Timeline for Is the Lie algebra-valued curvature two-form on a principal bundle P the curvature of a vector bundle over P?
Current License: CC BY-SA 3.0
17 events
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| Apr 22, 2012 at 20:17 | vote | accept | Spencer | ||
| Dec 2, 2011 at 18:11 | answer | added | Robert Bryant | timeline score: 10 | |
| Jul 12, 2011 at 19:51 | answer | added | Paul Siegel | timeline score: 6 | |
| Jun 28, 2011 at 22:19 | comment | added | Deane Yang | In particular, what you want to realize is that a connection $1$-form is always a $\mathfrak{g}$-valued $1$-form, whether the connection is on a principal $G$-bundle or on a vector bundle $E$ (with a right $G$-action). But even in the latter case $\mathfrak{g}$ is not the fiber of $E$. | |
| Jun 28, 2011 at 21:41 | comment | added | Deane Yang | Spencer, in the paragraph titled "The issue", you claim that a connection on a vector bundle $E$ is given by an $E$-valued 1-form. This is not the case. | |
| Jun 28, 2011 at 21:31 | comment | added | Spencer | @Deane If you're referring to the bit I think you're referring to then I am either still very confused or have failed to get my point across again: I have a connection on $P$ = principal connection, in the sense of smooth $G$- covariant choice of horizontal subspaces. This is equivalent to a one-form on $P$ with values in the lie algebra and some other properties. Then I discuss the possibility of another, different, connection in a vectore bundle $E \to P$. I was not supposed to suggest $\omega$ actually is said connection in $E$. | |
| Jun 28, 2011 at 21:10 | comment | added | Deane Yang | Your description of what a connection on a vector bundle $E$ is is incorrect. If you fix a trivialization of $E$, then the connection is not given by an $E$-valued $1$-form. It is given by an $Aut(E)$-valued $1$-form. For example, if you have a Riemannian manifold, the Levi-Civita connection is given, with respect to a trivialization of the tangent bundle, not by a vector-valued $1$-form but by a (skew-symmetric) matrix-valued $1$-form. | |
| Jun 28, 2011 at 18:13 | history | edited | Spencer | CC BY-SA 3.0 |
Re-worded the question because it wasn't coming across properly.
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| Jun 27, 2011 at 22:32 | answer | added | Vít Tuček | timeline score: 3 | |
| Jun 27, 2011 at 19:46 | comment | added | Theo Johnson-Freyd | Alternately, a connection is a $\mathfrak g$-valued 1-form on the principle bundle, satisfying some properties. Depending on your interests, a good place to read about the geometry might be Section 1 of Dan Freed's article "Classical Chern Simons Theory, Part 1" arxiv.org/abs/hep-th/9206021 . Freed defines the the curvature of a connection by working with the connection as a $\mathfrak g$-valued 1-form on the total space of the bundle; he defines it as $\Omega=d\omega+\frac12[\omega\wedge\omega]$, and then proves (or just quotes) various properties. | |
| Jun 27, 2011 at 19:36 | comment | added | Theo Johnson-Freyd | Small nitpick: A priori, the space of connections on a principle bundle is not even a vector space (what's the sum of two connections?). Rather, it is the space of sections of a bundle that is a torsor (=affine space) for the vector bundle whose sections are $\mathfrak g$-valued 1-forms. The point is that you can trivialize the bundle, at least locally, and this is chooses a base point for the space of connections. Certainly, for actual calculations, you always choose such a local trivialization, and then just check gauge-invariance. | |
| Jun 27, 2011 at 18:57 | comment | added | Spencer | @Spiro, Thanks for your comment. I think my trouble comes from the fact that from the point of view of the manifold $P$, $\omega$ is just a certain one-form with values in some vector space which happens to be $\mathfrak{g}$. I see how this object is a connection in the bundle $P \to M$ but what does this have to do with differrentiating $\mathfrak{g}$-valued forms on $P$? [@Jose Corrected, thanks.] | |
| Jun 27, 2011 at 18:41 | history | edited | Spencer | CC BY-SA 3.0 |
added 1 characters in body
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| Jun 27, 2011 at 18:00 | comment | added | Deane Yang | I sympathize with your struggles. I recommend initially working only with a principal $G$ bundle that is a sub-bundle of frames of a vector bundle. This requires assuming that $G$ is a subgroup of $GL(N)$, where $N$ is the rank of the vector bundle. After you understand this really well, it becomes much easier to understand and work with the general abstract case. | |
| Jun 27, 2011 at 17:58 | comment | added | José Figueroa-O'Farrill | Small nitpick: I think you mean "connection one-forms $A_i$" and not "curvature one-forms". Otherwise, let me echo Spiro's comment. | |
| Jun 27, 2011 at 17:49 | comment | added | Spiro Karigiannis | I don't understand. You already say that you have a connection $1$-form $\omega$. This is equivalent to an exterior covariant derivative. In order for one to be able to sensibly differentiate a $\mathfrak{g}$-valued form, one needs a connection. Try looking at Volume 1 of Kobayashi-Nomizu. Your questions should all be answered there. | |
| Jun 27, 2011 at 17:33 | history | asked | Spencer | CC BY-SA 3.0 |