## The seven types of graduate student applicant

Yes, it’s that time of year again.

1. Hide and Seek: Contacts you every day about the status of their application, then goes on radio silence the moment they receive an offer, never to be heard of again.
2. The No Chancer: Has an offer from Harvard, Princeton, and MIT, but still plans to attend the prospective student day because they fancy a three day holiday in (wherever your university is located).
3. The Copy & Paster: It has always been my dream to attend Michigan University, the best university in the world. Well, good luck with that.
4. The Nervous Nellie: Has some questions about the graduate program — a lot of questions. Wants you (that is, me) to answer detailed questions about everything from the reasonable (exact duties of a TA, particulars on graduate student stipend and health insurance) to the less so (graduation statistics and data for the last 10 years of graduates, upcoming schedule of faculty sabbaticals for the next three years, tips on the best place in Evanston to purchase toothpaste, etc.).
5. The Surprise: Never responds to any email query about whether they are interested in coming or whether they have offers from somewhere else, is completely discounted by the committee, but then ends up accepting on April 15.
6. The Googler: Makes an effort to look at the department website to customize their application, but gets it all wrong: I would really like to work with X, Y, and Z where X is a postdoc, Y has retired, and Z moved to a different institution two years ago.
7. The Unicorn: Actually accepts the offer well before April 15.

Tell me if I’ve missed anyone.

Posted in Mathematics, Rant | | 11 Comments

## Only Harvard Grads need apply

It’s hard to take articles in Slate too seriously, but I have to admit I was quite perplexed about the following article (with the concomitant research publication here).

The main thrust of the article seems to be as follows. A disproportionate number faculty at research universities in the US received PhDs from a small number of prestigious institutions, and hence (?) such hiring practices reflect profound social inequality. Is it just me, or does this appear to be utter bollocks? There is an obvious pair of hypotheses that would completely explain the data, namely:

2. Universities hire the strongest candidates they can, and admit the strongest graduate students they can.

Let’s examine these possibilities in the context of graduate school in mathematics. I have, on several occasions, been responsible for graduate admissions at my institution. I would say, on the whole, that prospective graduate students are among the most class conscious of anyone in academia. I would guess that, at least 75% of the time, a student will accept either the program that is the most highly ranked amongst those where they were admitted or a school within at most one or two places of their highest ranked option.

What about the second hypothesis? The worry here is that universities might view “undergraduate/graduate institution” as a proxy for “quality of candidate.” In my experience (being on hiring committees), this is utterly preposterous. I am not claiming that mathematical judgements are not a slippery thing — there are many variations which relate to matters of taste and inclination — but there are some reasonable objective criteria (GRE scores for graduate applications, publication record for job candidates) which would serve as a check against any implicit bias in this regard.

We here at Persiflage, however, are open to the idea that we may have missed something. So here are some other possibilities:

1. You are talking about Mathematics, a field for which it is easier to make reliable judgements about the quality of research, and a field for which there is a more pronounced spike in talent at the top of the scale. Is this true? I honestly don’t know. Perhaps whatever field it is that produces papers like the one under consideration is not something for which talent of any kind is an asset, and so there is no real difference between graduates from Harvard or from Podunk U. Less sarcastically, suppose (say) I compare the English department faculty at the top ranked place (taking from this list) Berkeley and compare it to a place also ranked in the top 25 schools but closer to the bottom of that list, say UIUC. Then, if I knew something, could I confidently say that one department is much better than the other?
2. You are talking about the experience of hiring/admitting students at a Group I university. Perhaps it is the case that, for lower ranked universities, there is insufficient expertise to hire on the basis of talent/output, and so PhD institution serves as a lazy way to evaluate the candidate. This seems to be a somewhat condescending argument, but it’s true that I don’t have any idea how hiring works at non-Group I universities. But surely the letters of recommendation would carry the most weight, and they would reflect the quality of research? At the very least, if you are going to claim this is what happens, you need to come up with a way to substantiate that claim.

Ultimately, I certainly don’t feel that I can rule out bias when it comes to hiring, but the fact that the paper under review uses “prestige” as a dirty word and doesn’t seem to acknowledge in any way that there is some correlation between prestige and quality of graduates is highly disturbing. Perhaps, as with this paper, the main goal is to substantiate the political beliefs of the authors rather than to undertake a serious academic inquiry. Still, even if the methodology is flawed, I would like people’s opinion on the conclusion.

Posted in Politics, Rant | Tagged , | 9 Comments

## Review of Buzzard-Gee

This is a review of the paper “Slopes of Modular Forms” submitted for publication in a Simons symposium proceedings volume.

tl;dr: This paper is a nice survey article on questions concerning the slopes of modular forms. Buzzard has given a (very explicit) conjecture which predicts the slopes of classical modular ($p$-stabilized) eigenforms of level prime to $p,$ at least under a certain regularity hypothesis. One consequence is that, under favourable circumstances, all the slopes are integers. The current paper describes the link between this and related problems to the $p$-adic Langlands program, as well as raising several further intriguing questions concerning the distributions of these slopes. The paper is well written, and is a welcome addition to the literature. I strongly recommend that this paper be accepted.

Review: Buzzard’s slope conjectures live somewhere in the world between 19th and 21st century mathematics. Suppose that one considers the space of over-convergent cusp forms of level $N = 1$ for $p = 2.$ Then, using nothing more than classical identities between modular functions, one may prove that the smallest eigenvalue of the compact operator $U_2$ is at most $\|2^3\|_2 = 1/8.$ On the other hand, it is now a “folklore” conjecture (Conjecture 4.1.1 of the paper under review) that, if $p$ is odd and $f$ is a classical modular form of level $\Gamma_0(N)$ prime to $p$, then the residual representation:

$\overline{\rho}: G_{\mathbf{Q}} \rightarrow \mathrm{GL}_2(k)$

is irreducible locally on the decomposition group at $p$ whenever the valuation of $a_p$ is not an integer. This problem seems to be a deep question in the p-adic Langlands program for $\mathrm{GL}_2(\mathbf{Q}_p).$ The two cases where this is known, $v(a_p) < 1$ and $v(a_p)$ sufficiently large, both require machinery from p-adic Hodge theory — in the former case, one needs the full local Langlands correspondence for $\mathrm{GL}_2(\mathbf{Q}_p).$

Comments: Here are some comments given in some order that bears little relation to the actual paper.

• I don’t like the table on page 6, in particular, because certain ranges of numbers are bunched together, the output looks a little strange. Can one improve this in some way? Perhaps finish at $3^{10}?$ Perhaps include only selected powers bigger than $3^{10}?$ Perhaps normalize for the length of the range?
• Corollary 5.1.2. Do you want to speculate on what happens in the reducible case? In some sense, in Buzzard’s conjecture, one doesn’t see the fact that the residual representations are globally reducible or not. On the other hand, weird stuff certainly happens for $p = 2,$ as previously mentioned here. What happens in the reducible case for $p = 3?$
• Conjecture 4.1.1 demands that $k$ is even, but that is a consequence of the level being of the form $\Gamma_0(N)$ — something which is noted immediately after the statement. So why include the condition in the statement of the conjecture? Also, perhaps it’s also worth remarking upon the case when $a_p$ is a unit.
• The authors (in Remark 4.1.3) point to the origins of this Conjecture 4.1.1 to around 2005. However, I feel like I remember some discussion of this conjecture in the Durham symposium of 2004. There were certainly hints of this conjecture on «la serviette de Kisin», upon which Mark gave a heuristic local argument for why the Eigencurve was proper — although the argument was slightly dodgy in that it collapsed if the napkin was rotated 90 degrees. (Of course, Mark was proved right when Hansheng Diao and Ruochuan Liu did indeed prove this result using local methods here.) Also, isn’t Conjecture 4.1.1 a consequence of Buzzard’s original conjecture as modified by Lisa Clay? Somehow it seems to me that what Remark 4.1.3 is referring to is the idea that Conjecture 4.1.1 is a consequence of a purely local conjecture, and refers to the period (2005?) when Breuil was formulating the first versions of the p-adic Langlands program.
• For Conjecture 4.2.1, wouldn’t it make more sense to normalize the valuation in terms of the coefficient field $\mathbf{Q}_p(\chi),$ so the statement once more becomes that $a_p$ has integral valuation?
• Why is the condition on Buzzard’s conjecture different when $p = 2?$ (I understand it has to be modified in order to have a chance of being true, but I am asking if there is any explanation for why this is necessary.)
• The authors remark (p.3) that it is not known whether there are infinitely many Buzzard irregular primes. Here is a short argument to prove that this is a consequence of standard conjectures of prime values of polynomials. We start with the observation that the first Buzzard irregular prime is $p =59,$ and that the offending representation:

$\rho_{Q \Delta} : G_{\mathbf{Q}} \rightarrow \mathrm{GL}_2(\mathbf{F}_{59})$

has exceptional image in the context of Serre and Swinnerton-Dyer (On l-adic representations and congruences for coefficients of modular forms, Antwerp III). Indeed, this particular example features prominently in that paper. I always thought this was not an entirely random coincidence, and since it seems relevant here, I thought I would finally bother to figure out what is going on. (For the next prime, $p = 79,$ the corresponding representation has image containing $\mathrm{SL}_2(\mathbf{F}_{79}),$ so it is somewhat of an accident.) The mod-59 representation above has projective image $S_4.$ Now suppose that $p \equiv 3 \mod 4$ is a prime such that $H/\mathbf{Q}$ is an $S_4$-extension which is unramified away from $p.$ Such a representation will give rise (following an argument of Tate) to a mod-$p$ representation of $\mathbf{Q}$ which is unramified away from $p.$ The congruence condition on $p$ implies that it will be odd, and hence modular, by Langlands-Tunnell. Now let us suppose, in addition, that $4$ divides the ramification index $e_p.$ Under this assumption, the representation cannot be locally reducible, because the ramification index of any power of the cyclotomic character divides $p-1 \equiv 2 \mod 4.$ Hence, if there are infinitely many such fields, there are infinitely many $\mathrm{SL}_2(\mathbf{Z})$-irregular primes. Consider the following fields studied by Darrin Doud (here):

$\displaystyle{K = \mathbf{Q}[u]/f(u), \quad f(u) = (u + x)^4 - p^* u},$

$\displaystyle{(-1)^{(p-1)/2} p = p^* = \frac{256 x^3 - y^2}{27}, \quad p^* \ne 1 + 4x}.$

Doud shows that $K$ has discriminant $(p^*)^3$ and Galois group $S_4,$ and it is easy to see that the splitting field $H/\mathbf{Q}$ has all the required properties needed above as long as $p \equiv 3 \mod 4.$ The formula relating tame ramification and the discriminant implies that $e_p(K/\mathbf{Q}) = 4.$ Standard conjectures now predict that there are infinitely many such primes of this form — if $y$ is odd, then $p^* < 0.$ The first few such primes which are $3 \mod 4$ are

$59, 107, 139, 283, \ldots$

which compares with the first few Buzzard irregular primes (taken from Buzzard’s paper):

$59, 79, 107, 131, 139, 151, 173 \ldots$

On the other hand, an unconditional proof by these means seems out of reach, because any modular $S_4$-extension unramified outside $p \equiv 3 \mod 4$ forces $\mathbf{Q}(\sqrt{-p})$ to have class number divisible by $3,$ and we don’t know if there exist infinitely many such primes.

## Random Photos

Lunt Hall, the Northwestern mathematics building,  recently underwent an upgrade of the fire alarm system. This includes introducing informative new signage, such as the following:

Another change is that the “internal” window to my office was boarded up. The window overlooked the internal stairwell and had previously been covered up with posters, but I always had the idea of removing them and using the window for office hours; students would look up at the window and ask questions, and I would offer them Delphic pronouncements in return. Alas, I will no longer be able to do this. Here’s a picture from the brief period between when the posters were removed and the window was boarded up; the photo is taken from the top of the stairs; the spot directly next to the window is a couple of feet lower:

Finally, I haven’t yet managed to get around to blogging about the Sarnak conference! So here at least is a picture, which is an action shot of me advocating (unsuccessfully, it turns out) for a certain paper to be accepted into Duke. (Pictured are Jonathan Wahl (Duke supremo), RLT, and me.)

Finally, here is an amusing box of free books I occasionally pass when I walk to work:

## H_2(Gamma_N(p),Z)

In this post (which is a follow-up to the last post), I wanted to compute the group $H_2(\Gamma_N(p),\mathbf{Z})$, where $\Gamma_N(p)$ is the congruence subgroup of $\mathrm{SL}_N(\mathbf{Z})$ for large enough $N$ and $p$ is prime. In fact, to make my life easier, I will also assume that $p > 3,$ and in addition, ignore $2$-torsion. The first problem is to compute the prime to $p$ torsion. By Charney’s theorem, this will come from the cohomology of the homotopy fibre $X$ of the map

$\displaystyle{ SK(\mathbf{Z}) \rightarrow SK(\mathbf{F}_p).}$

The relevant part of the Serre long exact sequence is, using classical computations of the first few K-groups of the integers together with Quillen’s computation of $K_*(\mathbf{F}_p),$

$\displaystyle{ 0 \rightarrow \pi_3(X) \rightarrow \mathbf{Z}/48 \mathbf{Z} \rightarrow \mathbf{Z}/(p^2 - 1) \mathbf{Z} \rightarrow \pi_2(X) \rightarrow \mathbf{Z}/2 \mathbf{Z} \rightarrow 0.}$

Here is where it is convenient to invert primes dividing $6;$ from Hurewicz theorem and Charney’s theorem we may deduce that, where $\sim$ denotes an equality up to a finite group of order dividing $48,$

$\displaystyle{ H_2(\Gamma_N(p),\mathbf{Z}[1/p]) \sim \mathbf{Z}/(p^2 - 1) \mathbf{Z}.}$

In order to deal with $3$-torsion, then we also have to show that the map $K_3(\mathbf{Z}) \otimes \mathbf{Z}/3 \mathbf{Z} \rightarrow K_3(\mathbf{F}_p)$ is injective for $p \ne 3.$ I have a sketch of this which I will omit from this discussion but it is not too hard (assuming Quillen-Lichtenbaum). It remains to compute the homology with coefficients in $\mathbf{Z}_p$. I previously computed that there was an isomorphism

$\displaystyle{ H_2(\Gamma_N(p),\mathbf{F}_p) = \wedge^2 \mathfrak{g} \oplus \mathbf{F}_p = H_2(G_N(p),\mathbf{F}_p) \oplus \mathbf{F}_p,}$

where $G_N = \mathrm{SL}_N(\mathbf{Z}_p)$ and $\mathfrak{g} = H_1(\Gamma_N(p),\mathbf{F}_p)$ is the adjoint representation.

Some facts concerning the cohomology of $G_N(p)$:

There are short exact sequences:

$\displaystyle{ 0 \rightarrow H_2(G_N(p),\mathbf{Z}_p)/p \rightarrow H_2(G_N(p),\mathbf{Z}/ p \mathbf{Z}) \rightarrow H_1(G_N(p),\mathbf{Z}_p)[p] \rightarrow 0,}$

$\displaystyle{ 0 \rightarrow H_2(G_N(p),\mathbf{Z}_p)/p^2 \rightarrow H_2(G_N(p),\mathbf{Z}/ p^2 \mathbf{Z}) \rightarrow H_1(G_N(p),\mathbf{Z}_p)[p^2] \rightarrow 0.}$

Since $H_1(G_N(p),\mathbf{Z}_p) = \mathfrak{g}$ is annihilated by $p,$ we may deduce that

$\displaystyle{ H_2(G_N(p),\mathbf{Z}_p)/p = H_2(G_N(p),\mathbf{Z}_p)/p^2}$

as long as

$\displaystyle{ |H_2(G_N(p),\mathbf{Z}/p^2 \mathbf{Z}) | = | H_2(G_N(p),\mathbf{Z}/ p \mathbf{Z})|.}$

Such an equality (for any group) is a claim about the Bockstein maps having a big an image as possible. Indeed, for any group $\Phi,$ there is an exact sequence:

$\displaystyle{ H_3(\Phi,{\mathbf{Z}}/p {\mathbf{Z}}) \rightarrow H_2(\Phi,{\mathbf{Z}}/p {\mathbf{Z}}) \rightarrow H_2(\Phi,{\mathbf{Z}}/p^2 {\mathbf{Z}}) \rightarrow H_2(\Phi,{\mathbf{Z}}/p {\mathbf{Z}}) \rightarrow H_1(\Phi,{\mathbf{Z}}/p {\mathbf{Z}})}$

The first and last maps here are the Bockstein maps $\beta_2$ and $\beta_1$. Since $p$ is odd, $\beta_1 \circ \beta_2 = 0$. On the other hand, we see that the orders of the cohomology groups with coefficients in $\mathbf{Z}/p \mathbf{Z}$ and $\mathbf{Z}/p^2 \mathbf{Z}$ will have the same order if and only if

$\displaystyle{ \ker(\beta_1) = \mathrm{im}(\beta_2).}$

Hence we have reduced to the following claim. Take the complex

$\displaystyle{ H_*(G_N(p),\mathbf{F}_p) = \wedge^* \mathfrak{g} }$

where the differentials are given by the Bockstein maps. Then we have to show that the cohomology of this complex vanishes in degree two. But what are the Bockstein map is in this case? Note that since $H_1(G_N(p),\mathbf{Z}_p) = \mathfrak{g}$ is annihilated by $p$, the Bockstein map $\beta_1$ will be a surjective map:

$\displaystyle{ \beta_1: \wedge^2 \mathfrak{g} \rightarrow \mathfrak{g}.}$

To compute this explicitly, recall that the isomorphism $H_1(G_N(p),{\mathbf{Z}}_p) = \mathfrak{g}$ comes from the identification of $\mathfrak{g}$ with $G_N(p)/G_N(p^2).$ Then, computation omitted due to laziness, we find that the Bockstein is precisely the Lie bracket. Moreover, since the (co-)homology is generated in degree one, the higher Bockstein maps can be computed from the first using the cup product formula. So the Bockstein complex above is, and I haven’t checked this because it must be true, the complex computing the mod-$p$ Lie algebra cohomology of $\mathfrak{g}$. And this cohomology vanishes in degrees one and two, so we are done. One consequence of this computation is that

$\displaystyle{ H_2(G_N(p),\mathbf{Z}_p) = H_2(G_N(p),\mathbf{Z}_p)/p}$

is annihilated by $p.$ Moreover, the last term can be identified with the kernel of the Lie bracket (Bockstein) on $H_2(G_N(p),\mathbf{F}_p) = \wedge^2 \mathfrak{g}.$

Returning to the main computation:

From the Hochschild–Serre spectral sequence and the computation of stable completed cohomology, one has an exact sequence:

$\displaystyle{ 0 \leftarrow H_2(G_N(p),\mathbf{Z}_p) \leftarrow H_2(\Gamma_N(p),\mathbf{Z}_p) \leftarrow \mathbf{Z}_p \leftarrow H_3(G_N(p),\mathbf{Z}_p).}$

From known results in characteristic zero, we immediately deduce that there is some $\alpha$ such that there is an exact sequence

$\displaystyle{ 0 \leftarrow H_2(G_N(p),\mathbf{Z}_p) \leftarrow H_2(\Gamma_N(p),\mathbf{Z}_p) \leftarrow \mathbf{Z}/p^{\alpha} \mathbf{Z} \leftarrow 0.}$

we also deduce that there is an exact sequence:

$\displaystyle{ 0 \leftarrow H_2(G_N(p),\mathbf{Z}/p^n \mathbf{Z}) \leftarrow H_2(\Gamma_N(p),\mathbf{Z}/p^n \mathbf{Z}) \leftarrow \mathbf{Z}/p^{\mathrm{min}(\alpha,n)} \mathbf{Z} \leftarrow 0,}$

There are spectral sequences:

$\displaystyle{ H_i(\mathrm{SL}_N(\mathbf{F}_p),H_j(\Gamma_N(p),A)) \Rightarrow H_{i+j}(\Gamma_N,A)}$

for $A = \mathbf{Z}/p^n \mathbf{Z}$ and $A = \mathbf{Z}_p.$
For both of these rings, we have

$\displaystyle{ H_1(\Gamma_N(p),A) = \mathfrak{g}, \qquad H_0(\Gamma_N(p),A) = A.}$

Moreover, for sufficiently large $N,$ we have

$\displaystyle{ H_i(\mathrm{SL}_N(\mathbf{F}_p),A) = 0,}$

this follows from and is equivalent to Quillen’s computation which implies that the $K$-groups of finite fields have order prime to $p.$ Since $H_2(\mathrm{SL}_N(\mathbf{Z}),\mathbf{Z}_p)$ is trivial for $p > 2,$ we deduce that

$\displaystyle{ H_0(\mathrm{SL}_N(\mathbf{F}_p),H_2(\Gamma_N(p),A)) = H_2(\mathrm{SL}_2(\mathbf{F}_p),\mathfrak{g}) = \mathbf{F}_p,}$

where the last equality was already used in my paper. The compatibility of the spectral sequence above for different $A$ implies that we also get an isomorphism

$\displaystyle{ H_0(\mathrm{SL}_N(\mathbf{F}_p),H_2(\Gamma_N(p),\mathbf{Z}_p)) = H_0(\mathrm{SL}_N(\mathbf{F}_p),H_2(\Gamma_N(p),\mathbf{Z}/p \mathbf{Z})).}$

On the other hand, the invariant class must be an element of order $p$ in

$\displaystyle{ \mathbf{Z}/p^{\alpha} \mathbf{Z};}$

and hence the reduction map

$\displaystyle{ \mathbf{Z}/p^{\alpha} \mathbf{Z} \rightarrow \mathbf{Z}/p \mathbf{Z}}$

sends an element of order $p$ to an element of order $p,$ and so $\alpha = 1.$

Putting things back together:

Assembling all the pieces, we see that we have proven the following:

Theorem: Let $p > 3,$ and let $N$ be sufficiently large. Let $\mathfrak{g}$ be the Lie algebra $\mathfrak{sl}_N$ over $\mathbf{F}_p.$ Then, up to a finite group of order dividing $48,$ we have

$\displaystyle{ \displaystyle{ H_2(\Gamma_N(p),\mathbf{Z}) \sim \mathbf{Z}/(p^2 - 1) \mathbf{Z} \oplus \mathbf{Z}/p \mathbf{Z} \oplus \ker \left( [\ , \ ] \wedge^2 \mathfrak{g} \rightarrow \mathfrak{g} \right)}.}$

Moreover, still with $p > 3,$ then up to a group of order dividing $16,$ we should have the same equality with $p^2 - 1$ replaced by $(p^2 - 1)/3.$

## Stable completed homology without Quillen-Lichtenbaum

Having just made (hopefully) the final revisions on my paper on stable completed cohomology groups, I wanted to record here a few remarks which didn’t otherwise make it into the paper.

The first is that, in addition to the result that $\widetilde{H}_2(\mathrm{SL},\mathbf{Z}_p) = \mathbf{Z}_p$ for $p > 2,$ one may also compute $\widetilde{H}_3(\mathrm{SL},\mathbf{Z}_p)$ for $p > 3.$ Namely:

$\displaystyle{\widetilde{H}_3(\mathrm{SL},\mathbf{Z}_p) = 0.}$

This result is proved in the paper up to a finite group, so the point here is the integral refinement. The computation of $\widetilde{H}_2$ comes from the Hurewicz isomorphism

$\displaystyle{\pi_2(SK(\mathbf{Z},\mathbf{Z}_p);\mathbf{Z}_p) \simeq \widetilde{H}^{\mathrm{cont}}_2(\mathbf{Z}_p)}.$

However, the Hurewicz theorem also gives an epimorphism

$\displaystyle{\pi_3(SK(\mathbf{Z},\mathbf{Z}_p);\mathbf{Z}_p) \rightarrow \widetilde{H}^{\mathrm{cont}}_3(\mathbf{Z}_p)},$

and one finds that the first group lives in an exact sequence

$H^2(\mathbf{Q}_p,\mathbf{Z}_p(2)) \rightarrow \pi_3(SK(\mathbf{Z},\mathbf{Z}_p);\mathbf{Z}_p) \rightarrow K_3(\mathbf{Z}) \otimes \mathbf{Z}_p$

Since both flanking groups vanish for $p > 3$, the middle group is zero, and the claim follows.

The second remark is that, throughout the paper, I assume the Quillen-Lichtenbaum conjecture, which is now a theorem due do Voevodsky and others. However, I must confess, I do not have the fine details of the argument at my fingertips. How much can one say without it? The answer is quite a lot. Due to work of Borel, Soulé, and Quillen (all of which is much more familiar to me, at least relatively speaking), we know that the $K$-groups of number fields are finitely generated abelian groups, we know their ranks, and we know that the Chern class maps to the appropriate Galois cohomology groups are surjective. Moreover, we understand $K_2(\mathscr{O}_F)$ completely in terms of Galois cohomology by work of Tate. (In this game, I am also giving up the results of Hesselholt and Masden on the $K$-theory of local fields, and instead using the results of Wagoner, which similarly give everything in very small degree and up to a finite group in higher degrees.)

In particular:

1. The computation of $H_2(\Gamma_N(p),\mathbf{F}_p)$ for large $N$, where $\Gamma_N(p)$ is the principal congruence subgroup of $\mathrm{GL}_N(\mathbf{Z})$, is unaffected. This also uses the computation of $K_3(\mathbf{Z})$ by Lee and Szczarba.
2. The identification of the completed $K$-groups with Galois cohomology groups still holds up to a finite group.
3. The computation of the rational stable completed homology groups$\widetilde{H}_*(\mathrm{SL},\mathbf{Z}_p) \otimes \mathbf{Q} = \mathbf{Q}[x_2,x_6,x_{10},x_{14}, \ldots ]$under the assumption that either $p$ is regular or $\zeta_p(3), \zeta_p(5), \zeta_p(7)$ etc. are all non-vanishing still holds.

Something that does require Quillen-Lichenbaum is the vanishing of the partially completed $K$-group for very regular primes.

Regarding the computation of the rational stable completed homology groups, the referee made a very interesting point (I will come back in a later post to the refeering of this paper and some other of my recent papers in a post on “what a great referee report should be”). I prove that the rational stable completed homology groups are the continuous homology of the homotopy fibre

$SK(\mathbf{Z},\mathbf{Z}_p) \rightarrow SK(\mathbf{Z}) \rightarrow SK(\mathbf{Z}_p)$

(The definition of $SK(\mathbf{Z},\mathbf{Z}_p)$ is just homotopy fibre of this map.) Now $SK(\mathbf{Z},\mathbf{Z}_p)$ is an infinite loop space, which under the assumption that $p$ is regular or on the non-vanishing of the $p$-adic zeta function at integral arguments, has the property that the homotopy groups with coefficients $\pi_n(SK(\mathbf{Z},\mathbf{Z}_p);\mathbf{Z}_p)$ are rationally non-zero in exactly degrees $2,6,10$, etc. The referee noted that the computation of rational stable completed homology should follow precisely from this description using the Milnor–Moore theorem, which shows that (for simply connected $H$-spaces) that the homology is (rationally) the universal enveloping algebra of the rational homotopy classes (and so, in particular, the Hurewicz map is rationally injective). One consequence is that the rational homotopy groups are precisely the primitive classes in rational homology. To orient the reader, this is exactly the theorem which allowed Borel to compute the rational $K$-groups of (rings of integers) of number fields from his computation of stable homology over $\mathbf{Q}$. Now I was a little worried about this, because the Milnor–Moore theorem does not literally apply, since one is comparing here homotopy groups with coefficients in $\mathbf{Z}_p$ and continuous homology (the latter is just the inverse limit of homology groups modulo $p^n$). However, having looked at the argument in Milnor–Moore and then having Paul Goerss explain it to me, the argument does indeed seem to simply work in this case. (Warning, this is a weaker statement than saying I checked the details.)

To be more precise, suppose that $G$ is a simply connected infinite loop space, and suppose that $G$ has the property that the groups $\pi_n(G;\mathbf{Z}/p^k)$ are finite for all $n$ and $k$, so $\pi_n(G;\mathbf{Z}_p)$ is the inverse limit of these groups. There is a pairing

$[,]: \quad \pi_r(G,\mathbf{Z}/p^k) \otimes \pi_s(G,\mathbf{Z}/p^k)\rightarrow \pi_{r+s}(G,\mathbf{Z}/p^k),$

which, after taking inverse limits in $k$ and tensoring with $\mathbf{Q}$, makes $\pi_*(G,\mathbf{Z}_p) \otimes \mathbf{Q}$ into a Lie algebra over $\mathbf{Q}_p$, then the Hurewicz map will induce an isomorphism

$U(\pi_*(G,\mathbf{Z}_p) \otimes \mathbf{Q}) \rightarrow H^{\mathrm{cont}}_*(G,\mathbf{Q}_p):= \projlim H_*(G,\mathbf{Z}/p^k) \otimes \mathbf{Q}$

of Hopf algebras. The key technical point required here is to define the appropriate pairing on homotopy groups with coefficients, which is done by Neisendorfer. (If $G$ is simply connected infinite loop space, one doesn’t have to worry about the issue of homotopy groups with coefficients in very low degree exhibiting certain pathologies.)

As another example of this, one can take $G = SK(\mathbf{Z}_p)$. In this case, the rational continuous homology reduces, by work of Lazard, to lie algebra cohomology, and gives an exterior algebra in odd degrees $> 1$. So $SK_n(\mathbf{Z}_p;\mathbf{Z}_p) \otimes \mathbf{Q}$ has dimension one in odd degrees $> 1$ and is zero for all even positive degrees. This is a result of Wagoner. In fact, Wagoner proves something slightly stronger, also capturing some information away from $p$. To do this, he also proves a version of the Milnor–Moore theorem, but his assumptions are more stringent than what we discuss above.

Posted in Mathematics | | 1 Comment

## Inverse Galois Problems II

David Zywina was in town today to talk about a follow up to his previous results mentioned previously on this blog. This time, he talked about his construction of Galois groups which were simple of orthogonal type, in particular, the simple groups

$\Omega(V) \subset \mathrm{SO}(V) \subset \mathrm{O}(V)$

where $V$ is a vector space $V$ over $\mathbf{F}_l$ of odd dimension at least five. The group $\Omega(V)$ here is a simple group of index two inside $\mathrm{SO}(V).$ In the special case when $n = 5,$ there is an exceptional isomorphism

$\Omega(V) \simeq \mathrm{PSp}_4(\mathbf{F}_l).$

In constrast to his constructions of number fields with Galois group $\mathrm{PSp}_2(\mathbf{F}_l),$ Zywina actually constructs a family of compatible families whose residual image is generically $\Omega(V).$ When David told me about this construction (scribbled on a piece of paper) in Frankfurt airport on the way back from Oberwolfach, I was troubled by something which I shall now explain. Without saying so much about the construction (you can read about it here), the compatible families of Galois representations of interest occur inside $H^2(X_T,\mathbf{Q}(1))$ for a carefully chosen family of non-isotrivial elliptic sufaces $X.$ As David explained in his talk today, the zero section and the fibres of bad reduction contribute a large Galois trivial summand to $H^{1,1},$ and the remaining piece is five dimensional. What disturbed me at the time was that this construction was surely liftable to a compatible family of four dimensional representations with generalized symplectic image. After all, Tate’s result on $H^2(G_{\mathbf{Q}},\mathbf{C}^{\times})$ guarantees that one can lift any projective representation with image in $\mathrm{PSp}_4(\mathbf{F}_l)$ to a genuine generalized symplectic representation. This representation should then come from a Siegel modular form, since all oddness conditions should be automatic. On the other hand, if you want a family of Galois representations giving rise to a family of Siegel modular forms, especially one for which the maximal difference between any two Hodge-Tate weights in $\wedge^2 W$ is two, then you expect that they have to come from a family of abelian surfaces, or at least abelian varieties $A$ of dimension $2n$ with endomorphisms by the ring of integers in a totally real field of degree $n.$ However, there is an obstruction to making this work — the corresponding Galois representations will have Hodge-Tate weights $[0,0,1,1],$ and they will have similitude character that is an even finite order character times the cyclotomic character. It’s easy to see that for such a family, the residual representations will (at least half the time) land in $\mathrm{PGSp}_4(\mathbf{F}_{l})$ and not in the simple index two subgroup, similar to what happens for modular forms of weight two. I thought at the time that I must have been making some group theory error, so after today’s talk we sorted out the details:

In the process of this computation, however, I realized what my error actually was. I was imagining that the original compatible family of Galois representations in $H^2$ had Hodge-Tate weights $[0,1,1,1,2],$ but they could equally have had Hodge-Tate weights $[0,0,1,2,2].$ And in this latter case, the Galois representation (up to twist) of the corresponding Siegel modular form in $\mathrm{GSp}_4$ will have Hodge-Tate weights $[-1,0,0,1].$ In particular, we are not looking for classical Siegel modular forms of low weight, but the nasty Siegel modular forms which do not contribute to holomorphic limits of discrete series and only occur in coherent cohomology via $H^1$ or $H^2.$ (A reference for this fact is George Boxer’s talk in Barbados.) And now everything makes sense! That is, if you have a family of Galois representations with Hodge Tate weights $[-1,0,0,1]$ and quadratic similitude character, then (with some good luck) you can really have projective representations which land in the right simple group for all but finitely many $l.$

A related point: when lifting projective representations using Tate’s theorem, one may have to increase the size of the residue field. In fact, when $\ell \equiv 3 \mod 4,$ it will not be possible to lift an odd $\mathrm{PSp}_4(\mathbf{F}_l)$ representation to one in $\mathrm{GSp}_4(\mathbf{F}_l)$ (there is an obstruction at infinity). Indeed, the natural lift is the group $\mathrm{Sp}_4(\mathbf{F}_l)$ together with a scalar matrix $I$ with $I^2 = -1.$ This suggests what the picture should be motivically: there should be an eight dimensional piece of $H^2(Y)$ (for some $Y)$ which admits an involution breaking the representation up into two four dimensional pieces, and these pieces will have coefficients in $\mathbf{Q}(\sqrt{-1}).$ Can one find such a $Y$ explicitly? This does remind me of the motivic lifting problems that Stefan Patrikis knows about.

From this analysis, it also becomes clearer why Zywina could find a family of compatible families with residual image $\Omega(V)$ when $\mathrm{dim}(V)$ is odd and at least five, but only isolated examples of compatible families with residual projective image $\mathrm{PSL}_2(\mathbf{F}_l) \simeq \Omega(V)$ with $\mathrm{dim}(V) = 3.$ In the latter case, the corresponding modular forms will be forced to have odd weight $k > 1,$ and so the Hodge-Tate weights will differ by at least two, and so Griffiths’ theorem implies that they should not deform in a family. On the other hand, if you want to look for Siegel modular forms which could possibly correspond to geometric families, and you want the similitude character to be an even power of the cyclotomic character times a finite character, then it is possible to escape the specter of Griffiths on your shoulder, but only barely — you will be pretty much forced to work with forms whose HT weights are $[-1,0,0,1].$ Of course, I’m not sure I can prove that any Siegel modular forms of this kind actually exist! (insisting the Mumford-Tate group is big, naturally). My proposal in the previous post to look for these representations using Siegel modular forms would also have only found sporadic compatible families, because to ensure computability and the determinant condition I suggested looking in weights where the Galois representation was regular and had HT weights something like $[0,1,3,4],$ — the gap being necessary to make the multiplier character a square of a Hodge-Tate character.

There is one check left on these musings (though I’m sure it must be correct), namely, that for the surface $X$ in 1.4 of Zywina’s paper, one should have

$h^{2,0}(X) = 2.$

Update: Here’s a proof of this statement:

Proof: The Hodge diamond of a minimal elliptic surface $\pi: X \rightarrow C$ was computed by Miranda, see IV.1.1 here; I’ll try to give a self contained argument. Let $\omega_E$ be the Euler characteristic of $\mathcal{O}_X.$ Let $L^{-1}$ be the bundle $L^{-1} = R^1 \pi_* \mathcal{O}_X$ on $X;$ it is a line bundle because the fibres are elliptic curves, so it makes sense to talk about $L.$ The bundle $L$ has positive degree if and only if the fibration is not isotrivial (this is not so hard, but let me give the proof of III.1.6 of Miranda as a reference); let us assume this is the case. From the Leray spectral sequence, there is an exact sequence

$0 \rightarrow H^1(C,\pi_* \mathcal{O}_X) \rightarrow H^1(X,\mathcal{O}_X) \rightarrow H^0(C,L^{-1}) \rightarrow H^2(C,\pi_* \mathcal{O}_X)$

Since $\pi_* \mathcal{O}_X = \mathcal{O}_C,$ the first term has dimension $g,$ the genus of $C.$ Since we are assuming that $L$ has positive degree, the third term is also zero, and hence the irregularity of a non-isotrivial elliptic surface is

$H^1(X,\mathcal{O}_X) = g.$

It follows that

$\chi(\mathcal{O}_X) = h^{0,0} - h^{1,0} + h^{2,0} = 1 - g + h^{2,0}.$

In our particular case, the genus of $C$ is zero. On the other hand, as noted in 2.4 of Zywina’s paper, the degree of the minimal discriminant is $12 \cdot \chi_E = 12 \cdot \chi(\mathcal{O}_X).$ In the example at hand, Zywina computes (see section 8) that $\chi_E = 3,$ and so

$h^{2,0} = 3 - (1-g) = 3 - 1 = 2.$

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