# why are all characters of the maximal torus in a Lie group weights?

Let $G$ be a compact connected Lie group, $T$ maximal torus, identified with $\mathbb{R}^n/\mathbb{Z}^n$, $X^*(T)$

the set of characters of $T$, naturally identified with $\mathbb{Z}^n$. Let next $\Phi$ denote the set of roots, corresponding to $T$, $$\Lambda=\{v\in \mathbb{R}^n: \forall \alpha\in \Phi,\ 2(v,\alpha)/(\alpha,\alpha)\in \mathbb{Z}\}$$ be the lattice of weights. Then $X^*(T)\subset \Lambda$.

What is the easiest and most elementary way to prove it?

It suffices to prove that for any $\lambda\in X^*(T)$ is such a character that for some finite-dimensional complex representation of $G$ there is a non-zero $T$-invariant subspace, on which $T$ acts by multiplying by $\lambda(\cdot)$. Is there any way to do it without appealing to infinite-dimensional representations (inducing $\lambda$ from $T$ to $G$ and some work after that)?

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"Easiest" depends on how you set things up: everything really hinges on how you want to identify $X^\ast(T)$ with $\mathbb Z^n$. It's probably cleanest if you don't work explicitly with $\mathbb Z^n$, but instead state everything in terms of Lie algebras and their duals. I personally like the setup given in Knapp, Lie Groups Beyond an Introduction, IV.6--7, which is fairly standard. In the end it all boils down to associating a copy of $SU(2)$ (or $\mathfrak{su}(2)$ or $\mathfrak{sl}_2(\mathbb C)$ ...) to each root $\alpha$, and then the integrality statement you're after ultimately follows from the fact that $$\exp 2 \pi i x = 1 \, \implies \, x \in \mathbb Z.$$

Edit: Here's an outline of the details. Let $\mathfrak t_0$ and $\mathfrak t$ denote, respectively, the real and complex Lie algebras of $T$ and set $$L = \{ \xi \in \mathfrak t_0 \mid \exp \lambda = 1 \}$$ for the kernel of the exponential map $\exp \colon \mathfrak t_0 \to T$. Also define $$L^\perp = \{ \lambda \in \mathfrak t^\ast \mid \lambda(L) \subset 2\pi i \mathbb Z \}.$$ (Admittedly, the $^\perp$ is a slight abuse of notation.) Then there is an isomorphism $$L^\perp \stackrel{\sim}{\longrightarrow} X^\ast(T)$$ given by sending $\lambda$ to the character $e^\lambda$ defined by $$e^\lambda(\exp \xi) = e^{\lambda(\xi)}, \qquad \xi \in \mathfrak t_0.$$ This is basically Proposition 4.58 in Knapp. So our objective now is to show that $L^\perp$ sits in the weight lattice $\Lambda$ --- in other words, we want to show that $$2\frac{(\lambda, \alpha)}{(\alpha,\alpha)} \in \mathbb Z$$ for all $\lambda \in L^\perp$ and $\alpha \in \Phi$ (Prop 4.59). To this end, let $\psi_\alpha \colon SU(2) \to G$ denote the root morphism corresponding to the root $\alpha \in \Phi$. This morphism has the property that $d \psi_\alpha(h) = 2 \alpha^\vee/(\alpha,\alpha)$, where $$h = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \in \mathfrak{su}(2).$$ Consequently, $$1 = \psi_\alpha(1) = \psi_\alpha(\exp (2\pi i h)) = \exp (2 \pi i \, d \psi_\alpha(h)),$$ that is, $$2 \pi i \, d \psi_\alpha(h) = 2 \pi i \cdot 2 \frac{\alpha^\vee}{(\alpha,\alpha)} \in L$$ whence $$\lambda(2 \pi i \cdot 2 \frac{\alpha^\vee}{(\alpha,\alpha)}) \in 2 \pi i \mathbb Z \iff 2\frac{(\lambda, \alpha)}{(\alpha,\alpha)} \in \mathbb Z,$$ as desired.

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So the fundamental theorem which is being used is that each $\mathfrak{su}(2)$ exponentiates to an $SU(2)$ or $PSU(2)$. I like this argument a lot. – David Speyer May 20 '11 at 10:58
thanks, that's perfect! – Fedor Petrov May 20 '11 at 18:37

To answer the last question here, it's certainly not necessary to introduce infinite-dimensional representations into the picture when studying compact Lie groups. On the other hand, I don't quite see the point of trying to relate in isolation the character group of a maximal torus to an abstractly defined "weight" lattice. This is best done within the standard context of finite dimensional representations of a compact Lie group, which of course takes a while to develop from scratch. Look for instance at a standard 1985 textbook by Brocker and tom Dieck Representations of Compact Lie Groups (Springer), where all of the classical theory is laid out systematically.

It's possible of course (as shown by Bourbaki) to treat roots and weights more abstractly without specific reference to Lie groups or Lie algebras, but this is rather artificial here. At any rate, it's risky to start out by identifying the character group of a maximal torus "somehow" with the standard lattice in $\mathbb{R}^n$, since the same could be done with the root lattice or (abstract) weight lattice. It's the placement of one lattice within another that counts.

It's important to have examples at hand, as a reminder that the character group of a maximal torus (say in a semisimple compact group) can vary anywhere between the root lattice (adjoint group) and full weight lattice (simply connected group).

P.S. Added observations: (1) You should assume your group is semisimple (having no central torus of positive dimension) if you want to compare $X(T)$ with the dual lattice of the root lattice, the latter being of finite index in the former. It's also helpful to recall from the general theory that the quotient of these lattices is isomorphic to the center of the simply connected covering group of your given group and to the fundamental group of the adjoint group. (2) As the answers here suggest, your question doesn't have any "easy" answer until you place it in one of the well-established textbook settings for studying structure and representations of compact Lie groups. There are various approaches using analysis, algebraic topology, algebraic geometry, Lie algebras, comparison with complex Lie groups, etc. Your question is mainly pedagogical, so you should first place it in one or more of these traditions.

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I'm not sure what I'm allowed to use here. Here is a proof, and I'll see whether or not you are happy with it.

Let $L$ be the lattice of those characters of $T$ which occur in finite dimensional representations of $G$. It is clearly a lattice because, if $\lambda$ occurs in $V$ and $\mu$ occurs in $U$, then $\lambda+\mu$ occurs in $V \otimes U$. As you say, it is enough to show that $L = X^*(T)$. Also, note that the root lattice is in $L$.

Let $\lambda$ be any element of $X^*(T)$. Let $s : T \to \mathbb{C}$ be the function $t \mapsto \sum_{w \in W} \lambda(w(t))$. (Implicitly using that the Weyl group is finite; am I allowed to assume this?) Then $s$ is a $W$-invariant function on $T$, and it is not zero since it is $|W|$ at the identity. Since every conjugacy class of $G$ intersects $T$, and does so in a $W$-orbit, we can extend $s$ to a conjugacy invariant function on $G$.

By the Peter-Weyl theorem, there must be some finite dimensional representation $V$ such that $\langle s, \chi_V \rangle \neq 0$. By the Weyl integral formula, $\langle s, \chi_V \rangle$ is an integral over $T$ of a product of three factors: $s$, which is a certain finite sum of characters of $T$; $\chi_V$, which is a certain finite sum of characters in $L$, and the Weyl integrand, which is a certain sum of characters in the root lattice. So the product of the last two terms is a sum of characters from $L$.

In order for this integrand to be nonzero, one of the characters in $s$ must be negative a character in the root lattice, so $-w(\lambda) \in L$ for some $w\in W$. Using that $L$ is $W$-stable, this shows that $\lambda \in L$.

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Did you switch from $V$ to $W$ at some point? – Keerthi Madapusi Pera May 19 '11 at 22:58
No, but I used W for two things: the Weyl group and the hypothetical representation in the second paragraph. I'll change the later use to $U$. – David Speyer May 19 '11 at 23:02
Thanks, I was just being dense. – Keerthi Madapusi Pera May 20 '11 at 2:52
Thanks, this is nice argument, the onliest problem why it is not appropriate for my students is that they do not know the proof of Peter-Weyl theorem, which requires some functional analysis (or it actually does not? the proof which I know requires some properties of compact operators.) – Fedor Petrov May 20 '11 at 18:39
Yes, I think there is no way of avoiding some difficult analysis there. That's the main reason I said I wasn't sure this was what you wanted. – David Speyer May 20 '11 at 19:22