# Monday, March 29 ## Maps of Double Complexes :::{.remark} Last time: we talked about hypercohomology. We're doing this so we can set up a Grothendieck spectral sequence. ::: :::{.definition title="Chain homotopies of double complexes"} Let $f, g:D\to E$ be two maps between double complexes. A **chain homotopy** from $f$ to $g$ consists of $s_{p, q}^h: D_{p, q} \to E_{p+1, q}$ and $s_{p, q}^v: D_{p, q} \to E_{p, q+1}$ for all $p, q$ satisfying the following conditions: 1. All of the possible maps $D_{p, q} \to E_{p, q}$ summed should be equal to $g-f$, i.e. $g-f = (d^h s^h + s^h d^h) + (d^v s^v + s^v d^v)$: \begin{tikzcd} &&&& {E_{p, q+1}} \\ \\ &&&& {E_{p, q}} && {E_{p+1, q}} \\ \\ {D_{p-1, q}} && {D_{p, q}} \\ \\ && {D_{p, q-1}} \arrow["{d^v}"{description}, draw={rgb,255:red,16;green,178;blue,32}, from=1-5, to=3-5] \arrow["{d^v}"{description}, draw={rgb,255:red,16;green,178;blue,32}, from=5-3, to=7-3] \arrow["{s_{p-1, q}^h}"{description, pos=0.2}, color={rgb,255:red,149;green,68;blue,55}, dashed, from=5-1, to=3-5] \arrow["{s^v_{p, q-1}}"{description, pos=0.9}, color={rgb,255:red,16;green,178;blue,32}, dashed, from=7-3, to=3-5] \arrow["{d^h}"{description}, draw={rgb,255:red,149;green,68;blue,55}, from=5-3, to=5-1] \arrow["{d^h}"{description}, draw={rgb,255:red,149;green,68;blue,55}, from=3-7, to=3-5] \arrow["{s_{p, q}^h}"{description, pos=0.2}, color={rgb,255:red,149;green,68;blue,55}, dashed, from=5-3, to=3-7] \arrow["{s^v_{p, q}}"{description, pos=0.9}, color={rgb,255:red,16;green,178;blue,32}, dashed, from=5-3, to=1-5] \arrow["{g-f}"{description}, curve={height=-30pt}, from=5-3, to=3-5] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsNixbMCw0LCJEX3twLTEsIHF9Il0sWzIsNCwiRF97cCwgcX0iXSxbMiw2LCJEX3twLCBxLTF9Il0sWzQsMiwiRV97cCwgcX0iXSxbNiwyLCJFX3twKzEsIHF9Il0sWzQsMCwiRV97cCwgcSsxfSJdLFs1LDMsImRediIsMSx7ImNvbG91ciI6WzEyNiw4NCwzOF19XSxbMSwyLCJkXnYiLDEseyJjb2xvdXIiOlsxMjYsODQsMzhdfV0sWzAsMywic197cC0xLCBxfV5oIiwxLHsibGFiZWxfcG9zaXRpb24iOjIwLCJjb2xvdXIiOls4LDQ2LDQwXSwic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fSxbOCw0Niw0MCwxXV0sWzIsMywic152X3twLCBxLTF9IiwxLHsibGFiZWxfcG9zaXRpb24iOjkwLCJjb2xvdXIiOlsxMjYsODQsMzhdLCJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19LFsxMjYsODQsMzgsMV1dLFsxLDAsImReaCIsMSx7ImNvbG91ciI6WzgsNDYsNDBdfV0sWzQsMywiZF5oIiwxLHsiY29sb3VyIjpbOCw0Niw0MF19XSxbMSw0LCJzX3twLCBxfV5oIiwxLHsibGFiZWxfcG9zaXRpb24iOjIwLCJjb2xvdXIiOls4LDQ2LDQwXSwic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fSxbOCw0Niw0MCwxXV0sWzEsNSwic152X3twLCBxfSIsMSx7ImxhYmVsX3Bvc2l0aW9uIjo5MCwiY29sb3VyIjpbMTI2LDg0LDM4XSwic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fSxbMTI2LDg0LDM4LDFdXSxbMSwzLCJnLWYiLDEseyJjdXJ2ZSI6LTV9XV0=) 2. The two rectangles below should be zero, i.e. $s^v d^h + d^h s^v = 0 = s^h d^v + d^v s^h$: \begin{tikzcd} && {E_{p-1, q+1}} && {E_{p, q+1}} \\ \\ &&&& {E_{p, q}} && {E_{p+1, q}} \\ \\ {D_{p-1, q}} && {D_{p, q}} &&&& {E_{p+1, q-1}} \\ \\ && {D_{p, q-1}} \arrow["{d^v}", from=1-5, to=3-5] \arrow["{d^v}", color={rgb,255:red,171;green,43;blue,43}, from=5-3, to=7-3] \arrow["{d^h}"', color={rgb,255:red,38;green,151;blue,38}, from=5-3, to=5-1] \arrow["{d^h}"', from=3-7, to=3-5] \arrow["{g-f}"{description}, from=5-3, to=3-5] \arrow["{d^v}"{description}, color={rgb,255:red,171;green,43;blue,43}, from=3-7, to=5-7] \arrow["{s^h_{p, q-1}}"', color={rgb,255:red,171;green,43;blue,43}, from=7-3, to=5-7] \arrow["{s^h_{p, q}}"', color={rgb,255:red,171;green,43;blue,43}, from=5-3, to=3-7] \arrow["{s^v_{p-1, q}}", color={rgb,255:red,38;green,151;blue,38}, from=5-1, to=1-3] \arrow["{d^h}"{description}, color={rgb,255:red,38;green,151;blue,38}, from=1-5, to=1-3] \arrow["{s^v_{p, q}}", color={rgb,255:red,38;green,151;blue,38}, from=5-3, to=1-5] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=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) ::: :::{.remark} The definition is set up so that $s^h + s^v: \Tot(D)_n \to \Tot(E)_{n+1}$ is a chain homotopy $\Totsum(D) \to \Totsum(E)$. ::: :::{.remark} Exercises 5.7.2 and 5.7.3 show: 1. If $f:A\to B$ is a chain map and $P\to A, Q\to B$ are CE resolutions, then there is a map of double complexes $\tilde f: P\to Q$ lifting $f$. 2. If $f, g: A\to B$ are chain homotopic, then $\tilde f, \tilde g$ are chain homotopic in the sense just defined. 3. Any two CE resolutions $P, P'$ of $A$ are chain homotopy equivalent, as are $\Totsum(F(P))$ and $\Totsum(F(P'))$ for any additive functor $F$. ::: :::{.remark} This last remark shouldn't be too hard to believe: chain homotopies are defined in terms of addition. ::: ## Hypercohomology :::{.definition title="Hyper Left-Derived Functors"} Let \( F : \cat{A} \to \cat{B} \) be a right-exact functor where $\cat{A}$ has enough projectives and $\cat{B}$ is cocomplete (closed under direct sums/coproducts). If $A \in \Ch(\cat{A})$ is a chain complex and $P\to A$ a CE resolution, define \[ \LL_i F(A) \da H_i \Totsum F(P): \Ch(\cat{A}) \to \cat{B} .\] If $f:A\to B$ is a chain map in $\Ch(\cat{A})$ and $\tilde f: P\to Q$ where $P, Q$ are CE resolutions of $A, B$ resp., define $L_iF(f)$ to be the map \[ H_i \Tot(F\tilde f) \to \LL_i F(B) .\] This yields a functor \[ \LL_i F: \Ch(\cat{A}) \to \cat{B} ,\] the **hyper left-derived functor** of $F$. ::: :::{.remark} Recall that chain homotopy yields a notion of equivalence, and chain homotopic maps induce the same map on homology. The same is true for double complexes. There is a lemma that shows a SES of double complexes induces a LES in homology. ::: :::{.proposition title="Convergence of spectral sequences and filtration comparison"} \envlist a. There is always a convergent spectral sequence \[ {}^{II} E^2_{p, q} (L_p F)(H_q(A)) \abuts \LL_{p+q} F(A) .\] b. If $A$ is bounded below complex, so there exists a $p_0$ such that $A_p=0$ for $p< p_0$, then there is another spectral sequence \[ {}^{I} E_{p, q}^2 = H_p L_q F(A) \abuts \LL_{p+q} F(A) .\] ::: :::{.proof title="of (a)"} These are the spectral sequences associated to the upper half-plane double complex $FP_{*, *}$. Recall that ${}^{II} E^2_{p, q} = H_p^v H_q^h (FP) \abuts H_{p+q} \Totsum FP \da \LL_{p+q} F(A)$. The filtration by rows is exhaustive since we are taking the direct sum, so any cycle or boundary is supported in some finite row. So what we want to show is that \[ {}^{II}E_{p, q}^2 (L_p F)(H_q A) = H_p^v H_q^h (FP) .\] The main claim is the following: $H_q^h(FP) = F H_q^h(P)$. Fix a row $p$ of the double complex so we can drop $p$ and $h$ from the notation. We have the following situation: \begin{tikzcd} \cdots && {P_{q-1}} && {P_{q}} && {P_{q+1}} && \cdots \\ \\ &&&& {Z_q} \\ \\ &&&& {B_q} \arrow[hook, from=5-5, to=3-5] \arrow[hook, from=3-5, to=1-5] \arrow["d"', from=1-9, to=1-7] \arrow["d"', from=1-7, to=1-5] \arrow["d"', from=1-5, to=1-3] \arrow["d"', from=1-3, to=1-1] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsNyxbMiwwLCJQX3txLTF9Il0sWzQsMCwiUF97cX0iXSxbNiwwLCJQX3txKzF9Il0sWzQsMiwiWl9xIl0sWzQsNCwiQl9xIl0sWzgsMCwiXFxjZG90cyJdLFswLDAsIlxcY2RvdHMiXSxbNCwzLCIiLDAseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9fX1dLFszLDEsIiIsMCx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV0sWzUsMiwiZCIsMl0sWzIsMSwiZCIsMl0sWzEsMCwiZCIsMl0sWzAsNiwiZCIsMl1d) We have a SES \[ 0 \to B_q \to Z_q \to H_q \to 0 ,\] which induces a LES \begin{tikzcd} \cdots && {L_2FH_q} && {L_1FH_q} \\ \\ {FB_q} && {FZ_q} && {FH_q} && 0 \arrow[from=3-1, to=3-3] \arrow[from=3-3, to=3-5] \arrow[from=3-5, to=3-7] \arrow[from=1-5, to=3-1, out=0, in=180] \arrow[from=1-3, to=1-5] \arrow[from=1-1, to=1-3] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsNyxbNiwyLCIwIl0sWzQsMiwiRkhfcSJdLFsyLDIsIkZaX3EiXSxbMCwyLCJGQl9xIl0sWzQsMCwiTF8xRkhfcSJdLFsyLDAsIkxfMkZIX3EiXSxbMCwwLCJcXGNkb3RzIl0sWzMsMl0sWzIsMV0sWzEsMF0sWzQsM10sWzUsNF0sWzYsNV1d) We have $L_1 FH_q = 0$, since in the CE resolution we assume that $H_q(P, d^h)$ is projective. The second SES we have is \[ 0 \to Z_q \to P_q \mapsvia{d} B_{q-1} \] inducing the LES \begin{tikzcd} \cdots && {L_2FP_q} && {L_1FB_{q-1}} \\ \\ {FZ_q} && {FP_q} && {FB_{q-1}} && 0 \arrow[from=3-1, to=3-3] \arrow[from=3-3, to=3-5] \arrow[from=3-5, to=3-7] \arrow[from=1-5, to=3-1, in=180, out=0] \arrow[from=1-3, to=1-5] \arrow[from=1-1, to=1-3] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsNyxbNiwyLCIwIl0sWzQsMiwiRkJfe3EtMX0iXSxbMiwyLCJGUF9xIl0sWzAsMiwiRlpfcSJdLFs0LDAsIkxfMUZCX3txLTF9Il0sWzIsMCwiTF8yRlBfcSJdLFswLDAsIlxcY2RvdHMiXSxbMywyXSxbMiwxXSxbMSwwXSxbNCwzXSxbNSw0XSxbNiw1XV0=) Here $L_i F B_{q-1} = 0$ since $B_{p-q}(P, d^h)$ was projective. Putting these together, we have \[ H_{q}(FP) = { \ker Fd : FP_q \to FP_{q-1} \over \im Fd : FP_{q+1} \to FP_{q} } \cong {FZ_q \over FB_q} \cong FH_q(P_{*, *}) .\] Now what is its vertical homology? The map $H_q(P_{*, *}) \to H_q(A)$ is a projective resolution, so apply $F$ to the source -- it's no longer exact, and you get $FH_q(P)$ from above, and taking homology yields the left-derived functors applied to the source. Thus \[ H_p^v FH_q^h(P) = L_p F( H_q (A)) ,\] and the left-hand side is equal to $H_p^v H_q^h (FP)$. ::: :::{.exercise title="Prove (b)"} Prove part (b) of the proposition. ::: :::{.remark} There is a cohomology variant of this: everything dualizes to $R^i F(A)$ for a left exact functor $F: \cat{A}\to \cat{B}$ where $A\in \Ch(\cat{A})$, $\cat{A}$ has enough injectives, and $B$ is complete. Using a right CE resolution $I^{*, *}$ of injective objects in $A$ yields an upper half-plane complex with $A^{*} \to I^{*,0}$ such that the induces maps on cohomology are themselves injective resolutions of $B^p(A^{*})$ and $H^p(A^{*})$. In this case \[ R^i F(A^{*}) = H^i \Totprod F(I^{*, *}) .\] We can prove dual version of all of the results about left hyper-derived functors, although there are some slight convergence issues to worry about due to the direct product. :::