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,
and more frogs:
Hopefully more animal pictures will come before a year passes!
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.)
While thinking about something else, I noticed recently the following result, which is certainly not new:
Let be a compact topological group [ADDITIONAL ASSUMPTION pointed out by Y. Choi: connected, Lie group], and let be a finite-dimensional irreducible unitary continuous representation of on a vector space . Then the natural representation of on decomposes as a direct sum of one-dimensional characters if and only if is of dimension .
One direction is clear: if has dimension one, then is simply the trivial one-dimensional representation. For the converse, here is an argument with character theory.
As a first step, note that if (of dimension , say) has this property, then in fact decomposes as a direct sum of distinct one-dimensional characters: indeed, the multiplicity of a character in is the same as
where is the probability Haar measure on , and since
by the orthogonality relations of characters. (Algebraically, this is just an application of Schur’s lemma).
Thus if we decompose into irreducible representations, we get
where the are distinct one-dimensional characters. We then know by orthogonality that
Now the last-integral is bounded by
(since ). Comparing, this means that there must be equality throughout in this estimate, which in turn implies that for all . Since is unitary of size , this implies that is scalar for all , 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 is nicely reminiscent of the Larsen alternative (see Section 6.3 of my notes on representation theory, for instance…)
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 of the -th Hermite polynomial , which is defined for integers by
This is a classical orthogonal polynomial (which implies in particular that all zeros of 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:
where is the first (real) zero of the function
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 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
which is only asymptotically weaker by a factor 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 are real and simple, we will use the easy facts that , and that is an even function for even, and an odd function for odd, and most importantly (since all other properties are rather generic!) that they satisfy the differential equation
The crucial lemma is the following result of Laguerre:
Let be a polynomial of degree . Let be a simple zero of , and let
Then if is any line or circle passing through and , either all zeros of are in , or both components of contain at least one zero of .
Before explaining the proof of this, let’s see how it gives the desired lower bound on the largest zero of . We apply Laguerre’s result with and . Using the differential equation, we obtain
Now consider the circle such that the segment is a diameter of .
Now note that is the smallest zero of (as we observed above, is either odd or even). We can not have : if that were the case, the unbounded component of the complement of the circle would not contain any zero, and neither would contain all zeros (since ), contradicting the conclusion of Laguerre's Lemma. Hence we get
and this implies
as claimed. (Note that if , one deduces easily that the inequality is strict, but there is equality for .)
Now for the proof of the Lemma. One defines a polynomial by
so that has degree and has zero set formed of the zeros of different from (since the latter is assumed to be simple). Using the definition, we have
We now compute the value at of the logarithmic derivative of , which is well-defined: we have
which becomes, by the above formulas and the definition of , the identity
where is a Möbius transformation.
Recalling that , this means that is the average of the . It is then elementary that for line , either is contained in , or intersects both components of the complement of . Now apply to this assertion: one gets that either is contained in , or intersects both components of the complement of . We are now done, after observing that the lines passing through are precisely the images under of the circles and lines passing through and through (because , and each line passes through in the projective line.)