Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 115 / 1

The equivalent formulations of orientability of a smooth manifold are:

A positive chart orients its coordinate frame. Conversely, smoothly oriented frames determine local positive coordinate volume forms. A partition of unity subordinate to an oriented atlas glues these positive forms: at each point they are positive multiples of one another, so their weighted sum cannot vanish. Finally, a nowhere-zero top form $\omega$ declares $(v_1,\dots ,v_n)$ positive exactly when $\omega (v_1,\dots ,v_n)>0$. These constructions are inverse at the level of orientations.

Let $\Omega$ orient $M\times \mathbb{R}$ and write $t$ for the coordinate on $\mathbb{R}$. Along $M\times \{0\}$, contraction with the transverse vector ${\partial }_t$ givesFor every basis of $T_{p}M$, adjoining ${\partial }_t$ gives a basis of $T_{(p,0)}(M\times \mathbb{R})$, so $\omega$ never vanishes. The top form $\omega$ orients $M$.

Assume $M\times N$ is oriented. Fix $q\in N$ and an ordered basis $(w_1,\dots ,w_n)$ of $T_{q}N$. Contracting the product orientation successively with the constant vectors $w_i$ along $M\times \{q\}$ produces a nowhere-zero top form on $M$. Hence $M$ is orientable. Fixing a point and a basis in $M$ gives an orientation of $N$ in the same way. This is the orientability of factors of a product manifold.