Let $G$ be a semisimple lie group, with Lie algebra $\mathfrak g$. For each representation $V$ of $G$, we can consider the bilinear form on $\mathfrak g$ that sends a pair of elements $(x,y)$ to $tr(xy|V)$.

What is the abelian group generated by these bilinear forms, including its embedding into the vector space of bilinear forms?

The bilinear form is certainly a $\mathbb Q$-linear combination of the Killing forms of the simple factors of $\mathfrak g$ - in other words, the bilinear forms coming from the adjoint representations. The coefficient of each Killing form is proportional to the trace of the corresponding Casimir operator on $V$.

This completely describes the $\mathbb Q$-vector space generated by these bilinear forms, but the abelian group is more difficult to characterize. Note that it depends on $G$, not just on $\mathfrak g$, as $G$ need not be simply-connected.

For $G=SL_n$, it's clear that the group is generated by the standard representation, or $\frac{1}{2n}$ times the Killing form. But for $SL_n/\mu_k$ with $k|n$, I can only figure out the answer for very small $k$ or $n$. For other simple groups, or for semisimple groups that do not factor as a product, it quickly gets very hard. One can reduce it to some combinatorics problem on the root lattice, but the combinatorics seems quite complicated in general.

  • $\begingroup$ Could you please elaborate on the definition of your quadratic form $tr(xy|V)$. Given a linear representation $V$ of $G$ do you mean for $X,Y \in \mathfrak{g}$ to have pairing given by $trace(e^X e^Y)$ ? $\endgroup$ – J. Martel May 5 '13 at 19:21
  • $\begingroup$ No. I mean $tr(XY)$, so that it is bilinear. $\endgroup$ – Will Sawin May 5 '13 at 19:47

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