Optional: A torus is contained in a maximal torus

Any torus is contained in a maximal torus

This video is optional.

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

Lemma:

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

Proof:

0.35 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 $T_{1}\subset T_{2}\subset T_{3}\subset\cdots$ each strictly included in the next.

2.05 We now pass to the level of Lie algebras: $\mathfrak{t}_{1}\subset\mathfrak{t}_{2}\subset\mathfrak{t}_{3}\subset\cdots.$ These are all subspaces of $\mathfrak{g}$ , 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 $\mathfrak{t}_{N+k}=\mathfrak{t}_{N}$ for all $k\geq 0$ .

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

Outlook

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

Remark:

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

Remark:

There is a more sophisticated way to think about $\lambda$ . We can think of $\lambda$ as the linear function $\lambda_{1}\theta_{1}+\cdots+\lambda_{n}\theta_{n}$ 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 $(i\theta_{1},\ldots,i\theta_{n})$ (just like how $\mathfrak{u}(1)=i\mathbf{R}$ ). Instead, the vector $(\theta_{1},\ldots,\theta_{n})$ lives in $i\mathfrak{t}\subset\mathfrak{t}\otimes\mathbf{C}$ . This means that $\lambda$ lives most naturally in $(i\mathfrak{t})^{*}$ (where $*$ means the dual space of linear functions). We will discuss this more formally when we talk about the Killing form.