be a simplex.\n\nThe converse is also true. | Yes |
6.5 Theorem. An \( n \) -dimensional polytope \( P = \operatorname{conv}\left\{ {{x}_{1},\ldots ,{x}_{v}}\right\} \) is simplicial if and only if, for a Gale diagram \( \bar{X} \) of \( X = \left( {\operatorname{vert}P}\right) \), the following condition holds: (1) For any hyperplane \( H \) in \( \operatorname{lin}\bar{X} \), with \( 0 \in H,0 \notin \) relint conv \( \left( {\bar{X} \cap H}\right) \) . | Proof. Suppose \( 0 \in \) relint conv \( \left( {\bar{X} \cap H}\right) \) for some hyperplane \( H \) . Since \( \dim P = n \) and \( P \) lies in a hyperplane of \( U \) which does not contain 0, we have \( \dim U = \operatorname{rank}X = n + 1 \), so, \( \operatorname{rank}\bar{X} = v - \operatorname{rank}X = v - n - 1 \) (Lemma 4.5).\n\n\n\n\n\nFigure 11a, b, c, d. (a) Double simplex. (b) Cyclic polytope \( C\left( {6,4}\right) \) . (c) Pyramid over a double simplex. (d) 2-Fold pyramid.\n\nThen, by Carathéodory’s theorem (I, Theorem 2.3), for \( k = \operatorname{rank}\left( {H \cap \bar{X}}\right) \leq \) \( v - n - 2 \), there exist \( k + 1 \leq v - n - 1 \) affinely independent elements \( {\bar{x}}_{1},\ldots ,{\bar{x}}_{k + 1} \in \bar{X} \cap H \) such that \( 0 \in \) relint conv \( \left\{ {{\bar{x}}_{1},\ldots ,{\bar{x}}_{k + 1}}\right\} \) . Therefore, \( \left\{ {{\bar{x}}_{1},\ldots ,{\bar{x}}_{k + 1}}\right\} \) is a coface of \( X \), and the corresponding (proper) face has at least \( v - \left( {v - n - 1}\right) = n + 1 \) vertices and cannot be a simplex.\n\nThe converse is also true. | Yes |
1.16 Lemma. Let \( F,{F}^{\prime } \) be cones in \( {\mathbb{R}}^{n} \) such that \( F \cdot {F}^{\prime } \) is defined. | Proof. (a) For \( x \in F \cdot {F}^{\prime } \), we may set\n\n\[ x = {\lambda }_{1}{y}_{1} + \cdots + {\lambda }_{k}{y}_{k} + {\lambda }_{k + 1}{y}_{k + 1} + \cdots + {\lambda }_{m}{y}_{m} \]\n\nwhere \( {y}_{1},\ldots ,{y}_{k} \in F,{y}_{k + 1},\ldots ,{y}_{m} \in {F}^{\prime } \), and \( {\lambda }_{1},\ldots ,{\lambda }_{m} \geq 0 \) . Then, for \( y \mathrel{\text{:=}} {\lambda }_{1}{y}_{1} + \cdots + {\lambda }_{k}{y}_{k},{y}^{\prime } \mathrel{\text{:=}} {\lambda }_{k + 1}{y}_{k + 1} + \cdots + {\lambda }_{m}{y}_{m} \), we have \( x = y + {y}^{\prime } \) . Hence, \( F \cdot {F}^{\prime } \subset F + {F}^{\prime } \) . The inclusion \( F + {F}^{\prime } \subset F \cdot {F}^{\prime } \) is trivially true, so (a) follows.\n\n(b) By definition, \( F \cdot {F}^{\prime } \) is a cone. Suppose \( \{ 0\} \) is not an apex of \( F \cdot {F}^{\prime } \) . Then, there exists \( x \neq 0 \) such that \( x \in F \cdot {F}^{\prime } \) and \( - x \in F \cdot {F}^{\prime } \) . By (a), we have \( x = y + {y}^{\prime } \) and \( - x = z + {z}^{\prime } \) where \( y, z \in F \) and \( {y}^{\prime },{z}^{\prime } \in {F}^{\prime } \) . Now \( 0 = x - x = y + z + {y}^{\prime } + {z}^{\prime } \) , so that \( y + z = - \left( {{y}^{\prime } + {z}^{\prime }}\right) \) . By the assumption \( \left( {\operatorname{lin}F}\right) \cap \left( {\operatorname{lin}{F}^{\prime }}\right) = \{ 0\} \), we obtain \( y + z = 0 = {y}^{\prime } + {z}^{\prime } \), and, hence, \( z = - y,{z}^{\prime } = - {y}^{\prime } \) . Since 0 is an apex of \( F \) and of \( {F}^{\prime } \), we find \( z = y = {z}^{\prime } = {y}^{\prime } = 0 \) and, therefore, \( x = 0 \), a contradiction.\n\n(c) Since, by (a) and (b), \( F \cdot {F}^{\prime } \) is a cone with apex 0, we can find a hyperplane \( H \) which does not pass through 0 and intersects \( F \cdot {F}^{\prime } \) in a polytope \( \left( {F \cdot {F}^{\prime }}\right) \cap H \neq \varnothing \) . By central projection from 0, we see that \( \left( {F \cdot {F}^{\prime }}\right) \cap H \) and \( \left( {F \cdot {F}^{\prime }}\right) \cap S \) are homeomorphic. It is readily seen that \( \left( {F \cdot {F}^{\prime }}\right) \cap H = \left( {F \cap H}\right) \cdot \left( {{F}^{\prime } \cap H}\right) \), so that (c) follows. | Yes |
1.16 Lemma. Let \( F,{F}^{\prime } \) be cones in \( {\mathbb{R}}^{n} \) such that \( F \cdot {F}^{\prime } \) is defined.\n\n(a) \( F \cdot {F}^{\prime } = F + {F}^{\prime } \) (vector sum).\n\n(b) \( F \cdot {F}^{\prime } \) is a cone with apex 0 .\n\n(c) If \( S \) is the unit sphere of \( {\mathbb{R}}^{n} \), then,\n\n\[ \left( {F \cap S}\right) \cdot \left( {{F}^{\prime } \cap S}\right) = \left( {F \cdot {F}^{\prime }}\right) \cap S \] | Proof. (a) For \( x \in F \cdot {F}^{\prime } \), we may set\n\n\[ x = {\lambda }_{1}{y}_{1} + \cdots + {\lambda }_{k}{y}_{k} + {\lambda }_{k + 1}{y}_{k + 1} + \cdots + {\lambda }_{m}{y}_{m} \]\n\nwhere \( {y}_{1},\ldots ,{y}_{k} \in F,{y}_{k + 1},\ldots ,{y}_{m} \in {F}^{\prime } \), and \( {\lambda }_{1},\ldots ,{\lambda }_{m} \geq 0 \) . Then, for \( y \mathrel{\text{:=}} {\lambda }_{1}{y}_{1} + \cdots + {\lambda }_{k}{y}_{k},{y}^{\prime } \mathrel{\text{:=}} {\lambda }_{k + 1}{y}_{k + 1} + \cdots + {\lambda }_{m}{y}_{m} \), we have \( x = y + {y}^{\prime } \) . Hence, \( F \cdot {F}^{\prime } \subset F + {F}^{\prime } \) . The inclusion \( F + {F}^{\prime } \subset F \cdot {F}^{\prime } \) is trivially true, so (a) follows.\n\n(b) By definition, \( F \cdot {F}^{\prime } \) is a cone. Suppose \( \{ 0\} \) is not an apex of \( F \cdot {F}^{\prime } \) . Then, there exists \( x \neq 0 \) such that \( x \in F \cdot {F}^{\prime } \) and \( - x \in F \cdot {F}^{\prime } \) . By (a), we have \( x = y + {y}^{\prime } \) and \( - x = z + {z}^{\prime } \) where \( y, z \in F \) and \( {y}^{\prime },{z}^{\prime } \in {F}^{\prime } \) . Now \( 0 = x - x = y + z + {y}^{\prime } + {z}^{\prime } \) , so that \( y + z = - \left( {{y}^{\prime } + {z}^{\prime }}\right) \) . By the assumption \( \left( {\operatorname{lin}F}\right) \cap \left( {\operatorname{lin}{F}^{\prime }}\right) = \{ 0\} \), we obtain \( y + z = 0 = {y}^{\prime } + {z}^{\prime } \), and, hence, \( z = - y,{z}^{\prime } = - {y}^{\prime } \) . Since 0 is an apex of \( F \) and of \( {F}^{\prime } \), we find \( z = y = {z}^{\prime } = {y}^{\prime } = 0 \) and, therefore, \( x = 0 \), a contradiction.\n\n(c) Since, by (a) and (b), \( F \cdot {F}^{\prime } \) is a cone with apex 0, we can find a hyperplane \( H \) which does not pass through 0 and intersects \( F \cdot {F}^{\prime } \) in a polytope \( \left( {F \cdot {F}^{\prime }}\right) \cap H \neq \varnothing \) . By central projection from 0, we see that \( \left( {F \cdot {F}^{\prime }}\right) \cap H \) and \( \left( {F \cdot {F}^{\prime }}\right) \cap S \) are homeomorphic. It is readily seen that \( \left( {F \cdot {F}^{\prime }}\right) \cap H = \left( {F \cap H}\right) \cdot \left( {{F}^{\prime } \cap H}\right) \), so that (c) follows. | Yes |
1.16 Lemma. Let \( F,{F}^{\prime } \) be cones in \( {\mathbb{R}}^{n} \) such that \( F \cdot {F}^{\prime } \) is defined.\n\n(a) \( F \cdot {F}^{\prime } = F + {F}^{\prime } \) (vector sum).\n\n(b) \( F \cdot {F}^{\prime } \) is a cone with apex 0 .\n\n(c) If \( S \) is the unit sphere of \( {\mathbb{R}}^{n} \), then,\n\n\[ \left( {F \cap S}\right) \cdot \left( {{F}^{\prime } \cap S}\right) = \left( {F \cdot {F}^{\prime }}\right) \cap S \] | Proof. (a) For \( x \in F \cdot {F}^{\prime } \), we may set\n\n\[ x = {\lambda }_{1}{y}_{1} + \cdots + {\lambda }_{k}{y}_{k} + {\lambda }_{k + 1}{y}_{k + 1} + \cdots + {\lambda }_{m}{y}_{m} \]\n\nwhere \( {y}_{1},\ldots ,{y}_{k} \in F,{y}_{k + 1},\ldots ,{y}_{m} \in {F}^{\prime } \), and \( {\lambda }_{1},\ldots ,{\lambda }_{m} \geq 0 \) . Then, for \( y \mathrel{\text{:=}} {\lambda }_{1}{y}_{1} + \cdots + {\lambda }_{k}{y}_{k},{y}^{\prime } \mathrel{\text{:=}} {\lambda }_{k + 1}{y}_{k + 1} + \cdots + {\lambda }_{m}{y}_{m} \), we have \( x = y + {y}^{\prime } \) . Hence, \( F \cdot {F}^{\prime } \subset F + {F}^{\prime } \) . The inclusion \( F + {F}^{\prime } \subset F \cdot {F}^{\prime } \) is trivially true, so (a) follows.\n\n(b) By definition, \( F \cdot {F}^{\prime } \) is a cone. Suppose \( \{ 0\} \) is not an apex of \( F \cdot {F}^{\prime } \) . Then, there exists \( x \neq 0 \) such that \( x \in F \cdot {F}^{\prime } \) and \( - x \in F \cdot {F}^{\prime } \) . By (a), we have \( x = y + {y}^{\prime } \) and \( - x = z + {z}^{\prime } \) where \( y, z \in F \) and \( {y}^{\prime },{z}^{\prime } \in {F}^{\prime } \) . Now \( 0 = x - x = y + z + {y}^{\prime } + {z}^{\prime } \) , so that \( y + z = - \left( {{y}^{\prime } + {z}^{\prime }}\right) \) . By the assumption \( \left( {\operatorname{lin}F}\right) \cap \left( {\operatorname{lin}{F}^{\prime }}\right) = \{ 0\} \), we obtain \( y + z = 0 = {y}^{\prime } + {z}^{\prime } \), and, hence, \( z = - y,{z}^{\prime } = - {y}^{\prime } \) . Since 0 is an apex of \( F \) and of \( {F}^{\prime } \), we find \( z = y = {z}^{\prime } = {y}^{\prime } = 0 \) and, therefore, \( x = 0 \), a contradiction.\n\n(c) Since, by (a) and (b), \( F \cdot {F}^{\prime } \) is a cone with apex 0, we can find a hyperplane \( H \) which does not pass through 0 and intersects \( F \cdot {F}^{\prime } \) in a polytope \( \left( {F \cdot {F}^{\prime }}\right) \cap H \neq \varnothing \) . By central projection from 0, we see that \( \left( {F \cdot {F}^{\prime }}\right) \cap H \) and \( \left( {F \cdot {F}^{\prime }}\right) \cap S \) are homeomorphic. It is readily seen that \( \left( {F \cdot {F}^{\prime }}\right) \cap H = \left( {F \cap H}\right) \cdot \left( {{F}^{\prime } \cap H}\right) \), so that (c) follows. | Yes |
Let \( P \) be a polytope and \( s\left( {p;F}\right) \mathcal{B}\left( P\right) \) a stellar subdivision of its boundary complex. Then, there exists a polytope \( {P}^{\prime } \) such that \[ s\left( {p;F}\right) \mathcal{B}\left( P\right) \approx \mathcal{B}\left( {P}^{\prime }\right) \] | We may assume \( \dim P = n \) . For \( \dim F = n - 1 \), we obtain a \( {P}^{\prime } \) by placing a sufficiently \ | No |
2.2 Theorem. Let \( P \) be a polytope and \( s\left( {p;F}\right) \mathcal{B}\left( P\right) \) a stellar subdivision of its boundary complex. Then, there exists a polytope \( {P}^{\prime } \) such that \n\n\[ \n s\left( {p;F}\right) \mathcal{B}\left( P\right) \approx \mathcal{B}\left( {P}^{\prime }\right) \n\] | Proof. We may assume \( \dim P = n \) . For \( \dim F = n - 1 \), we obtain a \( {P}^{\prime } \) by placing a sufficiently \ | No |
2.6 Theorem. Let \( \mathcal{C} \) be a cell complex and let \( {Z}_{1},\ldots ,{Z}_{r} \) be the 1-cells (in case \( \mathcal{C} \) is a complex of cones) or 0-cells otherwise (in some order). For \( {v}_{i} \in \) relint \( {Z}_{i}\left( \left\{ {v}_{i}\} = {Z}_{i}\text{in case of 0 -cells}), i = 1,\ldots, r\text{, we set}{\mathcal{C}}_{0} \mathrel{\text{:=}} \mathcal{C}\text{and}\right\} \right. \) \( {\mathcal{C}}_{i} = s\left( {{v}_{i};{Z}_{i}}\right) {\mathcal{C}}_{i - 1}, i = 1,\ldots, r \) . Then, \( {\mathcal{C}}_{r} \) is a simplicial complex which has the same 1-cells (in case of a complex of cones) or 0-cells (in case of a compact complex) as \( \mathcal{C} \) has. Furthermore, \( \left| {\mathcal{C}}_{r}\right| = \left| \mathcal{C}\right| \) . | Proof. By definition, \( s\left( {{v}_{i};{Z}_{i}}\right) \) does not add a 1-cell (in case of a cone complex) or a 0-cell (otherwise) to \( {\mathcal{C}}_{i - 1}, i = 1,\ldots, r \) . Suppose \( F \in {\mathcal{C}}_{r} \) is not a simplex cone or a simplex. Then, there exist \( {Z}_{j} \subset F,{Z}_{k} \subset F \) such that \( \operatorname{conv}\left( {{Z}_{j} \cup {Z}_{k}}\right) \) is not a cell of \( {\mathcal{C}}_{r} \) . But \( s\left( {{v}_{j};{Z}_{j}}\right) {\mathcal{C}}_{j - 1} \) contains \( \operatorname{conv}\left( {{Z}_{j} \cup {Z}_{k}}\right) \) (by definition), a contradiction. | Yes |
2.6 Theorem. Let \( \mathcal{C} \) be a cell complex and let \( {Z}_{1},\ldots ,{Z}_{r} \) be the 1-cells (in case \( \mathcal{C} \) is a complex of cones) or 0-cells otherwise (in some order). For \( {v}_{i} \in \) relint \( {Z}_{i}\left( \left\{ {v}_{i}\} = {Z}_{i}\text{in case of 0 -cells}), i = 1,\ldots, r\text{, we set}{\mathcal{C}}_{0} \mathrel{\text{:=}} \mathcal{C}\text{and}\right\} \right. \) \( {\mathcal{C}}_{i} = s\left( {{v}_{i};{Z}_{i}}\right) {\mathcal{C}}_{i - 1}, i = 1,\ldots, r \) . Then, \( {\mathcal{C}}_{r} \) is a simplicial complex which has the same 1-cells (in case of a complex of cones) or 0-cells (in case of a compact complex) as \( \mathcal{C} \) has. Furthermore, \( \left| {\mathcal{C}}_{r}\right| = \left| \mathcal{C}\right| \) . | Proof. By definition, \( s\left( {{v}_{i};{Z}_{i}}\right) \) does not add a 1-cell (in case of a cone complex) or a 0-cell (otherwise) to \( {\mathcal{C}}_{i - 1}, i = 1,\ldots, r \) . Suppose \( F \in {\mathcal{C}}_{r} \) is not a simplex cone or a simplex. Then, there exist \( {Z}_{j} \subset F,{Z}_{k} \subset F \) such that \( \operatorname{conv}\left( {{Z}_{j} \cup {Z}_{k}}\right) \) is not a cell of \( {\mathcal{C}}_{r} \) . But \( s\left( {{v}_{j};{Z}_{j}}\right) {\mathcal{C}}_{j - 1} \) contains \( \operatorname{conv}\left( {{Z}_{j} \cup {Z}_{k}}\right) \) (by definition), a contradiction. | Yes |
In \( {\mathbb{R}}^{2} \), let \( K \mathrel{\text{:=}} \operatorname{conv}\left\{ {2{e}_{1}, - 2{e}_{1},2{e}_{2}}\right\}, L \mathrel{\text{:=}} \operatorname{conv}\left\{ {{e}_{1}, - {e}_{1},{e}_{1} + }\right. \) \( \left. {2{e}_{2}, - {e}_{1} + 2{e}_{2}}\right\} \) . As is seen from Figure \( 4, d\left( {K, L}\right) = 1 \) . | Proof of Theorem 2.3.\n\n(1) and (3) are true by definition of \( d \) .\n\n(2) If \( d\left( {K, L}\right) = 0, K + 0 \cdot B \supset L \), and \( L + 0 \cdot B \supset K \) ; hence, \( K = L \) . Clearly, \( d\left( {K, K}\right) = 0 \) .\n\n(4) We set \( r \mathrel{\text{:=}} d\left( {K, M}\right), s \mathrel{\text{:=}} d\left( {M, L}\right) \), and \( t \mathrel{\text{:=}} d\left( {K, L}\right) \) . Then, \( K + {rB} \supset M \) , \( M + {rB} \supset K \) and \( M + {sB} \supset L, L + {sB} \supset M \) imply\n\n\[ K + \left( {r + s}\right) B \supset M + {sB} \supset L, \]\n\n\[ L + \left( {r + s}\right) B \supset M + {rB} \supset K, \]\n\nhence, \( r + s \geq t \) . | Yes |
Let \( S \mathrel{\text{:=}} \operatorname{conv}\left\{ {0,{e}_{1},{e}_{2},{e}_{3}}\right\} \) be a simplex in \( {\mathbb{R}}^{3} \), and \( {I}_{1} \mathrel{\text{:=}} \left\lbrack {0,{e}_{1}}\right\rbrack \) , \( {I}_{2} \mathrel{\text{:=}} \left\lbrack {0,{e}_{2}}\right\rbrack \) line segments. We wish to calculate \( V\left( {S,{I}_{1},{I}_{2}}\right) \) . | By Theorem 3.7,\n\n\[ \n{6V}\left( {S,{I}_{1},{I}_{2}}\right) = V\left( {S + {I}_{1} + {I}_{2}}\right) - V\left( {S + {I}_{1}}\right) - V\left( {S + {I}_{2}}\right) \n\]\n\n\[ \n- V\left( {{I}_{1} + {I}_{2}}\right) + V\left( S\right) + V\left( {I}_{1}\right) + V\left( {I}_{2}\right) . \n\]\n\nClearly \( V\left( {I}_{1}\right) = V\left( {I}_{2}\right) = V\left( {{I}_{1} + {I}_{2}}\right) = 0 \) since \( V\left( \cdot \right) \) denotes three-dimensional volume. For reasons of symmetry, \( V\left( {S + {I}_{1}}\right) = V\left( {S + {I}_{2}}\right) \) . We know \( V\left( S\right) = \frac{1}{6} \), so we must only calculate \( V\left( {S + {I}_{1}}\right) \) and \( V\left( {S + {I}_{1} + {I}_{2}}\right) \) . We obtain \( S + {I}_{1} \) from \( S \) by joining a triangular prism to it (see Figure 10) whose volume is evidently \( \frac{1}{2} \) . Therefore, \( V\left( {S + {I}_{1}}\right) = V\left( {S + {I}_{2}}\right) = \frac{2}{3} \). From Figure 10, it is readily seen that \( V\left( {S + {I}_{1} + {I}_{2}}\right) = 2\frac{1}{6} \). Therefore \( {6V}\left( {S,{I}_{1},{I}_{2}}\right) = 2\frac{1}{6} - \frac{4}{3} + \frac{1}{6} = 1 \), hence, \( V\left( {S,{I}_{1},{I}_{2}}\right) = \frac{1}{6} \). | Yes |
4.6 Lemma. Let \( P, Q \) be \( n \) -polytopes in \( {\mathbb{R}}^{n} \). Then, for any \( \varepsilon > 0 \), there exists a polytope \( {Q}^{\prime } \) such that (1) \( d\left( {Q,{Q}^{\prime }}\right) < \varepsilon \), and (2) \( P \) and \( {Q}^{\prime } \) are in skew position. | Proof. Let \( {G}_{1},\ldots ,{G}_{r} \) be those faces of \( Q \) for which faces \( {F}_{1},\ldots ,{F}_{r} \) of \( P \) exist, respectively, such that \( {U}_{{F}_{i}} \cap {U}_{{G}_{i}} \neq \{ 0\}, i = 1,\ldots, r \). Up to renumbering, let \( \dim {G}_{1} = \cdots = \dim {G}_{{r}_{1}} = : {k}_{1},\ldots ,\dim {G}_{{r}_{s}} = \cdots = \dim {G}_{r} = : {k}_{s} \) and \( {k}_{1} > \cdots > {k}_{s} \geq 1 \). From each \( {G}_{i} \), we choose a point \[ {q}_{i} \in \operatorname{relint}{G}_{i},\;i = 1,\ldots, r. \] Successively, we pull up \( {q}_{i} \) to a point \( {q}_{i}^{\prime } \) in the sense of III, Definition 2.3, where \( \begin{Vmatrix}{{q}_{i}^{\prime } - {q}_{i}}\end{Vmatrix} < \varepsilon \) can be assumed, \( i = 1,\ldots, r \). First, we pull up the \( {q}_{i} \) with \( \dim {G}_{i} = {k}_{1} \), then those \( {q}_{j} \) with \( \dim {G}_{j} = {k}_{2} \), and so on. Since \( {k}_{1} > \cdots > {k}_{s} \), in the \( i \) th pulling up the faces \( {G}_{i + 1},\ldots ,{G}_{r} \) are not affected. So, as a result of the \( r \) pulling ups, we obtain a polytope \( {Q}^{\prime } \) for which \( d\left( {Q,{Q}^{\prime }}\right) < \varepsilon \). If \( {G}_{j} \) is contained in a nonsharp \( N\left( F\right) \) -shadow boundary \( {\Gamma }_{N\left( F\right) }^{Q}(F = {F}_{j} \) or some other face \( F \) of \( P \) ) such that a translate of \( {G}_{j} \) is contained in \( {U}_{F} \), then, \( {\Gamma }_{N\left( F\right) }^{{Q}^{\prime }} \) no longer contains a point of \( {G}_{j} \), and, for \( {q}_{j} \in \) relint \( {G}_{j} \), the vertex \( {q}_{j}^{\prime } \) of \( {\widehat{\Theta }}_{F}^{{Q}^{\prime }} \) satisfies \( \left( {{U}_{F} + {q}_{j}^{\prime }}\right) \cap {Q}^{\prime } = \left\{ {q}_{j}^{\prime }\right\} \). This readily implies (2). | Yes |
4.13 Theorem. Let \( {K}_{1},\ldots ,{K}_{n} \) be convex bodies. Then,\n\n(a) \( V\left( {{K}_{1},\ldots ,{K}_{n}}\right) \geq 0 \) ,\n\n(b) \( V\left( {{K}_{1},\ldots ,{K}_{n}}\right) > 0 \) if and only if each \( {K}_{i} \) contains a line segment \( {I}_{i} = \) \( \left\lbrack {{a}_{i},{b}_{i}}\right\rbrack \) such that \( {b}_{1} - {a}_{1},\ldots ,{b}_{n} - {a}_{n} \) are linearly independent. | Proof.\n\n(a) If we replace each \( {K}_{i} \) by one of its points, \( {p}_{i}, i = 1,\ldots, n \), and apply Theorem 4.12, we obtain (a) from \( V\left( {{p}_{1},\ldots ,{p}_{n}}\right) = 0 \) (which follows, for example, by Theorem 3.7).\n\n(b) From Theorem 3.7, we obtain\n\n\[ n!V\left( {{I}_{1},\ldots ,{I}_{n}}\right) = V\left( {{I}_{1} + \cdots + {I}_{n}}\right) = \left| {\det \left( {{b}_{1} - {a}_{1},\ldots ,{b}_{n} - {a}_{n}}\right) }\right| ,\]\n\nand by (6), we know \( V\left( {{K}_{1},\ldots ,{K}_{n}}\right) \geq V\left( {{I}_{1},\ldots ,{I}_{n}}\right) \) . So, \( \mid \det \left( {{b}_{1} - }\right. \) \( \left. {{a}_{1},\ldots ,{b}_{n} - {a}_{n}}\right) \mid > 0 \) implies \( V\left( {{K}_{1},\ldots ,{K}_{n}}\right) > 0 \) .\n\nConversely, let \( V\left( {{K}_{1},\ldots ,{K}_{n}}\right) > 0 \), and suppose \( \mid \det \left( {{b}_{1} - {a}_{1},\ldots ,{b}_{n} - }\right. \) \( \left. \left. {a}_{n}\right) \right) \mid = 0 \) for arbitrary \( {a}_{i},{b}_{i} \in {K}_{i}, i = 1,\ldots, n \) . Then, the linear hull of \( \left( {{K}_{1} - {K}_{1}}\right) \cup \cdots \cup \left( {{K}_{n} - {K}_{n}}\right) \) has dimension, at most, \( n - 1 \), and, up to translations, \( {K}_{i} \subset H, i = 1,\ldots, n \) . Then, \( {K}_{{i}_{1}} + \cdots + {K}_{{i}_{r}} \subset H \) and \( V\left( {{K}_{{i}_{1}} + \cdots + {K}_{{i}_{r}}}\right) = 0 \) for arbitrary index sets \( \left\{ {{i}_{1},\ldots ,{i}_{r}}\right\} \subset \{ 1,\ldots, n\} \) . By Theorem 3.7, this implies \( V\left( {{K}_{1},\ldots ,{K}_{n}}\right) = 0 \), a contradiction. | Yes |
When does equality hold in (AF)? | Clearly a sufficient condition is \( K \) and \( L \) to be homothetic: If \( K = a + {tL} \) , \( t > 0 \), then, \( V\left( {K, L,\mathcal{C}}\right) = V\left( {a + {tL}, L,\mathcal{C}}\right) = {tV}\left( {L, L,\mathcal{C}}\right) \), so that equality in (AF) follows. However, homothety of \( K \) and \( L \) is not necessary.\n\nExample 1. In \( {\mathbb{R}}^{3} \), let \( {I}_{j} \mathrel{\text{:=}} \left\lbrack {{a}_{j},{b}_{j}}\right\rbrack, j = 1,2,3 \), be line segments such that \( {b}_{1} - {a}_{1},{b}_{2} - {a}_{2},{b}_{3} - {a}_{3} \) are linearly independent. Then, for \( K \mathrel{\text{:=}} {I}_{1} + {I}_{3} \) , \( L \mathrel{\text{:=}} {I}_{2} + {I}_{3},{K}_{1} \mathrel{\text{:=}} {I}_{1} + {I}_{2} \), an easy calculation shows that equality holds in (AF).\n\nExample 2. Let \( {K}_{1} \) be a 3-simplex in \( {\mathbb{R}}^{3} \), and let \( K, L \) be obtained from \( {K}_{1} \) by chopping off the vertices but leaving at least one point of each edge remaining (Figure 15). Then, \( \left( {2}^{\prime }\right) \) and equality in \( \left( {3}^{\prime }\right) \) are trivially true, so that equality in (AF) follows. | Yes |
Example 2. Let \( {K}_{1} \) be a 3-simplex in \( {\mathbb{R}}^{3} \), and let \( K, L \) be obtained from \( {K}_{1} \) by chopping off the vertices but leaving at least one point of each edge remaining (Figure 15). Then, \( \left( {2}^{\prime }\right) \) and equality in \( \left( {3}^{\prime }\right) \) are trivially true, so that equality in (AF) follows. | Proof. We can express (S) in terms of \( {\Theta }_{F}^{K},{\Theta }_{F}^{L} \) . Consider the set \( {\widehat{\Theta }}_{F}^{K} \) (see Definition 4.8). By the assumptions of the theorem and by using (1) in Lemma 5.4, we may suppose \( K, L \) to be strictly combinatorially isomorphic and both in skew position to \( P \) . Then \( {\widehat{\Theta }}_{F}^{K} \) is a polygonal arc. Both end points of \( {\widehat{\Theta }}_{F}^{K} \) are also end points of further local shadow boundaries. In fact, there is a one-to-one-correspondence between the set of local shadow boundaries and the set of edges of the dual polytope \( {P}^{ * } \) of \( P \) . It provides a homeomorphism between the union of all local shadow boundaries of \( K \) and the union of all (closed) edges of \( {P}^{ * } \) (which carries the edge graph of \( {P}^{ * } \) ). Since the edge graph of \( {P}^{ * } \) is connected, so is the union of all local shadow boundaries. Hence, from the definition of \( {\Theta }_{F}^{K} \) , (and, analoguously, \( {\Theta }_{F}^{L} \) ),(S) is equivalent to the following condition:\n\n\( \left( {\mathrm{S}}^{\prime }\right) \) For each \( \left( {n - 2}\right) \) -face \( F \) of \( P = {K}_{1} + \cdots + {K}_{n - 2} \) ,\n\n\[{\Theta }_{F}^{K} = {\Theta }_{F}^{L}\]\n\nIt even suffices to show the following:\n\n\( \left( {\mathrm{S}}^{\prime \prime }\right) \) Let \( F \) be an \( \left( {n - 2}\right) \) -face of \( P \), and let \( {a}_{K},{a}_{L} \) be parallel edges of \( {\Theta }_{F}^{K},{\Theta }_{F}^{L} \) , respectively, which do not contain 0 . Then, the lengths \( {\alpha }_{K},{\alpha }_{L} \) of \( {a}_{K},{a}_{L} \) , respectively, are equal.\n\nTo deduce \( \left( {\mathrm{S}}^{\prime \prime }\right) \) from equality in \( \left( \mathrm{{AF}}\right) \), we introduce the following operations. Let \( G = K \cap {H}_{K}\left( u\right) \) be a face of \( K \) . Also, let \( H \) be a hyperplane parallel to \( {H}_{K}\left( u\right) \) which separates vert \( G \) from (vert \( K \) ) \( \smallsetminus \) (vert \( G \) ), such that vert \( G \subset \) int \( {H}^{ + } \) , (vert \( K \) ) \( \smallsetminus \) (vert \( G \) ) \( \subset {H}^{ - } \) . Let \( \varepsilon \) be the distance of \( H \) and \( {H}_{K}\left( u\right) \) . Then, we set\n\n\[c\left( {u,\varepsilon }\right) K \mathrel{\text{:=}} K \cap {H}^{ - }\]\n\nand call \( c\left( {u,\varepsilon }\right) \) a cut operation (Figure 16).\n\nSuppose, in \( \left( {\mathrm{S}}^{\prime \prime }\right) \),\n\n(7)\n\n\[{\alpha }_{K} < {\alpha }_{L}\]\n\nLet \( {G}_{K},{G}_{L} \) be the inverse images of \( {a}_{K}, | Yes |
3.5 Lemma. If \( \sigma \) has an apex, then, the monoid \( \sigma \cap {\mathbb{Z}}^{n} \) has (up to renumbering) precisely one minimal system of generators. | Proof. Let \( {a}_{1},\ldots ,{a}_{t} \) and \( {b}_{1},\ldots ,{b}_{m} \) be different minimal systems of generators and \( {b}_{1} \notin \left\{ {{a}_{1},\ldots ,{a}_{t}}\right\} \), say. There exist linear combinations, say\n\n\[ \n{b}_{1} = \mathop{\sum }\limits_{{j = 1}}^{r}{\lambda }_{j}{a}_{j}\text{ for }{\lambda }_{j} \in {\mathbb{Z}}_{ \geq 0} \n\] \n\nand \n\n\[ \n{a}_{i} = \mathop{\sum }\limits_{{k = 1}}^{m}{\mu }_{ik}{b}_{k},\;{\mu }_{ik} \in {\mathbb{Z}}_{ \geq 0}\text{ for }i = 1,\ldots, t, \n\] \n\nthat yield a representation \n\n\[ \n{b}_{1} = \mathop{\sum }\limits_{{k = 1}}^{m}{\gamma }_{k}{b}_{k} \n\] \n\nwhere \( {\gamma }_{1} = \mathop{\sum }\limits_{{j = 1}}^{r}{\lambda }_{j}{\mu }_{j1} > 0 \), since the \( {b}_{k} \) ’s form a minimal system of generators as a monoid. On the other hand, \( {\gamma }_{1} \leq 1 \) ; otherwise, \( - {b}_{1} \in \sigma \), though \( \sigma \) has an apex. That implies \( {\gamma }_{1} = 1 \) and \( {\gamma }_{k} = 0 \) for \( k \geq 2 \), since \( \mathop{\sum }\limits_{{k = 2}}^{m}{\gamma }_{k}{b}_{k} = 0 \) . As a consequence, \( r = 1 \) and, thus, \( {b}_{1} = {a}_{1} \), a contradiction. | Yes |
For every monoid \( \sigma \cap {\mathbb{Z}}^{n} \), where the cone \( \sigma = \operatorname{pos}\left\{ {{a}_{1},\ldots ,{a}_{k}}\right\} \) is generated by primitive lattice vectors \( {a}_{j} \), is \( {S}_{0} \mathrel{\text{:=}} \left\{ {{\alpha }_{1}{a}_{1} + \cdots + {\alpha }_{k}{a}_{k} \mid {\alpha }_{1},\ldots ,{\alpha }_{k} \in {\mathbb{Z}}_{ \geq 0}}\right\} \) a submonoid? | As is seen from Example 1, \( {S}_{0} \) need not be equal to \( \sigma \cap {\mathbb{Z}}^{n} \). | No |
In \( {\mathbb{R}}^{3} \) let \( X = \left( {{a}_{1},\ldots ,{a}_{6}}\right) \) come from the vertices of a regular prism \( P \) with \( 0 \in \) int \( P \) ,(see Figure 8). Let the fan \( \sum \) be defined by the following facets:\n\n\[ \operatorname{pos}\left\{ {{a}_{1},{a}_{2},{a}_{3}}\right\} ,\operatorname{pos}\left\{ {{a}_{4},{a}_{5},{a}_{6}}\right\} ,\operatorname{pos}\left\{ {{a}_{1},{a}_{3},{a}_{4}}\right\} ,\operatorname{pos}\left\{ {{a}_{3},{a}_{4},{a}_{6}}\right\} ,\]\n\n\[ \operatorname{pos}\left\{ {{a}_{1},{a}_{2},{a}_{5}}\right\} ,\operatorname{pos}\left\{ {{a}_{1},{a}_{4},{a}_{5}}\right\} ,\operatorname{pos}\left\{ {{a}_{2},{a}_{3},{a}_{6}}\right\} ,\operatorname{pos}\left\{ {{a}_{2},{a}_{5},{a}_{6}}\right\} .\]\n\nThen, the affine relationships\n\n\[ {a}_{1} - {a}_{3} - {a}_{4} + {a}_{6} = 0 \]\n\nand\n\n\[ {a}_{1} - {a}_{2} - {a}_{4} + {a}_{5} = 0 \] | provide a basis for the space of affine dependences of \( X \) . A Gale transform \( {\bar{X}}_{\widehat{U}} \) is formed by the rows of the transposed of the matrix\n\n\[ \left( \begin{array}{rrrrrr} 1 & 0 & - 1 & - 1 & 0 & 1 \\ 1 & - 1 & 0 & - 1 & 1 & 0 \end{array}\right) \]\n\n(compare Figure 9). | Yes |
In \( {\mathbb{R}}^{3} \) choose a 3-simplex \( T \), a vertex \( v \) of \( T \), and a triangle \( \Delta \subset \) \( \left( {{\mathbb{R}}^{3} \oplus \mathbb{R}}\right) \smallsetminus \) aff \( T.P \mathrel{\text{:=}} \operatorname{conv}\left( {T \cup \Delta }\right) \) is then a polytope with 3 double-simplices and (in general) 4 simplices as facets. We denote the vertices of \( T \) by \( 1,2,3,4 = v \) (Figure 10), the vertices of \( \Delta \) (after projection) by 5,6,7. We consider a Schlegel diagram of \( P \), namely a central projection into \( T \ ). | We may place \( \Delta \) such that we obtain facets \( {F}_{1},\ldots ,{F}_{7} \), up to renumbering as follows:\n\n\[ \n{F}_{1} = \left\lbrack {1,2,4,5,6}\right\rbrack \n\]\n\n\[ \n{F}_{2} = \left\lbrack {2,3,4,6,7}\right\rbrack \n\]\n\n\[ \n{F}_{3} = \left\lbrack {1,3,4,5,7}\right\rbrack \n\]\n\n\[ \n{F}_{4} = \left\lbrack {2,5,6,7}\right\rbrack \n\]\n\n\[ \n{F}_{5} = \left\lbrack {4,5,6,7}\right\rbrack \n\]\n\n\[ \n{F}_{6} = \left\lbrack {1,2,3,5}\right\rbrack \n\]\n\n\[ \n{F}_{7} = \left\lbrack {2,3,5,7}\right\rbrack \text{.} \n\] | Yes |
4.12 Theorem (Oda’s criterion). A complete simplicial fan \( \sum \) is regular if and only if the following conditions are satisfied:\n\n(a) There exists in \( \sum \) at least one regular \( n \) -cone.\n\n(b) If \( \sigma = \operatorname{pos}\left\{ {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right\} ,{\sigma }^{\prime } = \operatorname{pos}\left\{ {{x}_{1}^{\prime },{x}_{2},\ldots ,{x}_{n}}\right\} \) are two adjacent \( n \) -cones, there exist integers \( {\alpha }_{2},\ldots ,{\alpha }_{n} \) such that\n\n\[ \n{x}_{1} + {x}_{1}^{\prime } + {\alpha }_{2}{x}_{2} + \cdots + {\alpha }_{n}{x}_{n} = 0.\n\] | Proof. Let \( \sum \) be regular, \( \det \sigma = \det \left( {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right) = 1 \) . Then, \( \det \left( {x}_{1}^{\prime }\right. \) , \( \left. {{x}_{2},\ldots ,{x}_{n}}\right) = - 1 \), and, hence,\n\n\[ \n\det \left( {{x}_{1} + {x}_{1}^{\prime },{x}_{2},\ldots ,{x}_{n}}\right) = 0.\n\]\n\nSince \( {x}_{2},\ldots ,{x}_{n} \) are linearly independent, we can write\n\n\[ \n{x}_{1} + {x}_{1}^{\prime } = - {\alpha }_{2}{x}_{2} - \cdots - {\alpha }_{n}{x}_{n}\n\]\n\nFrom\n\n\[ \n\det \left( {{x}_{1},{x}_{1}^{\prime },{x}_{3},\ldots ,{x}_{n}}\right) = \det \left( {{x}_{1}, - {\alpha }_{2}{x}_{2},{x}_{3},\ldots ,{x}_{n}}\right)\n\]\n\n\[ \n= - {\alpha }_{2}\det \left( {{x}_{1},\ldots ,{x}_{n}}\right) = - {\alpha }_{2},\n\]\n\nwe see that \( {\alpha }_{2} \) is integral. Similarly, \( {\alpha }_{3},\ldots ,{x}_{n} \) are shown to be integers. So (a) and (b) are true.\n\nConversely, let (a) and (b) be valid. For two adjacent \( n \) -cones \( \sigma = \) \( \operatorname{pos}\left\{ {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right\} ,{\sigma }^{\prime } \mathrel{\text{:=}} \operatorname{pos}\left\{ {{x}_{1}^{\prime },{x}_{2},\ldots ,{x}_{n}}\right\} \), we deduce from (b) that\n\n\[ \n\det \left( {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right) = - \det \left( {{x}_{1}^{\prime },{x}_{2},\ldots ,{x}_{n}}\right) .\n\]\n\nSince \( \sum \) is complete, we shall see that all \( n \) -cones have the same absolute value of det \( \sigma \), that is \( \left| {\det \sigma }\right| = : c \) . Consider the set of all \( n \) -cones of \( \sum \) with the same absolute value of \( \left| {\det \sigma }\right| \) . Suppose it does not cover \( {\mathbb{R}}^{n} \) . Then, there is a gap which contains at least one \( n \) -cone \( {\sigma }_{0} \), so the boundary of the gap contains at least one \( \left( {n - 1}\right) \) -side of an \( n \) -cone \( {\sigma }^{\prime } \) such that \( \left| {\det {\sigma }^{\prime }}\right| \neq \left| {\det {\sigma }_{0}}\right| \), contrary to the initial assumption.\n\nBy (a), therefore, all determinants of \( n \) -cones are \( \pm 1 \) . | Yes |
5.17 Theorem. Let \( \sum = \sum \left( P\right) \) be strongly polytopal. . (a) The polytope group \( \widetilde{\mathcal{G}} \) is the smallest group into which the semi-group of all polytopes strictly combinatorially isomorphic to \( P \) can be embedded. (b) Pic \( \sum \) can be generated by \( {f}_{n - 1}\left( P\right) - n - \lambda + 1 \) polytope elements strictly isomorphic to \( P \) where \( \lambda \) is defined according to Theorem 5.9. | Proof. (a) is a consequence of Theorem 5.15; (b) follows from Theorems 5.9 and 5.15. | No |
Example 7. If \( \sum \) consists of the cones into which \( {\mathbb{R}}^{n} \) is split by the coordinate hyperplanes, we can set \( {P}_{i} = \left\lbrack {0,{e}_{i}}\right\rbrack \) (line segments), \( i = 1,\ldots, n \), and we obtain a system of \( n \) polytope elements generating | \[ \text{Pic}\sum \cong {\mathbb{Z}}^{n}\text{.} \] | No |
Example 8. In the example of Figure 15, we may choose \( {P}_{1} \mathrel{\text{:=}} \left\lbrack {0,{e}_{1}}\right\rbrack ,{P}_{2} \mathrel{\text{:=}} \) \( \left\lbrack {0,{e}_{2}}\right\rbrack ,{P}_{3} \mathrel{\text{:=}} \operatorname{conv}\left\{ {{e}_{1},{e}_{2},{e}_{1} + {e}_{2}}\right\} \) . Therefore, | \[ \text{Pic}\sum \cong {\mathbb{Z}}^{3}\text{.} \] | Yes |
1.34 Lemma. Let \( \varphi : X \rightarrow Y \) be a morphism of affine varieties, and let \( {\varphi }^{ * } \) : \( {R}_{Y} \rightarrow {R}_{X} \) be the corresponding morphism of rings. Then,\n\n(a) \( {\varphi }^{ * } \) is an isomorphism if and only if \( \varphi \) is an isomorphism, and\n\n(b) \( {\varphi }^{ * } \) is surjective if and only if \( \varphi \) is a closed embedding. | Proof.\n\n(a) We need only show the \ | No |
1.38 Lemma. If \( Y \) is an affine algebraic variety, then, the dimension of \( Y \) is equal to the dimension of its coordinate ring \( {R}_{Y} \) . | Proof. The irreducible affine algebraic sets contained in \( Y \subset {\mathbb{C}}^{n} \) correspond to those prime ideals in \( R \mathrel{\text{:=}} \mathbb{C}\left\lbrack {{\xi }_{1},\ldots ,{\xi }_{n}}\right\rbrack \) which contain \( {i}_{Y} \) . These ideals are in one-to-one correspondence \( \varphi \) to prime ideals in \( {R}_{Y} = \mathbb{C}\left\lbrack {{\xi }_{1},\ldots ,{\xi }_{n}}\right\rbrack /{\mathrm{i}}_{Y} \) :\n\n\[ \begin{matrix} \varphi : R & \rightarrow & {R}_{Y} = R/{\mathfrak{i}}_{Y} \\ \mathfrak{p} & \mapsto & \varphi \left( \mathfrak{p}\right) . \end{matrix} \]\n\nSo, the above definitions imply the lemma. | Yes |
The largest possible \( n \) -dimensional cone is \( \sigma \mathrel{\text{:=}} {\mathbb{R}}^{n} \) . Then, viewed as a monoid, \( \sigma \cap {\mathbb{Z}}^{n} = {\mathbb{Z}}^{n} \) has generators \( {e}_{1},\ldots ,{e}_{n}, - {e}_{1},\ldots , - {e}_{n} \), so the associated algebra is \( {R}_{\sigma } = \mathbb{C}\left\lbrack {{z}_{1},\ldots ,{z}_{n},{z}_{1}^{-1},\ldots ,{z}_{n}^{-1}}\right\rbrack \) . The corresponding toric variety \( {X}_{\sigma } \) can be described in \( {\mathbb{C}}^{2n} \) with coordinates \( {\xi }_{1},\ldots ,{\xi }_{2n} \) as solution set of the equations | \[ {\xi }_{i}{\xi }_{n + i} = 1, i = 1,\ldots, n\text{.} \] Hence, \( {X}_{\sigma } = V\left( {{\xi }_{1}{\xi }_{n + 1} - 1,\ldots ,{\xi }_{n}{\xi }_{2n} - 1}\right) \) . For \( n = 1 \), we obtain a (complex) hyperbola with \( \left\{ {{\xi }_{1} = 0}\right\} \) and \( \left\{ {{\xi }_{2} = 0}\right\} \) as asymptotes (Figure 1); compare section 1, Example 7. We may also realize \( {X}_{\sigma } \) as the set of points \[ T \mathrel{\text{:=}} \left\{ {\left( {{z}_{1},\ldots ,{z}_{n}}\right) \in {\mathbb{C}}^{n} \mid {z}_{i} \neq 0, i = 1,\ldots, n}\right\} = {\left( \mathbb{C}\smallsetminus \{ 0\} \right) }^{n}, \] which is isomorphic to \( V\left( {{\xi }_{1}{\xi }_{n + 1} - 1,\ldots ,{\xi }_{n}{\xi }_{2n} - 1}\right) \) under the projection \( {\mathbb{C}}^{2n} \rightarrow {\mathbb{C}}^{n} \) . The inverse of the restricted projection \( V\left( {{z}_{1}{z}_{n + 1} - 1,\ldots ,{z}_{n}{z}_{2n} - 1}\right) \rightarrow T \) is given by \[ \left( {{z}_{1},\ldots ,{z}_{n}}\right) \mapsto \left( {{z}_{1},\ldots ,{z}_{n},{z}_{1}^{-1},\ldots ,{z}_{n}^{-1}}\right) . \] | Yes |
For \( \sigma \mathrel{\text{:=}} \operatorname{pos}\left( \left\{ {{e}_{1},{e}_{2}}\right\} \right) \), the monoid \( \sigma \cap {\mathbb{Z}}^{2} \) has linearly independent generators \( {e}_{1},{e}_{2} \) . Therefore, \( \mathfrak{a} = \mathfrak{o} \), the zero ideal, and \[ {X}_{\sigma } = {\mathbb{C}}^{2} \] | The same is true for each cone \[ \sigma = \operatorname{pos}\left( \left\{ {{e}_{1} + v{e}_{2},{e}_{2}}\right\} \right) ,\;\text{ for }v \in \mathbb{Z}. \] More generally, if \( \sigma = \operatorname{pos}\left( \left\{ {{a}_{1},\ldots ,{a}_{n}}\right\} \right) \) is a regular lattice cone in \( {\mathbb{R}}^{n} \), then, again, \( \mathfrak{a} = \mathfrak{o} \) ; hence, \( {X}_{\sigma } \) can by identified with the affine space \( {\mathbb{C}}^{n} \). | No |
For \( \sigma \mathrel{\text{:=}} \operatorname{pos}\left( \left\{ {{e}_{1},{e}_{1} + 2{e}_{2}}\right\} \right) \), the monoid \( \sigma \cap {\mathbb{Z}}^{2} \) is generated by \( {a}_{1} = {e}_{1},{a}_{2} = {e}_{1} + 2{e}_{2},{a}_{3} = {e}_{1} + {e}_{2} \) . There is a linear relationship | \[ {a}_{1} + {a}_{2} = 2{a}_{3} \] and, hence, a monomial equation \[ {u}_{1}{u}_{2} = {u}_{3}^{2} \] \( {X}_{\sigma } \) is a quadratic cone with \ | Yes |
Example 4. For the half-plane \( \sigma = \operatorname{pos}\left( \left\{ {{e}_{1}, - {e}_{1},{e}_{2}}\right\} \right) \) and fixed \( v \in {\mathbb{Z}}_{ \geq 0} \) the monoid \( \sigma \cap {\mathbb{Z}}^{2} \) has \( {e}_{1}, - {e}_{1}, b = v{e}_{1} + {e}_{2} \) as generators,(see Figure 3 for \( v = 2 \) ). In particular, \( {u}_{1} \mathrel{\text{:=}} {z}^{{e}_{1}} = {z}_{1},{u}_{2} \mathrel{\text{:=}} {z}^{-{e}_{1}} = {z}_{1}^{-1},{u}_{3} \mathrel{\text{:=}} {z}^{{e}_{2}} = {z}_{2} \) and \( {v}_{1} \mathrel{\text{:=}} {u}_{1} \) , \( {v}_{2} \mathrel{\text{:=}} {u}_{2},{v}_{3} \mathrel{\text{:=}} {z}^{v{e}_{1} + {e}_{2}} = {z}_{1}^{v}{z}_{2} \) are two coordinate systems, and | \[ {v}_{1} = {u}_{1}\;{u}_{1} = {v}_{1} \] \[ {v}_{2} = {u}_{2}\;{u}_{2} = {v}_{2} \] \[ {v}_{3} = {u}_{1}^{v}{u}_{3},\text{ and }\;{u}_{3} = {v}_{2}^{v}{v}_{3} \] are transformation formulae for the coordinates. | Yes |
Example 1. Let the projective plane \( {\mathbb{P}}^{2} = \left\{ {\left\lbrack {{\eta }_{0},{\eta }_{1},{\eta }_{2}}\right\rbrack \mid {\eta }_{i} \in \mathbb{C}}\right. \), not all \( {\eta }_{i} = \) \( 0\} \) be given, the homogeneous coordinates \( {\eta }_{0},{\eta }_{1},{\eta }_{2} \) being determined only up to a common multiple. | It is covered by three affine planes \( {A}_{0} \mathrel{\text{:=}} \left\{ {\left( {1,{\eta }_{1}{\eta }_{0}^{-1},{\eta }_{2}{\eta }_{0}^{-1}}\right) \mid {\eta }_{0} \neq 0}\right\} \) , \( {A}_{1} \mathrel{\text{:=}} \left\{ {\left( {{\eta }_{0}{\eta }_{1}^{-1},1,{\eta }_{2}{\eta }_{1}^{-1}}\right) \mid {\eta }_{1} \neq 0}\right\} \) and \( {A}_{2} \mathrel{\text{:=}} \left\{ {\left( {{\eta }_{0}{\eta }_{2}^{-1},{\eta }_{1}{\eta }_{2}^{-1},1}\right) \mid {\eta }_{2} \neq 0}\right\} \) . Setting \( {z}_{1} \mathrel{\text{:=}} {\eta }_{1}{\eta }_{0}^{-1},{z}_{2} \mathrel{\text{:=}} {\eta }_{2}{\eta }_{0}^{-1} \), we obtain \[ {A}_{0} = \left\{ \left( {{z}_{1},{z}_{2}}\right) \right\} ,{A}_{1} = \left\{ \left( {{z}_{1}^{-1},{z}_{2}{z}_{1}^{-1}}\right) \right\} \text{, and}{A}_{2} = \left\{ \left( {{z}_{2}^{-1},{z}_{1}{z}_{2}^{-1}}\right) \right\} \text{.} \] We find isomorphic coordinate rings, each representing an affine plane (compare Figure 4): \[ {\sigma }_{0} \mathrel{\text{:=}} {\mathbb{R}}_{ \geq 0}{e}_{1} + {\mathbb{R}}_{ \geq 0}{e}_{2},\;{\sigma }_{1} \mathrel{\text{:=}} {\mathbb{R}}_{ \geq 0}{e}_{2} + {\mathbb{R}}_{ \geq 0}\left( {-{e}_{1} - {e}_{2}}\right) , \] \[ {R}_{{\check{\sigma }}_{0}} = \mathbb{C}\left\lbrack {{z}_{1},{z}_{2}}\right\rbrack ,\;{R}_{{\check{\sigma }}_{1}} = \mathbb{C}\left\lbrack {{z}_{1}^{-1},{z}_{1}^{-1}{z}_{2}}\right\rbrack , \] \[ {X}_{{\check{\sigma }}_{0}} = {A}_{0},\;{X}_{{\check{\sigma }}_{1}} = {A}_{1}, \] \[ {\sigma }_{2} \mathrel{\text{:=}} {\mathbb{R}}_{ \geq 0}{e}_{1} + {\mathbb{R}}_{ \geq 0}\left( {-{e}_{1} - {e}_{2}}\right) \] \[ {R}_{{\check{\sigma }}_{2}} = \mathbb{C}\left\lbrack {{z}_{1}{z}_{2}^{-1},{z}_{2}^{-1}}\right\rbrack \] \[ {X}_{{\check{\sigma }}_{2}} = {A}_{2} \] | Yes |
Given \( {\mathbb{P}}^{1} \times {\mathbb{P}}^{1} = \left\{ {\left( {\left\lbrack {{\eta }_{0},{\eta }_{1}}\right\rbrack ,\left\lbrack {{\zeta }_{0},{\zeta }_{1}}\right\rbrack }\right) \mid \left( {{\eta }_{0},{\eta }_{1}}\right) \neq \left( {0,0}\right) }\right. \) , \( \left( {{\zeta }_{0},{\zeta }_{1}}\right) \neq \left( {0,0}\right) \} \) . | We may cover \( {\mathbb{P}}^{1} \times {\mathbb{P}}^{1} \) by four \ | No |
Example 3 Hirzebruch surfaces \( {\mathcal{H}}_{k} \) . We consider a hypersurface in \( {\mathbb{P}}^{1} \times {\mathbb{P}}^{2} = \) \( \left\{ {\left( {\left\lbrack {{\eta }_{0},{\eta }_{1}}\right\rbrack ,\left\lbrack {{\zeta }_{0},{\zeta }_{1},{\zeta }_{2}}\right\rbrack }\right) \mid \left( {{\eta }_{0},{\eta }_{1}}\right) \neq \left( {0,0}\right) ,\left( {{\zeta }_{0},{\zeta }_{1},{\zeta }_{2}}\right) \neq \left( {0,0,0}\right) }\right\} \) given by an equation\n\n\[ \n{\eta }_{0}^{k}{\zeta }_{0} = {\eta }_{1}^{k}{\zeta }_{1},\;k \in \mathbb{Z}.\n\]\n\nIt is called a Hirzebruch surface \( {\mathcal{H}}_{k} \) . | By a modification of the arguments in Example 2, we find four affine planes as charts whose gluing together depends on\n\n \n\n## FIGURE 5a, b.\n\n\( k \) . They are given by the spectra of the following rings (compare Figure 6):\n\n\[ \n{R}_{{\check{\sigma }}_{0}} = \mathbb{C}\left\lbrack {{z}^{{e}_{1}},{z}^{{e}_{2}}}\right\rbrack = \mathbb{C}\left\lbrack {{z}_{1},{z}_{2}}\right\rbrack \n\]\n\n\[ \n{R}_{{\breve{\sigma }}_{2}} = \mathbb{C}\left\lbrack {{z}^{-{e}_{2}},{z}^{{e}_{1} + k{e}_{2}}}\right\rbrack = \mathbb{C}\left\lbrack {{z}_{2}^{-1},{z}_{1}{z}_{2}^{k}}\right\rbrack \n\]\n\n\[ \n{R}_{{\check{\sigma }}_{1}} = \mathbb{C}\left\lbrack {{z}^{-{e}_{1}},{z}^{{e}_{2}}}\right\rbrack = \mathbb{C}\left\lbrack {{z}_{1}^{-1},{z}_{2}}\right\rbrack \n\]\n\n\[ \n{R}_{{\check{\sigma }}_{3}} = \mathbb{C}\left\lbrack {{z}^{-{e}_{1} - k{e}_{2}},{z}^{-{e}_{2}}}\right\rbrack = \mathbb{C}\left\lbrack {{z}_{1}^{-1}{z}_{2}^{-k},{z}_{2}^{-1}}\right\rbrack .\n\] | Yes |
Proposition 1.2.1. The number of vertices of odd degree in a graph is always even. | Proof. A graph on \( V \) has \( \frac{1}{2}\mathop{\sum }\limits_{{v \in V}}d\left( v\right) \) edges, so \( \sum d\left( v\right) \) is an even number. | Yes |
Proposition 1.2.2. Every graph \( G \) with at least one edge has a subgraph \( H \) with \( \delta \left( H\right) > \varepsilon \left( H\right) \geq \varepsilon \left( G\right) \) . | Proof. To construct \( H \) from \( G \), let us try to delete vertices of small degree one by one, until only vertices of large degree remain. Up to which degree \( d\left( v\right) \) can we afford to delete a vertex \( v \), without lowering \( \varepsilon \) ? Clearly, up to \( d\left( v\right) = \varepsilon \) : then the number of vertices decreases by 1 and the number of edges by at most \( \varepsilon \), so the overall ratio \( \varepsilon \) of edges to vertices will not decrease.\n\nFormally, we construct a sequence \( G = {G}_{0} \supseteq {G}_{1} \supseteq \ldots \) of induced subgraphs of \( G \) as follows. If \( {G}_{i} \) has a vertex \( {v}_{i} \) of degree \( d\left( {v}_{i}\right) \leq \varepsilon \left( {G}_{i}\right) \) , we let \( {G}_{i + 1} \mathrel{\text{:=}} {G}_{i} - {v}_{i} \) ; if not, we terminate our sequence and set \( H \mathrel{\text{:=}} {G}_{i} \) . By the choices of \( {v}_{i} \) we have \( \varepsilon \left( {G}_{i + 1}\right) \geq \varepsilon \left( {G}_{i}\right) \) for all \( i \), and hence \( \varepsilon \left( H\right) \geq \varepsilon \left( G\right) \).\n\nWhat else can we say about the graph \( H \) ? Since \( \varepsilon \left( {K}^{1}\right) = 0 < \varepsilon \left( G\right) \) , none of the graphs in our sequence is trivial, so in particular \( H \neq \varnothing \) . The fact that \( H \) has no vertex suitable for deletion thus implies \( \delta \left( H\right) > \varepsilon \left( H\right) \) , as claimed. | Yes |
Proposition 1.3.2. Every graph \( G \) containing a cycle satisfies \( g\left( G\right) \leq \) \( 2\operatorname{diam}G + 1 \) . | Proof. Let \( C \) be a shortest cycle in \( G \) . If \( g\left( G\right) \geq 2\operatorname{diam}G + 2 \), then \( C \) has two vertices whose distance in \( C \) is at least \( \operatorname{diam}G + 1 \) . In \( G \) , these vertices have a lesser distance; any shortest path \( P \) between them is therefore not a subgraph of \( C \) . Thus, \( P \) contains a \( C \) -path \( {xPy} \) . Together with the shorter of the two \( x - y \) paths in \( C \), this path \( {xPy} \) forms a shorter cycle than \( C \), a contradiction. | Yes |
Proposition 1.3.3. A graph \( G \) of radius at most \( k \) and maximum degree at most \( d \geq 3 \) has fewer than \( \frac{d}{d - 2}{\left( d - 1\right) }^{k} \) vertices. | Proof. Let \( z \) be a central vertex in \( G \), and let \( {D}_{i} \) denote the set of vertices of \( G \) at distance \( i \) from \( z \) . Then \( V\left( G\right) = \mathop{\bigcup }\limits_{{i = 0}}^{k}{D}_{i} \) . Clearly \( \left| {D}_{0}\right| = 1 \) and \( \left| {D}_{1}\right| \leq d \) . For \( i \geq 1 \) we have \( \left| {D}_{i + 1}\right| \leq \left( {d - 1}\right) \left| {D}_{i}\right| \), because every vertex in \( {D}_{i + 1} \) is a neighbour of a vertex in \( {D}_{i} \), and each vertex in \( {D}_{i} \) has at most \( d - 1 \) neighbours in \( {D}_{i + 1} \) (since it has another neighbour in \( \left. {D}_{i - 1}\right) \) . Thus \( \left| {D}_{i + 1}\right| \leq d{\left( d - 1\right) }^{i} \) for all \( i < k \) by induction, giving\n\n\[ \left| G\right| \leq 1 + d\mathop{\sum }\limits_{{i = 0}}^{{k - 1}}{\left( d - 1\right) }^{i} = 1 + \frac{d}{d - 2}\left( {{\left( d - 1\right) }^{k} - 1}\right) < \frac{d}{d - 2}{\left( d - 1\right) }^{k}. \] | Yes |
Theorem 1.3.4. (Alon, Hoory & Linial 2002)\n\nLet \( G \) be a graph. If \( d\left( G\right) \geq d \geq 2 \) and \( g\left( G\right) \geq g \in \mathbb{N} \) then \( \left| G\right| \geq {n}_{0}\left( {d, g}\right) \) . | 2.3.1] Corollary 1.3.5. If \( \delta \left( G\right) \geq 3 \) then \( g\left( G\right) < 2\log \left| G\right| \) .\n\nProof. If \( g \mathrel{\text{:=}} g\left( G\right) \) is even then\n\n\[ \n{n}_{0}\left( {3, g}\right) = 2\frac{{2}^{g/2} - 1}{2 - 1} = {2}^{g/2} + \left( {{2}^{g/2} - 2}\right) > {2}^{g/2},\n\]\n\nwhile if \( g \) is odd then\n\n\[ \n{n}_{0}\left( {3, g}\right) = 1 + 3\frac{{2}^{\left( {g - 1}\right) /2} - 1}{2 - 1} = \frac{3}{\sqrt{2}}{2}^{g/2} - 2 > {2}^{g/2}.\n\]\n\nAs \( \left| G\right| \geq {n}_{0}\left( {3, g}\right) \), the result follows. | No |
Proposition 1.4.1. The vertices of a connected graph \( G \) can always be enumerated, say as \( {v}_{1},\ldots ,{v}_{n} \), so that \( {G}_{i} \mathrel{\text{:=}} G\left\lbrack {{v}_{1},\ldots ,{v}_{i}}\right\rbrack \) is connected for every \( i \) . | Proof. Pick any vertex as \( {v}_{1} \), and assume inductively that \( {v}_{1},\ldots ,{v}_{i} \) have been chosen for some \( i < \left| G\right| \) . Now pick a vertex \( v \in G - {G}_{i} \) . As \( G \) is connected, it contains a \( v - {v}_{1} \) path \( P \) . Choose as \( {v}_{i + 1} \) the last vertex of \( P \) in \( G - {G}_{i} \) ; then \( {v}_{i + 1} \) has a neighbour in \( {G}_{i} \) . The connectedness of every \( {G}_{i} \) follows by induction on \( i \) . | Yes |
Corollary 1.5.2. The vertices of a tree can always be enumerated, say as \( {v}_{1},\ldots ,{v}_{n} \), so that every \( {v}_{i} \) with \( i \geq 2 \) has a unique neighbour in \( \left\{ {{v}_{1},\ldots ,{v}_{i - 1}}\right\} \) | Proof. Use the enumeration from Proposition 1.4.1. | No |
Corollary 1.5.4. If \( T \) is a tree and \( G \) is any graph with \( \delta \left( G\right) \geq \left| T\right| - 1 \) , then \( T \subseteq G \), i.e. \( G \) has a subgraph isomorphic to \( T \) . | Proof. Find a copy of \( T \) in \( G \) inductively along its vertex enumeration from Corollary 1.5.2. | No |
Lemma 1.5.5. Let \( T \) be a normal tree in \( G \) .\n\n(i) Any two vertices \( x, y \in T \) are separated in \( G \) by the set \( \lceil x\rceil \cap \lceil y\rceil \) .\n\n(ii) If \( S \subseteq V\left( T\right) = V\left( G\right) \) and \( S \) is down-closed, then the components of \( G - S \) are spanned by the sets \( \lfloor x\rfloor \) with \( x \) minimal in \( T - S \) . | Proof. (i) Let \( P \) be any \( x - y \) path in \( G \) . Since \( T \) is normal, the vertices of \( P \) in \( T \) form a sequence \( x = {t}_{1},\ldots ,{t}_{n} = y \) for which \( {t}_{i} \) and \( {t}_{i + 1} \) are always comparable in the tree oder of \( T \) . Consider a minimal such sequence of vertices in \( P \cap T \) . In this sequence we cannot have \( {t}_{i - 1} < {t}_{i} > {t}_{i + 1} \) for any \( i \), since \( {t}_{i - 1} \) and \( {t}_{i + 1} \) would then be comparable and deleting \( {t}_{i} \) would yield a smaller such sequence. So\n\n\[ x = {t}_{1} > \ldots > {t}_{k} < \ldots < {t}_{n} = y \]\n\nfor some \( k \in \{ 1,\ldots, n\} \) . As \( {t}_{k} \in \lceil x\rceil \cap \lceil y\rceil \cap V\left( P\right) \), the result follows.\n\n(ii) Since \( S \) is down-closed, the upper neighbours in \( T \) of any vertex of \( G - S \) are again in \( G - S \) (and clearly in the same component), so the components \( C \) of \( G - S \) are up-closed. As \( S \) is down-closed, minimal vertices of \( C \) are also minimal in \( G - S \) . By (i), this means that \( C \) has only one minimal vertex \( x \) and equals its up-closure \( \lfloor x\rfloor \) . | Yes |
Proposition 1.5.6. Every connected graph contains a normal spanning tree, with any specified vertex as its root. | Proof. Let \( G \) be a connected graph and \( r \in G \) any specified vertex. Let \( T \) be a maximal normal tree with root \( r \) in \( G \) ; we show that \( V\left( T\right) = V\left( G\right) \) . \n\nSuppose not, and let \( C \) be a component of \( G - T \) . As \( T \) is normal, \( N\left( C\right) \) is a chain in \( T \) . Let \( x \) be its greatest element, and let \( y \in C \) be adjacent to \( x \) . Let \( {T}^{\prime } \) be the tree obtained from \( T \) by joining \( y \) to \( x \) ; the tree-order of \( {T}^{\prime } \) then extends that of \( T \) . We shall derive a contradiction by showing that \( {T}^{\prime } \) is also normal in \( G \) . \n\nLet \( P \) be a \( {T}^{\prime } \) -path in \( G \) . If the ends of \( P \) both lie in \( T \), then they are comparable in the tree-order of \( T \) (and hence in that of \( {T}^{\prime } \) ), because then \( P \) is also a \( T \) -path and \( T \) is normal in \( G \) by assumption. If not, then \( y \) is one end of \( P \), so \( P \) lies in \( C \) except for its other end \( z \), which lies in \( N\left( C\right) \) . Then \( z \leq x \), by the choice of \( x \) . For our proof that \( y \) and \( z \) are comparable it thus suffices to show that \( x < y \), i.e. that \( x \in r{T}^{\prime }y \) . This, however, is clear since \( y \) is a leaf of \( {T}^{\prime } \) with neighbour \( x \) . | Yes |
Proposition 1.6.1. A graph is bipartite if and only if it contains no odd cycle. | (1.5.1) Proof. Let \( G = \left( {V, E}\right) \) be a graph without odd cycles; we show that \( G \) is bipartite. Clearly a graph is bipartite if all its components are bipartite or trivial, so we may assume that \( G \) is connected. Let \( T \) be a spanning tree in \( G \), pick a root \( r \in T \), and denote the associated tree-order on \( V \) by \( { \leq }_{T} \) . For each \( v \in V \), the unique path \( {rTv} \) has odd or even length. This defines a bipartition of \( V \) ; we show that \( G \) is bipartite with this partition.\n\n\n\nFig. 1.6.3. The cycle \( {C}_{e} \) in \( T + e \)\n\nLet \( e = {xy} \) be an edge of \( G \) . If \( e \in T \), with \( x{ < }_{T}y \) say, then \( {rTy} = {rTxy} \) and so \( x \) and \( y \) lie in different partition classes. If \( e \notin T \) then \( {C}_{e} \mathrel{\text{:=}} {xTy} + e \) is a cycle (Fig. 1.6.3), and by the case treated already the vertices along \( {xTy} \) alternate between the two classes. Since \( {C}_{e} \) is even by assumption, \( x \) and \( y \) again lie in different classes. | Yes |
Proposition 1.7.1. \( G \) is an \( {MX} \) if and only if \( X \) can be obtained from \( G \) by a series of edge contractions, i.e. if and only if there are graphs \( {G}_{0},\ldots ,{G}_{n} \) and edges \( {e}_{i} \in {G}_{i} \) such that \( {G}_{0} = G,{G}_{n} \simeq X \), and \( {G}_{i + 1} = {G}_{i}/{e}_{i} \) for all \( i < n \) . | Proof. Induction on \( \left| G\right| - \left| X\right| \) . | No |
A connected graph is Eulerian if and only if every vertex has even degree. | Proof. The degree condition is clearly necessary: a vertex appearing \( k \) times in an Euler tour (or \( k + 1 \) times, if it is the starting and finishing vertex and as such counted twice) must have degree \( {2k} \) .\n\nConversely, let \( G \) be a connected graph with all degrees even, and let\n\n\[ W = {v}_{0}{e}_{0}\ldots {e}_{\ell - 1}{v}_{\ell } \]\n\nbe a longest walk in \( G \) using no edge more than once. Since \( W \) cannot be extended, it already contains all the edges at \( {v}_{\ell } \). By assumption, the number of such edges is even. Hence \( {v}_{\ell } = {v}_{0} \), so \( W \) is a closed walk.\n\nSuppose \( W \) is not an Euler tour. Then \( G \) has an edge \( e \) outside \( W \) but incident with a vertex of \( W \), say \( e = u{v}_{i} \). (Here we use the connectedness of \( G \), as in the proof of Proposition 1.4.1.) Then the walk\n\n\[ {ue}{v}_{i}{e}_{i}\ldots {e}_{\ell - 1}{v}_{\ell }{e}_{0}\ldots {e}_{i - 1}{v}_{i} \]\n\n\nis longer than \( W \), a contradiction. | Yes |
Proposition 1.9.1. The induced cycles in \( G \) generate its entire cycle space. | Proof. By definition of \( \mathcal{C}\left( G\right) \) it suffices to show that the induced cycles in \( G \) generate every cycle \( C \subseteq G \) with a chord \( e \) . This follows at once by induction on \( \left| C\right| \) : the two cycles in \( C + e \) that have \( e \) but no other edge in common are shorter than \( C \), and their symmetric difference is precisely \( C \) . | Yes |
Proposition 1.9.2. The following assertions are equivalent for edge sets \( F \subseteq E \) :\n\n(i) \( F \in \mathcal{C}\left( G\right) \) ;\n\n(ii) \( F \) is a disjoint union of (edge sets of) cycles in \( G \) ;\n\n(iii) All vertex degrees of the graph \( \left( {V, F}\right) \) are even. | Proof. Since cycles have even degrees and taking symmetric differences preserves this,(i) \( \rightarrow \) (iii) follows by induction on the number of cycles used to generate \( F \) . The implication (iii) \( \rightarrow \) (ii) follows by induction on \( \left| F\right| \) : if \( F \neq \varnothing \) then \( \left( {V, F}\right) \) contains a cycle \( C \), whose edges we delete for the induction step. The implication (ii) \( \rightarrow \) (i) is immediate from the definition of \( \mathcal{C}\left( G\right) \) . | Yes |
Proposition 1.9.3. Together with \( \varnothing \), the cuts in \( G \) form a subspace \( {\mathcal{C}}^{ * } \) of \( \mathcal{E}\left( G\right) \) . This space is generated by cuts of the form \( E\left( v\right) \) . | Proof. Let \( {\mathcal{C}}^{ * } \) denote the set of all cuts in \( G \), together with \( \varnothing \) . To prove that \( {\mathcal{C}}^{ * } \) is a subspace, we show that for all \( D,{D}^{\prime } \in {\mathcal{C}}^{ * } \) also \( D + {D}^{\prime } \) \( \left( { = D - {D}^{\prime }}\right) \) lies in \( {\mathcal{C}}^{ * } \) . Since \( D + D = \varnothing \in {\mathcal{C}}^{ * } \) and \( D + \varnothing = D \in {\mathcal{C}}^{ * } \) , we may assume that \( D \) and \( {D}^{\prime } \) are distinct and non-empty. Let \( \left\{ {{V}_{1},{V}_{2}}\right\} \) and \( \left\{ {{V}_{1}^{\prime },{V}_{2}^{\prime }}\right\} \) be the corresponding partitions of \( V \) . Then \( D + {D}^{\prime } \) consists of all the edges that cross one of these partitions but not the other (Fig. 1.9.1). But these are precisely the edges between \( \left( {{V}_{1} \cap {V}_{1}^{\prime }}\right) \cup \left( {{V}_{2} \cap {V}_{2}^{\prime }}\right) \) and \( \left( {{V}_{1} \cap {V}_{2}^{\prime }}\right) \cup \left( {{V}_{2} \cap {V}_{1}^{\prime }}\right) \), and by \( D \neq {D}^{\prime } \) these two sets form another partition of \( V \) . Hence \( D + {D}^{\prime } \in {\mathcal{C}}^{ * } \), and \( {\mathcal{C}}^{ * } \) is indeed a subspace of \( \mathcal{E}\left( G\right) \) . | Yes |
Lemma 1.9.4. Every cut is a disjoint union of bonds. | Proof. Consider first a connected graph \( H = \left( {V, E}\right) \), a connected subgraph \( C \subseteq H \), and a component \( D \) of \( H - C \) . Then \( H - D \), too, is connected (Fig. 1.9.2), so the edges between \( D \) and \( H - D \) form a minimal cut. By the choice of \( D \), this cut is precisely the set \( E\left( {C, D}\right) \) of all \( C - D \) edges in \( H \) .\n\n\n\nFig. 1.9.2. \( H - D \) is connected, and \( E\left( {C, D}\right) \) a minimal cut\n\n\n\nTo prove the lemma, let a cut in an arbitrary graph \( G = \left( {V, E}\right) \) be given, with partition \( \left\{ {{V}_{1},{V}_{2}}\right\} \) of \( V \) say. Consider a component \( C \) of \( G\left\lbrack {V}_{1}\right\rbrack \), and let \( H \) be the component of \( G \) containing \( C \) . Then \( E\left( {C,{V}_{2}}\right) = E\left( {C, H - C}\right) \) is the disjoint union of the edge sets \( E\left( {C, D}\right) \) over all the components \( D \) of \( H - C \) . By our earlier considerations these sets are minimal cuts in \( H \), and hence bonds in \( G \) . Now the disjoint union of all these edge sets \( E\left( {C,{V}_{2}}\right) \), taken over all the components \( C \) of \( G\left\lbrack {V}_{1}\right\rbrack \), is precisely our cut \( E\left( {{V}_{1},{V}_{2}}\right) \) . | Yes |
Theorem 1.9.5. The cycle space \( \mathcal{C} \) and the cut space \( {\mathcal{C}}^{ * } \) of any graph satisfy\n\n\[ \mathcal{C} = {\mathcal{C}}^{* \bot }\;\text{ and }\;{\mathcal{C}}^{ * } = {\mathcal{C}}^{ \bot }.\] | Proof. (See also Exercise 30.) Let us consider a graph \( G = \left( {V, E}\right) \) . Clearly, any cycle in \( G \) has an even number of edges in each cut. This implies \( \mathcal{C} \subseteq {\mathcal{C}}^{* \bot } \) .\n\nConversely, recall from Proposition 1.9.2 that for every edge set \( F \notin \mathcal{C} \) there exists a vertex \( v \) incident with an odd number of edges in \( F \) . Then \( \langle E\left( v\right), F\rangle = 1 \), so \( E\left( v\right) \in {\mathcal{C}}^{ * } \) implies \( F \notin {\mathcal{C}}^{* \bot } \) . This completes the proof of \( \mathcal{C} = {\mathcal{C}}^{* \bot } \) .\n\nTo prove \( {\mathcal{C}}^{ * } = {\mathcal{C}}^{ \bot } \), it now suffices to show \( {\mathcal{C}}^{ * } = {\left( {\mathcal{C}}^{* \bot }\right) }^{ \bot } \) . Here \( {\mathcal{C}}^{ * } \subseteq {\left( {\mathcal{C}}^{* \bot }\right) }^{ \bot } \) follows directly from the definition of \( \bot \) . But \( {\mathcal{C}}^{ * } \) has the same dimension as \( {\left( {\mathcal{C}}^{* \bot }\right) }^{ \bot } \), since \( \left( \dagger \right) \) implies\n\n\[ \dim {\mathcal{C}}^{ * } + \dim {\mathcal{C}}^{* \bot } = m = \dim {\mathcal{C}}^{* \bot } + \dim {\left( {\mathcal{C}}^{* \bot }\right) }^{ \bot }.\]\n\nHence \( {\mathcal{C}}^{ * } = {\left( {\mathcal{C}}^{* \bot }\right) }^{ \bot } \) as claimed. | No |
Theorem 1.9.6. Let \( G \) be a connected graph and \( T \subseteq G \) a spanning tree. Then the corresponding fundamental cycles and cuts form a basis of \( \mathcal{C}\left( G\right) \) and of \( {\mathcal{C}}^{ * }\left( G\right) \), respectively. If \( G \) has \( n \) vertices and \( m \) edges, then\n\n\[ \dim \mathcal{C}\left( G\right) = m - n + 1\;\text{ and }\;\dim {\mathcal{C}}^{ * }\left( G\right) = n - 1. \] | Proof. Since an edge \( e \in T \) lies in \( {D}_{e} \) but not in \( {D}_{{e}^{\prime }} \) for any \( {e}^{\prime } \neq e \), the cut\n\n\( \left( {1.5.3}\right) \)\n\n\( {D}_{e} \) cannot be generated by other fundamental cuts. The fundamental cuts therefore form a linearly independent subset of \( {\mathcal{C}}^{ * } \), of size \( n - 1 \) (Corollary 1.5.3). Similarly, an edge \( e \in E \smallsetminus E\left( T\right) \) lies on \( {C}_{e} \) but not on any other fundamental cycle; so the fundamental cycles form a linearly independent subset of \( \mathcal{C} \), of size \( m - n + 1 \) . Thus,\n\n\[ \dim {\mathcal{C}}^{ * } \geq n - 1\;\text{ and }\;\dim \mathcal{C} \geq m - n + 1. \]\n\nBut\n\n\[ \dim {\mathcal{C}}^{ * } + \dim \mathcal{C} = m = \left( {n - 1}\right) + \left( {m - n + 1}\right) \]\n\nby Theorem 1.9.5 and \( \left( \dagger \right) \), so the two inequalities above can hold only with equality. Hence the sets of fundamental cuts and cycles are maximal as linearly independent subsets of \( {\mathcal{C}}^{ * } \) and \( \mathcal{C} \), and hence are bases. | Yes |
Corollary 2.1.3. If \( G \) is \( k \) -regular with \( k \geq 1 \), then \( G \) has a 1 -factor. | Proof. If \( G \) is \( k \) -regular, then clearly \( \left| A\right| = \left| B\right| \) ; it thus suffices to show by Theorem 2.1.2 that \( G \) contains a matching of \( A \) . Now every set \( S \subseteq A \) is joined to \( N\left( S\right) \) by a total of \( k\left| S\right| \) edges, and these are among the \( k\left| {N\left( S\right) }\right| \) edges of \( G \) incident with \( N\left( S\right) \) . Therefore \( k\left| S\right| \leq k\left| {N\left( S\right) }\right| \), so \( G \) does indeed satisfy the marriage condition. | Yes |
For every set of preferences, \( G \) has a stable matching. | Proof. Call a matching \( M \) in \( G \) better than a matching \( {M}^{\prime } \neq M \) if \( M \) makes the vertices in \( B \) happier than \( {M}^{\prime } \) does, that is, if every vertex \( b \) in an edge \( {f}^{\prime } \in {M}^{\prime } \) is incident also with some \( f \in M \) such that \( {f}^{\prime }{ \leq }_{b}f \) . Given a matching \( M \), call a vertex \( a \in A \) acceptable to \( b \in B \) if \( e = {ab} \in \) \( E \smallsetminus M \) and any edge \( f \in M \) at \( b \) satisfies \( f{ < }_{b}e \) . Call \( a \in A \) happy with \( M \) if \( a \) is unmatched or its matching edge \( f \in M \) satisfies \( f{ > }_{a}e \) for all edges \( e = {ab} \) such that \( a \) is acceptable to \( b \) .\n\nStarting with the empty matching, let us construct a sequence of matchings that each keep all the vertices in \( A \) happy. Given such a matching \( M \), consider a vertex \( a \in A \) that is unmatched but acceptable to some \( b \in B \) . (If no such \( a \) exists, terminate the sequence.) Add to \( M \) the \( { \leq }_{a} \) -maximal edge \( {ab} \) such that \( a \) is acceptable to \( b \), and discard from \( M \) any other edge at \( b \) .\n\nClearly, each matching in our sequence is better than the previous, and it is easy to check inductively that they all keep the vertices in \( A \) happy. So the sequence continues until it terminates with a matching \( M \) such that every unmatched vertex in \( A \) is inacceptable to all its neighbours in \( B \) . As every matched vertex in \( A \) is happy with \( M \), this matching is stable. | Yes |
A graph \( G \) has a 1-factor if and only if \( q\left( {G - S}\right) \leq \left| S\right| \) for all \( S \subseteq V\left( G\right) \) . | Proof. Let \( G = \left( {V, E}\right) \) be a graph without a 1-factor. Our task is to\nbad set find a bad set \( S \subseteq V \), one that violates Tutte’s condition.\n\nWe may assume that \( G \) is edge-maximal without a 1-factor. Indeed, if \( {G}^{\prime } \) is obtained from \( G \) by adding edges and \( S \subseteq V \) is bad for \( {G}^{\prime } \), then \( S \) is also bad for \( G \) : any odd component of \( {G}^{\prime } - S \) is the union of components of \( G - S \), and one of these must again be odd.\n\nWhat does \( G \) look like? Clearly, if \( G \) contains a bad set \( S \) then, by its edge-maximality and the trivial forward implication of the theorem,\n\nall the components of \( G - S \) are complete and every vertex\n\n\( \left( *\right) \)\n\n\( s \in S \) is adjacent to all the vertices of \( G - s \) .\n\nBut also conversely, if a set \( S \subseteq V \) satisfies \( \left( *\right) \) then either \( S \) or the empty set must be bad: if \( S \) is not bad we can join the odd components of \( G - S \) disjointly to \( S \) and pair up all the remaining vertices - unless \( \left| G\right| \) is odd, in which case \( \varnothing \) is bad.\n\nSo it suffices to prove that \( G \) has a set \( S \) of vertices satisfying \( \left( *\right) \) . \( S \) Let \( S \) be the set of vertices that are adjacent to every other vertex. If this set \( S \) does not satisfy \( \left( *\right) \), then some component of \( G - S \) has non- \( a, b, c \) adjacent vertices \( a,{a}^{\prime } \) . Let \( a, b, c \) be the first three vertices on a shortest \( a - {a}^{\prime } \) path in this component; then \( {ab},{bc} \in E \) but \( {ac} \notin E \) . Since \( b \notin S \) , \( d \) there is a vertex \( d \in V \) such that \( {bd} \notin E \) . By the maximality of \( G \), there \( {M}_{1},{M}_{2} \) is a matching \( {M}_{1} \) of \( V \) in \( G + {ac} \), and a matching \( {M}_{2} \) of \( V \) in \( G + {bd} \) .\n\nLet \( P = d\ldots v \) be a maximal path in \( G \) starting at \( d \) with an edge from \( {M}_{1} \) and containing alternately edges from \( {M}_{1} \) and \( {M}_{2} \) (Fig. 2.2.2). If the last edge of \( P \) lies in \( {M}_{1} \), then \( v = b \), since otherwise we could continue \( P \) . Let us then set \( C \mathrel{\text{:=}} P + {bd} \) . If the last edge of \( P \) lies in \( {M}_{2} \) , then by the maximality of \( P \) the \( {M}_{1} \) -edge at \( v \) must be \( {ac} \), so \( v \in \{ a, c\} \) ; then let \( C \) be the cycle \( {dPvbd} \) . In each case, \( C \) is an even cycle with every other edge in \( {M}_{2} \), and whose only edge not in \( E \) is \( {bd} \) . Replacing in \( {M}_{2} \) its edges on \( C \) with the edges of \( C - {M}_{2} \), we obtain a matching of \( V \) contained in \( E \), a contradiction. | Yes |
Every bridgeless cubic graph has a 1-factor. | Proof. We show that any bridgeless cubic graph \( G \) satisfies Tutte’s condition. Let \( S \subseteq V\left( G\right) \) be given, and consider an odd component \( C \) of \( G - S \) . Since \( G \) is cubic, the degrees (in \( G \) ) of the vertices in \( C \) sum to an odd number, but only an even part of this sum arises from edges of \( C \) . So \( G \) has an odd number of \( S - C \) edges, and therefore has at least 3 such edges (since \( G \) has no bridge). The total number of edges between \( S \) and \( G - S \) thus is at least \( {3q}\left( {G - S}\right) \) . But it is also at most \( 3\left| S\right| \), because \( G \) is cubic. Hence \( q\left( {G - S}\right) \leq \left| S\right| \), as required. | Yes |
Lemma 2.3.1. Let \( k \in \mathbb{N} \), and let \( H \) be a cubic multigraph. If \( \left| H\right| \geq {s}_{k} \) , then \( H \) contains \( k \) disjoint cycles. | Proof. We apply induction on \( k \) . For \( k \leq 1 \) the assertion is trivial, so let \( k \geq 2 \) be given for the induction step. Let \( C \) be a shortest cycle in \( H \) .\n\nWe first show that \( H - C \) contains a subdivision of a cubic multigraph \( {H}^{\prime } \) with \( \left| {H}^{\prime }\right| \geq \left| H\right| - 2\left| C\right| \) . Let \( m \) be the number of edges between \( C \) and \( H - C \) . Since \( H \) is cubic and \( d\left( C\right) = 2 \), we have \( m \leq \left| C\right| \) . We now consider bipartitions \( \left\{ {{V}_{1},{V}_{2}}\right\} \) of \( V\left( H\right) \), beginning with \( {V}_{1} \mathrel{\text{:=}} V\left( C\right) \) . If \( H\left\lbrack {V}_{2}\right\rbrack \) has a vertex of degree at most 1 we move this vertex to \( {V}_{1} \) , obtaining a new partition \( \left\{ {{V}_{1},{V}_{2}}\right\} \) crossed by fewer edges. Suppose we can perform a sequence of \( n \) such moves, but no more. Then the resulting partition \( \left\{ {{V}_{1},{V}_{2}}\right\} \) is crossed by at most \( m - n \) edges. And \( H\left\lbrack {V}_{2}\right\rbrack \) has at most \( m - n \) vertices of degree less than 3, because each of these is incident with a cut edge. These vertices have degree exactly 2 in \( H\left\lbrack {V}_{2}\right\rbrack \), since we could not move them to \( {V}_{1} \) . Let \( {H}^{\prime } \) be the cubic multigraph obtained from \( H\left\lbrack {V}_{2}\right\rbrack \) by suppressing these vertices. Then\n\n\[ \left| {H}^{\prime }\right| \geq \left| H\right| - \left| C\right| - n - \left( {m - n}\right) \geq \left| H\right| - 2\left| C\right| \]\n\nas desired.\n\nTo complete the proof, it suffices to show that \( \left| {H}^{\prime }\right| \geq {s}_{k - 1} \) . Since \( \left| C\right| \leq 2\log \left| H\right| \) by Corollary 1.3.5 (or by \( \left| H\right| \geq {s}_{k} \), if \( \left| C\right| = g\left( H\right) \leq 2 \) ), and \( \left| H\right| \geq {s}_{k} \geq 6 \), we have\n\n\[ \left| {H}^{\prime }\right| \geq \left| H\right| - 2\left| C\right| \geq \left| H\right| - 4\log \left| H\right| \geq {s}_{k} - 4\log {s}_{k}. \]\n\n(In the last inequality we use that the function \( x \mapsto x - 4\log x \) increases for \( x \geq 6 \) .)\n\nIt thus remains to show that \( {s}_{k} - 4\log {s}_{k} \geq {s}_{k - 1} \) . For \( k = 2 \) this is clear, so we assume that \( k \geq 3 \) . Then \( {r}_{k} \leq 4\log k \) (which is obvious for \( k \geq 4 \), while the case of \( k = 3 \) has to be calculated), and hence\n\n\[ {s}_{k} - 4\log {s}_{k} = 4\left( {k - 1}\right) {r}_{k} + 4\log k + 4\log \log k + {16} \]\n\n\[ - \left( {8 + 4\log k + 4\log {r}_{k}}\right) \]\n\n\[ \geq {s}_{k - 1} + 4\log \log k + 8 - 4\log \left( {4\log k}\right) \]\n\n\[ = {s}_{k - 1}\text{.} \] | Yes |
There is a function \( f : \mathbb{N} \rightarrow \mathbb{R} \) such that, given any \( k \in \mathbb{N} \), every graph contains either \( k \) disjoint cycles or a set of at most \( f\left( k\right) \) vertices meeting all its cycles. | We show the result for \( f\left( k\right) \mathrel{\text{:=}} {s}_{k} + k - 1 \) . Let \( k \) be given, and let \( G \) be any graph. We may assume that \( G \) contains a cycle, and so it has a maximal subgraph \( H \) in which every vertex has degree 2 or 3 . Let \( U \) be its set of degree 3 vertices.\n\nLet \( \mathcal{C} \) be the set of all cycles in \( G \) that avoid \( U \) and meet \( H \) in exactly one vertex. Let \( Z \subseteq V\left( H\right) \smallsetminus U \) be the set of those vertices. For each \( z \in Z \) pick a cycle \( {C}_{z} \in \mathcal{C} \) that meets \( H \) in \( z \), and put \( {\mathcal{C}}^{\prime } \mathrel{\text{:=}} \left\{ {{C}_{z} \mid z \in Z}\right\} \) . By the maximality of \( H \), the cycles in \( {\mathcal{C}}^{\prime } \) are disjoint.\n\nLet \( \mathcal{D} \) be the set of the 2-regular components of \( H \) that avoid \( Z \) . Then \( {\mathcal{C}}^{\prime } \cup \mathcal{D} \) is another set of disjoint cycles. If \( \left| {{\mathcal{C}}^{\prime } \cup \mathcal{D}}\right| \geq k \), we are done. Otherwise we can add to \( Z \) one vertex from each cycle in \( \mathcal{D} \) to obtain a set \( X \) of at most \( k - 1 \) vertices that meets all the cycles in \( \mathcal{C} \) and all the 2-regular components of \( H \) . Now consider any cycle of \( G \) that avoids \( X \) . By the maximality of \( H \) it meets \( H \) . But it is not a component of \( H \), it does not lie in \( \mathcal{C} \), and it does not contain an \( H \) -path between distinct vertices outside \( U \) (by the maximality of \( H \) ). So this cycle meets \( U \).\n\nWe have shown that every cycle in \( G \) meets \( X \cup U \) . As \( \left| X\right| \leq k - 1 \) , it thus suffices to show that \( \left| U\right| < {s}_{k} \) unless \( H \) contains \( k \) disjoint cycles. But this follows from Lemma 2.3.1 applied to the graph obtained from \( H \) by suppressing its vertices of degree 2 . | Yes |
A multigraph contains \( k \) edge-disjoint spanning trees if and only if for every partition \( P \) of its vertex set it has at least \( k\left( {\left| P\right| - 1}\right) \) cross-edges. | For the proof of Theorem 2.4.1, let a multigraph \( G = \left( {V, E}\right) \) and \( G = \left( {V, E}\right) \) \( k \in \mathbb{N} \) be given. Let \( \mathcal{F} \) be the set of all \( k \) -tuples \( F = \left( {{F}_{1},\ldots ,{F}_{k}}\right) \) of edge-disjoint spanning forests in \( G \) with the maximum total number of edges, i.e. such that \( \parallel F\parallel \mathrel{\text{:=}} \left| {E\left\lbrack F\right\rbrack }\right| \) with \( E\left\lbrack F\right\rbrack \mathrel{\text{:=}} E\left( {F}_{1}\right) \cup \ldots \cup E\left( {F}_{k}\right) \) is as large as possible. If \( F = \left( {{F}_{1},\ldots ,{F}_{k}}\right) \in \mathcal{F} \) and \( e \in E \smallsetminus E\left\lbrack F\right\rbrack \), then every \( {F}_{i} + e \) contains a cycle \( \left( {i = 1,\ldots, k}\right) \) : otherwise we could replace \( {F}_{i} \) by \( {F}_{i} + e \) in \( F \) and obtain a contradiction to the maximality of \( \parallel F\parallel \) . Let us consider an edge \( {e}^{\prime } \neq e \) of this cycle, for some fixed \( i \) . Putting \( {F}_{i}^{\prime } \mathrel{\text{:=}} {F}_{i} + e - {e}^{\prime } \) , and \( {F}_{j}^{\prime } \mathrel{\text{:=}} {F}_{j} \) for all \( j \neq i \), we see that \( {F}^{\prime } \mathrel{\text{:=}} \left( {{F}_{1}^{\prime },\ldots ,{F}_{k}^{\prime }}\right) \) is again in \( \mathcal{F} \) ; we say that \( {F}^{\prime } \) has been obtained from \( F \) by the replacement of the edge \( {e}^{\prime } \) with \( e \) . Note that the component of \( {F}_{i} \) containing \( {e}^{\prime } \) keeps its vertex set when it changes into a component of \( {F}_{i}^{\prime } \) . Hence for every path \( x\ldots y \subseteq {F}_{i}^{\prime } \) there is a unique path \( x{F}_{i}y \) in \( {F}_{i} \) ; this will be used later. We now consider a fixed \( k \) -tuple \( {F}^{0} = \left( {{F}_{1}^{0},\ldots ,{F}_{k}^{0}}\right) \in \mathcal{F} \) . The set \( {F}^{0} \) of all \( k \) -tuples in \( \mathcal{F} \) that can be obtained from \( {F}^{0} \) by a series of edge replacements will be denoted by \( {\mathcal{F}}^{0} \) . Finally, we let \[ {E}^{0} \mathrel{\text{:=}} \mathop{\bigcup }\limits_{{F \in {\mathcal{F}}^{0}}}\left( {E \smallsetminus E\left\lbrack F\right\rbrack }\right) \] and \( {G}^{0} \mathrel{\text{:=}} \left( {V,{E}^{0}}\right) \) . Lemma 2.4.3. For every \( {e}^{0} \in E \smallsetminus E\left\lbrack {F}^{0}\right\rbrack \) there exists a set \( U \subseteq V \) that is connected in every \( {F}_{i}^{0}\left( {i = 1,\ldots, k}\right) \) and contains the ends of \( {e}^{0} \) . Proof. As \( {F}^{0} \in {\mathcal{F}}^{0} \), we have \( {e}^{0} \in {E}^{0} \) ; let \( {C}^{0} \) be the component of \( {G}^{0} \) containing \( {e}^{0} \) . We shall prove the assertion for \( U \mathrel{\text{:=}} V\left( {C}^{0}\right) \) . | Yes |
Corollary 2.4.2. Every \( {2k} \) -edge-connected multigraph \( G \) has \( k \) edge-disjoint spanning trees. | Proof. Every set in a vertex partition of \( G \) is joined to other partition sets by at least \( {2k} \) edges. Hence, for any partition into \( r \) sets, \( G \) has at least \( \frac{1}{2}\mathop{\sum }\limits_{{i = 1}}^{r}{2k} = {kr} \) cross-edges. The assertion thus follows from Theorem 2.4.1. | Yes |
Lemma 2.4.3. For every \( {e}^{0} \in E \smallsetminus E\left\lbrack {F}^{0}\right\rbrack \) there exists a set \( U \subseteq V \) that is connected in every \( {F}_{i}^{0}\left( {i = 1,\ldots, k}\right) \) and contains the ends of \( {e}^{0} \) . | Proof. As \( {F}^{0} \in {\mathcal{F}}^{0} \), we have \( {e}^{0} \in {E}^{0} \) ; let \( {C}^{0} \) be the component of \( {G}^{0} \) \( {C}^{0} \) containing \( {e}^{0} \) . We shall prove the assertion for \( U \mathrel{\text{:=}} V\left( {C}^{0}\right) \) . \n\nLet \( i \in \{ 1,\ldots, k\} \) be given; we have to show that \( U \) is connected in \( {F}_{i}^{0} \) . To this end, we first prove the following: \n\nLet \( F = \left( {{F}_{1},\ldots ,{F}_{k}}\right) \in {\mathcal{F}}^{0} \), and let \( \left( {{F}_{1}^{\prime },\ldots ,{F}_{k}^{\prime }}\right) \) have been \n\nobtained from \( F \) by the replacement of an edge of \( {F}_{i} \) . If \n\n(1) \n\n\( x, y \) are the ends of a path in \( {F}_{i}^{\prime } \cap {C}^{0} \), then also \( x{F}_{i}y \subseteq {C}^{0} \) . \n\nLet \( e = {vw} \) be the new edge in \( E\left( {F}_{i}^{\prime }\right) \smallsetminus E\left\lbrack F\right\rbrack \) ; this is the only edge of \( {F}_{i}^{\prime } \) not lying in \( {F}_{i} \) . We assume that \( e \in x{F}_{i}^{\prime }y \) : otherwise we would have \( x{F}_{i}y = x{F}_{i}^{\prime }y \) and nothing to show. It suffices to show that \( v{F}_{i}w \subseteq {C}^{0} \) : then \( \left( {x{F}_{i}^{\prime }y - e}\right) \cup v{F}_{i}w \) is a connected subgraph of \( {F}_{i} \cap {C}^{0} \) that contains \( x, y \), and hence also \( x{F}_{i}y \) . Let \( {e}^{\prime } \) be any edge of \( v{F}_{i}w \) . Since we could replace \( {e}^{\prime } \) in \( F \in {\mathcal{F}}^{0} \) by \( e \) and obtain an element of \( {\mathcal{F}}^{0} \) not containing \( {e}^{\prime } \), we have \( {e}^{\prime } \in {E}^{0} \) . Thus \( v{F}_{i}w \subseteq {G}^{0} \), and hence \( v{F}_{i}w \subseteq {C}^{0} \) since \( v, w \in x{F}_{i}^{\prime }y \subseteq {C}^{0} \) . This proves (1). \n\nIn order to prove that \( U = V\left( {C}^{0}\right) \) is connected in \( {F}_{i}^{0} \) we show that, for every edge \( {xy} \in {C}^{0} \), the path \( x{F}_{i}^{0}y \) exists and lies in \( {C}^{0} \) . As \( {C}^{0} \) is connected, the union of all these paths will then be a connected spanning subgraph of \( {F}_{i}^{0}\left\lbrack U\right\rbrack \) . \n\nSo let \( e = {xy} \in {C}^{0} \) be given. As \( e \in {E}^{0} \), there exist an \( s \in \mathbb{N} \) and \( k \) -tuples \( {F}^{r} = \left( {{F}_{1}^{r},\ldots ,{F}_{k}^{r}}\right) \) for \( r = 1,\ldots, s \) such that each \( {F}^{r} \) is obtained from \( {F}^{r - 1} \) by edge replacement and \( e \in E \smallsetminus E\left\lbrack {F}^{s}\right\rbrack \) . Setting \( F \mathrel{\text{:=}} {F}^{s} \) in (1), we may think of \( e \) as a path of length 1 in \( {F}_{i}^{\prime } \cap {C}^{0} \) . Successive applications of (1) to \( F = {F}^{s},\ldots ,{F}^{0} \) then give \( x{F}_{i}^{0}y \subseteq {C}^{0} \) as desired. | Yes |
Theorem 2.4.4. (Nash-Williams 1964)\n\n\( A \) multigraph \( G = \left( {V, E}\right) \) can be partitioned into at most \( k \) forests if and only if \( \parallel G\left\lbrack U\right\rbrack \parallel \leq k\left( {\left| U\right| - 1}\right) \) for every non-empty set \( U \subseteq V \) . | Proof. The forward implication was shown above. Conversely, we show\n\n\( \left( {1.5.3}\right) \)\n\nthat every \( k \) -tuple \( F = \left( {{F}_{1},\ldots ,{F}_{k}}\right) \in \mathcal{F} \) partitions \( G \), i.e. that \( E\left\lbrack F\right\rbrack = \) \( E \) . If not, let \( e \in E \smallsetminus E\left\lbrack F\right\rbrack \) . By Lemma 2.4.3, there exists a set \( U \subseteq V \) that is connected in every \( {F}_{i} \) and contains the ends of \( e \) . Then \( G\left\lbrack U\right\rbrack \) contains \( \left| U\right| - 1 \) edges from each \( {F}_{i} \), and in addition the edge \( e \) . Thus \( \parallel G\left\lbrack U\right\rbrack \parallel > k\left( {\left| U\right| - 1}\right) \), contrary to our assumption. | Yes |
Every directed graph \( G \) has a path cover \( \mathcal{P} \) and an independent set \( \left\{ {{v}_{P} \mid P \in \mathcal{P}}\right\} \) of vertices such that \( {v}_{P} \in P \) for every \( P \in \mathcal{P} \) . | We prove by induction on \( \left| G\right| \) that for every path cover \( \mathcal{P} = \) \( {P}_{i} \) \( \left\{ {{P}_{1},\ldots ,{P}_{m}}\right\} \) of \( G \) with \( \operatorname{ter}\left( \mathcal{P}\right) \) minimal there is a set \( \left\{ {{v}_{P} \mid P \in \mathcal{P}}\right\} \) as \( {v}_{i} \) claimed. For each \( i \), let \( {v}_{i} \) denote the last vertex of \( {P}_{i} \) .\n\nIf \( \operatorname{ter}\left( \mathcal{P}\right) = \left\{ {{v}_{1},\ldots ,{v}_{m}}\right\} \) is independent there is nothing more to show, so we assume that \( G \) has an edge from \( {v}_{2} \) to \( {v}_{1} \) . Since \( {P}_{2}{v}_{2}{v}_{1} \) is again a path, the minimality of \( \operatorname{ter}\left( \mathcal{P}\right) \) implies that \( {v}_{1} \) is not the only vertex of \( {P}_{1} \) ; let \( v \) be the vertex preceding \( {v}_{1} \) on \( {P}_{1} \) . Then \( {\mathcal{P}}^{\prime } \mathrel{\text{:=}} \) \( {\mathcal{P}}^{\prime },{G}^{\prime } \) \( \left\{ {{P}_{1}v,{P}_{2},\ldots ,{P}_{m}}\right\} \) is a path cover of \( {G}^{\prime } \mathrel{\text{:=}} G - {v}_{1} \) (Fig. 2.5.1). Clearly, any independent set of representatives for \( {\mathcal{P}}^{\prime } \) in \( {G}^{\prime } \) will also work for \( \mathcal{P} \) in \( G \), so all we have to check is that we may apply the induction hypothesis to \( {\mathcal{P}}^{\prime } \) . It thus remains to show that \( \operatorname{ter}\left( {\mathcal{P}}^{\prime }\right) = \left\{ {v,{v}_{2},\ldots ,{v}_{m}}\right\} \) is minimal among the sets of last vertices of path covers of \( {G}^{\prime } \) .\n\nSuppose then that \( {G}^{\prime } \) has a path cover \( {\mathcal{P}}^{\prime \prime } \) with \( \operatorname{ter}\left( {\mathcal{P}}^{\prime \prime }\right) \subsetneqq \operatorname{ter}\left( {\mathcal{P}}^{\prime }\right) \) . If a path \( P \in {\mathcal{P}}^{\prime \prime } \) ends in \( v \), we may replace \( P \) in \( {\mathcal{P}}^{\prime \prime } \) by \( {Pv}{v}_{1} \) to obtain a path cover of \( G \) whose set of last vertices is a proper subset of \( \operatorname{ter}\left( \mathcal{P}\right) \) , contradicting the choice of \( \mathcal{P} \) . If a path \( P \in {\mathcal{P}}^{\prime \prime } \) ends in \( {v}_{2} \) (but none in \( v \) ), we similarly replace \( P \) in \( {\mathcal{P}}^{\prime \prime } \) by \( P{v}_{2}{v}_{1} \) to obtain a contradiction to the minimality of \( \operatorname{ter}\left( \mathcal{P}\right) \) . Hence \( \operatorname{ter}\left( {\mathcal{P}}^{\prime \prime }\right) \subseteq \left\{ {{v}_{3},\ldots ,{v}_{m}}\right\} \) . But now \( {\mathcal{P}}^{\prime \prime } \) and the trivial path \( \left\{ {v}_{1}\right\} \) together form a path cover of \( G \) that contradicts the minimality of \( \operatorname{ter}\left( \mathcal{P}\right) \) . | Yes |
In every finite partially ordered set \( \left( {P, \leq }\right) \), the minimum number of chains with union \( P \) is equal to the maximum cardinality of an antichain in \( P \) . | If \( A \) is an antichain in \( P \) of maximum cardinality, then clearly \( P \) cannot be covered by fewer than \( \left| A\right| \) chains. The fact that \( \left| A\right| \) chains will suffice follows from Theorem 2.5.1 applied to the directed graph on \( P \) with the edge set \( \{ \left( {x, y}\right) \mid x < y\} \) . | No |
Proposition 3.1.3. A graph is 2-connected if and only if it can be constructed from a cycle by successively adding \( H \) -paths to graphs \( H \) already constructed (Fig. 3.1.2). | Proof. Clearly, every graph constructed as described is 2-connected. Conversely, let a 2-connected graph \( G \) be given. Then \( G \) contains a cycle, and hence has a maximal subgraph \( H \) constructible as above. Since any edge \( {xy} \in E\left( G\right) \smallsetminus E\left( H\right) \) with \( x, y \in H \) would define an \( H \) - path, \( H \) is an induced subgraph of \( G \) . Thus if \( H \neq G \), then by the connectedness of \( G \) there is an edge \( {vw} \) with \( v \in G - H \) and \( w \in H \) . As \( G \) is 2-connected, \( G - w \) contains a \( v - H \) path \( P \) . Then \( {wvP} \) is an \( H \) -path in \( G \), and \( H \cup {wvP} \) is a constructible subgraph of \( G \) larger than \( H \) . This contradicts the maximality of \( H \) . | Yes |
A graph \( G \) is 3-connected if and only if there exists a sequence \( {G}_{0},\ldots ,{G}_{n} \) of graphs with the following properties:\n\n(i) \( {G}_{0} = {K}^{4} \) and \( {G}_{n} = G \) ;\n\n(ii) \( {G}_{i + 1} \) has an edge \( {xy} \) with \( d\left( x\right), d\left( y\right) \geq 3 \) and \( {G}_{i} = {G}_{i + 1}/{xy} \), for every \( i < n \) . | If \( G \) is 3-connected, a sequence as in the theorem exists by Lemma 3.2.1. Note that all the graphs in this sequence are 3-connected.\n\nConversely, let \( {G}_{0},\ldots ,{G}_{n} \) be a sequence of graphs as stated; we show that if \( {G}_{i} = {G}_{i + 1}/{xy} \) is 3-connected then so is \( {G}_{i + 1} \), for every \( i < n \) . Suppose not, let \( S \) be a separator of at most 2 vertices in \( {G}_{i + 1} \), and let \( {C}_{1},{C}_{2} \) be two components of \( {G}_{i + 1} - S \) . As \( x \) and \( y \) are adjacent, we may \( {C}_{1},{C}_{2} \) assume that \( \{ x, y\} \cap V\left( {C}_{1}\right) = \varnothing \) (Fig. 3.2.2). Then \( {C}_{2} \) contains neither\n\n\n\nFig. 3.2.2. The position of \( {xy} \in {G}_{i + 1} \) in the proof of Theorem 3.2.2\n\nboth vertices \( x, y \) nor a vertex \( v \notin \{ x, y\} \) : otherwise \( {v}_{xy} \) or \( v \) would be separated from \( {C}_{1} \) in \( {G}_{i} \) by at most two vertices, a contradiction. But now \( {C}_{2} \) contains only one vertex: either \( x \) or \( y \) . This contradicts our assumption of \( d\left( x\right), d\left( y\right) \geq 3 \) . | Yes |
Let \( G = \left( {V, E}\right) \) be a graph and \( A, B \subseteq V \) . Then the minimum number of vertices separating \( A \) from \( B \) in \( G \) is equal to the maximum number of disjoint \( A - B \) paths in \( G \) . | We apply induction on \( \parallel G\parallel \) . If \( G \) has no edge, then \( \left| {A \cap B}\right| = k \) and we have \( k \) trivial \( A - B \) paths. So we assume that \( G \) has an edge \( e = {xy} \) . If \( G \) has no \( k \) disjoint \( A - B \) paths, then neither does \( G/e \) ; here, we count the contracted vertex \( {v}_{e} \) as an element of \( A \) (resp. \( B \) ) in \( G/e \) if in \( G \) at least one of \( x, y \) lies in \( A \) (resp. \( B \) ). By the induction hypothesis, \( G/e \) contains an \( A - B \) separator \( Y \) of fewer than \( k \) vertices. Among these must be the vertex \( {v}_{e} \), since otherwise \( Y \subseteq V \) would be an \( A - B \) separator in \( G \) . Then \( X \mathrel{\text{:=}} \left( {Y \smallsetminus \left\{ {v}_{e}\right\} }\right) \cup \{ x, y\} \) is an \( A - B \) separator in \( G \) of exactly \( k \) vertices.\n\nWe now consider the graph \( G - e \) . Since \( x, y \in X \), every \( A - X \) separator in \( G - e \) is also an \( A - B \) separator in \( G \) and hence contains at least \( k \) vertices. So by induction there are \( k \) disjoint \( A - X \) paths in \( G - e \) , and similarly there are \( k \) disjoint \( X - B \) paths in \( G - e \) . As \( X \) separates \( A \) from \( B \), these two path systems do not meet outside \( X \), and can thus be combined to \( k \) disjoint \( A - B \) paths. | Yes |
Lemma 3.3.2. If an alternating walk \( W \) as above ends in \( B \smallsetminus V\left\lbrack \mathcal{P}\right\rbrack \) , then \( G \) contains a set of disjoint \( A - B \) paths exceeding \( \mathcal{P} \) . | Proof. We may assume that \( W \) has only its first vertex in \( A \smallsetminus V\left\lbrack \mathcal{P}\right\rbrack \) and only its last vertex in \( B \smallsetminus V\left\lbrack \mathcal{P}\right\rbrack \) . Let \( H \) be the graph on \( V\left( G\right) \) whose edge set is the symmetric difference of \( E\left\lbrack \mathcal{P}\right\rbrack \) with \( \left\{ {{e}_{0},\ldots ,{e}_{n - 1}}\right\} \) . In \( H \) , the ends of the paths in \( \mathcal{P} \) and of \( W \) have degree 1 (or 0, if the path or \( W \) is trivial), and all other vertices have degree 0 or 2 . For each vertex \( a \in \left( {A \cap V\left\lbrack \mathcal{P}\right\rbrack }\right) \cup \left\{ {x}_{0}\right\} \), therefore, the component of \( H \) containing \( a \) is a path, \( P = {v}_{0}\ldots {v}_{k} \) say, which starts in \( a \) and ends in \( A \) or \( B \) . Using conditions (i) and (iii), one easily shows by induction on \( i = 0,\ldots, k - 1 \) that \( P \) traverses each of its edges \( e = {v}_{i}{v}_{i + 1} \) in the forward direction with respect to \( \mathcal{P} \) or \( W \) . (Formally: if \( e \in {P}^{\prime } \) with \( \left. {{P}^{\prime } \in \mathcal{P}\text{, then }{v}_{i} \in {P}^{\prime }{\mathring{v}}_{i + 1}\text{; if }e = {e}_{j} \in W\text{, then }{v}_{i} = {x}_{j}\text{ and }{v}_{i + 1} = {x}_{j + 1}\text{. }}\right) \) Hence, \( P \) is an \( A - B \) path. Similarly, for every \( b \in \left( {B \cap V\left\lbrack \mathcal{P}\right\rbrack }\right) \cup \left\{ {x}_{n}\right\} \) there is an \( A - B \) path in \( H \) that ends in \( b \) . The set of \( A - B \) paths in \( H \) therefore exceeds \( \mathcal{P} \). | Yes |
Corollary 3.3.5. Let \( a \) and \( b \) be two distinct vertices of \( G \) .\n\n(i) If \( {ab} \notin E \), then the minimum number of vertices \( \neq a, b \) separating \( a \) from \( b \) in \( G \) is equal to the maximum number of independent \( a - b \) paths in \( G \) .\n\n(ii) The minimum number of edges separating \( a \) from \( b \) in \( G \) is equal to the maximum number of edge-disjoint \( a - b \) paths in \( G \) . | Proof. (i) Apply Theorem 3.3.1 with \( A \mathrel{\text{:=}} N\left( a\right) \) and \( B \mathrel{\text{:=}} N\left( b\right) \) .\n\n(ii) Apply Theorem 3.3.1 to the line graph of \( G \), with \( A \mathrel{\text{:=}} E\left( a\right) \) and \( B \mathrel{\text{:=}} E\left( b\right) \) . | Yes |
Theorem 3.3.6. (Global Version of Menger's Theorem) \( \\left\\lbrack \\begin{array}{l} {4.2.7} \\ {6.6.1} \\ {9.4.2} \\end{array}\\right\\rbrack \) (i) A graph is \( k \) -connected if and only if it contains \( k \) independent paths between any two vertices.\n\n(ii) A graph is \( k \) -edge-connected if and only if it contains \( k \) edge-disjoint paths between any two vertices. | Proof. (i) If a graph \( G \) contains \( k \) independent paths between any two vertices, then \( \\left| G\\right| > k \) and \( G \) cannot be separated by fewer than \( k \) vertices; thus, \( G \) is \( k \) -connected.\n\nConversely, suppose that \( G \) is \( k \) -connected (and, in particular, has more than \( k \) vertices) but contains vertices \( a, b \) not linked by \( k \) independent paths. By Corollary 3.3.5 (i), \( a \) and \( b \) are adjacent; let \( {G}^{\\prime } \\mathrel{\\text{:=}} G - {ab} \) . Then \( {G}^{\\prime } \) contains at most \( k - 2 \) independent \( a - b \) paths. By Corollary 3.3.5 (i), we can separate \( a \) and \( b \) in \( {G}^{\\prime } \) by a set \( X \) of at most \( k - 2 \) vertices. As \( \\left| G\\right| > k \), there is at least one further vertex \( v \\notin X \\cup \\{ a, b\\} \) in \( G \) . Now \( X \) separates \( v \) in \( {G}^{\\prime } \) from either \( a \) or \( b \) -say, from \( a \) . But then \( X \\cup \\{ b\\} \) is a set of at most \( k - 1 \) vertices separating \( v \) from \( a \) in \( G \) , contradicting the \( k \) -connectedness of \( G \).\n\n(ii) follows straight from Corollary 3.3.5 (ii). | Yes |
Corollary 3.4.2. Given a graph \( G \) with an induced subgraph \( H \), there are at least \( \frac{1}{2}{\kappa }_{G}\left( H\right) \) independent \( H \) -paths and at least \( \frac{1}{2}{\lambda }_{G}\left( H\right) \) edge-disjoint \( H \) -paths in \( G \) . | Proof. To prove the first assertion, let \( k \) be the maximum number of independent \( H \) -paths in \( G \) . By Theorem 3.4.1, there are sets \( X \subseteq V\left( {G - H}\right) \) and \( F \subseteq E\left( {G - H - X}\right) \) with\n\n\[ k = \left| X\right| + \mathop{\sum }\limits_{{C \in {\mathcal{C}}_{F}}}\left\lfloor {\frac{1}{2}\left| {\partial C}\right| }\right\rfloor \]\n\nsuch that every \( H \) -path in \( G \) has a vertex in \( X \) or an edge in \( F \) . For every \( C \in {\mathcal{C}}_{F} \) with \( \partial C \neq \varnothing \), pick a vertex \( v \in \partial C \) and let \( {Y}_{C} \mathrel{\text{:=}} \partial C \smallsetminus \{ v\} \) ; if \( \partial C = \varnothing \), let \( {Y}_{C} \mathrel{\text{:=}} \varnothing \) . Then \( \left\lfloor {\frac{1}{2}\left| {\partial C}\right| }\right\rfloor \geq \frac{1}{2}\left| {Y}_{C}\right| \) for all \( C \in {\mathcal{C}}_{F} \) . Moreover, for \( Y \mathrel{\text{:=}} \mathop{\bigcup }\limits_{{C \in {\mathcal{C}}_{F}}}{Y}_{C} \) every \( H \) -path has a vertex in \( X \cup Y \) . Hence\n\n\[ k \geq \left| X\right| + \mathop{\sum }\limits_{{C \in {\mathcal{C}}_{F}}}\frac{1}{2}\left| {Y}_{C}\right| \geq \frac{1}{2}\left| {X \cup Y}\right| \geq \frac{1}{2}{\kappa }_{G}\left( H\right) \]\n\nas claimed.\n\nThe second assertion follows from the first by considering the line graph of \( G \) (Exercise 18). | Yes |
Lemma 3.5.1. There is a function \( h : \mathbb{N} \rightarrow \mathbb{N} \) such that every graph of average degree at least \( h\left( r\right) \) contains \( {K}^{r} \) as a topological minor, for every \( r \in \mathbb{N} \) . | Proof. For \( r \leq 2 \), the assertion holds with \( h\left( r\right) = 1 \) ; we now assume that \( r \geq 3 \) . We show by induction on \( m = r,\ldots ,\left( \begin{array}{l} r \\ 2 \end{array}\right) \) that every graph \( G \) with average degree \( d\left( G\right) \geq {2}^{m} \) has a topological minor \( X \) with \( r \) vertices and \( m \) edges; for \( m = \left( \begin{array}{l} r \\ 2 \end{array}\right) \) this implies the assertion with \( h\left( r\right) = {2}^{\left( \begin{array}{l} r \\ 2 \end{array}\right) } \) .\n\nIf \( m = r \) then, by Propositions 1.2.2 and 1.3.1, \( G \) contains a cycle of length at least \( \varepsilon \left( G\right) + 1 \geq {2}^{r - 1} + 1 \geq r + 1 \), and the assertion follows with \( X = {C}^{r} \) .\n\nNow let \( r < m \leq \left( \begin{array}{l} r \\ 2 \end{array}\right) \), and assume the assertion holds for smaller \( m \) . Let \( G \) with \( d\left( G\right) \geq {2}^{m} \) be given; thus, \( \varepsilon \left( G\right) \geq {2}^{m - 1} \) . Since \( G \) has a component \( C \) with \( \varepsilon \left( C\right) \geq \varepsilon \left( G\right) \), we may assume that \( G \) is connected. Consider a maximal set \( U \subseteq V\left( G\right) \) such that \( U \) is connected in \( G \) and \( \varepsilon \left( {G/U}\right) \geq {2}^{m - 1} \) ; such a set \( U \) exists, because \( G \) itself has the form \( G/U \) with \( \left| U\right| = 1 \) . Since \( G \) is connected, we have \( N\left( U\right) \neq \varnothing \) .\n\nLet \( H \mathrel{\text{:=}} G\left\lbrack {N\left( U\right) }\right\rbrack \) . If \( H \) has a vertex \( v \) of degree \( {d}_{H}\left( v\right) < {2}^{m - 1} \), we may add it to \( U \) and obtain a contradiction to the maximality of \( U \) : when we contract the edge \( v{v}_{U} \) in \( G/U \), we lose one vertex and \( {d}_{H}\left( v\right) + 1 \leq \) \( {2}^{m - 1} \) edges, so \( \varepsilon \) will still be at least \( {2}^{m - 1} \) . Therefore \( d\left( H\right) \geq \delta \left( H\right) \geq \) \( {2}^{m - 1} \) . By the induction hypothesis, \( H \) contains a \( {TY} \) with \( \left| Y\right| = r \) and \( \parallel Y\parallel = m - 1 \) . Let \( x, y \) be two branch vertices of this \( {TY} \) that are non-adjacent in \( Y \) . Since \( x \) and \( y \) lie in \( N\left( U\right) \) and \( U \) is connected in \( G \) , \( G \) contains an \( x - y \) path whose inner vertices lie in \( U \) . Adding this path to the \( {TY} \), we obtain the desired \( {TX} \) . | Yes |
There is a function \( f : \mathbb{N} \rightarrow \mathbb{N} \) such that every \( f\left( k\right) \) -connected graph is \( k \) -linked, for all \( k \in \mathbb{N} \) . | Proof. We prove the assertion for \( f\left( k\right) = h\left( {3k}\right) + {2k} \), where \( h \) is a \( G \) function as in Lemma 3.5.1. Let \( G \) be an \( f\left( k\right) \) -connected graph. Then \( K \) \( d\left( G\right) \geq \delta \left( G\right) \geq \kappa \left( G\right) \geq h\left( {3k}\right) \) ; choose \( K = T{K}^{3k} \subseteq G \) as in Lemma 3.5.1, \( U \) and let \( U \) denote its set of branch vertices.\n\n\( {s}_{i},{t}_{i} \) For the proof that \( G \) is \( k \) -linked, let distinct vertices \( {s}_{1},\ldots ,{s}_{k} \) and \( {t}_{1},\ldots ,{t}_{k} \) be given. By definition of \( f\left( k\right) \), we have \( \kappa \left( G\right) \geq {2k} \) . Hence by Menger’s theorem (3.3.1), \( G \) contains disjoint paths \( {P}_{1},\ldots ,{P}_{k} \) , \( {P}_{i},{Q}_{i} \) \( {Q}_{1},\ldots ,{Q}_{k} \), such that each \( {P}_{i} \) starts in \( {s}_{i} \), each \( {Q}_{i} \) starts in \( {t}_{i} \), and all\n\n\( \mathcal{P} \) these paths end in \( U \) but have no inner vertices in \( U \) . Let the set \( \mathcal{P} \) of these paths be chosen so that their total number of edges outside \( E\left( K\right) \) is as small as possible.\n\nLet \( {u}_{1},\ldots ,{u}_{k} \) be those \( k \) vertices in \( U \) that are not an end of a path in \( \mathcal{P} \) . For each \( i = 1,\ldots, k \), let \( {L}_{i} \) be the \( U \) -path in \( K \) (i.e., the subdivided edge of the \( {K}^{3k} \) ) from \( {u}_{i} \) to the end of \( {P}_{i} \) in \( U \), and let \( {v}_{i} \) be the first vertex of \( {L}_{i} \) on any path \( P \in \mathcal{P} \) . By definition of \( \mathcal{P}, P \) has no more edges outside \( E\left( K\right) \) than \( P{v}_{i}{L}_{i}{u}_{i} \) does, so \( {v}_{i}P = {v}_{i}{L}_{i} \) and hence \( P = {P}_{i} \) (Fig. 3.5.1). Similarly, if \( {M}_{i} \) denotes the \( U \) -path in \( K \) from \( {u}_{i} \) to the end of \( {Q}_{i} \) in \( U \), and \( {w}_{i} \) denotes the first vertex of \( {M}_{i} \) on any path in \( \mathcal{P} \), then this path is \( {Q}_{i} \) . Then the paths \( {s}_{i}{P}_{i}{v}_{i}{L}_{i}{u}_{i}{M}_{i}{w}_{i}{Q}_{i}{t}_{i} \) are disjoint for different \( i \) and show that \( G \) is \( k \) -linked. | Yes |
Lemma 4.2.1. Let \( G \) be a plane graph, \( f \in F\left( G\right) \) a face, and \( H \subseteq G \) a subgraph.\n\n(i) \( H \) has a face \( {f}^{\prime } \) containing \( f \) .\n\n(ii) If the frontier of \( f \) lies in \( H \), then \( {f}^{\prime } = f \) . | Proof. (i) Clearly, the points in \( f \) are equivalent also in \( {\mathbb{R}}^{2} \smallsetminus H \) ; let \( {f}^{\prime } \) be the equivalence class of \( {\mathbb{R}}^{2} \smallsetminus H \) containing them.\n\n(ii) Recall from Section 4.1 that any arc between \( f \) and \( {f}^{\prime } \smallsetminus f \) meets the frontier \( X \) of \( f \) . If \( {f}^{\prime } \smallsetminus f \neq \varnothing \) then there is such an arc inside \( {f}^{\prime } \) , whose points in \( X \) do not lie in \( H \) . Hence \( X \nsubseteq H \) . | Yes |
Lemma 4.2.2. Let \( G \) be a plane graph and \( e \) an edge of \( G \). \( \begin{array}{r} \left\lbrack {4.5.1}\right\rbrack \\ \left\lbrack {4.5.2}\right\rbrack \\ \left\lbrack {12.5.4}\right\rbrack \end{array} \) (i) If \( X \) is the frontier of a face of \( G \), then either \( e \subseteq X \) or \( X \cap \overset{ \circ }{e} = \varnothing \). (ii) If \( e \) lies on a cycle \( C \subseteq G \), then \( e \) lies on the frontier of exactly two faces of \( G \), and these are contained in distinct faces of \( C \). (iii) If \( e \) lies on no cycle, then \( e \) lies on the frontier of exactly one face of \( G \). | Proof. We prove all three assertions together. Let us start by considering one point \( {x}_{0} \in e \). We show that \( {x}_{0} \) lies on the frontier of either exactly two faces or exactly one, according as \( e \) lies on a cycle in \( G \) or not. We then show that every other point in \( e \) lies on the frontier of exactly the same faces as \( {x}_{0} \). Then the endpoints of \( e \) will also lie on the frontier of these faces - simply because every neighbourhood of an endpoint of \( e \) is also the neighbourhood of an inner point of \( e \).\n\n\( G \) is the union of finitely many straight line segments; we may assume that any two of these intersect in at most one point. Around every point \( x \in \overset{ \circ }{e} \) we can find an open disc \( {D}_{x} \), with centre \( x \), which meets \( {D}_{x} \) only those (one or two) straight line segments that contain \( x \).\n\nLet us pick an inner point \( {x}_{0} \) from a straight line segment \( S \subseteq e \). \( {x}_{0}, S \) Then \( {D}_{{x}_{0}} \cap G = {D}_{{x}_{0}} \cap S \), so \( {D}_{{x}_{0}} \smallsetminus G \) is the union of two open half-discs. Since these half-discs do not meet \( G \), they each lie in a face of \( G \). Let us denote these faces by \( {f}_{1} \) and \( {f}_{2} \); they are the only faces of \( G \) with \( {x}_{0} \) \( {f}_{1},{f}_{2} \) on their frontier, and they may coincide (Fig. 4.2.1).\n\nIf \( e \) lies on a cycle \( C \subseteq G \), then \( {D}_{{x}_{0}} \) meets both faces of \( C \) (Theorem 4.1.1). Since \( {f}_{1} \) and \( {f}_{2} \) are contained in faces of \( C \) by Lemma 4.2.1, this implies \( {f}_{1} \neq {f}_{2} \). If \( e \) does not lie on any cycle, then \( e \) is a bridge and thus links two disjoint point sets \( {X}_{1},{X}_{2} \) as in Lemma 4.1.3, with \( {X}_{1} \cup {X}_{2} = G \smallsetminus \overset{ \circ }{e} \). Clearly, \( {f}_{1} \cup \overset{ \circ }{e} \cup {f}_{2} \) is the subset of a face \( f \) of \( G - e \). By Lemma 4.1.3, \( f \smallsetminus \overset{ \circ }{e} \) is a face of \( G \). But \( f \smallsetminus \overset{ \circ }{e} \) contains \( {f}_{1} \) and \( {f}_{2} \) by definition of \( f \), so \( {f}_{1} = f \smallsetminus \mathring{e} = {f}_{2} \) since \( {f}_{1},{f}_{2} \) and \( f \) are all faces of \( G \).\n\nNow consider any other point \( {x}_{1} \in \overset{ \circ }{e} \). Let \( P \) be the arc from \( {x}_{0} \) to \( P \) \( {x}_{1} \) contained in \( e \). Since \( P \) is compact, finitely many of the discs \( {D}_{x} \) \( {D}_{0},\ldots ,{D}_{n} \) with \( x \in P \) cover \( P \). Let us enumerate these discs as \( {D}_{0},\ldots ,{D}_{n} \) in the natural order of their centres along \( P \) ; adding \( {D}_{{x}_{0}} \) or \( {D}_{{x}_{1}} \) as necessary, we may assume that \( {D}_{0} = {D}_{{x}_{0}} | Yes |
Corollary 4.2.3. The frontier of a face is always the point set of a subgraph. | The subgraph of \( G \) whose point set is the frontier of a face \( f \) is said to bound \( f \) and is called its boundary; we denote it by \( G\left\lbrack f\right\rbrack \) . A face is said to be incident with the vertices and edges of its boundary. By Lemma 4.2.1 (ii), every face of \( G \) is also a face of its boundary; we shall use this fact frequently in the proofs to come. | No |
Proposition 4.2.4. A plane forest has exactly one face. | Proof. Use induction on the number of edges and Lemma 4.1.3. | No |
Lemma 4.2.5. If a plane graph has different faces with the same boundary, then the graph is a cycle. | Proof. Let \( G \) be a plane graph, and let \( H \subseteq G \) be the boundary of\n\n(4.1.1)\n\ndistinct faces \( {f}_{1},{f}_{2} \) of \( G \) . Since \( {f}_{1} \) and \( {f}_{2} \) are also faces of \( H \), Proposition 4.2.4 implies that \( H \) contains a cycle \( C \) . By Lemma 4.2.2 (ii), \( {f}_{1} \) and \( {f}_{2} \) are contained in different faces of \( C \) . Since \( {f}_{1} \) and \( {f}_{2} \) both have all of \( H \) as boundary, this implies that \( H = C \) : any further vertex or edge of \( H \) would lie in one of the faces of \( C \) and hence not on the boundary of the other. Thus, \( {f}_{1} \) and \( {f}_{2} \) are distinct faces of \( C \) . As \( C \) has only two faces, it follows that \( {f}_{1} \cup C \cup {f}_{2} = {\mathbb{R}}^{2} \) and hence \( G = C \) . | Yes |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.