Timeline for Gauss-Bonnet invariant Ω: explicit intrinsic expression for Π in Ω=dΠ?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 5, 2015 at 13:27 | vote | accept | duetosymmetry | ||
| Feb 4, 2015 at 12:23 | answer | added | Robert Bryant | timeline score: 11 | |
| Feb 4, 2015 at 2:45 | comment | added | Robert Bryant | @duetosymmetry: I think I see a source of terminological confusion. Apparently, you are interpreting the functions $u_i$ that Chern introduces in §2 of his paper as components of an arbitrary vector field. Admittedly, Chern calls them the 'components of the vector $\frak{v}$', but this is misleading; what he really means is that the $u_i$ are part of a local coordinate system on the unit sphere bundle, which he calls $M$. Thus, it's misleading to say that his construction is 'in terms of an arbitrary vector field'; what he really means is that $\Pi$ is going to be constructed on $M$, not $R$. | |
| Feb 3, 2015 at 16:26 | comment | added | Robert Bryant | @duetosymmetry: Chern 1944 gives the formula for $\Pi$ explicitly as a sum of $n/2$ terms, and it involves factorial coefficients and factors of $\pi^{n/2}$. It doesn't make sense for me to type it out here, because you have Chern's paper in front of you. See equation $(24)$ and look back at $(15)$ for the definition of the individual $\Phi_m$. One could, of course, pull $\Pi$ back to the $\mathrm{SO}(m)$ frame bundle, where it would become a polynomial with constant coefficients in the connection and curvature forms, but Chern doesn't do it that way; he wants it on the sphere bundle. | |
| Feb 3, 2015 at 16:18 | history | edited | duetosymmetry | CC BY-SA 3.0 |
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| Feb 3, 2015 at 16:12 | comment | added | duetosymmetry | @RobertBryant yes, I did not say it 'depends on', just that it is 'in terms of'—but it indeed must be independent. Since it is independent, what is the explicit expression without reference to the vector? | |
| Feb 3, 2015 at 16:02 | comment | added | Robert Bryant | Chern's 1944 construction of $\Pi$ does not depend on a choice of vector field. He defines $\Pi$ as a canonical $n{-}1$ form on $M^{2n-1}=\mathsf{S}(R)$, the unit sphere bundle of the oriented Riemannian $n$-manifold $R^n$ (where $n$ is even) and shows that $\mathrm{d}\Pi=\pi^*\Omega$, where $\Omega$ is the Gauss-Bonnet form on $R^n$ and $\pi:M\to R$ is the base-point mapping. The choice of a vector field only comes into play at the end, when he uses it to get a section $V$ of the bundle $\pi:M\to R$ away from a single point, where he then uses other properties of $\Pi$ to prove the theorem. | |
| Feb 3, 2015 at 15:27 | comment | added | duetosymmetry | What about for dimension 4 or higher? | |
| Feb 3, 2015 at 14:56 | comment | added | Ben McKay | For surfaces, $\Pi$ is the connection 1-form. | |
| Feb 3, 2015 at 14:44 | history | asked | duetosymmetry | CC BY-SA 3.0 |