# Why are all compact Lie groups matrix Lie groups?

I've seen it said that this is a consequence of the Peter-Weyl theorem (here), but I don't know how to do this and have been unsuccessful in finding a reference.

• Hint: the compact group $G$ acts faithfully on $L^2(G)$ and this action is completely reducible with finite dimensional irreducible factors. Jun 14, 2020 at 6:00
• Right, so if we look at the kernel over all finite-dimensional representations, this is trivial. But I don't see how that implies that there is a finite-dimensional representation with trivial kernel, or why it should be smooth. In particular, $\Pi_{S} \mathbb Z_2$ is compact for any indexing set $S$, but taking $S$ to be larger than the cardinality of every $GL_n(\mathbb C)$, we have a compact group with no faithful finite dim representations. Jun 14, 2020 at 6:12
• You can argue as follows. Start with some irrep $V_1$. The action of $G$ has some kernel $K_1$. Since the action of $G$ on $L^2(G)$ is faithful, pick a nonzero element $k_1 \in K_1$ and it will act nontrivially on some other irrep $V_2$. The action of $G$ on $V_1 \oplus V_2$ has some kernel $K_2$ strictly contained in $K_1$ (since it doesn't contain $k_1$). Keep going like this... Jun 14, 2020 at 6:35
• @AshwinTrisal also every continuous homomorphism between Lie groups is smooth: math.stackexchange.com/questions/2858695/… Jun 14, 2020 at 8:40

Lemma. Let $$G$$ be a compact Lie group and $$... be a chain of proper compact (necessarily Lie) subgroups. Then this sequence is necessarily finite.
Proof. Dimension in this sequence can drop only finitely many times, hence, WLOG, for each $$i$$, $$G_i< G_{i-1}$$ is a codimension 0 subgroup; hence $$G_i$$ is open in $$G_{i-1}$$; hence (by compactness) is of finite index in $$G_{i-1}$$. However, the index $$G_n (if finite) is at most the number of connected components of $$G$$, which has to be finite by compactness of $$G$$ (and since $$G$$ is a manifold!). Thus, the sequence eventually terminates. qed
Let $$\lambda: G\to L^2(G)$$ be the left-regular representation; it is a faithful representation. By the P-W theorem, this representation splits as a direct sum of irreducible finite-dimensional factors $$V_\alpha, \alpha\in A$$. Take $$\beta_1\in A$$ such that $$\lambda_1: G\to GL(V_{\beta_1})$$ (the projection of $$\lambda$$) is a nontrivial representation and let $$G_1 be the kernel of $$\lambda_1$$. Clearly, $$G_1$$ is a closed subgroup of $$G$$. If $$G_1=\{1\}$$, we are done. Otherwise, there exists $$\beta_2\in B$$ such that $$G\to GL(V_{\beta_2})$$ is nontrivial on $$G_1$$. Let $$G_2< G_1$$ denote the kernel of $$G\to GL(V_{\beta_1}\oplus V_{\beta_2}).$$
Continue inductively. According to Lemma, this sequence of subgroups of $$G$$ eventually terminates and we obtain a faithful representation $$G\to GL(\oplus_{i=1}^n V_{\beta_i}).$$ As for smoothness of this representation, see this link given by freakish.
Remark. I proved the lemma using the theorem that closed subgroups of a Lie group are Lie subgroups. In fact, one can avoid appealing to this theorem and to the lemma in full generality: In our setting, each $$G_i< G_{i-1}$$ is the kernel of a continuous matrix representation of $$G_{i-1}$$ where we can assume (inductively) that $$G_{i-1}$$ is a compact Lie group. There is an elementary argument (given by José Carlos Santos in his answer here) that such a representation is necessarily smooth, hence, its kernel is a Lie subgroup.