Is there a generalization of Pfaffians?
$begingroup$
For an skew-symmetric matrix $A$ (meaning $A^T=-A$), the Pfaffian is defined by the equation $(text{Pf},A)^2=det A$. It is my understanding that this is defined for anti-symmetric matrices because it is known that the determinant of an anti-symmetric matrix is always a square of a polynomial in the entries of the matrix.
Now, skew-symmetry is sufficient to prove that the determinant is a square of a polynomial, but it is not necessary. The simplest example is the $2ntimes 2n$ matrix $A=a I_{2n}$ with $ainmathbb{C}$ and $I_k$ the $ktimes k$ identity matrix. The determinant is $det A = a^{2n} = (a^n)^2$. Of course, for $aneq 0$, $A$ is not skew-symmetric.
I have a few questions about this.
- Is there a generalization of a Pfaffian for any matrix whose determinant is a square of a polynomial?
- Is there a characterization (or some known set of properties) of matrices whose determinants are squares of polynomials?
- (Edit) Are there any known necessary and sufficient conditions for a matrix to have its determinant be the square of a polynomial (aside from skew-symmetry being sufficient)?
(Edit 2) For those who are curious, these questions arise from a problem from physics I am working on. I have a certain class of matrices whose characteristic polynomials (which arise as the determinant of a non-skew-symmetric matrix) appear to be the squares of Chebyshev polynomials. If I could prove that these characteristic polynomials must be squares of polynomials (using properties of the matrix) then I may be able to use some of the properties attributed to Pfaffians (or the proper generalization to non-skew-symmetric matrices) to confirm that they are indeed squared Chebyshev polynomials.
(Edit 3) To be as concrete as possible, I am looking for any information (e.g., answers to questions 1-3) on the set
$${Ainmathcal{M}_n(mathbb{C}): det A = p({a_{ij}})^2text{ with }ptext{ a polynomial} }$$
where $mathcal{M}_n(mathbb{C})$ is the set of $ntimes n$ complex matrices and $a_{ij}$ is the $i,j$'th entry of $A$.
linear-algebra determinant pfaffian
asked Nov 27 '18 at 19:26
UglyMousanova19UglyMousanova19
364
$endgroup$
|
show 11 more comments
$begingroup$
For an skew-symmetric matrix $A$ (meaning $A^T=-A$), the Pfaffian is defined by the equation $(text{Pf},A)^2=det A$. It is my understanding that this is defined for anti-symmetric matrices because it is known that the determinant of an anti-symmetric matrix is always a square of a polynomial in the entries of the matrix.
Now, skew-symmetry is sufficient to prove that the determinant is a square of a polynomial, but it is not necessary. The simplest example is the $2ntimes 2n$ matrix $A=a I_{2n}$ with $ainmathbb{C}$ and $I_k$ the $ktimes k$ identity matrix. The determinant is $det A = a^{2n} = (a^n)^2$. Of course, for $aneq 0$, $A$ is not skew-symmetric.
I have a few questions about this.
- Is there a generalization of a Pfaffian for any matrix whose determinant is a square of a polynomial?
- Is there a characterization (or some known set of properties) of matrices whose determinants are squares of polynomials?
- (Edit) Are there any known necessary and sufficient conditions for a matrix to have its determinant be the square of a polynomial (aside from skew-symmetry being sufficient)?
(Edit 2) For those who are curious, these questions arise from a problem from physics I am working on. I have a certain class of matrices whose characteristic polynomials (which arise as the determinant of a non-skew-symmetric matrix) appear to be the squares of Chebyshev polynomials. If I could prove that these characteristic polynomials must be squares of polynomials (using properties of the matrix) then I may be able to use some of the properties attributed to Pfaffians (or the proper generalization to non-skew-symmetric matrices) to confirm that they are indeed squared Chebyshev polynomials.
(Edit 3) To be as concrete as possible, I am looking for any information (e.g., answers to questions 1-3) on the set
$${Ainmathcal{M}_n(mathbb{C}): det A = p({a_{ij}})^2text{ with }ptext{ a polynomial} }$$
where $mathcal{M}_n(mathbb{C})$ is the set of $ntimes n$ complex matrices and $a_{ij}$ is the $i,j$'th entry of $A$.
linear-algebra determinant pfaffian
asked Nov 27 '18 at 19:26
UglyMousanova19UglyMousanova19
364
$endgroup$
1
$begingroup$
Determinants can be squares for various random reasons. Are you asking about certain families of matrices whose determinants are squares? Otherwise I'd say it's a rather vague question.
$endgroup$
– darij grinberg
Nov 27 '18 at 19:44
1
$begingroup$
I don't know if there exist an extension of Pfaffians to matrices other than the antisymmetric ones. But if you are interested into these polynomials, here is a paper that enlarges the point of view, in particular by using exterior algebra : kurims.kyoto-u.ac.jp/~kyodo/kokyuroku/contents/pdf/1302-14.pdf
$endgroup$
– Jean Marie
Nov 27 '18 at 19:59
1
$begingroup$
@JeanMarie Thanks for the response. I have actually read quite a bit on Pfaffians in the context of Soliton theory (in particular the book "The Direct Method in Soliton Theory" by Hirota - same author as that paper you linked). They have many nice properties that could be very useful, but are defined only for anti-symmetric matrices (and rely on this fact quite heavily it seems).
$endgroup$
– UglyMousanova19
Nov 27 '18 at 23:21
1
$begingroup$
Maybe a way to rephrase your question to appeal to those who deem it vague: Let $R$ be the ring of polynomials in matrix entries, so $detin R$. Let us call an ideal $Isubseteq R$ Pfaffian if $det$ becomes a square in $R/I$ and denote by $J$ the intersection of all Pfaffian ideals. What is $J$? It characterizes the largest subvariety of our matrix space where we can define something like a Pfaffian. And also a question: Can you name a Pfaffian ideal that does not contain the vanishing ideal of all skew-symmetric matrices?
$endgroup$
– Jesko Hüttenhain
Nov 27 '18 at 23:25
1
$begingroup$
@JeskoHüttenhain: What about the ideal defining $operatorname{SL}_nleft(Kright)$? That would be Pfaffian, too.
$endgroup$
– darij grinberg
Nov 28 '18 at 3:34
|
show 11 more comments
$begingroup$
For an skew-symmetric matrix $A$ (meaning $A^T=-A$), the Pfaffian is defined by the equation $(text{Pf},A)^2=det A$. It is my understanding that this is defined for anti-symmetric matrices because it is known that the determinant of an anti-symmetric matrix is always a square of a polynomial in the entries of the matrix.
Now, skew-symmetry is sufficient to prove that the determinant is a square of a polynomial, but it is not necessary. The simplest example is the $2ntimes 2n$ matrix $A=a I_{2n}$ with $ainmathbb{C}$ and $I_k$ the $ktimes k$ identity matrix. The determinant is $det A = a^{2n} = (a^n)^2$. Of course, for $aneq 0$, $A$ is not skew-symmetric.
I have a few questions about this.
- Is there a generalization of a Pfaffian for any matrix whose determinant is a square of a polynomial?
- Is there a characterization (or some known set of properties) of matrices whose determinants are squares of polynomials?
- (Edit) Are there any known necessary and sufficient conditions for a matrix to have its determinant be the square of a polynomial (aside from skew-symmetry being sufficient)?
(Edit 2) For those who are curious, these questions arise from a problem from physics I am working on. I have a certain class of matrices whose characteristic polynomials (which arise as the determinant of a non-skew-symmetric matrix) appear to be the squares of Chebyshev polynomials. If I could prove that these characteristic polynomials must be squares of polynomials (using properties of the matrix) then I may be able to use some of the properties attributed to Pfaffians (or the proper generalization to non-skew-symmetric matrices) to confirm that they are indeed squared Chebyshev polynomials.
(Edit 3) To be as concrete as possible, I am looking for any information (e.g., answers to questions 1-3) on the set
$${Ainmathcal{M}_n(mathbb{C}): det A = p({a_{ij}})^2text{ with }ptext{ a polynomial} }$$
where $mathcal{M}_n(mathbb{C})$ is the set of $ntimes n$ complex matrices and $a_{ij}$ is the $i,j$'th entry of $A$.
linear-algebra determinant pfaffian
asked Nov 27 '18 at 19:26
UglyMousanova19UglyMousanova19
364
$endgroup$
For an skew-symmetric matrix $A$ (meaning $A^T=-A$), the Pfaffian is defined by the equation $(text{Pf},A)^2=det A$. It is my understanding that this is defined for anti-symmetric matrices because it is known that the determinant of an anti-symmetric matrix is always a square of a polynomial in the entries of the matrix.
Now, skew-symmetry is sufficient to prove that the determinant is a square of a polynomial, but it is not necessary. The simplest example is the $2ntimes 2n$ matrix $A=a I_{2n}$ with $ainmathbb{C}$ and $I_k$ the $ktimes k$ identity matrix. The determinant is $det A = a^{2n} = (a^n)^2$. Of course, for $aneq 0$, $A$ is not skew-symmetric.
I have a few questions about this.
- Is there a generalization of a Pfaffian for any matrix whose determinant is a square of a polynomial?
- Is there a characterization (or some known set of properties) of matrices whose determinants are squares of polynomials?
- (Edit) Are there any known necessary and sufficient conditions for a matrix to have its determinant be the square of a polynomial (aside from skew-symmetry being sufficient)?
(Edit 2) For those who are curious, these questions arise from a problem from physics I am working on. I have a certain class of matrices whose characteristic polynomials (which arise as the determinant of a non-skew-symmetric matrix) appear to be the squares of Chebyshev polynomials. If I could prove that these characteristic polynomials must be squares of polynomials (using properties of the matrix) then I may be able to use some of the properties attributed to Pfaffians (or the proper generalization to non-skew-symmetric matrices) to confirm that they are indeed squared Chebyshev polynomials.
(Edit 3) To be as concrete as possible, I am looking for any information (e.g., answers to questions 1-3) on the set
$${Ainmathcal{M}_n(mathbb{C}): det A = p({a_{ij}})^2text{ with }ptext{ a polynomial} }$$
where $mathcal{M}_n(mathbb{C})$ is the set of $ntimes n$ complex matrices and $a_{ij}$ is the $i,j$'th entry of $A$.
linear-algebra determinant pfaffian
linear-algebra determinant pfaffian
asked Nov 27 '18 at 19:26
UglyMousanova19UglyMousanova19
364
asked Nov 27 '18 at 19:26
UglyMousanova19UglyMousanova19
364
edited Nov 28 '18 at 20:26
UglyMousanova19
asked Nov 27 '18 at 19:26
UglyMousanova19UglyMousanova19
364
asked Nov 27 '18 at 19:26
UglyMousanova19UglyMousanova19
364
asked Nov 27 '18 at 19:26
UglyMousanova19UglyMousanova19
364
364
1
$begingroup$
Determinants can be squares for various random reasons. Are you asking about certain families of matrices whose determinants are squares? Otherwise I'd say it's a rather vague question.
$endgroup$
– darij grinberg
Nov 27 '18 at 19:44
1
$begingroup$
I don't know if there exist an extension of Pfaffians to matrices other than the antisymmetric ones. But if you are interested into these polynomials, here is a paper that enlarges the point of view, in particular by using exterior algebra : kurims.kyoto-u.ac.jp/~kyodo/kokyuroku/contents/pdf/1302-14.pdf
$endgroup$
– Jean Marie
Nov 27 '18 at 19:59
1
$begingroup$
@JeanMarie Thanks for the response. I have actually read quite a bit on Pfaffians in the context of Soliton theory (in particular the book "The Direct Method in Soliton Theory" by Hirota - same author as that paper you linked). They have many nice properties that could be very useful, but are defined only for anti-symmetric matrices (and rely on this fact quite heavily it seems).
$endgroup$
– UglyMousanova19
Nov 27 '18 at 23:21
1
$begingroup$
Maybe a way to rephrase your question to appeal to those who deem it vague: Let $R$ be the ring of polynomials in matrix entries, so $detin R$. Let us call an ideal $Isubseteq R$ Pfaffian if $det$ becomes a square in $R/I$ and denote by $J$ the intersection of all Pfaffian ideals. What is $J$? It characterizes the largest subvariety of our matrix space where we can define something like a Pfaffian. And also a question: Can you name a Pfaffian ideal that does not contain the vanishing ideal of all skew-symmetric matrices?
$endgroup$
– Jesko Hüttenhain
Nov 27 '18 at 23:25
1
$begingroup$
@JeskoHüttenhain: What about the ideal defining $operatorname{SL}_nleft(Kright)$? That would be Pfaffian, too.
$endgroup$
– darij grinberg
Nov 28 '18 at 3:34
|
show 11 more comments
1
$begingroup$
Determinants can be squares for various random reasons. Are you asking about certain families of matrices whose determinants are squares? Otherwise I'd say it's a rather vague question.
$endgroup$
– darij grinberg
Nov 27 '18 at 19:44
1
$begingroup$
I don't know if there exist an extension of Pfaffians to matrices other than the antisymmetric ones. But if you are interested into these polynomials, here is a paper that enlarges the point of view, in particular by using exterior algebra : kurims.kyoto-u.ac.jp/~kyodo/kokyuroku/contents/pdf/1302-14.pdf
$endgroup$
– Jean Marie
Nov 27 '18 at 19:59
1
$begingroup$
@JeanMarie Thanks for the response. I have actually read quite a bit on Pfaffians in the context of Soliton theory (in particular the book "The Direct Method in Soliton Theory" by Hirota - same author as that paper you linked). They have many nice properties that could be very useful, but are defined only for anti-symmetric matrices (and rely on this fact quite heavily it seems).
$endgroup$
– UglyMousanova19
Nov 27 '18 at 23:21
1
$begingroup$
Maybe a way to rephrase your question to appeal to those who deem it vague: Let $R$ be the ring of polynomials in matrix entries, so $detin R$. Let us call an ideal $Isubseteq R$ Pfaffian if $det$ becomes a square in $R/I$ and denote by $J$ the intersection of all Pfaffian ideals. What is $J$? It characterizes the largest subvariety of our matrix space where we can define something like a Pfaffian. And also a question: Can you name a Pfaffian ideal that does not contain the vanishing ideal of all skew-symmetric matrices?
$endgroup$
– Jesko Hüttenhain
Nov 27 '18 at 23:25
1
$begingroup$
@JeskoHüttenhain: What about the ideal defining $operatorname{SL}_nleft(Kright)$? That would be Pfaffian, too.
$endgroup$
– darij grinberg
Nov 28 '18 at 3:34
1
1
$begingroup$
Determinants can be squares for various random reasons. Are you asking about certain families of matrices whose determinants are squares? Otherwise I'd say it's a rather vague question.
$endgroup$
– darij grinberg
Nov 27 '18 at 19:44
$begingroup$
Determinants can be squares for various random reasons. Are you asking about certain families of matrices whose determinants are squares? Otherwise I'd say it's a rather vague question.
$endgroup$
– darij grinberg
Nov 27 '18 at 19:44
1
1
$begingroup$
I don't know if there exist an extension of Pfaffians to matrices other than the antisymmetric ones. But if you are interested into these polynomials, here is a paper that enlarges the point of view, in particular by using exterior algebra : kurims.kyoto-u.ac.jp/~kyodo/kokyuroku/contents/pdf/1302-14.pdf
$endgroup$
– Jean Marie
Nov 27 '18 at 19:59
$begingroup$
I don't know if there exist an extension of Pfaffians to matrices other than the antisymmetric ones. But if you are interested into these polynomials, here is a paper that enlarges the point of view, in particular by using exterior algebra : kurims.kyoto-u.ac.jp/~kyodo/kokyuroku/contents/pdf/1302-14.pdf
$endgroup$
– Jean Marie
Nov 27 '18 at 19:59
1
1
$begingroup$
@JeanMarie Thanks for the response. I have actually read quite a bit on Pfaffians in the context of Soliton theory (in particular the book "The Direct Method in Soliton Theory" by Hirota - same author as that paper you linked). They have many nice properties that could be very useful, but are defined only for anti-symmetric matrices (and rely on this fact quite heavily it seems).
$endgroup$
– UglyMousanova19
Nov 27 '18 at 23:21
$begingroup$
@JeanMarie Thanks for the response. I have actually read quite a bit on Pfaffians in the context of Soliton theory (in particular the book "The Direct Method in Soliton Theory" by Hirota - same author as that paper you linked). They have many nice properties that could be very useful, but are defined only for anti-symmetric matrices (and rely on this fact quite heavily it seems).
$endgroup$
– UglyMousanova19
Nov 27 '18 at 23:21
1
1
$begingroup$
Maybe a way to rephrase your question to appeal to those who deem it vague: Let $R$ be the ring of polynomials in matrix entries, so $detin R$. Let us call an ideal $Isubseteq R$ Pfaffian if $det$ becomes a square in $R/I$ and denote by $J$ the intersection of all Pfaffian ideals. What is $J$? It characterizes the largest subvariety of our matrix space where we can define something like a Pfaffian. And also a question: Can you name a Pfaffian ideal that does not contain the vanishing ideal of all skew-symmetric matrices?
$endgroup$
– Jesko Hüttenhain
Nov 27 '18 at 23:25
$begingroup$
Maybe a way to rephrase your question to appeal to those who deem it vague: Let $R$ be the ring of polynomials in matrix entries, so $detin R$. Let us call an ideal $Isubseteq R$ Pfaffian if $det$ becomes a square in $R/I$ and denote by $J$ the intersection of all Pfaffian ideals. What is $J$? It characterizes the largest subvariety of our matrix space where we can define something like a Pfaffian. And also a question: Can you name a Pfaffian ideal that does not contain the vanishing ideal of all skew-symmetric matrices?
$endgroup$
– Jesko Hüttenhain
Nov 27 '18 at 23:25
1
1
$begingroup$
@JeskoHüttenhain: What about the ideal defining $operatorname{SL}_nleft(Kright)$? That would be Pfaffian, too.
$endgroup$
– darij grinberg
Nov 28 '18 at 3:34
$begingroup$
@JeskoHüttenhain: What about the ideal defining $operatorname{SL}_nleft(Kright)$? That would be Pfaffian, too.
$endgroup$
– darij grinberg
Nov 28 '18 at 3:34
|
show 11 more comments
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1
$begingroup$
Determinants can be squares for various random reasons. Are you asking about certain families of matrices whose determinants are squares? Otherwise I'd say it's a rather vague question.
$endgroup$
– darij grinberg
Nov 27 '18 at 19:44
1
$begingroup$
I don't know if there exist an extension of Pfaffians to matrices other than the antisymmetric ones. But if you are interested into these polynomials, here is a paper that enlarges the point of view, in particular by using exterior algebra : kurims.kyoto-u.ac.jp/~kyodo/kokyuroku/contents/pdf/1302-14.pdf
$endgroup$
– Jean Marie
Nov 27 '18 at 19:59
1
$begingroup$
@JeanMarie Thanks for the response. I have actually read quite a bit on Pfaffians in the context of Soliton theory (in particular the book "The Direct Method in Soliton Theory" by Hirota - same author as that paper you linked). They have many nice properties that could be very useful, but are defined only for anti-symmetric matrices (and rely on this fact quite heavily it seems).
$endgroup$
– UglyMousanova19
Nov 27 '18 at 23:21
1
$begingroup$
Maybe a way to rephrase your question to appeal to those who deem it vague: Let $R$ be the ring of polynomials in matrix entries, so $detin R$. Let us call an ideal $Isubseteq R$ Pfaffian if $det$ becomes a square in $R/I$ and denote by $J$ the intersection of all Pfaffian ideals. What is $J$? It characterizes the largest subvariety of our matrix space where we can define something like a Pfaffian. And also a question: Can you name a Pfaffian ideal that does not contain the vanishing ideal of all skew-symmetric matrices?
$endgroup$
– Jesko Hüttenhain
Nov 27 '18 at 23:25
1
$begingroup$
@JeskoHüttenhain: What about the ideal defining $operatorname{SL}_nleft(Kright)$? That would be Pfaffian, too.
$endgroup$
– darij grinberg
Nov 28 '18 at 3:34