Timeline for Gauss-Bonnet invariant Ω: explicit intrinsic expression for Π in Ω=dΠ?
Current License: CC BY-SA 3.0
8 events
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| Feb 9, 2015 at 17:04 | comment | added | Robert Bryant | @duetosymmetry: Ah, I see what's bothering you. Yes, of course, there is an $\mathrm{SO}(n)$-invariant expression on $B$, say $\tilde\Phi$ whose exterior derivative is $\tilde\Omega$, but it is useless for proving the Gauss-Bonnet theorem via Chern's method because $\tilde\Phi$ does not descend to $M$. Chern's point is that $\tilde \Pi$ does descend to $M$, which provides the key link between the Gauss-Bonnet integral and the sum of indices of a vector field on $R$ with isolated zeros because $\tilde \Pi$ pulls back to each fiber of $M\to R$ to be the volume form on that fiber. | |
| Feb 9, 2015 at 15:28 | comment | added | duetosymmetry | this is exactly what confuses me. Shouldn't there exist an SO($n$) invariant expression, like there does for the Chern-Simons form? | |
| Feb 9, 2015 at 12:42 | comment | added | Robert Bryant | @duetosymmetry: (Responding to your first comment) Actually, $\mathrm{SO}(n)$-invariance does not hold for any of the terms in $\tilde \Pi$. For example, note that, when $n=4$, that second term only involves the $\omega_{1k}$ for $k=2,3,4$, and not all $6$ of the $\omega_{ij}$, as it would have to if it were fully $\mathrm{SO}(4)$-invariant. In fact, the terms in $\tilde \Pi$ are only $\mathrm{SO}(n{-}1)$-invariant, which makes sense because the fibers of $\pi_1:B\to M$ are $\mathrm{SO}(n{-}1)$-orbits. If you are still doubtful, when I have time, I'll put in a more explicit explanation. | |
| Feb 8, 2015 at 20:22 | comment | added | duetosymmetry | This latter expression clearly has a frame SO(n) invariance. However, I can't for the life of me deduce what frame-independent expression leads to the first term (which contains $\omega\wedge\omega\wedge\omega$). Do you have any hits for how to proceed to find such an expression for the first term? | |
| Feb 8, 2015 at 20:17 | comment | added | duetosymmetry | Thanks, Robert, for putting in the effort to write this didactic answer. It really is helpful, and gets almost all the way where I wanted to go. I know fully well that none of the expressions depend on an arbitrary choice of coordinate system. This can be easily seen for the SO(n) invariance of the frame indices in $\tilde{\Omega}$. However, in the examples, this is not transparent. For $n=2$ it's clear that we can take $\tilde{\Pi}\propto\epsilon^{ij}\omega_{ij}$. For $n=4$, it's also clear that we can write the rightmost term as $\tilde{\Pi}\supset\epsilon^{ijkl}\omega_{ij}\wedge\Omega_{ij}$ | |
| Feb 5, 2015 at 13:27 | vote | accept | duetosymmetry | ||
| Feb 5, 2015 at 12:35 | history | edited | Robert Bryant | CC BY-SA 3.0 |
added some explanatory sentences that may help clarify things for the reader
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| Feb 4, 2015 at 12:23 | history | answered | Robert Bryant | CC BY-SA 3.0 |