Orientability on smooth manifolds

In my lecture, we have introduced orientable manifolds $$M$$. An orientation on $$M$$ is a class of oriented atlases, an oriented atlas consists of charts, s.t. dthe transition maps are orientation-preserving.

But now we also have the term of an orientation of the tangent space and I don't really understand what this means and how it is induced by the orientation on the manifold itself.

Can somebody explain this?

• A map $f:M\rightarrow N$ is orientation preserving if $\det Df_p>0$ for all $p\in M$. Of course $\det Df$ only makes sense in charts, I hope it is clear what this means. Also take a look at this math.stackexchange.com/questions/139215/… Jul 5, 2019 at 20:38

One definition (#1) of the orientation of a manifold $$X$$ is a choice of orientation on $$T_xX$$ for all $$x \in X$$ which varies continuously with $$x$$. So an orientation on a manifold by definition corresponds to choosing orientations on $$T_xX$$ continuously for all $$x \in X$$.

Aside on orientations of vector spaces

(#1') Let $$V$$ be a real $$n$$-dimensional vector space, and $$(v_1, \dots, v_n)$$ and $$(v_1', \dots, v_n')$$ be bases of $$V$$. Then $$v_i'= \sum_{j=1}^n A_{ij}v_j$$ for $$A \in \mathrm{GL}(n, \mathbb{R})$$. Now define an equivalence relation on bases of $$V$$ by $$(v_1, \dots, v_n) \sim (v_1', \dots, v_n')$$ if and only if $$\det A > 0$$. Then an orientation of $$V$$ is a choice of equivalence class $$[ v_1, \dots, v_n ]$$.

(#2') There is a second notion of orientations of vector spaces. Let $$V$$ be as above. Then $$\bigwedge^n V \cong \mathbb{R}$$ and we have $$0 \neq v_1 \wedge \cdots \wedge v_n \in \bigwedge^n V$$. By what we've said, $$\bigwedge^n V \setminus 0 \cong (-\infty, 0) \cup (0, \infty)$$ has two connected components.

It is a fact that if $$(v_1, \dots, v_n)$$ and $$(v_1', \dots, v_n')$$ are bases of $$V$$ with $$v_i'= \sum_{j=1}^n A_{ij}v_j$$ then $$v_1' \wedge \cdots \wedge v_n' = \det A \cdot v_1 \wedge \cdots \wedge v_n .$$

So define an equivalence relation $$(v_1, \dots, v_n) \sim (v_1', \dots, v_n')$$ if and only if $$v_1' \wedge \cdots \wedge v_n'$$ and $$v_1 \wedge \cdots \wedge v_n$$ lie in the same connected component of $$\bigwedge^n V \setminus 0$$.

You may ask how the definition of orientation of a manifold that I've given above and the definition you've given in terms of charts are equivalent.

Firstly, as we've said in (#2') above, one can interpret an orientation on a vector space $$V$$ as choosing a connected component of $$\bigwedge^n V \setminus 0$$. Applying this definition of orientations of vector spaces to the vector spaces $$T_xX$$ gives a second definition for the orientation of a manifold:

#2: An orientation of an $$n$$-manifold $$X$$ is an equivalence class $$[ \omega ]$$ of non-vanishing top forms $$\omega \in \Omega^n(X)$$ where $$\omega$$ is equivalent to $$\omega'$$ if and only if $$\omega'=f \circ \omega$$ for a smooth function $$f : X \to (0 , \infty)$$.

We finally come to the third definition (your definition) of the orientation of a manifold:

#3: We first define oriented charts. Let $$X$$ be an $$n$$-manifold with orientation $$[ \omega ]$$. Let $$(U, \phi)$$ be a chart on $$X$$. We call $$(U, \phi)$$ oriented if $$\phi^*(\omega) = f \cdot dx_1 \wedge \cdots \wedge dx_n$$ where $$f : U \to \mathbb{R}$$ and $$f >0$$. we can find an atlas $$\mathcal{A} = \{ (U_i, \phi_i) \mid i \in I \}$$ for $$X$$ consisting of only oriented charts. Call $$\mathcal{A}$$ an oriented atlas. For two such charts on $$X$$ with local coordinates $$(x_1, \dots, x_n)$$, $$(y_1, \dots, y_n)$$ we have $$\det \left( \frac{ \partial y_i}{\partial x_j} \right)_{i, j=1}^n > 0$$ on the overlap. Then we can define an oriented manifold to be a manifold with oriented atlas.

Hopefully this chain of definitions sufficiently links together the way of thinking about orientations of manifolds via charts and the way of thinking about orientations of manifolds as orientations of their tangent spaces.

• Thank you for the answer! I think my biggest problem was that I completely forgot about the idea of orientation on a vector space. I have one question left on the first definition, you say "choice of orientation on $T_xM$" but here it is not clear to me which kind of orientation you mean. Jul 6, 2019 at 9:34
• You're welcome; I'll add discussion about orientations of vector spaces later today when I get back Jul 6, 2019 at 12:30
• Edited to include aside on vector space orientations. Jul 6, 2019 at 23:35
• Ok, thanks! That's a really good explanation! Jul 8, 2019 at 16:16

If a vector space has dimension n, then the space of n-linear alternating functions, denoted $$Alt^n$$(V) on V is one-dimensional. Let $$\omega \not=$$ 0 be such a form. If we view this space as a module over $$C^{\infty}$$ functions on V, we have two classes $$\mathbb{R}_+\omega$$ and $$\mathbb{R}_-\omega$$. These classes are the orientations on V. Given an n-dimensional manifold M, we say that a n-form $$\omega$$ is an assignment from p $$\in$$ M to $$\omega_p$$ $$\in$$ $$Alt^n(T_pM)$$. If $$\omega_p$$ $$\not= 0$$ for every p, we say M is oriented.