I asked this in math.stackexchange, but it disappeared from the "main list" almost immediately, so I hope it will be appropriate as a separate question in MO.

For a given function $f\in C(G)$ on a compact group $G$ its Fourier transform is defined as the family of operators
$$ \widehat{f}_\sigma=\int_Gf(t)\cdot\sigma(t^{-1}) \ \text{d}\ t,\quad \widehat{f}_\sigma:X_\sigma\to X_\sigma $$ where $\sigma:G\to B(X_\sigma)$ runs over the dual object $\widehat{G}$ and $\int...\text{d}\ t$ means the integral with respect to the normed Haar measure (i.e. $\int 1\ \text{d}\ t=1$). It is known that in the space $L_2(G)$ the following equality holds: $$ f(t)=\sum_{\sigma\in\widehat{G}}\dim X_\sigma\cdot\text{tr}\Big(\widehat{f}_\sigma\circ\sigma(t)\Big),\quad t\in G. $$ I wonder if this can be interpreted as an equality in $C(G)$ in some specific sense (or as a pointwise equality)? As an example, in the case $G=\mathbb{T}$ the Fejér theorem states that the arithmetical means of the partial sums of the series converge to $f$ in $C(G)$ (and hence pointwisely). Is it possible that someting similar is true for all compact groups?

P.S. As far as I understand (Marc Palm mentions this here), the situation becomes simpler if we replace $C(G)$ by $C^\infty(G)$. I would be grateful if somebody could clarify details or give references.

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    $\begingroup$ Some of the comments here mathoverflow.net/questions/162856 may be relevant: the problem is that unless the Fourier series converges absolutely (which forces you to be in $A(G)$) one has to specify some summation method over the index set $\widehat{G}$. But maybe there is something like the Fejer kernel, at least for compact Lie groups $\endgroup$
    – Yemon Choi
    May 18, 2014 at 15:32
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    $\begingroup$ You should probably also add a link on your MSE question pointing here (I personally think the question should stay here on MO, but others may disagree.) $\endgroup$
    – Yemon Choi
    May 18, 2014 at 15:33
  • $\begingroup$ I gave the link to MSE. Or do you mean this link: mathoverflow.net/questions/162856/…? $\endgroup$ May 18, 2014 at 15:37
  • $\begingroup$ Yes, I think that there must (probably) be a tricky summation method like the Fejér one. $\endgroup$ May 18, 2014 at 15:38
  • $\begingroup$ Sorry, I meant that on your MSE question math.stackexchange.com/questions/800347 you should add a link that points here $\endgroup$
    – Yemon Choi
    May 18, 2014 at 15:39

2 Answers 2


One basic fact is that (up to an irrelevant constant depending on the normalization of the measure), the ratio of sup-norm to $L^2$ norm in a finite-dimensional space of functions on a compact topological group $K$ is bounded by the square root of the dimension of that space, and that estimate is sharp. (E.g., this is proven in some notes http://www.math.umn.edu/~garrett/m/mfms/notes_2013-14/09_spheres.pdf and is well-known...)

In particular, for many non-abelian topological groups, this sup-norm grows, which makes pointwise convergence very difficult.

On the other hand, for compact Lie groups, a polynomial bound is easily provable for these dimensions, so sufficient differentiability implies very good pointwise convergence.

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    $\begingroup$ Marc Palm refers to a Harish-Chandra paper here mathoverflow.net/questions/162856/…. Paul, do you mean these results or something else? $\endgroup$ May 21, 2014 at 10:02
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    $\begingroup$ @SergeiAkbarov, in comparison to the H-C result, there is a simpler result applicable to compact Lie groups: use the lemma comparing sup norm and L^2 norm, the decomposition of biregular representation, a crude estimate on the dimension of the irreducible with a given highest weight, and just sufficient differentiability with respect to Casimir, to prove very good pointwise convergence for sufficiently differentiable functions on compact Lie groups. $\endgroup$ May 21, 2014 at 15:12
  • $\begingroup$ Hm... I had a hope that there is a paper that I can refer to... :) OK, Paul, thank you. $\endgroup$ May 22, 2014 at 10:23
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    $\begingroup$ @SergeiAkbarov, it occurred to me that there are some papers in the mid-1970s in conferences in Paris and nearby that address such things... can't remember the particulars, but I dimly recall that some appear in (relatively early) SLN's. $\endgroup$ Jan 13, 2015 at 20:58

The case of point wise absolute convergence for connected compact Lie groups and sufficiently smooth functions requires no deep analysis, I thought I'd write it out here. In general, the problem is basically Fourier analysis on a $\mathrm{Rank}(G)$ dimensional torus together with $|\Phi_{+}|$ summations by parts, where Rank is the dimension of the maximal torus and $\Phi_{+}$ is the positive roots.

Let $G$ be a connected compact Lie group, $T$ a maximal torus, $M$ the character lattice of the torus, $W$ the Weyl group and $\Phi$ the root system. Let $\Phi_{+}$ be the positive root system, let $M_+$ be the chamber of dominant weights and let $\rho = \frac{1}{2} \sum_{\alpha\in \Phi_{+}} \alpha$. For $\lambda \in M_{+}$, let $V_{\lambda}$ be the corresponding representation and let $d_{\lambda}$ be $\dim V_{\lambda}$. Let $\pi_{\lambda}: G \to V_{\lambda}$ be the representation map and let $\chi_{\lambda}(g) = \mathrm{Tr}\ \pi_{\lambda}(g)$ be the character. We fix a positive definite Hermitian inner product on each $V_{\lambda}$ so that $\pi(g)$ is unitary. We normalize Haar measure so that $\int_G 1 = 1$.

Let $f$ be a continuous function on $G$. Clearly, it is enough to prove pointwise convergence at the identity. Define $$\bar{f}(x) = \int_{g \in G} f(gxg^{-1}),$$ so $\bar{f}$ is a class function.

Lemma Pointwise convergence of $f$ at $1$ is equivalent to pointwise convergence of $\bar{f}$ at $1$ (with the same ordering of irreps).

Proof Choose $v_1$, $v_2$, ..., $v_{d_{\lambda}}$ an orthonormal basis of $V_{\lambda}$. The Fourier series of $f$ at $x \in G$ is, by definition, $$\sum_{\lambda \in M_{+}} d_{\lambda} \sum_{i=1}^{d_{\lambda}} \sum_{j=1}^{d_{\lambda}} \langle v_i,\ \pi_{\lambda}(x) v_j \rangle \int_{g \in G} f(g) \langle v_j,\ \pi_{\lambda}(g) v_i \rangle .$$ (Some $g$ and $g^{-1}$'s may be switched here.) Taking $x$ to be the identity, we have $\langle v_i, v_j \rangle = \delta_{ij}$ so the sum simplifies to $$\sum_{\lambda \in M_{+}} d_{\lambda} \int_{g \in G} f(g) \sum_{i=1}^{d_{\lambda}} \langle v_i,\ \pi_{\lambda}(g) v_i \rangle = \sum_{\lambda \in M_{+}} d_{\lambda} \int_{g \in G} f(g) \chi_{\lambda}(g).$$ Since $\chi_{\lambda}$ is a class function, the integrals are unchanged by replacing $f$ with the class function $\bar{f}$. $\square$.

From now on, we assume that $f$ is a class function. So $f$ is determined by its restriction to $T$, which is $W$ invariant. Until I say otherwise, assume that $f$ is sufficiently smooth that the Fourier coefficients $\int_{\theta \in T} e^{\mu}(\theta) f(\theta)$ decay faster $|\mu|^{-N}$, where $N$ is some integer depending on $G$ which you could extract from the proof. Noting that $d_{\lambda} = O(|\lambda|^K)$ for some other constant $K$, this will make all sums absolutely convergent, so we can rearrange at will.

For $\alpha \in M$, let $e^{\alpha}$ be corresponding character of the torus. We will also sometimes abuse notation by writing things like $e_{\alpha/2}$, which is only defined on the double cover of the torus; our total expressions will be well defined. Let $\Delta :T \to \mathbb{C}$ be the Weyl denominator: $$\Delta(\theta) = \prod_{\alpha \in \Phi_{+}} \left( e^{\alpha/2}(\theta) - e^{- \alpha/2}(\theta) \right).$$

Using the Weyl integration formula to convert integrals of class functions $G$ to integrals over $T$ and the Weyl character formula to express $\chi_{\lambda}$;, we want to show $$f(1) = \mbox{constant} \cdot \sum_{\lambda \in M_{+}} d_{\lambda} \int_{\theta \in T} \Delta(\theta)^2 \frac{\sum_{w \in W} (-1)^w e^{w(\lambda+\rho)}(\theta)}{\Delta(\theta)} f(\theta)$$ $$= \mbox{constant} \cdot \sum_{\lambda \in M_{+}} d_{\lambda} \int_{\theta \in T} \Delta(\theta) \sum_{w \in W} (-1)^w e^{ w(\lambda+\rho)}(\theta) f(\theta) $$ $$= \mbox{constant} \cdot \sum_{\lambda \in M_{+}} \sum_{w \in W} (-1)^w d_{\lambda} \int_T e^{w(\lambda+\rho)}(\theta) \Delta(\theta) f(\theta). $$ (Here the constant absorbs $2 \pi$'s, signs and $|W|$ in some combination that isn't worth keeping track of.)

For $\mu \in M$, define $d(\mu) = \prod_{\alpha \in \Phi_{+}} \langle \mu, \alpha \rangle$. If $\mu = \lambda + \rho$, then $d(\mu) = d_{\lambda}$ by the Weyl dimension formula. If $\mu = w(\lambda+\rho)$ for $\lambda$ dominant, then $d(\mu) = (-1)^w d_{\lambda}$ by noting that $d$ is clearly $W$-antisymmetric. If $\mu$ is not of the form $w(\lambda+\rho)$ for any dominant $\lambda$ and any $w$, then $\mu$ is on one of the hyperplanes $\langle \ , \alpha \rangle =0$, so $d(\mu)=0$. In short, we can make the change of variable $\mu=w(\lambda+\rho)$ and write $$\mbox{constant} \cdot \sum_{\mu \in M+\rho} d(\mu) \int_T \Delta(\theta) f(\theta) e^{\mu}(\theta).$$

I pause to record what we are doing for $SU(2)$. For $f$ an even, $2 \pi$ periodic function, we are trying to show that $$f(0) = \frac{-1}{4 \pi} \sum_{n=-\infty}^{\infty} n \int_{\theta=0}^{2 \pi} f(\theta) (e^{i \theta} - e^{- i \theta}) \cdot e^{i n \theta} d \theta. \quad (\ast)$$ Assuming that the Fourier cofficients are small enough to rearrange this sum, we can regroup as $$\frac{-1}{4 \pi} \sum_{m=-\infty}^{\infty} \int_{\theta=0}^{2 \pi} f(\theta) ((m-1) - (m+1)) e^{i m \theta} \quad (\ast \ast)$$ which is clearly the standard Fourier tranform. Note, though, that the rate of decay of Fourier coefficients which justifies the rearrangement is more stringent than that which makes the Fourier sum absolutely convergent.

We claim that the same thing happens in general. Expanding the product for $\Delta$, we can rewrite the sum as

$$\sum_{\nu \in M} \int_T f(\theta) \sum_{\epsilon: \Phi_{+} \to \{ \pm 1 \}} \prod_{\alpha \in \Phi_{+}} \epsilon(\alpha) \cdot d\left( \nu - \sum_{\alpha \in \Phi_{+}} \epsilon(\alpha) \alpha \right) e^{\nu}(\theta)$$

Now, $d(x)$ is a polynomial of degree $|\Phi^{+}|$, and we have applied $|\Phi|$ partial difference operators to it. So $\sum_{\epsilon: \Phi_{+} \to \{ \pm 1 \}} \prod \epsilon(\alpha) d(x- (1/2) \sum \epsilon(\alpha) \alpha)$ is a constant, and I leave it as a challenge for the reader to compute that it is $|W|$. So our sum collapses to $$\mbox{constant} \cdot \sum_{\nu \in M} f(\theta) e^{\nu}(\theta).$$ Because the world is just, the constant works out.

Stated more conceptually, $\Delta(\theta) e^{\nu}(\theta) f(\theta)$ is a $|\Phi_{+}|$-fold partial difference of $\int_{\theta} e^{\nu}(\theta) f(\theta)$, and $d(\nu)$ is a degree $|\Phi_{+}|$ polynomial. If we can justify summing by parts enough times, we should be able to turn $$\sum_{\nu} \int_{\theta} \Delta(\theta) e^{\nu}(\theta) f(\theta) \cdot d(\nu)$$ into $$\sum_{\nu} \int_{\theta} e^{\nu}(\theta) f(\theta) \cdot 1.$$

I'm pretty sure that I learned somewhere that, if $\sum A_i (B_{i+1} - B_i)$ is $(C,k)$ Cesaro summable, then the summation by parts $\sum (A_{i-1} - A_i) B_i$ is $(C,k+1)$ Ceasro summable. Taking $A_n=n$ and $B_n = \int_{\theta=0}^{2 \pi} f(\theta) e^{i n \theta} d \theta$, we see that $(\ast)$ should be $(C,2)$ Cesaro summable.

Unfortunately, on higher dimensional torii, the question of what the analogue of Cesaro summation is a mess.

  • $\begingroup$ Sorry if I've missed something obvious, but when you talk about pointwise convergence, what ordering are you putting on your index set? Or are you talking about pointwise unordered summability? $\endgroup$
    – Yemon Choi
    May 21, 2014 at 17:45
  • $\begingroup$ For most of the answer, I have a function smooth enough that the sums are absolutely convergent so I don't need to say. Then, when I start talking about Cesaro summability, I only make a precise statement for $SU(2)$, where I order the irreps by dimension. But I need to do some editing; let me see if I can clean it up. $\endgroup$ May 21, 2014 at 17:54

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