Bilinear map on semisimple Lie group I am looking at Proposition 2.3.2 in Pressley and Segal's book "Loop Groups" (on page 14). The statement is:
If $\mathfrak{g}$ is the Lie algebra of a compact semisimple group $G$ then any $\mathbb{C}$-linear $G$-invariant map $B:\mathfrak{g}_{\mathbb{C}} \times \mathfrak{g}_{\mathbb{C}}\to\mathbb{C}$ is necessarily symmetric.
The proof first assumes $G$ is simple, which is pretty easy by Schur's lemma. Then, they decompose $\mathfrak{g} = \mathfrak{g}_1 \oplus ...\oplus \mathfrak{g}_k$ according to the adjoint representation.
They then claim "The factors $\mathfrak{g}_i$ are obviously all non-isomorphic as representations of $G$"
I don't understand the claim here. If I take $G$ to be simple, then $G\times G$ is semisimple, and its Lie algebra splits into a direct sum of isomorphic representations, no?
 A: I admit I've been confused by this too, but no, the two "summands" of the adjoint representation of $G \times G$, i.e. its restrictions to the first and second summand of $\mathfrak g \oplus \mathfrak g$, respectively, are actually not isomorphic to each other (as $G\times G$-representations).
For starters, notice that the kernel of the first one is $Z(G) \times G$, whereas the kernel of the second is $G \times Z(G)$. Sure, these are isomorphic. But isomorphic representations have the same kernel (not just isomorphic ones).
From here you'll realize that the two representations treat the two factors of the group very differently from each other, so much that they are not isomorphic as representations of $G \times G$.
Relatedly, I used to think: Is there not an obvious equivalence given by the flip? But no, note that the "flip" map $\mathfrak g \oplus \mathfrak g$, $(x,y) \mapsto (y,x)$ is not equivariant under the action of $G \times G$. (It would be equivariant for the action of the diagonal, i.e. of just $G$. Of course, seen this way as representations just of $G$, the two reps are isomorphic. But not as reps of $G \times G$.)
To tie this back to the point in your source where you were confused: If $G = G_1 \times \dots G_k$ with Lie algebra $\mathfrak g_1 \oplus \dots \oplus\mathfrak g_k$, then distinct summands $\mathfrak g_i$ are mutually not isomorphic as $G = G_1 \times \dots G_k$-representations; even if for some pairs $i \neq j$ we have $G_i \simeq G_j$, which admittedly would make $\mathfrak g_i \simeq \mathfrak g_j$ but only as $G_i$- (or $G_j$-)representation, not as representation of the full group $G$.
A: An invariant (under $\frak{g}$) $\langle \cdot, \cdot \rangle$ means
$$\langle [X,Y], Z\rangle + \langle Y, [X,Z]\rangle = 0$$
If $X$, $Y$ are in $\frak{g}_{\alpha}$, $Z \in \frak{g}_{\beta}$, $\alpha\ne \beta$, then $[X,Z]=0$, and so $\langle [X,Y], Z\rangle = 0$. We conclude that $$\langle [\frak{g}_{\alpha}, \frak{g}_{\alpha}], \frak{g}_{\beta}\rangle=0$$
Now, if $\frak{g}$ is semisimple then its factors $\frak{g}_{\alpha}$ are simple, so $[\frak{g}_{\alpha}, \frak{g}_{\alpha}]= \frak{g}_{\alpha}$. We conclude
$$\langle \frak{g}_{\alpha}, \frak{g}_{\beta}\rangle=0$$
Therefore, an invariant form on $\frak{g}$ is the direct sum of invariant forms on $\frak{g}_{\alpha}$.
Assume now $\frak{g}$ simple, $V$ an irreducible f.d. representation of $\frak{g}$. Then there exists at most one ( up to multiplication by scalars) $\frak{g}$-invariant form on $V$ . Indeed, such a form is equivalent to  a $\frak{g}$-map from $V$ to $V^{*}$, and the dimension of this space is at most $1$-dimensional.
Now, for $\frak{g}$ simple, the Killing form is symmetric, invariant and non-degenerate ( so non-zero)( non-deg. equivalent to semi-simple -- Theorem of Cartan). Therefore: every invariant form on $\frak{g}$ is symmetric.
The same will happen for $\frak{g}$ semi-simple.
