Is there a generalization of Pfaffians?

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 $$2n\times 2n$$ matrix $$A=a I_{2n}$$ with $$a\in\mathbb{C}$$ and $$I_k$$ the $$k\times k$$ identity matrix. The determinant is $$\det A = a^{2n} = (a^n)^2$$. Of course, for $$a\neq 0$$, $$A$$ is not skew-symmetric.

1. Is there a generalization of a Pfaffian for any matrix whose determinant is a square of a polynomial?
2. Is there a characterization (or some known set of properties) of matrices whose determinants are squares of polynomials?
3. (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 $$\{A\in\mathcal{M}_n(\mathbb{C}): \det A = p(\{a_{ij}\})^2\text{ with }p\text{ a polynomial} \}$$ where $$\mathcal{M}_n(\mathbb{C})$$ is the set of $$n\times n$$ complex matrices and $$a_{ij}$$ is the $$i,j$$'th entry of $$A$$.

• 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. Nov 27, 2018 at 19:44
• 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 Nov 27, 2018 at 19:59
• @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). Nov 27, 2018 at 23:21
• 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 $\det\in R$. Let us call an ideal $I\subseteq R$ 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? Nov 27, 2018 at 23:25
• @JeskoHüttenhain: What about the ideal defining $\operatorname{SL}_n\left(K\right)$? That would be Pfaffian, too. Nov 28, 2018 at 3:34

Only a partial answer. The problem with defining the pfaffian of a matrix whose determinant is the square of the polynomial is that the sign of the pfaffian may be not well defined. For example, one may have two matrices $$A$$ and $$B$$ such that $$det A=det B=(polynomial)^2$$ but $$pf A=-pf B$$. A possible approach is to think in terms of unitary transformations and equivalence classes.
Let $$A$$ be a $$2n\times 2n$$ matric, not necessarily antisymmetric. Let $$\mathcal U(A)$$ be the set of all matrices $$B=U A U^\dagger$$ unitarily equivalent to $$A$$. Now consider the subset $$\bar{\mathcal U}(A)\subset\mathcal U(A)$$ of matrices which are antisymmetric. It is clear that all matrices in $${\mathcal U}$$ have the same determinant, and all matrices in $$\bar{\mathcal U}$$ have the same pfaffian. Therefore one can define the pfaffian of any $$A\in{\mathcal U}$$ as the pfaffian of any $$B\in\bar{\mathcal U}$$. In short, one can define
$$pf A=pf (U A U^\dagger)$$
if there exists a unitary matrix $$U$$ such that $$U A U^\dagger$$ is antisymmetric. (The unitary matrix does not need to be uniquely defined, of course).