I have some trouble understanding a proposition in Leung's paper "Symplectic Structures on Gauge Theory" published in Commun. Math. Phys. 193, 47 – 67 (1998).

I expose here the setup for my question: Let $\mathcal{A}$ be the set of unitary connections on a holomorphic vector bundle $E\rightarrow X$. We pull back $E$ to the product $\mathbb{E}\rightarrow \mathcal{A}\times X$ and we call that bundle the universal vector bundle on $\mathcal{A}\times X$. It is endowed with a canonical connection $\mathbb{A}$ whose curvature is described here Canonical connection on $\mathcal{A}\times X$.

This curvature $\mathbb{A}^2$ belongs to the space $\Omega^2(\mathcal{A}\times X,End(\mathbb{E}))$ that can be decomposed using the decomposition (on the level of forms) $$\Omega^2(\mathcal{A}\times X)=\Omega^2(\mathcal{A})\oplus \Omega^1(\mathcal{A})\otimes \Omega^1(X)\oplus \Omega^2(X)$$ Now for the middle component it is seemingly commonly accepted to write it (in Leung's paper $\mathbb{F}^{1,1}$) $$\mathbb{F}^{1,1}_{x,A}(v,B)=B(v)$$ where $v\in T_xX$ and $B\in T_A\mathcal{A}$.

Here is where I start missing something: The curvature is a 2-form, that is we have $$\mathbb{F}_{x,A}(v\oplus A,w\oplus B)=F_A(v,w)+A(w)-B(v)$$ for $v,w\in T_xX$ and $A,B\in T_A\mathcal{A}$, and so $$\mathbb{F}^{1,1}_{x,A}(v\oplus A,w\oplus B)=A(w)-B(v)$$ So the expression above for $\mathbb{F}^{1,1}$ is inacurate.

This problem makes things worse later in proposition 1 where we consider powers of $\mathbb{F}^{1,1}$ in the expression of the index of the elliptic family $\mathcal{A}$ over $\Gamma(X,E)$. As a matter of fact it uses the fact that $$(\mathbb{F}^{1,1})^k(B_1,...,B_k)=B_1\wedge...\wedge B_k$$ which does not make sense to me.

Any help would be much appreciated

  • $\begingroup$ Up to a sign, the "commonly accepted" formula follows from your equation (after "and so") by setting $A = B = 0$. This second equation can be recovered by skewsymmetrization, so both formulas are equivalent. $\endgroup$ Dec 14 '20 at 17:32
  • $\begingroup$ @BertramArnold I understand that they are related but it is not enough to get the proper expression for the powers $\endgroup$
    – BinAcker
    Dec 15 '20 at 10:53
  • $\begingroup$ When you write $\mathbb F_{x,A}$ then $(x,A)$ is your basepoint, isn't it? Moreover, $A$ then is a connection, and $B$ is an $\mathfrak{u}(n)$-valued endomorphism. So it would be better in the third display to use $A$ only once on each side and replace the other $A$ by, say, $C$. The signs in that formula are surely just a matter of convention. $\endgroup$ Dec 15 '20 at 13:58
  • $\begingroup$ @SebastianGoette you're right I used A for two purposes, should have taken another letter. $\endgroup$
    – BinAcker
    Dec 15 '20 at 20:26

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