# What corresponds to the operation of taking traces in of the Fourier transformation on a finite group?

I have a question about the Fourier transfomation on a finite non-comutative group. I hope that it is a known fact in the Representation Theory but I cannot find it written explicitly in textbooks.

Let $$G$$ be a finite non-abelian group, $$\hat{?}:L_2(G)\to \hat L_2(\hat G)$$ be its Fourier transformation and $$\check{?}:\hat L_2(\hat G)\to L_2(G)$$ be the inverse Fourier transformation. The space $$L_2(G)$$ carries the standard norm induced by the inner product $$\langle x,y\rangle=\sum_{g\in G}x(g)\cdot\overline{y(g)}$$. The norm on $$\hat L_2(\hat G)$$ is normalized so that the Fourier transformations $$\hat?$$ is an isometry.

Let $$P:\hat L_2(\hat G)\to \hat L_2(\hat G)$$ be the function assigning to a sequence of matrices $$(M_\alpha)_{\alpha\in\hat G}$$ in $$\hat L_2(\hat G)$$ the sequence $$(\frac 1{\dim(\alpha)}Tr(M_\alpha)\cdot\mathrm{Id}_{\alpha})_{\alpha\in\hat G}$$ of normalized traces of those matrices multiplied by the identity matrices. So, $$P$$ is a linear projection of $$\hat L_2(\hat G)$$ onto its $$|\hat G|$$-dimensional subspace.

Question 1. What is the norm of a sequence $$(M_\alpha)_{\alpha\in \hat G}\in P(\hat L_2(\hat G))$$ in the Hilbert space $$\hat L_2(\hat G)$$? Is it equal to $$(\frac1{|\hat G|}\sum_{\alpha\in\hat G}|\frac1{\dim(\alpha)}Tr(M_\alpha)|^2)^{1/2}$$?

I am interested in describing the projector $$A=\check{?}\circ P\circ \hat{?}:L_2(G)\to L_2(G)$$. It should assign to each function $$f\in L_2(G)$$ some class function on $$G$$.

Question 2. Is $$A$$ the averaging of $$f$$ over conjugacy classes? If yes, where can I find a reference to this fact?

• In the non abelian case by $\hat{L}_2(\hat G)$ do you just mean the direct product of matrix algebras coming from the Wedderburn decomposition? – Benjamin Steinberg Apr 7 at 11:53
• Exercise 6.2 of Serre’s book on rep theory answers your first question. Your formula is off. You should multiply by the dimension not divide – Benjamin Steinberg Apr 7 at 12:01
• @BenjaminSteinberg But the formula in Ex 6.2 is for the inverse Fourrier transform... And in the definition of norm one should impose some multipliers to make to Fourier transformation an isometry. As far as I understand, the Fourrier transformation of the characteristic function of the singleton at the unit of the group is the sequence of identity matrices and this sequence should have norm one if we want that the Fourier transformation to be an isometry. To obtain this 1 from traces one should divide by the dimension, not multiply. That was my philosophy of writing exactly such formula. – Taras Banakh Apr 7 at 12:16
• @BenjaminSteinberg The answer to your first comment is yes: by $\hat L_2(\hat G)$ I understand the image of the Fourier transformation, i.e., the direct sum of endomorphism algebras corresponding to irreducible representations of $G$, – Taras Banakh Apr 7 at 12:19
• I stand by my second comment. The sum of the dimensions squared is the order of the group so the formula I propose gives the sequence of identity matrices norm 1 and yours does not. – Benjamin Steinberg Apr 7 at 12:28

Exercise 6.2 in Serre's book on representation theory of finite groups can be interpreted as saying that for functions $$f,h$$ on $$G$$ one has

$$\sum_{g\in G}\overline{f(g)}{h(g)} = \dfrac{1}{|G|}\sum_{\rho \in \hat G}\dim\rho\cdot \mathrm{Tr}(\hat f(\rho)^*\hat h(\rho))$$ and so the correct formula for the norm should be $$\|(M_{\rho})_{\rho\in \hat G}\|^2 = \dfrac{1}{|G|}\sum_{\rho \in \hat G}\dim\rho\cdot \mathrm{Tr}(M^*_{\rho}M_\rho))$$ where $$*$$ denotes the Hermitian adjoint. See Teras and CECCHERINI-SILBERSTEIN et al for this version. Serre uses a slightly different bilnear form that agrees with this one on characters.

For the second question you are correct I will use the second orthogonality relations which says that $$\sum_{\rho\in \hat G}\overline{\mathrm{Tr}(\rho(g))}\mathrm{Tr}(\rho(h))$$ is $$0$$ unless $$g,h$$ are conjugate, in which case it is $$\dfrac{|G|}{|\mathrm{Cl}(g)|}$$ where $$\mathrm{Cl}(g)$$ is the conjugacy class of $$g$$.

Now $$P\hat f$$ is the sequence whose $$\rho$$-component is $$\sum_{h\in G}\frac{f(h)}{\dim \rho}\mathrm{Tr}(\rho(h))I$$.

The Fourier inversion theorem says that if $$k$$ is the function with $$\hat k=P\hat f$$, then $$k(g) =\frac{1}{|G|}\sum_{\rho\in \hat G}\dim \rho\cdot \mathrm{Tr}(\rho(g)^*\hat k(\rho))$$ $$=\frac{1}{|G|}\sum_{h\in G}f(h)\sum_{\rho\in \hat G}\overline{\mathrm{Tr}(\rho(g))}\mathrm{Tr}(\rho(h))=\sum_{h\sim g}\dfrac{f(h)}{|\mathrm{Cl}(g)|}$$ where $$\sim$$ is conjugacy by the second orthogonality relations, which is the average value of $$f$$ on the conjugacy class of $$g$$.

• Now I see that your (=Serre's) formula also yields 1 for the sequence of unit matrices. So, my heuristic was wrong. Thank you for clarification. – Taras Banakh Apr 7 at 12:47
• I'm assuming you are using unitary representations – Benjamin Steinberg Apr 8 at 12:00
• Yes, I am using the most standard representations for which the theory is developed. – Taras Banakh Apr 8 at 12:46