# Talk 2: Symplectic L-functions and their topological analogues 


> Abstract: The topology of the symplectic group enters into many different areas of mathematics. After discussing a couple of “classical” manifestations of this, I will explain a new one, in the theory of L-functions, as well as a purely topological analogue of the statement. I am not going to assume any familiarity with the theory of L-functions for the talk. 
Joint work with Amina Abdurrahman.


## Symplectic Spaces

:::{.remark}
Note: for $L$ a field,
\[
\squares L \da \ts{a^2 \st a\in L}
.\]

:::

:::{.remark}
Recall that any symplectic vector space is isomorphic to $\qty{\RR^{2g}, J\da \matt{0}{\id_g}{- \id_g}{0}}$.
Write 
\[
\SP(V) = \ts{g: V\to V\st \inp{gx}{gy} = \inp{x}{y}} = \ts{A\in \GL_{2g}(\RR) \st A^tJA = J}
.\]
Note 
\[
\SP_2(\RR) = \ts{M\in \GL_2(\RR) \st \det M = 1}
,\]
and $\SP_{2g}(\RR)$ is connected but $\SP_{2g}(\RR) \homotopic \RR^2\smz$.
Let $p: G\to \SP_{2g}(\RR)$ be the universal cover; since the base is a topological group, picking any $p\inv(1)$ yields an essentially unique group structure on $G$ by analytically continuing the group law.
In fact $G$ is a Lie group and $\ker p \cong \ZZ \in Z(G)$ is central, so there is a SES
\[
\ZZ\to G\to \SP_{2g}(\RR)
.\]
Note that there are not faithful finite-dimensional reps of $G$, but there are infinite-dimensional reps important to quantization in physics.
:::

## Surfaces

:::{.remark}
A theorem of Meyer on surfaces: let $\Sigma$ be a (compact closed oriented smooth) surface and $\rho: \pi_1 \Sigma\to \SP_{2g}\RR$; this corresponds to a local system, so form the twisted cohomology $H^1(\Sigma; \rho)$.
Using the cup product and symplectic pairing, one can produce a *symmetric* pairing
\[
L: H^1(\Sigma;\rho)\tensorpower{\RR}{2}\to \RR
.\]
There is an isomorphism $(H^1, L)\cong (\RR^{p+q}, \diag_{p}(-1) \oplus \diag_q(1))$, so $\signature(\Sigma, \rho) \da \signature L \da p-q$ is an interesting invariant.
:::

:::{.remark}
Recall that 4-manifolds $M$ also admit a signature on $H^2(M; \RR)$.
Let $M\to\Sigma$ be surface bundle over a surface -- if $M$ is a product, $\signature M = 0$.
Chern-Hirzebruch-Serre: there is a monodromy morphism $\pi_1\Sigma\to \SP_{2g}\RR$ which is trivial if $\signature M = 0$.
In fact $\signature(M) = \signature(\Sigma, \rho)$, so this situation naturally arises for fibered 4-manifolds.
:::

:::{.remark}
Recall $\pi_1 \Sigma_g = \gens{a_i,b_i \st \prod_i [a_i, b_i] = e}$.
Consider trying to form a lift:

\begin{tikzcd}
	&& G \\
	\\
	{\pi_1\Sigma} && {\SP_{2g}(\RR)}
	\arrow[from=3-1, to=3-3]
	\arrow[from=1-3, to=3-3]
	\arrow[dashed, from=3-1, to=1-3]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsMyxbMCwyLCJcXHBpXzFcXFNpZ21hIl0sWzIsMiwiXFxTUF97Mmd9KFxcUlIpIl0sWzIsMCwiRyJdLFswLDFdLFsyLDFdLFswLDIsIiIsMix7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dXQ==)

Note $m\da \prod [\tilde\rho a_i, \tilde\rho b_i] \in \ker p \cong \ZZ$, and it turns out that
\[
\signature(\Sigma, \rho) = p-q = 4m
.\]
The fact that $\ker p \in Z(G)$ was essential in making sure this doesn't depend on choices.
Note that this only determines $m$ up to sign!
:::

:::{.remark}
Toward generalizing, the above central extension determines a class $b\in H^2(\SP_{2g}(\RR); \ZZ)$ and $p^* b\in H^2( \Sigma; \ZZ)$.
Using the pairing against the fundamental class $[\Sigma]$ yields
\[
\signature(\Sigma, \rho) = 4 \int_\Sigma p^* b
.\]

When replacing $\RR$ with $L$, replace $\ZZ$ by $W_L$, the Witt group of quadratic forms over $L$.
:::

# 3-manifolds

:::{.remark}
Consider now varying the situation in a 1-dimensional family -- take $\Sigma\to M\to S^1$ a surface bundle over a circle with fibers $\Sigma_t$, each yielding $\rho_t: \pi_1\Sigma_t \to \SP_{2g}(\RR)$ for $t\in S^1$.
Each $t$ yields a quadratic vector space $H^1(\Sigma_t,\rho_t)$ of signature $(p, q)$.
Monodromy yields an element $m\in \Aut H^1(\Sigma_0, \rho_0) \subseteq \Orth_{p, q}$.
If $p,q>0$ then $\size\pi_0\Orth_{p, q} = 4$ -- which connected component does this land in?
How to separate the components: there is a determinant map
\[
\det: \Orth_{p, q} \to \ts{\pm 1}
.\]
There is a *spinor norm*
\[
\spinornorm: \Orth_{p, q} &\to \ts{\pm 1} = \RR\units/\squares\RR\units \\
\mathrm{reflection}_v &\mapsto \inp{v}{v}
.\]
This works with $\RR$ replaced by $L$, using the fact that $\Orth_{p, q}$ is generated by reflections.
For $(V, L)$ a symmetric space, one gets
\[
\det: \Orth(V) &\to \ts{\pm 1}
\]
and 
\[
\spinornorm: \Orth(V) &\to L\units/\squares L\units
.\]
:::

:::{.theorem title="?"}
It turns out that 

- $m\in \Orth(H^1(\Sigma_0, \rho_0))$,
- $\det m = 1$, and
- $\spinornorm(m) = \int_M p^* c\in L\units/\squares L\units$.

Note that there is a Soulé etale Chern class 
\[
m\in H^3(\SP_{2g}(L), L\units/\squares L\units)
.\]
Pulling this back and integrating yields
\[
p^* c\in H^3(M; L\units/\squares L\units) \mapsvia{\int_M} L\units/\squares L\units
.\]

:::

:::{.proof title="?"}
Re-interplay $\spinornorm(m)$ as a quantity $\RT(M, \rho)$ where $\RT$ is the *Reidemeister torsion*, which makes sense for any $(M, \rho)$ (not just those fibered over $S^1$).
This is a bordism invariant, i.e. if $M\sim N$ are bordant then $\RT(M,\rho) = \RT(N,\rho')$.
Thus there exists some formula of the desired type, where $c$ is unknown -- the trick is to compute enough examples to determine $c$.
:::

## Symplectic $L\dash$functions

:::{.remark}
For $X\in\smooth\proj\Var\slice{\FF_q}$ and $\rho: \pi_1 X\to \SP_{2g} k$ for $k$ finite containing $\sqrt{q}$, there is an associated $L\dash$function $L(X, \rho; T)\in k(T)$ where $T\approx q^{-s}$.
There is a functional equation relating $T\mapstofrom {1\over qT}$.
Evaluate at the center of symmetry $T = {1\over\sqrt q}$ to define $L(X, \rho) \da L(X, \rho; 1/\sqrt q)$.
:::

:::{.theorem title="?"}
Suppose $L(X, \rho) \neq 0$; then modulo squares
\[
L(X, \rho) = \int_X \rho^* c \qquad \in k\units/\squares k\units
,\]
which is a spinor norm of Frobenius (see Zassenhaus).
This requires some conditions, e.g. $\rho\mid_{\pi_1 X}$ is surjective, congruences on $\size k$ and $q\mod 8$, and $\gcd(q,\size \SP_{2g}(k)) = 1$ (which may not be necessary).
:::

:::{.remark}
Interpretation: $L(X,\rho) = ab^2$ where $a$ is simple and $b$ is complicated.
BSD gives a conjectural formula for this which includes a lot of squares.
So there is a cohomological obstruction to the existence of a square root of $L(X, \rho)$.
:::

:::{.remark}
On the proof: try to pass validity from one known example $(X,\rho)$ to other examples $(X', \rho')$ with $X\slice{\FF_q}, X'\slice{\FF_q'}$ and images in $\SP_{2g}(k), \SP_{2g}(k')$ respectively.

- Pass from $q$ to $q'$ using a moduli space of pairs $(X,\rho)$ where a topological theorem controls the fiber over $\CC$.
- Pass from $k$ to $k'$ using *compatible local systems* from NT.

:::

:::{.remark}
Final comparison:

- For 2-manifolds, $\signature(\Sigma, \rho) = \int_\Sigma p^* b$.
- For 3-manifolds, $\RT(M,\rho) = \int_M p^* c$.
- Symplectic $L\dash$functions: ?
:::