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Let $\pi$ be a group, $G$ a compact lie group with lie algebra $g$, $A:\pi\to G$ a representation which composes with the adjoint map to give $g$ a $\pi$-module structure. I want to construct a Kuranishi map $K:H^1(\pi;g)\to H^2(\pi;g)$ (defined in a nbd of $0$) so that the space of conjugacy classes of representations near $A$ look locally like $K^{-1}(0)/G_A$, where $G_A$ is the stabilizer of the conjugation action of $A$. I want to do this in the bar complex, which has cochains $C^p(\pi;g)=funct(\pi^p,g)$, and differentials $\partial^i$.

All of this is standard, except one step: I can define the map $K:H^1(\pi;g)\to C^2(\pi;g)/image(\partial^1)$ but I don't see how to project to $H^2(\pi;g)$ without increasing the zero set, or to prove the image already lies in $H^2(\pi;g)$.

More detail: I start with the "curvature map" $F:C^1(\pi;g)\to C^2(\pi,g)$ defined (near 0) by $$F(a)(x,y)=\log(\exp(a(x))A(x)\exp(a(y))A(x)^{-1})-a(xy)$$ which locally cuts out the representations $\pi\to G$ near $A$ (i.e. $x\to \exp(a(x))A(x)$ for $a\in F^{-1}(0)$), so that $dF_0=\partial^1$. I reduce to produce $K$ as above in the usual way by splitting $C^2=image \ dF_0\oplus coker \ dF_0$, project to $image\ dF_0$ and apply the implicit function theorem. The CBH formula shows that $F(a)=\partial^1 a + \tfrac{1}{2}[a,a] +$ 3-fold and higher lie brackets.

I know the argument with curvature and differential forms. The trick there is to use the Bianchi identity and simple estimates to adjust for the fact that the complex $0\to \Omega^0\to \Omega^1\to \Omega^2\to 0$ is not elliptic at $\Omega^2$ on a manifold of dimension $>2$.

Q: How is it done in this situation? What is the substitute for the Bianchi identity argument?

Note that working with diffl forms is not as sharp since $H^2(M;g)$ might be bigger than $H^2(\pi;g)$. One cheat is to choose a high dim'l manifold $M$ with the right fundamental group and with $H^2(M;g)=H^2(\pi;g)$, but surely there is a purely algebraic argument with the bar complex I'm not seeing.

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