Q stringlengths 4 3.96k | A stringlengths 1 3k | Result stringclasses 4
values |
|---|---|---|
Corollary 5.3.4. If \( {C}_{\phi } \) is a composition operator and \( \phi \left( a\right) = a \) for some \( a \in \mathbb{D} \), then\n\n\[ \n{\Pi }_{0}\left( {C}_{\phi }\right) \subset \{ 1\} \cup \mathop{\bigcup }\limits_{{k = 1}}^{\infty }\left\{ {\left( {\phi }^{\prime }\left( a\right) \right) }^{k}\right\} \sub... | Proof. This corollary is an immediate consequence of Corollary 5.3.2 and Theorem 5.3.3. | No |
Corollary 5.3.5. If \( {C}_{\phi } \) is a compact composition operator satisfying \( \phi \left( a\right) = a \) for some \( a \in \mathbb{D} \), then\n\n\[ \sigma \left( {C}_{\phi }\right) = \{ 0\} \cup \{ 1\} \cup \mathop{\bigcup }\limits_{{k = 1}}^{\infty }\left\{ {\left( {\phi }^{\prime }\left( a\right) \right) }^... | Proof. This follows immediately from the previous corollary and the Fredholm alternative. | No |
Theorem 5.4.1. Let \( \phi \) be an automorphism of the unit disk other than the identity automorphism. Then \( \phi \) has at most two fixed points in the complex plane. Either \( \phi \) has one fixed point inside \( \mathbb{D} \) and does not fix any point in \( {S}^{1} \) or \( \phi \) has all of its fixed points i... | Proof. Note that the equation \( \phi \left( z\right) = z \) is either a linear or a quadratic equation and thus has at most two roots. Hence \( \phi \) has at most two fixed points in the complex plane.\n\nLet \( \phi \left( z\right) = \lambda \frac{a - z}{1 - \bar{a}z} \) . If \( \phi \left( 0\right) = 0 \), then \( ... | Yes |
Theorem 5.4.3. If \( \phi \) is a disk automorphism satisfying \( \phi \left( a\right) = a \) for some \( a \in \mathbb{D} \), then the spectrum of \( {C}_{\phi } \) is the closure of \[ \left\{ {{\left( {\phi }^{\prime }\left( a\right) \right) }^{n} : n = 0,1,2,\ldots }\right\} . \] | Proof. Recall that if \( \psi = {\phi }_{a} \circ \phi \circ {\phi }_{a} \), then \( \psi \left( 0\right) = 0 \) (see the proof of Theorem 5.2.3). Now, \( \psi \) is a disk automorphism so it has the form \( \psi \left( z\right) = \mu \frac{b - z}{1 - \bar{b}z} \) for some \( b \) in \( \mathbb{D} \) and \( \mu \) of m... | Yes |
Lemma 5.4.4. Let \( \phi \) be a parabolic disk automorphism satisfying \( \phi \left( 1\right) = 1 \) . If \( \Gamma \left( z\right) = i\frac{1 + z}{1 - z} \) and the function \( F \) is defined by \( F = \Gamma \circ \phi \circ {\Gamma }^{-1} \), then there is a real number \( \beta \) such that \( F\left( z\right) =... | Proof. Note that the function \( \Gamma \) is a conformal mapping of the open unit disk onto the open upper half-plane. Moreover, \( \Gamma \) maps \( {S}^{1} \) bijectively onto \( \mathbb{R} \cup \{ \infty \} \) . Notice that \( \Gamma \left( 1\right) = \infty \) . Clearly, \( F \) is an automorphism of the upper hal... | Yes |
Theorem 5.4.6. If \( \phi \) is a parabolic disk automorphism, then\n\n\[ \sigma \left( {C}_{\phi }\right) = {\Pi }_{0}\left( {C}_{\phi }\right) = {S}^{1}. \] | Proof. By the previous theorem, it suffices to show that \( {S}^{1} \subset {\Pi }_{0}\left( {C}_{\phi }\right) \) . As in the proof of the previous theorem, we can and do assume that \( \phi \left( 1\right) = 1 \) and define \( \Gamma \) and \( F \) as in Lemma 5.4.4. In particular, \( F\left( z\right) = z + \beta \) ... | Yes |
Lemma 5.4.7. Let \( \phi \) be a hyperbolic disk automorphism satisfying \( \phi \left( 1\right) = 1 \) and \( \phi \left( {-1}\right) = - 1 \) . Then there exists a real number a between -1 and 1 such that\n\n\[ \phi \left( z\right) = \frac{z - a}{1 - {az}} \]\n\nfor all \( z \in \mathbb{D} \) . Moreover, \( {\phi }^{... | Proof. Since \( \phi \) is a disk automorphism, there exists an \( a \) in \( \mathbb{D} \) and a \( \lambda \) of modulus 1 such that\n\n\[ \phi \left( z\right) = \lambda \frac{a - z}{1 - \bar{a}z}. \]\n\nNow \( \phi \left( 1\right) = 1 \) implies that \( \lambda \left( {a - 1}\right) = 1 - \bar{a} \) and \( \phi \lef... | Yes |
Lemma 5.4.8. Let \( \phi \) be a hyperbolic disk automorphism satisfying \( \phi \left( 1\right) = 1 \) and \( \phi \left( {-1}\right) = - 1 \) . If \( \Gamma \left( z\right) = i\frac{1 + z}{1 - z} \) and the function \( G \) is defined by \( G = \) \( \Gamma \circ \phi \circ {\Gamma }^{-1} \), then there is a \( \gamm... | Proof. Note that \( G \) is an automorphism of the upper half-plane and that \( G \) maps the extended real line to the extended real line. Also\n\n\[ G\left( 0\right) = \Gamma \left( {\phi \left( {-1}\right) }\right) = \Gamma \left( {-1}\right) = 0, \]\n\nand\n\n\[ G\left( \infty \right) = \Gamma \left( {\phi \left( 1... | Yes |
Proposition 1.1. We have\n\n\[ \n{A}_{m, n}\left( q\right) = \mathop{\sum }\limits_{{i \geq 0}}p\left( {i; \leq n; \leq m}\right) {q}^{i} = {\left\lbrack \begin{matrix} m + n \\ m \end{matrix}\right\rbrack }_{q}. \n\] | Proof. We show that \( {A}_{m, n}\left( q\right) \) and \( {\left\lbrack \begin{matrix} m + n \\ m \end{matrix}\right\rbrack }_{q} \) satisfy the same initial conditions and the same recurrence. For \( m = 0 \) or \( n = 0 \) we have \( {A}_{0, n}\left( q\right) = {A}_{m,0}\left( q\right) = 1 \) since in this case we g... | Yes |
Corollary 1.2. We have\n\n\\[ \n\\left( {1 + {xq}}\\right) \\left( {1 + x{q}^{2}}\\right) \\cdots \\left( {1 + x{q}^{n}}\\right) = \\mathop{\\sum }\\limits_{{k = 0}}^{n}{\\left\\lbrack \\begin{array}{l} n \\ k \\end{array}\\right\\rbrack }_{q}{q}^{\\left( \\begin{matrix} k + 1 \\ 2 \\end{matrix}\\right) }{x}^{k}.\n\\] | Proof. Expanding the left-hand side we obtain\n\n\\[ \n\\left( {1 + {xq}}\\right) \\cdots \\left( {1 + x{q}^{n}}\\right) = \\mathop{\\sum }\\limits_{{k = 0}}^{n}{b}_{k}\\left( q\\right) {x}^{k},\n\\]\n\nwhere\n\n\\[ \n{b}_{k}\\left( q\\right) = \\mathop{\\sum }\\limits_{{\\lambda \\in {\\operatorname{Par}}_{d}\\left( {... | Yes |
Corollary 1.4.\n\n\\[ \n\\mathop{\\sum }\\limits_{{\\sigma \\in S\\left( n\\right) }}{q}^{\\operatorname{inv}\\left( \\sigma \\right) } = \\frac{\\left( {1 - {q}^{n}}\\right) \\left( {1 - {q}^{n - 1}}\\right) \\cdots \\left( {1 - q}\\right) }{{\\left( 1 - q\\right) }^{n}}. \n\\] | Example. For \\( n = 3 \\), the inversion numbers are \\( {I}_{3,0} = 1,{I}_{3,1} = \\) \\( {I}_{3,2} = 2,{I}_{3,3} = 1 \\) . Thus\n\n\\[ \n\\mathop{\\sum }\\limits_{{\\sigma \\in S\\left( 3\\right) }}{q}^{\\operatorname{inv}\\left( \\sigma \\right) } = {q}^{3} + 2{q}^{2} + {2q} + 1 \n\\]\n\nin agreement with\n\n\\[ \n... | Yes |
Proposition 2.1. A sequence \( \left( {{A}_{i}\left( z\right) }\right) \) of series converges if and only if for any \( n,\deg \left( {{A}_{i + 1}\left( z\right) - {A}_{i}\left( z\right) }\right) > n \) for all but finitely many \( i \) . We write this, for short, as \( \lim \deg \left( {{A}_{i + 1}\left( z\right) - {A... | Proof. Suppose \( \left( {{A}_{i}\left( z\right) }\right) \) converges to \( F\left( z\right) \) . Then for fixed \( k,{a}_{i, k} = \) \( {f}_{k} \) for \( i \geq {i}_{k} \), and so \( {a}_{i + 1, k} - {a}_{i, k} = 0 \) for \( i \geq {i}_{k} \) . Now let \( n \) be arbitrary. Then for \( k = 0,1,\ldots, n \) ,\n\n\[ \n... | Yes |
Corollary 2.2. The sum \( \mathop{\sum }\limits_{{k \geq 0}}{A}_{k}\left( z\right) \) exists if and only if \( \lim \deg {A}_{k} \) \( = \infty \) . | Another way to state this is that \( \mathop{\sum }\limits_{{k \geq 0}}{A}_{k}\left( z\right) \) exists if and only if for any \( n \), there are only finitely many non-zero \( n \) -th coefficients among the \( {A}_{k}\left( z\right) \) . So our notion of convergence agrees with the natural way to define the sum \( F\... | Yes |
Proposition 2.4. If \( \mathop{\prod }\limits_{{k \geq 1}}{A}_{k}\left( z\right) \) and \( \mathop{\prod }\limits_{{k \geq 1}}{B}_{k}\left( z\right) \) are admissible products, then so is \( \mathop{\prod }\limits_{{k \geq 1}}{A}_{k}\left( z\right) {B}_{k}\left( z\right) \), and we have\n\n\[ \mathop{\prod }\limits_{{k... | Proof. From (3) we immediately infer that \( \mathop{\prod }\limits_{{k \geq 1}}{A}_{k}\left( z\right) {B}_{k}\left( z\right) \) is again admissible. Furthermore, the factors that make a non-trivial contribution to \( \left\lbrack {z}^{n}\right\rbrack \) are the same on both sides, so we have in reality reduced the mul... | Yes |
Proposition 2.5. Let \( {p}_{n}^{\left( \alpha \right) }\left( x\right) = \mathop{\sum }\limits_{{k = 0}}^{n}{c}_{n, k}{p}_{k}^{\left( \beta \right) }\left( x\right) \) with \( \alpha = \left( {{a}_{1},{a}_{2}}\right. \) , \( \ldots ),\beta = \left( {{b}_{1},{b}_{2},\ldots }\right) \) . Then\n\n\[ \n{c}_{0,0} = 1,{c}_{... | Proof. To simplify the notation we set \( {p}_{n}\left( x\right) = {p}_{n}^{\left( \alpha \right) }\left( x\right) ,{q}_{n}\left( x\right) = \) \( {p}_{n}^{\left( \beta \right) }\left( x\right) \) ; thus \( {p}_{n}\left( x\right) = \mathop{\sum }\limits_{{k = 0}}^{n}{c}_{n, k}{q}_{k}\left( x\right) \).\n\nClearly, \( {... | Yes |
Lemma 3.2. Let \( f, g : {\mathbb{N}}_{0} \rightarrow \mathbb{C} \) and \( h : {\mathbb{N}}_{0} \rightarrow \mathbb{C} \) be defined by\n\n\[ h\left( \left| X\right| \right) = \mathop{\sum }\limits_{\left( S, T\right) }f\left( \left| S\right| \right) g\left( \left| T\right| \right) \]\n\nwhere \( \left( {S, T}\right) \... | Proof. Let \( \left| X\right| = n \) . There are \( \left( \begin{array}{l} n \\ k \end{array}\right) \) such pairs \( \left( {S, T}\right) \) with \( \left| S\right| = k \) , \( \left| T\right| = n - k \) ; hence\n\n\[ h\left( n\right) = \mathop{\sum }\limits_{{k = 0}}^{n}\left( \begin{array}{l} n \\ k \end{array}\rig... | Yes |
Theorem 3.3 (Composition Formula). Let \( f, g : {\mathbb{N}}_{0} \rightarrow \mathbb{C} \) with \( f\left( 0\right) = \) 0, and let \( h : {\mathbb{N}}_{0} \rightarrow \mathbb{C} \) be defined through\n\n\[ h\left( \left| X\right| \right) = \mathop{\sum }\limits_{{k \geq 1}}\mathop{\sum }\limits_{{\left\{ {{B}_{1},\ld... | Proof. Let \( \left| X\right| = n \), and denote by \( {h}_{k}\left( n\right) \) the inner sum of the righthand side of (3) for fixed \( k \) . Since the blocks \( {B}_{i} \) are nonempty (and therefore distinct), they can be permuted in \( k \) ! ways. Hence we get by (2) (note that \( f\left( 0\right) = 0 \) )\n\n\[ ... | Yes |
Corollary 3.4 (Exponential Formula). Let \( f : {\mathbb{N}}_{0} \rightarrow \mathbb{C} \) with \( f\left( 0\right) = 0 \) and \( h : {\mathbb{N}}_{0} \rightarrow \mathbb{C} \) be defined by\n\n\[ h\left( \left| X\right| \right) = \mathop{\sum }\limits_{{k \geq 1}}\mathop{\sum }\limits_{{\left\{ {{B}_{1},\ldots ,{B}_{k... | Example. In how many ways \( h\left( n\right) \) can we decompose an \( n \) -set into nonempty blocks and choose a linear order on each block? Here \( g\left( n\right) = 1 \) for all \( n, f\left( n\right) = n! \) ; hence \( \widehat{G}\left( z\right) = {e}^{z},\widehat{F}\left( z\right) = \mathop{\sum }\limits_{{n \g... | No |
Theorem 3.5. Let \( f, g : {\mathbb{N}}_{0} \rightarrow \mathbb{C} \) with \( f\left( 0\right) = 0 \) and \( h : {\mathbb{N}}_{0} \rightarrow \mathbb{C} \) be defined by\n\n\[ h\left( \left| X\right| \right) = \mathop{\sum }\limits_{{\sigma \in S\left( X\right) }}f\left( \left| {C}_{1}\right| \right) \cdots f\left( \le... | Proof. There are \( \left( {j - 1}\right) \) ! ways to make a \( j \) -set into a cyclic permutation. Hence we may write\n\n\[ h\left( \left| X\right| \right) = \mathop{\sum }\limits_{{\left\{ {{B}_{1},\ldots ,{B}_{k}}\right\} \in \prod \left( X\right) }}\left( {\left| {B}_{1}\right| - 1}\right) !f\left( \left| {B}_{1}... | Yes |
Corollary 3.9 (Lagrange). Let \( H\left( z\right) = {zG}\left( z\right), G\left( 0\right) \neq 0 \), and \( {H}^{\langle - 1\rangle }\left( z\right) \) the compositional inverse, \( {H}^{\langle - 1\rangle }\left( 0\right) = 0 \) . Then\n\n\[ \left\lbrack {z}^{n}\right\rbrack {H}^{\langle - 1\rangle }\left( z\right) = ... | Proof. Note that \( F\left( z\right) = {H}^{\langle - 1\rangle }\left( z\right) \) satisfies the equation\n\n\[ F\left( z\right) = z{G}^{-1}\left( {F\left( z\right) }\right) \]\n\nsince\n\n\[ H\left( {F\left( z\right) }\right) = F\left( z\right) G\left( {F\left( z\right) }\right) = z{G}^{-1}\left( {F\left( z\right) }\r... | Yes |
Proposition 4.1. Let \( {P}_{i} \) be recurrence operators for \( F\left( {n, k}\right) \) . If\n\n\[ S\left( {N, n}\right) = \mathop{\sum }\limits_{{i = 1}}^{t}{A}_{i}\left( {N, K, n, k}\right) {P}_{i}\left( {N, K, n, k}\right) + \left( {K - I}\right) C\left( {N, K, n, k}\right) \]\n\ndoes not depend on \( K, k \), th... | Example. The simplest case arises when there is a single recurrence operator \( P \) that does not depend on \( k \) . For the binomial coefficients \( F\left( {n, k}\right) = \left( \begin{array}{l} n \\ k \end{array}\right) \), we have the recurrence operator from above\n\n\[ P = {NK} - I - K = \left( {K - I}\right) ... | Yes |
Lemma 4.2. The maximality of the degree of \( p\left( n\right) \) implies \( \gcd (q\left( n\right) , \( r\left( {n + j}\right) ) = 1 \) for all \( j \geq 0 \) . | Proof. Suppose to the contrary that \( g\left( n\right) = \gcd \left( {q\left( n\right), r\left( {n + j}\right) }\right) \neq \) 1 for some \( j > 0 \) . Then \( g\left( {n - j}\right) \mid r\left( n\right) \) and\n\n\[ \n\frac{q\left( n\right) }{r\left( n\right) } = \frac{g\left( n\right) }{g\left( {n - 1}\right) } \c... | Yes |
Lemma 4.3. \( \left( {S\left( n\right) }\right) \) is a \( {CF} \) if and only if \( f\left( n\right) \) is a rational function. | Proof. If \( f\left( n\right) \) is a quotient of polynomials, then \( \left( {S\left( n\right) }\right) \) is clearly a closed form since \( \left( {a\left( n\right) }\right) \) is. Suppose, conversely, that \( \left( {S\left( n\right) }\right) \) is a CF. Then\n\n\[ f\left( n\right) = \frac{p\left( n\right) S\left( {... | Yes |
Theorem 5.1. Let \( E \) be a finite set, and \( f, g : {2}^{E} \rightarrow K \) functions into a field \( K \) of characteristic 0 . Then\n\n\[ f\left( A\right) = \mathop{\sum }\limits_{{T \supseteq A}}g\left( T\right) \left( {\forall A}\right) \Leftrightarrow g\left( A\right) = \mathop{\sum }\limits_{{T \supseteq A}}... | Proof. Assume the equality on the left-hand side. Then\n\n\[ \mathop{\sum }\limits_{{T \supseteq A}}{\left( -1\right) }^{\left| T\right| - \left| A\right| }f\left( T\right) = \mathop{\sum }\limits_{{T \supseteq A}}{\left( -1\right) }^{\left| T\right| - \left| A\right| }\mathop{\sum }\limits_{{U \supseteq T}}g\left( U\r... | Yes |
Corollary 5.2. Let \( X \) be a universe, \( E = \\left\\{ {{e}_{1},\\ldots ,{e}_{n}}\\right\\} \) a set of properties, and \( {N}_{p} \) the number of elements in \( X \) that possess precisely \( p \) properties. Then\n\n\[ \n{N}_{p} = \\mathop{\\sum }\\limits_{{k = p}}^{n}{\\left( -1\\right) }^{k - p}\\left( \\begin... | Proof. We have by (10),\n\n\[ \n{N}_{p} = \\mathop{\\sum }\\limits_{{A : \\left| A\\right| = p}}{N}_{ = A} = \\mathop{\\sum }\\limits_{{A : \\left| A\\right| = p}}\\mathop{\\sum }\\limits_{{T \\supseteq A}}{\\left( -1\\right) }^{\\left| T\\right| - \\left| A\\right| }{N}_{ \\supseteq T} \n\]\n\n\[ \n= \\mathop{\\sum }\... | Yes |
Theorem 5.5 (Möbius Inversion). Let \( P \) be a locally finite poset, and \( f, g : P \rightarrow K \) . Then\n\n(i) Inversion from below:\n\n\[ f\left( a\right) = \mathop{\sum }\limits_{{x \leq a}}g\left( x\right) \left( {\forall a \in P}\right) \Leftrightarrow \]\n\n\[ g\left( a\right) = \mathop{\sum }\limits_{{x \l... | Proof. Set \( \bar{f}\left( {0, a}\right) = f\left( a\right) \), and \( \bar{f}\left( {x, y}\right) = 0 \) for \( x \neq 0 \), and similarly \( \bar{g}\left( {0, a}\right) = g\left( a\right) ,\bar{g}\left( {x, y}\right) = 0 \) for \( x \neq 0 \) . The left-hand side in (i) is then equivalent to \( \bar{f} = \bar{g} * \... | Yes |
Theorem 5.7 (Involution Principle). Let \( f : S \rightarrow T \) be a sign-preserving bijection between two signed sets \( S \) and \( T \), and let \( \varphi \) and \( \psi \) be alternating involutions on \( S \) and \( T \), respectively, with \( {\operatorname{Fix}}_{\varphi }S \subseteq {S}^{ + },{\operatorname{... | Proof. Since \( \left| {{\operatorname{Fix}}_{\varphi }S}\right| = \left| {S}^{ + }\right| - \left| {S}^{ - }\right| = \left| {T}^{ + }\right| - \left| {T}^{ - }\right| = \left| {{\operatorname{Fix}}_{\psi }T}\right| \), it remains to verify (18). We have the two involutions \( \varphi \) and \( \psi \), and we may reg... | Yes |
Lemma 6.1. Let \( G \) act on \( X \) . Then for any \( x \in X \) , \[ \left| {M\left( x\right) }\right| = \frac{\left| G\right| }{\left| {G}_{x}\right| }.\] | Proof. We have \( M\left( x\right) = \{ \bar{g}x : g \in G\} \) . Now, \[ \bar{g}x = \bar{h}x \Leftrightarrow {\bar{g}}^{-1}\bar{h}x = x \Leftrightarrow {g}^{-1}h \in {G}_{x} \Leftrightarrow h = {ga}, a \in {G}_{x}. \] This means that for every \( g \in G \), exactly \( \left| {G}_{x}\right| \) group elements, namely a... | Yes |
Lemma 6.2 (Burnside-Frobenius). Let the group \( G \) act on \( X \), and let \( \mathcal{M} \) be the set of patterns. Then\n\n\[ \left| \mathcal{M}\right| = \frac{1}{\left| G\right| }\mathop{\sum }\limits_{{g \in G}}\left| {X}_{g}\right| \] | Proof. Let \( y \in M\left( x\right) \), or equivalently \( M\left( y\right) = M\left( x\right) \) . Then by (2), \( \left| {G}_{x}\right| = \left| {G}_{y}\right| \), and therefore\n\n\[ \left| G\right| = \left| {M\left( x\right) }\right| \left| {G}_{\mathcal{Y}}\right| = \mathop{\sum }\limits_{{\mathcal{Y} \in M\left(... | Yes |
Theorem 6.4. Let \( N \) and \( R \) be sets, \( G \) a group that acts on \( N \) , \( \mathcal{F} \) a \( G \) -closed class, and \( {x}_{j}\left( {j \in R}\right) \) variables. Then\n\n\[ w\left( {\mathcal{F};G}\right) = \mathop{\sum }\limits_{{M \in {\mathcal{M}}_{\mathcal{F}}}}w\left( M\right) = \frac{1}{\left| G\... | Proof. Let \( M \) be a pattern of \( \mathcal{F} \) under \( G \) . If we apply \( G \) to \( M \), then, of course, we obtain only one pattern, namely \( M \) . By the lemma of Burnside-Frobenius,\n\n\[ 1 = \frac{1}{\left| G\right| }\mathop{\sum }\limits_{{g \in G}}\left| {M}_{g}\right| \]\n\nwhere \( {M}_{g} = \{ f ... | Yes |
Theorem 6.5 (Pólya-Redfield). Let \( N \) and \( R \) be sets, \( \left| N\right| = n,\left| R\right| = \) \( r \), and \( G \) a group acting on \( N,{x}_{j}\left( {j \in R}\right) \) variables. Then\n\n\[ w\left( {\operatorname{Map}\left( {N, R}\right) ;G}\right) = \mathop{\sum }\limits_{{M \in \mathcal{M}}}w\left( M... | In particular,\n\n\[ \left| \mathcal{M}\right| = Z\left( {G;r, r,\ldots, r}\right) . \] | Yes |
Corollary 6.6. Let \( {m}_{k} \) be the number of patterns of weight \( {x}^{k} \) in \( \operatorname{Map}\left( {N, R}\right) \) under \( G \), where all \( {m}_{k} \) are assumed to be finite. Then\n\n\[ w\left( {\operatorname{Map}\left( {N, R}\right) ;G}\right) = \mathop{\sum }\limits_{{k \geq 0}}{m}_{k}{x}^{k} \]\... | Proof. As in problem \( \mathrm{C},\mathop{\sum }\limits_{{a \in R}}w{\left( a\right) }^{k} = R\left( {x}^{k}\right) \), and the result follows. | No |
Let \( R\left( x\right) = \mathop{\sum }\limits_{{n \geq 0}}{r}_{n}{x}^{n} \) be a generating function. Then\n\n\[ \mathop{\sum }\limits_{{n \geq 0}}Z\left( {S\left( n\right) ;R\left( x\right), R\left( {x}^{2}\right) ,\ldots, R\left( {x}^{n}\right) }\right) {y}^{n} = \exp \left( {\mathop{\sum }\limits_{{k \geq 1}}R\lef... | Example. We want to determine the generating function \( U\left( x\right) = \) \( \mathop{\sum }\limits_{{n \geq 1}}{u}_{n}{x}^{n} \), where \( {u}_{n} \) is the number of isomorphism classes of rooted trees with \( n \) vertices. Two rooted trees \( \left( {T, v}\right) \) and \( \left( {{T}^{\prime },{v}^{\prime }}\r... | Yes |
Proposition 6.8. Let \( N \) be an \( n \) -set, \( R = \{ 0,1\}, G \) a group acting on \( N \), and \( \mathcal{F} \subseteq \operatorname{Map}\left( {N, R}\right) \) a \( G \) -closed class. Then with \( h : 0 \leftrightarrow 1 \) in \( R \) , \[ w\left( {{\mathcal{F}}^{c};G}\right) = \mathop{\sum }\limits_{{M \in {... | Proof. Consider the action of \( G \) on \( {\mathcal{F}}^{c} \) defined by \[ f \rightarrow {f}^{\prime } = h \circ f \circ \bar{g}\;\left( {g \in G}\right) . \] Now, \( {f}^{\prime } = h \circ f \circ \bar{g} \) implies \( {f}^{\prime } = f \circ \overline{{k}_{f}g} \) . Conversely, if \( {f}^{\prime } = f \circ \bar... | Yes |
Corollary 6.9. The weight enumerator of the self-complementary patterns of \( \operatorname{Map}\left( {N, R}\right) \) is given by\n\n\[ w\left( {\operatorname{Map}{\left( N, R\right) }^{c};G}\right) = Z\left( {G;{\lambda }_{1},\ldots ,{\lambda }_{n}}\right) ,\] \n\nwith\n\n\[ {\lambda }_{k} = \left\{ \begin{array}{ll... | In particular,\n\n\[ \left| {\operatorname{Map}{\left( N, R\right) }^{c}}\right| = Z\left( {G;0,2,0,2,\ldots }\right) . \] | No |
Proposition 6.10. We have\n\n\[ \mathop{\sum }\limits_{{k = 0}}^{\left( \begin{array}{l} n \\ 2 \end{array}\right) }{g}_{n, k}{x}^{k} = Z\left( {\bar{S}\left( n\right) ;1 + x,1 + {x}^{2},\ldots ,1 + {x}^{\left( \begin{matrix} n \\ 2 \end{matrix}\right) }}\right) ,\] \n\nand in particular,\n\n\[ g\left( n\right) = Z\lef... | Thus our task consists in computing the cycle index \( Z\left( {\bar{S}\left( n\right) }\right) \) . Let \( \sigma \in S\left( n\right) \) have type \( t\left( \sigma \right) = {1}^{{c}_{1}}{2}^{{c}_{2}}\ldots {n}^{{c}_{n}} \), and consider a cycle \( A = \) \( \left( {{a}_{1},\ldots ,{a}_{k}}\right) \) of \( \sigma \)... | Yes |
Corollary 6.11. The number of non-isomorphic self-complementary graphs on \( n \) vertices is given by \[ Z\left( {\bar{S}\left( n\right) ;0,2,0,2,\ldots }\right) \text{.} \] | In the example \( n = 4 \) we have \( Z\left( {\bar{S}\left( 4\right) ;0,2,0,2,0,2}\right) = \frac{1}{24}\left\lbrack {6 \cdot 2 \cdot 2}\right\rbrack = \) 1, with the path \( \bullet \bullet \bullet \bullet \) being the only such graph. | No |
Corollary 6.12. Let \( G = \left( {V, E}\right) \) be a graph, and \( {g}_{k}\left( G\right) \) the number of non-isomorphic subgraphs of \( G \) with \( k \) edges. Then \[ \mathop{\sum }\limits_{{k \geq 0}}{g}_{k}\left( G\right) {x}^{k} = Z\left( {{\bar{S}}_{G};1 + x,1 + {x}^{2},\ldots }\right) . \] | The edge group consists of the four permutations id, (5)(12)(34), (5)(13)(24), (5)(14)(23) (the Klein 4-group), and we compute \[ Z\left( {{\bar{S}}_{G};{z}_{1},\ldots ,{z}_{5}}\right) = \frac{1}{4}\left\lbrack {{z}_{1}^{5} + 3{z}_{1}{z}_{2}^{2}}\right\rbrack . \] Thus \[ \mathop{\sum }\limits_{{k = 0}}^{5}{g}_{k}\left... | Yes |
Theorem 6.15 (de Bruijn). Let \( \mathcal{F} \subseteq \operatorname{Map}\left( {N, R}\right) \) be a \( G, H \) -closed class. Then\n\n\[ w\left( {\mathcal{F};G, H}\right) = \frac{1}{\left| H\right| }\mathop{\sum }\limits_{{h \in H}}w\left( {\mathcal{F}\left( h\right) ;G}\right) \]\n\n\[ = \frac{1}{\left| G\right| \le... | Proof. Let \( M \in {\mathcal{M}}_{\mathcal{F}} \) . The \( G, H \) -pattern \( M \) splits into disjoint \( G \) - patterns, which we denote by \( {fG} \) . Regarding them as new elements, we obtain by the same argument as in the proof of Theorem 6.4,\n\n\[ w\left( M\right) = \frac{1}{\left| H\right| }\mathop{\sum }\l... | Yes |
Corollary 6.16. Let \( G \leq S\left( N\right), H \leq S\left( R\right) ,\left| N\right| = n,\left| R\right| = r \), and \( {P}_{1},\ldots ,{P}_{m} \) the \( H \) -orbits of \( R \) . Any \( h \in H \) decomposes uniquely into a product \( h = {h}_{1}\cdots {h}_{m} \) with \( {h}_{i} \in S\left( {P}_{i}\right) \) . We ... | \[ w\left( {\operatorname{Map}\left( {N, R}\right) ;G, H}\right) = \frac{1}{\left| H\right| }\mathop{\sum }\limits_{{h \in H}}Z\left( {G;{\lambda }_{1}\left( h\right) ,\ldots ,{\lambda }_{n}\left( h\right) }\right) ,\] where \[ {\lambda }_{k}\left( h\right) = \mathop{\sum }\limits_{{i = 1}}^{m}\left( {\mathop{\sum }\li... | Yes |
Corollary 6.17. For \( G \leq S\left( N\right), H \leq S\left( R\right) \), we have\n\n\[ \left| {\mathcal{M}}_{\text{Inj }}\right| = \frac{1}{\left| G\right| \left| H\right| }\mathop{\sum }\limits_{{g \in G, h \in H}}\mathop{\prod }\limits_{{k = 1}}^{n}{k}^{{c}_{k}\left( g\right) }\left( {{c}_{k}\left( g\right) }\righ... | Formula (11) can be conveniently described by the so-called cap product of Redfield. Let \( {z}_{1}^{{a}_{1}}{z}_{2}^{{a}_{2}}\cdots {z}_{n}^{{a}_{n}} \) be a monomial appearing in the cycle index \( Z\left( G\right) \), and \( {z}_{1}^{{b}_{1}}{z}_{2}^{{b}_{2}}\cdots {z}_{r}^{{b}_{r}} \) one of \( Z\left( H\right) \) ... | Yes |
Lemma 7.1. Let \( A = {A}^{\sigma ,\tau } = \left( {a}_{n, k}\right) \), then we have for all \( m, n \geq 0 \) ,\n\n\[ \mathop{\sum }\limits_{{k \geq 0}}{a}_{m, k}{a}_{n, k}{T}_{k} = {B}_{m + n}\;\left( { = {a}_{m + n,0}}\right) . \] | Proof. We use induction on \( m \) . For \( m = 0 \) ,\n\n\[ \mathop{\sum }\limits_{{k \geq 0}}{a}_{0, k}{a}_{n, k}{T}_{k} = {a}_{0,0}{a}_{n,0} = {B}_{n}. \]\n\nAssume that (4) holds for \( m - 1 \) and all \( n \) . By (1),\n\n\[ \mathop{\sum }\limits_{k}{a}_{m, k}{a}_{n, k}{T}_{k} = \mathop{\sum }\limits_{{k \geq 0}}... | Yes |
A lower triangular matrix A with main diagonal 1 is a Catalan matrix if and only if \( {AT}{A}^{T} = H \) for some diagonal matrix \( T = \left( {{T}_{i}{\delta }_{i, j}}\right) \) with \( {T}_{0} = 1,{T}_{k} \neq 0 \) for all \( k \), and some Hankel matrix \( H = \left( {B}_{i + j}\right) \) . If \( {AT}{A}^{T} = H \... | Proof. We have already seen one direction, so assume \( {AT}{A}^{T} = H \) ,\n\nthat is,\n\n\[ \mathop{\sum }\limits_{{k \geq 0}}{a}_{m, k}{a}_{n, k}{T}_{k} = {B}_{m + n}\;\text{ for all }m, n. \]\n\n(6)\n\nFirst we note that \( m = 0 \) in (6) implies \( {a}_{n,0} = {B}_{n} \) . Next we claim that \( {AT}{A}^{T} = H \... | Yes |
Lemma 7.3. The sequence \( \left( {{p}_{n}\left( x\right) }\right) \) is an OPS with \( L \) and \( {\lambda }_{n} \) as in (10) if and only if\n\n\[ L\left( {{x}^{k}{p}_{n}\left( x\right) }\right) = \left\{ \begin{array}{ll} {\lambda }_{n} & k = n\;\left( {{\lambda }_{0} = 1}\right) , \\ 0 & k < n. \end{array}\right. ... | Proof. Since \( \left( {{p}_{n}\left( x\right) }\right) \) constitutes a basis of \( \mathbb{C}\left\lbrack x\right\rbrack \), we have \( {x}^{k} = \mathop{\sum }\limits_{{i = 0}}^{k}{c}_{k, i}{p}_{i}\left( x\right) \) with \( {c}_{k, k} = 1 \), and hence\n\n\[ L\left( {{x}^{k}{p}_{n}\left( x\right) }\right) = \mathop{... | Yes |
Theorem 7.4 (Favard). A polynomial sequence \( \left( {{p}_{n}\left( x\right) }\right) \) is an OPS if and only if \( \left( {{p}_{n}\left( x\right) }\right) \) satisfies recurrence (8) for some pair of sequences \( \sigma = \left( {s}_{k}\right) ,\tau = \left( {t}_{k}\right) \) . | Proof. Suppose \( \left( {{p}_{n}\left( x\right) }\right) \) satisfies (8), where \( A = {A}^{\sigma ,\tau }, U = {A}^{-1} = \) \( \left( {u}_{n, k}\right) ,{p}_{n}\left( x\right) = \mathop{\sum }\limits_{{k = 0}}^{n}{u}_{n, k}{x}^{k} \), and where \( T = \left( {{T}_{i}{\delta }_{i, j}}\right), H = \left( {B}_{i + j}\... | Yes |
Theorem 7.5. A sequence \( \left( {B}_{n}\right) \) is a Catalan sequence if and only if \( \det {H}_{n} \neq 0 \) for all \( n \), where \( {H}_{n} = {\left( {B}_{i + j}\right) }_{i, j = 0}^{n} \) is the Hankel matrix. | Proof. We have already seen one half. Suppose then \( \det {H}_{n} \neq 0 \) for all \( n \) . We define the sequence \( \left( {{p}_{n}\left( x\right) }\right) \) by \( {p}_{0}\left( x\right) = 1 \), and for \( n \geq 1 \n\n\[ \n\begin{array}{r} \det \left( \begin{matrix} {B}_{0} & {B}_{1} & \ldots & {B}_{n} \\ {B}_{1... | Yes |
Lemma 7.7 (Jacobi). Let \( M \) be an \( n \times n \) -matrix, \( n \geq 2 \). Then \[ \det M \cdot \det {M}_{1, n;1, n} = \det {M}_{1,1} \cdot \det {M}_{n, n} - \det {M}_{1, n} \cdot \det {M}_{n,1}, \] where \( {M}_{1, n;1, n} \) is the submatrix with rows \( 1, n \) and columns \( 1, n \) deleted, where we set \( \d... | Proof. We regard the entries of \( M \) as variables. To shorten the notation, let us set \( \left| M\right| = \det M \). Consider the matrix \[ {M}^{ * } = \left( \begin{matrix} \left| {M}_{1,1}\right| & 0 & 0\ldots 0 & {\left( -1\right) }^{n + 1}\left| {M}_{n,1}\right| \\ - \left| {M}_{1,2}\right| & 1 & 0\ldots 0 & {... | Yes |
Corollary 7.8. Let \( \left( {B}_{n}\right) \) be a sequence, and \( {d}_{n}^{\left( k\right) } = \det {H}_{n}^{\left( k\right) } \) . Then\n\n\[ \n{d}_{n}^{\left( k\right) }{d}_{n - 2}^{\left( k + 2\right) } = {d}_{n - 1}^{\left( k\right) }{d}_{n - 1}^{\left( k + 2\right) } - {\left( {d}_{n - 1}^{\left( k + 1\right) }... | Example. We know for the ordinary Catalan numbers \( {C}_{n} \) that \( {d}_{n}^{\left( 0\right) } = {d}_{n}^{\left( 1\right) } = 1 \) . Hence (8) yields for \( k = 2 \) ,\n\n\[ \n{d}_{n}^{\left( 2\right) } = {d}_{n - 1}^{\left( 2\right) } + 1\n\]\n\nand thus \( {d}_{n}^{\left( 2\right) } = \det {H}_{n}^{\left( 2\right... | Yes |
Lemma 7.9. Suppose \( \left( {{A}_{k}\left( z\right) }\right) \) and \( \left( {{C}_{k}\left( z\right) }\right) \) with \( {C}_{k}\left( 0\right) = 1 \) are connected as in (5). Then \( \left( {{A}_{k}\left( z\right) }\right) \) satisfies (4) if and only if\n\n\[ \n{C}_{k}\left( z\right) = \frac{1}{1 - {s}_{k}z - {t}_{... | Proof. Suppose (4) holds for \( \left( {{A}_{k}\left( z\right) }\right) \) . Then\n\n\[ \n\begin{cases} \frac{{A}_{0} - 1}{z} & = {s}_{0}{A}_{0} + {t}_{1}{A}_{1}, \\ \frac{{A}_{k}}{z} & = {A}_{k - 1} + {s}_{k}{A}_{k} + {t}_{k + 1}{A}_{k + 1}\;\left( {k \geq 1}\right) . \end{cases}\n\]\n\nHence by (5),\n\n\[ \n{C}_{0} =... | Yes |
Lemma 7.10. If \( A \leftrightarrow \left( {B, F}\right) \), then we have\n\n\[ U \leftrightarrow \left( {\frac{1}{B\left( {F}^{\langle - 1\rangle }\right) },{F}^{\langle - 1\rangle }}\right) ,\]\n\nwhere \( {F}^{\langle - 1\rangle } \) is the compositional inverse of \( F \) . | Proof. We have to show that\n\n\[ {U}_{k}\left( z\right) = \frac{1}{B\left( {{F}^{\langle - 1\rangle }\left( z\right) }\right) }\frac{{\left( {F}^{\langle - 1\rangle }\left( z\right) \right) }^{k}}{{Q}_{k}}.\]\n\nFrom \( {AU} = I \) follows\n\n\[ {u}_{0, k}{A}_{0}\left( z\right) + {u}_{1, k}{A}_{1}\left( z\right) + \cd... | Yes |
Proposition 7.11. Let \( A \leftrightarrow \left( {B, F}\right) \), and \( \left( {{p}_{n}\left( x\right) }\right) \) the corresponding OPS. Then we have\n\n\[ \mathop{\sum }\limits_{{n \geq 0}}{p}_{n}\left( x\right) \frac{{z}^{n}}{{Q}_{n}} = \frac{1}{B\left( {{F}^{\langle - 1\rangle }\left( z\right) }\right) }\mathop{... | Proof. Let us define the infinite column-vectors \( Z = \left( \frac{{z}^{n}}{{Q}_{n}}\right) \) and \( X = \left( {x}^{n}\right) \). We obtain\n\n\[ {Z}^{T}\left( {UX}\right) = \left( {1,\frac{z}{{Q}_{1}},\frac{{z}^{2}}{{Q}_{2}},\ldots }\right) \left( \begin{matrix} {p}_{0}\left( x\right) \\ {p}_{1}\left( x\right) \\ ... | Yes |
Lemma 7.12. Let \( H\left( z\right) = \mathop{\sum }\limits_{{n \geq 0}}{h}_{n}\frac{{z}^{n}}{{Q}_{n}} \) . (i) We have \( {h}_{n} = {\left. H\left( \Delta \right) {x}^{n}\right| }_{x = 0} \) for all \( n \) . (ii) If \( p\left( x\right) = \mathop{\sum }\limits_{{i = 0}}^{k}{p}_{i}{x}^{i} \), then \( {\left. H\left( \D... | Proof. Recall the notation \( \left\lbrack \begin{array}{l} n \\ m \end{array}\right\rbrack = \frac{{Q}_{n}}{{Q}_{m}{Q}_{n - m}} \) . Now \( \Delta {x}^{n} = {q}_{n}{x}^{n - 1} \) , and thus \( {\Delta }^{m}{x}^{n} = {q}_{n}{q}_{n - 1}\cdots {q}_{n - m + 1}{x}^{n - m} \), or \[ \frac{{\Delta }^{m}{x}^{n}}{{Q}_{m}} = \l... | Yes |
Lemma 7.13. Let \( A = \left( {a}_{n, k}\right) \) be invertible, and \( {A}_{k}\left( z\right) = \mathop{\sum }\limits_{{n \geq 0}}{a}_{n, k}\frac{{z}^{n}}{{Q}_{n}} \) . If for two polynomials \( p\left( x\right), q\left( x\right) \), we have \( {\left. {A}_{k}\left( \Delta \right) p\left( x\right) \right| }_{x = 0} =... | Proof. The previous lemma shows that\n\n\[ \n{\left. {A}_{k}\left( \Delta \right) p\left( x\right) \right| }_{x = 0} = \mathop{\sum }\limits_{i}{p}_{i}{a}_{i, k},{\left. {A}_{k}\left( \Delta \right) q\left( x\right) \right| }_{x = 0} = \mathop{\sum }\limits_{i}{q}_{i}{a}_{i, k}. \n\]\n\nIn other words, \( {p}^{T}A = {q... | Yes |
Lemma 7.14. Let \( A \) be invertible, and \( \left( {{p}_{n}\left( x\right) }\right) \) the polynomial sequence corresponding to \( U = {A}^{-1},{p}_{n}\left( x\right) = \mathop{\sum }\limits_{{i = 0}}^{n}{u}_{n, i}{x}^{i} \) . Then we have\n\n\[ \n{\left. {A}_{k}\left( \Delta \right) {p}_{n}\left( x\right) \right| }_... | Proof. By Lemma 7.12 and \( U = {A}^{-1} \) ,\n\n\[ \n{\left. {A}_{k}\left( \Delta \right) {p}_{n}\left( x\right) \right| }_{x = 0} = \mathop{\sum }\limits_{i}{a}_{i, k}{u}_{n, i} = {\delta }_{n, k}.\n\] | Yes |
Proposition 7.15. Let \( A \) be invertible, and \( \left( {{p}_{n}\left( x\right) }\right) \) the polynomial sequence corresponding to \( U = {A}^{-1} \) . For an arbitrary generating function \( H\left( z\right) = \mathop{\sum }\limits_{{n \geq 0}}{h}_{n}\frac{{z}^{n}}{{Q}_{n}} \) , | \[ H\left( z\right) = \mathop{\sum }\limits_{{k \geq 0}}\left( {\left. H\left( \Delta \right) {p}_{k}\left( x\right) \right| }_{x = 0}\right) {A}_{k}\left( z\right) . \] | Yes |
Proposition 7.16. Let \( A \leftrightarrow \left( {B, F}\right) \), and \( \left( {{p}_{n}\left( x\right) }\right) \) the corresponding orthogonal polynomial system. Then we have\n\n\[ F\left( \Delta \right) {p}_{n}\left( x\right) = {q}_{n}{p}_{n - 1}\left( x\right) . \] | Proof. By (18),\n\n\[ {\left. {A}_{k}\left( \Delta \right) \left( F\left( \Delta \right) {p}_{n}\left( x\right) \right) \right| }_{x = 0} = {\left. B\left( \Delta \right) \frac{F{\left( \Delta \right) }^{k + 1}}{{Q}_{k}}{p}_{n}\left( x\right) \right| }_{x = 0} \]\n\n\[ = {\left. {q}_{k + 1}{A}_{k + 1}\left( \Delta \rig... | Yes |
Proposition 7.17. Suppose \( A \leftrightarrow \left( {B, F}\right) \) . Then\n\n\[{\widetilde{p}}_{n}\left( x\right) = B\left( \Delta \right) {p}_{n}\left( x\right) \text{ for all }n.\]\n\n(20) | Proof. From (18) we get \( {\left. \frac{F{\left( \Delta \right) }^{k}}{{Q}_{k}}{\widetilde{p}}_{n}\left( x\right) \right| }_{x = 0} = {\delta }_{n, k} \), and on the other hand, again by (18),\n\n\[{\left. \frac{F{\left( \Delta \right) }^{k}}{{Q}_{k}}\left( B\left( \Delta \right) {p}_{n}\left( x\right) \right) \right|... | Yes |
Proposition 7.18. Let \( A \leftrightarrow \left( {B, F}\right) ,\widetilde{A} \leftrightarrow \left( {1, F}\right) \) in the exponential calculus, with the associated polynomial sequences \( \left( {{p}_{n}\left( x\right) }\right) \) and \( \left( {{\widetilde{p}}_{n}\left( x\right) }\right) \) . Then we have\n\n\[ \n... | Proof. By the expansion theorem applied to \( \widetilde{A} \) and \( H\left( z\right) = {e}^{yz} \) and (22) we obtain\n\n\[ \n{e}^{yz} = {\left. \mathop{\sum }\limits_{{k \geq 0}}{e}^{yD}{\widetilde{p}}_{k}\left( x\right) \right| }_{x = 0}\frac{F{\left( z\right) }^{k}}{k!} = \mathop{\sum }\limits_{{k \geq 0}}{\wideti... | Yes |
Consider \( \sigma = \left( {{s}_{k} = a + k}\right) \) , \( \tau = \left( {{t}_{k} = {ak}}\right) \) . In our previous notation we have \( s = 1, a = b, u = 0 \) , which means that we have to solve the system\n\n\[ \left\{ \begin{array}{l} {F}^{\prime } = 1 + F, \\ {B}^{\prime } = {aB}\left( {1 + F}\right) . \end{arra... | Clearly \( F\left( z\right) = {e}^{z} - 1 \), and \( \frac{{B}^{\prime }}{B} = a\left( {1 + F}\right) = a{e}^{z} \) implies \( B\left( z\right) = \) \( {e}^{a\left( {{e}^{z} - 1}\right) } \), since \( B\left( 0\right) = 1 \) . We have already seen this function in Section 3.3; it is the exponential generating function ... | Yes |
Consider \( \sigma = \left( {{s}_{k} = a + {2k}}\right) \) , \( \tau = \left( {{t}_{k} = k\left( {a + k - 1}\right) }\right) \), that is, \( s = 2, u = 1, a = b \) . The differential equations (13) are\n\n\[ \left\{ \begin{array}{l} {F}^{\prime } = 1 + {2F} + {F}^{2} = {\left( 1 + F\right) }^{2} \\ {B}^{\prime } = {aB}... | One easily computes \( F\left( z\right) = \frac{z}{1 - z} \) and \( {F}^{\langle - 1\rangle }\left( z\right) = \frac{z}{1 + z} \) . Furthermore, \( {\left( \log B\right) }^{\prime } = \frac{{B}^{\prime }}{B} = a\left( {1 + F}\right) = \frac{a}{1 - z} \), and thus\n\n\[ B\left( z\right) = \frac{1}{{\left( 1 - z\right) }... | Yes |
Example 3. Hermite Polynomials. Our final example is \( \sigma \equiv a \) , \( \tau = \left( {{t}_{k} = {bk}}\right) \) . Here \( s = u = 0 \), and system (13) reads\n\n\[ \n{F}^{\prime } = 1,\;{B}^{\prime } = B\left( {a + {bF}}\right) .\n\] | We obtain \( F\left( z\right) = z \), and \( {\left( \log B\right) }^{\prime } = \frac{{B}^{\prime }}{B} = a + {bz} \) ; thus \( B\left( z\right) = {e}^{{az} + \frac{b{z}^{2}}{2}} \) . In particular, for \( a = b = 1 \) we obtain the exponential generating function of the involution numbers \( {i}_{n} \) considered in ... | Yes |
Lemma 7.19. Suppose \( A = {A}^{\sigma ,\tau } = \left( {a}_{n, k}\right) \) . (i) \( \widetilde{A} = {PA} \) is the Catalan matrix corresponding to the sequences \( \widetilde{\sigma } \equiv \sigma + 1 = \left( {{s}_{0} + 1,{s}_{1} + 1,{s}_{2} + 1,\ldots }\right) ,\widetilde{\tau } = \tau . \) (ii) \( \bar{A} = \left... | We know that the \( \ell \) -th power of \( P \) has as \( \left( {n, k}\right) \) -entry \( \left( \begin{array}{l} n \\ k \end{array}\right) {\ell }^{n - k},\ell \in \) \( \mathbb{Z} \) (see Exercise 2.38). Thus if \( \widetilde{A} = {\widetilde{A}}^{\sigma + \ell ,\tau } \), then \( \widetilde{A} = {P}^{\ell }A \), ... | No |
Proposition 7.21. We have\n\n\[ \n{B}_{n} = \mathop{\sum }\limits_{{P \in \Pi \left( n\right) }}W\left( P\right) \n\] | Proof. We have to check the recurrence\n\n\[ \n\mathop{\sum }\limits_{{P \in \Pi \left( {n + 1}\right) }}W\left( P\right) = a\mathop{\sum }\limits_{{P \in \Pi \left( n\right) }}W\left( P\right) + b\mathop{\sum }\limits_{{k = 0}}^{{n - 1}}\left( \begin{array}{l} n \\ k \end{array}\right) {s}^{n - k - 1}\mathop{\sum }\li... | Yes |
Proposition 7.22. We have\n\n\[ \n{B}_{n} = \mathop{\sum }\limits_{{\pi \in \bar{S}\left( n\right) }}W\left( \pi \right) \n\] | The proof of recurrence (6) for \( \mathop{\sum }\limits_{{\pi \in \bar{S}\left( n\right) }}W\left( \pi \right) \) is analogous, by considering separately the cases \( {\pi }_{n + 1} = n + 1 \) and \( {\pi }_{n + 1} \neq n + 1 \), and is left to the exercises. | No |
Proposition 8.1. Let \( f\left( {{x}_{1},\ldots ,{x}_{n}}\right) \) be an alternating polynomial of degree \( d \) . Then\n\n\[ \frac{f\left( {{x}_{1},\ldots ,{x}_{n}}\right) }{\mathop{\prod }\limits_{{1 \leq i < j \leq n}}\left( {{x}_{i} - {x}_{j}}\right) } \]\n\n(2)\n\nis a symmetric polynomial of degree \( d - \left... | Proof. Since both numerator and denominator are alternating, the quotient is symmetric, and the degree is clearly \( d - \left( \begin{array}{l} n \\ 2 \end{array}\right) \) . So all we have to show is that the quotient in (2) is a polynomial.\n\nConsider \( f\left( {{x}_{1},\ldots ,{x}_{n}}\right) \) as a polynomial i... | Yes |
Proposition 8.2. Let \( \lambda = {\lambda }_{1}{\lambda }_{2}\ldots {\lambda }_{r} \in \operatorname{Par}\left( m\right) \) . Then\n\n\[ \n{e}_{\lambda } = \mathop{\sum }\limits_{{\mu \in \operatorname{Par}\left( m\right) }}{M}_{\lambda \mu }{m}_{\mu }\n\]\n\n(2)\n\nwhere \( {M}_{\lambda \mu } \) is the number of 0,1-... | Proof. We have\n\n\[ \n{e}_{\lambda } = {e}_{{\lambda }_{1}}{e}_{{\lambda }_{2}}\cdots = \mathop{\sum }\limits_{{{i}_{1} < \cdots < {i}_{{\lambda }_{1}}}}{x}_{{i}_{1}}\cdots {x}_{{i}_{{\lambda }_{1}}}\mathop{\sum }\limits_{{{j}_{1} < \cdots < {j}_{{\lambda }_{2}}}}{x}_{{j}_{1}}\cdots {x}_{{j}_{{\lambda }_{2}}}\cdots .\... | Yes |
Theorem 8.4. The set \( \left\{ {{e}_{\lambda }\left( x\right) : \lambda \in \operatorname{Par}\left( m\right) }\right\} \) is a basis for \( {\Lambda }^{m}\left( X\right) \) . | Proof. We have to show that any monomial function \( {m}_{\mu } \) is in the span \( \left\langle {{e}_{\lambda } : \lambda \in \operatorname{Par}\left( m\right) }\right\rangle \) . To accomplish this we first define a total ordering on the set \( \operatorname{Par}\left( m\right) \) . Set\n\n\[ \lambda < \mu : \Leftri... | Yes |
Theorem 8.6. The set \( \left\{ {{h}_{\lambda }\left( x\right) : \lambda \in \operatorname{Par}\left( m\right) }\right\} \) is a basis for \( {\Lambda }^{m}\left( X\right) \) . | Proof. Since \( \left\{ {{e}_{\lambda } : \lambda \in \operatorname{Par}\left( m\right) }\right\} \) is a basis, it suffices to show that any \( {e}_{\lambda } \) is in the span \( \left\langle {{h}_{\mu } : \mu \in \operatorname{Par}\left( m\right) }\right\rangle \) . First we note from (1) and (3) that\n\n\[ \left( {... | Yes |
Theorem 8.7. Let \( \lambda = {\lambda }_{1}\ldots {\lambda }_{n} \in \) Par with possible 0’s at the end. Then\n\n\[ \n{s}_{\lambda }\left( {{x}_{1},\ldots ,{x}_{n}}\right) = \det {\left( {h}_{{\lambda }_{i} - i + j}\right) }_{i, j = 1}^{n}, \]\n\nwhere we set \( {h}_{k} = 0 \) for \( k < 0 \) . | Proof. Let \( {e}_{j}^{\left( \ell \right) } \) be the \( j \) -th elementary symmetric function on \( X = \) \( \left\{ {{x}_{1},\ldots ,{x}_{n}}\right\} \smallsetminus {x}_{\ell } \) . By (1) and (3) of the previous section,\n\n\[ \n\mathop{\sum }\limits_{{k \geq 0}}{h}_{k}{z}^{k} \cdot \mathop{\sum }\limits_{{j = 0}... | Yes |
Corollary 8.9. Let \( \lambda \in \operatorname{Par}\left( n\right) ,\lambda = {\lambda }_{1}\ldots {\lambda }_{n} \), with possible 0 ’s at the end. Then \[ {s}_{\lambda }\left( {{x}_{1},\ldots ,{x}_{n}}\right) = \mathop{\sum }\limits_{{\mu \in \operatorname{Par}\left( n\right) }}{K}_{\lambda \mu }{m}_{\mu }, \] where... | Proof. A monomial \( {x}^{T} = {x}_{1}^{{\mu }_{1}}\cdots {x}_{n}^{{\mu }_{n}} \) appears as often as there are SSTs of type \( \mu \) . | Yes |
Theorem 8.10. The set \( \left\{ {{s}_{\lambda } : \lambda \in \operatorname{Par}\left( n\right) }\right\} \) is a basis for \( {\Lambda }^{n}\left( X\right), X = \) \( \left\{ {{x}_{1},\ldots ,{x}_{n}}\right\} \) . | Proof. We proceed as in the proof of Theorem 8.4. This time we choose the so-called dominance order on \( \operatorname{Par}\left( n\right) \) . Let \( \lambda = {\lambda }_{1}\ldots {\lambda }_{n} \) , \( \mu = {\mu }_{1}\ldots {\mu }_{n} \) in \( \operatorname{Par}\left( n\right) \) with possible 0 ’s at the end. The... | Yes |
Corollary 8.12 (MacMahon). We have\n\n\\[ \n\\mathop{\\sum }\\limits_{{n \\geq 0}}{pp}\\left( n\\right) {q}^{n} = \\mathop{\\prod }\\limits_{{i \\geq 1}}\\frac{1}{{\\left( 1 - {q}^{i}\\right) }^{i}}. \n\\] | Proof. All we have to do is to let \\( r, s \\), and \\( t \\) go to infinity in (8). Let \\( n = i + j + k - 1 \\) or \\( n + 1 = i + j + k \\), that is, \\( n + 1 \\) is an ordered partition into three parts. We know from Section 1.5 that there are precisely \\( \\left( \\begin{array}{l} n \\\\ 2 \\end{array}\\right)... | Yes |
Lemma 8.13. If \( T \) is an SST, then so is \( T \leftarrow k \) . | Proof. The rows in \( T \leftarrow k \) are obviously monotone, and rule (1) implies that the columns are strictly increasing. | No |
Lemma 8.14. Suppose we perform \( T \leftarrow k \), and then \( \left( {T \leftarrow k}\right) \leftarrow \ell \) with \( \ell \geq k \). Then the insertion path of \( \ell \) runs strictly to the right of the insertion of \( k \). | Example. For \( \left( {T \leftarrow 3}\right) \leftarrow 3 \) in the example above we get\n\n\[ \text{1223366} \]\n\n\[ \text{23345} \]\n\n\[ \left( {T \leftarrow 3}\right) \leftarrow 3 : \;{44677} \]\n\n\[ \text{6778} \]\n\n\[ \text{7 8} \] | No |
Theorem 8.15. The RSK algorithm gives a bijection between the matrices \( A \) over \( {\mathbb{N}}_{0} \) (with finitely many nonzero elements) and the ordered pairs \( \left( {P, Q}\right) \) of semistandard tableaux of the same shape. Furthermore, type \( \left( P\right) = \operatorname{col}\left( A\right) \), type ... | Proof. Suppose \( A\overset{\mathrm{{RSK}}}{ \rightarrow }\left( {P, Q}\right) \) . By Lemma 8.13, \( P \) is an SST, and \( Q \) has by construction the same shape as \( P \) . So we have to show that \( Q \) is also an SST. Since in the \( 2 \times n \) -scheme the first row \( {i}_{1} \leq \cdots \leq {i}_{n} \) is ... | Yes |
Corollary 8.16 (Cauchy). Let \( X = \\left\\{ {{x}_{1},{x}_{2},\\ldots }\\right\\}, Y = \\left\\{ {{y}_{1},{y}_{2},\\ldots }\\right\\} \) . Then \[ \\mathop{\\prod }\\limits_{{i, j}}\\frac{1}{1 - {x}_{i}{y}_{j}} = \\mathop{\\sum }\\limits_{{\\lambda \\in \\operatorname{Par}}}{s}_{\\lambda }\\left( x\\right) {s}_{\\lamb... | Proof. We write \[ \\mathop{\\prod }\\limits_{{i, j}}\\frac{1}{1 - {x}_{i}{y}_{j}} = \\mathop{\\prod }\\limits_{{i, j}}\\left( {\\mathop{\\sum }\\limits_{{{a}_{ji} \\geq 0}}{\\left( {x}_{i}{y}_{j}\\right) }^{{a}_{ji}}}\\right) . \] Expanding the right-hand side of (3), we see that a monomial \( {x}^{\\alpha }{y}^{\\bet... | Yes |
Corollary 8.17. Let \( \mu, v \in \operatorname{Par}\left( n\right) \) . Then\n\n\[ \mathop{\sum }\limits_{{\lambda \in \operatorname{Par}\left( n\right) }}{K}_{\lambda \mu }{K}_{\lambda v} = {N}_{\mu v} \]\n\nwhere \( {K}_{\lambda ,\mu },{N}_{\mu \nu } \) are defined as in the previous sections. | Proof. According to Exercise 8.14,\n\n\[ \mathop{\prod }\limits_{{i, j}}\frac{1}{1 - {x}_{i}{y}_{j}} = \mathop{\sum }\limits_{{\mu ,\nu \in \operatorname{Par}}}{N}_{\mu \nu }{m}_{\mu }\left( x\right) {m}_{\nu }\left( y\right) \]\n\nSo for \( \mu, v \in \operatorname{Par}\left( n\right) \) the coefficient of \( {x}^{\mu... | Yes |
Lemma 8.18. Suppose \( \alpha = {a}_{1}\ldots {a}_{n} \) and \( \beta = {b}_{1}\ldots {b}_{n} \) have associated tableaux \( {P}_{\alpha } \) and \( {P}_{\beta } \) . If the order permutations are identical, \( {\pi }_{\alpha } = {\pi }_{\beta } \), then \( {P}_{\beta } \) arises from \( {P}_{\alpha } \) by replacing e... | Proof. All we have to notice is that at every stage of the algorithm the insertion path of \( {a}_{t + 1} \) in \( {P}_{\alpha }\left( t\right) \) to produce \( {P}_{\alpha }\left( {t + 1}\right) \) is identical to the insertion path of \( {b}_{t + 1} \) in \( {P}_{\beta }\left( t\right) \) . But this is clear by rule ... | Yes |
If the RSK algorithm observes the symmetry property (5) for all permutation matrices, then it holds in general. | Proof. Suppose \( A\overset{\mathrm{{RSK}}}{ \rightarrow }\left( {P, Q}\right) ,\widetilde{A} \rightarrow \left( {\widetilde{P},\widetilde{Q}}\right) \) . Then by (6) and (7),\n\n\n\nand we are done. | No |
Lemma 8.21. Given \( \pi \in S\left( n\right) \), then \( {fs}\left( a\right) = {is}\left( a\right) \) for all \( a \in \) \( \{ 1,\ldots, n\} \) . | Proof. Property (iii) immediately implies \( {fs}\left( a\right) \leq {is}\left( a\right) \) . Now let \( {b}_{1} < {b}_{2} < \cdots < {b}_{{is}\left( a\right) } = a \) be a longest increasing subword ending in \( a \) . We clearly have \( {is}\left( {b}_{k}\right) = k,1 \leq k \leq {is}\left( a\right) \), and therefor... | Yes |
Corollary 8.25. We have\n\n\[ \frac{1}{\mathop{\prod }\limits_{{i \geq 1}}\left( {1 - {x}_{i}}\right) \mathop{\prod }\limits_{{1 \leq i < j}}\left( {1 - {x}_{i}{x}_{j}}\right) } = \mathop{\sum }\limits_{{\lambda \in \operatorname{Par}}}{s}_{\lambda }\left( x\right) . \] | Proof. The product on the left is\n\n\[ \left( {\mathop{\sum }\limits_{{k \geq 0}}{x}_{1}^{k}}\right) \left( {\mathop{\sum }\limits_{{k \geq 0}}{x}_{2}^{k}}\right) \cdots \left( {\mathop{\sum }\limits_{{k \geq 0}}{\left( {x}_{1}{x}_{2}\right) }^{k}}\right) \cdots .\n\nIt follows that the coefficient of \( {x}^{\alpha }... | Yes |
Theorem 8.26. We have\n\na. \( \mathop{\sum }\limits_{{\lambda \in \operatorname{Par}\left( n\right) }}f{\left( \lambda \right) }^{2} = n! \) ,\n\nb. \( \mathop{\sum }\limits_{{\lambda \in \operatorname{Par}\left( n\right) }}f\left( \lambda \right) = {i}_{n} \) (number of involutions). | Looking at the example above we obtain \( f\left( 4\right) = 1, f\left( {31}\right) = 3 \) , \( f\left( {22}\right) = 2, f\left( {211}\right) = 3, f\left( {1111}\right) = 1 \), which agrees with \( {1}^{2} + {3}^{2} + \) \( {2}^{2} + {3}^{2} + {1}^{2} = {24} \) and \( 1 + 3 + 2 + 3 + 1 = {10} = {i}_{4} \) . Note also t... | Yes |
Proposition 8.27. Let \( \pi \overset{\text{ ins }}{ \rightarrow }P \), and \( \lambda = {\lambda }_{1}\ldots {\lambda }_{r} \) the shape of \( P \) . Then a. \( {is}\left( \pi \right) = {\lambda }_{1} \) , b. \( \# \{ \pi \in S\left( n\right) : \) is \( \left( \pi \right) = t\} = \mathop{\sum }\limits_{{\lambda \in \o... | It is natural to look at decreasing subwords as well. Let \( {ds}\left( \pi \right) \) be the length of a longest decreasing subword of \( \pi \) . We suspect that \( {ds}\left( \pi \right) = {\lambda }_{1}^{ * } = \) number of rows of \( \lambda \), and this is indeed the case. If \( \pi = {\pi }_{1}\ldots {\pi }_{n} ... | No |
Lemma 8.28. Let \( T \) be an SST with distinct entries, and \( k \neq \ell \) two numbers not contained in \( T \) . Then\n\n\[ \ell \rightarrow \left( {T \leftarrow k}\right) = \left( {\ell \rightarrow T}\right) \leftarrow k. \]\n\n(1)\n\nIn other words, the operations \( \leftarrow \) and \( \rightarrow \) commute. | Proof. Let \( {W}_{1} : k < {x}_{1} < \cdots < {x}_{r} \) be the insertion path when \( k \) is inserted in \( T \) . Thus \( {x}_{1} \) is bumped by \( k,{x}_{2} \) is bumped by \( {x}_{1} \), and so on. The element \( {x}_{i} \) is therefore in row \( i \) of \( T \), and in row \( i + 1 \) of \( T \leftarrow k \) . ... | Yes |
Lemma 8.29. Let \( w = {w}_{1}\ldots {w}_{n} \) be a sequence of distinct numbers, \( {w}^{ * } = {w}_{n}\ldots {w}_{1} \) the reversed sequence. Denote by \( T\left( {{w}_{1}\ldots {w}_{n}}\right) \) the tableau obtained by rowwise insertion of \( {w}_{1},{w}_{2},\ldots ,{w}_{n} \), and by \( {T}^{\prime }\left( {{w}_... | Proof. For \( n = 1 \) there is nothing to prove, and for \( n = 2 \) the assertion is easily checked. We proceed by induction on \( n \) . By the previous lemma and induction,\n\n\[ T\left( {{w}_{1}\ldots {w}_{n}}\right) = T\left( {{w}_{1}\ldots {w}_{n - 1}}\right) \leftarrow {w}_{n} = {T}^{\prime }\left( {{w}_{1}\ldo... | Yes |
Proposition 8.30. Let \( \pi = {\pi }_{1}\ldots {\pi }_{n} \in S\left( n\right) \) with \( \pi \overset{\text{ins }}{ \rightarrow }P \) . Then \( {\pi }^{ * }\overset{ins}{ \rightarrow }{P}^{ * } \) for the reversed permutation \( {\pi }^{ * } = {\pi }_{n}\ldots {\pi }_{1} \) . | Proof. By the lemma, \( P = T\left( {{\pi }_{1}\ldots {\pi }_{n}}\right) = {T}^{\prime }\left( {{\pi }_{1}\ldots {\pi }_{n}}\right) \), where \( {\pi }^{ * } = {\pi }_{n}\ldots {\pi }_{1} \) is inserted columnwise. Hence \( {\pi }^{ * }\overset{ins}{ \rightarrow }{P}^{ * } \) when \( {\pi }^{ * } \) is inserted rowwise... | Yes |
Proposition 9.1. Suppose \( e = \{ u, v\} \) is not a loop of \( G \) . Then \[ \chi \left( {G;\lambda }\right) = \chi \left( {G \smallsetminus e;\lambda }\right) - \chi \left( {G/e;\lambda }\right) . \] | Proof. Look at the \( \lambda \) -colorings of \( G \smallsetminus e \) . The colorings \( c \) with \( c\left( u\right) \neq c\left( v\right) \) correspond bijectively to the colorings of \( G \), and those with \( c\left( u\right) = c\left( v\right) \) to the colorings of \( G/e \) . The result follows. | Yes |
Proposition 9.2. Let \( G = \left( {V, E}\right) \) be a loopless graph. Then \( \chi \left( {G;\lambda }\right) = \) \( {a}_{0}{\lambda }^{n} + {a}_{1}{\lambda }^{n - 1} + \cdots + {a}_{n} \) is a polynomial in \( \lambda \) with coefficients in \( \mathbb{Z} \) . Furthermore,\n\n(i) \( \chi \left( {G;\lambda }\right)... | Proof. We use induction on \( \left| E\right| \) . If \( G \) has no edges, then \( \chi \left( {G;\lambda }\right) = \) \( {\lambda }^{n} \) . Otherwise, pick an edge \( e \in E \) . Then\n\n\[ \chi \left( {G;\lambda }\right) = \chi \left( {G \smallsetminus e;\lambda }\right) - \chi \left( {G/e;\lambda }\right) ,\]\n\... | No |
Theorem 9.5. Let \( f \) be a chromatic invariant with \( A, B,\alpha ,\beta \) as above. Then for all graphs \( G = \left( {V, E}\right) \) , \[ f\left( G\right) = {\alpha }^{\left| E\right| - \left| V\right| + k\left( G\right) }{\beta }^{\left| V\right| - k\left( G\right) }T\left( {G;\frac{A}{\beta },\frac{B}{\alpha ... | Proof. Formula (10) is certainly true for edgeless graphs. If \( G \) consists only of bridges and loops, say \( \ell \) bridges and \( m \) loops, then \( f\left( G\right) = {A}^{\ell }{B}^{m} \) by (ii) above. Now, \( \left| V\right| - k\left( G\right) = \ell ,\left| E\right| - \left| V\right| + k\left( G\right) = m ... | Yes |
Proposition 9.6. We have\n\n\\[ \n\\chi \\left( {G;\\lambda }\\right) = {\\left( -1\\right) }^{\\left| V\\right| - k\\left( G\\right) }{\\lambda }^{k\\left( G\\right) }T\\left( {G;1 - \\lambda ,0}\\right) .\n\\] \n\n(11) | For \\( {K}_{4}^{ - } \\) we obtain from \\( T\\left( {{K}_{4}^{ - };x, y}\\right) = {x}^{3} + 2{x}^{2} + x + {2xy} + y + {y}^{2} \\) ,\n\n\\( \\chi \\left( {{K}_{4}^{ - };\\lambda }\\right) = - \\lambda \\left\\lbrack {{\\left( 1 - \\lambda \\right) }^{3} + 2{\\left( 1 - \\lambda \\right) }^{2} + \\left( {1 - \\lambda... | No |
Let \( F\left( {G;A}\right) \) be the number of \( A \) -flows; then\n\n\[ F\left( {G;A}\right) = {\left( -1\right) }^{\left| E\right| - \left| V\right| + k\left( G\right) }T\left( {G;0,1 - \left| A\right| }\right) . | Now this implies the somewhat unexpected result that the number of flows depends only on the size \( \left| A\right| \), but not on the structure of the group \( A \), nor on the orientation. We may thus give the following definition:\n\nDefinition. The flow polynomial \( F\left( {G;\lambda }\right) \) is the polynomia... | Yes |
Proposition 9.8. We have for \( G = \\left( {V, E}\\right) \), \[ {ac}\\left( G\\right) = T\\left( {G;2,0}\\right) = {\\left( -1\\right) }^{\\left| V\\right| }\\chi \\left( {G; - 1}\\right) = \\left| {\\chi \\left( {G; - 1}\\right) }\\right| . \] | Proof. With (11) the Recipe Theorem 9.5 will imply (16), once we show that for \( e \) not a loop or bridge, \[ {ac}\\left( G\\right) = {ac}\\left( {G \\smallsetminus e}\\right) + {ac}\\left( {G/e}\\right) . \] But this is easy. Let \( e = \\{ u, v\\} \), and consider an arbitrary acyclic orientation \( O \) of \( G \\... | Yes |
Corollary 9.11. We have\n\n\[ \nQ\left( {H;x}\right) = \mathop{\sum }\limits_{{T \subseteq \{ 1,\ldots, n\} }}{\left( x - 1\right) }^{\left| T\right| - \operatorname{rk}\left( {A}_{T}\right) }, \n\]\n\n(5)\n\nwith \( \operatorname{rk}\left( {A}_{\varnothing }\right) = 0 \) by definition. | It follows from recurrence (3) that \( Q\left( {H;x}\right) \) is a polynomial with nonnegative integer coefficients, which in view of the expression (4) is rather surprising. Furthermore, (3) implies\n\n\[ \nQ\left( {{H}^{ab};x}\right) = Q\left( {{H}^{ab} \smallsetminus b;x}\right) + Q\left( {H \smallsetminus a;x}\rig... | No |
Corollary 9.13. We have\n\n\( Q\left( {H;1}\right) = \# \) induced subgraphs of \( H \) with an odd number of\n\nperfect matchings (including the empty set). | Since a forest has at most one perfect matching; this yields in particular the following: | No |
Proposition 9.15. Let \( G \) be a 2-in 2-out graph, and \( H \) any of its interlace graphs. Then for the number \( e\left( G\right) \) of Eulerian cycles, and for \( {e}_{k}\left( G\right) \), we have\na. \( e\left( G\right) = Q\left( {H;1}\right) \),\nb. \( e\left( {G;x}\right) \mathrel{\text{:=}} \mathop{\sum }\lim... | Proof. For an interlace graph \( H \) of \( G \) define \( f\left( H\right) = e\left( G\right) \) . We clearly have \( f\left( {{H}_{1}\dot{ \cup }{H}_{2}}\right) = f\left( {H}_{1}\right) f\left( {H}_{2}\right) \), and the recurrence holds because of (2). It remains to consider \( H = {K}_{1} \), with\n\n![8ae744ce-978... | Yes |
Corollary 9.18. We have\n\n\[ P\\left( {G;2}\\right) = \\left\\{ \\begin{matrix} {2}^{\\left| V\\right| } & \\text{ if }G\\text{ is Eulerian,} \\\\ 0 & \\text{ otherwise. } \\end{matrix}\\right. \] | Proof. An admissible 2-valuation must assume alternate numbers around a black face of \( \\widetilde{G} \) . Hence every black face must have an even number of boundary edges, or equivalently \( G \) must be Eulerian. Since there are two possibilities for each black face, the result follows. | Yes |
Theorem 9.19. Suppose \( G \) is a connected plane 3-regular graph; then\n\n\[ P\left( {G;3}\right) = \# 3\\text{-edge-colorings.} \]\n | Proof. The following picture shows the whole proof. Relate a 3- edge-coloring \( g \) of \( G \) to a 3-valuation \( \\bar{g} \) of \( \\widetilde{G} \) as in the figure:\n\n\n\nIt is easily seen that \( \\bar{g} \) ... | No |
Lemma 9.21. Suppose the weightings \( W \) and \( {W}^{\prime } \) differ by an additive constant, that is, \( {\alpha }^{\prime } = \alpha + m,{\beta }^{\prime } = \beta + m,{y}^{\prime } = y + m \) . Then \( S\left( {\widetilde{G},{W}^{\prime }; - 2}\right) = S\left( {\widetilde{G}, W; - 2}\right) . | Example. Take \( W = {W}_{0,1, - 1} \) and \( {W}^{\prime } = {W}_{1,2,0} \) . The first weighting gives the Penrose polynomial, while the second is covered by the previous theorem. The lemma implies\n\n\[ \nP\left( {G; - 2}\right) = S\left( {\widetilde{G},{W}_{0,1, - 1}; - 2}\right) = S\left( {\widetilde{G},{W}_{1,2,0... | No |
Corollary 9.23. The number of Eulerian cycles in the 2-in 2-out graph \( {\widetilde{G}}^{c} \) equals the number of spanning trees of \( G \) . | Consider, finally, the evaluation \( \alpha = \beta = 0,\gamma = 1 \) . That is, we take the unique transition system \( q \) consisting only of crossing transitions. Subtracting 1, we get the weighting \( \overline{\alpha } = \overline{\beta } = - 1,\overline{y} = 0 \) . Lemma 9.21 and (9) imply\n\n\[ S\left( {\wideti... | Yes |
Corollary 9.24. Let \( G = \left( {V, E, F}\right) \) be a connected plane graph, and \( \widetilde{G} \) its medial graph. Then\n\n\[ T\left( {G; - 1, - 1}\right) = {\left( -1\right) }^{\left| E\right| }{\left( -2\right) }^{c\left( q\right) - 1}, \]\n\nwhere \( q \) is the all-crossing transition system of \( \widetil... | Example. We have computed the Tutte polynomial for the graph \( {K}_{4}^{ - } \) in Section 9.1, obtaining \( T\left( {{K}_{4}^{ - }; - 1, - 1}\right) = 2 \) . Thus \( c\left( q\right) = 2 \), and the figure shows the corresponding decomposition into two cycles. | No |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.