# Friday, April 02 ## Review: The Lyndon-Hochschild-Serre Spectral Sequence :::{.remark} We're trying to prove the Lyndon-Hochschild-Serre spectral sequence for $H\normal G$. ::: :::{.lemma title="?"} Let $H\normal G$ and $A\in\modsleft{G}$ with \[ \rho: G\to { G \over H} .\] Then $A_H, A^H$ are in \( \modsleft{{G \over H}} \) and $(\wait)^H$ (respectively $(\wait)_H$) are right (respectively left) adjoin to \[ \phi^\#: \modsleft{G \over H} \to \modsleft{G} .\] ::: :::{.theorem title="Lyndon-Hochschild-Serre Spectral Sequence"} Let $H\normal G$ and $A\in \mods{G}$, then there exist two $Q1$ spectral sequences: \[ E_{p, q}^2 &= H_p\qty{ {G \over H}, H_q(H;A)} \abuts H_{p+q}(G; A) \\ E^{p, q}_2 &= H^p\qty{ { G \over H}, H_q(H;A)} \abuts H^{p+q}(G; A) .\] ::: :::{.proof title="?"} We want to write this as a composition of functors: \begin{tikzcd} {\mods{G}} && {\mods{G/H}} && {} \\ \\ && \Ab \arrow["{(\wait)^H}", from=1-1, to=1-3] \arrow["{(\wait)^{G\over H}}", from=1-3, to=3-3] \arrow["{(\wait)^G}"', dashed, from=1-1, to=3-3] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCJcXG1vZHN7R30iXSxbMiwwLCJcXG1vZHN7Ry9IfSJdLFsyLDIsIlxcQWIiXSxbNCwwXSxbMCwxLCIoXFx3YWl0KV5IIl0sWzEsMiwiKFxcd2FpdClee0dcXG92ZXIgSH0iXSxbMCwyLCIoXFx3YWl0KV5HIiwyLHsic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV1d) We can write \[ (A^H)^{G/H} &= \ts{ a\in A \st ha = a \forall h\in H } \\ &= \ts{ a\in A^H \st \bar{g} a =a \forall \bar{g}\in G/H} \\ &= \ts{ a\in A \st \alpha= a \forall g\in G } \\ &= A^G .\] By the lemma, $(\wait)^H$ is right adjoint to $\rho^{\#}$, which is exact. By prop 2.3.10, it sends injectives to injectives, and injectives are $F\dash$acyclic for $F(\wait) = (\wait)^{G \over H}$. So this is a valid setup for the Grothendieck spectral sequence. ::: ## Application: Bootstrapping Homology of Cyclic Groups :::{.example title="?"} Let $C_m$ be cyclic of order $m$, and suppose we have the results from section 6.2: 1. If $m$ is odd, \[ H_q(C_m; \ZZ) = \begin{cases} \ZZ & q=0 \\ \ZZ/m & q \text{ odd} \\ 0 & q\text{ even}. \end{cases} \] 2. If $H\leq Z(G)$ and $A$ is a trivial \(G\dash\)module, then $G/H \actson H_*(H; A)$ trivially as well. [^note_iso_star] 3. If $A$ is a trivial $C_2\dash$module and let $\times 2:A\to A$ be multiplication, then \[ H_p(C_2; A) = \begin{cases} A & p = 0 \\ \coker(\times 2) = A/2A & p \text{ odd} \\ \ker(\times s) = \ts{ a\in A \st 2a = 0 } & p \text{ even}. \end{cases} \] Note that the previous fact was a special case of multiplication by $m$. Using the SES \[ 0 \to C_m \to C_{2m} \to C_2 \to 0 ,\] we can use the LHS spectral sequence to compute \[ E_{p, q}^2 = H_p( C_2; H_q(C_m; \ZZ)) \abuts H_{p+q}(C_{2m}; \ZZ) .\] Let $A = H_q(C_m; \ZZ)$, then by fact (2) we'll get a trivial $C_2\dash$module, and we can then use fact (3). - For $q=0$ we have \[ E_{p, 0}^2 &= H_p(C_2; \ZZ) \\ &= \begin{cases} \ZZ & p=0 \\ \ZZ/2 & p \text{ odd} \\ 0 & p \text{ even} \end{cases} && \text{by (3)} .\] - For $p=0$ we have \[ E_{0, q}^2 &= H_q(C_m; \ZZ) \\ &= \begin{cases} \ZZ & p=0 \\ \ZZ/m & p \text{ odd} \\ 0 & p \text{ even}. \end{cases} \] - For $q>0$ odd and $p>0$ odd, note that $\ZZ/m \mapsvia{\times 2} \ZZ/m$ is a bijection for odd $m$, so \[ E_{p, q}^2 = H_p(C_2; \ZZ/m) = 0 && \text{since } { \ZZ/m \over 2\ZZ/m} = 0 .\] - For $q>0$ odd and $p>0$ even, \[ E_{p, q}^2 = H_p(C_2; \ZZ/m) = 0 .\] - For $q>0$ even and $p>0$, \[ H_q(C_m; \ZZ) = 0 \implies E_{p, q}^2 = 0 .\] Thus the $E_2$ page of the LHS spectral sequence looks like the following, where there is only one possible nontrivial differential which is forced to be zero: \begin{tikzcd} q \\ & \bullet \\ && \vdots \\ 5 && {\ZZ/m} & \bullet &&&&&& {E_2} \\ 4 && \bullet & \bullet & \bullet \\ 3 && {\ZZ/m} & \bullet & \bullet & \bullet \\ 2 && \bullet & \bullet & \bullet & \bullet & \bullet \\ 1 && {\ZZ/m} & \bullet & \bullet & \bullet & \bullet & \bullet \\ 0 && \ZZ & {\ZZ/2} & \bullet & {\ZZ/2} & \bullet & {\ZZ/2} & \bullet & \cdots \\ \bullet &&&&&&&&&&& \bullet \\ & \bullet & 0 & 1 & 2 & 3 & 4 & 5 &&&& p \arrow[dashed, no head, from=10-1, to=10-12] \arrow["{d=0}"', color={rgb,255:red,92;green,92;blue,214}, from=9-5, to=8-3] \arrow[dashed, no head, from=2-2, to=11-2] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=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) Note that each diagonal only has (at most) two nonzero terms along the axes, and so we'll get a 2-term filtration. Recall that in general we get \( \ts{ F_i H_n }_{i=1}^n \) where $F_{\leq -1} H_n =0$ and $F_{\geq n}H_n = H_n$. Here $E_{0, n}^{\infty }$ comes from $F_{-1}, F_0$ and $E_{n, 0}^{\infty }$ comes from $F_{n-1}, F_n$. So we have \[ H_0(C_{2m}; \ZZ) &= \ZZ \\ H_n(C_{2m}; \ZZ) &= 0 \text{ for $n$ even} .\] For $n$ odd, we get a SES \[ 0 \to \ZZ/m \to H_n(C_{2m}; \ZZ) \to \ZZ/2 \to 0 .\] Letting $B\in \Ab$ be the middle term, its order is $2m$, the product of the two outer elements. By Cauchy's theorem, since $2\divides \# B$, there is an element $y\in B$ of order 2. So send the generator of $\ZZ/2$ to $y$ to form the splitting. Thus \[ B\cong \ZZ/m \oplus \ZZ/2 \cong \ZZ/m \times \ZZ/2 \cong \ZZ/2m ,\] where we've now used the $\gcd(2, m) = 1$. So \[ H_n(C_{2m}; \ZZ) = \begin{cases} \ZZ & n=0 \\ \ZZ/2m & n\text{ even} \\ 0 & n \text{ odd}. \end{cases} \] ::: :::{.question} Can you get the group homology of any cyclic group this way? Similar formulas likely hold, see section 6.2. ::: ## Restriction and Inflation :::{.remark} The exact sequence of low degree terms in the cohomological LHS spectral sequence are of the form \begin{tikzcd} 0 && {H^1(G/H; A^H)} && {H^1(G; A)} && {H^1(H; A)} \\ \\ && {H^2(G/H; A^H)} && {H^2(G; A)} \arrow[from=1-1, to=1-3] \arrow["{\text{inflation}}", from=1-3, to=1-5] \arrow["{\text{restriction}}", from=1-5, to=1-7] \arrow["{d_2}"', from=1-7, to=3-3, out=0, in=180] \arrow["{\text{inflation}}", from=3-3, to=3-5] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsNixbMCwwLCIwIl0sWzIsMCwiSF4xKEcvSDsgQV5IKSJdLFs0LDAsIkheMShHOyBBKSJdLFs2LDAsIkheMShIOyBBKSJdLFsyLDIsIkheMihHL0g7IEFeSCkiXSxbNCwyLCJIXjIoRzsgQSkiXSxbMCwxXSxbMSwyLCJcXHRleHR7aW5mbGF0aW9ufSJdLFsyLDMsIlxcdGV4dHtyZXN0cmljdGlvbn0iXSxbMyw0LCJkXzIiLDJdLFs0LDUsIlxcdGV4dHtpbmZsYXRpb259Il1d) Note that these maps have particular name, **inflation** and **restriction**. ::: :::{.remark} We thought of homology as a functor of the module $A$, but here we see it's varying. Can this be thought of as a functor of the group instead? Setup: let $\rho: H\to G$ be a group morphism, then recall that any \(G\dash\)module becomes an \(H\dash\)module by composition with $\rho$, which yields an exact functor \[ \rho^{\#}: \mods{G} \to \mods{H} .\] Letting $A\in\mods{G}$, set - $T_n(A) \da H_n(G; A)$ - $T^n(A) \da H^n(G; A)$ - $S_n(A) \da H_n(G; \rho^{\#} A)$ - $S^n(A) \da H^n(G; \rho^{\#} A)$ ::: [^note_iso_star]: Note that this can be phrased in terms of the image of the functor lying in trivial modules.