# Tuesday, August 31 :::{.remark} Today: a systematic way to compute group cohomology by taking standard resolution. For a fixed group $G$, we want to resolve $\ZZ$ by free $\ZZ[G]\dash$modules, so take a simplicial resolution \begin{tikzcd} \cdots \stackar{7}[r] & G\cartpower{3} \stackar{5}[r] & G^\cartpower{2} \stackar{3}[r] & G \end{tikzcd} Taking free $\ZZ\dash$modules yields \begin{tikzcd} \cdots \stackar{7}[r] & \ZZ[G\cartpower{3}] \stackar{5}[r] & \ZZ[G \cartpower{2}] \stackar{3}[r] & \ZZ[G] \end{tikzcd} Note that this is a simplicial set whose realization is $EG$. ::: :::{.proposition title="?"} $C_\bullet(G)$ is exact, and $\ZZ[G\cartpower{n}]$ is free in $\mods{\ZZ[G]}$ where $G\actson G\cartpower{n}$ diagonally and this extends linearly. ::: :::{.proof title="?"} $\ZZ[G \cartpower{n}]$ is a free $\ZZ[G]\dash$module, using that $\ts{(1, g_1, \cdots, g_{n-1}) \st g_k \in G}$ is a free basis, since these are representatives for $G\dash$orbits on $G\cartpower{n}$. That this is an exact complex will follow from a nullhomotopy $h: \ZZ[G\cartpower{n-1}] \to \ZZ[G\cartpower{n}]$ so that $hd + dh = \id$. Take the map $h(g_1, \cdots, g_n) = (e, g_1, \cdots, g_n)$, then \[ (hd)(g_1, \cdots, g_n) &= h \sum (-1)^i (g_1, \cdots, \hat g_i, \cdots, g_n)\\ &= \sum (-1)^i (e, g_1, \cdots, \hat g_i, \cdots, g_n) .\] and \[ (dh)(g_1, \cdots, g_n) &= d (e, g_1, \cdots, g_n) \\ &= (g_1,\cdots, g_n) - \sum (-1)^i (e, g_1, \cdots, \hat g_i, \cdots, g_n) ,\] and adding these two cancels the two summed terms and yields the identity. Then just recall from homological algebra that $x \in \ker d$ implies $x = hdx + dhx = dhx$, so $x\in \im d$, so this makes the complex exact. ::: :::{.corollary title="?"} For $G \in \Grp$ discrete and $M\in \mods{G}$, \[ H^*(G; M) &= H^*( \Homcx_{\ZZ[G]}(C_\bullet(G), M) ) \\ H_*(G; M) &= H^*( M \tensor_{\ZZ[G]} C_\bullet(G) ) .\] ::: :::{.remark} Can we find a smaller way to represent this? Note that \[ \ZZ[G^{\times n}] = \bigoplus_{(g_1, \cdots, g_n) \in G^{n-1}} \ZZ[G](1, g_1, \cdots, g_{n-1}) ,\] and there is a free/forgetful adjunction between modules and sets that yields \[ \Hom_{\ZZ[G]}( \ZZ[G \cartpower{n} ], M ) \cong \Hom_{\Set}(G\cartpower{n-1 }, M) .\] ::: :::{.definition title="Reduced Complex"} For $G\in \Grp$ discrete and $M\in \mods{G}$, set \[ \tilde C^r(G; M) \da \Hom_{\Set}(G \cartpower{r}, M) .\] The boundary maps are given by \[ \delta: \tilde C^0(G, M) &\to \tilde C^1(G, M) \\ \delta f( \sigma) &= \sigma f(\wait) - f(\wait) \\ \\ \delta: \tilde C^1(G, M) &\to \tilde C^2(G, M) \\ \delta f( \sigma, \tau) &= \sigma f(\tau) - f(\sigma \tau) + f(\sigma) \\ \\ \delta: \tilde C^2(G, M) &\to \tilde C^3(G, M) \\ \delta f( \sigma, \tau, \rho) &= \sigma f(\tau, \rho) - f(\sigma\tau, \rho) + f(\sigma, \tau\rho) - f(\sigma, \tau) .\] The pattern is multiply by $\sigma$ on the outside, cycle through multiplying it to each argument, and for the last term leave $\sigma$ off. ::: :::{.remark} Punchline: in principle, group cohomology is computable -- however, the complex is quite large and not practical for large groups. ::: ## Some Formal Properties :::{.proposition title="Spectral Sequences"} For $H\normal G$ and $M\in \mods{G}$, the **Hochschild-Serre spectral sequence** reads \[ E_2^{p, q} = H^p( G/H; H^q(H; M)) \abuts H^{p+q}(G; M) .\] ::: :::{.remark} This is useful for inducting on the lengths of composition series, since e.g. for solvable groups one can take $G/H$ to be cyclic and $H$ a smaller solvable group. ::: :::{.proposition title="Inflation/Restriction Exact Sequence"} This spectral sequence induces an **inflation/restriction exact sequence** \begin{tikzcd} 0 \\ {H^1\qty{ {G \over H}; M^H}} && {H^1\qty{G; M}} && {H^1\qty{H; M}^{G\over H}} \\ \\ {H^2\qty{ {G \over H}; M^H}} && {H^2\qty{G; M}} \arrow[from=1-1, to=2-1] \arrow[from=2-1, to=2-3] \arrow[from=2-3, to=2-5] \arrow[from=2-5, to=4-1, out=0, in=180] \arrow[from=4-1, to=4-3] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsNixbMCwxLCJIXjFcXHF0eXsge0cgXFxvdmVyIEh9OyBNXkh9Il0sWzIsMSwiSF4xXFxxdHl7RzsgTX0iXSxbNCwxLCJIXjFcXHF0eXtIOyBNfV57R1xcb3ZlciBIfSJdLFswLDMsIkheMlxccXR5eyB7RyBcXG92ZXIgSH07IE1eSH0iXSxbMiwzLCJIXjJcXHF0eXtHOyBNfSJdLFswLDAsIjAiXSxbNSwwXSxbMCwxXSxbMSwyXSxbMiwzXSxbMyw0XV0=) ::: :::{.remark} This comes from the bottom-left corner of the HS spectral sequence, which is a general principle for first quadrant spectral sequences. Note that the $G/H$ action comes from $G\actson H$ by conjugation, which yields a $G\dash$action on $H^*$, and since $H$ acts trivially on $H^*(H; M)$ (since e.g. $M^H$ has a trivial action), this action factors through $G/H$. ::: ## Forms, Torsors, and $H^1$ :::{.definition title="Forms/descent, a pseudo-definition"} Let $X\slice{k}$ be an object (e.g. a variety, a group scheme, a variety with extra structure), then a **form** of $X$ over $k$ is an object $X'\slice{k}$ with an isomorphism $X'\slice{k^s} \mapsvia{\sim} X$ (i.e. a **descent** of $X$). ::: :::{.example title="?"} For $X \da \PP^n\slice{k^s}$ then a form of $X\slice{k}$ is a Severi-Brauer variety, for example a smooth conic. ::: :::{.example title="Severi Brauers"} Let $E$ be a genus 1 curve, then $E$ is a form for its Jacobian $\Jac(E)$, i.e. it becomes isomorphic to its Jacobian if it has a rational point. Not every curve has such a point, so they only become isomorphic after base changing to a separable closure. Note that $\Jac(E) \actson E$ by addition of divisors (since Jacobians have degree zero, curves have divisors of degree 1, and adding them yields a degree 1 divisor). It is in fact a torsor. ::: :::{.example title="?"} If $L\slice{k}$ is a finite separable extension then $L$ is a form of $(k^s)\cartpower{n}$. ::: :::{.example title="?"} The groups $\SO(p, q)\slice{\RR}$, the matrices preserving a quadratic form \[ h_{p, q}\da \diag(1, \cdots, 1, -1, \cdots, -1) \] with $p$ copies of 1 and $q$ copies of $-1$, and these are all forms of $\SO(p+q)\slice{\CC}$. ::: :::{.proposition title="?"} Suppose $X\slice{k}$ is some object (e.g. a variety, then forms of $X_{k^s}$ over $k$ are canonically in bijection with $H^1_{\Gal}(k; \Aut(X_{k^s}))$ (recalling that this was defined as a direct limit). Note that this automorphism group may be nonabelian, which we still need to define. ::: :::{.proof title="?"} Suppose $\Aut(X_{k^s})$ is abelian, then we'll show the following stronger claim: :::{.claim} $X'_L \mapsvia{\sim} X_L$ since there is a bijection \[ \correspond{ \text{Forms of } X_{k^s} \\ \text{split by } L \slice{k} } &\mapstofrom H^1_\Gal(L\slice k; \Aut(X_L)) .\] ::: :::{.proof title="?"} Recall that \[ H^1(L\slice k; \Aut(X_L)) = H^1( \tilde C^\bullet(\Gal(L/k)); \Aut(X_L) ) .\] Given $X'\slice k$ split by $L$, we want a map $\Gal(L/k) \to \Aut(X_L)$. Choose an isomorphism $X'_L \mapsvia{\sim} X_L$, noting that Galois acts on the LHS since it's defined over $k$, which will be different from the natural action on the right-hand side. So we can take a map \[ f: \Gal(L/k) \to \Aut(X'/L) \mapsvia{\sim} \Aut(X_L) ,\] although this is not generally a homomorphism. Instead, $f(\sigma \tau) = f(\sigma) f(\tau)^\sigma$, a **crossed homomorphism** which involves acting on the coefficients of defining equations (which come from $L$). This says that $f\in \ker \delta$, the differential for $\tilde C^\bullet$. So we now have a map from forms split by $L$ to $H^1(\Gal(L/k), \Aut(X_L))$, and we'll show it's injective and surjective. \ **Injectivity:** Suppose $X', X''$ are isomorphic forms of $X$, so we have an isomorphism defined over $k$ of the form $X_L' \mapsvia{\sim} X_L''$. :::{.exercise title="?"} This changes $f$ by an element of the form $\delta(g)$ for $g\in \Aut(X_L)$. ::: **Surjectivity:** Given a crossed homomorphism $f: \Gal(L/k) \to \Aut(X_L)$, we want to produce a form of $X\slice k$ mapping to it. This is the hardest part of the argument! Suppose $X\slice k$ is a variety. First suppose $X\in \Aff\Var$, so $X = \spec R$ and $\Gal(L\slice k)\actson_f R_L = R \tensor_k L$, which is only an $L\dash$semilinear action. Then $X' = \spec(R_L)^{\Gal(L/k)}$, and the claim is that $X_L' \cong X_L$. The proof of this is **Galois descent**, i.e. there is an equivalence of tensor categories \[ \adjunction{(\wait) \tensor L} {(\wait)^{\Gal(L\slice k)} } {\mods{k}^{\tensor} }{\mods{L}^{\tensor} + \text{ a semilinear action of } \Gal(L\slice k) } \] Now for general $X$, one reduces to the case of affines. One can alternatively prove Galois descent without reference to affine varieties. ::: :::