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Consider a principal bundle $P\rightarrow M$ and the associated vector bundle $P\times_{\rho}V$ over $M$ such that $(p,v)=(pg^{-1},\rho(g)v)$. The connection on the principal bundle $A$ defines a covariant derivative on $P\times_{\rho}V$ as follows.

View a section $s$ on $P\times_{\rho}V$ as a $G$-equivariant map $s^{P}:P\rightarrow V$. This means $S^{P}(pg^{-1})=\rho(g)s^{P}(p)$. This defines a $G$-equivariant homomorphism $s^{P}_{*}$ from $TP$ to $V$.

Let $x\in M$ and $v\in TM_{x}$, $H_{A}$ be the horizontal vector bundle in $TP$ defined by $A$. Then for any $p$ be the inverse image of $x$, there is a unique horizontal vector $v_{A}\in H_{A}|_{p}$ being the horizontal lift of $v$ such that $\pi_{*}v_{A}=v$. The covariant derivative $\nabla_{A}$ sends section $v$ to the equivalence of the pair $(p,s^{P}_{*}v_{A})$. A routine exercise showed this is not dependent on the choice of $p$.

Taubes claim in his book Differential Geometry that locally we can write the covariderive $\nabla_{A}$ on $P\times_{\rho}V$ as $$x\rightarrow (x,ds_{U}+\rho_{*}(a_{U})s_{U})$$where $\rho_{*}:\mathcal{G}\rightarrow End(V)$ is the differential of $\rho$ at the identity. I do not understand how he derived this identity given the above definition of $\nabla_{A}$. Even if we use the canonical identification $$\phi_{U}^{*}H_{A}=(x,-g^{-1}a_{U}(v)g)\in TU\oplus \mathcal{G}$$ where $\phi_{U}^{*}$ is the local trivialization. I still do not know how to derive the desired identity. Note in particular the connection 1-form on $P$ is given by $$g^{-1}dg+g^{-1}a_{U}g$$ where $a_{U}:U\rightarrow \mathcal{G}\otimes T^{*}M$. However I do not know how to put all these formulas together.

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1 Answer 1

Actually more is true: globally we can write $\nabla s = ds^P + \rho_*(a) s^P \in \Omega^1(P)\otimes \mathfrak g$ where $a$ is the (global) connection 1-form on $P$. Notice that if $v_A$ is horizontal then by definition $a(v_A) = 0$ so that $\nabla s (v_A) = ds^P(v_A) = s^P_* v_A$, agreeing with what you wrote.

Now you may ask then why we didn't just define $\nabla s$ as $d s^P$. The point is that we want $\nabla s$ to Lie in $\Omega^1(M) \otimes \mathfrak g$, where we view $\Omega^1(M) \subset \Omega^1(P)$ by the pullback of the projection. This identifies $\Omega^1(M)$ with those forms in $\Omega^1(P)$ which are right invariant and horizontal (i.e. they annihilate $\ker \pi_*$). The problem is that $ds^P$ is not horizontal: let $X^*$ be the vector field on $P$ given by $p \mapsto p\exp(tX)$ for $X \in \mathfrak g$. Then $$ ds^P(X^*_p) = \frac{d}{dt}\vert_{t=0} s(p\exp(tX)) = \frac{d}{dt}\vert_{t=0} \rho(\exp(-tX)) s(p) = -(\rho_* X) s(p). $$ However, this is cancelled by the $\rho_* a$ since $a(X^*) = X$ by definition of a connection form.

To answer the question in the title, the curvature 2-form is just $\rho_* \Omega$ where $\Omega \in \Omega^2(M)\otimes \mathfrak g$ is the curvature form of the connection on $P$.

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well done. One question remains - you explained the motivation, not the derivation of the formula. Can you tell me how to derive it? –  Bombyx mori Oct 16 '12 at 13:14

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