Connection Transformation Formula; Degree 3 Cech Cohomology

Question

While reading through Brylinski, as in all of my posts, I am trying to understand the following equation:

$ g_* \tilde{\theta} = \tilde{\theta} - g^{-1} dg$

Setting

I have a principal $B$-bundle, $Q$, over my space M, with connection $\theta$ (values in the Lie Algebra $\mathfrak{b}$ of the lie group $B$), and a central group extension by $\mathbb{C}^*$,

$$ 0 \to \mathbb{C}^* \to \tilde{B} \to B \to 0$$ where the map $p:\tilde{B} \to B$ is a principal $\mathbb{C}^*$ fibration, yielding an exact sequence of sheaves of groups:

$$ 0 \to \underline{\mathbb{C}}_M^* \to \underline{\tilde{B}}_M \to \underline{B}_M \to 0$$ where $M$ is a paracompact space.

Long story short, I restrict my attention of the space to an open set $ U \subset M$, small enough so that I can "lift" the bundle to a $\tilde{B}$-bundle, $\tilde{Q}$, and the connection to a $\tilde{\mathfrak{b}}$-valued connection $\tilde{\theta}$ on the bundle $\tilde{Q}$. The lifted bundle must satisy the condition that we have a bundle isomorphism $f: \tilde{Q}/\mathbb{C}^* \tilde{\to} Q_{U} $. The lifted connection must satisfy $$q \circ \tilde{\theta} = f^*\theta$$ as 1-forms with values in $\mathfrak{b} = \tilde{\mathfrak{b}}/\mathbb{C}$; where $q$ is simply the map which quotients out by $\mathbb{C}$.

Finally, $g$ is simply a $\mathbb{C}^*$-valued function, which can be thought of as a bundle isomorphism given our local picture is a trivialization. By $g_*$ we simply mean the pullback of $g^{-1}$.

Warning: Matrices Beware!

I know that this formula is true for vector bundles, or when the group automorphisms are represented by invertible matrices; I have read that literature already. Unless you are prepared to help me understand why this situation is most certainly in the land of finite vector bundles or matrix groups, such an answer would be redundant. The formula also makes sense up to the fact that the transformed connection and original connection must differ by a complex-valued form, which is exactly what this formula prescribes.

Answer 1

The equality follows directly from the definition of a connection, and is independent of the context of lifting structure groups, or degree three cohomology.

Recall that a connection on a principal $G$-bundle $\pi:P \to M$ is a 1-form $\omega \in \Omega^1(P,\mathfrak{g})$ such that $$ p_2^{\ast}\omega = Ad_g^{-1} (p_1^{\ast}\omega) + g^{\ast}\theta $$ over $P \times_M P$, where $p_1,p_2: P \times_M P \to P$ are the two projections, and $g: P \times_M P \to G$ is the difference map defined by $g(p,p')\cdot p=p'$. Here $g^\ast\theta=g^{-1}dg$, whatever notation is preferred.

Now, if $b: P \to G$ is a smooth map, it induces a map $$ \tilde b: P \to P \times_M P: p \mapsto (p\cdot b(p),p). $$ Pullback of above defining equation along $\tilde b$ produces $$ \omega = Ad_b(b_{\ast}\omega)+b^*\theta. $$ If $b$ is central, this is your equation.

This works of course also, if $b:M \to G$ is defined on the base (you didn't say clearly where $b$ is defined). Then use $b' := b \circ \pi$ instead.

Answer 2

Let $P \to M$ be a $G$ principal bundle with connection $\theta$ and let $g:M \to Z/G)$ a function with values in the center of $G$. Let $\omega $ be the left-invariant Maurer-Cartan form on $G$. Then $g$ induces, by left-multiplication, as bundle automorphism of $P$ and the formula $g^{\ast}\theta - \theta = g^{\ast} \omega$ holds. The identity $\omega = g^{-1} dg$ holds for linear groups.

Proof: Assume $P=M \times G$. You can write the connection as $\theta=pr_{G}^{\ast} \omega + pr_{M}^{\ast} \eta$, where $\omega$ is the left-invariant Maurer-Cartan form on $G$ and $\eta$ is a $\mathfrak{g}$-valued form on $M$. A $G$-valued function $g:M \to G$ induces a map $\mu_g :P \to P$ by right-multiplication. Then $\mu_{g}^{\ast} \theta - \theta= \mu_{g}^{\ast} (pr_{G}^{\ast} \omega )- pr_{G}^{\ast} \omega$, because $pr_M \circ \mu_g = pr_M$.

Furthermore, $pr_G \circ \mu_g (m,h) = hg(m) $, so $pr_G \circ \mu_g$ is the product of the two $G$-valued functions $pr_G$ and $g \circ pr_M$ on $P$.

Next you have to invoke the fact: if $f_0,f_1: P \to G$ are two functions, then $(f_0 f_1)^{\ast} \omega = Ad (f_1)^{-1}f_{0}^{\ast} \omega + f_{1}^{\ast} \omega$. This is because $(f_0 f_1)^{-1} d(f_0 f_1 )= f_{1}^{-1} f_{0}^{-1} df_0 f_1 + f_{1}^{-1} df_1=Ad (f_1)^{-1} f_{0}^{\ast} \omega + f_{1}^{\ast} \omega$. This is for linear groups, and holds in general as any Lie group is isogenous to a linear one.

Therefore $\mu_{g}^{\ast} (pr_{G}^{\ast} \omega )=pr_{G}^{\ast} \omega + g^{\ast} \omega$, as $g$ was assumed to be central.

Answer 3

Since my comment came out TeX-scrambled, I repeat it here:

Let $\mu:G\times G\to G$ be the multiplication and let $x.y=\mu(x,y)=\mu^y(x)=\mu_x(y)$; this is to fix notation for right and left transport on G. Then, for $f:M\to G$, the notation $f^{−1}.df$ is shorthand for $$x\mapsto T(\mu_{f(x)^{-1}}).T_x f: T_x M \to T_{f(x)} G \to T_eG = \mathfrak g$$ This notation also works in infinite dimensions.