Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $M$ be a connected orientable smooth manifold.

Is it true that $M$ must have only 2 orientations?

If yes, why?

share|cite|improve this question
See this question for more details. The answer is yes. Why just two orientations? Because you are picking out compatibility based on Jacobian determinants being positive. 2 choices since "positive, negative" is a list of 2 things. Or another way to look at it: Because $\mathrm{GL}_n(\mathbb{R})$ has two connected components (matrices with positive determinants and those with negative determinants). – Bill Cook Feb 15 '12 at 19:11
Alternatively, two orientations agree on an open set and disagree on an open set. – Mariano Suárez-Alvarez Feb 15 '12 at 19:35
@MarianoSuárez-Alvarez sorry I didn't understand what you mean – Jr. Feb 15 '12 at 20:08
@Jr.: What he means is that if I have two arbitrary orientations, then on any given open set, they either agree or they disagree. – Zhen Lin Feb 15 '12 at 20:12

I am also trying this question and again like you do, I am only given very basic definitions of orientations using atlas. Let me try first and see if this works, and give me your comments! (By the way I am reading sharpe's book)

Let $M$ be a connected orientable smooth manifold.

We say that an atlas $(U_{i},\phi_{i})$ is oriented if the jacobian of the transformation between two charts at the intersection is positive. (i.e. $\det\mbox{Jac}\phi_{j}\circ\phi_{i}^{-1}>0$)

Let us say that two oriented atlas $(U_{i},\phi_{i})$ and $(V_{j},\psi_{j})$ are equivalent if the union forms an oriented atlas for $M$.


Let $(U_{i},\phi_{i})$ and $(V_{j},\psi_{j})$ be oriented atlases such that its union forms an atlas for $M$. If there exists $p\in U_{\alpha}\cap V_{\beta}\subseteq M$ such that the determinant of the Jacobian $\psi_{\beta}\circ\phi_{\alpha}^{-1}$ is positive, then both atlases are equivalent.


We will use the fact that $M$ is connected! We define two subsets as below

$S_{1}=\{p\in M: \exists U_{i},V_{j}\mbox{ such that }\det\mbox{Jac}(\psi_{j}\circ\phi_{i}^{-1})|_{p}>0\}$

$S_{2}=\{p\in M: \exists U_{i},V_{j}\mbox{ such that }\det\mbox{Jac}(\psi_{j}\circ\phi_{i}^{-1})|_{p}<0\}$

Note that we don't have $0$ as a case because $\psi_{j}\circ\phi_{i}^{-1}$ is a diffeomorphism on the intersection.

Now we have $S_{1}\cap S_{2}=\emptyset$ because suppose we have $p\in S_{1}\cap S_{2}$, then $p\in U_{i}\cap V_{j}$ and $p\in U_{k}\cap V_{l}$ such that $\det\mbox{Jac}(\psi_{j}\circ\phi_{i}^{-1})>0$ and $\det\mbox{Jac}(\psi_{l}\circ\phi_{k}^{-1})<0$. But we will run into problems because while $\det\mbox{Jac}(\psi_{j}\circ\phi_{i}^{-1})>0$, we also have


On the other hand, $S_{1}\cup S_{2}=M$, because if $p$ lies in $M$, it must go into some $U_{i}$ and $V_{k}$ and so the determinant of the jacobian is either $>0$ or $<0$.

Now it is easy to see that both $S_{1}$ and $S_{2}$ are open because taking jacobian, and then the determinant are all continuous functions and both $(0,\infty)$ and $(-\infty, 0)$ are open in $\mathbb{R}$, (so the pre-image is open in the intersection $U_{i}\cap V_{j}$ and hence is also open in $M$.

But there exists a point which both atlas agree (or compatible), this means $S_{1}$ is not empty, and so $S_{2}$ is empty because $M$ is connected.

In a same manner, if both oriented atlases disagree on one point, they disagree everywhere.

Attempted proof to your question

Since $M$ is orientable, we have already found one oriented atlas $(U_{i},\phi_{i})$.

We take the same atlas and give a negative orientation. This can be easily done by post composing $\phi_{i}$ with the operation $L:(x_{1},\ldots,x_{n})\mapsto (-x_{1},\ldots, x_{n})$ to get $L\circ\phi_{i}$.

These are the only two orientations by our previous claim because if we choose another oriented atlas $(W_{k},\lambda_{k})$ for $M$, just choose one point $p\in U_{i}\cap W_{k}$. Then because the determinant of the jacobian is either positive or negative, therefore $(W_{k},\lambda_{k})$ must agree with either $(U_{i},\phi_{i})$ or $(U_{i},L\circ\phi_{i})$.


share|cite|improve this answer

I think that this question can be easily answered if one analyzes the definitions carefully. Let me recall the following relevant definitions:

Definition 1 An $n$-form on a smooth $n$-manifold $M$ is a function $\Omega$ that assigns to each $p\in M$ an alternating $n$-tensor $\Omega_p$ on the tangent space $T_p(M)$. The $n$-form $\Omega$ on $M$ is said to be smooth if $\Omega(X_1,\dots,X_n)$ is a smooth function on $U$ whenever $X_1,\dots,X_n$ are smooth vector fields on any open subset $U$ of $M$.

Exercise 1: Prove that the set of alternating $n$-tensors on an $n$-dimensional vector space $V$ is itself a vector space of dimension $1$.

Exercise 2: The definition of smoothness of an $n$-form on a smooth $n$-manifold $M$ given in Definition 1 is "coordinate-free" (a desirable situation in differential geometry). Prove that this definition is equivalent to the following one using local coordinates on the smooth manifold $M$:

Definition 2 The $n$-form $\Omega$ on the smooth manifold $M$ is smooth if and only if $\Omega(E_1,\dots,E_n)$ is a smooth function on $U$ whenever $(U,\phi)$ is a coordinate neighborhood of $M$ and $E_1,\dots,E_n$ are the corresponding coordinate frames on $U$ (i.e., $E_i=\phi_{*}^{-1}\left(\frac{\partial}{\partial x^i}\right)$ for all $1\leq i\leq n$).

Definition 3 A smooth $n$-manifold $M$ is said to be orientable if there is a smooth $n$-form $\Omega$ on $M$ which is nowhere zero on $M$. In this case, $\Omega$ is said to be an orientation of $M$. An oriented smooth manifold $M$ is a smooth manifold $M$ equipped with an orientation of $M$. Two orientations $\Omega_1$ and $\Omega_2$ of $M$ are said to be equivalent, written $\Omega_1\equiv \Omega_2$, if there is a positive smooth function $f$ on $M$ such that $f\Omega_1=\Omega_2$.

Lemma The relation $\Omega_1\equiv \Omega_2$ is an equivalence relation on the set of orientations of $M$. If the smooth manifold $M$ is connected, then there are exactly two equivalence classes.

Proof. The reflexivity of the relation follows from the smoothness of the constant function $1$ on $M$, the symmetry follows from the smoothness of $\frac{1}{f}$ on $M$ when $f$ is a positive smooth function on $M$, and the transitivity follows from the smoothness of the product $fg$ on $M$ when $f$ and $g$ are positive smooth functions on $M$.

Let $\Omega_1$ and $\Omega_2$ be orientations of $M$. Let us define a function $f$ on $M$ by the rule $f=\frac{\Omega_2}{\Omega_1}$ (this definition is valid because of Exercise 1 and the fact that $\Omega_1\neq 0$ on $M$).

Exercise 3: Prove that $f$ is smooth on $M$.

Note that $f\neq 0$ on $M$ because $\Omega_2\neq 0$ on $M$. Since $M$ is connected, it follows that either $f$ is positive on $M$ or negative on $M$. We conclude that either $\Omega_2\equiv \Omega_1$ or $\Omega_2\equiv -\Omega_1$. Hence there are at most two equivalence classes. However, we clearly cannot have $\Omega_1\equiv -\Omega_1$. The proof is complete.

I mention the following definition because it is very important in the formulation of Stokes' theorem and in the study of Lie groups (to mention a couple of famous examples).

Definition 4 A volume element on a smooth oriented connected manifold $M$ is a nowhere vanishing smooth $n$-form on $M$ which belongs to the equivalence class of $M$ determined by the orientation of $M$.

I hope this helps!

share|cite|improve this answer
thank you for the answer, but the book I'm reading assumes no previous knowledge in differential forms.The definition of orientable manifold is a manifold with a differentiable structure such that the differential of the change of coordinates has positive determinant. – Jr. Feb 16 '12 at 3:42
@Jr.: You should show that your definition and this definition are equivalent. This is the better definition, in my opinion, since it manifestly permits only two orientations on each connected component! – Zhen Lin Feb 16 '12 at 7:28
@Jr. I think my answer is self-contained in the sense that I have defined "$n$-form" and "smooth $n$-form" assuming only a knowledge of such concepts as "differentiable structure on a manifold", "smooth function" and "smooth vector field" which are fairly basic. The term "alternating tensor" might be unfamiliar but can be easily looked up. Also, it is easy to prove the equivalence between the definition that you have noted in your comment and the definition that I have used of orientability. Indeed, if you are familiar with partitions of unity, then it should be a straightforward exercise. – Amitesh Datta Feb 16 '12 at 7:31
Awesome! Your proof is clear and intuitive. Thanks. – uhbif19 Jul 13 '15 at 19:53

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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