It is known that the Dirac $\delta(U)$ distribution can be expanded in representations for compact groups.
For example concerning the $U(1)$ $$ \delta(\phi)= \frac{1}{2\pi}\sum_{n} e^{in\phi}$$ Or another example is $SU(2)$ where the $\delta$ is expanded over the group like $$\delta(U) = \sum_{j} (2j+1) TrD^{(j)}(U)$$ where $D^{(j)}(U)$ are the the Wigner matrices.
There exist a similar expansion-like formula for the $\delta$ over the quantum group $SU(2)_{q}$ ?

  • $\begingroup$ Could you introduce us to what you call "The $\delta(u)$" ? Has it a name ? $\endgroup$
    – Jean Marie
    Sep 19, 2020 at 8:43
  • $\begingroup$ Thanks for the note, I have edited the post hoping now it is more clear. $\endgroup$
    – Pipe
    Sep 19, 2020 at 10:37
  • $\begingroup$ Thanks. Isn't it more a "Dirac comb" than a "single Dirac" ? $\endgroup$
    – Jean Marie
    Sep 19, 2020 at 12:45
  • 1
    $\begingroup$ @JeanMarie, certainly in the case of "Dirac on the circle", it pulls back to Dirac comb on the line. But for non-abelian compact groups there's nothing obviously comparable, so I think it's reasonable to call it "Dirac on X" when X is a smooth manifold. $\endgroup$ Mar 16, 2021 at 19:14

1 Answer 1


Honestly, I've never worked with Wigner matrices, so I am not sure if I understand your question/motivation correctly and therefore whether my answer will be anyhow useful. But let's try:

First of all, I think that the object you are refering to as Dirac distribution is what is in the quantum group framework called the counit and denoted by $\varepsilon$. Classically, it is the functional $C(G)\to\mathbb{C}$ mapping $f\mapsto f(e)$, where $e$ is the unit element of the group. Since with quantum groups, we work already by definition with functions rather than points, the counit actually stands kind of somewhere in the centre of the whole story, so I am not sure if it useful to try to expand it somehow, but maybe yes, I don't know.

Your examples are written in a slightly informal language, so let's formalize the first one, to make sure we understand each other (as I said, I have no practice with the second one). So, take $G=U(1)$ to be the circle. Let's denote by $\phi\in G$ its elements, where $\phi$ denotes the angle on the circle. This group has the fundamental representation $\rho\colon\phi\mapsto e^{i\phi}$ and all other irreducibles are given by $\mathop{\rm Irr}G=\{\rho^k\mid k\in\mathbb{Z}\}$. Since all the irreducibles are one-dimensional (which is because $G$ is commutative), we actually have that those are just functions $G\to\mathbb{C}$, so $\rho^k\in C(G)$ and they map $\phi\mapsto e^{ik\phi}$. Those functions even form a basis of $C(G)$ (or $L^2(G)$ or whatever) by expanding to the Fourier series. Now each of those functions induce canonically a functional $\omega_k\in C(G)^*$ mapping $\omega_k(f)=\langle\rho^k,f\rangle={1\over 2\pi}\int e^{ik\phi}f(\phi)\,{\rm d}\phi$. In particular $\omega_k(\rho^l)=\delta_{kl}$. Now, you claim that $$\varepsilon=\sum_{k\in\mathbb{Z}}\omega_k.$$

Now, how this extends to the quantum group case: Let $G$ be a compact matrix quantum group with fundamental representation $u$. Denote by $O(G)\subset C(G)$ the coordinate algebra, that is, the $*$-algebra generated by the entries $u_{ij}$ of the fundamental representation. Denote by $\mathop{\rm Irr}G$ again the set of irreducibles of $G$ (up to equivalence). For a given $\alpha\in\mathop{\rm Irr}G$, we denote by $u^\alpha$ the corresponding representation and by $n_\alpha$ its size. It holds again that the entries of the irreducibles, that is the set $\{u_{ij}^\alpha\}$ wiht $\alpha\in\mathop{\rm Irr}G$, $i,j=1,...,n_\alpha$ forms a vector space basis of $O(G)$. Therefore, any functional $\omega\in O(G)^*$ is determined by the numbers $\omega^\alpha_{ij}:=\omega(u_{ij}^\alpha)$. Consequently, we can write $$O(G)^*\simeq\prod_{\alpha\in\mathop{\rm Irr}G}M_{n_\alpha}(\mathbb{C})$$ Now, we can take kind of the identity matrix for each of the factor. More precisely, for every $\alpha\in\mathop{\rm Irr}G$, define $\omega_\alpha\in O(G)^*$ by $\omega_\alpha(u_{ij}^\beta)=\delta_{ij}\delta_{\alpha\beta}$. Then finally, we can write $$\varepsilon=\sum_{\alpha\in\mathop{\rm Irr}G}\omega_\alpha.$$ (Now, as I am thinking of it, I am not sure, what topology you should choose in order to make this work. But at least pointwise it's true.)

I am not sure, however, if this is what you are looking for.

EDIT: As I understand it, the Wigner matrices are the irreducible representations of $SU(2)$. That is, $D^{(\alpha)}=u^\alpha$ here. So you could write something like $\omega_\alpha\approx\mathop{\rm Tr} D^{(\alpha)}$ and you get (almost) exactly the formula you've mentioned. But formally this is incorrect, because you should use the dual forms $\omega_\alpha$ instead of the representations $D^{(\alpha)}$ (and I suppose that the $(2j+1)$ factor is probably some normalization that appears when passing to this dual).

Nevertheless, I think that if you were really seriously working with quantum groups, then instead of those sums with questionable meaning, you would probably just use the simple equality $\varepsilon(u_{ij}^{\alpha})=\delta_{ij}$ (and maybe the fact that those $u_{ij}^\alpha$ form a basis of your algebra). But it depends on what you are working on I guess.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .