# Showing the positivity of the determinant of $\mathfrak{sp}(n)$ without making use of diagonalization

Let $$\mathfrak{sp}(n)$$ be the lie algebra of compact symplectic group $$\mathrm{SP}(n)$$, regarded as a compact form of $$\mathfrak{sp}(2n,\mathbb{C})$$, so we can talk about its (complex) determinant.

Let $$M\in \mathfrak{sp}(n)$$, then $$M$$ has purely imaginary eigenvalues $$(ix_1,ix_2,\dots,ix_n,-ix_1,-ix_2,\dots,-ix_n)$$, so $$\det(M)=(x_1x_2\cdots x_n)^2\geq 0.$$

My question is

Is there a coordinate independent way to show that every element of $$\mathfrak{sp}(n)$$ has nonnegative determinant?

I would want an argument without using eigenvalues, nor anything that cannot be expressed as a function of the matrix entries.

1. I will like to know if there is some geometric arguments.

I also want to understand the algebra behind. For $$n=2$$, I have tried expanding $$\det(M)$$, but I cannot find a way to express it as a sum of non-negative terms.

A useful way to show the positivity of an algebraic expression is to write it as a sum of terms, and each term is either a norm square, or can be shown to be non-negative by a direct application of the Cauchy-Schwartz inequality. For example, we know $$\mathrm{tr}(A^4)\geq 0$$ because $$\mathrm{tr}(A^4)=||A^2||^2$$.

Of course we have $$\det(M)=(x_1x_2\cdots x_n)^2$$, but the problem $$(x_1x_2\cdots x_n)$$ is not expressible by $$M$$.

1. Can $$\det(M)$$ be expressed a sum of such non-negative terms? If yes, what are they? If not, what are the extra ingredients we need to show the positivity apart from Cauchy Schwartz or completing squares?

• Because some now-deleted answers were confused about this, might be worth recording that the matrices $M \in \mathfrak{sp}(n)$ are those which satisfy $\Omega M = -M^T\Omega$ where $\Omega$ is some fixed non-singular skew-symmetric matrix, e.g. $\Omega = \begin{pmatrix} 0 & I \\ -I & 0 \end{pmatrix}$. Feb 25 at 17:05

Here is a proof of $$\det M\geq 0$$ for $$M\in \mathfrak{sp}(n)$$ based on the lemma that every complex matrix is consimilar to a real matrix.

Acknowledgment: In what follows I was helped by feedback I received at MSE.

By construction, the $$2n\times 2n$$ complex matrix $$M\in \mathfrak{sp}(n)$$ is skew-Hermitian and Hamiltonian, which means that it has the $$n\times n$$ block decomposition $$M=\begin{pmatrix} A&B\\ C&-A^T\end{pmatrix},\;\;\text{with}\;\;A=-A^\ast,\;\;B=B^T=-C^\ast=-\bar{C}.$$ Here $$M^T$$ denotes the transpose, $$\bar{M}$$ the complex conjugate, and $$M^\ast$$ the conjugate transpose.

By continuity of the determinant it is sufficient to consider $$\det A\neq 0$$. Then Schur's determinant identity gives

$$\det M=\det(-AA^T-ACA^{-1}B)=\det(A\bar{A}+A\bar{B}A^{-1}B)$$ $$\qquad=\det(A\bar{A})\det(1+\bar{A}^{-1}\bar{B}A^{-1}B)$$ $$\qquad=|\det A|^2\det(1+\bar{X}X),\;\;\text{with}\;\;X=A^{-1}B.$$

Now I apply the consimilarity lemma, to write $$X=SR\bar{S}^{-1}$$ with $$R$$ a real matrix. This gives $$\det M=|\det A|^2\det(1+\bar{S}R^2\bar{S}^{-1})=|\det A|^2\det(1+R^2)$$ $$\qquad=|\det A|^2\det(1+iR)\det(1-iR)$$ $$\quad=|\det A|^2|\det(1+iR)|^2\geq 0.$$

• Ultimately this seems to use eigenvalues in a way that's pretty similar to what the OP did. Feb 26 at 1:22
• For $X=\begin{pmatrix} a & b \\ c & d \end{pmatrix}$, $|b|^4\det(1+X\overline{X})=||b|^2-\overline{b}^2 \det(X)|^2+|b|^2|a\overline{b}+b\overline{d}|^2$. So it is possible to write the determinant as a sum of norm square in the case $n=2$. Feb 26 at 3:47
• @SamHopkins --- I worked a bit more on the proof, I think I have now removed any reliance on eigenvalues. Feb 26 at 12:43
• @CarloBeenakker Can $R$ be expressed using $X$? I think you proof is using the fact that $X\overline{X}$ is conjugate to the square of a real matrix, which is not so much different from a proof using eigenvalues Feb 27 at 1:05
• a proof along these lines using eigenvalues would rely on the fact that the negative eigenvalues of $X\bar{X}$ have multiplicity two, which is not what I am using here (although indeed this property can be derived from the consimilarity of $X$ with a real matrix). Feb 27 at 7:42