# proof of basic fact that torus actions are diagonalizable

Suppose a torus $T=(\mathbb{C}^\ast)^n$ acts on a finite dimensional vector space $W$, and define for $m \in M$ ($M$ is the character lattice of $T$) the eigenspace $W_m$ by $$W_m = \{w \in W \mid t\cdot w = \chi^m(t)w \text{ for all }t \in T \}$$ i.e. for $w \in W_m$ is a simultaneous eigenvector for all $t \in T$, with eigenvalue $\chi^m(t)$ depending on $t \in T$. Then it is a famous fact $$W=\underset{m \in M} \bigoplus W_m$$ Can someone provide a somewhat self-contained proof of this result? I don't know much about the theory of algebraic groups.

$\def\C{\mathbb{C}}$ My answer uses results from James Humphreys' Linear Algebraic Groups. I'll cite the pages which have the results I'm going to use because the proofs he gives are entirely self-contained and I would mostly be transcribing them otherwise.

The key idea is that the torus $(\mathbb{C}^{*})^{n}$ consists of commuting semisimple elements and that homomorphisms of algebraic groups preserve semisimplicity. This is all you need to know if you don't want to read the very long answer that follows.

For an algebraic group $G$ and an element $x \in G$, we say that $x$ is semisimple if the action of $x$ on the coordinate ring $\mathbb{C}[G]$ defined by $(x \cdot f(g)) = f(g\cdot x)$ can be diagonalized. Now, the coordinate ring of $(\mathbb{C}^{*})^{n}$ is $\mathbb{C}[x_{1}^{\pm}, \ldots, x_{n}^{\pm}].$ Now, if $r = (r_{1}, \ldots, r_{n})$ is an element of $(\mathbb{C}^{*})^{n}$ and $g$ is an arbitrary element of the same, then for any $a_{1}, \ldots, a_{n}$,

$$(r\cdot x_{1}^{a_{1}}\cdots x_{n}^{a_{n}})(g) = r_{1}^{a_{1}}\cdots r_{n}^{a_{n}}(x_{1}^{a_{1}}\cdots x_{n}^{a_{n}})(g).$$ Hence $$r \cdot x_{1}^{a_{1}} \cdots x_{n}^{a_{n}} = r_{1}^{a_{1}}\cdots r_{n}^{a_{n}} (x_{1}^{a_{1}}\cdots x_{n}^{a_{n}}).$$

Hence, each of the monomials in the coordinate ring of $(\mathbb{C}^{*})^{n}$ is an eigenvector for $r$. Thus, every element in $(\mathbb{C}^{*})^{n}$ acts diagonally on the coordinate ring and is hence semisimple.

The next result is found in page 99 of Humphreys, where he proves (among other things) that homomorphisms of algebraic groups preserve semisimplicty. That is, if $\rho: G \rightarrow G'$ is a morphism of algebraic groups and if $g \in G$ is semisimple, then $\rho(g)$ is semisimple in $G'.$

So now, if we have a representation $V$ of $(\C^{*})^n$, then we have a homomorphism $\rho: (\C^{*})^{n} \rightarrow GL(V)$ whose image consists entirely of semisimple elements. So now, we show that if an element $g \in GL(V)$ is semisimple then it acts diagonally on $V$. This will then imply that each element of $\rho((\C^{*})^{n})$ acts diagonally on $V$ as and hence the entire action can be simultaneously diagonalized since the operators also commute with each other.

To finish the proof, we use the multiplicative Jordan decomposition. This can be found on page 96 of Humphreys (and many other sources I would imagine.) This says that for any element $g \in GL(V)$, there exist $g_{s}, g_{u} \in GL(V)$ such that

1. $g, g_{s}, g_{u}$ commute with any linear map $V\rightarrow V$ that commutes with $g$.
2. $g_{s}$ is diagonalizable.
3. $g_{u}$ is unipotent i.e. there exists some $m$ such that $(g_{u} - 1)^{m} = 0.$

So now, since $g_{s}$ is diagonalizable, it must act semisimply on $\C[GL(V)]$. This is because we let $x_{i, j}$ be coordinates for $GL(V) \cong M(n \times n)$ in the basis such that $g_{s}$ is diagonal, then $\C[GL(V)] = \C[X_{1,1}, \ldots, x_{n, n}]_{\det}$ and the elements of the form

$$\frac{x_{1, 1}^{a_{1, 1}}\cdots x_{n, n}^{a_{n, n}}}{\det^{l}}$$

form a spanning set in $\C[GL(V)]$ of eigenvectors for $g_{s}$ and hence for $g_{s}^{-1}.$ As $g$ also acts diagonally on $\C[GL(V)]$, and commutes with $g_{s}^{-1}$, the action of $g$ and $g_{s}^{-1}$ on $\C[GL(V)]$ must be simulataneously diagonalizable. Hence, $gg_{s}^{-1} = g_{u}$ must also act diagonally on $\C[GL(V)]$.

But $g_{u} - 1$ acts nilpotently on $\C[GL(V)]$ because its $m$th power is $0$. Hence, the action of $g_{u}$ in $\C[GL(V)]$ must have eigenvalue $1$ and is hence the identity. Since the representation of an algebraic group in its coordinate ring is faithful, this implies that $g_{u} = 1$ and hence $g = g_{s}$ is diagonalizable in $GL(V)$. This is what we needed.

• This is useful, thank you. I had one question. I know that a finite set of commuting diagonalizable linear operators can be simultaneously diagonalized. But $(\mathbb{C}^\ast)^n$ has infinitely many elements, so what is the argument that they can all be simultaneously diagonilized? – ykm Jun 11 '14 at 4:57
• $(\mathbb{C}^{*})^{n}$ is the Zariski closure of the subgroup generated by finitely many elements. If these finitely many elements are diagonalized, so is the subgroup they generate and hence so is its Zariski closure as the set of diagonal matrices is Zariski closed. – Siddharth Venkatesh Jun 11 '14 at 8:08