# L^2 basis of class functions on a compact Lie group that are point-wise small

Consider first the torus group $\mathbb{T}^k$. A natural $L^2$ basis is given by the 1-dimensional complex representations: $(\theta_1, \ldots, \theta_k) \mapsto e^{i \sum_j c_j \theta_j}$ for integral $c_j$'s. This basis has also the nice property that each element in it is point-wise bounded by $1$ in absolute value. In general, an $L^2$ function takes the form $g = \sum_n q_n f_n$ where $f_n$ are the basis functions described above. $g$ can have arbitrary bad $L^\infty$ bound, since the only requirement is that $\sum q_n^2 = 1$, but $\sum_n q_n$ can be $\infty$.

Now my question is, for the unitary or orthogonal group, what is the smallest uniform $L^\infty$ bound one can achieve on an orthogonal basis on the space of class functions? For the unitary groups, any orthogonal basis of class functions forms an orthogonal basis in the space of symmetric functions of the eigenvalues under the inner product such that the Schur polynomials are orthonormal. So the question can be phrased in terms of symmetric functions. For the orthogonal groups, one orthogonal (not normalized) basis of class functions comes from $\prod_{i=1}^n (Tr X^i)^{\alpha_i}$ where for some odd $i$, $\alpha_i$ is odd (I am not entirely sure about this, but see this paper theorem 4).

Edit: what about all $L^2$ functions in general? Can one get a uniform $L^\infty$ bound on the basis of size $e^{\mathcal{O}(n)}$ for $SO(n)$ or $U(n)$?

• So, unlike the torus, you don't want the basis of characters, since the best pointwise bound you can get for $Tr(\pi(g))$ is $\dim\pi$. For finite groups $G$, you have the basis of normalized characteristic functions of conjugacy classes $C$, which is not good either since it gives you a bound of $\sqrt{|G|/|C|}$. Jan 4, 2012 at 6:09
• @Alain: yes you are right. I guess my stated basis class functions for $SO(n)$ are not really orthogonal. Jan 4, 2012 at 6:26

There is a sharp comparison of sup norm to $L^2$ norm on compact topological groups $K$. (I saw a proof of the version of this for orthogonal groups or spheres in Stein-Weiss "Fourier Analysis on Euclidean Spaces", but the argument succeeds generally. E.g., section 7 of http://www.math.umn.edu/~garrett/m/mfms/notes_c/spheres_I.pdf)

The assertion (as alluded-to by @Alain Valette) is that the sup of the ratio of sup norm to $L^2$ norm is $\sqrt{dim V}$ if the total measure is normalized to 1, where $V$ is a (e.g.) left $K$-stable collection of functions on $K$. This estimate is sharp, in that it is attained.

Indeed, the argument does not depend on the space of functions $V$ being an irreducible repn, and it is easy to do tangible experiments on products of circles to see the phenomenon manifest.

Edit: (Thx @Denis... for noting broken link.) And, yes, the Stein-Weiss trick does not refer to basis elements, but to a subspace, so does not answer the first, original question exactly. Only the question-at-the-end. However, on another hand, the pointwise interaction of orthonormal basis elements is illuminated by the inevitability of that estimate.

• As far as I can understand (the link doesn't work for me), you consider the ratio on a linear subspace, which is not the same as the question where one wants the ratio on elements of a fixed basis. E.g. for a finite non-abelian group $G$, you can find an orthonormal basis of the space of functions on $G$ where every function has constant modulus $1$. (Just pick-up a bijection with $\mathbf{Z}/|G|\mathbf{Z}$ and transfer the characters from the latter.) Of course it's not clear what one can hope to do with such a basis... Jan 5, 2012 at 6:53
• @Denis: I understand your construction for finite groups now. Can you elaborate on your method for Lie groups? Certainly one can't map $SO(n)$ diffeomorphically onto the torus. Jan 5, 2012 at 19:13

I think for $U(n)$ and similar connected Lie groups you can construct orthonormal bases of the $L^2$ space of class functions which are everywhere of modulus $1$: use the Weyl integration formula to represent the space of conjugacy classes as $[0,\pi]^n$ with an explicit measure, and define functions $f(x)=\varphi(\psi(x))$ where

(1) $\varphi$ runs over an orthonormal basis of $[0,\pi]^n$ with respect to Lebesgue measure, all of which satisfy $|\varphi|=1$ (which can be done using characters);

(2) $\psi$ is a fixed function so that the change of variable formula shows that $\varphi\mapsto f$ is an isometry from the $L^2$-space with respect to Lebesgue measure to the class-function space.

For finite groups, similarly, one can construct an orthonormal basis with maximal $L^{\infty}$-norm equal to $1$ (though usually not with constant modulus).

• It's not entirely clear to me why this would work. Consider $\varphi_1 (\theta) = e^{i \sum_j a_j \theta_j}$ and $\varphi_2 = e^{i \sum_j b_j \theta_j}$ that are orthogonal on the torus, i.e., $\langle \varphi_1, \varphi_2 \rangle =0$. Let $Jac(\theta)$ be the Jacobian factor for going from the Haar measure on the torus to that on $U(n)$. Then we basically need $int e^{i\sum_j (a_j - b_j) \psi(\theta)} Jac(\psi(\theta)) d(\psi(\theta))$ to vanish for some $\psi$. But unless $\psi$ is linear, the exponential factor doesn't resemble anything that would vanish under integration. Jan 4, 2012 at 20:09
• I guess my question is why would composition by $\psi$ yield an isometry between the two function spaces? Jan 4, 2012 at 20:12
• Also if your argument works, then the modulus must be constant even on finite groups right? Otherwise the $L^2$ norm would be strictly less than 1. Jan 4, 2012 at 20:13