Timeline for Totally non parallelizable manifold
Current License: CC BY-SA 3.0
26 events
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| Feb 22, 2014 at 12:53 | comment | added | Ali Taghavi | @IgorBelegradek I apologize for my misunderestanding. Now I underestand you consider $TE$ as a vector bundle on $E$ not as a vector bundle on $TM$.Thanks again for your answer. | |
| Feb 20, 2014 at 17:37 | comment | added | Ali Taghavi | @IgorBelegradek sorry there is software problem in my computer system(For chat).may be continue in this comment page? | |
| Feb 19, 2014 at 20:20 | comment | added | Igor Belegradek | let us continue this discussion in chat | |
| Feb 19, 2014 at 20:16 | comment | added | Ali Taghavi | I learnt it from chapter 11 (Connection) of "Lecture on differentiable manifolds" by Siavash Shahshahani. | |
| Feb 19, 2014 at 19:58 | comment | added | Ali Taghavi | ....because$Dp^{-1}(TU)=Tp^{-1}(U)\simeq T(U\times \mathbb{R}^{n})\simeq TU\times T\mathbb{R}^{n} \simeq TU\times \mathbb{R}^{2n}$. So $(TE,TM,Dp)$ is trivial on $TU$ and the dimension of fibre is equal to 2n. (no matter what dim(M) is) | |
| Feb 19, 2014 at 19:44 | comment | added | Ali Taghavi | @IgorBelegradek In order to remove misunderestanding, lets review the proof of the fact 'IF (E,M,p) is a n-bundle then (TE,TM,Dp) is a 2n-bundle, where(It is not important what is dim of M). Assume that E is trivial over$U\subset M$. Then $TE$ is trivial on $TU$, because | |
| Feb 19, 2014 at 15:50 | comment | added | Igor Belegradek | @AliTaghavi: Sorry, I still do not understand what you mean. You seem to confuse the bundle $TE$ and its total space. The total space of $TE$ has dimension $2(n+k)$. The bundle $TE$ restricted to any subset of $E$ has fiber dimension $n+k$. You could consider $TM$ as a submanifold of $TE$ but not of $E$, so it make sense to talk about the normal bundle of the submanifold $TM$ in the manifold $TE$, but not of the restriction of the bundle $TE$ to the subset $TM$. When you get the terminology straight it would be easier for me to see the issue. | |
| Feb 19, 2014 at 15:01 | comment | added | Ali Taghavi | @IgorBelegradek assume that $E$ is a n dimensional vector bundle on a k dimensional manifold $M$. Then TE is a 2n dimensional vectoe bundle on $TM$.(Note that we do NOT consider $TE$ as a vector bundle on $E$. But we consider $TE$ as a vector bundle on $TM$. So $TE$ is a 2n dimensional bundle on $TM$(Independent of k!)No my question: How the 2n-bundle TE (restricted to the zero section of TM) is isomorphic to n+k bundle $E\oplus TM$? I think that some thing is missing in your answer. | |
| Feb 18, 2014 at 23:36 | comment | added | Igor Belegradek | Can you delete your earlier comment, and write the one with all dimensions correct? | |
| Feb 18, 2014 at 23:32 | comment | added | Igor Belegradek | @AliTaghavi: I do not follow what you wrote: if $E$ is $\mathbb R^n$-bundle over $k$-manifold, then $E$ has dimension $n+k$, so $TE$ is an $\mathbb R^{n+k}$-bundle. | |
| Feb 18, 2014 at 16:59 | comment | added | Ali Taghavi | @IgorBelegradek In my previous comment I have to write "n+k" not "n+2k". | |
| Feb 18, 2014 at 8:19 | comment | added | Ali Taghavi | @IgorBelegradek Just now I realized that perhaps something is missing in your answer. If $E$ is a n_vector bundle on a k-manifold $M$ then $TE$ (and its restriction to every subset) is a 2n- bundle. How a 2n-bundle can be isomorphic to a n+2k bundle $E\oplus TM$? Am I missing something? | |
| Dec 13, 2013 at 10:52 | comment | added | Ali Taghavi | @Igor perhaps this is an alternative proof for parallelizability of $TS^{2}$:According to your statements it is sufficient to prove $TS^{2} \oplus TS^{2} $ is a trivial bundle on $S^{2}. But this a consequence of the fact "the complexification of tangent bundle of spheres is trivial", this is said in Sawans paper "f.g projective module and vector bundles" But what is a proof for this statement of Swan's paper? | |
| Dec 13, 2013 at 0:06 | vote | accept | Ali Taghavi | ||
| Dec 13, 2013 at 0:03 | comment | added | Igor Belegradek | Ali: because an open manifold is homotopy equivalent to a complex of lower dimension. Michael: there is no criterion of parallelizability in terms of characteristic classes. Ali: $CP^2$ has nonzero first Pontryagin class. Actually, what you ask are good exercises. You should do them yourself. | |
| Dec 12, 2013 at 23:58 | comment | added | Ali Taghavi | @Igor another question: why the tangent bundle of $\mathbb{C}P^{2}$ is a non torsion element in real K theory? | |
| Dec 12, 2013 at 23:43 | comment | added | Michael | @IgorBelegradek, by criterion I meant something computable, such as a certain characteristic class vanishing. It cannot be Euler class = 0 because of $S^5$, for example. | |
| Dec 12, 2013 at 23:41 | comment | added | Ali Taghavi | @Igor Why for open manifold, stable parallelizable imply parallelizable? | |
| Dec 12, 2013 at 23:39 | comment | added | Igor Belegradek | @Ali: yes, $TS^2$ is parallelizable. In general, an open manifold is parallelizable iff it is stably parallelizable. As I explained $T(TS^2)$ is $2TS^2$ in $K$-theory, but $TS^2$ is stably trivial, and hence so is $2TS^2$. Hence $S^2$ answers your second question. | |
| Dec 12, 2013 at 23:38 | comment | added | Ali Taghavi | @Igor Is $TS^{2}$ ot $T^{2} S^{2}$ a counterexample to the last part of my question? | |
| Dec 12, 2013 at 23:33 | comment | added | Igor Belegradek | @Michael: criterion? By definition, a manifold is parallelizable iff its tangent bundle is trivial. | |
| Dec 12, 2013 at 23:30 | comment | added | Ali Taghavi | @Igor thank you very much for your beautiful answer. you said "think of S^2" do you mean TS^2 is parallelizable? | |
| Dec 12, 2013 at 23:21 | comment | added | Michael | I see. In particular, Stiefel-Whitney of $TTM$ would be 0, therefore $TTM$ is always orientable. Could you remind me the criterion of parallelizability? | |
| Dec 12, 2013 at 23:01 | comment | added | Igor Belegradek | @Michael, I never claimed that $TE(\xi)$ is parallelizable iff its $0$-section is (this is false, think of $S^2$). The point is that a rank $k$ vector bundle $\xi$ over $B$ is a homotopy class of maps $B\to BO(k)$, so if $B$ deformation retracts to $S$, then $\xi$ is completely determined (up to isomorphism) by its restriction to $S$. | |
| Dec 12, 2013 at 22:51 | comment | added | Michael | Igor, could you explain something in your answer? First, why is it valid to restrict $TE(\xi)$ to the 0-section of $\xi$ for the purpose of answering the question, i.e. how one would show that $TE(\xi)$ parallelizable iff the 0-section is? The reason I am asking is that it's possible apriori that parallelizability is not stable property relative to direct sum with a trivial bundle. | |
| Dec 12, 2013 at 21:42 | history | answered | Igor Belegradek | CC BY-SA 3.0 |