# A Primate’s Post

About a year ago I started watching videos of a course on stress called Stress and Your Body, produced by The Great Courses, and given by Robert Sapolsky. I would say it is mid-level, in that the lectures are far from technical, and yet quite a bit of the neurobiology is explained. For some reason  I didn’t take the viewing seriously until a couple of weeks ago. But then I decided I really wanted to know about stress, and watched a lecture every day, until I have now finally finished the course.

And it was brilliant. Saplosky is fun. For example, he tells us that he really hates Wagner’s music, “can’t stand him!”, in order to give an example of psychological stress for him specifically. And then he goes on to talk about the sympathetic and parasympathetic nervous systems and the specific biological effects of his stress – even more fun.

The one thing that I keep finding more evidence for is that the brain is essentially a prediction-reaction box. It just predicts all the time, and reacts in a way that it predicts will make for better predictions later. As in, there isn’t necessarily a “good” state to want – other than the ability to predict itself. But maybe I keep “finding” this since I’m already predicting it to be true…

Figuring out that I really liked this professor, I decided to also listen to his (audio*)book ‘A Primate’s Memoir: A Neuroscientist’s Unconventional Life Among the Baboons‘. In this instance there was no long neglect, and I finished listening in two weeks (mostly during walks and car drives). I highly recommend the book, but only as a fun memoir of a neuroscientist who worked in Africa. It is not a book on Sapolsky’s work, and by far most of the chapters are not about his baboon troop. It does not reflect an objective view of Africa or anything like that. I say this since some bad reviews on Amazon expected the opposite.

Most of all, I like that the book has a large number of characters. Second to that, is the contrast between the cultures (the many cultures – I’m not implying that there are only two). On the one hand I feel I know more about Kenya now (maybe Kenya of twenty years ago). On the other, I realise that if anything, I only know of Sapolsky’s Kenya. But I’m satisfied with that for now. Finally, listening to long descriptions of behaviour of baboons, with clear personalities, friendships, rivalries, is quite the eye opener.

* I’ll note that the audiobook, narrated by Mike Chamberlane, has a good pace and fun accentuation. I don’t know if this is really how it was, but I enjoyed his soft and slow rendering of the locals speech.

# A Master’s Problem

I’d like to tell you about my masters thesis problem, given to me by my advisor David Soudry. It is essentially about translating the Descent construction of Ginzburg-Rallis-Soudry to the case of finite fields. Their work, over local and global fields, has had a major impact on the field of Langlands Functoriality. Here’s one of their main results, in its simplest form possible, as they have proved in greater generality:

Theorem (Ginzburg-Rallis-Soudry ’99). Let $\pi$ be an irreducible, supercuspidal, self-dual representation of $GL_{2n}(\mathbb{Q}_p)$ such that $L^s(\pi,\Lambda^2,s)$has a pole at $0$. Then there exists an explicit construction of a representation $\sigma_\pi$ of $\widetilde{Sp}_{2n}(\mathbb{Q}_p)$ with the right (according to the Langlands conjectures) gamma factor.

One of the main reasons why descent is so powerful is that it can give us lots of explicit information on endoscopic representations, without using proxies such trace formulas or converse theorems.

It is not hard to figure out what the theorem should be over finite fields. In this post I will explain, and prove a bit of, the smallest case.

Fix an odd prime power $q$. For convenience (mine mostly) we will present everything with matrices. Let

$J = \begin{pmatrix} &&&1\\ &&1&\\ &1&&\\ 1&&& \end{pmatrix}$

and let $\hat{G}:=SO_4(\mathbb{F}_q)=\{g\in GL_5(\mathbb{F}_q)\ |\ {}^t gJg=J\}$. We have the usual parabolic subgroup with Levi subgroup isomorphic to $G:=GL_2(\mathbb{F}_q)$:

$P = \{\begin{pmatrix} g & M \\ & J {}^tg^{-1} J \end{pmatrix}\ |\ g\in GL_2(\mathbb{F}_q),\ M\in M_{2}(\mathbb{F}_q),\ {}^t(Jg^{-1}M)=-Jg^{-1}M \}$.

We will also need a specific vector, and my advisor’s favourite is $v= {}^t(0,1,1,0)$. Then it is easy to see that the subgroup of $SO_4(\mathbb{F}_q)$ fixing $v$ (* I like to multiply from the left) is isomorphic to $H:=SO_3(\mathbb{F}_q)$. Denote by $\textup{Irr}_{sd}(G)$ the irreducible self dual representations (well, isomorphism classes) of $G$.

Definition For $\tau\in \textup{Irr}_{sd}(G)$, extend to $P$, induce to $\hat{G}$, and take the component of least dimension, which we’ll call $\pi(\tau)$. The descent of $\tau$ is the restriction $\textup{res}_H^{\hat{G}}\ \pi(\tau)\in \textup{Reps}(H)$.

The part where we “take the component of least dimension” is analogous to taking the residue of Eisenstein series in the original Ginzburg-Rallis-Soudry construction. With this definition, we have the following result:

Theorem The map $\tau\mapsto \textup{res}_H^{\hat{G}}\ \pi(\tau)$ is a bijection between the $\textup{Irr}_{sd}(G)$ and $\textup{Irr}(H)$, that preserves dimension. Moreover, for each such $\tau$ there is a $s\in Sp_2(\mathbb{F}_q)\le G$, such that $\tau$ is in the Lusztig family of $s$, and then $\textup{res}_H^{\hat{G}}\ \pi(\tau)$ is in the Lusztig family of $s$, considered as an element of the dual of $H$, which is isomorphic to $Sp_2(\mathbb{F}_q)$.

It is easy to compute the descent of the trivial representation. Indeed, the induction to $\hat{G}$ will contain the trivial representation, and so $\pi(\tau)=1$, giving that the descent is the trivial representation of $H$. Great!

Let’s compute one general case, that of “principal series” in general position. These are the representation of the form $\tau=\textup{Ind}_B^G(\theta\times\theta^{-1})$, where $\theta^2\ne1$. The $\theta^{-1}$ comes from the self-dual condition. Note that the existence of such $\theta$ implies that $q>3$. So we’ll assume that. ($B$ is a Borel subgroup.)

At this point I want to start using Deligne-Lusztig theory. So $\tau=R_{T_{1^2}}^G(\theta\times\theta^{-1})$. The induction, by transitivity of Deligne-Lusztig inductions, is $\rho:=R_{T_{1^2}}^{\hat{G}}(\theta\times\theta^{-1})$. Here $T_{1^2}$ is the obvious torus rationally isomorphic to $\mathbb{G}_m^2$, in $G$ or $\hat{G}$ depending on the context.

The element $J$, which can thought of as in the Weyl group of $\hat{G}$, fixes $T_{1^2}$, and also $\theta\times\theta^{-1}$. It is to go over the four Weyl elements and see that $1,J$ are the only such elements. So $(\rho,\rho)=2$, by a basic result in Deligne-Lusztig theory, Theorem 6.8 in [DL].

We also have the virtual character $\rho':=-R_{T_{2}}^{\hat{G}}(\theta\circ\textup{det})$. By the same argument as above, $(\rho',\rho')=2$. Going to a quadratic extension, the two tori become rationally conjugate, and it is a simple computation to show that $(\theta\times\theta^{-1})\circ\textup{Norm}$ and $\theta\circ\textup{det}\circ\textup{Norm}$ are conjugate. By standard results we get that $\rho$ and $\rho$ must share at least one component. Since also $(\rho,\rho')=0$, we must have that

$\rho=W+\pi,\ \rho'=W-\pi$

for two irreducible proper characters $W, \pi$. But we can also calculate the dimension of $\rho$ and $\rho'$, using Theorem 7.1 of [DL]:

$\rho(1)=\frac{|\hat{G}|_{p'}}{|T_{1^2}^F|}=\frac{(q^2-1)^2}{(q-1)^2} = (q+1)^2$
$\rho'(1)=\frac{|\hat{G}|_{p'}}{|T_{2}^F|}=\frac{(q^2-1)^2}{q^2-1} = q^2-1$

Since the dimension of $\rho'$ is positive, $\pi$ must have dimension smaller than $W$, and we get

$\pi=\frac{\rho-\rho'}{2}$
$\pi(1)=\frac{\rho(1)-\rho'(1)}{2}=q+1$

Alright! Let’s finish with a computation of the character at another semi-simple element. This means that there is an $a\in \mathbb{F}_q^\times$ such that $a^2\ne 1$. Choose such, and let

$g=\begin{pmatrix} a&&& \\ &1&& \\ &&1& \\ &&&a^{-1} \end{pmatrix}\in H$

It is easy to see that $g$ is not conjugate to any element of $T_{2}$, so by the character formula for Deligne-Lusztig characters, Theorem 4.2 in [DL], we get

$\pi(g)=\frac{1}{2}R_{T_{1^2}}^{\hat{G}}(\theta\times\theta^{-1})(g)=\theta(a)+\theta^{-1}(a)$

Looking at the character table of $H\cong PGL_2(\mathbb{F}_q)$, for example here, we can see that there is exactly one representation of dimension $q+1$ that can have the above character value. Working with Deligne-Lusztig characters, this representation’s character must be

$\textup{res}_H^{\hat{G}}\ \pi(R_{T^2}^G(\theta\times\theta^{-1}))=R_{T_{1^2}}^H(\theta\times\theta^{-1}).$

Which is exactly what one would expect.

My thesis work is to prove the analogue of the above result to the general case, but only for those $\tau$ that descend to cuspidal representations that are associated to the Coxeter torus. The strategy of the proof works for all irreducible self-dual representations that are not unipotent, and I am preparing an article on this.

The computation of descent for unipotent characters is a bit weird, and I’m not sure what the correct result even is. I will sometime soon write a post on computations with $GL_4$, in which I will also define descent in general.

—————————

[DL] P. Deligne, G. Lusztig, Representations of reductive groups over finite fields, Ann. of Math. (2) 103, no. 1, 103?161. (1976).

# Deterministic Square Rooting

A thought on deterministic square root extraction in finite fields had crossed my mind. In a recent paper by Tsz Wo Sze, “On taking square roots without quadratic nonresidues over finite fields”, Mathematics of Computation, volume 80, number 275, essentially, an algorithm is given that works in time polynomial in $\sum_{q|p-1}q$, where the field we are working in is $\mathbb{F}_p$.

The only other deterministic square root algorithm is Schoof’s (that I’m aware of). As Tsz Wo Sze comments, if we could keep taking square roots: $-1, \sqrt{-1}, \sqrt[4]{-1}, \dots$, we would end up with a non-residue, which can be used in the Tonelli-Shanks algorithm. So, I was thinking, why not try the obvious generalisation of Schoof’s algorithm using the hyperelliptic curves $C_s$:

$y^2=x^{2^s}+1$

where $2^s || p-1$, so that the Frobenius endomorphism can be expressed in $End(C_s)\cong \mathbb{Z}[\zeta_{2^s}]$, and some algebra gives us an explicit $\sqrt[2^s]{-1}$. We see that if $s$ is bounded by a polynomial function in $\log \log{p}$, we should have a polynomial time deterministic algorithm for square roots in $\mathbb{F}_p$. I don’t know, maybe there’s some kind of catch.

# Large 3-Ranks of Class Groups

Here’s an idea I once had for generating class groups of imaginary quadratic fields with large 3-ranked class group, maybe you can make something of it:

In Rankin Soleng’s paper “Homomorphisms From the Group of Rational Points On Elliptic Curves to Class Groups of Quadratic Number Fields”, J. Number Theory Volume 46, Issue 2, Feburary 1994, homomorphisms from a subgroup of rational points on an elliptic curve to certain class groups are defined.

Say $E$ is an elliptic curve given by the Weierstrass equation

$y^2 = x^3+a_2x^2+a_4x+a_6$

$a_6<0$, and $P=(x_0,y_0) \in E(\mathbb{Q})$ a point that satisfies a small technical condition called primitive, then we send $P$ to
$[(x_0,y_0-\sqrt{a_6})] \in Cl(\mathbb{Q}(\sqrt{a_6}))$
The primitive points form a subgroup of finite index in $E(\mathbb{Q})$.

Soleng uses these homomorphisms to generate families of imaginary quadratic number fields with non-trivial 3-ranked class group, just by using elliptic curves with with non-trivial 3-torsion.

My idea is then to try to build more than one homomorphism into the same class group. For any $n \in \mathbb{Z}$, that is negative enough, we can shift the elliptic curve to get a new Weierstrass equation

$y^2 = (x+n)^3+a_2(x+n)^2+a_4(x+n)+a_6 = x^3+\ldots+(n^3+a_2n^2+a_4n+a_6)$

So the new $a_6$ is the polynomial evaluated at $n$. If this value is a square times the old value, we should get another map from $E$ to the same class group. This is equivalent to $n$ being the $x$ coordinate of a point on the $a_6$-quadratic twist of $E$. Note that this quadratic twist always has the point $(0,1)$. Multiples of this point give candidate values for $n$, of course only finitely many integral.

If we have an elliptic curve, with non-trivial 3-torsion, and with many of these different homomorphisms into the same class group, we should expect the class group to have a large 3-rank.

I tried for a few days to get this to work and gave up after a while.