TITLE: Is there a Peter-Weyl-Theorem over the real numbers?
QUESTION [8 upvotes]: I'd like to know: Is there a Peter-Weyl-like Theorem over the real numbers? If so, can you provide a reference?
With "over the real numbers" I mean that the representations considered act on vector spaces over the real numbers.
Best,
Leon

REPLY [2 votes]: I think that the Peter-Weyl theorem can be formulated for complex and real representations simultaneously using the ideas from the book by Bröcker and tom Dieck "Representations of compact Lie groups", Chapter III, Section 6, Exercise 4.
Let $\mathbb{K}$ be either $\mathbb{C}$ or $\mathbb{R}$ and let $\lambda$ be an irreducible representation of a compact topological group $G$ in a finite-dimensional $\mathbb{K}$-linear space $E$ with $G$-invariant inner product. Let $D(E)$ be the set of linear operators in $E$ that intertwine $\lambda$ with itself. If $\mathbb{K}=\mathbb{C}$, then $D(E)=\mathbb{C}$, while if $\mathbb{K}=\mathbb{R}$, then $D(E)$ may take three values, $\mathbb{R}$, $\mathbb{C}$, and $\mathbb{H}$, the skew field of quaternions.
Let $E^*$ be the linear space, dual to $E$, that is, the set of $\mathbb{K}$-linear functionals over $E$. It is a right $D(U)$-linear space via composition. On the other hand, $E$ is a left $D(U)$-linear space via evaluation. The tensor product $E^*\otimes_{D(U)}E$ is defined. A simple bookkeeping shows that the dimension of this $\mathbb{K}$-linear space is equal to $\frac{(\dim E)^2}{\dim D(E)}$. If $\mathbb{K}=\mathbb{C}$, then the denominator is equal to $1$, if $\mathbb{K}=\mathbb{R}$, then it may take three different values: $1$, $2$, and $4$.
The map that maps the tensor product $x^*\otimes y$ to the continuous function $f\colon G\to\mathbb{K}$, $f(g)=x^*(\lambda(g^{-1})x)$, can be extended by linearity to all of $E^*\otimes_{D(U)}E$. The image of this map is a finite-dimensional subspace of $C(G)$, call it $E_{\lambda}$. It is exactly the subspace grnerated by the matrix entries of the representation $\lambda$.
The Peter--Weyl theorem becomes:

If $\lambda_1$ and $\lambda_2$ are not equivalent, then the intersection of the spaces $E_{\lambda_1}$ and $E_{\lambda_2}$ is equal to $\{0\}$.

The set of finite $\mathbb{K}$-linear combinations of the elements of the spaces $E_{\lambda}$ is dense in both $C(G)$ and $L^2(G)$.

The space $E_{\lambda}$ carries the direct sum of $\frac{\dim E}{\dim D(E)}$ copies of the representation $\lambda$, an the Hilbert direct sum of all these representations is equivalent to the left regular representation of $G$ in $L^2(G)$.