Let $E$ be a real oriented vector bundle of rank $k$ over a differentiable manifold $M$ of dimension $n$. By ''orientable'', I mean that the structure group of the bundle can be reduced to $GL^+(k,\mathbb{R})$.

We have (since $E$ is orientable) a Thom class $[\Phi] \in H^k_{cv}(E)$ ($*$) which restricts to a generator of $H^k_c(F) \cong H^k_c(\mathbb{R}^k) \cong \mathbb{R}$ for every fiber $F$, since $\int_F i^*(\Phi) = 1$ for every fiber $F$, where $i:F \to E$ is the inclusion.

I'd like to know if the existence of such a cohomology class guarantees that $E$ is orientable. More precisely, I'm wondering if the following holds:

Claim: Let $E$ be a real vector bundle of rank $k$ over $M$. There exists a closed form $\Phi \in \Omega^k_{cv}(E)$ such that $\int_F i^*(\Phi) = 1$ for every fiber $F$ (where $i: F \to E$ is the inclusion) if and only if $E$ is orientable.

If the claim held, then we could define an orientation of a vector bundle as a choice of such a form $\Phi$.

The reason (I may be wrong, of course) I think it should hold is that $\int_F i^*(\Phi) = 1$ implies $[i^*\Phi]$ is a generator of $H^k_c(F)$, and gives a ''sort of'' (can this be formalized?) orientation of $F$ for every fiber $F$. Since $\Phi$ is defined globally and is continuous, this choice of orientation is consistent, and then defines an orientation for $E$.

$(*)$ $\Phi \in \Omega^k_{cv}(E)$ is a closed differential form with compact vertical support. See for example Bott-Tu's Differential forms in Algebraic Topology.


1 Answer 1


Sure. It suffices to orient each fiber $E_x$ (in a locally compatible way, of course). A basis $(e_1)_x,\dots,(e_k)_x$ for $E_x$ will be declared positive iff $\Phi(z(x))((e_1)_x,\dots,(e_k)_x)>0$, where $z(x)$ is the point of the zero section of $E$ corresponding to $x$. (The Thom class has far more information in it than just orientation!)

EDIT: If you start with an arbitrary $\Phi$ with vertical compact support, as opposed to specifically the Thom class, you can use it to define a form $\tilde\Phi$ supported in a neighborhood of the zero section as follows. Fix once and for all a ball $B=B(0,r)\subset\Bbb R^k$. Let $\pi\colon E\to M$ be the bundle projection. Cover $M$ by open sets $U_i$ over which we have trivializations $g_i\colon U_i\times\Bbb R^k\to \pi^{-1}(U_i)$, and suppose $g_i^*\Phi$ is supported in $U_i\times V_i$. Choose a diffeomorphism $\psi_i\colon B\to V_i$ so that $(g_i^*\Phi)(\psi_i(0))\ne 0$. Set $\phi_i = \text{id}\times\psi_i$ and $\omega_i = (g_i^{-1})^*(g_i\circ\phi_i)^*\Phi$. Let $\{\rho_i\}$ be a partition of unity subordinate to $\{U_i\}$ and set $\tilde\Phi = \sum(\rho_i\circ\pi)\omega_i$.

COMMENT: This form is most likely no longer closed, but still integrates to $1$ on each fiber, and, by construction, is nonzero along the zero section.

  • $\begingroup$ Thanks for the answer! There's a detail I'm not sure about though. We know that $\int_{E_x} i^*\Phi = 1$ and that $i^*\Phi$ has compact support in $E_x$, but we could still have $\Phi(z(x)) = 0$, right? In that case, how do we define a positive basis in $E_x$? $\endgroup$
    – Mauro
    May 27, 2018 at 23:35
  • $\begingroup$ If you look at the construction, the form is a bump function supported in a neighborhood of the zero section. $\endgroup$ May 27, 2018 at 23:59
  • $\begingroup$ Yes, when we assume $\Phi$ is the Thom class I can see that. However, in the claim I made in the original question I only assumed $\Phi$ is a form with compact vertical support, but isn't necessarily supported in a neighbourhood of the zero section (I'm sure this is just a minor detail). $\endgroup$
    – Mauro
    May 28, 2018 at 0:15
  • $\begingroup$ @Mauro: See my edit. This should satisfy your curiosity. :) $\endgroup$ May 28, 2018 at 1:06
  • $\begingroup$ Thanks! It definitely did, it was just a detail after all! $\endgroup$
    – Mauro
    May 28, 2018 at 1:59

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