# Plenary: Akshay Venkatesh :::{.remark} A fairy tale: see [Langlands Elephant](https://www.youtube.com/watch?v=8yRNQIkiudQ). The point of this talk is to see the entire elephant! ::: :::{.remark} In the analogy between number fields and 3-manifolds, automorphic forms are on the number field side -- what is the manifold analog? Some history: - Mazur 63/64, in conversations with Artin: $\spec \ZZ/p\ZZ \to \spec \ZZ$ is like a knot in a (simply connected) 3-manifold. - Weil 49: Weil conjectures, there should be an "algebraic" cohomology theory $H^*(\wait)$ for $X\slice k$ for $k = \bar k$ where $H^*(X(\CC))$ recovers singular cohomology. - Artin/Grothendieck: for finite coefficients, realized this using **étale cohomology** $H^*_\et(X)$. - Tate 62, Poitou 61: looked at Galois cohomology, showed $H_\et^*(\spec \ZZ)$ (or other number rings) has a "Poincare duality" $H^i \mapstofrom H^{3-i}$, making it look like a 3-manifold. Recovers results in class field theory. ::: :::{.example title="Varieties with automorphisms"} Let $X = V(x^3+y^3+z^3) \subseteq \PP^2\slice \CC$; there are automorphisms - $x\mapsto y, x\mapsto z$, etc - $x\mapsto \bar x, y\mapsto \bar y$, etc - $x,y,z \mapsto \sigma(x), \sigma(y), \sigma(z)$ for $\sigma\in S_3$ These should all act on $H^*(X)$ for any such $H^*$, but conjugation is quite discontinuous. ::: :::{.example title="Manifolds vs schemes"} Let $X \in \Mfd$, how does one compute $H^*_\sing(X; \ZZ)$? Reduce to a linear-algebraic problem by triangulating and forming a chain complex of simplices. However, for $X \da \spec R, R\da \ZZ\invert{2}$, \[ H^1_\et(X; C_2) &= R\units/\squares{R\units}= \ts{\pm 1,\pm 2} \cong C_2\cartpower{2} \\ H^2_\et(X; C_2) &= \Quat\Alg\slice R = \ts{ \Mat_{2\times 2}(R), R[i,j,k]/\gens{i^2=j^2=k^2=-1} } .\] so $H^2$ classifies quaternion algebras over $R$. So computing this is very different to the case of manifolds! Also note that these are not dual on-the-nose, since $H^1,H^2$ have different orders. ::: :::{.remark} Comparing duality for number rings vs 3-manifolds. For us, a number ring will be - $\OO_{K, S}$ for $K$ a number field, where $S$ is a finite number of primes we can invert (the **$S\dash$integers of $K$**), or - functions on a smooth curve over $\FF_q$. For simplicity, we'll take $R = \ZZ\invert{p}, X = \spec R$. For $M\in\Ab\Grp$ a $p\dash$torsion group (where we need the order to be invertible in $R$), we'll consider \[ H^i(X; M) \da H^i_\et(\spec \ZZ\invert p; M) .\] There is a LES involving $M\dual \da \Hom_\Grp(M, S^1) = \Hom_\Grp(M, \mu_{p^\infty})$ where $\mu_{p^\infty}$ are the $p\dash$power roots of 1: \begin{tikzcd} &&&& {H^{i-1}(\cdots)} \\ \\ {H^{3-i}(\ZZ\invert p; M\dual)\dual} && {H^i(\ZZ\invert p; M)} && {H^i(\QQpadic; M)} \\ \\ {H^{i+1}(\cdots)} \arrow[from=3-1, to=3-3] \arrow[from=3-3, to=3-5] \arrow[from=1-5, to=3-1] \arrow[from=3-5, to=5-1] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsNSxbMiwyLCJIXmkoXFxaWlxcaW52ZXJ0IHA7IE0pIl0sWzQsMiwiSF5pKFxcUVFwYWRpYzsgTSkiXSxbMCwyLCJIXnszLWl9KFxcWlpcXGludmVydCBwOyBNXFxkdWFsKVxcZHVhbCJdLFs0LDAsIkhee2ktMX0oXFxjZG90cykiXSxbMCw0LCJIXntpKzF9KFxcY2RvdHMpIl0sWzIsMF0sWzAsMV0sWzMsMl0sWzEsNF1d) On the other hand, let $X\in \Mfd^3$ be a manifold with boundary, then there is a LES \begin{tikzcd} &&&& {H^{i-1}(\cdots)} \\ \\ {H^i(X, \bd X; M) \cong H^{3-i}(X; M\dual)\dual} && {H^i(X; M)} && {H^i(\bd X; M)} \\ \\ {H^{i+1}(\cdots)} \arrow[from=3-1, to=3-3] \arrow[from=3-3, to=3-5] \arrow[from=1-5, to=3-1] \arrow[from=3-5, to=5-1] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsNSxbMiwyLCJIXmkoWDsgTSkiXSxbNCwyLCJIXmkoXFxiZCBYOyBNKSJdLFswLDIsIkheaShYLCBcXGJkIFg7IE0pIFxcY29uZyBIXnszLWl9KFg7IE1cXGR1YWwpXFxkdWFsIl0sWzQsMCwiSF57aS0xfShcXGNkb3RzKSJdLFswLDQsIkhee2krMX0oXFxjZG90cykiXSxbMiwwXSxbMCwxXSxbMywyXSxbMSw0XV0=) So $X \approx \ZZ\invert p$ is a (nonorientable!) 3-manifold in this analogy, with $\bd X \approx \QQpadic$, which is now like a 2-manifold. The nonorientable assumption here is related to the need to twist the Galois actions on the scheme side. How to realize this: delete a tubular neighborhood of the knot, then - $\bd X = \bd \nu(\spec \ZZ/p\ZZ) \approx \QQpadic$. - $\spec \ZZpadic \approx \nu$, - $\bd \spec \ZZpadic\approx \bd \spec \ZZ\invert{p} \sim \spec \QQpadic$. ![](figures/2022-03-05_11-44-31.png) ::: :::{.remark} \envlist - The 3-dimensional objects: $\spec R$ for $R = \ZZ, \ZZ\invert p, \ZZ\adjoin{\sqrt 2}$, or $\FF_p(t)$, $\ZZpadic$, or projective smooth curves over $\FF_p$. - The 2-dimensional objects: $\QQpadic, \FF_p\fls{t}$, or projective smooth curves over $\bar \FF_p$. ::: :::{.remark} Relating to automorphic forms: for $G=\SL_2$, - For $\ZZ$, we study the vector space $\mca_\ZZ = \ts{\text{functions on } \dcosetl{G(\ZZ)}{G(\RR)}}$. - For $\ZZ\invert p$, we instead look at $\mca_{\ZZ\invert p}= \ts{\text{functions on } \dcosetl{G(\ZZ\invert p)}{ G(\RR) \times G(\QQpadic)} }$. We would like some association that works similarly: \[ \Mfd^3 &\to \Vect\slice k \\ M &\mapsto \mca_M .\] ::: :::{.example title="?"} A non-example is $M\mapsto H_\sing^*(M; \CC)$, which behaves nothing like $\ZZ\mapsto \mca_\ZZ$. E.g. it has the wrong type of functoriality: the map $\ZZ\to \ZZ\adjoin{\sqrt 2}$ is like a branched double cover, but for manifolds there are maps both ways and here it is difficult to find wrong-way maps. Moreover the corresponding spaces of automorphic forms $\mca_{\ZZ}, \mca_{\ZZ\invert p}$ differ by far more than just a part coming from $\QQpadic$. Note also that $H^*(M \union N) \cong H^*(M) \oplus H^*(N)$, but for automorphic forms we have $\mca_{\ZZ \oplus \ZZ} \cong \mca_\ZZ \tensor \mca_\ZZ$, where the source is like a degenerate quadratic extension. ::: :::{.remark} A heavily studied piece of the analogy: \[ \txt{\text{Automorphic forms}}\mapstofrom \TQFT ,\] coming from the [Kapustin-Witten 2006](https://arxiv.org/abs/hep-th/0604151), where a (4-dimensional) TQFT is a essentially a monoidal functor: \[ \TQFT^3 = [(\Bord^3, \disjoint) \to (\Vect\slice k, \tensor_k)] .\]. > See [Atiyah TQFT](https://www.math.ru.nl/~mueger/TQFT/At.pdf), section 2. :::