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\author{Gabriel C. Drummond-Cole}
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\title{Microlocal Categories\\Dmitry Tamarkin}
\begin{document}\maketitle
[Came in late]

Talking about $S\in D_{>0}(G\times\rR^D\times \rR^D\times \rR)$.
$Sp(2D)$ acts on $T^*\rR^D$.

We have $S|_{e\times\rR^D\times \rR^D\times \rR}=\zZ\Delta\boxtimes \zZ_{t\ge 0}[D]$.  We think of $SO(D)$ as sitting inside $Sp(2D)$ and this lifts to the universal cover $\widetilde{Sp}(2D)$ so we can talk about $S|_{SO(D)\times\rR^D\times \rR^D\times \rR}$.

$\Sigma\in D_{>0}(SO(D)\times\rR^D\times \rR^D\times \rR)$, $\Sigma=\zZ_K \boxtimes \zZ_{t\ge 0}[D]$ for $K=\{g,x_1,x_2|x_2=gx_1\}\subset SO(D)\times \rR^D\times \rR^D$.  

Claim: $S\cong \Sigma\otimes \Ll$ for some local system $\Ll\in Loc(SO(D))$ with $\Ll|e\cong \zZ$. 

You can define a functor that will give you this $\Ll$.  Let me call the group $SO(D)$ by $H$ and the space $\rR^D$ by $E$, and then I need a convolution
$D(H\times E\times E\times\rR)\stackrel{\circ}{\times}D(H\times E\times E\times R)\stackrel{\circ}{\to} D(H\times H\times E\times E\times \rR)$, where this is convolution of the second $E$ of the first factor and the first $E$ of the second factor.  You also have an embedding $D(H\times E\times E\times \rR)\stackrel{i}{\to} D(H\times H\times E\times E\times R)$ where $h$ embeds as $h,h^{-1}$.  

Let me also define $\tilde{\Sigma}=\zZ_{\tilde{K}}\boxtimes \zZ_{t\ge 0}$ where $\tilde{K}$ is the opposite: $\{g,x_1,x_2|x_1=g(x_2)$.  

Now we can look at $i^{-1}(S\circ \tilde{\Sigma})$.  From the singular support, you can see that it is a locally constant sheaf on $H\times \Delta_E\times\rR_{\ge 0}$.  Restricting on this diagonal, where it is supported, it is locally constant.  Since $\Delta_E\times\rR_{\ge 0}$ is contractible, you will get a local system on $H$.  If you do the same thing with $\Sigma$ you will get the inverse system.  Therefore, since we know this one, we can get our original sheaf $S$ by applying the inverse functor.  Call this one $G(S)$ and $F(S)=i^{-1}(S\circ \Sigma)$  so then $F\circ G\cong \id$, and $S\cong F(\Ll\boxtimes \zZ_\Delta\boxtimes \zZ_{t\ge 0})$.  Which local system is this?  We can prove that the fiber at the unit is $\zZ$ since $\Ll\otimes \Sigma|_e\cong S_e$.  There are two of these, the trivial one and the twisted one.  If we restrict to $SO(2)$ we can see this.  So look at $\Ll|_{SO(2)}$ where $SO(2)\subset SO(D)\subset \widetilde{Sp}(2D)$.  You could just as easily inject in $SU(2)$ in $SU(D)$ in $\widetilde{Sp}(2D)$.  So we might as well assume that $D=2$.  Now $SU(2)\subset\widetilde{Sp}(4)$ is a homotopy equivalence, by accident, so now we need to look at $SU(2)\times E\times E\times \rR$.  Now we really need to construct the sheaf here and see what happens when we restrict to $SO(2)$.  We're stuck, it's really a concrete computation.

We have $\rR^4$ with coordinates $(q_1,q_2,p_1,p_2)$.  It's useful to introduce the coordinates $Q=q_1+iq_2$ and $P=p_1-ip_2$, and then $\alpha=\re (PdQ)$.  If you have a matrix $\displaystyle\left(\begin{array}{cc}a&b\\-\bar{b}&\bar{a}\end{array}\right)$ where $|a|^2+|b|^2=1$, then \[
\left(\begin{array}{c}\tilde{Q}\\\tilde{P}\end{array}\right)=\left(\begin{array}{cc}a&b\\-\bar{b}&\bar{a}\end{array}\right)\left(\begin{array}{c}Q\\
P\end{array}\right).\]
Now we will quantize using generating functions.  We need to do it on two charts separately.  One, $U_1$, will be $a\ne 0$ and the other $U_2$ will be $b\ne 0$.  They will mismatch by a local system.  You need to choose a pair of independent coordinates.  If $a\ne 0$ you can choose $\tilde{Q}$ and $P$ (you will have $\tilde{Q}=aQ+bP$ and $P=\frac{\tilde{P}+\bar{b}Q}{\bar{a}}$). You need to compute $\re(\tilde{P}d\tilde{Q}+Q dP)$ which is necessarily $dS(\tilde{Q},P)$.  This can be computed, we have $S(\tilde{Q},P)=\re\frac{2P\tilde{Q}-bP^2-\bar{b}\tilde{Q}^2}{2a}$

If $b\ne 0$ you can use $Q$ and $\tilde{Q}$ [!].  You have another formula and you define
$d\Sigma=\re(\tilde{P}d\tilde{Q}+PdQ)$ and $\Sigma=\re\frac{\bar{a}\tilde{Q}^2+aQ^2-2Q\tilde{Q}}{2b}$.

Now let us see what kind of sheaves we can obtain from these functions and what we get on the overlap.

The idea is that first I can consider $t+S(\tilde{Q},P)\ge 0$ with $\rR(\tilde{q}_1,\tilde{q}_2,p_1,p_2)\times \rR$.  We need to do a Fourier transform.  If you take the convolution with $t-PQ\ge 0$, overall it will look as $t+S(\tilde{Q},P)-PQ\ge 0$.  Take the constant sheaf on this set in $\rR^6_{Q\tilde{Q}P}\times \rR$, and then you push forward along $P$ to $\rR^4_{Q\tilde{Q}}\times \rR$.  I forgot the group variable, you should also multiply all factors by $U_1\subset SU(2)$ where $a\ne 0$.

Now we should compare it to the other quantization $\zZ_{t+\Sigma(Q,\tilde{Q})\ge 0}$.  Call these $\Ss_1$ and $\Ss_2$.

[Quantization of a group is a sheaf on $G\times E\times E\times \rR$, and if you restrict to an element you should get a transformation of $E$.  An object on $E\times E\times \rR$ gives you an endofunctor on $E\times \rR$.   It's a way of encoding an action of $G$ on $E\times \rR$.]

Now we need to compare two objects.  Both can be restricted to the overlap $U_1\cap U_2$.  We will see they differ by a local system.  You remove two big circles from $S^3$, which are linked but do not intersect.  You need to complete the square, if $a$ and $b$ are nonzero, and you get the following formula.

\[S(\tilde{Q},P)-PQ=\re\left\{
-\frac{b}{2a}\Pi^2+\Sigma(Q,\tilde{Q})
\right\}
\]
where $\Pi=(P+\frac{a}{b}-\frac{1}{b}\tilde{Q})^2$.

This overlap is $S^1$ times a disk.  If $a\ne 0$, maybe you can just, what you need to do, if $b\ne 0$, you can instead of projecting along $P$, you can project along $\Pi$.  So restricting to the intersection the first sheaf is 
$R\Pi_!\zZ_{\left\{\re\left(t-\frac{b}{2a}\Pi^2+\Sigma(Q,\tilde{Q})\right)\ge 0\right\}}$.  We can focus to where we project, first restrict yourself to $R\Pi^0_!(\zZ_{t-\re\frac{b}{2a}\Pi^2\ge 0})$ and we can look at $U_1\cap U_2\times \rR^2_\Pi\times \rR_t\stackrel{\Pi^0}{\to}{\rR}$.

I don't have time, this is a good exercise, you have $U_2\stackrel{Arg\frac{a}{b}}{\to }S^1$ and the pushforward is $\alpha^*\Ll_{S^1}\boxtimes \zZ_{t\ge 0}$.  So $\Ss_1\cong \Ss_2\otimes \alpha^*\Ll_{S^1}$.

How can you glue this with one sheaf?  You need to choos it to be $\Ss_1$ on one chart and $\Ss_2$ on the other chart or vice versa.  Then you need to restrict it to $SO(2)$, and you realize this as $\left(\begin{array}{cc}e^{i\phi}&0\\0&e^{-i\phi}\end{array}\right)$.  You have two choices because we don't know which one to shift.  The neighborhood of $1$, when $b\ne 0$, we should have the unchanged sheaf.  So we should change $\Ss_1$ to $\Ss_1\times \Ll$, and we need to restrict to $SO(2)$, and if you restrict you will get what I said.  This can be computed, and you will get the formula $\Ll_{S^1}\otimes \re \zZ_{t+\re(P(\tilde{Q}e^{-i\phi}-Q))>0}$.  

I completely forgot the homological shift.  They will work out together.  This one, you can prove that this is $t$ plus the Fourier transformation, and then you project on $P$ you'll get the delta function on the difference $\tilde{Q}e^{-i\phi}-Q$.
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