# Symmetric matrices as a module over the skewsymmetric ones

I'm trying to understand the Cartan decomposition of a semisimple Lie algebra, $\mathfrak g=\mathfrak k \oplus \mathfrak p$, where $[\mathfrak k,\mathfrak p] \subseteq \mathfrak p$, cf. the wikipedia article on Cartan decomposition.

I posted the following question on math.stackexchange.com, where Darij suggested to repost the question here as an answer is not completely obvious, I suppose.

Let $\mathfrak {so}_{n}$ denote the skew-symmetric complex $n \times n$-matrices and let $M$ denote the symmetric $n \times n$-matrices of trace 0.

Then $M$ is a module over the Lie algebra $\mathfrak {so}_n$ (this comes from the Cartan decomposition of $\mathfrak {sl}_n$).

What is the decomposition of $M$ into irreducible $\mathfrak {so}_n$-modules?

The standard representation of $\mathfrak {so}_n$ has dimension $n$, the adjoint representation has dimension $\frac 1 2 n \cdot (n-1)$ and there are two spin representations of small dimension. But I don't see a way how these, together with trivial representations, should add up to the dimension of $M$, which is $\frac 1 2 n \cdot (n+1)-1$.

• Are we in characteristic zero? If so, then it's irreducible, at least for $n>2$. I don't understand your last paragraph, though. Most irreps are not direct sum of the adjoint, the standard and the spinorial reps. The representation ring is generated multiplicatively by the standard and the spinorial reps. Indeed, the one you want is in the tensor product of the standard rep with itself. This question is not really MO level, though, my all due to respect to darij grinberg. – José Figueroa-O'Farrill Feb 21 '11 at 11:12
• José: Is there any easy reason why it is irreducible, short of combing for highest weights? – darij grinberg Feb 21 '11 at 12:19
• Sorry, I meant to ask: If it is irreducible, what is the highest weight then? Of course, if it is irreducible, it won't decompose as a non-trivial direct sum. – Guntram Feb 21 '11 at 12:33
• If you think in terms of the tensor calculus, it's "obvious", whatever that means. If the standard rep has highest weight [1,0,...,0] then this is [2,0,...,0], I believe. – José Figueroa-O'Farrill Feb 21 '11 at 15:37

To elaborate on Jose's comment above, to obtain $M$, one can take the symmetric square of the standard representation of $\mathfrak{so}_n$ and subtract off the trivial summand leaving an irreducible component $M$. Since the highest weight of the symmetric square is twice the highest weight of the standard representation, and this highest weight is obviously not part of the trivial summand, it follows that it is the highest weight of the irreducible piece $M$.

• Sorry, but why is $M$ irreducible? We know that it has only one highest weight vector for the weight [2,0,0,...,0], and we have killed one highest weight vector for the weight [1,0,0,...,0], but is there a good reason we do not have more [1,0,0,...,0] HWVs or even [1,1,0,0,...,0] HWVs? (Disclaimer: I know almost no Lie algebra rep. theory beyound sl.) – darij grinberg Feb 21 '11 at 16:17
• I guess one (overly simplistic perhaps) way to think about it is that in the symmetric matrices which Guntram used to describe $M$, there are no 'obvious' subspaces that are invariant under the $\mathfrak{so}_n$ action, except for the trace part. Another way to see it without having to look for invariant subspaces is to use the Weyl Dimension Formula applied to the irreducible module with highest weight twice the weight of the standard rep, call this module $N$. As $N$ is an irreducible component of the symmetric square and dimension 1 less than the symmetric square, $M\subset N$, hence $M=N$ – ARupinski Feb 21 '11 at 16:36

I'm still not totally sure that this question is appropriate to the site, but since it seems to have generated some activity, perhaps I should expand on my cryptic comment concerning the irreducibility in terms of tensors.

The Lie group $SO(n)$ is the subgroup of the general linear group $GL(n)$ of $\mathbb{R}^n$ which preserves the "dot" product and the associated volume form.

Suppose now that $V$ is an irreducible representation of $GL(n)$. It will decompose into irreducible representations of $SO(n)$ as $V = V_1 \oplus \cdots \oplus V_n$, where $n\geq 1$. The projector $V \to V_i$ is $SO(n)$-invariant (as an element in $\mathrm{Hom}(V,V_i)$) and hence it has to be built of the "dot" product and the volume form, these being the elementary $SO(n)$-invariant tensors.

Let $V$ denote the standard $n$-dimensional representation of $SO(n)$. It is also irreducible under $GL(n)$. The tensor square of $V$ decomposes into two $GL(n)$ irreducible representations: $$V \otimes V = S^2V \oplus \Lambda^2 V$$ Under $SO(n)$, $\Lambda^2 V$ is the adjoint representation, which is irreducible if $\mathrm{dim} V \neq 4$. For $\mathrm{dim} V = 4$, the volume element defines an $SO(n)$-equivariant endomorphism of $\Lambda^2$ which decomposes it into two three-dimensional irreducible representations. This is nothing but the fact that $SO(4)$ is not simple.

How about $S^2V$? Using the dot product we can take the "trace" of a symmetric matrix. This gives an $SO(n)$-equivariant map $S^2 V \to \mathbb{R}$, whose kernel is the representation in the OP's question. It is irreducible because using the available invariant tensors one cannot extract any further components.

Now the above, as written, is not a proof. To show that the representation is irreducible, it is probably best to resort to roots and weights. However for small representations, the sort of argument above gives a good feel for their (ir)reducibility.