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Suppose $G$ is a group and $V$ an irreducible representation of $G$. One has that $V\otimes V\cong \Lambda^2(V)\oplus Sym^2(V)$. It is well-known that if the trivial representation appears as a subrepresentation of $\Lambda^2(V)$ then $V$ is of quaternionic type; while if the trivial representation appears as a subrepresentation of $Sym^2(V)$ then $V$ is a of real type. From this approach, it is clear that the trivial representation cannot appear in both $\Lambda^2(V)$ and $Sym^2(V)$.

What I am curious about is as follows:

Question: Is there is some (relatively easy) way to see why the trivial representation cannot appear in both $\Lambda^2(V)$ and $Sym^2(V)$ without introducing the machinery of real/quaternionic types?

As a bit of motivation, if one looks at other subrepresentations, then for example if $G = G_2$ and $V_n$ is an $n$-dimensional irreducible representation of $G_2$, then $V_{64}$ appears as a subrepresentation of both $\Lambda^2(V_{27})$ and $Sym^2(V_{27})$. In particular it is possible for the intertwining number of $\Lambda^2(V)$ and $Sym^2(V)$ to be nonzero, but by the real vs. quaternionic characterization, the trivial representation is somehow special in that it cannot contribute to the intertwining number.

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up vote 6 down vote accepted

The trivial representation appears in $\wedge^2 V$ if and only if the representation $V^{\ast}$ has a $G$-invariant alternating bilinear form (because $\wedge^2 V\cong\wedge^2\left(\left(V^{\ast}\right)^{\ast}\right)$ is isomorphic to the $G$-module of all alternating bilinear forms on $V^{\ast}$, and $G$-invariant forms correspond to $G$-fixed elements).

ctrl+c & ctrl+v:

The trivial representation appears in $\mathrm{Sym}^2 V$ if and only if the representation $V^{\ast}$ has a $G$-invariant symmetric bilinear form (because $\mathrm{Sym}^2 V\cong\mathrm{Sym}^2\left(\left(V^{\ast}\right)^{\ast}\right)$ is isomorphic to the $G$-module of all symmetric bilinear forms on $V^{\ast}$, and $G$-invariant forms correspond to $G$-fixed elements).

So we have to prove that for an irreducible representation $V$, the representation $V^{\ast}$ cannot have both a nontrivial $G$-invariant symmetric bilinear form and a nontrivial $G$-invariant alternating bilinear form. More generally, an irreducible representation $W$ of $G$ cannot have two linearly independent $G$-invariant bilinear forms. In fact, a bilinear form on the $k$-vector space $W$ can be seen as a homomorphism $W\to W^{\ast}$, and the bilinear form is $G$-invariant if and only if this homomorphism is $G$-equivariant. But Schur's lemma yields that there cannot be two linearly independent $G$-equivariant homomorphisms from $W$ to $W^{\ast}$, since both $W$ and $W^{\ast}$ are irreducible representations.

So much for the case when the ground field is algebraically closed (which is probably your case). In the general case, I think the assertion is not true, though I don't know a counterexample right out of my head.

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Counterexample for the nonalgebraically closed case: $k=\mathbb{R}$, $G = \mathbb{Z}/n$ for $n>2$, $V=\mathbb{R}^2$ with $G$ acting by rotation by $2 \pi/n$. Rotation of the plane preserves the standard (symmetric) inner product and the standard (anti-symmetric) area form. – David Speyer May 13 '11 at 19:39
Very nice counterexample! – darij grinberg May 13 '11 at 19:40
So it still reduces to the fact that the representation is self-dual, but at least your approach avoids introducing the equivalence with real/quaternionic types. – ARupinski May 13 '11 at 20:06
@ARupinski: ... and is also valid over arbitrary fields. – Qiaochu Yuan May 14 '11 at 8:40

This is essentially what Darij wrote, but without mentioning the bilinear forms. (I had written it out before reading far enough into Darij's post to see that he was really doing the same thing, after the part about bilinear forms.) Think of $V\otimes V$ as $\text{Hom}(V^*,V)$. An occurrence of the trivial representation in $V\otimes V$ thus amounts to a $G$-equivariant linear map from $V^*$ to $V$. Since $V$ and therefore also $V^*$ are irreducible, Schur's lemma says that the space of such maps has dimension either 1 (iff $V$ and $V^*$ are isomorphic) or 0. So there's at most one occurrence of the trivial representation in $V\otimes V$.

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So to answer the question implicit in the last part of the OP ("why is the trivial representation special"), the trivial representation is special because $\text{Hom}_G(1, V)$ represents the functor sending a representation to its $G$-invariant subspace, and this functor sends $\text{Hom}(A, B)$ to $\text{Hom}_G(A, B)$. – Qiaochu Yuan May 14 '11 at 8:39
By the way, doesn't the identification of $V \otimes V$ with $\text{Hom}(V^{\ast}, V)$ only hold if $V$ is finite-dimensional? If $V$ is infinite-dimensional I think the latter is much larger... – Qiaochu Yuan May 14 '11 at 8:42
I was assuming finite-dimensionality, but now I see that there's no finiteness in the question. – Andreas Blass May 14 '11 at 17:28
Still assuming finite-dimensionality, it might be worth pointing out that the argument gives more than was asserted. If $V$ and $W$ are any two irreducible representations of $G$, then the number of copies of the trivial representation in $V \otimes W$ is 1 if $V$ and $W$ are dual to each other and 0 otherwise. – Andreas Blass May 14 '11 at 17:30
If you take unitary representations on infinite dimensional Hilbert spaces you can reformulate some of this stuff. In this case, $V\otimes V^{*}$ is isomorphic to Hilbert-Schmidt operators from $V$ to $V,$ and has a copy of the trivial representation if and only if $V$ is finite dimensional. Also, one can reformulate Schur's Lemma to say that a unitary representation is irreducible if and only if there are no nontrivial bounded operators commuting with $\pi(g)$ for all $g\in G.$ (Think of commuting operators as $G$-modular maps). This requires some tools though, namely the spectral theorem. – Benjamin Hayes May 17 '11 at 19:35

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