Is it Possible to Apply the Finite Field Method to Hyperplane Arrangements in $mathbb{R}^n$?
$begingroup$
I have a question regarding the finite field method to compute the characteristic polynomial of a hyperplane arrangement. I am using the book "Enumerative Combinatorics, Volume 1" second edition by Richard P. Stanley. This method may be found in chapter 3.11.4.
I will begin by stating a problem I am interested solving using the finite field method. Then I will state the theorem and tell you might thoughts on this. In the bottom of this post, you may find my questions.
I'm interested in solving 3.114 (b) in the book by Stanley:
For an arrangement $mathcal{A}$ below (in $mathbb{R}^n$) show that the characteristic polynomials are as indicated. $x_i=x_j$ for $1leq i<jleq n$ and $sum_{i=1}^n x_i=0$. Then
$$chi_{mathcal{A}}(x)=(x-1)^2(x-2)(x-3)cdots (x-n+1).$$
If I am allowed to use the finite field method, I have a solution to this problem that gives me the correct polynomial. However, by reading the theorem (I will state it down below), I'm not sure why I would be allowed to use it.
The theorem only says something about $mathbb{Q}^n$ and not $mathbb{R}^n$.
Let us now state the theorem.
Theorem:
Let $mathcal{A}$ be an arrangement in $mathbb{Q}^n$, and suppose that $L(mathcal{A})cong L(mathcal{A}_q)$ for some prime power $q$. Then
$$
chi_{mathcal{A}}(q)=#Big ( mathbb{F}_q - bigcup_{Hinmathcal{A}_q} H Big )=q^n-#bigcup_{Hinmathcal{A}_q} H.
$$
To me, it really seems as we need to work in the vector space $mathbb{Q}^n$ to apply this theorem. This theorem does not seem to tell us that we can do this in the vector space $mathbb{R}^n$.
I still have some hope that I might be able to apply the above theorem for two reasons.
(1) Just to quote Stanley in the first paragraph when he starts to talk about the finite field method
In this subsection we will describe a method based on finite fields
for computing the characteristic polynomial of an arrangement
defined over $mathbb{Q}$.
It seems to me that he wants to tell the reader that the finite field method works if your hyperplanes are defined over $mathbb{Q}$. After doing some research what "defined over $mathbb{Q}$" means, it seems as this means the hyperplanes have rational coefficients. Surely, in the problem I want to solve the hyperplanes does, indeed, have rational coefficients.
(2) In example 3.11.11 in the book, Stanley uses the finite field method to find the characteristic polynomial of the braid arrangement $mathcal{B}_n$. This is an arrangement in $mathbb{K}^n$ with hyperplanes $x_i-x_j=0$ for $1leq i<jleq n$.
He hasn't said anything about which field $mathbb{K}$ we consider. Thus, I guess, we could for instance let $mathbb{K}=mathbb{R}$? Notice the polynomials are defined over $mathbb{Q}$.
It is very likely that I am misinterpreting something. Perhaps I am misinterpreting the theorem or the example?
I would be really happy if someone could tell me why we can apply the theorem when we have a hyperplane arrangement in $mathbb{R}^n$ (if it is possible) and if not perhaps tell me what's happening in example 3.11.11.
Thanks for taking your time!
combinatorics polynomials finite-fields
asked Jan 12 at 21:50
JoeJoe
4214
$endgroup$
add a comment |
$begingroup$
I have a question regarding the finite field method to compute the characteristic polynomial of a hyperplane arrangement. I am using the book "Enumerative Combinatorics, Volume 1" second edition by Richard P. Stanley. This method may be found in chapter 3.11.4.
I will begin by stating a problem I am interested solving using the finite field method. Then I will state the theorem and tell you might thoughts on this. In the bottom of this post, you may find my questions.
I'm interested in solving 3.114 (b) in the book by Stanley:
For an arrangement $mathcal{A}$ below (in $mathbb{R}^n$) show that the characteristic polynomials are as indicated. $x_i=x_j$ for $1leq i<jleq n$ and $sum_{i=1}^n x_i=0$. Then
$$chi_{mathcal{A}}(x)=(x-1)^2(x-2)(x-3)cdots (x-n+1).$$
If I am allowed to use the finite field method, I have a solution to this problem that gives me the correct polynomial. However, by reading the theorem (I will state it down below), I'm not sure why I would be allowed to use it.
The theorem only says something about $mathbb{Q}^n$ and not $mathbb{R}^n$.
Let us now state the theorem.
Theorem:
Let $mathcal{A}$ be an arrangement in $mathbb{Q}^n$, and suppose that $L(mathcal{A})cong L(mathcal{A}_q)$ for some prime power $q$. Then
$$
chi_{mathcal{A}}(q)=#Big ( mathbb{F}_q - bigcup_{Hinmathcal{A}_q} H Big )=q^n-#bigcup_{Hinmathcal{A}_q} H.
$$
To me, it really seems as we need to work in the vector space $mathbb{Q}^n$ to apply this theorem. This theorem does not seem to tell us that we can do this in the vector space $mathbb{R}^n$.
I still have some hope that I might be able to apply the above theorem for two reasons.
(1) Just to quote Stanley in the first paragraph when he starts to talk about the finite field method
In this subsection we will describe a method based on finite fields
for computing the characteristic polynomial of an arrangement
defined over $mathbb{Q}$.
It seems to me that he wants to tell the reader that the finite field method works if your hyperplanes are defined over $mathbb{Q}$. After doing some research what "defined over $mathbb{Q}$" means, it seems as this means the hyperplanes have rational coefficients. Surely, in the problem I want to solve the hyperplanes does, indeed, have rational coefficients.
(2) In example 3.11.11 in the book, Stanley uses the finite field method to find the characteristic polynomial of the braid arrangement $mathcal{B}_n$. This is an arrangement in $mathbb{K}^n$ with hyperplanes $x_i-x_j=0$ for $1leq i<jleq n$.
He hasn't said anything about which field $mathbb{K}$ we consider. Thus, I guess, we could for instance let $mathbb{K}=mathbb{R}$? Notice the polynomials are defined over $mathbb{Q}$.
It is very likely that I am misinterpreting something. Perhaps I am misinterpreting the theorem or the example?
I would be really happy if someone could tell me why we can apply the theorem when we have a hyperplane arrangement in $mathbb{R}^n$ (if it is possible) and if not perhaps tell me what's happening in example 3.11.11.
Thanks for taking your time!
combinatorics polynomials finite-fields
asked Jan 12 at 21:50
JoeJoe
4214
$endgroup$
1
$begingroup$
$Lleft(mathcal{A}right)$ does not change if you extend your base field. I don't know of a particularly slick way to explain why, except of "everything boils down to solving systems of linear equations, and the Gaussian elimination algorithm does not care if the field suddenly grows larger".
$endgroup$
– darij grinberg
Jan 12 at 23:17
$begingroup$
@darijgrinberg, thank you for your answer! :) I think I will get it (why it works in $mathbb{R}^n$), if I spend an hour or so to think through your answer.
$endgroup$
– Joe
Jan 12 at 23:25
1
$begingroup$
Actually a great first step is to convince yourself of the following: Write $mathcal{A}$ as $left(H_iright)_{i in I}$ (where $I$ is an indexing set and the $H_i$ are the hyperplanes). Then, the intersection lattice $Lleft(mathcal{A}right)$ is uniquely determined if you know the dimensions $dim left(bigcaplimits_{i in J} H_iright)$ for all subsets $J$ of $I$. Now all you need is to show that the latter dimensions don't change when the base field grows.
$endgroup$
– darij grinberg
Jan 12 at 23:27
add a comment |
$begingroup$
I have a question regarding the finite field method to compute the characteristic polynomial of a hyperplane arrangement. I am using the book "Enumerative Combinatorics, Volume 1" second edition by Richard P. Stanley. This method may be found in chapter 3.11.4.
I will begin by stating a problem I am interested solving using the finite field method. Then I will state the theorem and tell you might thoughts on this. In the bottom of this post, you may find my questions.
I'm interested in solving 3.114 (b) in the book by Stanley:
For an arrangement $mathcal{A}$ below (in $mathbb{R}^n$) show that the characteristic polynomials are as indicated. $x_i=x_j$ for $1leq i<jleq n$ and $sum_{i=1}^n x_i=0$. Then
$$chi_{mathcal{A}}(x)=(x-1)^2(x-2)(x-3)cdots (x-n+1).$$
If I am allowed to use the finite field method, I have a solution to this problem that gives me the correct polynomial. However, by reading the theorem (I will state it down below), I'm not sure why I would be allowed to use it.
The theorem only says something about $mathbb{Q}^n$ and not $mathbb{R}^n$.
Let us now state the theorem.
Theorem:
Let $mathcal{A}$ be an arrangement in $mathbb{Q}^n$, and suppose that $L(mathcal{A})cong L(mathcal{A}_q)$ for some prime power $q$. Then
$$
chi_{mathcal{A}}(q)=#Big ( mathbb{F}_q - bigcup_{Hinmathcal{A}_q} H Big )=q^n-#bigcup_{Hinmathcal{A}_q} H.
$$
To me, it really seems as we need to work in the vector space $mathbb{Q}^n$ to apply this theorem. This theorem does not seem to tell us that we can do this in the vector space $mathbb{R}^n$.
I still have some hope that I might be able to apply the above theorem for two reasons.
(1) Just to quote Stanley in the first paragraph when he starts to talk about the finite field method
In this subsection we will describe a method based on finite fields
for computing the characteristic polynomial of an arrangement
defined over $mathbb{Q}$.
It seems to me that he wants to tell the reader that the finite field method works if your hyperplanes are defined over $mathbb{Q}$. After doing some research what "defined over $mathbb{Q}$" means, it seems as this means the hyperplanes have rational coefficients. Surely, in the problem I want to solve the hyperplanes does, indeed, have rational coefficients.
(2) In example 3.11.11 in the book, Stanley uses the finite field method to find the characteristic polynomial of the braid arrangement $mathcal{B}_n$. This is an arrangement in $mathbb{K}^n$ with hyperplanes $x_i-x_j=0$ for $1leq i<jleq n$.
He hasn't said anything about which field $mathbb{K}$ we consider. Thus, I guess, we could for instance let $mathbb{K}=mathbb{R}$? Notice the polynomials are defined over $mathbb{Q}$.
It is very likely that I am misinterpreting something. Perhaps I am misinterpreting the theorem or the example?
I would be really happy if someone could tell me why we can apply the theorem when we have a hyperplane arrangement in $mathbb{R}^n$ (if it is possible) and if not perhaps tell me what's happening in example 3.11.11.
Thanks for taking your time!
combinatorics polynomials finite-fields
asked Jan 12 at 21:50
JoeJoe
4214
$endgroup$
I have a question regarding the finite field method to compute the characteristic polynomial of a hyperplane arrangement. I am using the book "Enumerative Combinatorics, Volume 1" second edition by Richard P. Stanley. This method may be found in chapter 3.11.4.
I will begin by stating a problem I am interested solving using the finite field method. Then I will state the theorem and tell you might thoughts on this. In the bottom of this post, you may find my questions.
I'm interested in solving 3.114 (b) in the book by Stanley:
For an arrangement $mathcal{A}$ below (in $mathbb{R}^n$) show that the characteristic polynomials are as indicated. $x_i=x_j$ for $1leq i<jleq n$ and $sum_{i=1}^n x_i=0$. Then
$$chi_{mathcal{A}}(x)=(x-1)^2(x-2)(x-3)cdots (x-n+1).$$
If I am allowed to use the finite field method, I have a solution to this problem that gives me the correct polynomial. However, by reading the theorem (I will state it down below), I'm not sure why I would be allowed to use it.
The theorem only says something about $mathbb{Q}^n$ and not $mathbb{R}^n$.
Let us now state the theorem.
Theorem:
Let $mathcal{A}$ be an arrangement in $mathbb{Q}^n$, and suppose that $L(mathcal{A})cong L(mathcal{A}_q)$ for some prime power $q$. Then
$$
chi_{mathcal{A}}(q)=#Big ( mathbb{F}_q - bigcup_{Hinmathcal{A}_q} H Big )=q^n-#bigcup_{Hinmathcal{A}_q} H.
$$
To me, it really seems as we need to work in the vector space $mathbb{Q}^n$ to apply this theorem. This theorem does not seem to tell us that we can do this in the vector space $mathbb{R}^n$.
I still have some hope that I might be able to apply the above theorem for two reasons.
(1) Just to quote Stanley in the first paragraph when he starts to talk about the finite field method
In this subsection we will describe a method based on finite fields
for computing the characteristic polynomial of an arrangement
defined over $mathbb{Q}$.
It seems to me that he wants to tell the reader that the finite field method works if your hyperplanes are defined over $mathbb{Q}$. After doing some research what "defined over $mathbb{Q}$" means, it seems as this means the hyperplanes have rational coefficients. Surely, in the problem I want to solve the hyperplanes does, indeed, have rational coefficients.
(2) In example 3.11.11 in the book, Stanley uses the finite field method to find the characteristic polynomial of the braid arrangement $mathcal{B}_n$. This is an arrangement in $mathbb{K}^n$ with hyperplanes $x_i-x_j=0$ for $1leq i<jleq n$.
He hasn't said anything about which field $mathbb{K}$ we consider. Thus, I guess, we could for instance let $mathbb{K}=mathbb{R}$? Notice the polynomials are defined over $mathbb{Q}$.
It is very likely that I am misinterpreting something. Perhaps I am misinterpreting the theorem or the example?
I would be really happy if someone could tell me why we can apply the theorem when we have a hyperplane arrangement in $mathbb{R}^n$ (if it is possible) and if not perhaps tell me what's happening in example 3.11.11.
Thanks for taking your time!
combinatorics polynomials finite-fields
combinatorics polynomials finite-fields
asked Jan 12 at 21:50
JoeJoe
4214
asked Jan 12 at 21:50
JoeJoe
4214
edited Jan 12 at 23:04
Joe
asked Jan 12 at 21:50
JoeJoe
4214
asked Jan 12 at 21:50
JoeJoe
4214
asked Jan 12 at 21:50
JoeJoe
4214
4214
1
$begingroup$
$Lleft(mathcal{A}right)$ does not change if you extend your base field. I don't know of a particularly slick way to explain why, except of "everything boils down to solving systems of linear equations, and the Gaussian elimination algorithm does not care if the field suddenly grows larger".
$endgroup$
– darij grinberg
Jan 12 at 23:17
$begingroup$
@darijgrinberg, thank you for your answer! :) I think I will get it (why it works in $mathbb{R}^n$), if I spend an hour or so to think through your answer.
$endgroup$
– Joe
Jan 12 at 23:25
1
$begingroup$
Actually a great first step is to convince yourself of the following: Write $mathcal{A}$ as $left(H_iright)_{i in I}$ (where $I$ is an indexing set and the $H_i$ are the hyperplanes). Then, the intersection lattice $Lleft(mathcal{A}right)$ is uniquely determined if you know the dimensions $dim left(bigcaplimits_{i in J} H_iright)$ for all subsets $J$ of $I$. Now all you need is to show that the latter dimensions don't change when the base field grows.
$endgroup$
– darij grinberg
Jan 12 at 23:27
add a comment |
1
$begingroup$
$Lleft(mathcal{A}right)$ does not change if you extend your base field. I don't know of a particularly slick way to explain why, except of "everything boils down to solving systems of linear equations, and the Gaussian elimination algorithm does not care if the field suddenly grows larger".
$endgroup$
– darij grinberg
Jan 12 at 23:17
$begingroup$
@darijgrinberg, thank you for your answer! :) I think I will get it (why it works in $mathbb{R}^n$), if I spend an hour or so to think through your answer.
$endgroup$
– Joe
Jan 12 at 23:25
1
$begingroup$
Actually a great first step is to convince yourself of the following: Write $mathcal{A}$ as $left(H_iright)_{i in I}$ (where $I$ is an indexing set and the $H_i$ are the hyperplanes). Then, the intersection lattice $Lleft(mathcal{A}right)$ is uniquely determined if you know the dimensions $dim left(bigcaplimits_{i in J} H_iright)$ for all subsets $J$ of $I$. Now all you need is to show that the latter dimensions don't change when the base field grows.
$endgroup$
– darij grinberg
Jan 12 at 23:27
1
1
$begingroup$
$Lleft(mathcal{A}right)$ does not change if you extend your base field. I don't know of a particularly slick way to explain why, except of "everything boils down to solving systems of linear equations, and the Gaussian elimination algorithm does not care if the field suddenly grows larger".
$endgroup$
– darij grinberg
Jan 12 at 23:17
$begingroup$
$Lleft(mathcal{A}right)$ does not change if you extend your base field. I don't know of a particularly slick way to explain why, except of "everything boils down to solving systems of linear equations, and the Gaussian elimination algorithm does not care if the field suddenly grows larger".
$endgroup$
– darij grinberg
Jan 12 at 23:17
$begingroup$
@darijgrinberg, thank you for your answer! :) I think I will get it (why it works in $mathbb{R}^n$), if I spend an hour or so to think through your answer.
$endgroup$
– Joe
Jan 12 at 23:25
$begingroup$
@darijgrinberg, thank you for your answer! :) I think I will get it (why it works in $mathbb{R}^n$), if I spend an hour or so to think through your answer.
$endgroup$
– Joe
Jan 12 at 23:25
1
1
$begingroup$
Actually a great first step is to convince yourself of the following: Write $mathcal{A}$ as $left(H_iright)_{i in I}$ (where $I$ is an indexing set and the $H_i$ are the hyperplanes). Then, the intersection lattice $Lleft(mathcal{A}right)$ is uniquely determined if you know the dimensions $dim left(bigcaplimits_{i in J} H_iright)$ for all subsets $J$ of $I$. Now all you need is to show that the latter dimensions don't change when the base field grows.
$endgroup$
– darij grinberg
Jan 12 at 23:27
$begingroup$
Actually a great first step is to convince yourself of the following: Write $mathcal{A}$ as $left(H_iright)_{i in I}$ (where $I$ is an indexing set and the $H_i$ are the hyperplanes). Then, the intersection lattice $Lleft(mathcal{A}right)$ is uniquely determined if you know the dimensions $dim left(bigcaplimits_{i in J} H_iright)$ for all subsets $J$ of $I$. Now all you need is to show that the latter dimensions don't change when the base field grows.
$endgroup$
– darij grinberg
Jan 12 at 23:27
add a comment |
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$begingroup$
$Lleft(mathcal{A}right)$ does not change if you extend your base field. I don't know of a particularly slick way to explain why, except of "everything boils down to solving systems of linear equations, and the Gaussian elimination algorithm does not care if the field suddenly grows larger".
$endgroup$
– darij grinberg
Jan 12 at 23:17
$begingroup$
@darijgrinberg, thank you for your answer! :) I think I will get it (why it works in $mathbb{R}^n$), if I spend an hour or so to think through your answer.
$endgroup$
– Joe
Jan 12 at 23:25
1
$begingroup$
Actually a great first step is to convince yourself of the following: Write $mathcal{A}$ as $left(H_iright)_{i in I}$ (where $I$ is an indexing set and the $H_i$ are the hyperplanes). Then, the intersection lattice $Lleft(mathcal{A}right)$ is uniquely determined if you know the dimensions $dim left(bigcaplimits_{i in J} H_iright)$ for all subsets $J$ of $I$. Now all you need is to show that the latter dimensions don't change when the base field grows.
$endgroup$
– darij grinberg
Jan 12 at 23:27