# 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 $\mathrm{tr}(\tilde{\rho}(h)^2) = 0$.

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.

• 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
• Some comments from Brian Conrad: For your MO question, rather than use an argument specific to SL_n, it seems simpler to argue more structurally in terms of the Lie algebra: for a connected linear algebraic group G over a field of characteristic 0, in any finite-dimensional G-representation the G-stable subspaces are the same as Lie(G)-stable subspaces (see section 7 of Borel's book on linear algebraic groups). So the complete reducibility of the representation theory for a given G is equivalent to the same for its Lie algebra, and for Lie algebras in char. 0 the complete reducibility always – Abhinav Kumar Aug 20 '11 at 1:36
• ... holds when the Lie algebra is semisimple (which means by definition that its maximal solvable Lie ideal vanishes, and can be tested in various ways -- non-degeneracy of Killing form, consideration of weight spaces under action of a torus, etc.). Rather generally, this is how one proves that for any connected semisimple group G over a field k of char. 0, the linear repns of G are completely reducible: one shows Lie(G) must be semisimple. Then with a bit more one gets the same conclusion for any reductive k-group (possibly disconnected). – Abhinav Kumar Aug 20 '11 at 1:37