# Vector bundle orientable if and only if determinant bundle is (Bott and Tu Prop. 11.4)

Let $$E$$ be a vector bundle over a smooth manifold $$M$$. By definition of the determinant bundle, it's clear that if $$E$$ is orientable then $$\det E$$ is as well, since $$\det E$$ is defined to be trivialized on the same cover as $$E$$ with transition functions which are the determinants of the transition functions of $$E$$. Bott and Tu Proposition 11.4 claims that the converse is true as well, but doesn't seem to provide any further justification. The fact that $$\det E$$ is orientable just means there exists a trivialization of it whose transition maps are positive numbers in $$GL(1, \mathbf{R})$$. For example, $$\det E$$ could be trivialized on all of $$M$$. There's no reason why $$E$$ needs to be trivialized on the same cover, so the existence of a trivializing cover for $$\det E$$ which is orientable on its own doesn't seem to imply the orientability of $$E$$. So is this really sufficient to deduce that $$\det E$$ orientable implies $$E$$ is? Probably I have completely misunderstood this.

Fix trivializations $$(\varphi_i)$$ of $$\det E$$ on an open cover $$(U_i)$$ of $$M$$ such that the transition maps are positive. Now fix trivializations $$(\psi_j)$$ of $$E$$ on an open cover $$(V_j)$$ such each $$V_j$$ is connected. Note that we can turn each $$\psi_j$$ into a trivialization $$\det \psi_j$$ of $$\det E$$ on $$V_j$$. For each $$p\in V_j$$, we can then consider the transition map between $$\det\psi_j$$ and $$\varphi_i$$ for any $$i$$ such that $$p\in U_i$$, and the sign of this transition map is independent of the choice of $$U_i$$ containing $$p$$. The sign of this transition map is thus a well-defined locally constant function on all of $$V_j$$, and thus is constant since $$V_j$$ is connected. Modifying $$\psi_j$$ if necessary (compose with a reflection), we may assume the transition map is positive.
I now claim that these trivializations $$\psi_j$$ (modified so that $$\det\psi_j$$ always has positive transition maps with each $$\varphi_i$$) witness the orientability of $$E$$. Indeed, given $$p\in V_j\cap V_k$$, the transition map between $$\psi_j$$ and $$\psi_k$$ at $$p$$ has positive determinant iff the transition map between $$\det\psi_j$$ and $$\det\psi_k$$ is positive. But we know that both $$\det\psi_j$$ and $$\det\psi_k$$ have positive transition maps to $$\varphi_i$$ for any $$i$$ such that $$p\in U_i$$, and so they must also have positive transition maps to each other.