Occurence of trivial representation in a tensor square. 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.
 A: 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.
A: 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 $V\otimes V$.
