Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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$


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.

share|improve this question
Could someone please tell me what I can do to make this post more answerable? –  cheyne Nov 27 '12 at 15:05
Well, ask a question. I know this material (I work with bundle gerbes, among other things), and I can't figure out what you want. Do you want an explanation of the equation at top, or what? Your bolded sentence in the last paragraph doesn't make sense to me. Perhaps provide a reference to where in Brylinski you are stuck. –  David Roberts Dec 6 '12 at 22:30
Is it the Maurer-Cartan form $g^{-1}dg$ that you don't understand? –  David Roberts Dec 6 '12 at 22:30
Thank you, David. I apologize if I wasn't clearer, the first two lines of my post (namely the grey box) were "my question": namely, I can't verify the equation in the grey box in the beginning (Page 174, eq(4-22) in Brylisnki). I understand a bit about the Maurer-Cartan form, but not that I get $g^{-1}dg$ when my lie group is not a matrix goup. Any help or answers are greatly appreciated. –  cheyne Dec 6 '12 at 22:55
$g^{-1}dg$ is an abuse of notation, really it means the pullback by $g$ of the Maurer-Cartan form on $\mathbb{C}^*$. This latter is a 1-form defined locally by $d\log g$ having picked some branch for the logarithm on $\mathbb{C} - L$ where $L$ is some ray (including 0). Alternatively, view $\mathbb{C}^*$ as the matrix group $GL(1,\mathbb{C})$ and apply what you know. –  David Roberts Dec 6 '12 at 23:55
add comment

3 Answers

up vote 2 down vote accepted

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.

share|improve this answer
By $f$, you mean $b$, right? –  cheyne Dec 7 '12 at 18:03
@Cheyne: right, that was a typo. I've edited my answer. –  Konrad Waldorf Dec 8 '12 at 13:31
add comment

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.

share|improve this answer
Thank you very much for this, Johannes. I am working through the details at the moment. Firstly, I think the main thing I was missing was that any connection can be decomposed where one piece is the Maurer Cartan form; this is obvious to me now (thanks!). Secondly, I am working through your equations after "Next you have to invoke the fact" and I am just making sure that when you talk about "linear groups" you are allowing for infinite dimensions? –  cheyne Dec 7 '12 at 0:46
Also, I don't understand your "definition" of $f^*\omega$ = f^{-1} \omega (df)$, when $f$ is $G$-valued and $\omega$ is $\mathfrak{g}$ - valued; unless you implicitly mean an adjoint map induced by $f$ on the Lie-Algebras? –  cheyne Dec 7 '12 at 1:03
Let $\mu:G\x G\to G$ 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 `$$x\mapsto T(\mu_{f(x)^{-1}).T_x f: T_xM \to T_{f(x)} G \to T_eG = \mathfrak g$$' This also works in infinite dimensions. –  Peter Michor Dec 7 '12 at 8:11
add comment

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.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.