# nonabelian reciprocity law

I heard the following relation in a talk by Peter Scholze. Could someone explain "in a simple way" what is the precise relation between the polynomial $x^4-7x^2-3x+1$ and the integral homology of the orbit space $\mathbb{H}^3/\Gamma$ where:

1. $\mathbb{H}^3$ is the hyperbolic 3-space.
2. $\Gamma\subset SL_{2}(\mathbb{Z}[i])$ such that for any $(\gamma_{r,s})_{r,s\in\{1,2\}}=\gamma\in \Gamma$ we have $\gamma_{1,2}\equiv 0 \textrm{ mod } 183$ and $\gamma_{2,2}\equiv 1 \textrm{ mod } 3$
• Notice that the discriminant of your polynomial is $3^2 \cdot 61^2$, and $183 = 3 \cdot 61$. Commented Mar 4, 2017 at 16:59
• @user94041: Yes, that's on the first shown slide of the linked video ... Commented Mar 6, 2017 at 14:57

Will Sawin's answer is perfectly correct, but I wanted to add some further perspective.

What Peter Scholze has proven is a profound generalization of (one half) of Class Field Theory. The particular example you state is simply a special case of a much more general result.

To set the stage, let me recall some of the set-up from class field theory. Let $F$ be a number field. For simplicity of exposition, suppose that the class group of $F$ is equal to one. Let $\mathfrak{q}$ be a prime of $F$, and suppose that $\mathfrak{q}$ has norm $q$. Now ask the question: does $F$ admit an abelian extension of degree $p$ which is only ramified at the prime $\mathfrak{q}$?

Class Field Theory provides an answer to this question. The answer is that one should consider a certain quotient of the Idele Class group $F^{\times} \backslash \mathbf{A}^{\times}_F$ depending on $\mathfrak{q}$, namely, the quotient

$$X_{\mathfrak{q}}:=F^{\times} \backslash \mathbf{A}^{\times}/U_{\mathfrak{q}},$$

where $U = \prod U_v$ is a compact open of the finite adeles such that $U_v = \mathcal{O}^{\times}_{F,v}$ for all primes $v$ except for $\mathfrak{q}$, and for $v = \mathfrak{q}$, $U_v$ is subgroup of $\mathcal{O}^{\times}_{F,\mathfrak{q}}$ consisting of elements congruent to $1 \mod \mathfrak{q}$. This quotient space is, geometrically, a disconnected union of a product of circles. What does class field theory say? It says that the zeroth homology group:

$$H_0(X_{\mathfrak{q}},\mathbf{Z})$$

completely determines (and is determined by) the abelian extensions of $F$ which are unramified everywhere away from $\mathfrak{q}$ and have degree prime to $q$. In particular, we have a geometric object whose homology relates directly to Galois Representations. Now it turns out that one can give a quite precise description of the connected components of $X_{\mathfrak{q}}$ in terms of the quotient of $(\mathcal{O}_F/\mathfrak{q})^{\times}$ by the global units, which leads to an explicit computational way to determine the existence of abelian extensions. Here's an explicit example.

Let $F = \mathbf{Q}(\sqrt{2})$, and $\epsilon = \sqrt{2} - 1$. Let $\mathfrak{q}$ be a prime of $\mathcal{O}_F$ whose norm is prime and congruent to $1 \mod 3$. Then there exists an abelian extension of $F$ of degree $3$ unramified outside $\mathfrak{q}$ if and only if $\epsilon$ is a cube modulo $\mathfrak{q}$.

OK, now let's turn to $2$-dimensional representations. Again, let's fix a field $F$ and a prime $\mathfrak{q}$. This time, consider the space

$$X_{\mathfrak{q}} = \mathrm{GL}_2(F) \backslash \mathrm{GL}_2(\mathbf{A})/U_{\mathfrak{q}},$$

where now $U_{\mathfrak{q}}$ is a compact open subgroup whose level reflects the choice of prime $\mathfrak{q}$. It turns out that the homology of this space is related to the existence of two dimensional Galois representations. It's usually traditional (convenient) to also take the quotient by a further compact subgroup $K$ of $\mathrm{GL}_2(\mathbf{R})$ as well, and also the $\mathbf{Q}$-split part of the center.

So what kind of Galois representations do these homology classes relate to, at least conjecturally? Well, one has to be quite careful in specifying the precise degrees, which depend somewhat delicately on the signature of the field $F$. But let's consider the case when $F = \mathbf{Q}$, in which case the objects above are none other than the classical modular curves. In this case, the cohomology of modular curves corresponds, via the Eichler-Shimura isomorphism, to classical modular forms of weight two. And then, by the combined efforts of many people (Shimura, Deligne, Langlands, etc), we expect that these then give rise (taking into account the Hecke operators) to Galois representations:

$$\rho:\mathrm{Gal}(\overline{\mathbf{Q}}/\mathbf{Q}) \rightarrow \mathrm{GL}_2(\overline{\mathbf{Q}}_p).$$

What is surprising at first is that these Galois representaitons have infinite image, and determining (even conjecturally) precisely what the image should be requires the entire theory of integral p-adic Hodge Theory, local-Langlands, and so on. The actual construction of these Galois representations crucially uses the fact that the objects $X_{\mathfrak{q}}$ we have constructed above are actually algebraic varieties over number fields (in this case curves), so we can use the powerful methods of etale cohomology developed by Grothendieck and his school to construct Galois representations. This is usually viewed as one half of the problem of "Reciprocity" in the Langlands program: going from automorphic representations (in this case in their avatar as cohomology classes, which requires the representations to be of a special kind) to Galois Representations. One should view this as the analog in class field theory of constructing a non-trivial abelian extension (say the Hilbert class field) starting with the class group. Naturally, there is another half of the problem which is the converse, and here the profound work of Wiles is relevant.

To get back to Scholze's talk, however, requires a slight detour. Already for the case $F = \mathbf{Q}$ above, one could consider the cohomology not with coefficients in characteristic zero, but in characteristic $p$. In this case, there is a generalization of Reciprocity which associates (in a precise bijective way) such eigenclasses under Hecke operators to representations:

$$\rho:\mathrm{Gal}(\overline{\mathbf{Q}}/\mathbf{Q}) \rightarrow \mathrm{GL}_2(\overline{\mathbf{F}}_p).$$

(again with some technical conditions on what representations can occur). This amounts to a famous conjecture of Serre, proved spectacularly by Khare and Wintenberger, surely one of the great theorems in the subject since Wiles. In this case, the arguments crucially use the following observation: If one takes the mod-p cohomology of an algebraic curve (or orientable topological surface), then such classes always lift to characteristic zero. By Serre's conjecture, this implies a correpsonding statement on the Galois side which is incredibly profound. However, at least one has tools to study this quesiton, since there is a direct link between mod-p torsion classes in $H_1$ and classes in characteristic zero.

Now let us take $F$ to be an imaginary quadratic field, say $\mathbf{Q}(i)$. In this case, the corresponding $X_{\mathfrak{q}}$ turn out to be a disjoint union of spaces of the shape

$$\Gamma \backslash \mathbf{H}^3,$$

where $\mathbf{H}^3$ is hyperbolic three space, and $\Gamma$ is a congruence subgroup of $\mathrm{GL}_2(\mathbf{Z}[i])$. This means (since the cover is contratible) that the homology groups of this space are just the homology groups of $\Gamma$. Once again, Langlands reciprocity predicts, from any cohomology class which is a Hecke eigenform, a corresponding Galois representation:

$$\rho:\mathrm{Gal}(\overline{\mathbf{Q}}/F) \rightarrow \mathrm{GL}_2(\overline{\mathbf{Q}}_p)$$

with particular properties. Now it's much harder to see why this might be true, since the hyperbolic three manifold no longer has any association with algebraic geometry. One method is to use the fact that characteristic zero classes in homology can often be computed by de Rham cohomology, and hence by harmonic differentials, and hence by automorphic forms. Then one might hope to move the automorphic forms around between different groups (using Langlands base change or variations thereof), and move to a situation where one has access to geometry. This was done (in this case) by Harris-Soudry-Taylor, who found the Galois representations.

The REAL issue, however, is when one now considers mod-p cohomology classes. There is no reason why (and it is often false) a mod-p cohomology class of $\Gamma$ (or on the corrresponding arithmetic three manifold) should lift to characteristic zero. Hence this is not directly related to Langlands reciprocity. That is we HAVE GONE BEYOND THE LANGLANDS PROGRAM AS WE KNOW IT. However, an amazing generalization of this reciprocity (due to Ash and others) says that the something analogous happens in this case too, namely, given a mod-p torsion eigenclass in cohomology, one obtains a mod-p Galois representation. This is an outrageous conjecture, which was completely open until Peter Scholze stunningly proved it (in many cases) in 2013.

It's interesting to ask about the precise relationship between cohomology and Galois representations, which was your (great!) question. Will Sawin has explained what is going on in this case. But it is worth noting that the recipe (going from eigenalues of Hecke operators to the trace of Frobenius elements) is a completely general feature of these mod-p Langlands reciprocity conjectures. To recall the precise statment:

$$\mathrm{Trace}(\rho(\mathrm{Frob}_{v})) = a_v \mod p,$$

where $a_v$ is the eigenvalue of $T_v$. In this case, the prime $p$ si equal to three, and so representation of interest has the shape:

$$\mathrm{Gal}(\overline{\mathbf{Q}}/F) \rightarrow \mathrm{GL}_2(\overline{\mathbf{F}}_3).$$

The composite of this map to $\mathrm{PGL}_2(\overline{\mathbf{F}}_3)$ turns out the land in $\mathrm{PGL}_2(\mathbf{F}_3) = S_4$ and turns out to have image $A_4$, and moreover is the natural representation of the Galois group of the given polynomial.

Let's now use this to understand solutions to the polynomial $x^4−7x^2−3x+1$ modulo $q$, for a prime $q = N(\mathfrak{q})$ of the Gaussian integers. Certainly, since the polynomial has rational coefficients, it has a solution modulo $q$ if and only if it has a solution modulo $\mathfrak{q}$, since $\mathbf{Z}/q = \mathbf{Z}[i]/\mathfrak{q}$. It turns out that one can tell if any polynomial of degree $n$ has a root modulo $q$ in terms of the corresponding Frobenius element considered as a conjugacy class inside $S_n$, where the Galois group acts on the roots in the obvious way. In this case, the Galois group is $A_4$ acting on the roots in the obvious way to identify the group with $A_4$. The group $A_4$ only has four conjugacy classes. In general, a polynomial doesn't have a root if and only if the corresponding element of $S_n$ has no fixed points. In $A_4$, this means it is conjugate to $(12)(34)$. There are four roots if Frobenius is trivial, and one root of Frobenius is conjugate to $(123)$ or its inverse. In the first, second, and third cases, the corresponding elements of $A_4 = \mathrm{SL}_2(\mathbf{F}_3)$ are

$$\left(\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right), \quad \left(\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right), \quad \left(\begin{matrix} 0 & 1 \\ -1 & 1 \end{matrix} \right).$$

Any lift of the first matrix to $\mathrm{GL}$ has trace zero, and any lift of the other two matrices do not, thus the trace of $\rho(\mathrm{Frob}_v)$ will be zero modulo three exactly when the quartic has a rational solution modulo $q$. But now, the magic happens, and Scholze's theorem associates these traces to the Hecke eigenvalues, and thus, remarkably, one can deduce that the Hecke operator $T_{\mathfrak{q}}$ on this eigenclass is non-trivial modulo $3$ exactly when the quartic has a solution modulo $q$.

Of course, one should really think of this example not in isolation but as part of the vast web of interlacing conjectures in the modern Langlands program. Still, it's nice to have explicit examples!

(There are other interesting features of this example, namely that the $A_4$ extension comes from the restriction of an even representation of $\mathbf{Q}$, but we do not discuss those here.)

• I really enjoyed reading this! I'm a bit of a novice at this stuff, so perhaps you'd excuse my naive question: When you said "mod $p$ cohomology lifts/does not lift to characteristic 0", I assume you're talking about singular cohomology? Ie, you're asking if every cohomology class in $H^i(X,\mathbb{F}_p)$ is the image of a class in $H^i(X,\mathbb{Z})$? (How does one get this map? - for homology the map comes from the universal coeff. theorem) Or, were you talking about a different type of cohomology? Commented Mar 10, 2017 at 4:40
• Dear @Amy, I am indeed talking about the map $H^i(X,\mathbf{Z}) \rightarrow H^i(X,\mathbf{F}_p)$ (but one can similarly consider the map $H_i(X,\mathbf{Z}) \rightarrow H_i(X,\mathbf{F}_p)$). Even though Cohomology is contravariant in spaces, it is covariant in the coefficients. The cohomology is computed from the homology of the co-chain complex; think of a cohomology class as represented by a class in $\mathrm{Hom}(C,\mathbf{Z})$ for some $C$; this maps naturally to $\mathrm{Hom}(C,\mathbf{F}_p)$ and this induces a map on cohomology. I'm not sure if this answers your question or not. Commented Mar 10, 2017 at 4:56
• Ah, of course! Here I assume by "$C$" you meant some chain group with coefficients in... $\mathbb{Z}$? Do you know what the cokernel is of the reduction mod $p$ map is measured by? (In the homology case it's a certain Tor group) Commented Mar 10, 2017 at 5:29
• The cokernel of $H^i(X,\mathbf{Z}) \rightarrow H^i(X,\mathbf{F}_p)$ is $H^{i+1}(X,\mathbf{Z})[p]$; the cokernel of $H_i(X,\mathbf{Z}) \rightarrow H_i(X,\mathbf{F}_p)$ is $H_{i-1}(X,\mathbf{Z})[p]$. Here $M[p]$ denotes the $p$-torsion of $M$. Commented Mar 10, 2017 at 15:34

Peter Scholze explains the same example in more detail in this talk https://youtu.be/cFdm0B9KLcQ?t=48m and gives even more detail here https://youtu.be/gI3Ta73yVuo?t=23m55s.

In brief, the relationship is that for every prime $\mathfrak p$ of $\mathbb Z[i]$ not dividing $183$, the polynomial has a root mod $\mathfrak p$ if and only if the eigenvalue of the $\mathfrak p$th Hecke operator on a certain 3-torsion class in that integral homology group is nonzero.

The $p$th Hecke operator comes from the subgroup $\Gamma_0(\mathfrak p)$ of $\Gamma$ which has two equivalent descriptions - as the subgroup of $\Gamma$ consisting of elements where $\gamma_{1,2} \equiv 0\mod \mathfrak p$ and the subgroup consisting of elements where $\gamma_{2,1}\equiv 0 \mod \mathfrak p$. The isomorphism between these two subgroups comes from multiplying $\gamma_{1,2}$ by $\mathfrak p$ and dividing $\gamma_{2,1}$ by $\mathfrak p$.

This isomorphism means there are two distinct injections $\Gamma_0 (\mathfrak p) \to \Gamma$, hence two distinct covering maps maps $\mathbb H^3/\Gamma_0(\mathfrak p) \to \mathbb H^3/\Gamma$. Pulling back a homology class along the first map (possible because the map is a finite covering) and pushing it forward along the second map produces an operator $T_\mathfrak p: H_1( \mathbb H^3/\Gamma,\mathbb Z) \to H_1( \mathbb H^3/\Gamma,\mathbb Z)$.

Figuiredo located a $3$-torsion class in this group that is an eigenvector of all these operators, and Scholze's result implies that the eigenvalue of $T_\mathfrak p$ on this class is nonzero if and only if the polynomial has a root mod $\mathfrak p$.

• In other words, the correct answer to the question "is there some explanation of this remark in this video?" is "yes; it is explained in great detail in the very same video." Now explain to me why the question has 22 upvotes and then I will understand this website. Commented Mar 9, 2017 at 22:33
• @PoundSterling I don't know if I'm the one who's supposed to explain this to you, but I'm guessing most of the people who upvoted it did not watch the video, and just thought the question was interesting. Commented Mar 9, 2017 at 23:47
• that is one theory. Another is that most of the people who upvoted the question were about as lazy as the OP, and thought it sounded sexy simply because it had Peter Scholze's name in it. And most of those people will not be particularly interested in reading the details of your answer. Commented Mar 10, 2017 at 0:19
• @PoundSterling Often users upvote a question simply to encourage a "good answer", instead of expressing that it is a "good question". It is also often the case with popular-but-advanced mathematics (or popular mathematicians) that people hope to find a conceptual or easy exposition, some sort of quick route to understanding. One thing worth repeating is that this forum is community moderated, therefore we can improve questions and answers by asking for particular edits. For example, we could have asked the OP to clarify what specifically he wasn't satisfied with in the videos, etc. Commented Mar 10, 2017 at 1:07