It's time for another episode of *Wild Spectral Sequences*! I haven't written about this in a long time! Truthfully, I didn't want to prove any more diagram lemmas, even though I still think the spectral sequence approach is much more fun than the diagram chasing approach. Today however, I'm going to talk a little about the Lyndon-Hochschild-Serre (LHS) spectral sequence for group cohomology applied to the first cohomology of tori.

A **torus** $T$ over a field $F$ is an algebraic group such that $T_{\overline{F}}$ is isomorphic to $\G_{m,\overline{F}}^n$. In other words, it's a form of $\G_m^n$ for some $n$. The cohomology I'm talking about is the Galois cohomology group $H^1(F,T)$. Where does this group show up? If $G$ acts on an $F$-variety $V$, and $T\subset G$ is an $F$-torus that is the stabiliser of some $v\in V(F)$, then the "stable class" $G(\overline{F})v\cap V(F)$ of $v$ in $V(F)$ may be different than the usual orbit $G(F)v$.

The $G(F)$-orbits in the stable class of $v$ is then parametrised by the abelian group

Specifically, if $gv$ for $g\in G(\overline{F})$ represents a $G(F)$-conjugacy class in the stable class of $v$ then the corresponding element of $D$ is $\sigma\mapsto g^{-1}\sigma(g)$. It's a rather fascinating problem in itself to figure out the differences between $G(F)$-conjugacy and $G(\overline{F})$-conjugacy, and I'll probably give examples of calculations later. Moreover, this difference between the two types of conjugacy is extremely useful in the comparison of trace formula and establishing cases of Langlands functoriality through the fundamental lemma. More on this later.

## Where Are the Spectral Sequences?

Wait, isn't this post supposed to be about spectral sequences? It is! Sometimes, you actually want to *compute* $H^1(F,T)$ explicitly! So, how to you do it? Of course, there's some finite Galois extension $E/F$ for which $T$ splits into a bunch of copies of $\G_m$. So you might think that because of Hilbert's theorem 90, you should be able to compute $H^1(F,T)$ as the *finite* Galois cohomology group $H^1(E/F,T)$.

It's true! And, did you know you can prove this using spectral sequences? Of course you did, because this is an episode of *Wild Spectral Sequences* after all.

## The Proof

The actual proof is kind of simple, but it illustrates:

**The Lyndon-Hochschild-Serre Spectral Sequence**

$$E^{pq}_2 = H^p(G/H,H^q(H,A))\Rightarrow H^{p + q}(G,A)$$.

Here, $G$ is a group, $A$ is a $G$-module, and $H\leq G$ is a normal subgroup. Finally, $H^q(H,A)$ actually has to be considered as a $G/H$-module, whose action of $G/H$ may be nontrivial, even if $G$ acts trivially on $A$! Crazy! How do we apply this to show that we can compute $H^1(F,T)$ as $H^1(E/F,T)?$ Well, we know that $H^1(F,T) \cong \varinjlim_{U} H^1(\Gamma/U,T(\overline{F})^U)$ where $U$ runs over all the open normal subgroups of $\Gamma = \mathrm{Gal}(\overline{F}/F)$. In other words, we can compute $H^1(F,T)$ as the direct limit of cohomology groups for the finite Galois extensions. We can take the cofinal set in this direct limit where all the extensions contain $E$.

So, we have to prove the following: suppose that $M/E/F$ are Galois extensions. Then the inflation map $H^1(E/F,T)\to H^1(M/F,T)$ is an isomorphism. Luckily, this map will appear in the spectral sequence. To use the spectral sequence, put $G = \mathrm{Gal}(M/F)$ and $H\leq G$ be the subgroup $\mathrm{Gal}(M/E)$. Then $G/H$ is naturally isomorphic to $\mathrm{E/F}$. Put $A = T(M)$. Then, the spectral sequence tells us we have a filtration of $F^1H\to H^1(M/F,T(M))$.

The piece $F^1H \cong E^{1,0}_\infty = E^{1,0}_2 = H^1(E/F,T(E))$. The theory of the spectral sequence tells us that the map $H^1(E/F,T(E))\to H^1(M/F,T(M))$ is the inflation map, which is the same map that is in the direct limit. So far, so good. But now, the spectral sequence also tells us that the quotient $H^1(M/F,T(M))/H^1(E/F,T(E))$ is isomorphic to $E_\infty^{0,1}$, which is in turn the kernel of the map $E_2^{0,1}\to E_2^{2,0}$. Since we don't want to think to hard about what this map actually is, checking whether $E_2^{0,1} = 0$ is a good first strategy.

But $E_2^{0,1} = H^0(E/F,H^1(M/E,T(M)))$. The point is now, as a $\mathrm{Gal}(M/E)$-module, $T(M)$ is just a bunch of copies of $M^\times$, with the action of $\mathrm{Gal}(M/E)$ being induced by the action on $M$. This of course is not true if $T(M)$ is considered as a $\mathrm{Gal}(M/F)$-module! Hence Hilbert's theorem 90 gives us $E_2^{0,1} = 0$.

**Conclusion**: the inflation map $H^1(E/F,T(E))\to H^1(M/F,T(M))$ appearing in the direct limit is an isomomorphism, and hence $H^1(F,T)\cong H^1(E/F,T)$.

*Notice*: no snakes were harmed in the testing of this spectral sequence.