# Smooth atlas oriented iff transition maps have positive Jacobian determinant

First of all, my definition: An orientation of an $m$-dimensional smooth (connected) manifold $M$ is a choice of orientations for all tangent spaces $T_p M$, $p \in M$, such that there exists an atlas $\{(U_\alpha, h_\alpha)\}_{\alpha \in A}$ such that the maps $$T_p h_\alpha : T_pM \to T_{h_\alpha(p)} \mathbb R^m \cong \mathbb R^m$$ are orientation preserving for all $p \in M$ and all $\alpha \in A$.

I want to show that an atlas is oriented iff for all tranisition maps $h_{\alpha \beta} := h_\beta \circ h_\alpha^{-1}$ the Jacobian matrix $D h_{\alpha \beta} = D (h_\beta \circ h_\alpha^{-1})$ has positive determinant, i.e. $\det D h_{\alpha\beta} > 0$.

I have struggles with both directions. If I assume that my atlas is (positive) oriented, then for every tangent space the change of basis has positive determinant, by definition. How does that imply that $\det D h_{\alpha\beta} > 0$? Similarly, if all transition maps have a positive Jacobian determinant, how do I get to $\det T_ph_\alpha$ for all $p$ and $\alpha$?

I'd be very happy for some tips!

• Does the below help? Jul 16 '17 at 8:42

The maps $T_p h_{\alpha}$ are precisely the differentials of $h_{\alpha}: U_{\alpha} \to h_{\alpha}(U_{\alpha})$. Now what does it mean for this map to be smooth on $U_{\alpha} \subset M$? Precisely that, $h_{\alpha} \circ h_{\beta}^{-1}:h_{\beta}(U_{\beta}) \subset \mathbb{R}^m \to \mathbb{R}^m$ is smooth where $U_{\beta} \subset U_{\alpha}$. You just needed to recall that in order to differentiate, we need coordinates for the domain. To get these coordinates we use the inverse of a local chart. This is an example of what it means for $f: M \to \mathbb{R}$ to be smooth. Although never said, in calculus we've always used this association of $f$ with $f \circ \phi^{-1}$, but it was hidden due to the fact that we used the identity map for our local chart.

• I see your point. The differential of a map $h_\alpha$ is locally given by the Jacobian of $h_\alpha \circ h_\beta^{-1}$, so I think if $\det T_p h_\alpha > 0$, then the Jacobian determinant should also be positive. But how do I conclude $\det T_p h_\alpha > 0$ from the definition? I'm having a hard time understanding this "orientation preserving" property.
– user462006
Jul 16 '17 at 9:35
• There is too much notation and I think that is confusing you. Here we have, \begin{align*} &h_{\alpha} \circ h_{\beta}^{-1}: h_{\beta}(U) \to h_{\alpha}(U) \\ \\ & \Rightarrow T_p h_{\alpha} = D(h_{\alpha} \circ h_{\beta}^{-1})(p): T_ph_{\beta}(U) \to T_q h_{\alpha}(U) \end{align*} where $q =( h_{\alpha} \circ h_{\beta}^{-1})(p)$. Hence, it is a linear map from $\mathbb{R}^n$ to $\mathbb{R}^n$. We have a definition for linear maps $T$ between euclidean spaces which are orientation preserving, namely $\textbf{det}(T)>0$. Jul 16 '17 at 9:59
• Aaaah, now I get it. The other direction works similar. Thank you very much! (Edit: And you are right, there is way too much notation. That is one thing I do not like about this topic, I hope this gets better. :D)
– user462006
Jul 16 '17 at 11:10
• I was encouraged to make up my own when reading differential geometry. Most authors do the same because they know too how the notation differs amongst themselves. I'm just glad I was able to help. Jul 16 '17 at 15:05