# Linear reductivity of $SL_n$ in char $0$: proof in Mukai's book

I'm reading through Mukai's excellent book "Introduction to Invariants and Moduli", and am stuck on a proof in Chapter 4. He's proving that $G = SL_n$ over a field $k$ of characteristic $0$ is linearly reductive, i.e. for every epimorphism $V \rightarrow W$ of representations of $G$, the induced map on invariants $V^G \rightarrow W^G$ is also surjective. Let $\rho$ be a representation of $SL_n$ and let $\tilde{\rho}$ be the induced representation on the Lie algebra and the distribution algebra at the identity. Let $\Omega$ be the Casimir element/operator. Let $T$ be the torus of diagonal matrices in $SL_n$ and let $\frak{h}$ be its Lie algebra.

Mukai reduces the proof of linear reductivity to the following assertion: if $\mathrm{tr}( \tilde{\rho}(\Omega)) = 0$ we must also have $\mathrm{tr}(\tilde{\rho}(h)) = 0$ for all $h \in \frak{h}$. He then says: we will do this just for $SL_2$; the general case is similar. For $SL_2$ we have $\frak{h}$ is one-dimensional spanned by multiples of the root $h = \epsilon_1 - \epsilon_2$, and by explicit calculation $\mathrm{tr}( \tilde{\rho}(\Omega)) = \mathrm{tr}(\tilde{\rho}(h)^2)$. So the assertion is immediate. But I don't quite see how the "similar" proof for $SL_n$ works. It would be great if someone could explain this to me!

In Mukai's approach to this proof (in particular his Prop 4.49), care is needed typographically, all squaring must come before taking the trace. Note that having $\mathrm{tr}(\tilde{\rho}(h)) = 0$ for all h does not imply $\tilde\rho$ is trivial (eg 2 dim rep of SL_2), the condition we need is $\mathrm{tr}(\tilde{\rho}(h)^2) = 0$.
If it's a proof of Mukai's Prop 4.49 you're after, again lets WLOG k=ℂ. By Weyl's unitary trick, for $h\in \frak h$ the expression$$\int_{SU_n}\mathrm{Ad}(k)h^2dk$$ is $SU_n$-invariant, hence a scalar multiple of the Casimir. And we can see it's nonzero by taking it's trace on some nontrivial representation.
• Hi Peter, thanks! You're right about $\textrm{tr}(\tilde{\rho}(h)^2)$ of course, so I changed that above. Weyl's unitary trick does work, but I also found another answer (a purely algebraic one) in Goodman-Wallach Section 3.3.3. The idea is to first reduce it to linear reductivity of the Lie algebra (see Brian Conrad's comments from an email below) and then to use Casimir to show that an extension of two highest weight representations of the Lie algebra is a direct sum, and finally to bootstrap/induct using dimension. – Abhinav Kumar Aug 20 '11 at 1:35