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I have been working on the following problem.problem.

I have spent plenty of time trying to solve it myself. I am, however, unable to prove one small step in the argument. Beneath you can find my attempt. Step 2 (showing that $D$ is a principal derivation) is not so clear to me. I guess there must be some direct argument that says that $H^1(G, k^{m \times n}) = 0$, rather than doing explicit calculations.

Any help or advice is appreciated. Thanks.


1) We first show that $D: G \to k^{m \times n}$ is a derivation. For this we check that $D(gh)=gD(h) + D(g)$.

\begin{eqnarray*} gD(h) + D(g) &=& g \cdot [ C(h) B^{-1}(h) ] + C(g)B^{-1}(g) \\ &=& A(g)C(h)B^{-1}(h)B^{-1}(g) + C(g)B^{-1}(g) \\ &=& A(g)C(h)B^{-1}(gh) + C(g)B^{-1}(g) \\ &=& [A(g)C(h) + C(g)B^{-1}(g) B(gh) ] B^{-1}(gh) \\ &=& [A(g)C(h) + C(g)B^{-1}(g) B(g) B(h) ] B^{-1}(gh) \\ &=& [A(g)C(h) + C(g) B(h) ] B^{-1}(gh) \\ &=& C(gh)B^{-1}(gh) \\ &=& D(gh) \end{eqnarray*} The second last step can be understood by looking at the matrix representation of $\rho$. When one uses the fact that $\rho(gh)=\rho(g)\rho(h)$ and looks explicitly at the matrix multiplications one sees the desired identity in the right-top corner.

2) I don't see why $D$ is principal...

3) First bring $\begin{pmatrix} 1 & M \\ 0 & 1 \end{pmatrix}^{-1}$ to the right hand side. Next work out the left and right-hand matrix multiplications explicitly: \begin{eqnarray*} \begin{pmatrix} A(g) & C(g) + M B(g) \\ 0 & B(g) \end{pmatrix} &=& \begin{pmatrix} 1 & M \\ 0 & 1 \end{pmatrix} \cdot \begin{pmatrix} A(g) & C(g) \\ 0 & B(g) \end{pmatrix} \\ &=& \begin{pmatrix} A(g) & 0 \\ 0 & B(g) \end{pmatrix} \cdot \begin{pmatrix} 1 & M \\ 0 & 1 \end{pmatrix} \\ &=& \begin{pmatrix} A(g) & A(g) M \\ 0 & B(g) \end{pmatrix} \end{eqnarray*}

Taking the identity of the right-top corners, we get that $C(g) + M B(g) = A(g)M$. This is the condition that must hold such that we have

$$\begin{pmatrix} 1 & M \\ 0 & 1 \end{pmatrix} \cdot \begin{pmatrix} A(g) & C(g) \\ 0 & B(g) \end{pmatrix} \cdot \begin{pmatrix} 1 & M \\ 0 & 1 \end{pmatrix}^{-1} = \begin{pmatrix} A(g) & 0 \\ 0 & B(g) \end{pmatrix}$$ But this condition is satisfied whenever $D$ is a principal derivation. When this is the case, there exists an $M$ such that $$C(g)B(g)^{-1} = D(g)= g \cdot M - M = A(g) M B(g)^{-1} - M$$ or $$C(g) = A(g) M - M B(g)$$ or $$C(g) + M B(g) = A(g)M$$ which was the desired result.

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You are apparently expected to prove (or know?) that $H^1\left(G, M\right) = 0$ whenever $G$ is a finite group, $k$ is a field of characteristic $0$ (or of positive characteristic coprime to $\left|G\right|$) and $M$ is a $k\left[G\right]$-module. Do you know the notions of restriction and corestriction, as well as the fact that $\mathrm{Cor}^G_H\circ\mathrm{Res}^G_H$ is multiplication by $\left[G:H\right]$ on $H^r\left(G,M\right)$ for every $r$ ? –  darij grinberg Jan 25 '13 at 22:46
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