It helped me to figure out a version of the Peter-Weyl Theorem for real representations (which has all the properties except that the multiplicity of the irreps can be smaller than their dimension).

We have used this to prove a “Wigner-Eckart Theorem” for equivariant convolution kernels in deep learning, which allows to parameterize these kernels for arbitrary compact groups and their homogeneous spaces:

https://arxiv.org/abs/2010.10952

Originally, the Wigner-Eckart Theorem is used to figure out the degrees of freedom of spherical tensor operators:

]]>I think you rather mean spectral subspaces, and the argument may be like this: suppose that the spectrum of is not a point, then it can be decomposed as a union of two non-empty Borel subsets. Consider the orthogonal projector onto the spectral subspace corresponding to one of them; it is a Borel function of , so it commutes with all operators of the representation as well as (because 1. polynomials in commute with the representation operators, 2. continuous functions can be approximated by polynomials, and 3. indicator functions can be approximated by continuous functions). Now this implies that the corresponding spectral subspace is -invariant, so the representation is reducible.

Or maybe I am missing something and there is a simpler proof?

*[“Eigenspace” changed to “invariant space” – T.]*

Ah, I see now where things are hiding. Lemma 2 is true in the real case as well, but when applied to equivariant maps of an irrep to itself only gives an orthogonal map on , not a constant multiple of the identity. In the complex case the only unitary equivariant maps on are multiples of the identity because of the spectral theorem (any non-trivial eigenspace would break irreducibility), but in the real case one can have non-trivial orthogonal equivariant maps (e.g. consider the tautological orthogonal rep of on , which is irreducible but has all the rotations as equivariant orthogonal maps).

]]>thank you for the correction!

In the real case, since we’re considering self-adjoint operators I don’t see the problem you’re stating. Assume for simplicity that we already knew that all irreducible representations were finite-dimensional. Then is self-adjoint, i.e. symmetric, and thus diagonalizable over . Can’t we proceed from there as in your proof and obtain the same result?

In general, I’m wondering how far we can take this and what goes wrong about Peter-Weyl in the real case. As far as I can see, the only thing that goes wrong is that we can’t conclude that the multiplicity of each irreducible representation equals its dimension, but the basic decomposition of the regular representation into irreducible representations should hold.

Best wishes,

Leon

Sorry. “invariant subspace” should be “eigenspace” here. The problem in the real case is that eigenspaces can be complex even when the operators are real and orthogonal (consider for instance a rotation matrix).

]]>thank you for this helpful right-up! I have two questions on Lemma 2, Schur’s Lemma for unitary representations:

1.: “Observe that any closed invariant subspace of {T^* T} is {G}-invariant”

How can we see this? If {X} is closed and invariant under {T^* T} and we look at {g \in G}, we can deduce that {\rho(g)X} is closed and invariant under {T^* T} as well, but I don’t see how this helps us in showing that it is a subset of {X}. Am I missing something?

2.: Does this Lemma also work if we look at representations over the real numbers, instead of the complex numbers, and replace “unitary” by “orthogonal”? After all, the constant multiple seems to be a real number, and I don’t see what else might go wrong.

]]>Yes, it is in the part on non abelian Fourier analysis, after equation 1 :

“This is an isomorphism not only of Hilbert spaces, but of the left-action of {G}. Indeed, it is an isomorphism of the bi-action of {G \times G} on both the left and right of both {L^2(G)} and {HS(V_\xi)}, defined by…”

I suppose, after now looking at the comment of F. Vom Ende, that it is due to a change that you made in the function on G that you associate to an operator on a representation space (the i_ξ).

It’s a minor point anyway, and more importantly, thank you for sharing your insights on your blog.

*[Action should be fixed now, thanks – T.>]*

Can you be more precise (e.g., by pointing out a specific line in the blog post that may require clarification or correction)?

]]>Are you sure that with your convention of Fourier inverse i_ξ , it entertwinnes (L,R) with the action on operators as you defined? I think it works when the Fourier inverse maps T in HS(V_ξ ) to the function on G : Tr(T ξ(g^(-1) ) .

Best Regards ]]>

*[Corrected, thanks – T.]*