# E. Kowalski’s blog

Comments on mathematics, mostly.

## The Spring menagerie

I think readers can legitimately complain that not only have I not added a new post for a long time, but more schockingly, my last animal-related one goes back more than one year. So, to celebrate the recent belated aperçus of spring in Zürich and around, here are some pictures:

The first two are cheating, since they come from the Masoala Hall — but the first one illustrates the beautiful views from the very new canopy walk:

while the second is a rarely-seen lizard

Next comes a well-camouflaged bird, this one from a park in Graz

and another one from the aforementioned canopy

after which come a frog,

a snail,

and more frogs:

Hopefully more animal pictures will come before a year passes!

Written by Kowalski

May 6th, 2013 at 9:24 pm

## A missing word

From the blog of the rare books collection of the ETH Library, I just learnt that the word for the study and classification of grape species that I was looking for is “ampelography” (ampélographie in French).

(The relevance of this word to my daily life is that the computers on my home network are named after grapes; red grapes are reserved for desktops and white for laptops.)

Written by Kowalski

April 21st, 2013 at 6:48 pm

## Another exercise with characters

While thinking about something else, I noticed recently the following result, which is certainly not new:

Let $G$ be a compact topological group [ADDITIONAL ASSUMPTION pointed out by Y. Choi: connected, Lie group], and let $\rho$ be a finite-dimensional irreducible unitary continuous representation of $G$ on a vector space $V$. Then the natural representation $\pi$ of $G$ on $\mathrm{End}(V)$ decomposes as a direct sum of one-dimensional characters if and only if $\rho$ is of dimension $1$.

One direction is clear: if $\rho$ has dimension one, then $\pi$ is simply the trivial one-dimensional representation. For the converse, here is an argument with character theory.

As a first step, note that if $\rho$ (of dimension $d\geq 1$, say) has this property, then in fact $\pi$ decomposes as a direct sum of distinct one-dimensional characters: indeed, the multiplicity of a character $\chi$ in $\pi$ is the same as
$n_{\chi}=\int_{G}\chi(x)\mathrm{Tr}(\pi(g))dg,$
where $dg$ is the probability Haar measure on $G$, and since
$\mathrm{Tr}(\pi(g))=|\mathrm{Tr}(\rho(g))|^2,$
we get
$n_{\chi}\leq \int_{G}\mathrm{Tr}(\pi(g))dg=1$
by the orthogonality relations of characters. (Algebraically, this is just an application of Schur’s lemma).

Thus if we decompose $\pi$ into irreducible representations, we get
$\pi=\bigoplus_{1\leq i\leq d^2} \chi_i,$
where the $\chi_i$ are distinct one-dimensional characters. We then know by orthogonality that
$d^2=\int_{G} |\mathrm{Tr}(\pi(g))|^2 dg=\int_{G} |\mathrm{Tr}(\rho(g))|^4 dg.$

Now the last-integral is bounded by
$\int_{G} |\mathrm{Tr}(\rho(g))|^4 dg\leq \mathrm{Max}_{g}|\mathrm{Tr}(\rho(g))|^2 \times \int_G|\mathrm{Tr}(\rho(g))|^2dg\leq d^2,$
(since $|\mathrm{Tr}(\rho(g))|\leq d$). Comparing, this means that there must be equality throughout in this estimate, which in turn implies that $|\mathrm{Tr}(\rho(g))|=d$ for all $g\in G$. Since $\rho(g)$ is unitary of size $d$, this implies that $\rho(g)$ is scalar for all $g$, and since it is assumed to be irreducible, it is in fact one-dimensional.

I see two interesting points in this argument: (1) is there a purely algebraic proof of the last part? I haven’t thought very hard about this yet, but it would be nice to have one; (2) the appearance of the fourth moment of $\rho$ is nicely reminiscent of the Larsen alternative (see Section 6.3 of my notes on representation theory, for instance…)

Written by Kowalski

March 22nd, 2013 at 10:19 am

Posted in Exercise,Mathematics

## Zeros of Hermite polynomials

In my paper with É. Fouvry and Ph. Michel where we find upper bounds for the number of certain sheaves on the affine line over a finite field with bounded ramification, the combinatorial part of the argument involves spherical codes and the method of Kabatjanski and Levenshtein, and turns out to depend on the rather recondite question of knowing a lower bound on the size of the largest zero $x_n$ of the $n$-th Hermite polynomial $H_n$, which is defined for integers $n\geq 1$ by
$H_n(x)=(-1)^n e^{x^2} \frac{d^n}{dx^n}e^{x^2}.$

This is a classical orthogonal polynomial (which implies in particular that all zeros of $H_n$ are real and simple). The standard reference for such questions seems to still be Szegö’s book, in which one can read the following rather remarkable asymptotic formula:
$x_n=\sqrt{2n}-\frac{i_1}{\sqrt[3]{6}}\frac{1}{(2n)^{1/6}}+o(n^{-1/6})$
where $i_1=3.3721\ldots>0$ is the first (real) zero of the function
$\mathrm{A}(x)=\frac{\pi}{3}\sqrt{\frac{x}{3}}\Bigl\{J_{1/3}\Bigl(2\Bigl(\frac{x}{3}\Bigr)^{3/2}\Bigr)+J_{-1/3}\Bigl(2\Bigl(\frac{x}{3}\Bigr)^{3/2}\Bigr)\Bigr\}$
which is a close cousin of the Airy function (see formula (6.32.8) in Szegö’s book, noting that he observes the Peano paragraphing rule, according to which section 6.32 comes before 6.4).

(Incidentally, if — like me — you tend to trust any random PDF you download to check a formula like that, you might end up with a version containing a typo: the cube root of $6$ is, in some printings, replaced by a square root…)

Szegö references work of a number of people (Zernike, Hahn. Korous, Bottema, Van Veen and Spencer), and sketches a proof based on ideas of Sturm on comparison of solutions of two differential equations.

As it happens, it is better for our purposes to have explicit inequalities, and there is an elementary proof of the estimate
$x_n\geq\sqrt{\frac{n-1}{2}},$
which is only asymptotically weaker by a factor $2$ from the previous formula. This is also explained by Szegö, and since the argument is rather cute and short, I will give a sketch of it.

Besides the fact that the zeros of $H_n$ are real and simple, we will use the easy facts that $\deg(H_n)=n$, and that $H_n$ is an even function for $n$ even, and an odd function for $n$ odd, and most importantly (since all other properties are rather generic!) that they satisfy the differential equation
$y''-2xy'+2ny=0.$

The crucial lemma is the following result of Laguerre:

Let $P\in \mathbf{C}[X]$ be a polynomial of degree $n\geq 1$. Let $z_0$ be a simple zero of $P$, and let
$w_0=z_0-2(n-1)\frac{P'(z_0)}{P''(z_0)}.$
Then if $T\subset \mathbf{C}$ is any line or circle passing through $z_0$ and $w_0$, either all zeros of $P$ are in $T$, or both components of $\mathbf{C}-T$ contain at least one zero of $P$.

Before explaining the proof of this, let’s see how it gives the desired lower bound on the largest zero $x_n$ of $H_n$. We apply Laguerre’s result with $P=H_n$ and $z_0=x_n$. Using the differential equation, we obtain
$w_0=x_n-\frac{n-1}{x_n}.$
Now consider the circle $T$ such that the segment $[w_0,z_0]$ is a diameter of $T$.

Now note that $-x_n$ is the smallest zero of $H_n$ (as we observed above, $H_n$ is either odd or even). We can not have $w_0<-x_n$: if that were the case, the unbounded component of the complement of the circle $T$ would not contain any zero, and neither would $T$ contain all zeros (since $-x_n\notin T$), contradicting the conclusion of Laguerre's Lemma. Hence we get $-x_n\leq w_0=x_n-\frac{n-1}{x_n},$
and this implies
$x_n\geq \sqrt{\frac{n-1}{2}},$
as claimed. (Note that if $n\geq 3$, one deduces easily that the inequality is strict, but there is equality for $n=2$.)

Now for the proof of the Lemma. One defines a polynomial $Q$ by
$P=(X-z_0)Q,$
so that $Q$ has degree $n-1$ and has zero set $Z$ formed of the zeros of $P$ different from $z_0$ (since the latter is assumed to be simple). Using the definition, we have
$Q'(z_0)=P'(z_0),\quad\quad Q''(z_0)=\frac{1}{2}P''(z_0).$
We now compute the value at $z_0$ of the logarithmic derivative of $Q$, which is well-defined: we have
$\frac{Q'}{Q}=\sum_{\alpha\in Z}\frac{1}{X-\alpha},$
hence
$\frac{Q'}{Q}(z_0)=\sum_{\alpha\in Z}\frac{1}{z_0-\alpha},$
which becomes, by the above formulas and the definition of $w_0$, the identity
$\frac{1}{z_0-w_0}=\frac{1}{n-1}\sum_{\alpha\in Z}\frac{1}{z_0-\alpha},$
or equivalently
$\gamma(w_0)=\frac{1}{n-1}\sum_{\alpha\in Z}{\gamma(\alpha)},$
where $\gamma(z)=1/(z_0-z)$ is a Möbius transformation.

Recalling that $|Z|=n-1$, this means that $\gamma(w_0)$ is the average of the $\gamma(\alpha)$. It is then elementary that for line $L$, either $\gamma(Z)$ is contained in $L$, or $\gamma(Z)$ intersects both components of the complement of $L$. Now apply $\gamma^{-1}$ to this assertion: one gets that either $Z$ is contained in $\gamma^{-1}(L)$, or $Z$ intersects both components of the complement of $\gamma^{-1}(L)$. We are now done, after observing that the lines passing through $\gamma(w_0)$ are precisely the images under $\gamma$ of the circles and lines passing through $w_0$ and through $z_0$ (because $\gamma(z_0)=\infty$, and each line passes through $\infty$ in the projective line.)

Written by Kowalski

February 20th, 2013 at 10:43 pm

Posted in Exercise,Mathematics