Mixing
Show that $χ_G(i − 1) = 0$
$begingroup$
Can you give me a hint for this exercise about Poset (Partially ordered set)?
Let $χ_G(t)$ be the characteristic polynomial of the graphical
arrangement $A_G$. Suppose that $χ_G(i) = 0$, where $i in
> mathbb{N}$, $i > 1$. Show that $χ_G(i − 1) = 0$
By Enumerative Combinatorics from Stanley, chapter 3.11, I know that
the characteristic polynomial $χ_A(x)$ of the arrangement $A$ is $χ_A(x) = sum_{t in L(A)} mu(hat0,t)x^{dim(t)}$.
And if $V = {v_1, . . . , v_p}$, then define the graphical arrangement $A_G$ to be the arrangement in $R^p$ with hyperplanes $x_i = x_j$, where $v_i$ and $v_j$ are adjacent vertices of $G$. The characteristic polynomial $χ_G(t)$ of the graphical arrangement $A_G$ is also the chromatic polynomial of the graph $G$. The value of $χ_G(t)$ at a positive integer $t$ equals the number of ways to color the vertices of the graph $G$ in $t$ colors so that all neighboring pairs of vertices have different colors.
If we have that $χ_G(i) = 0 Rightarrow χ_A(i)=sum_{t in L(A)} mu(hat0,t)x^{dim(t)}$= 0$
I have to proceeding by backward induction?
order-theory
edited Jan 13 at 18:24
greedoid
42.4k1153105
asked Nov 26 '18 at 1:14
Vlllbbb1777Vlllbbb1777
113
$endgroup$
add a comment |
$begingroup$
Can you give me a hint for this exercise about Poset (Partially ordered set)?
Let $χ_G(t)$ be the characteristic polynomial of the graphical
arrangement $A_G$. Suppose that $χ_G(i) = 0$, where $i in
> mathbb{N}$, $i > 1$. Show that $χ_G(i − 1) = 0$
By Enumerative Combinatorics from Stanley, chapter 3.11, I know that
the characteristic polynomial $χ_A(x)$ of the arrangement $A$ is $χ_A(x) = sum_{t in L(A)} mu(hat0,t)x^{dim(t)}$.
And if $V = {v_1, . . . , v_p}$, then define the graphical arrangement $A_G$ to be the arrangement in $R^p$ with hyperplanes $x_i = x_j$, where $v_i$ and $v_j$ are adjacent vertices of $G$. The characteristic polynomial $χ_G(t)$ of the graphical arrangement $A_G$ is also the chromatic polynomial of the graph $G$. The value of $χ_G(t)$ at a positive integer $t$ equals the number of ways to color the vertices of the graph $G$ in $t$ colors so that all neighboring pairs of vertices have different colors.
If we have that $χ_G(i) = 0 Rightarrow χ_A(i)=sum_{t in L(A)} mu(hat0,t)x^{dim(t)}$= 0$
I have to proceeding by backward induction?
order-theory
edited Jan 13 at 18:24
greedoid
42.4k1153105
asked Nov 26 '18 at 1:14
Vlllbbb1777Vlllbbb1777
113
$endgroup$
$begingroup$
I don't understand why this was downvoted; on the face of it, it looks like a high quality question.
$endgroup$
– Shaun
Nov 26 '18 at 1:25
add a comment |
$begingroup$
Can you give me a hint for this exercise about Poset (Partially ordered set)?
Let $χ_G(t)$ be the characteristic polynomial of the graphical
arrangement $A_G$. Suppose that $χ_G(i) = 0$, where $i in
> mathbb{N}$, $i > 1$. Show that $χ_G(i − 1) = 0$
By Enumerative Combinatorics from Stanley, chapter 3.11, I know that
the characteristic polynomial $χ_A(x)$ of the arrangement $A$ is $χ_A(x) = sum_{t in L(A)} mu(hat0,t)x^{dim(t)}$.
And if $V = {v_1, . . . , v_p}$, then define the graphical arrangement $A_G$ to be the arrangement in $R^p$ with hyperplanes $x_i = x_j$, where $v_i$ and $v_j$ are adjacent vertices of $G$. The characteristic polynomial $χ_G(t)$ of the graphical arrangement $A_G$ is also the chromatic polynomial of the graph $G$. The value of $χ_G(t)$ at a positive integer $t$ equals the number of ways to color the vertices of the graph $G$ in $t$ colors so that all neighboring pairs of vertices have different colors.
If we have that $χ_G(i) = 0 Rightarrow χ_A(i)=sum_{t in L(A)} mu(hat0,t)x^{dim(t)}$= 0$
I have to proceeding by backward induction?
order-theory
edited Jan 13 at 18:24
greedoid
42.4k1153105
asked Nov 26 '18 at 1:14
Vlllbbb1777Vlllbbb1777
113
$endgroup$
Can you give me a hint for this exercise about Poset (Partially ordered set)?
Let $χ_G(t)$ be the characteristic polynomial of the graphical
arrangement $A_G$. Suppose that $χ_G(i) = 0$, where $i in
> mathbb{N}$, $i > 1$. Show that $χ_G(i − 1) = 0$
By Enumerative Combinatorics from Stanley, chapter 3.11, I know that
the characteristic polynomial $χ_A(x)$ of the arrangement $A$ is $χ_A(x) = sum_{t in L(A)} mu(hat0,t)x^{dim(t)}$.
And if $V = {v_1, . . . , v_p}$, then define the graphical arrangement $A_G$ to be the arrangement in $R^p$ with hyperplanes $x_i = x_j$, where $v_i$ and $v_j$ are adjacent vertices of $G$. The characteristic polynomial $χ_G(t)$ of the graphical arrangement $A_G$ is also the chromatic polynomial of the graph $G$. The value of $χ_G(t)$ at a positive integer $t$ equals the number of ways to color the vertices of the graph $G$ in $t$ colors so that all neighboring pairs of vertices have different colors.
If we have that $χ_G(i) = 0 Rightarrow χ_A(i)=sum_{t in L(A)} mu(hat0,t)x^{dim(t)}$= 0$
I have to proceeding by backward induction?
order-theory
order-theory
edited Jan 13 at 18:24
greedoid
42.4k1153105
asked Nov 26 '18 at 1:14
Vlllbbb1777Vlllbbb1777
113
edited Jan 13 at 18:24
greedoid
42.4k1153105
asked Nov 26 '18 at 1:14
Vlllbbb1777Vlllbbb1777
113
edited Jan 13 at 18:24
greedoid
42.4k1153105
edited Jan 13 at 18:24
greedoid
42.4k1153105
edited Jan 13 at 18:24
greedoid
42.4k1153105
42.4k1153105
asked Nov 26 '18 at 1:14
Vlllbbb1777Vlllbbb1777
113
asked Nov 26 '18 at 1:14
Vlllbbb1777Vlllbbb1777
113
asked Nov 26 '18 at 1:14
Vlllbbb1777Vlllbbb1777
113
113
$begingroup$
I don't understand why this was downvoted; on the face of it, it looks like a high quality question.
$endgroup$
– Shaun
Nov 26 '18 at 1:25
add a comment |
$begingroup$
I don't understand why this was downvoted; on the face of it, it looks like a high quality question.
$endgroup$
– Shaun
Nov 26 '18 at 1:25
$begingroup$
I don't understand why this was downvoted; on the face of it, it looks like a high quality question.
$endgroup$
– Shaun
Nov 26 '18 at 1:25
$begingroup$
I don't understand why this was downvoted; on the face of it, it looks like a high quality question.
$endgroup$
– Shaun
Nov 26 '18 at 1:25
add a comment |
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$begingroup$
I don't understand why this was downvoted; on the face of it, it looks like a high quality question.
$endgroup$
– Shaun
Nov 26 '18 at 1:25