Fundamental Groups of Compact Lie Groups

We show that the size of the fundamental group of a compact connected Lie group is controlled by the center.

I passed my qualifying exam on Friday, and one of the questions on my exam used the following very nice result.

Theorem 1. Let \(G\) be a connected compact Lie group. The group \(\pi_1(G)\) is finitely generated, abelian, and of rank \(\dim Z(G)\).

Example 2. If \(Z(G)\) is finite, then \(\pi_1(G)\) is finite. In fact, in Corollary 59, we will compute \(\pi_1(G)\) when that \(G\) has trivial center, which is the most interesting case.

Example 3. With \(G=S^1\), we see that \(G\) is abelian and thus equal to its center, so \(\dim Z(G)=1\). Similarly, \(\pi_1(G)\) is isomorphic to \(\mathbb Z\).

Remark 4. It follows from the so-called polar decomposition that any connected Lie group is homotopy equivalent to a compact Lie group. We will not discuss this further.

This post is an excuse to tour many topics in Lie theory. Most notably, we will use Lie’s fundamental theorems, highest weight theory, and the Peter–Weyl theorem. We will provide sketches of the theory for the latter two tools because the ideas in highest weight theory are useful, and the Peter–Weyl theorem is not a typical addition to a first course on Lie theory. We will not say anything interesting about Lie’s fundamental theorems because I don’t understand them.

The Center

The dimension is controlled by the connected component, so a good place to start would be by understanding \(Z(G)^\circ\). Because it is not very hard, and we are at the beginning of the blog post, we will argue that \(Z(G)^\circ\) is actually a closed Lie subgroup.

Lemma 5. Let \(G\) be a connected Lie group. Then the center \(Z(G)\) is a closed Lie subgroup with Lie algebra \(\mathfrak z(\mathfrak g)\).

Recall that the Lie algebra is the tangent space at the identity.

Proof. We claim that the center is the kernel of the adjoint representation \(\operatorname{Ad}_\bullet\colon G\to\operatorname{GL}(\mathfrak g)\), which then completes the proof upon taking the differential. On one hand, \(Z(G)\) commutes with everything in \(G\), so its adjoint operator is trivial, so \(Z(G)\subseteq\ker\operatorname{Ad}_\bullet\). On the other hand, if a given \(g\in G\) has \(\operatorname{Ad}_gX=X\) for all \(X\in\mathfrak g\), then it follows that \(\operatorname{Ad}_g(\exp X)=\exp X\) for all \(X\in\mathfrak g\) by formalism of the exponential. Thus, \(g\) commutes with an open neighborhood of \(1\in G\), which is enough to achieve \(g\in Z(G)\) because \(G\) is connected. \(\blacksquare\)

Lemma 6. Let \(G\) be a Lie group. Then the connected component of the identity, denoted \(G^\circ\), is a normal closed Lie subgroup.

Proof. Let \(\pi_0(G)\) denote the set of connected components of \(G\). We claim that we can give \(\pi_0(G)\) a group structure by taking the quotient of the group structure on \(G\). This will complete the proof because then \(G^\circ\) is the kernel of the projection \(G\to\pi_0(G)\), where \(\pi_0(G)\) is now a zero-dimensional Lie group.

To prove the claim, the main issue is showing that the group multiplication is well-defined. Accordingly, suppose \(g\) and \(g’\) live in the same connected component of \(G\), and suppose the same for \(h\) and \(h’\). Then there is a path from \(g\) to \(g’\), and the same is true for \(h\) and \(h’\). Multiplying the paths together pointwise provides a path from \(gh\) to \(g’h’\), so they are in the same connected component. \(\blacksquare\)

We now explain the structure of \(Z(G)^\circ\).

Proposition 7. Let \(Z\) be a compact, connected, abelian Lie group of dimension \(d\). Then \(Z\) is isomorphic to \(\left(S^1\right)^d\).

Proof. Because \(Z\) is abelian, its Lie algebra \(\mathfrak z\) is also abelian. Thus, by the Baker–Campbell–Hausdorff formula (or other means), one sees that \(\exp\colon\mathfrak z\to Z\) is a homomorphism and local diffeomorphism. Because \(Z\) is connected, we conclude that \(\exp\) is surjective, and we know that its kernel \(K\) must be a discrete subgroup of \(\mathfrak z\). Because \(\mathfrak z\) is some Euclidean space of dimension \(d\), we know \(K\) is free and finitely generated of some rank \(r\le d\), so there is an identification \(\mathfrak z\cong\mathbb R^d\) for which \(K\cong\mathbb Z^r\). Then \[Z\cong\mathbb R^d/\mathbb Z^r.\] The only way for \(Z\) to be compact is to have \(r=d\), and the result follows. \(\blacksquare\)

The Fundamental Group

To ground ourselves, we pick up a few facts about the fundamental group of a Lie group.

Lemma 8. Let \(M\) be a smooth manifold. Then \(\pi_1(M)\) is finitely generated.

Sketch. This is a fairly technical argument in differential geometry, so we will merely sketch it. It is enough to show that \(M\) is homotopic to a locally finite countable graph \(G\). To this end, one covers \(M\) by countably many relatively compact open balls in such a way that each ball only intersects finitely many others. Then the vertex set of the graph consists of the centers of all the balls, and two vertices are connected if and only if the balls intersect. To show that \(M\) and \(G\) are homotopic, choose any path through \(M\), and keep track of which balls the path intersects; then one can deform the path to \(G\). \(\blacksquare\)

Proposition 9. Let \(G\) be a Lie group. Then its universal cover \(G_{\mathrm{sc}}\) is also a Lie group.

Proof. The proposition is used to check that \(G_{\mathrm{sc}}\) is actually a manifold: \(G_{\mathrm{sc}}\) is locally homeomorphic to \(G\), so \(G_{\mathrm{sc}}\) is Hausdorff and locally Euclidean, and Lemma 8 implies that the covering has at most countable degree, so \(G_{\mathrm{sc}}\) is still second countable.

It remains to provide the group structure. This is done via “path-lifting”: choose some point \(1\in G_{\mathrm{sc}}\) lying above \(1\in G\) to be our identity. Then, for example, because \(G_{\mathrm{sc}}\times G_{\mathrm{sc}}\) is simply connected, the composite \[G_{\mathrm{sc}}\times G_{\mathrm{sc}}\twoheadrightarrow G\times G\stackrel m\to G\] lifts uniquely to \(G_{\mathrm{sc}}\) in such a way that \(m(1,1)=1\). Similarly, one can lift the inverse map \(i\colon G\to G\) to an inverse map \(i\colon G_{\mathrm{sc}}\to G_{\mathrm{sc}}\). The uniqueness of the liftings implies that we actually have a group structure. For example, for the associativity check, one sees that the two maps \(G_{\mathrm{sc}}\times G_{\mathrm{sc}}\times G_{\mathrm{sc}}\to G_{\mathrm{sc}}\) must be equal because they lift the same projection \(G\times G\times G\to G\). \(\blacksquare\)

Remark 10. In fact, the construction of the group structure on \(G_{\mathrm{sc}}\) shows that the projection \(G_{\mathrm{sc}}\to G\) is a group homomorphism.

We now explain, in rough terms, what the center has to do with the fundamental group.

Lemma 11. Let \(H\) be a normal closed Lie subgroup of a Lie group \(G\), and let \(G/H\) be the quotient. Then the quotient map \(\pi\colon G\to G/H\) is a fiber bundle with fiber \(H\).

Sketch. In the stated generality, this is mostly differential geometry related to the construction of the manifold structure on the quotient \(G/H\), so we will be fairly sketchy. Fix some \(g\in G\), and we would like to show that the fiber of \(G\to G/H\) around \(g\) is homeomorphic to \(H\). By translation, we may as well assume \(g=1\). Now, an open neighborhood of \(1\in G\) looks like an open neighborhood of \(0\in \mathfrak g\), and an open neighborhood of \(1\in G/H\) looks like an open neighborhood of \(0\in\mathfrak g/\mathfrak h\), so the pre-image in \(\mathfrak g\) is \(\mathfrak h\) times this open neighborhood.

Having given a general sketch, we remark that we will only use this result in the case that \(H\subseteq G\) is discrete. Then one can show directly that \(G\to G/H\) is a covering space, and the result follows. \(\blacksquare\)

Proposition 12. Let \(G\) be a connected Lie group, and let \(\pi\colon G_{\mathrm{sc}}\to G\) be its universal cover. Let \(Z\) be the kernel of \(\pi\).

1. The subgroup \(Z\subseteq G_{\mathrm{sc}}\) is discrete and central.
2. One has \(Z\cong\pi_1(G)\).

Proof. For the first part, note that \(\dim Z=\dim G_{\mathrm{sc}}-\dim G=0\), so \(Z\) is successfully discrete. It remains to show that \(Z\) is central. By the proof of Lemma 5, it is enough to show that any \(z\in Z\) acts trivially on \(\mathfrak g\). Well, \(\operatorname{Ad}_\bullet\colon G_{\mathrm{sc}}\to\operatorname{GL}(\mathfrak g)\) is the unique lifting of \(\operatorname{Ad}_\bullet\colon G\to\operatorname{GL}(\mathfrak g)\), so \(\operatorname{Ad}_z\) factors through \(G\) and hence must be trivial.

For the second part, note that \[1\to Z\to G_{\mathrm{sc}}\to G\to 1\] is a fiber bundle by Lemma 11, hence a Serre fibration, so we may take the long exact sequence in homotopy to get \[\underbrace{\pi_1(G_{\mathrm{sc}})}_0\to\pi_1(G)\to\pi_0(Z)\to\underbrace{\pi_0(G_{\mathrm{sc}})}_1,\] as desired. \(\blacksquare\)

Remark 13. It is possible to avoid the use of the homotopy fiber sequence (and even the annoying calculation in Lemma 11) and instead prove the second part directly by path-lifting. We have chosen the current approach because it seems more natural.

Finite Center Implies Semisimple

We will shortly reduce the proof of Theorem 1 to the case where \(Z(G)\) is finite. Let’s start by explaining why this case is interesting: it implies that \(\mathfrak g_{\mathbb C}\) is semisimple.

Proposition 14. Let \(G\) be a compact Lie group with Lie algebra \(\mathfrak g\). Then \(\mathfrak g_{\mathbb C}\) is reductive.

Proof. By inductively taking orthogonal complements of ideals, it is enough to exhibit a \(\mathfrak g\)-invariant positive-definite Hermitian inner product on \(\mathfrak g_{\mathbb C}\). Indeed, this implies that \(\mathfrak g_{\mathbb C}\) is a direct sum of abelian ideals and simple Lie algebras, so \(\mathfrak g_{\mathbb C}\) is reductive.

We exhibit the required Hermitian inner product using Weyl’s unitary trick. Start with any positive-definite Hermitian inner product \(\langle-,-\rangle_0\) on \(\mathfrak g_{\mathbb C}\), and then define the inner product \(\langle-,-\rangle\) on \(\mathfrak g_{\mathbb C}\) by \[\langle X,Y\rangle:=\int_{G}\langle gX,gY\rangle_0\,dg,\] where \(dg\) is a choice of right Haar measure on \(G\). (For example, \(dg\) can be chosen as the unique up to scalar right-invariant top-degree differential form on \(G\); for further explanation, see the section on the Peter–Weyl theorem below.) This integral is well-defined because the integrand is continuous, and \(G\) has finite measure (because it is compact). The pairing \(\langle-,-\rangle\) is also easily checked to be Hermitian, bilinear, and positive-definite. Lastly, \(\langle-,-\rangle\) is \(G\)-invariant, so taking the derivative of \[\langle\exp(X)Y,Z\rangle=\langle Y,\exp(-X)Z\rangle\] shows that \(\langle-,-\rangle\) is \(\mathfrak g\)-invariant as well. \(\blacksquare\)

Corollary 15. Let \(G\) be a compact Lie group with Lie algebra \(\mathfrak g\). If \(Z(G)\) is finite, then \(\mathfrak g_{\mathbb C}\) is semisimple.

Proof. By the proof of Proposition 14, \(\mathfrak g_{\mathbb C}\) is a direct sum of abelian ideals with simple Lie algebras. The center of a simple Lie algebra is trivial, so we conclude that \(\mathfrak g_{\mathbb C}\) is a direct sum of the center \(\mathfrak z(\mathfrak g)\) Because \(\mathfrak g_{\mathbb C}\) is reductive, it is a direct sum of the center \(\mathfrak z(\mathfrak g)_{\mathbb C}\) and a semisimple Lie algebra: indeed, it is enough to show that the short exact sequence \[0\to\mathfrak z(\mathfrak g)_{\mathbb C}\to\mathfrak g_{\mathbb C}\to\mathfrak g_{\mathbb C}/\mathfrak z(\mathfrak g)_{\mathbb C}\to0\] splits as \(\mathfrak g_{\mathbb C}\)-representations (because subrepresentations of \(\mathfrak g_{\mathbb C}\) are just ideals), which is true because the adjoint action by \(\mathfrak g_{\mathbb C}\) factors through the semismiple Lie algebra \(\mathfrak g_{\mathbb C}/\mathfrak z(\mathfrak g)_{\mathbb C}\). But now \(Z(G)\) is discete, so \(\mathfrak z(\mathfrak g)\) is trivial by Lemma 5, so we are done! \(\blacksquare\)

Remark 16. In fact, any complex reductive Lie algebra is a direct sum of the center and a semisimple Lie algebra. Of course, this is a corollary of the Levi decomposition, but this case is easier to show than the full Levi decomposition. The usual way to show this is to use the complete reducibility of representations of complex semisimple Lie algebras to deduce that the short exact sequence \[0\to\mathfrak z(\mathfrak g)_{\mathbb C}\to\mathfrak g_{\mathbb C}\to\mathfrak g_{\mathbb C}/\mathfrak z(\mathfrak g)_{\mathbb C}\to0\] splits as representations of \(\mathfrak g_{\mathbb C}/\mathfrak z(\mathfrak g)_{\mathbb C}\) and hence as representations of \(\mathfrak g\).

Reduction to Finite Center

We are now ready to reduce our argument to the finite center case. We will use Lie’s (first and second) fundamental theorems.

Theorem 17 (Lie). Let \(G\) be a Lie group. Then taking the Lie algebra provides a bijection between connected closed Lie subgroups \(H\subseteq G\) and Lie subalgebras \(\mathfrak h\subseteq\mathfrak g\).

Proof. We only discuss uniqueness because existence is difficult. For uniqueness, simply note that a Lie group is generated as a group by an open neighborhood of the identity, which in turn can be discovered by taking the exponential of its Lie algebra. \(\blacksquare\)

Theorem 18 (Lie). Let \(G\) and \(H\) be Lie groups with Lie algebras \(\mathfrak g\) and \(\mathfrak h\), respectively. If \(G\) is connected, the differential map \[\operatorname{Hom}(G,H)\to\operatorname{Hom}(\mathfrak g,\mathfrak h)\] is injective. If \(G\) is also simply connected, the differential map is bijective.

Proof. We only discuss the first part because the second part is rather difficult. Suppose \(\varphi\colon G\to H\) induces the zero map \(d\varphi\colon\mathfrak g\to\mathfrak h\). Now, \(\exp(d\varphi(X))=\varphi(\exp(X))\) by properties of the exponential, so it follows that \(\varphi\) is trivial on an open neighborhood of \(1\in G\). Thus, \(\varphi\) is trivial because \(G\) is connected. \(\blacksquare\)

Here is our application.

Lemma 19. To prove Theorem 1, it is enough to prove that \(\pi_1(G)\) is finite for any compact connected Lie group \(G\) with finite center.

Proof. Let \(G\) be a connected compact Lie group, and let \(G_{\mathrm{sc}}\) be its universal cover. The main point is to compute \(G_{\mathrm{sc}}\). By the argument of Corollary 15, we may write \(\mathfrak g=\mathfrak z\oplus\mathfrak g’\), where \(\mathfrak z\) is the center, and \(\mathfrak g’\) is some semisimple Lie algebra. Let \(G’\subseteq G\) be the closed Lie subgroup of Theorem 17 cut out by \(\mathfrak g’\), and let \(G’_{\mathrm{sc}}\) be its universal cover. Then \(G_{\mathrm{sc}}\) and \(\mathbb R^{\dim\mathfrak z}\times G’_{\mathrm{sc}}\) are both simply connected Lie algebras with the same Lie algebra, so they must be isomorphic by Theorem 18!

The previous paragraph provides us with a short exact sequence \[1\to Z\to\mathbb R^{\dim\mathfrak z}\times G’_{\mathrm{sc}}\to G\to1.\] By Propositions 9 and 12, we know that \(Z\) is isomorphic to \(\pi_1(G)\) and a finitely generated abelian group, so we are interested in showing that \(\operatorname{rank}Z=\dim\mathfrak z\).

• We claim that \(\operatorname{pr}_1Z\) has rank \(\dim\mathfrak z\). Indeed, note that \(G\) projects onto \(\mathbb R^{\dim\mathfrak z}/\operatorname{pr}_1Z\), so the latter is a compact connected abelian Lie group. Thus, the claim follows by the argument of Proposition 7.
• We claim that \(\operatorname{pr}_2Z\) is finite. In fact, we claim that \(Z(G’_{\mathrm{sc}})\) is finite. Because \(\mathfrak g’\) is semisimple, it has trivial center, so \(Z(G’_{\mathrm{sc}})\) is discrete, so it is enough to show that \(G’_{\mathrm{sc}}\) is compact. Thus, it is enough to show that \(G’_{\mathrm{sc}}\to G’\) is a finite cover, which is equivalent to \(\pi_1(G’)\) being finite. This holds by hypothesis because \(Z(G’)\subseteq G’\) is discrete in a compact space.

The above two observations now complete the proof: the former lower-bounds the rank, and the latter upper-bounds the rank because \(Z\subseteq\operatorname{pr}_1Z\times\operatorname{pr}_2Z\). \(\blacksquare\)

In fact, it is now not too hard to even reduce to the case where \(Z(G)\) is trivial.

Lemma 20. To prove Theorem 1, it is enough to prove that \(\pi_1(G)\) is finite for any compact connected Lie group \(G\) with trivial center.

Proof. By Lemma 19, we reduce to the case where \(G\) has finite center. Now, consider the short exact sequence \[1\to Z(G)\to G\to G/Z(G)\to1.\] Arguing as in Proposition 12, this is a fiber bundle by Lemma 11, hence a Serre fibration, so the long exact sequence in homotopy yields the exact sequence \[\pi_1(Z(G))\to\pi_1(G)\to\pi_1(G/Z(G)).\] The left group is trivial, and the right group is finite by hypothesis, so the middle group is also finite. \(\blacksquare\)

Remark 21. As before, it is not too hard to avoid using the full power of the long exact sequence (and Lemma 11) in this case. For example, simply note that the universal cover of \(G/Z(G)\) has finite index over \(G/Z(G)\) because its fundamental group is finite, so the same is true for the cover \(G\) of \(G/Z(G)\).

Highest Weight Theory

As a first intermission, we discuss the structure theory of complex semisimple Lie algebras. The hurried reader who already knows something about highest weight theory is encouraged to skip this section except to read the statement of Corollary 37 and skim the proof if desired.

Our goal is to say something about highest weight theory. Let’s begin by recalling the root decomposition.

Definition 22. Fix a complex Lie algebra \(\mathfrak g\). Then \(X\in\mathfrak g\) is semisimple if and only if \(\operatorname{ad}_X\colon\mathfrak g\to\mathfrak g\) is a semisimple operator. A Cartan subalgebra \(\mathfrak h\) is a maximal subalgebra of \(\mathfrak g\) with the property that \(\mathfrak h\) is abelian and has only semisimple elements.

Because a Cartan subalgebra is abelian, all the operators commute with each other, so they can be simultaneously diagonalized.

Definition 23. Fix a complex semisimple Lie algebra \(\mathfrak g\) with Cartan subalgebra \(\mathfrak h\). Then we define the root system \(\Phi\subseteq\mathfrak h^*\) to be the finite set of nonzero eigenvalues of the adjoint action of \(\mathfrak h\) on \(\mathfrak g\).

Theorem 24 (Root decomposition). Let \(\mathfrak g\) be a complex semisimple Lie algebra with Cartan subalgebra \(\mathfrak h\), and let \((-,-)\) be a non-degenerate \(\mathfrak g\)-invariant bilinear form on \(\mathfrak g\).

1. The action of \(\mathfrak h\) on \(\mathfrak g\) diagonalizes as \[\mathfrak g=\mathfrak h\oplus\bigoplus_{\alpha\in\Phi}\mathfrak g_\alpha.\]
2. We have \(\dim\mathfrak g_\alpha=1\) for each \(\alpha\in\Phi\).
3. The set \(\Phi\) spans \(\mathfrak h^*\).
4. For any \(\alpha,\beta\in\Phi\), the number \(2(\alpha,\beta)/(\alpha,\alpha)\) is an integer.
5. The set \(\Phi\) is preserved by reflections: for each \(\alpha\in\Phi\), the reflection \(s_\alpha(\lambda)=\lambda-2(\lambda,\alpha)/(\alpha,\alpha)\) preserves \(\Phi\).

Proof. We omit this proof. No part of this theorem is particularly hard, but the argument is very long, and it uses tools that we will not have a use for in the sequel, so we will not introduce them. \(\blacksquare\)

Definition 25. Fix a complex semisimple Lie algebra \(\mathfrak g\) with Cartan subalgebra \(\mathfrak h\), and let \(\Phi\subseteq\mathfrak h^*\) be the root system. For each \(\alpha\in\Phi\), define the coroot \(\alpha^\lor\in\mathfrak h\) to be the unique vector so that \[\lambda(\alpha^\lor)=2\cdot\frac{(\lambda,\alpha)}{(\alpha,\alpha)}\] for each \(\lambda\in\mathfrak h^*\). We may also denote \(\alpha^\lor\) by \(h_\alpha\).

Our discussion of highest weight theory will need to know something about the combinatorics of root systems.

Definition 26. Fix a complex semisimple Lie algebra \(\mathfrak g\) with Cartan subalgebra \(\mathfrak h\), and let \(\Phi\subseteq\mathfrak h^*\) be the root system. For generic \(t\in\mathfrak h\), define the set of positive roots to be \[\Phi^+:=\{\alpha\in\Phi:\alpha(t)>0\}.\] We define the set of negative roots \(\Phi^-\) similarly. (Here, “generic” means that \(\alpha(t)\ne0\) for all \(\alpha\in\Phi\).) The choice of \(\Phi^+\subseteq\Phi\) makes \(\Phi\) polarized.

Definition 27. Fix a complex semisimple Lie algebra \(\mathfrak g\) with Cartan subalgebra \(\mathfrak h\), and let \(\Phi\subseteq\mathfrak h^*\) be the polarized root system. A positive root is simple if it cannot be written as a sum of more than one positive root.

Remark 28. It turns out that the set of simple roots is a basis for \(\mathfrak h^*\). An inductive argument shows that the positive roots can be written as \(\mathbb Z_{\ge0}\)-linear combinations of the simple roots.

We are now ready to turn to highest weight theory. The goal of highest weight theory is to describe the irreducible finite-dimensional representations of a complex semisimple Lie algebra. We start with some easier observations.

Lemma 29. Let \(\mathfrak g\) be a complex semisimple Lie algebra with Cartan subalgebra \(\mathfrak h\). Then any finite-dimensional irreducible representation \(V\) of \(\mathfrak g\) admits an \(\mathfrak h\)-eigenspace decomposition \[V=\bigoplus_{\lambda\in\mathfrak h^*}V[\lambda].\]

Proof. Let \(V’\) be the subspace of \(V\) on which \(\mathfrak h\) diagonalizes. Certainly \(V’\) is nonzero, so it is enough to show that \(V’\) is preserved by \(\mathfrak g\). Certainly it is preserved by \(\mathfrak h\), so by the root decomposition, it is enough to be preserved by the root spaces \(\mathfrak g_\alpha\). Well, any \(e_\alpha\in\mathfrak g_\alpha\) has \([h,e_\alpha]=\alpha(h)e_\alpha\) by definition of \(\mathfrak g_\alpha\), so \(e_\alpha\) maps \(V[\lambda]\) to \(V[\lambda+\alpha]\). \(\blacksquare\)

Thus, it will be interesting to understand the eigenvalues appearing in a representation. Traditionally, these are called weights. It will turn out that the weights are controlled by the largest one.

Definition 30. Let \(\mathfrak g\) be a complex semisimple Lie algebra with Cartan subalgebra \(\mathfrak h\), and let \(\Phi\) be the polarized root system. For a representation \(V\) of \(\mathfrak g\), a vector \(v\in V\) is singular if and only if it is an eigenvector for \(\mathfrak h\), and \(\mathfrak g_\alpha v=0\) for all \(\alpha\in\Phi^+\).

Lemma 31. Let \(\mathfrak g\) be a complex semisimple Lie algebra with Cartan subalgebra \(\mathfrak h\), and let \(\Phi\) be the polarized root system. Then any finite-dimensional irreducible representation \(V\) of \(\mathfrak g\) admits a nonzero singular vector.

Proof. Let \(\{\alpha_1,\ldots,\alpha_r\}\) be the simple roots of the polarized root system. Choose \(\lambda\in\mathfrak h^*\) to be the eigenvalue of \(V\) maximizing the sum of coefficients with respect to the basis of simple roots. As in the proof of Lemma 29, we see that \([\mathfrak g_\alpha,V[\lambda]]\subseteq V[\lambda+\alpha]\), but \(V[\lambda+\alpha]=0\) for any positive root \(\alpha\) by construction of \(\lambda\). Thus, any nonzero vector in \(V[\lambda]\) will suffice. \(\blacksquare\)

Approximately speaking, we can hope to understand the highest weight from a singular vector, and then the rest of the vector space is generated from that vector.

The remainder of highest weight theory relies on the technique of “reduction to \(\mathfrak{sl}_2(\mathbb C)\).” Of course, we are not going to give a complete account of the theory, but we can say something concrete about \(\mathfrak{sl}_2(\mathbb C)\).

Proposition 32. The finite-dimensional irreducible representations of \(\mathfrak{sl}_2(\mathbb C)\) are the symmetric powers of the standard representation \(\mathbb C^2\).

Proof. We will show that every finite-dimensional irreducible representation \(V\) is a symmetric power of \(\mathbb C^2\). Define \[(e,f,h):=\left\{\begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix},\begin{bmatrix}0 & 0 \\ 1 & 0\end{bmatrix},\begin{bmatrix}1 & 0 \\ 0 & -1\end{bmatrix}\right\}.\] For example, \(h\) spans the diagonal matrices in \(\mathfrak{sl}_2(\mathbb C)\), which is a Cartan subalgebra. By the earlier lemmas, we know that \(V\) admits a weight decomposition into eigenspaces of \(\mathfrak h\), and \(V\) admits a singular vector \(v_\lambda\) of some eigenvalue \(\lambda\), which we think of as the eigenvalue of \(h\).

Now, using that \(ev_\lambda=0\), one calculates that for each nonnegative integer \(m\), we have \[e^mf^mv_\lambda=m!\lambda(\lambda-1)\cdots(\lambda-(m-1))v_\lambda.\] In particular, \(f^m\in V[\lambda-2m]\) is a nonzero vector unless \(\lambda\) is a nonnegative integer less than \(m\). Because \(V\) is finite-dimensional, we conclude that \(\lambda\) must be a nonnegative integer. By comparing eigenvalues and eigenspaces of \(h\), one finds that \(V\) is the \(\lambda\)th symmetric power of \(\mathbb C^2\). \(\blacksquare\)

What is remarkable is the integrality of the eigenvalue.

Definition 33. Let \(\mathfrak g\) be a complex semisimple Lie algebra with Cartan subalgebra \(\mathfrak h\), and let \(\Phi\) be the polarized root system. A weight \(\lambda\in\mathfrak h^*\) is integral if and only if \(\alpha^\lor(\lambda)\in\mathbb Z\) for all coroots \(\alpha^\lor\). It is dominant if and only if \(\alpha^\lor(\lambda)\ge0\) for all positive roots \(\alpha\).

Lemma 34. Let \(\mathfrak g\) be a complex semisimple Lie algebra with Cartan subalgebra \(\mathfrak h\), and let \(\Phi\) be the polarized root system. Let \(V\) be a finite-dimensional irreducible representation of \(\mathfrak g\). Then all eigenvalues of \(V\) are integral.

Proof. We use “reduction to \(\mathfrak{sl}_2(\mathbb C)\).” For each positive root \(\alpha\), define the subspace \[\mathfrak{sl}_2(\mathbb C)_\alpha:=\mathfrak g_\alpha\oplus\mathfrak g_{-\alpha}\oplus\mathbb C\alpha^\lor.\] This subspace turns out to be isomorphic to \(\mathfrak{sl}_2(\mathbb C)\) (but we will not argue this). Now, view \(V\) as a representation of \(\mathfrak{sl}_2(\mathbb C)_\alpha\). Then all the eigenvalues with respect to \(\alpha^\lor\) must be integral by Proposition 32, so the result follows. \(\blacksquare\)

Lemma 35. Let \(\mathfrak g\) be a complex semisimple Lie algebra with Cartan subalgebra \(\mathfrak h\), and let \(\Phi\) be the polarized root system. Let \(V\) be a finite-dimensional irreducible representation of \(\mathfrak g\), and let \(v_\lambda\in V[\lambda]\) be a nonzero singular vector. Then \(\lambda\) is dominant.

Proof. Again, we reduce to \(\mathfrak{sl}_2(\mathbb C)_\alpha\): the subrepresentation generated by \(v_\lambda\) must be finite-dimensional, so it follows that the eigenvalue \(\alpha^\lor(\lambda)\) must be a nonnegative integer by the argument of Proposition 32. \(\blacksquare\)

Theorem 36 (Highest weight). Let \(\mathfrak g\) be a complex semisimple Lie algebra with Cartan subalgebra \(\mathfrak h\), and let \(\Phi\) be the polarized root system. For each dominant integral weight \(\lambda\), there is a unique irreducible finite-dimensional representation \(L_\lambda\) with a nonzero singular vector of weight \(\lambda\).

Sketch. We sketch the construction of \(L_\lambda\), but we will not run any check on it. We start with the “universal” module with a nonzero singular vector \(v_\lambda\) of weight \(\lambda\), which is the \(\mathfrak g\)-module \(M_\lambda\) generated by \(v_\lambda\) with the relations \[\begin{cases}hv_\lambda=\lambda(h)v_\lambda & \text{for all }h\in\mathfrak h, \\\mathfrak g_\alpha v_\lambda=0 & \text{for all }\alpha\in\Phi^+.\end{cases}\] In particular, one finds that \(M_\lambda\) admits an eigenspace decomposition, and its eigenvalues are of the form \(\lambda-\mu\), where \(\mu\) is a \(\mathbb Z\)-linear combination of positive roots. It turns out that \(M_\lambda\) admits a maximal nonzero quotient \(L_\lambda\) (given by taking the quotient by the sum of all nonzero submodules), and \(L_\lambda\) is finite-dimensional when \(\lambda\) is dominant integral.

This completes the construction. Let’s at least sketch the most difficult check on \(L_\lambda\), which is that it is finite-dimensional. If \(\lambda\) is integral dominant, then \(L_\lambda\) features the relation \[f_\alpha^{\lambda(\alpha^\lor)}v_\lambda=0\] for all \(f_\alpha\in\mathfrak g_{-\alpha}\), where \(\alpha\) is a positive root. (Roughly speaking, one shows this by reduction to \(\mathfrak{sl}_2(\mathbb C)_\alpha\).) It now turns out that the quotient of \(M_\lambda\) by these relations is finite-dimensional and isomorphic to \(L_\lambda\). \(\blacksquare\)

Corollary 37. Let \(\mathfrak g\) be a complex semisimple Lie algebra with Cartan subalgebra \(\mathfrak h\), and let \(\Phi\) be the polarized root system. Fix a finite-dimensional representation \(V\) of \(\mathfrak g\). For an eigenvalue \(\mu\) of \(V\), one has \[L_{\mu+\lambda}\subseteq L_\lambda\otimes V\] for \(\lambda\) chosen so that \(\lambda(\alpha^\lor)\) is large enough for all positive coroots \(\alpha^\lor\).

Proof. By irreducibility, it is enough to give a nonzero map \(L_{\mu+\lambda}\to L_\lambda\otimes V\). Because \(L_{\mu+\lambda}\) is the unique nonzero finite-dimensional quotient of \(M_{\mu+\lambda}\), it is equivalent to provide a nonzero map \(M_{\mu+\lambda}\to L_\lambda\otimes V\). By construction of \(M_{\mu+\lambda}\) as being generated by a nonzero singular vector of weight \(\mu+\lambda\), it is enough to provide a map \((\mu+\lambda)\to L_\lambda\otimes V\), where \((\mu+\lambda)\) is being viewed as a character on \(\mathfrak b_+:=\mathfrak h\oplus\bigoplus_{\alpha>0}\mathfrak g_\alpha\). It is then equivalent to provide a map \(L_\lambda^*(\mu+\lambda)\to V\).

Now, by applying a dual construction as \(M_\lambda\), one finds that \(L_\lambda^*\) is the maximal nonzero quotient of the module \(\overline M_{-\lambda}\) generated by a “lowest weight vector” \(v_{-\lambda}\) of weight \(-\lambda\), given the conditions \(\mathfrak g_{-\alpha} v_{-\lambda}=0\) for all \(\alpha\in\Phi^+\). Thus by rearranging, one sees that \(\overline M_{-\lambda}\) is generated by \(v_{-\lambda}\) as a representation of \(\mathfrak b\). Furthermore, by the proof of Theorem 36, we see that \(L_\lambda^*\) is cut out from \(\overline M_{-\lambda}\) by the relations \[e_\alpha^{\lambda(\alpha^\lor)}v_{-\lambda}=0\] for all \(e_\alpha\in\mathfrak g_\alpha\), where \(\alpha\in\Phi^+\).

Thus, providing a \(\mathfrak b\)-invariant map \(L_\lambda^*(\mu+\lambda)\) is equivalent to providing a candidate vector \(v\in V\) of weight \(-\lambda+\mu+\lambda\) which also satisfies the condition \(e_\alpha^{\lambda(\alpha^\lor)}v=0\) as in the previous paragraph. Accordingly, just choose any nonzero vector \(v\in V[\mu]\), and then take \(\lambda\) large enough to satisfy the second condition. \(\blacksquare\)

Remark 38. This argument actually shows that \[\dim\operatorname{Hom}_{\mathfrak g}(L_{\mu+\lambda},L_\lambda\otimes V)=\dim V[\mu]\] for \(\lambda\) “sufficiently large” with respect to \(V\) (and \(\mu\)).

The Central Character

We now return to our story. Let \(G\) be a compact Lie group with trivial center, and we want to show that \(\pi_1(G)\) is finite. This basically amounts to bounding the degree of the finite covers of \(G\), which by Proposition 12 amounts to controlling their centers. One expects to be able to control groups by their representations, so we will content ourselves with understanding how the center of these finite covers act on representations.

Proposition 39 (Schur). Let \(\varphi\colon V\to V\) be an endomorphism of an irreducible representation of a Lie group \(G\). Then \(\varphi\) is a scalar.

Proof. As a linear operator, \(\varphi\) admits an eigenvalue \(\lambda\in\mathbb C\). Thus, \(\varphi-\lambda\operatorname{id}_V\) has a kernel, which is a nontrivial subrepresentation of \(V\), so it must be all of \(V\), so \(\varphi=\lambda\operatorname{id}_V\). \(\blacksquare\)

Thus, to each element \(z\) in the center and each irreducible representation \(V\), we receive a scalar given by the action of \(z\) on \(V\). Let’s organize these scalars. It is convenient to pass to the simply connected cover of \(G\), which we do not yet know is compact.

Definition 40. Fix a simply connected Lie group \(G\) with semisimple Lie algebra \(\mathfrak g_{\mathbb C}\) with Cartan subalgebra \(\mathfrak h\). For each dominant integral weight \(\lambda\in\mathfrak h^*\), define \(\chi_\lambda\colon Z(G)\to\mathbb C^\times\) to be the character by which \(Z(G)\) acts on \(L_\lambda\).

Remark 41. For this definition to be well-defined, we need to know that the representation \(\mathfrak g\to\mathfrak{gl}(L_\lambda)\) lifts to \(G\), which holds by Theorem 18.

Notation 42. Fix a complex semisimple Lie algebra \(\mathfrak g\) with Cartan subalgebra \(\mathfrak h\). Let \(P\subseteq\mathfrak h^*\) be the weight lattice of integral weights, and let \(P^+\subseteq P\) be the subset of dominant integral weights. Lastly, let \(Q\subseteq\mathfrak h^*\) be the root lattice generated by \(\Phi\).

Remark 43. Theorem 24 implies that \(Q\subseteq P\) and that \(Q\) spans \(\mathfrak h^*\). For example, this implies that \([P:Q]<\infty\).

Lemma 44. Fix a simply connected Lie group \(G\) with semisimple Lie algebra \(\mathfrak g_{\mathbb C}\) with Cartan subalgebra \(\mathfrak h\). Then \(\chi_\bullet\colon P^+\to\operatorname{Hom}(Z(G),\mathbb C^\times)\) extends uniquely to a homomorphism on \(P\).

Proof. Because any element of \(P\) can be written as a difference of dominant weights, the extension is certainly unique. To be well-defined, it is enough (after some rearrangement) to check that \(\chi_\lambda\chi_\mu=\chi_{\lambda+\mu}\) for any dominant integral \(\lambda\) and \(\mu\).

Well, note that \(L_\lambda\otimes L_\mu\) has a singular vector of weight \(\lambda+\mu\). Thus, by the proof of Theorem 36, it receives a map from \(M_{\lambda+\mu}\), which because the image is finite-dimensional descends to \(L_{\lambda+\mu}\). Thus, \(L_{\lambda+\mu}\subseteq L_\lambda\otimes L_\mu\)! Considering the scalar action of \(Z(G)\) on these representations completes the proof. \(\blacksquare\)

Lemma 45. Let \(G\) be a compact Lie group with trivial center, and let \(G_{\mathrm{sc}}\) be its universal cover. Then \(\chi_\bullet\colon P\to\operatorname{Hom}(Z(G_{\mathrm{sc}}),\mathbb C^\times)\) contains \(Q\) in its kernel.

Proof. We use Corollary 37. Fix notation as in Lemma 44. Fix a root \(\alpha\), and we want to show that \(\chi_\alpha=1\). It is enough to show that \(\chi_{\lambda+\alpha}=\chi_\lambda\) for any \(\lambda\in P\). Well, let’s use Corollary 37 with \(V\) equal to the adjoint representation \(\mathfrak g_{\mathbb C}\). Then \(\mathfrak g[\alpha]\ne0\), so for \(\lambda\) “large enough,” we know that \[L_{\alpha+\lambda}\subseteq L_\lambda\otimes\mathfrak g_{\mathbb C}.\] Now, \(\mathfrak g\) is already a representation of \(G\), so \(Z(G_{\mathrm{sc}})\) acts trivially on \(\mathfrak g_{\mathbb C}\). (Namely, \(Z(G_{\mathrm{sc}})\) is contained in the kernel of \(G_{\mathrm{sc}}\to G\) because \(G\) has trivial center.) Thus, examining the action of \(Z(G)_{\mathrm{sc}}\) on these representations yields \(\chi_{\alpha+\lambda}=\chi_\lambda\), as required. \(\blacksquare\)

Thus, we have produced a map \[P/Q\to\operatorname{Hom}(Z(G_{\mathrm{sc}}),\mathbb C^\times).\] This map will turn out to be an isomorphism, thereby giving the needed finiteness.

The Peter–Weyl Theorem

As a second intermission, we say something about the representation theory of compact groups; we will be rather sketchy. Once again, the hurried reader is encouraged to skip this section except for reading the statements of Corollaries 53 and 55.

Recall that a choice of basis of \(\operatorname{Lie}G\) parallelizes a compact Lie group by giving it a frame of left-invariant vector fields. Taking the top exterior power then produces a left-invariant top-degree differential form, which by standard results in differential geometry, provides a left-invariant measure on \(G\). Because \(\land^{\dim G}\Omega_G\) is a line bundle, this left-invariant differential form is unique up to scalar.

For sanity, we note the following.

Lemma 46. Let \(G\) be a connected compact Lie group. Then any top-degree left-invariant differential form \(\omega\) on \(G\) is also right-invariant.

Proof. For brevity, set \(\mathfrak g:=\operatorname{Lie}G\) and \(n:=\dim G\). Because \(\omega\) is left-invariant already, being right-invariant is equivalent to being invariant under conjugation. Thus, we would like to show that \(\operatorname{Ad}_g\omega=\omega\) for all \(g\in G\). It is equivalent to ask for \(\operatorname{Ad}_g\omega_1\) to equal \(\omega_1\) as elements of \(T_1G\), which is \(\land^{\dim G}\mathfrak g\). Thus, we need to study the character \[\det\operatorname{Ad}_\bullet\colon G\to\operatorname{GL}(\mathbb R).\] Because \(G\) is compact and connected, the image is a compact and connected subgroup of \(\mathbb R^\times\), but the only possible such subgroup is \(\{1\}\), so this character is trivial! \(\blacksquare\)

Remark 47. It is still true that the left-invariant measure on \(G\) is right-invariant without the connectivity assumption. The moral is that the measure only depends on the global section \(\left|\omega\right|\) instead of the actual differential form \(\omega\).

Having access to a measure allows us to consider function spaces like \(L^2(G)\).

Notation 48. Let \(G\) be a compact Lie group. Then define \(C(G)^{\mathrm{fin}}\) to be the image of the span of all matrix coefficients of all finite-dimensional continuous representations of \(G\) in \(C(G)\). Here, a matrix coefficient of a representation \(V\) is a function of the form \(g\mapsto\ell(gv)\) for some \(v\in V\) and \(\ell\in V^*\).

Remark 49. It is not too hard to show that \(C(G)^{\mathrm{fin}}\) equals the set of all vectors \(f\) for which \[\dim\operatorname{span}\{Rg(f):g\in G\}<\infty,\] where the action \(R\colon G\to C(G)\) is given by right translation. (Such vectors in a representation are called \(G\)-finite.) The same applies for left translation. This explains the notation \(^{\mathrm{fin}}\).

Lemma 50. Let \(G\) be a compact Lie group. Then the matrix coefficient map defines a \((G\times G)\)-invariant isomorphism \[\bigoplus_{V\in\operatorname{IrRep}G}(V\otimes V^*)\to C(G)^{\mathrm{fin}}.\] The sum on the left-hand side varies over finite-dimensional irreducible representations of \(G\).

Proof. Representations of \(G\) are completely reducible by Weyl’s unitary trick, so the natural inclusion \[\bigoplus_{V\in\operatorname{IrRep}G}\left(V\otimes\operatorname{Hom}_G(V,C(G)^{\mathrm{fin}}\right)\to C(G)^{\mathrm{fin}}\] is an isomorphism; here, \(G\) is acting on \(C(G)\) by right translation. Thus, it remains to compute these multiplicity spaces. It is equivalent to compute the multiplicity spaces \(\operatorname{Hom}_G(V,C(G))\). Well, there is certainly a map \(V^*\to\operatorname{Hom}_G(V,C(G))\) given by sending \(\ell\in V^*\) to the morphism \(v\mapsto\ell(gv)\). Conversely, there is a map \(\operatorname{Hom}_G(V,C(G))\to V^*\) given by evaluating at \(1\in G\). One can check that these are inverse maps and that the resulting isomorphism unwinds into the lemma. \(\blacksquare\)

Theorem 51 (Peter–Weyl). Let \(G\) be a compact Lie group. Then the matrix coefficient map provides a \((G\times G)\)-invariant isomorphism \[\widehat{\bigoplus_{V\in\operatorname{IrRep}G}}(V\otimes V^*)\to L^2(G).\] Here, the left-hand side is a Hilbert space direct sum.

Sketch. The matrix coefficient map sends \(v\otimes\ell\in V\otimes V^*\) to the function \(g\mapsto\ell(gv)\). The usual orthogonality of matrix coefficients arguments show that the various maps \(V\otimes V^*\to L^2(G)\) are isometric embeddings, and all of their images are isomorphic.

Thus, the main issue is showing that \(C(G)^{\mathrm{fin}}\) is a dense subspace. This requires a trick. For continuous and compactly supported test functions \(\varphi\) on \(G\), the convolution operator \(R\varphi\) on \(L^2(G)\) defined by \[R\varphi(f)(x):=\int_G\varphi(g)f(xg)\,dg\] is Hilbert–Schmidt. Choosing \(\varphi\) from a suitable approximate identity allows us to use the Spectral theorem for Hilbert–Schmidt operators to diagonalize a dense subspace of \(L^2(G)\) into finite-dimensional \(G\)-invariant subspaces. \(\blacksquare\)

Corollary 52. Let \(G\) be a compact Lie group. Then \(G\) admits a finite-dimensional faithful representation.

Proof. The idea is to filter \(L^2(G)\) using the Peter–Weyl theorem. Enumerate the irreducible representations of \(G\) by \(\{V_i\}_i\), and define \(W_n\) to be \(V_1\oplus\cdots\oplus V_n\) for each \(n\). Then \[\ker W_1\supseteq\ker W_2\supseteq\ker W_3\supseteq\cdots.\] This is a descending sequence of closed Lie subgroups of \(G\), so eventually the dimension must stabilize, which causes the connected component of the identity to stabilize. But there are only finitely many connected components by compactness, so eventually the subgroups must actually stabilize. Let \(K\) be the stabilization; say \(K=\ker W_N\). Then \(K\) acts trivially on \(L^2(G)\) by Theorem 52, so \(K\) is trivial. Thus, \(W_N\) is the desired faithful representation. \(\blacksquare\)

Corollary 53. Let \(G\) be a compact Lie group. Then \(C(G)^{\mathrm{fin}}\) is dense in \(C(G)\), where \(C(G)\) is equipped with the supremum norm.

Proof. We use the Stone–Weierstrass theorem. Note that \(C(G)^{\mathrm{fin}}\) is a subalgebra under convolution (using Remark 49), preserved by complex conjugation (by taking matrix coefficients of conjugates of representations), and it separates points on \(G\) (by Corollary 29), so the result follows. \(\blacksquare\)

Remark 54. One can more or less reverse the logic of Corollary 53 to show that any subalgebra \(A\subseteq C(G)^{\mathrm{fin}}\) which is preserved by complex conjugation and separates points must equal \(C(G)^{\mathrm{fin}}\). Indeed, \(A\) admits only \(G\)-finite vectors, so we just have to compute the multiplicity spaces \(\operatorname{Hom}_G(V,A)\) where \(V\) is irreducible. But \(A\) is dense in \(C(G)\) by the Stone–Weierstrass theorem, so \[\operatorname{Hom}_G(V,A)\subseteq\operatorname{Hom}_G(V,C(G))=\operatorname{Hom}_G\left(V,C(G)^{\mathrm{fin}}\right)\] is a dense subspace, which must then be an equality!

Corollary 55. Let \(G\) be a compact Lie group, and let \(V\) be a finite-dimensional faithful representation. Then \(V\) and \(V^*\) \(\otimes\)-generate the category of finite-dimensional representations of \(G\).

Proof. Now, any finite-dimensional representation can be found inside \(C(G)^{\mathrm{fin}}\), so it is enough to show that the matrix coefficients of \(V\) generate \(C(G)^{\mathrm{fin}}\). (Convolutions of matrix coefficients can be found inside tensor powers.) Well, let \(A\) be the generated subalgebra of \(C(G)^{\mathrm{fin}}\), and we will use Remark 54. Certainly \(A\) separates points because \(V\) is faithful.

It thus remains to check that \(A\) is preserved by complex conjugation, for which we need a trick. It is enough to show that the complex conjugates of the matrix coefficients of \(V\) can be found in \(V^*\). Let \(\rho_V\colon G\to\operatorname{GL}(V)\) be the structure morphism. By Weyl’s unitary trick (as in Proposition 14), we may assume that the \(G\)-action on \(V\) is unitary. But then for each \(g\in G\), \[\rho_V(g)^\dagger=\rho_V(g)^{-1}.\] But now the action of \(g\in G\) on \(V^*\) is given by the action of \(g^{-1}\in G\) on \(V\), so we are done! \(\blacksquare\)

Completion of the Proof

Let’s return to the earlier setting, where \(G\) is a compact group with trivial center. As promised, let’s show that our central character homomorphism \(\chi_\bullet\) is an isomorphism.

Lemma 56. Let \(G\) be a compact Lie group with trivial center, and let \(G_{\mathrm{sc}}\) be its universal cover. Then \(\chi_\bullet\colon P/Q\to\operatorname{Hom}(Z(G_{\mathrm{sc}}),\mathbb C^\times)\) is injective.

Proof. This is an application of Corollary 55. By adding large weights in \(Q\) to push into \(P^+\), it is enough to show that if \(\lambda\in P^+\) satisfies \(\chi_\lambda=1\), then \(\lambda\in Q\).

The key point is that if \(\chi_\lambda=1\), then the representation \(G_{\mathrm{sc}}\to\operatorname{GL}(L_\lambda)\) descends to \(G_{\mathrm{sc}}/Z(G_{\mathrm{sc}})\). Observe that this latter quotient is \(G\): on one hand, because \(G\) has trivial center, the kernel of \(G_{\mathrm{sc}}\to G\) contains \(Z(G_{\mathrm{sc}})\). On the other hand, Proposition 12 implies that the kernel of \(G_{\mathrm{sc}}\to G\) is contained in \(Z(G_{\mathrm{sc}})\).

We now complete the proof. We know that \(L_\lambda\) descends to a representation of \(G\). Because \(G\) has trivial center, the adjoint representation \(\mathfrak g_{\mathbb C}\) of \(G\) is faithful by the proof of Lemma 5. Thus, by Corollary 55 (!), \(\mathfrak g_{\mathbb C}\) and its dual—which happens to be itself, though we don’t need to know this—generate the category of finite-dimensional representations of \(G\). In particular, \(L_\lambda\) can be found in some \(\otimes\)-power of \(\mathfrak g_{\mathbb C}\) (and its dual). Thus, the weights of \(L_\lambda\) can be found in the \(\mathbb Z\)-span of the weights of \(\mathfrak g_{\mathbb C}\), which is \(Q\). \(\blacksquare\)

Lemma 57. Let \(G\) be a compact Lie group with finite center. For any character \(\chi\colon Z(G)\to\mathbb C^\times\), there is a finite-dimensional irreducible representation \(V\) for which \(\chi\) is the central character of \(\chi_V\) acting on \(V\).

Proof. This is an application of Corollary 53. Indeed, by Corollary 53, for any \(\varepsilon>0\), we can find matrix coefficients \(\{g\mapsto\ell_i(gv_i)\}\) of irreducible representation \(\{V_i\}\) and coefficients \(\{c_i\}\) so that \[\left|\chi(g)-\sum_ic_i\ell_i(gv_i)\right|<\varepsilon\] for all \(g\in Z(G)\). (Note that \(Z(G)\) is finite and hence discrete.) By adjusting the coefficients, this is actually saying that \[\left|\chi(g)-\sum_ic_i\chi_i(g)\right|<\varepsilon\] for all \(g\in Z(G)\), where \(\chi_i\) is the central character of \(V_i\).

Now, recall that the characters of \(Z(G)\) form a basis for the space \(\operatorname{Mor}(Z(G),\mathbb C)\). The approximation in the previous paragraph means that the central characters of irreducible representations of \(G\) span a dense subspace of \(\operatorname{Mor}(Z(G),\mathbb C)\), which must just be the full space because it is finite-dimensional. Thus, the central characters must exhaust all characters: otherwise, they couldn’t span! \(\blacksquare\)

Proposition 58. Let \(G\) be a compact Lie group with trivial center, and let \(G_{\mathrm{sc}}\) be its universal cover. Then \(\chi_\bullet\colon P/Q\to\operatorname{Hom}(Z(G_{\mathrm{sc}}),\mathbb C^\times)\) is an isomorphism.

Proof. Injectivity follows from Lemma 56, so it only remins to achieve surjectivity, which is the role of Lemma 57. It is enough to show that \(Z(G_{\mathrm{sc}})\) has size bounded by the size of \(P/Q\).

To use Lemma 57, we need to pass to compact groups, but \(G_{\mathrm{sc}}\) is not yet known to be compact, so we require a trick. Because \(Z(G_{\mathrm{sc}})\) is already a finitely generated abelian group, it is enough to show that any finite-index subgroup has index at most \([P:Q]\). Well, let \(Z\subseteq Z(G_{\mathrm{sc}})\) be a finite-index subgroup, and then \(G’=G_{\mathrm{sc}}/Z\) is a finite cover of \(G\) and hence compact.

We now apply Lemma 57, which tells us that any character \(\chi\) of \(Z(G’)\) arises from an irreducible representation of \(G’\). Such a representation differentiates to a representation of \(\mathfrak g_{\mathbb C}\), where it continues to be irreducible: any factorization over \(\mathfrak g_{\mathbb C}\) amounts to the data of a nonzero non-identity idempotent which is \(\mathfrak g_{\mathbb C}\)-invariant and hence \(G’\)-invariant by exponentiation. Thus, we may write \(\chi=\chi_\lambda\) for some \(\lambda\in P^+\). We conclude that \[\chi_\bullet\colon P/Q\to\operatorname{Hom}(Z(G_{\mathrm{sc}})/Z,\mathbb C^\times)\] is surjective, so the proposition follows. \(\blacksquare\)

Corollary 59. Let \(G\) be a compact Lie group with trivial center. Then \[\pi_1(G)\cong\operatorname{Hom}(P/Q,\mathbb C^\times).\] Here, \(P\) and \(Q\) are the weight and root lattices of the Lie algebra \(\mathfrak g_{\mathbb C}\), which is semisimple.

Proof. By Proposition 12, \(\pi_1(G)\) is isomorphic to the kernel of the projection \(G_{\mathrm{sc}}\to G\) from the universal cover, which is \(Z(G_{\mathrm{sc}})\) by the proof of Lemma 56. The dual of \(Z(G_{\mathrm{sc}})\) is isomorphic to \(P/Q\) by Proposition 58, so we are done upon undoing the dual. \(\blacksquare\)

Proof of Theorem 1. By Lemma 20, we reduce to the case where \(G\) has trivial center. This follows from Corollary 59 and Remark 43. \(\blacksquare\)