Timeline for Are all splittings of the normal bundle to a submanifold locally isomorphic?
Current License: CC BY-SA 3.0
7 events
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| Feb 1, 2019 at 17:25 | comment | added | Georges Elencwajg | Of course given some supplementary structure on a manifold such as,say, a Riemannian metric we can identify the quotient bundle with a subbundle , but conceptually it is better to consider the normal bundle as a quotient of the restricted tangent bundle. | |
| Feb 1, 2019 at 17:18 | comment | added | Georges Elencwajg | The only natural definition of the normal bundle to $S$ in $M$ is, just as Michael writes, $N=\frac {TM\vert S}{TS}$. This is the definition adopted by all algebraic geometers and complex analysts. There is NO natural way to see that bundle $N$ as a subbundle of $TM\vert S$: don't we strictly forbid our students to confuse a quotient vector space with a subvector space? (To be continued) . | |
| Aug 29, 2017 at 13:11 | comment | added | Robert Bryant | @MichaelBächtold: Of course, 'normal' is already a misleading descriptor for this concept, since there is no inner product involved. A better term might have been a 'transversal subbundle' or a 'complementary subbundle', while TM/TS could have been called the 'tangential quotient bundle' or some such. However, we can't fix all of these misleading terminologies. | |
| Aug 29, 2017 at 12:26 | comment | added | Michael Bächtold | Sorry for my sloppy use of terminology. I had considered using "choice of normal bundle", but in many places the quotient $TM/TS$ is called the normal bundle, and there's no choice involved. I ended up talking about the splitting of $TM/TS$, but what I meant was the splitting of the short exact sequence. | |
| Aug 29, 2017 at 11:12 | history | edited | Robert Bryant | CC BY-SA 3.0 |
added 644 characters in body
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| Aug 29, 2017 at 11:03 | vote | accept | Michael Bächtold | ||
| Aug 29, 2017 at 11:01 | history | answered | Robert Bryant | CC BY-SA 3.0 |