Any nontrivial torus is contained in a maximal torus (so that, in particular, $G$ contains a maximal torus).

# Optional: A torus is contained in a maximal torus

## Any torus is contained in a maximal torus

This video is optional.

We have so far proved that any compact matrix group $G$ contains a nontrivial torus. We will now show that:

Suppose, for a contradiction, that we have a torus ${T}_{1}$ not contained in a maximal torus. Then ${T}_{1}$ cannot itself be maximal (it's contained in itself), so it must be contained in a strictly bigger torus ${T}_{2}$ , and ${T}_{2}$ is also not allowed to be maximal. Therefore ${T}_{2}$ is contained in a strictly bigger non-maximal torus${T}_{3}$ . Continuing in this manner, I can construct a sequence of nested tori $${T}_{1}\subset {T}_{2}\subset {T}_{3}\subset \mathrm{\cdots}$$ each strictly included in the next.

We now pass to the level of Lie algebras: $${\U0001d531}_{1}\subset {\U0001d531}_{2}\subset {\U0001d531}_{3}\subset \mathrm{\cdots}.$$ These are all subspaces of $\U0001d524$ , which is a finite-dimensional vector space. Therefore this sequence of Lie algebras must "stabilise" at some point, that is there exists an $N$ such that ${\U0001d531}_{N+k}={\U0001d531}_{N}$ for all $k\ge 0$ .

We saw earlier that a torus is exp of its Lie algebra, so ${T}_{k}=\mathrm{exp}({\U0001d531}_{k})$ for all $k$ . But then ${T}_{N+k}={T}_{N}$ for all $k\ge 0$ , and this is a contradiction because ${T}_{N}\subset {T}_{N+1}$ is supposed to be a strict inclusion.

## Outlook

Now we have found a maximal torus $t:U{(1)}^{n}\to G$ ($T=t(U{(1)}^{n})$ ), we will use it in the following way. Given a (smooth, complex) representation $R:G\to GL(V)$ , we can restrict to get a representation ${R|}_{T}:T\to GL(V)$ of our maximal torus. Because $T\cong U{(1)}^{n}$ , we deduce that $V=\oplus {W}_{\lambda}$ where $${W}_{\lambda}=\{v\in V:R(t({e}^{i{\theta}_{1}},\mathrm{\cdots},{e}^{i{\theta}_{n}}))v={e}^{i({\lambda}_{1}{\theta}_{1}+\mathrm{\cdots}+{\lambda}_{n}{\theta}_{n})}v\}.$$ Here, $\lambda =({\lambda}_{1},\mathrm{\dots},{\lambda}_{n})$ .

The bigger our torus, the larger $n$ is, so the more integers ${\lambda}_{i}$ we have to split up our weight spaces.

There is a more sophisticated way to think about $\lambda $ . We can think of $\lambda $ as the linear function ${\lambda}_{1}{\theta}_{1}+\mathrm{\cdots}+{\lambda}_{n}{\theta}_{n}$ of the $\theta $ s. The $\theta $ s are coordinates on a certain vector space. This is not quite the Lie algebra of $T$ , because the Lie algebra of $T$ consists of $n$ -tuples of imaginary numbers $(i{\theta}_{1},\mathrm{\dots},i{\theta}_{n})$ (just like how $\U0001d532(1)=i\mathbf{R}$ ). Instead, the vector $({\theta}_{1},\mathrm{\dots},{\theta}_{n})$ lives in $i\U0001d531\subset \U0001d531\otimes \mathbf{C}$ . This means that $\lambda $ lives most naturally in ${(i\U0001d531)}^{*}$ (where $*$ means the dual space of linear functions). We will discuss this more formally when we talk about the Killing form.