# nowhere vanishing vector field on a manifold

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 $\dim E^c=1$.

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 $L\subset TM$?

• 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

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).

• It seems that the Stiefel-Whitney class was designed for my question (^-^). All answers here are great! – Pengfei Nov 25 '10 at 12:00
• 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.

• 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
• 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

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.

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

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