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.
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asked Jun 14, 2020 at 5:37
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1$\begingroup$ Hint: the compact group $G$ acts faithfully on $L^2(G)$ and this action is completely reducible with finite dimensional irreducible factors. $\endgroup$– Moishe KohanJun 14, 2020 at 6:00
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1$\begingroup$ 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. $\endgroup$– Ashwin TrisalJun 14, 2020 at 6:12
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2$\begingroup$ 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... $\endgroup$– Qiaochu YuanJun 14, 2020 at 6:35
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1$\begingroup$ @AshwinTrisal also every continuous homomorphism between Lie groups is smooth: math.stackexchange.com/questions/2858695/… $\endgroup$– freakishJun 14, 2020 at 8:40
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1 Answer
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The key difference between compact Lie groups and general compact (Hausdorff) topological groups is a Noetherian property:
Lemma. Let $G$ be a compact Lie group and $$ ... <G_2 < G_1< G=G_0 $$ 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 $G_n<G$ (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
With this in mind:
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 $G_1<G$ 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.
answered Jun 14, 2020 at 14:29