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!



Linear reductivity of $SL_n$ in char $0$: proof in Mukai's book
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 $\mathrm{tr}(\tilde{\rho}(h)^2) = 0$.


I don't know how the 'similar' proof works either. Instead I'd approach it by saying WLOG k=ℂ (Lefschetz principle!). Then use Weyl's unitary trick to reduce the proof of linear reductivity to the case of representations of compact groups.


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.


