This might be forced to migrate to math.SE but let me still risk it.

The spaces $\operatorname{S}^2(k^n)$ and $\Lambda^2(k^{n+1})$ from the title have equal dimensions. Is there a *natural* isomorphism between them?

To make the question more MOish - choosing a basis, elements of $\operatorname{S}^2(V)$ can be identified with symmetric matrices, so this does have a Jordan algebra structure. On the other hand, if $V$ has even dimension $2k$, the adjoint representation of the Lie algebra $\mathfrak{sp}(2k)$ can be identified with $\operatorname{S}^2(V)$, so that the latter has a Lie algebra structure. On the third hand, regardless of the dimension, the adjoint representation of $\mathfrak{so}(m)$ can be identified with $\Lambda^2(k^m)$, so the latter also has a Lie algebra structure. And on the fourth hand, some isomorphism as above would give it also some Jordan algebra structure.

So all in all we seem to get a Lie algebra structure and a Jordan algebra structure on each of these spaces. This more MOish question then is: understanding by "natural" an isomorphism interchanging these Lie and Jordan structures in some way - is there a nice description of such an isomorphism?

Slightly more generally, for some involutive automorphism of $k$ one might ask for similar isomorphisms with "symmetric" and "exterior" replaced by "Hermitian" and "skew-Hermitian".

I am tagging this with reference-request, surely this must be in the literature. I tried The Book of Involutions, but could not find it there. Maybe I did not look hard enough, don't know.

I also tried to look for it here on MO; the closest I could find is Symmetric matrices as a module over the skewsymmetric ones but it is not what I need...

**Slightly later:**

As user44191 notes in the comment below, there is a more general question about $\operatorname{S}^i(k^n)\cong\Lambda^i(k^{n+i-1})$ (pertaining to $\left(\binom ni\right)=\binom{n+i-1}i$ and the stars and bars combinatorics), although what algebra structures might be involved in this case is not clear to me.

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