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.

- 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 $M$.

- 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!