Let $G$ be a compact group. I learned the version of the Peter-Weyl theorem which says: the matrix coefficients of $G$ are dense in $L^2(G)$. Call this Peter-Weyl I.

Apparently there is another version which states: For any $g \in G$ there exists a finite dimensional unitary representation $(\pi,V)$ such that $\pi(g) \neq I$ (identity). Call this Peter-Weyl II.

Can one prove Peter-Weyl II using Peter-Weyl I? A short slick proof is what I'm looking for, of course.

  • $\begingroup$ Do you mean compact Lie group? $\endgroup$ – user99914 Oct 14 '15 at 4:04
  • $\begingroup$ Sure. I don't think it matters to much at the end of the day. $\endgroup$ – nigel Oct 14 '15 at 4:07
  • $\begingroup$ @JohnMa I just want the group to satisfy whatever conditions the Peter-Weyl theorems require. $\endgroup$ – nigel Oct 14 '15 at 4:19

Let $H\subseteq G$ be the intersection of the kernels of all the finite-dimensional unitary representations of $G$; we wish to show $H=\{1\}$. Let $q:G\to G/H$ be the quotient map. For any integrable function $f:G/H\to\mathbb{C}$, $f\circ q$ is integrable and we have $\int_G f\circ q=\int_{G/H} f$, where both integrals are with respect to the Haar measures (this can be proven for $f$ continuous by uniqueness of the Haar measure on $G/H$, and then extends to all of $L^1(G/H)$ since continuous functions are dense). In particular, composition with $q$ defines an isometry $q^*:L^2(G/H)\to L^2(G)$. Furthermore, every matrix coefficient of $G$ is in the image of $q^*$, because if $f$ is a matrix coefficient, then $f(g)=f(hg)$ for all $h\in H$, $g\in G$ by definition of $H$.

Let us now assume Peter-Weyl I. The image of $q^*$ is a closed subspace containing every matrix coefficient, so it must be all of $L^2(G)$. But if $H$ is nontrivial, there is a continuous function $f$ on $G$ which is not constant on $H$, and it is easy to see that such a function cannot be in the image of $q^*$. Thus $H$ must be trivial, proving Peter-Weyl II.


This answer is actually using a stronger version of Peter Weyl I, which state that the matrix coefficients are even dense in $C(G)$, the space of continuous functions. (So that might not answer your question)

First we show that $G$ must have a faithful representation.

To show that, let $g\in G\setminus \{e\}$ and $f$ be continuous on $G$ so that $f(g) \neq f(e)$. Then by Peter Weyl I, there is a matrix coefficient $u$ so that $u(g) \neq u(e)$. If this $u$ comes from the representation $\phi_1 : G \to GL(V)$, then we see that $g\notin \ker(\phi)$. Let $K_1$ be this kernel. $K_1$ is a closed subgroup of $G$ which is strictly smaller then $G$.

We are done if $K_1 = \{e\}$. If not, then we find $g_2\in K_1\setminus \{e\}$ and similarly construct a representation $\phi_2$ on $G$ so that the kernel $\tilde K_2$ has the property that $K_2:= \cap K_1\neq K_1$.

Then inductively we find a descending sequence of closed subgroups

$$G \supsetneqq K_1 \supsetneqq K_2 \supsetneqq K_3 \supsetneqq \cdots $$

But this chain must stability by dimension consideration and the fact that $G$ is compact. Thus $K_m = \{e\}$ for some $m$. Then the representation

$$\Phi = \phi_1 \oplus \cdots \oplus \phi_m$$

is injective: if $g\in G$ and $\Phi(g) = 0$, then $g\in K_i$ for all $i=1, 2\cdots, m$. In particular $g = e$.

Now the claimed results (Peter Weyl II) follows from the fact that every representation of a compact Lie group admits a $G$-invariant metric. You may check more in the book representation of compact Lie groups, where I learnt the above argument from this book.

  • $\begingroup$ I greatly appreciate that you typed this out for me. However, I was thinking there would be a shorter argument, and one which had a more analytic than Lie-Theoretic feel. $\endgroup$ – nigel Oct 14 '15 at 4:40
  • $\begingroup$ I don't think your second sentence is as easy as you make it sound. For instance, it is easy to see that for any point $g\in U(1)$, the set of continuous functions $f$ on $U(1)$ such that $f(g)=f(e)$ is actually dense in $L^2(U(1))$. In any case, once you have your second sentence, you're actually done, because you've proven that finite-dimensional representations of $G$ separate points. $\endgroup$ – Eric Wofsey Oct 14 '15 at 4:43
  • $\begingroup$ @EricWofsey : My bad, I was thinking about the $C^0$ version of the Peter Weyl I, I don't even notice that. $\endgroup$ – user99914 Oct 14 '15 at 4:47

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