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I am wondering if there are necessary and sufficient conditions under which an one-dimensional subbundle of $TM$ has a nowhere vanishing vector field.

More precisely let $M$ be a compact smooth manifold.

a. When dose there exist a one-dimensional (smooth or continuous) subbundle $L\subset TM$?

b. If $L\subset TM$ is a continuous/smooth line subbundle of $TM$, does there exist a nowhere vanishing continuous/smooth section $X:M\to L$? If so, the euler characteristic of $L$ should be zero.

This is related to the partially hyperbolic system $f:M\to M$ and $TM=E^s\oplus E^c\oplus E^u$. I am curious if there is a 'center flow' if $\dim E^c=1$.


To Ryan: Am I right to say the following about your answers:

  1. If there exists an 1-dimensional subbundle $L$ of $TM$, then $\chi(M)=0$. This is independent of the case whether $M$ is orientiable or not.

  2. If $M$ is orientable, then there always exists an orientable 1-dimensional subbundle $L$ of $TM$.

Another question is, when is an 1-dimensional subbundle $L\subset TM$ orientable? Is it sufficient to assume that $M$ is orientable?


Thank you all. I did not formulate some questions properly. What I really mean is:

I. For a given line bundle $L\subset TM$, what is the obstruction for $L$ beging orientable? (or equivalently trivial according to Georges)

For example let $L_{\mathbb{C}}$ be a complex line bundle over a complex manifold $M$, if the top Chern class $c_1(L_{\mathbb{C}})$ does not vanish, then $L_{\mathbb{C}}$ can not be trivial. Is there some similar results in the real case?

II. Is there an example such that $M$ is orientable and has a non-orientable line bundle $L\subset TM$?

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1  
Regarding your question (1), yes. That's what I'm saying. The covering space argument is one way to approach this. Re (2) that is an if and only if $\chi M =0$, where I use $\chi M$ as shorthand for the Euler characteristic. This is of course assuming $M$ is connected. –  Ryan Budney Nov 24 '10 at 18:08

5 Answers 5

up vote 3 down vote accepted

A bundle is orientable if and only if its first Stiefel-Whitney class is 0 (one can see the first Stiefel-Whitney class as the function $w_1: H_1(M)\rightarrow \mathbb{Z}_2$ which associate to a loop the sign of the determinant of the monodromy).

As mentionned by Ryan, if a line bundle is non-orientable then there is a two sheeted cover of $M$ which orients $L$ (the covering correspond exactly to the index two subgroup $ker(w_1)$) this implies that if an oriented bundle admits a 1-dimensionnal sub-bundle the its Euler class has to be $0$ (regardless if the bundle is the tangent bundle or not).

Finally one can easily see that $T(T^2)$ is the sum of two non-trivial line bundle:

The canonical line bundle $\gamma$ on $\mathbb{R}P^1$ is non trivial but $\gamma\oplus\gamma^*$ is (it is oriented, 2 dimensionnal and admit a section given by the trace map). Pulling back this bundle by the projection $T^2\rightarrow S^1$ you get a trivial bundle (hence the tangent bundle) on $T^2$ written as a sum of two non trivial line bundle ($w_1$ is non zero on each summand).

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It seems that the Stiefel-Whitney class was designed for my question (^-^). All answers here are great! –  Pengfei Nov 25 '10 at 12:00
3  
Stiefel-Whitney classes are designed for a more general question than the one you ask -- the issue of constructing linearly-independent sections of vector bundles. Your questions have answers that don't need "big machines" like that, but of course they are convieniently answered by big machines. –  Ryan Budney Nov 25 '10 at 16:04
    
I agree that Stiefel-Whitney classes in general are designed for far more general problem. However in my answer I only talked about the first Stiefel-Whitney class which, correct me if I'm wrong, is specifically designed to address the orientability of vector bundle and is easily defined. –  Noz Nov 25 '10 at 16:27

If there is a 1-dimensional sub-bundle of $TM$, if it was an orientable bundle you'd have $\chi M = 0$. Consider the case it's non-orientable. Then there would be a 2-sheeted connected "orientation cover" of $M$, $\tilde M \to M$ such that the 1-dimensional sub-bundle of $M$ pulls-back to a trivial 1-dimensional sub-bundle of $\tilde M$. So $\chi \tilde M = 0$, but since Euler characteristic is multiplicative under covers, $\chi M = 0$. So if $TM$ is orientable, this implies there is a non-zero vector field on $TM$ by obstruction theory.

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Isn't $TM$ always orientable? –  Dylan Wilson Nov 24 '10 at 3:44
    
Specifically: it seems like if you pull back any atlas of $M$ you'll get an oriented atlas of $TM$ since the differentials of the transition maps will be triangular block matrices with the two matrices on the diagonal being the same. –  Dylan Wilson Nov 24 '10 at 3:48
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As a manifold $TM$ is always orientable, yes. But I was thinking of orientability in the bundle sense, ie $TM$ orientable if and only if $M$ is orientable. –  Ryan Budney Nov 24 '10 at 4:35
    
To Ryan: I added a comment in the question since I could change lines there. –  Pengfei Nov 24 '10 at 8:59

Dear Pengfei, about your supplementary questions:

a) "when is an 1-dimensional subbundle L⊂TM orientable ?" A line bundle is orientable if and only if it is trivial (independently of it being or not a subbundle of the tangent bundle). To prove the non-trivial implication, endow your orientable line bundle with a Riemann metric and choose in every fibre the unique vector of length one which has the correct orientation.This gives you a nowhere zero section .

b) "If M is orientable, then there always exists an orientable 1-dimensional subbundle L of TM " No.The tangent bundle $TS^2$of the two-sphere $S^2$ has no line subbundle at all. Indeed that subbundle would be trivial since line bundles are classified by $H^1( S^2, \mathbb Z/(2))$, which is zero . But the trivial bundle is not a subbundle of the tangent bundle since "you can't comb a sphere".

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Here is a cool proof that if there is a nonvanishing vector field then the Euler characteristic must be zero:

Suppose $X$ is a nonvanishing vector field. Let $f_t$ be the corresponding time $t$ flow. Then for some small $\varepsilon > 0$, the flow $f_\varepsilon$ has no fixed points (to show this, we must use the compactness of the manifold $M$). So by the Lefschetz fixed point theorem, we have $$0 = \sum_i (-1)^i \operatorname{Tr}(f_{\varepsilon,\ast} : H_i(M) \to H_i(M)). $$ But since $f_\varepsilon$ is homotopic to $f_0 = \operatorname{Id}$, we have that the RHS is equal to $$\sum_i (-1)^i \operatorname{Tr}(\operatorname{Id}:H_i(M) \to H_i(M)) = \sum_i (-1)^i \operatorname{rk}H_i(M),$$ which is the Euler characteristic.

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WOW! I like this! Then the 'orientable cover' argument applies if the line bundle is not orientable. –  Pengfei Nov 25 '10 at 5:49

To construct a non orientable line bundle on the torus $\mathbb{T}^2$ you can proceed as follows:

Consider a planar Reeb foliation of the anulus (see this drawing) and then glue the boundary circles obtaining a one dimensional non singular foliation of the two dimensional torus. Notice that it cannot be orientable since in one direction, when you follow the holonomy, you return with the wrong orientation (notice also, as explained above, that if you glue two such foliations, i.e. you consider the double cover, you get an orientable foliation).

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Yes! This example is fantastic. Both the non-orientability and its orientable cover can be realized directly. –  Pengfei Nov 25 '10 at 10:59

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