# Coordinate free isomorphism between $d+1$-dimensional antisymmetric rank $2$ tensors and $d$-dimensional symmetric rank $2$ tensors

The space of $(d+1)$-dimensional antisymmetric matrices has the same dimension as the space of $d$-dimensional symmetric matrices, $\frac12d(d+1)$. There are isomorphisms between the two spaces, e.g. for $d=2$

$$\begin{pmatrix} a & b\\ b & c \end{pmatrix} \leftrightarrow \begin{pmatrix} 0 & a & b\\ -a & 0 & c\\ -b & -c & 0 \end{pmatrix}.$$

This isomorphism is obviously only defined after we have chosen basis vectors.

My question: given a space of $(d+1)$-dimensional antisymmetric rank $2$ tensors and a space of $d$-dimensional symmetric rank $2$ tensors, what is the least that we can get away with specifying in order to provide an isomorphism between them? In the above example we needed bases. Ideally it would be desirable to have something coordinate free, but my intuition tells me that this may not be possible, in particular because they transform differently.

• Suggestion: If you think of your symmetric matrix as a $2$-form: $\sum_i a_i x_i^2 + \sum_{i<j} a_{ij} x_ix_j$ then an isomorphism is given by sending the quadratic form $x_i^2$ to $dx_0 \wedge dx_i$, and the quadratic form $x_i x_j$ (for $i<j$) to $dx_i \wedge dx_j$. Can that be interpreted in a natural geometric or algebraic way? – Ryan Budney May 4 '17 at 18:02

It seems that you are asking for some kind of isomorphism between $S^2(\mathbb{F}^d)$ and $\Lambda^2(\mathbb{F}^{d+1})$ that would 'have the greatest symmetry', in the sense that it would commute with some group action on $\mathbb{F}^d$ and $\mathbb{F}^{d+1}$, the bigger the better, so that it would be as independent of a choice of basis as possible.

Ryan's suggestion of a basis-dependent identification has a kind of symmetry based on the symmetric group $S_d$, but it isn't induced by a corresponding symmetry of the underlying vector spaces $\mathbb{F}^d$ and $\mathbb{F}^{d+1}$ because, if you permute the bases of these spaces, you'll see that the $x_ix_j$ permute among themselves, but the $\mathrm{d}x_i\wedge\mathrm{d}x_j$ permute and change signs at the same time, so it's not really based on the geometry of the underlying vector spaces.

It seems that what you want to find a subgroup $H\subset\mathrm{GL}(d,\mathbb{F})\times \mathrm{GL}(d{+}1,\mathbb{F})$ such that $S^2(\mathbb{F}^d)$ and $\Lambda^2(\mathbb{F}^{d+1})$ would be isomorphic as $H$-modules, and you'd want $H$ to be maximal with this property. It is an interesting questions as to what the maximal such subgroups $H$ could look like.

Here is an example of the kind of thing that you might consider as an answer to your question:

For a $2$-dimensional vector space $V$, there is a natural isomorphism $$\Lambda^2\bigl(S^2(V)\bigr) = S^2(V)\otimes\Lambda^2(V),$$ i.e., this isomorphism is $\mathrm{GL}(V)$-equivariant, and this provides an isomorphism between the skew-symmetric $2$-forms on $S^2(V^*)\simeq \mathbb{F}^3$ and the quadratic forms on $V^*\simeq \mathbb{F}^2$, twisted by the determinant $\Lambda^2(V)$, a $1$-dimensional vector space.

You can get rid of this twisting if you fix a volume form on $V$ and consider only volume perserving automorphisms, i.e., $\mathrm{SL}(V)$. Alternatively, you can consider $H = \mathrm{SL}(V)\times \mathbb{F}^\times$ and let the $\mathrm{SL}(V)$ factor act on $S^2(V)$ and $V$ as above, but let an invertible element $\lambda\in F^\times$ act on $V\times S^2(V)$ as $(\lambda,\lambda)$ (i.e., with equal weights, instead of as $(\lambda,\lambda^2)$), and this will give an $H$ acting on $V\times S^2(V)$ for which $S^2(V)$ and $\Lambda^2\bigl(S^2(V)\bigr)$ are isomorphic as $H$-modules on the nose (with no $\Lambda^2(V)$-twisting).

This generalizes to the case of all $d$, because, by the Clebsch-Gordan formulae, there is an isomorphism of $\mathrm{GL}(V)$-modules $$S^2\bigl(S^{d-1}(V)\bigr)\otimes\Lambda^2(V) = \Lambda^2\bigl(S^{d}(V)\bigr).$$ Since $S^{d-1}(V)\simeq\mathbb{F}^d$ and $S^{d}(V)\simeq\mathbb{F}^{d+1}$, this seems to correspond to the kind of thing you are thinking about. (As before, you can get rid of the $\Lambda^2(V)$-factor by restricting to $\mathrm{SL}(V)$ and instead making the scalars act with equal weights on $S^{d-1}(V)$ and $S^{d}(V)$, instead of with their 'natural' weights $d{-}1$ and $d$. The vector spaces $S^{d-1}(V)$ and $S^{d}(V)$ remain irreducible under this action.)

Note, however, that when $k\ge2$, the module $S^2\bigl(S^{k}(V)\bigr)$ is no longer an irreducible $\mathrm{SL}(V)$ module, instead, we have $$S^2\bigl(S^{k}(V)\bigr)\simeq \Lambda^2\bigl(S^{k+1}(V)\bigr) \simeq \bigoplus_{0\le j\le k/2} S^{2k-4j}(V).$$

A good test case would be $d=3$. I suspect that, in this case, the scalar-modified embedding as above of $H=\mathrm{GL}(V)$ into $$\mathrm{GL}\bigl(S^2(V)\bigr)\times \mathrm{GL}\bigl(S^3(V)\bigr) \simeq \mathrm{GL}(3,\mathbb{F})\times \mathrm{GL}(4,\mathbb{F})$$ makes $\mathrm{GL}(V)$ into a maximal subgroup of $\mathrm{GL}(3,\mathbb{F})\times \mathrm{GL}(4,\mathbb{F})$ for which $S^2(\mathbb{F}^3)$ and $\Lambda^2(\mathbb{F}^4)$ are isomorphic as $H$-modules, and I can show that any such $H$ that is a connected Lie group that contains a simple Lie subgroup must be this one, up to isomorphism. What happens in higher dimensions, I don't know.