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Let $V$ is a vector bundle or sphere bundle over smooth manifolds $M$. We can get an pullback bundle over $V$ by $p^*TM$ , $p: V \rightarrow M$ . If we give the inner product on each fiber of $p^*TM$ , is there a differetial form on $V$ such that it pullback by a section of $V$ is the Euler form ? I know two famous examples, frist is Mathai-Quillen Thom form, second is Causs-Bonnet-Chern type formula on Finsler manifolds. Is there some other examples?

Edit: May be my expression is not exactly. I restate it now. The first, I express why I have this question. If $M$ is a Riemannian manifold, we can use the connection of $TM$ to construct a differetial form that is Pfaffian, it's integration on $M$ is Euler characteristic. This is Gauss-Bonnet-Chern formula. If $M$ is a Finsler manifold, we must use pullback bundle $p^*TM$ on sphere bundle, then use the preserves connection of $p^*TM$ can get a Gauss-Bonnet-Chern type formula, that is a differential form on sphere bundle and pullback by a section with isolated singular can get a "Euler form"(I mean a differetial form, it's integration on $M$ is Euler characteristic) on $M$. About this you can read "A Gauss-Bonnet-Chern formula for Finsler manifolds" by Z.Shen http://www.math.iupui.edu/~zshen/Research/preprintindex.html We also can use torsion-free connection of $p^*TM$ to get a Gauss-Bonnet-Chern type formula, you can read On the Gauss–Bonnet Formula in Riemann–Finsler Geometry by B.Lackey. If we think Mathai-Quillen Thom form, we use pullback bundle $p^*TM$ on $TM$, and the preserves connection of $p^*TM$ can use to construct a differential form on $TM$ we need. I think the metric on $p^*TM$ is the thing we must use, and we needn't restrict on $TM$ or $SM$, we can use another vector bundle or sphere bundle to get a pullback bundle on it.

The second, I turn to my question. If $V$ is a vector bundle or sphere bundle over smooth manifolds $M$. $p^*TM$ is a pullback bundle on $V$, I want to use the connection of this pullback bundle to construct a differential form on $V$ and use a section to pullback it, then get a "Euler form".

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  • $\begingroup$ By "the" Euler form, I think you mean a canonical differential form representing the Euler class, such as the Pfaffian of the curvature. And I think you mean the Euler form of the tangent bundle $TM$. $\endgroup$
    – Guangbo Xu
    Oct 9, 2010 at 20:14
  • $\begingroup$ Yes. I mean the differential form representing the Euler class. Here I mean Euler form of the tangent bundle TM. $\endgroup$
    – Chen
    Oct 10, 2010 at 2:52
  • $\begingroup$ Then what has the inner product on $p^*TM$ to do with those forms? The question itself has nothing to do with the pullback bundle $p^*TM$. I guess you already have a Riemannian metric on $M$ so you can say "the Euler form" of $TM$, which is the Pfaffian of the curvature. If you don't specify a differential form, the answer is trivial in the cohomology level. $\endgroup$
    – Guangbo Xu
    Oct 10, 2010 at 16:49
  • $\begingroup$ May be I use this word "the Euler form" is not exactly, I mean a differential form on $M$, it's integration on $M$ is the Euler characteristic of the manifold. $M$ also can have Finsler metric. You say it is trivial in the cohomology level, can you say it more concretely? Thanks $\endgroup$
    – Chen
    Oct 13, 2010 at 5:15
  • $\begingroup$ Sorry I didn't make it clear. If $V$ has nothing to do with $TM$, then if $e$ is an Euler form of $TM$, then $p^*e$ is a differential form on $V$, and then for any section $s$ of $V$, obviously $s^* p^* e= (ps)^* e=e$, by tautology. But this is not much useful. In case $V=TM$ or the sphere bundle of $TM$, we have Thom forms or angular forms respectively. The pull-back of the former gives an Euler form since it can be viewed as a definition of Euler class. For the latter(sphere bundle), if you have a section, then the Euler characteristic is zero. $\endgroup$
    – Guangbo Xu
    Oct 13, 2010 at 20:20

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