Sign in the product of the LHS spectral sequence - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T15:19:32Z http://mathoverflow.net/feeds/question/85291 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/85291/sign-in-the-product-of-the-lhs-spectral-sequence Sign in the product of the LHS spectral sequence TJ 2012-01-09T21:31:03Z 2012-01-10T18:27:59Z <p>Given an extension of groups $$1 \to H \to G \to Q \to 1,$$ there is a spectral sequence $$E^{ip}_2(M) = H^i(Q,H^p(H,M)) \Rightarrow H^{i+p}(G,M).$$ I understand that the composition of the cup products for $Q$ and $H$ defines a pairing $$E_2^{ip}(M) \otimes E_2^{jq}(N)\hspace{180pt}$$ $$\begin{array}{cl} = &amp; H^i(Q,H^p(H,M) \otimes H^j(Q,H^q(H,M) \newline \xrightarrow[]{\cup_Q} &amp; H^{i+j}(Q,H^p(H,M) \otimes H^q(H,N)) \newline \xrightarrow[]{\cup_H^\ast} &amp; H^{i+j}(Q,H^{p+q}(H,M \otimes N)) \newline = &amp; E_2^{i+j,p+q}(M \otimes N). \end{array}$$ But according to section 7.3 in L. Evens' book (The Cohomology of Groups), the sign $$(-1)^{pj}$$ is needed in order to make this pairing a proper product in the spectral sequence. </p> <p><em>Question:</em> Where does this sign come from ? </p> http://mathoverflow.net/questions/85291/sign-in-the-product-of-the-lhs-spectral-sequence/85294#85294 Answer by Chris Gerig for Sign in the product of the LHS spectral sequence Chris Gerig 2012-01-09T22:22:37Z 2012-01-09T22:22:37Z <p>It comes in when comparing the definition of that spectral sequence (for $E_2$) to the normal cup product:<br> You take your projective resolutions $X$ and $Y$ associated to the groups $G$ and $G/H=Q$, and take diagonal maps $X\rightarrow X\otimes X$ and $Y\rightarrow Y\otimes Y$. These induce a map $Hom(Y\otimes Y,Hom(X\otimes X,M\otimes N))\rightarrow Hom(Y,Hom(X,M\otimes N))$, for appropriate module homomorphism groups. But to define the multiplicative structure on the relevent double complexes for $E$, you obtain the desired pairing by tacking on a product map to this above map:<br> $Hom(Y,Hom(X,M))\otimes Hom(Y,Hom(X,N))\rightarrow Hom(Y\otimes Y,Hom(X\otimes X,M\otimes N))$, which is defined in an obvious way <em>yet it involves a sign</em>. Now when you compare this $E_2$-product to the orindary cup product for $G/H$ cohomology, that sign $(-1)^{pj}$ appears.</p> http://mathoverflow.net/questions/85291/sign-in-the-product-of-the-lhs-spectral-sequence/85307#85307 Answer by Ralph for Sign in the product of the LHS spectral sequence Ralph 2012-01-10T01:09:18Z 2012-01-10T18:27:59Z <p>Let $X \to k$ resp. $Y \to k$ be projective resolutions of $k$ over $kG$ resp. $kQ$. In short, the reason for the sign is the twist $$T : X \otimes Y \to Y \otimes X,\; x \otimes y \mapsto (-1)^{ij} \cdot y \otimes x\quad,\quad x \in X_i, y \in Y_j.$$</p> <p>In detail: First note that if $U \to k$ is a projective resolution of $k$ over $kG$ and $\Delta: U \to U \otimes U$ is a diagonal approximation, then (using Evens' sign conventions) cup-product $$Hom_G(U,M) \otimes Hom_G(U,N) \to Hom_G(U,M\otimes N)$$ is given as follows: Let $\Delta_{ij}: U_{i+j} \to U_i \otimes U_j$ be the $(i,j)$-component of $\Delta$. If $\Delta_{ij}(u) = u_i \otimes u_j$ then $$(f \cup g)(u) = f(u_i) \otimes g(u_j).\hspace{80pt}(\ast)$$ Now let $\Delta_X: X \to X \otimes X$ and $\Delta_Y: Y \to Y \otimes Y$ be diagonal approximations. $X \otimes Y \to k$ is a projective resolution for $G$ (diagonal operation) and $$\Delta: X \otimes Y \xrightarrow{\Delta_X \otimes \Delta_Y} X \otimes X \otimes Y \otimes Y \xrightarrow{\;T\;} (X\otimes Y) \otimes (X \otimes Y)$$ is a diagonal approximation. </p> <p>Let $x \in X_{i+j}$ and $y \in Y_{p+q}$. If $\Delta_{i,j}(x) = x_i \otimes x_j$ and $\Delta_{p,q}(y) = y_p \otimes y_q$ then $$\Delta_{i+p,j+q}(x \otimes y) = T(x_i \otimes x_j \otimes y_p \otimes y_q) =(-1)^{jp}(x_i \otimes y_p) \otimes (x_j \otimes y_q).$$ </p> <p>Hence we have for the cup-product $$\cup: Hom_G(X \otimes Y,M) \otimes Hom_G(X \otimes Y,N) \to Hom_G(X \otimes Y,M \otimes N)$$ $$(f \cup g) (x \otimes y) = (-1)^{jp} f(x_i \otimes y_p) \otimes g(x_j \otimes y_q).\quad\quad(1)$$</p> <p>One can check that this product preserves the usual filtration and hence defines a product on the spectral sequence that is compatible with the product in the cohomology of $G$. </p> <p>Since your pairing is a composition of the cup-product on $Q$ and on $H$, it can be seen from $(\ast)$ that no sign occurs there. Furthermore, using the isomorphism $$Hom_G(X \otimes Y,-) \cong Hom_Q(Y,Hom_H(X,-))$$ of complexes, one obtains by definition checking - use $(\ast)$ - that your pairing is given on cochain level by $$Hom_G(X \otimes Y,M) \otimes Hom_G(X \otimes Y,N) \to Hom_G(X \otimes Y,M \otimes N)$$ $$(f \cup g) (x \otimes y) = f(x_i \otimes y_p) \otimes g(x_j \otimes y_q).$$ This shows that the two products differ by the sign $(-1)^{jp}$. </p> <p><strong>Remark:</strong> The sign depends on the definition of the differential in the $Hom$-cocomplex (that influences $(\ast)$). See for example the paper [Hochschild, Serre: Cohomology of Group Extensions, Trans. Amer. Math. Soc. 74(1),1953, pp. 110-134] Chap. II, Theorem 3 for a different sign. </p> <p><strong>Added:</strong> To explicate this remark, consider the sign convention in Brown's group cohomology book. He uses $$\delta: Hom_G(U_n,M) \to Hom_G(U_{n+1},M),\; f \mapsto (-1)^{n+1}f \circ d_{n+1}.$$ This leads to the cup product $$(f \cup g)(u) = (-1)^{ij} \cdot f(u_i) \otimes g(u_j)\hspace{80pt}(\ast')$$ Thus we obtain in place of $(1)$: $$(f \cup g)(x \otimes y) = (-1)^{jp} (-1)^{(i+p)(j+q)} f(x_i \otimes y_p) \otimes g(x_j \otimes y_q)\quad\quad (1')$$ Let $F \in Hom_H(Y_p,Hom_Q(X_i,M))$ correspond to $f \in Hom_G( (X\otimes Y)_{i+p},M)$ and $G \in Hom_H(Y_q,Hom_Q(X_j,M))$ correspond to $g \in Hom_G( (X\otimes Y)_{j+q},N)$. Then by $(\ast')$: $$(F \cup_Q G)(y) = (-1)^{pq} \cdot F(y_p) \otimes F(y_q).$$ Thus we obtain for your pairing $$(F \cup G)(y)(x) = (-1)^{pq} \big (F(y_p) \cup_H F(y_q) \big )(x) = (-1)^{pq}(-1)^{ij} F(y_p)(x_i) \otimes F(y_q)(x_j).$$ Comparing with $(1')$ shows that the two products differ by the sign $(-1)^{iq}$, in accordance with Hochschild-Serre. </p>