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I am reading a physics paper (hep-th/9204083 v1 above eq (4.15)) where there is a nice formula that reads

$$ \delta(U-\mathbb{1}) = \sum\limits_{r} \dim(r)\, \mathrm{tr}_r(U) $$ Here $\:U\:$ is a group element, $\:1\:$ is the unit group element, $\:r\:$ are the representations.

I would like to know if this formula has a name (for example something like Weyl's formula), if I can get any intuitive argument why it should be true.

I am not sure the formula applies to $\:U(1)\:$ group. The simplest case maybe would be $\:SU(2)\:$ and already there this formula seems quite complicated. I would like to verify the formula for the simplest possible case.

Is this formula a generalized Fourier transform?

Andreas
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  • Andreas, we have MathJax active on the site which means that you can mark up mathematics in a language equivalent to LaTeX's math mode by enclosing it in single dollar signs ($) for inline math or double dollar signs ($$) for block set equations. That is writing $\delta(U-\mathbb{1}) = \sum_r \dim(r) \mathrm{tr}_r(U)$ will mark up as $\delta(U-\mathbb{1}) = \sum_r \dim(r) \mathrm{tr}_r(U)$. – dmckee --- ex-moderator kitten Dec 16 '16 at 05:55
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    Which physics paper? – Qmechanic Dec 16 '16 at 08:57
  • This definitely has to do with Fourier analysis, and its generalisations to functions of group elements. This must have something to do with the fact that $f(g) = \sum_{\xi \in \hat{G}} \dim(V_\xi) \langle \hat{f}(\xi), \rho(g)\rangle$ where $\hat{G}$ is the Pontryagin dual of a fixed compact group, $\rho : G \to U(H)$ is a unitary representation and $V \subset H$ is a closed invariant subspace. – JamalS Dec 16 '16 at 11:43
  • For further context, and to make this thread more useful for future visitors, you should reference the paper where you found this result. – Emilio Pisanty Dec 17 '16 at 16:40
  • hep-th/9204083 v1 above eq (4.15). – Andreas Dec 17 '16 at 23:42

2 Answers2

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Yes it is. You are writing down the projection to the origin of the group through its character expansion, indeed a generalization of the Fourier transform, as you guessed.

A good book would cover it all, such as Vilenkin's classic, Ch I, §4. I'll use the physicist's informal seat- of-the-pants language and avoid mathematese, which probably got you stranded in the first place.

Your formula is the Fourier expansion for U(1), whose reps are all one dimensional, $e^{ir\theta}$, and the group-invariant Haar measure is trivial/flat, ; so you get the δ(θ-0), having projected to the origin of the group.

The point is $\mathrm{tr}_r(U)\equiv \chi_r(U)$ is the character of the rep, a class function, that is it is the same for the entire equivalent class containing U, by the similarity invariance of the trace.

The character of the identity, then, is just $\chi_r(\mathbb{1}) = \dim(r)$, the dimension of the rep. From the orthogonality of the characters outlined in the linked WP article, you then see you may represent U as a direct sum of this U in every rep, with its character in that rep as a coefficient, so, then, just like the Fourier expansion.

So you are writing the completeness relation of your group with the origin of the group as the destination. Note however, in what you write, that the right hand side is a class function, so the sloppy l.h.s. must be interpreted as acting on class functions and dialing their angles to the origin, e.g. $\int dU \delta(U-\mathbb{1}) \chi_s (U)= \dim(s) $.

For SU(2), the simplest case, this is not that recondite. There is only one equivalence class--recall a similarity transformation can represent any group element as a rotation $R_3$ around some Euler axis arrived at by this similarity transformation: so you are just looking at $e^{i\theta J_3}$. The characters then are just Chebyshev polynomials of the second kind, a Dirichlet kernel, $$ \chi_s(\theta)=\frac{\sin ((2s+1)\theta/2)}{\sin (\theta/2)}, $$ the obvious sum of the 2s +1 exponentials in the trace of $R_3$. The Haar measure is $$ \int_0^{4\pi} d\theta \frac{\sin^2 (\theta/2)}{2\pi}. $$

You are then summing $$ (n+1)\frac{e^{i(n+1)\theta/2} - e^{-i(n+1)\theta/2}}{e^{i\theta/2} - e^{-i\theta/2}} $$ over all positive integers n=2s+1. (Cheap and dirty evaluations may end up yielding 0/0. My favorite formal wisecrack might be defining $S(\theta)\equiv \sum_{n=1}^{\infty}e^{in\theta/2}$, evaluating $\partial_\theta (S(\theta)+S(-\theta))$, whence $-\partial_\theta \delta(\theta/2)/\sin(\theta/2)$, given the Fourier kernel of the periodic δ, the Dirac comb, but no need to take it too seriously!). Large ns will net you periodic δs, as per the Dirichlet kernel WP article, but the point is the smaller n terms conspire to interfere destructively.

Still, technicalities aside, the projection is bound to work out. Check out that $$ \int_0^{4\pi} d\theta \frac{\sin^2 (\theta/2)}{2\pi} \frac{(-\partial_\theta \delta(\theta/2))}{2\sin(\theta/2)} ~ \chi_7 (\theta)= 15, $$ after the integration by parts. You may find practical examples of this technique in our paper here.


Note added in mathematese, since the OP appears interested in the structure of characters providing a complete orthonormal set for class functions. Fulton and Harris, p 440, summarize the Peter-Weyl theorem, which generalizes Fourier analysis to compact groups: "The characters of irreducible representations span a dense subspace of the space of continuous class functions." So the character completeness relation projects the equivalence class of U to the identity, an equivalence class of its own.

Here is a link to the stuff in Vilenkin & Klimyk 1991, not as good as the 68 classic, in my parochial view...

Finally, the WP section on D matrices' orthogonality illustrates completeness and the identity limit for SU(2), reviewed by Schwinger in eqn (3.96) of his monumental 1952 review.

Cosmas Zachos
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  • Goodness. I normally like to think I understand stuff in (the neighbourhood of) this area, but this answer is some tough going. Is Vilenkin's book a good introduction to the topic? What other good texts are there? – Emilio Pisanty Dec 17 '16 at 16:38
  • Vilenkin rules supreme... puts most others to shame. Early pencil&paper lattice gauge theorists used such a lot: there is no substitute for practical use & reliance... so, Creutz's book on lattice gauge theory? Apologies if answer is stressful... Maybe you could improve some wording to make it friendlier? Never sure what's missing... – Cosmas Zachos Dec 17 '16 at 17:36
  • Creutz Ch 8 group integration skips mathematese and runs into the trenches. Skip the physics, if not interested, but the machinery is there in full functionality... – Cosmas Zachos Dec 17 '16 at 17:43
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Combining Zachos formula

$$\int dU \delta(U-1) \chi_r(U) = \chi_r(1) = dim(r)$$

with

$$\delta(U-1) = \sum_s dim(s) \chi_s(U)$$

leads me to

$$\sum_s dim(s) \int dU \chi_s(U)\chi_r(U) = dim(r)$$

In other words, we should have

$$\int dU \chi_s(U) \chi_r(U) = \delta_{r,s}$$

This might be easier to prove than the original formula.

Andreas
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  • Yes, for finite groups this is the "rows" orthogonality, as @ACuriousMind reminds you, the starting point in my discussion; whereas the "columns" orthogonality evokes what you wrote in your question, only you particularize it to the identity. Witten's (4.12) is the analog of the completeness relation of orthogonal polynomials, the resolution of the identity, $\delta(U-W) = \sum_r \chi_r ^*(W)\chi_r(U)$. – Cosmas Zachos Dec 18 '16 at 13:02
  • Maybe then I should correct it as follows, – Andreas Dec 18 '16 at 23:00
  • My conclusion seems to be that also – Andreas Dec 18 '16 at 23:15
  • My conclusion seems to be that also $\int dU \chi_s^*(U) \chi_r(U) = \delta_{r,s}$ should hold. – Andreas Dec 18 '16 at 23:16