Here is my view of orientability on a vector space $V$ of dimension $m>0$: let $I(V)$ be the set of linear isomorphisms from $V$ to $\mathbb{R}^m$. Given $\rho,\sigma\in{I(V)}$, we get a linear automorphism $\sigma\circ\rho^{-1}:\mathbb{R}^m\rightarrow\mathbb{R}^m$ with $\det(\sigma\circ\rho^{-1})\not=0$. We write $\sigma\sim\rho$ if $\det(\sigma\circ\rho^{-1})>0$, which defines an equivalence relation on $I(V)$. We define $\mathrm{Or}(V)=I(V)/\sim$, so that $|\mathrm{Or}(V)|=2$, and define an orientation on $V$ to be an element of $\mathrm{Or}(V)$. An oriented vector space is then a (finite dimensional) vector space equipped with an orientation.

Now, a basis of an oriented vector space $V$ determines an element of $I(V)$ which sends this basis to the standard basis of $\mathbb{R}^m$. We refer to this basis as positively oriented (with respect to our orientation on $V$) if this map lies in the equivalence class of our orientation, and negatively oriented otherwise.

For now I am considering an $m$-manifold embedded in $\mathbb{R}^p$. Then an orientation on such a manifold is as assignment of an orientation to each tangent space $T_x{M}$, such that there is an atlas of charts $\phi_\alpha:U_\alpha\rightarrow{V_\alpha}\subseteq\mathbb{R}^m$ such that for all $\alpha$ and all $x\in{U_\alpha}$, the map $d_x\phi_\alpha:T_x{M}\rightarrow\mathbb{R}^m$ lies in the orientation class of the orientation in $T_x{M}$. We refer to $M$ as an oriented manifold, and say $M$ is orientable if it admits an orientation.

Now, I'm aware that the following question has been asked before, but all the answers I've found have used concepts I haven't come across yet, such as differential forms, pullbacks etc. The problem is to show that if $M$ is an orientable manifold and $f:M\rightarrow{N}$ is a diffeomorphism between manifolds (where $N$ is also an $m$-manifold embedded in $\mathbb{R}^q$), then $N$ is also orientable. I've tried using the fact that $\{\phi_\alpha\circ{f}^{-1}\}_\alpha$ will be an atlas for $N$ if $\{\phi_\alpha\}$ is an altas for $M$, but am getting nowhere. I'm thinking the right way to go about it is a basis approach?


If $g_i : {\bf R}^n\rightarrow M $ is orientable chart for $M$, then for any two, $$ {\rm det}\ d(g_2^{-1}\circ g_1 ) > 0 $$ since $M$ is orientable

So $h_i:=f\circ g_i : {\bf R}^n\rightarrow N $ is chart so that $$ {\rm det}\ d( h_2^{-1}\circ h_1) = {\rm det}\ d(g_2^{-1}\circ f^{-1} \circ f \circ g_1)= {\rm det}\ d(g_2^{-1}\circ g_1) >0 $$

  • 1
    $\begingroup$ Surely $g_i$ and $h_i$ don't have to be any chart - an orientation on a manifold only requires the existence of some chart with the above properties? $\endgroup$
    – jl2
    Mar 28 '16 at 10:21
  • $\begingroup$ You are right I fix it $\endgroup$
    – HK Lee
    Mar 28 '16 at 10:26
  • 2
    $\begingroup$ Thank you - so this relies on the fact that a manifold is orientable if and only if there is an atlas that preserves orientation at the 'crossover' points? $\endgroup$
    – jl2
    Mar 28 '16 at 10:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.