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In Knapp's book Representation Theory of Semisimple Groups: An Overview Based on Examples, he proves the following theorem of Weyl: If $G$ is a compact connected semisimple Lie group, then $\pi_1(G)$ is finite. I am confused about a detail in the proof, which I now explain.

First, Knapp shows (by exploiting compactness of $G$) that $\pi_1(G)$ is a finitely generated abelian group.

Let $\widetilde{G}$ be the universal cover of $G$, so that $G = \widetilde{G}/Z$ for a discrete subgroup $Z \subset Z(\widetilde{G})$ isomorphic to $\pi_1(G)$. Knapp assumes for the sake of contradiction that $Z$ is infinite. Then (since it is a finitely generated abelian group, so by the structure theory...) there exist subgroups $Z_1 \subset Z$ of arbitrarily large finite index.

Now for some notation and recall of facts. Let $\mathfrak{g}$ be the Lie algebra of $G$ (and fix a Cartan $\mathfrak{h}$ in $\mathfrak{g} \otimes \mathbf{C}$), and let $d$ be the index of the root lattice in the weight lattice for $\mathfrak{g} \otimes \mathbf{C}$ (in particular $d$ is a finite number). To make my notation (which comes from Fulton--Harris, which I studied much more carefully) consistent with Knapp, my "weight lattice" is what Knapp calls the "algebraically integral forms" --- it is the set of $\lambda \in \mathfrak{h}^*$ such that $\langle \lambda, \alpha^\vee \rangle \in \mathbf{Z}$ for all roots $\alpha$. Given a compact connected Lie group $G' = \widetilde{G}/Z'$ with Lie algebra $\mathfrak{g}$, we can consider the lattice $\Lambda_{G'}$ of ``analytically integral weights for $G'$'', namely the $\lambda \in \mathfrak{h}^*$ such that $\lambda(X) = 1$ for all $X \in \mathfrak{h} \cap \mathfrak{g}$ with the property that $\exp_{G'}(X) = 1$. It is a fact that $\Lambda_{G'}$ contains the root lattice $\Lambda_R$ and is contained in the weight lattice $\Lambda_W$(this is Knapp (4.15)). Moreover, if $Z(\widetilde{G}) \supset Z'' \supset Z'$ and $G'' = \widetilde{G}/Z''$, then the number of sheets in the covering map $G' \to G''$ is $[\Lambda_{G'} : \Lambda_{G''}]$ (this is Knapp 4.25).

It seems to me that the rest of the proof of Weyl's theorem is a direct consequence of the above paragraph: choose $Z_1$ so that $[Z : Z_1] > d$. Since $[Z : Z_1]$ is the number of sheets in the covering of compact connected groups $\widetilde{G}/Z_1 \to \widetilde{G}/Z = G$, by the above paragraph we have $[\Lambda_{\widetilde{G}/Z_1} : \Lambda_{\widetilde{G}/Z}] > d$. This contradicts the inclusions $\Lambda_R \subset \Lambda_{\widetilde{G}/Z} \subset \Lambda_{\widetilde{G}/Z_1} \subset \Lambda_W$ and the fact that $[\Lambda_W : \Lambda_R] = d$.

However, the proof in Knapp doesn't quite finish here: he concludes (somehow) from this that $\widetilde{G}/Z_1$ is not a linear group, which contradicts the fact that it is compact (which is because $[Z : Z_1]$ is finite and $G$ is compact). Why is this extra step necessary, and why is it clear from this that $\widetilde{G}/Z_1$ is not linear ?

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    $\begingroup$ In case you are interested: there is a formula for for $\pi_1(G)$ where $G$ is a connected semisimple compact group $G$. Ideed, after passing to the complexification, you may assume that $G$ is a connected complex semisimple groups Lie. Then a formula for $\pi_1(G)$ is given in Chapter 3, Theorem 2.9 on page 92 of Gorbatsevich, Onishchik, and Vinberg, Structure of Lie groups and Lie Algebras, Onishchik and Vinberg (eds.), Lie Groups and Lie Algebras III, Encyclopaedia of Mathematical Sciences, vol. 41, Springer-Verlag 1994. $\endgroup$ Commented Apr 18, 2023 at 13:30
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    $\begingroup$ The formula for a complex semisimple group $G$ can be written as follows: Let $T\subset G$ be a maximal torus. Let $\rho\colon \widetilde G\to G$ be the universal covering. Write $\widetilde T=\rho^{-1}(T)\subset\widetilde G$. Then $\pi_1(G)\simeq X_*(T)/\rho_*X_*(\widetilde T)$ where $X_*$ denotes the cocharacter group: $X_*(T)={\rm Hom}({\Bbb C}^\times, T)$ (we consider only polynomial homomorphisms). $\endgroup$ Commented Apr 18, 2023 at 13:42

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It seems to me that you are right and that the proof does not need the detour to linear Lie groups. I think the reason why Knapp writes the proof of his Theorem 4.26 in this way is that throughout paragraph IV.5 he assumes that G is a compact linear Lie group. Strictly speaking, Knapp shows the necessary results about lattices only for such groups.

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  • $\begingroup$ Okay ! That makes sense, since (for example at the beginning of section 2) the fact that G is linear is used to understand the Cartan and how it sits inside the complexification. If you use Fulton--Harris' perspective, I suppose the difficulty gets shifted into the appendices of the book and into the brief discussion on compact forms. $\endgroup$
    – babu_babu
    Commented Apr 30, 2023 at 19:49

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