# How to find the orientation of a manifold

I do not quite understand why the orientation of a manifold is defined to be a smooth nowhere vanishing differential form. For example, $$S^1$$, the unit circle, is an orientable manifold. Intuitively, I would like to pick counter-clockwise as my positive direction, but how does this correspond to a differential form? I just cannot associate vectors that define the orientation in the usual sense with differential forms. Can someone explain this to me a little bit and (hopefully) give some examples? Thanks a lot in advance!

• Of course you can. Vectors $v$ that point in the correct direction at $p$ are ones for which $\omega(p)(v) > 0$. In the case of the circle, we typically take $\omega = -y\,dx + x\,dy$ (pulled back to the circle). Apr 8, 2021 at 20:26
• Hello professor, I found your lectures on Youtube very helpful! Could you elaborate just a bit on the part "pulled back to the circle"? Apr 8, 2021 at 21:38
• That is a $1$-form on $\Bbb R^2$, but you can restrict it to any submanifold of $\Bbb R^2$. If you parametrize the circle in the obvious way, by $f(\theta) = (\cos(\theta),\sin\theta)$, then you should check that $f^*\omega = d\theta$ gives a nowhere-zero $1$-form on the circle that gives you the counterclockwise orientation you desire. Apr 8, 2021 at 21:43
• I see! Thank you very much for your help! Apr 8, 2021 at 21:50

Picking the counter-clockwise direction means that with each point of $$S^1$$, you associate a tangential vector pointing in that direction. This smells like a global section through the tangent bundle, whereas differential forms are sections through the co-tangential bundle - why the extra "co"? Well, you will get difficulties when trying to generalize this to $$S^2$$: The generalization would amount to picking a basis (that we want to declare as positively oriented) of the two-dimensional tangential space in each point and such that this choice depends smoothly on the point. But you can't do this, says the Hairy Ball Theorem.
Actually, instead of picking an example of a positively oriented base of the tangential space for every point, we should look for a method that can decide whether a base is positively or negatively oriented. That is, for $$n$$ vector fields $$X_1,\ldots, X_n$$, we want to compute some $$\alpha(X_1,\ldots, X_n)$$ such that the result is positive for positively oriented bases, negative for negatively oriented bases (and zero in case of linear dependnce). As each $$X_i$$ is in $$TM$$, this $$\alpha$$ is a map $$TM^n\to \Bbb R$$. To make life easy, we of course want $$\alpha$$ to be linear in each component (i.e., multi-linear). Moreover, swapping $$X_i\leftrightarrow X_j$$ shall result in a sign change; in particular, if $$X_i=X_j$$ with $$i\ne j$$, then the result shall be zero. This means that $$\alpha$$ is an alternating multi-linear form $$\bigwedge^nTM\to \Bbb R$$, in other words, a global section through the $$n$$th outer power of the co-tangential bundle - also known as a differential $$n$$-form.