Sign in the product of the LHS spectral sequence

2012-01-09T21:31:03Z

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 $$ E_2^{ip}(M) \otimes E_2^{jq}(N)\hspace{180pt}$$ $$\begin{array}{cl} = & H^i(Q,H^p(H,M) \otimes H^j(Q,H^q(H,M) \newline \xrightarrow[]{\cup_Q} & H^{i+j}(Q,H^p(H,M) \otimes H^q(H,N)) \newline \xrightarrow[]{\cup_H^\ast} & H^{i+j}(Q,H^{p+q}(H,M \otimes N)) \newline = & E_2^{i+j,p+q}(M \otimes N). \end{array}$$ 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.

Question: Where does this sign come from ?

Answer by Chris Gerig for Sign in the product of the LHS spectral sequence

2012-01-09T22:22:37Z

It comes in when comparing the definition of that spectral sequence (for $E_2$) to the normal cup product:
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:
$Hom(Y,Hom(X,M))\otimes Hom(Y,Hom(X,N))\rightarrow Hom(Y\otimes Y,Hom(X\otimes X,M\otimes N))$, which is defined in an obvious way yet it involves a sign. Now when you compare this $E_2$-product to the orindary cup product for $G/H$ cohomology, that sign $(-1)^{pj}$ appears.

Answer by Ralph for Sign in the product of the LHS spectral sequence

2012-01-10T01:09:18Z

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 $$T : X \otimes Y \to Y \otimes X,\; x \otimes y \mapsto (-1)^{ij} \cdot y \otimes x\quad,\quad x \in X_i, y \in Y_j.$$

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.

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 $$\Delta_{i+p,j+q}(x \otimes y) = T(x_i \otimes x_j \otimes y_p \otimes y_q) =(-1)^{jp}(x_i \otimes y_p) \otimes (x_j \otimes y_q).$$

Hence we have for the cup-product $$\cup: Hom_G(X \otimes Y,M) \otimes Hom_G(X \otimes Y,N) \to Hom_G(X \otimes Y,M \otimes N)$$ $$(f \cup g) (x \otimes y) = (-1)^{jp} f(x_i \otimes y_p) \otimes g(x_j \otimes y_q).\quad\quad(1)$$

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 $G$.

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 $(-1)^{jp}$.

Remark: 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.

Added: 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.

Sign in the product of the LHS spectral sequence - MathOverflow

Sign in the product of the LHS spectral sequence

Answer by Chris Gerig for Sign in the product of the LHS spectral sequence

Answer by Ralph for Sign in the product of the LHS spectral sequence