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-Schwarz 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 Schwarz or completing squares?

Thanks in advanced!

  • $\begingroup$ 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. $\Omega = \begin{pmatrix} 0 & I \\ -I & 0 \end{pmatrix}$. $\endgroup$ Feb 25, 2021 at 17:05

1 Answer 1


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 $$\det M=|\det A|^2\det(1+\bar{S}R^2\bar{S}^{-1})=|\det A|^2\det(1+R^2)$$ $$\qquad=|\det A|^2\det(1+iR)\det(1-iR)$$ $$\quad=|\det A|^2|\det(1+iR)|^2\geq 0.$$

  • $\begingroup$ Ultimately this seems to use eigenvalues in a way that's pretty similar to what the OP did. $\endgroup$ Feb 26, 2021 at 1:22
  • 1
    $\begingroup$ 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 $n=2$. $\endgroup$ Feb 26, 2021 at 3:47
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    $\begingroup$ @SamHopkins --- I worked a bit more on the proof, I think I have now removed any reliance on eigenvalues. $\endgroup$ Feb 26, 2021 at 12:43
  • $\begingroup$ @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 $\endgroup$ Feb 27, 2021 at 1:05
  • $\begingroup$ 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). $\endgroup$ Feb 27, 2021 at 7:42

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