3
$\begingroup$

Let $M$ be a closed manifold, $\nabla_1,~\nabla_2$ be two euclidean connections on $TM$, by the classical Chern Simons formula, we know, there is an transgression form(Chern-Simons) $\tilde e(TM,\nabla_1,\nabla_2)$ such that $$d\tilde e(TM,\nabla_1,\nabla_2)=e(TM,\nabla_2)-e(TM,\nabla_2),$$ where $e(TM,\nabla_i)$ denotes the Euler class associated with the connection $\nabla_i$.

Q: If $\dim M$ is odd, can we say $\tilde e(TM,\nabla_1,\nabla_2)=0$ in cohomology sense, i.e. $\tilde e(TM, \nabla_1,\nabla_2)$ is an exact form.

  • 3
    Isn't the Euler class for odd-dimensional manifolds always 0 in $\mathbb{R}$? I don't understand the others but if your formula for the transgression form is correct, then it just follows from the vanishing of the Euler classes. – 2017-02-15
  • 0
    @MoisheCohen Yes. But, I mean the Chern-Simons class, on the odd-dim closed manifold, the Euler class is zero, so Chern-Simons is closed. Then, can we say it is exact ? I do not follow – 2017-02-20
  • 0
    Sorry, I made a mistake, $\tilde e$ is the Chern-Simons class, i.e. $d\tilde e=e_1-e_0$. – 2017-02-20

0 Answers 0