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

I'm looking for a (comprehensible) reference for the Frolicher-Nijenhuis bracket, hopefully more down to earth than Michor's books and different from Saunders's book on Jets.

I'm interested in it as this bracket seems the appropriate means to define the curvature of a general Ehresmann connection on a bundle.

Alternatively it would be really great to have a reference (again different from Michor) that defines connections, curvature and holonomy on general bundles, later specializing to principal bundles, vector bundles and maybe even to the riemannian setting.

share|improve this question
    
I'd be interested in any reference for general Ehresmann connections on fibre bundles -- I only know this theory for principal bundles and their associated fibre bundles. –  José Figueroa-O'Farrill Mar 4 '10 at 0:06
    
Wikipedia isn't too bad... en.wikipedia.org/wiki/Frölicher–Nijenhuis_bracket The basic properties are probably pretty easy to prove. –  Kevin H. Lin Mar 4 '10 at 4:21
2  
The wikipedia link doesn't seem to be working right. How do I do it correctly? –  Kevin H. Lin Mar 4 '10 at 4:25
    
The only reference I know of is this (and of course the book by Michor alone, which is more or less the same) emis.de/monographs/KSM/index.html The only problem is that it may be a little bit too much sophisticated for me. In Jeffrey Lee's book (Manifolds and Differential Geometry) Ehresmann connections are introduced, but he switches right away to vector bundles and so, for example, he doesn't define curvature in the general setting. –  babubba Mar 4 '10 at 7:48
    
Actually the wiki link to Ehresmann connection contains what I was wondering about. I was expecting a different definition, somehow, but it is indeed very similar to the case of principal and associated bundles. The links is en.wikipedia.org/wiki/Ehresmann_connection , by the way. –  José Figueroa-O'Farrill Mar 4 '10 at 18:14

1 Answer 1

up vote 5 down vote accepted

I also only learned about Frolicher-Nijenhuis brackets from Saunders' book on jets but I doubt that there is any other authorative reference besides Michor's book and the original papers. I don't know if this is what the original poster intended, but here's how I understand the link between FN and curvature. Generally, I tend to stay away from the full generality of the FN bracket, and only use those (axiomatic) properties that I need...

Let $\pi: E \rightarrow M$ be a fiber bundle. An Ehresmann connection is a subbundle $H$ of $TE$ such that $H \oplus VE = TE$, where $VE$ is the vertical bundle. Denote the projector from $TE$ onto $H$ by $h$. Saunders (and many other authors) define Ehresmann connections directly in terms of bundle maps $h$, since they are easier to work with, but this doesn't matter.

Now, to introduce curvature, we would like to formalize the idea that curvature is the failure of "parallel transport" to commute, where of course we haven't define parallel transport properly. However, it is only a small step of the imagination to guess that this must be related to the integrability of $H$, i.e.whether $[h(X), h(Y)] \in H$ for arbitrary vector fields $X, Y$. The failure of two horizontal vector fields to be horizontal again is measured by the expression $$ R(X,Y) = [h(X), h(Y)] - h([h(X), h(Y)]), $$ and in fact this is nothing but Saunders' definition 3.5.13 of curvature. Note that $R(X, Y) = 0$ if either $X$ or $Y$ is in $VE$.

The Frohlicher-Nijenhuis bracket of two linear bundle maps is in general quite complicated, but if you look at proposition 3.4.15 in Saunders, you get that for a linear bundle map $h: TE \rightarrow TE$, $$ [h, h] (X,Y) = 2(h([X,Y]) + [h(X), h(Y)] - h([h(X),Y]) - h([X,h(Y)])). $$ Now note that the right-hand side is zero as soon as either $X$ or $Y$ is in $VE$, just as for $R$. If both $X$ and $Y$ are horizontal, we can apply the above formula to $X = h(X)$ and $Y = h(Y)$, and conclude that $[h, h] (X, Y) = 2( [h(X), h(Y)] - h([h(X), h(Y)]))$ (where we use the identity $h \circ h = h$ liberally), and so $$ R = \frac{1}{2} [h,h]. $$

If you have a principal fiber bundle, $h$ is related to the connection one-form $\mathcal{A}$, and the above formula gives the curvature of $\mathcal{A}$ in terms of the covariant differential of $\mathcal{A}$. For a symplectic connection, something similar happens.

Edit: here's how I think it works for principal fiber bundles. Take a connection one-form $A: TE \to \mathfrak{g}$, and let $\sigma: \mathfrak{g} \rightarrow VE$ be the infinitesimal generator of the $G$-action. The composition $v := \sigma \circ \mathcal{A}$ is then the vertical projector of the connection and $h := 1 - v$ is the horizontal one. Now plug this expression for $h$ into the formula for the curvature: $$ [h, h] = [1, 1] - [\sigma \circ \mathcal{A}, 1] - [1, \sigma \circ \mathcal{A}] + [\sigma \circ \mathcal{A}, \sigma \circ \mathcal{A}]. $$ The term $[1,1]$ is zero, and you can use Saunders' proposition 3.4.15 to show that term 2 and 3 vanish as. The last term can be written as (again using S3.4.15) as $$ [\sigma \circ \mathcal{A}, \sigma \circ \mathcal{A}] (X, Y) = 2( (\sigma \circ \mathcal{A})^2([X, Y]) + [ (\sigma \circ \mathcal{A})(X), (\sigma \circ \mathcal{A})(Y)] - (\sigma \circ \mathcal{A})([(\sigma \circ \mathcal{A})(X), Y)]) - (\sigma \circ \mathcal{A})([X, (\sigma \circ \mathcal{A})(Y)])). $$

Now, show that this vanishes whenever $X$ or $Y$ is vertical, so that we can take $X$ and $Y$ to be horizontal. In that case, the above simplifies to $$ [\sigma \circ \mathcal{A}, \sigma \circ \mathcal{A}] (X, Y) = 2(\sigma \circ \mathcal{A})([X, Y]) $$ or $$ R(X, Y) = (\sigma \circ \mathcal{A}) ([X^h, X^h]) $$ where $X^h$ represents the horizontal part of $X$. But $\mathcal{A} ([X^h, X^h])$ is just the negative of the curvature (as a two-form with values in $\mathfrak{g}$) of $\mathcal{A}$, so that $$ R(X, Y) = -\sigma ( \mathcal{B}(X, Y) ). $$

share|improve this answer
    
Thanks. Would you mind being more specific in the principal bundle case? –  babubba Mar 4 '10 at 14:42
1  
Sure. The idea is that in the principal bundle case, the connection one-form is related to the vertical projector of the connection. Hence, by subtracting A from the identity, you get h above. At this point, I would probably just use the coordinate expression of [h, h] and try to observe its relationship with DA, but let me see if I can find a nice intrinsic proof. –  jvkersch Mar 5 '10 at 7:32

Your Answer

 
discard

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.