# E. Kowalski's blog

18.05.2014

### Is the Kierkegaardian idea true? and other queries

Filed under: Language,Literature,Mathematics @ 20:22

In February, I was invited to give talks in Bristol and Oxford, and I spent the night after the second talk in a guest room of Worcester College. While looking at the brochure explaining the history of this college, I noticed that a previous guest had left a cryptic inscription, which I took a photograph of:

Kierkegaard

Can anybody make a guess of what is the first line? It is

What is the **** historical perspective?

but I can’t read the missing fourth word!

And what is the Kierkergaardian idea, really?

11.05.2014

### The discrete spectrum is discrete

Filed under: Exercise,Mathematics @ 19:06

No, this post is not an exercise in tautological reasoning: the point is that the word “discrete” is relatively overloaded. In the theory of automorphic forms, “discrete spectrum” (or “spectre discret”) is the same as “cuspidal spectrum”, and refers to those automorphic representations (of a given group $G$ over a given global field $F$) which are realized as closed subspaces of the relevant representation space on $L^2(G(\mathbf{Q})\backslash G(\mathbf{A}_F))$, by opposition with the “continuous spectrum”, whose components may fail to be actual subrepresentations.

However, we can also think of automorphic representations as points in the unitary dual of the relevant adélic group $G(\mathbf{A}_F)$, and this has a natural topology (the Fell topology), for which it then makes sense to ask whether the set of all cuspidal automorphic representations is discrete or not.

The two notions might well be unrelated: for instance, if we take the group $\mathrm{SL}_2(\mathbf{R})$ and the direct sum over $t\in\mathbf{Q}$ of the principal series with parameter $t$, we obtain a representation containing only “discrete” spectrum, but parameterized by a “non-discrete” subset of the unitary dual.

It is nevertheless true that the discrete spectrum is discrete in the unitary dual, in the automorphic case. I am sure this is well-known. (Unless the topology on the unitary dual is much weirder than I expect; my expertise in this respect is quite limited, but I remember asking A. Venkatesh who told me that the pathologies of the unitary dual topology are basically irrelevant as far as the automorphic case is concerned). I think this must be well-known, but I don’t remember seeing it mentioned anywhere (hence this post…)

Here is the argument, at least over number fields, and for $\mathrm{GL}_n$. Let $\pi$ be a given cuspidal representation. The unitary dual topology on the adélic group is, if I understand right, the restricted direct product topology with respect to the unramified spectrum. So a neighborhood of $\pi$ is determined by a finite set $S$ of places and corresponding neighborhoods $U_s$ of $\pi_s$ for $s\in S$. We want to find such a neighborhood in which $\pi$ is the unique automorphic cuspidal representation.

First, we can fix a neighborhood $U_{\infty}$ of the archimedean (Langlands) parameters that is relatively compact. Next, we note (or claim…) that for a finite place $v$, the exponent of the conductor is locally constant, so we get neighborhoods $U_v$ of $\pi_v$, with $U_v$ the unramified spectrum when $\pi$ is unramified at $v$, such that all automorphic representations in
$U=U_{\infty}\times \prod_{v} U_v$
have arithmetic conductor equal to that of $\pi$. With the archimedean condition, it follows that the Iwaniec-Sarnak analytic conductor of cuspidal representations in $U$ is bounded.

However, a basic “height-like” property is that there are only finitely many cuspidal representations with bounded analytic conductor (this is proved by Michel-Venkatesh in Section 2.6.5 of this paper, and a different proof based on spherical codes, as suggested by Venkatesh, is due to Brumley). Thus $U$ almost isolates $\pi$, in the sense that only finitely many other cuspidal representations can lie in $U$. Denote by $X$ this set of representations.

Now, the unitary dual is (or should be…) at least minimally separated so that, for any two cuspidal representations $\pi_1$ and $\pi_2$, there is an open set which contains $\pi_1$ and not $\pi_2$, say $V_{\pi_1,\pi_2}$. Then
$U'=U\cap \bigcap_{\rho\in X-\{\pi\}} V_{\pi,\rho}$
is an open neighborhood of $\pi$ which only contains the cuspidal representation $\pi$

03.05.2014

### More conjugation shenanigans

Filed under: Exercise,Mathematics @ 16:46

After I wrote my last post on the condition $\xi H\xi =H$ in a group, I had a sudden doubt concerning the case in which this arose: there we assume that we have a coset $T=\xi H$ such that $g^2\in H$ for all $g\in T$. I claimed that this implies $\xi H \xi =H$, but really the argument I wrote just means that $\xi H \xi \subset H$: for all $g\in H$, we get $\xi g \xi g \in H$, hence $\xi g \xi \in H$. But what about the other inclusion?

I had in mind a case where the groups involved are finite, or close enough, so that the reverse inclusion is indeed obviously true. But it is amusing to see that in fact what I wrote is correct in all cases: if $H$ is a subgroup of an arbitrary group $G$ and $\xi\in H$ satisfies $\xi H \xi\subset H$, then in fact $\xi H \xi = H$: taking the inverse of the inclusion gives $\xi^{-1} H\xi^{-1}=\xi^{-1}H^{-1}\xi^{-1}\subset H^{-1}=H.$

I find this interesting because, when it comes to the normalizer, the analogue of this fact is not true: the condition $\xi H\xi^{-1}\subset H$ is not, in general, equivalent with $\xi H\xi^{-1}=H$.

(I found this explained in an exercise in Bourbaki, Algèbre, Chapitre I, p. 134, Exercice 27, or page 146 of the English edition; here is a simple case of the counterexample from that exercise: consider the group $G$ of permutations of $\mathbf{Z}$; consider the subgroup $H$ which is the pointwise stabilizer of $\mathbf{N}=\{0,1,2,\ldots\}$, and the element $\xi\in G$ which is just
$\xi(x)=x+1.$

Then we have $\xi H\xi^{-1}\subset H$, because the left-hand side is the pointwise stabilizer of $\xi(\mathbf{N})$ which is a subset of $\mathbf{N}$. But $\xi H\xi^{-1}$ is not equal to $H$, because $\xi^{-1} H\xi$ is the pointwise stabilizer of $\xi^{-1}(\mathbf{N})$, and there are elements in $G$ which fix $\mathbf{N}$ but not $\xi^{-1}(\mathbf{N})=\{-1,0,1,\ldots\}$.)

It seems natural to ask: what exactly is the set $X$ of all $(a,b)\in\mathbf{Z}^2$ such that $\xi^a H\xi^b\subset H$ is equivalent to $\xi^a H \xi^b=H$? This set $X$ contains the line $(n,n)$ for $n\in\mathbf{Z}$, and also the coordinates axes $(n,0)$ and $(0,n)$ (since we then deal with cosets of $H$).

In fact, one can determine $X$: it is the complement of the line $(n,-n)$, for $n\not=0$ (which corresponds to conjugation). Indeed, suppose for instance that $a+b\geq 1$. Let $H$ and $\xi$ be such that $\xi^a H \xi^b\subset H$.

We can write
$-a=(a+b-1)a-a(a+b),\quad\quad -b=(a+b-1)b-b(a+b),$
and
$\xi^{-a}H\xi^{-b}=\xi^{(a+b-1)a} (\xi^{a+b})^{-a}H (\xi^{a+b})^{-b} \xi^{(a+b-1)b}.$
Since $\xi^{a+b}\in H$, and since $a+b-1\geq 0$ is the common exponent of $\xi^a$ and $\xi^b$ at the two extremities of the right-hand side, it follows that this right-hand side is contained in $H$.

A similar argument works for $a+b\leq -1$, using
$-a=(-a-b-1)a+a(a+b),\quad\quad -b=(-a-b-1)b+b(a+b),$
where again it is crucial that the coefficient $(-a-b-1)$ appears on both sides, and that it is $\geq 0$.

Since we already know that $(1,-1)\notin X$, and in fact that $(-n,n)\notin X$ for $n\not=0$ (the same setting as the counterexample above works, because the $\xi$ we used is an $n$-th power for every $n\not=0$), we have therefore determined the set $X$

01.05.2014

### Normalizers everywhere

Filed under: Exercise,Mathematics @ 13:05

In working on a paper, I found myself in the amusing but unusual situation of having a group $G$, a subgroup $H$ and an element $\xi\in G$ such that
$\xi H\xi =H.$

This certainly can happen: the two obvious cases are when $H=G$, or when $\xi$ is an involution that happens to be in the normalizer $N(H)$ of $H$.

In fact the general case is just a tweak of this last case: we have $\xi H\xi=H$ if and only if $\xi^2\in H$ and $\xi\in N(H)$, or in other words, if $\xi$ belongs to the normalizer, and is an involution modulo $H$.

This is of course easy to check. I then asked myself: what about the possibility that
$\xi^a H\xi^b = H$,
where $a$ and $b$ are arbitrary integers? Can one classify when this happens? The answer is another simple exercise (that I will probably use when I teach Algeba I next semester): this is the case if and only if $\xi^{a+b}\in H$ and $\xi^{(a,b)}\in N(H)$, where $(a,b)$ is the gcd of $a$ and $b$. In particular, for all pairs $(a,b)$ where $a$ and $b$ are coprime, the condition above implies that $\xi$ belongs to the normalizer of $H$.

Here is the brief argument: having fixed $\xi$, let
$M=\{(\alpha,\beta)\in\mathbf{Z}^2\,\mid\, \xi^{\alpha}H\xi^{\beta} =H\}.$
This set is easily seen to be a subgroup of $\mathbf{Z}^2$. Furthermore, note that $(\alpha,\beta)\in M$ implies that $\xi^{\alpha+\beta}\in H$, which in turns means that $(\alpha+\beta,0)\in M$ and $(0,\alpha+\beta)\in M$.

Hence if $(a,b)\in M$, we get
$(a,-a)=(a,b)-(0,a+b)\in M,\quad\quad (b,-b)=(a+b,0)-(a,b)\in M$,
so that
$M\cap (1,-1)\mathbf{Z}$
contains $(a,-a)\mathbf{Z}\cap (b,-b)\mathbf{Z}$, which is just
$(d,-d)\mathbf{Z},$
where $d=(a,b)$. But $(d,-d)\in M$ means exactly that $\xi^{d}\in N(H)$.

Thus we have got the first implication. Conversely, the conclusion means exactly that
$(a+b,0)\in M,\quad (d,-d)\in M.$
But then
$(a,b)=(a+b,0)-(b,-b)=(a+b,0)-b/d (d,-d)\in M$
shows that $\xi^a H\xi^b=H$.

To finish, how did I get to this situation? This can arise quite naturally as follows: one has a collection $X$ of representations $\rho$ of a fixed group $\Gamma$, and an action of a group latex $G$ on these representations $X$ (action up to isomorphism really).
For a given representation $\rho_0$, we can then define a group
$H=\{g\in G\,\mid\, g\cdot \rho_0\simeq \rho_0\}$
and also a subset
$T=\{g\in G\,\mid\, g\cdot \rho_0\simeq D(\rho_0)\},$
where $D(\cdot)$ denotes the contragredient representation.

It can be that $T$ is empty, but let us assume it is not. Then $T$ has two properties: (1) it is a coset of $H$ (because $H$ acts on $T$ simply transitively); (2) we have $g^2\in H$ for all $g\in T$ (because the contragredient of the contragredient is the representation itself).

This means that, for some $\xi\in G$, we have $T=\xi H$, and furthermore
$\xi H\xi=H$
since $\xi g\xi g\in H$ for all $g\in H$.

By the previous discussion, we therefore get a good handle on the structure of $T$: either it is empty, or it is of the form $\xi H$ for some $\xi\in G$ such that $\xi^2\in H$ and $\xi$ normalizes $H$ in $G$. In particular, if $H$ is trivial (which happens often), either $T$ is empty, or it consists of a single element $\xi$ which is an involution of $G$.

25.03.2014

### All Hail the distinguished achievement professor!

Filed under: Mathematics @ 13:55

Mr. Quomodocumque is probably too modest to mention it himself, so let me be the first mathematics blogger to congratulate Jordan Ellenberg on becoming a Vilas Distinguished Achievement Professor! Which hopefully comes with a lot of free time to visit Switzerland…

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