# Friday, March 26 ## 5.6: Two Spectral Sequences on Total Complexes :::{.remark} Recall that we had two filtrations on a total complex: the first was fixing a vertical line and replacing everything to the right with zeros, which was given by $^{I}E_{p, q}^0 = F_p(\Tot)/ F_{p-1}(\Tot) = C_{p, q}$. Taking homology with the vertical differentials yielded $^{I}E_{p, q}^1 = H_q^v(C_{p,*})$, and $^{I} E_{p, q}^2 = H_p^h H_q^v(C_{*, *})$. Applying the classical convergence theorem when this is 1st quadrant yields some spectral sequence with these as the pages which converges to $H_{p+q}(\Tot(C))$. ::: :::{.definition title="The second filtration"} We'll define a filtration by rows: let $^{II}F_n \Tot(C)$ be the total complex of the double complex \[ ({}^{II} \tau_{\leq n} C)_{p, q} &= \begin{cases} C_{p, q} & p, q\leq \\ 0 & p, q > n. \end{cases} \] This is the complex gotten by replacing everything below the $n$th row with zeros. We define the 0th page \[ {}^{II} E^{0}_{p, q} = { {}^{II} F_p \Tot(C)_{p+q} \over {}^{II} F_{p-1} \Tot(C)_{p+q} } = C_{q, p} ,\] which follows from the fact that we are modding out a full diagonal by a diagonal with one fewer elements: \begin{tikzcd} & \ddots & \vdots & \vdots & \vdots & \vdots & \ddots \\ & \cdots & 0 & 0 & 0 & 0 & \cdots \\ p &&&&&& {} & {} \\ & \cdots & \textcolor{rgb,255:red,214;green,92;blue,92}{\bullet} & \bullet & \bullet & \bullet & \cdots \\ & \cdots & \bullet & \textcolor{rgb,255:red,92;green,92;blue,214}{\bullet} & \bullet & \bullet & \cdots \\ & \cdots & \bullet & \bullet & \bullet & \bullet & \cdots \\ & \cdots & \bullet & \bullet & \bullet & \bullet & \ddots \\ & \ddots & \vdots & \vdots & \vdots & \ddots & \ddots \\ && q \arrow["{F_p}"', shift right=3, color={rgb,255:red,214;green,92;blue,92}, squiggly, no head, from=7-6, to=4-3] \arrow["{F_{p-1}}", shift left=3, color={rgb,255:red,92;green,92;blue,214}, squiggly, no head, from=7-6, to=5-4] \arrow[dashed, no head, from=3-1, to=3-8] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=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) ::: :::{.warnings} Note the switched order! ::: :::{.remark} Note that the differential is \[ d^0: E^0_{p, q} &\to E^0_{p, q-1} \\ = d^h: C_{q, p} &\to C_{q-1, p} .\] We similarly have ${}^II E_{p, q}^I = H_q^h(C_{*, p})$, again noting the switched indices, with differential \[ d^1: E^1_{p, q} &\to E_{p-1, q}^1 \\ =H^h(C_{q, p}) &\to H^h(C_{*, p-1}) \] which comes from the original differential inducing a map on horizontal homology. Then ${}^{II} E^2_{p, q} = H_p^v H_q^h(C)$. ::: :::{.remark} Note that transposing everything about the line $p=q$ interchanges filtrations $I$ and $II$, and thus the two spectral sequences ${}^{I}E_{p, q} \mapstofrom {}^{II} E_{q, p}$. Using that first quadrant sequences are canonically bounded, we can apply the classical convergence theorem to ${}^{II} E$ to obtain \[ {}^{II}E_{p, q}^2 \abuts H_{p+q}( \Tot(C) ) .\] Transposing sends $QIV$ to $QII$ and thus ${}^{II} E \abuts H_{p+q}\Totsum(C)$. Note that this does not guarantee anything about $\Totprod(C)$. ::: :::{.remark} In particular, if we have a $QI$ double complex, both filtrations converge to the homology of the total complex. ::: ## Application: Balancing Tor :::{.remark} Our proof in 2.7 that $\Tor_*^R(A, B)$ could be computed either by a projective resolution $\complex{P}\surjects A$ or a projective resolution $\complex{Q}\surjects B$ was a disguised spectral sequence argument. So we'll go recover it using the actual spectral sequence. ::: :::{.remark} We have a $QI$ double complex $C$ given by $C_{p, q} \da (P\tensor Q)_{p, q} = P_p\tensor Q_q$, and we now have two spectral sequences converging to $H_*(\Tot(P\tensor Q))$. Taking the first filtration, we can write \[ H_q^v(\Tot(C)) = H_q(P_p \tensor Q_q) = P_p \tensor H_q(Q) .\] Using that $P$ is an exact complex, and noting that we delete the augmentation when taking homology, we have \[ H_1^v(\Tot(C)) = \begin{cases} 0 & q>0 \\ P_p\tensor B & q=0. \end{cases} \] Thus \[ E^2_{p, q} = \begin{cases} H_p^h(P_* \tensor B) & q=0 \\ 0 & 1>0, \end{cases} \] meaning that this collapses at $E^2$ and we have \[ H_p (\Tot(P\tensor Q) ) \cong L_p(\wait \tensor B)(A) \da \Tor^R_p(A, B) .\] Now consider taking the second filtration, which yields \[ {}^{II} E_{p, q}^1 = H_q^h( P_q \tensor Q_p) = H_q(P_*) \tensor Q_p = \begin{cases} A_\tensor Q_p & q=0 \\ 0 & q>0. \end{cases} \] The second pages comes from taking the vertical homology, so \[ {}^{II}E_{p, q}^2 = H_p^v H_q^h(P_q \tensor Q_p) = \begin{cases} H^v_p(A\tensor Q) & q=0 \\ 0 & q>0. \end{cases} ,\] which is $L_p(A\tensor \wait)(B)$ in $q=0$. Since ${}^{II}E_{p, q}^2 \abuts H_{p+q}(\Tot(P\tensor Q)) = L_p(\wait \tensor B)(A)$, and we thus have \[ L_p(A\tensor \wait)(B) \cong L_p(\wait \tensor B)(A) .\] ::: :::{.remark} See the this section of Weibel for other applications in the exercises: the Kunneth formula, the Universal Coefficient Theorem, and the Acyclic Assembly Lemma. ::: ## Hypercohomology :::{.remark} We'd like to compute derived functors acting on chain complexes instead of just objects. ::: :::{.definition title="Cartan-Eilenberg Resolutions"} Let $\cat{A}$ be an abelian category with enough projectives and let $\complex{A} \in \Ch(\cat{A})$. A (left) **Cartan-Eilenberg resolution** (a CE resolution) $P_{*, *}$ of $A_*$ is an upper half-plane complex (so $P_{p, q} = 0$ when $q<0$) of projective objects and an augmentation chain map $P_{*, 0} \mapsvia{\eps} A_*$ such that 1. If $A_p=0$ then the entire column $P_{p, *}$ is zero. 2. The augmentation induces maps on boundaries and in homology which are projective resolutions in $\cat{A}$: \[ B_p(P, d^h) &\mapsvia{B_p(\eps)} B_p(A) \\ H_p(P, d^h) &\mapsvia{H_p(\eps)} H_p(A) .\] ::: :::{.remark} So we have the following situation \begin{tikzcd} {q:} & \cdots && {P_{p+1, q}} && {P_{p, q}} && {P_{p-1, q}} && \cdots \\ &&& \vdots && \vdots && \vdots \\ & \cdots && {P_{p+1, 1}} && {P_{p, 1}} & {} & {P_{p-1, 1}} && \cdots \\ \\ & \cdots && {P_{p+1, 0}} && {P_{p, 0}} && {P_{p-1, 0}} && \cdots \\ {} &&&&&&&&&& {} \\ & \cdots && {A_{p-1}} && {A_{p}} && {A_{p-1}} && \cdots \arrow[from=3-8, to=3-6] \arrow[from=7-8, to=7-6] \arrow[from=7-6, to=7-4] \arrow[from=5-8, to=5-6] \arrow[from=5-6, to=5-4] \arrow[from=3-6, to=3-4] \arrow[from=3-4, to=5-4] \arrow[from=5-4, to=7-4] \arrow[from=3-6, to=5-6] \arrow[from=5-6, to=7-6] \arrow[from=3-8, to=5-8] \arrow[from=5-8, to=7-8] \arrow[from=3-10, to=3-8] \arrow[from=5-10, to=5-8] \arrow[from=7-10, to=7-8] \arrow[from=7-4, to=7-2] \arrow[from=5-4, to=5-2] \arrow[from=3-4, to=3-2] \arrow[from=1-10, to=1-8] \arrow[from=1-8, to=1-6] \arrow[from=1-6, to=1-4] \arrow[from=1-4, to=1-2] \arrow[color={rgb,255:red,92;green,92;blue,214}, dotted, no head, from=6-1, to=6-11] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=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) The situation in row $q$ will be: \begin{tikzcd} \cdots && {P_{p+1, q}} && {P_{p, q}} && {P_{p-1, q}} && \cdots \\ \\ &&&& {Z_p(P, d^h)} \\ &&&&&& {H_p(P, d^h)_q} \\ &&&& {B_p(P, d^h)} \arrow[from=1-9, to=1-7] \arrow[from=1-7, to=1-5] \arrow[from=1-5, to=1-3] \arrow[from=1-3, to=1-1] \arrow[hook, from=3-5, to=1-5] \arrow[hook, from=5-5, to=3-5] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsOCxbNiwwLCJQX3twLTEsIHF9Il0sWzQsMCwiUF97cCwgcX0iXSxbMiwwLCJQX3twKzEsIHF9Il0sWzgsMCwiXFxjZG90cyJdLFswLDAsIlxcY2RvdHMiXSxbNCwyLCJaX3AoUCwgZF5oKSJdLFs0LDQsIkJfcChQLCBkXmgpIl0sWzYsMywiSF9wKFAsIGReaClfcSJdLFszLDBdLFswLDFdLFsxLDJdLFsyLDRdLFs1LDEsIiIsMCx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV0sWzYsNSwiIiwwLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiaG9vayIsInNpZGUiOiJ0b3AifX19XV0=) Here when we take the homology of the complex along the rows $p$, we'll obtain \[ H_q(P, d^h) = {Z_p(P, d^h)_q \over B_p(P, d^h)_q} ,\] and since the induces maps preserve cycles and boundaries, we get induced maps on homology. Exercise 5.7.1 shows that $P_{p, *} \mapsvia{\eps} A_p$ will be a projective resolution in $\cat{A}$ and so $Z_p(P, d^h)_* \to Z_p(A)$. ::: :::{.lemma title="?"} Every $A_*$ has a CE resolution $P_{*, *} \mapsvia{\eps} A$. ::: :::{.proof title="?"} Choose a levelwise resolution and use the horseshoe lemma: \begin{tikzcd} 0 && {B_p(A)} && {Z_p(A)} && {H_p(A)} && 0 \\ \\ 0 && {P^B_{p, *}} && \textcolor{rgb,255:red,92;green,92;blue,214}{P^Z_{p, *}} && {P^H_{p, *}} && 0 \arrow[from=1-1, to=1-3] \arrow[from=1-3, to=1-5] \arrow[from=1-5, to=1-7] \arrow[from=1-7, to=1-9] \arrow[from=3-3, to=1-3] \arrow[from=3-7, to=1-7] \arrow[from=3-1, to=3-3] \arrow[draw={rgb,255:red,92;green,92;blue,214}, dashed, from=3-3, to=3-5] \arrow[draw={rgb,255:red,92;green,92;blue,214}, dashed, from=3-5, to=3-7] \arrow[from=3-7, to=3-9] \arrow[draw={rgb,255:red,92;green,92;blue,214}, dashed, from=3-5, to=1-5] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsMTAsWzAsMCwiMCJdLFsyLDAsIkJfcChBKSJdLFs0LDAsIlpfcChBKSJdLFs2LDAsIkhfcChBKSJdLFs4LDAsIjAiXSxbNiwyLCJQXkhfe3AsICp9Il0sWzQsMiwiUF5aX3twLCAqfSIsWzI0MCw2MCw2MCwxXV0sWzIsMiwiUF5CX3twLCAqfSJdLFswLDIsIjAiXSxbOCwyLCIwIl0sWzAsMV0sWzEsMl0sWzIsM10sWzMsNF0sWzcsMV0sWzUsM10sWzgsN10sWzcsNiwiIiwxLHsiY29sb3VyIjpbMjQwLDYwLDYwXSwic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV0sWzYsNSwiIiwxLHsiY29sb3VyIjpbMjQwLDYwLDYwXSwic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV0sWzUsOV0sWzYsMiwiIiwxLHsiY29sb3VyIjpbMjQwLDYwLDYwXSwic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV1d) Recall that this involved a direct sum construction. Now do a similar thing for the following SES: \begin{tikzcd} 0 && {Z_p(A)} && {A_p} && {B_{p-1}(A)} && 0 \\ \\ 0 && {P^Z_{p, *}} && \textcolor{rgb,255:red,92;green,92;blue,214}{P^A_{p, *}} && {P^B_{p-1, *}} && 0 \arrow[from=1-1, to=1-3] \arrow[from=1-3, to=1-5] \arrow["{d_p}", from=1-5, to=1-7] \arrow[from=1-7, to=1-9] \arrow[from=3-3, to=1-3] \arrow[from=3-7, to=1-7] \arrow[from=3-1, to=3-3] \arrow[draw={rgb,255:red,92;green,92;blue,214}, dashed, from=3-3, to=3-5] \arrow["{\tilde{d_p}}", draw={rgb,255:red,92;green,92;blue,214}, dashed, from=3-5, to=3-7] \arrow[from=3-7, to=3-9] \arrow[draw={rgb,255:red,92;green,92;blue,214}, dashed, from=3-5, to=1-5] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=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) We use the fact that we have the two side resolutions from the previous step. So set $P_{p, q} \da P_{p, q}^A$ assembled into a double complex using the sign trick: $d^v \da (-1)^p d$ where we used the differential $d$ from $P_{p, *}^A$. We can now define \[ d^h: P^A_{p+1, *} \mapsvia{\tilde d_{p+1} } P_{p, *}^B \injects P_{p, *}^Z \injects P_{p, *}^A .\] One then checks that $B_p(\eps)$ and $H_p(\eps)$ are indeed projective resolutions. :::