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:



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> > **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? 
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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.



Occurence of trivial representation in a tensor square.
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.


