# $e^{xy}dx \wedge dy$: determine the $1$-form that it induces on $S^1$ and check if the obtained $1$-form respects or not the induced orientation

Consider the $2$-form $e^{xy}dx \wedge dy$ on $\mathbb{R}^2$.

Determine the $1$-form that it induces on $S^1$, viewed as the boundary of $B_2$.

Check if the obtained $1$-form respects or not the induced orientation.

Honestly I don't know where to begin, some help would be appreciated.

Somehow the answer for the first part is $\frac{e^{xy}}{y}dx$ or $-\frac{e^{xy}}{x}dy$.

The induced orientation of $S^1$ is given by $dx$ (this I know).

• I edited the question, I found an answer in someone's notes but I have no clue why is that – Leafar Jun 8 '15 at 20:07
• I'm following these notes: people.maths.ox.ac.uk/hitchin/hitchinnotes/manifolds2012.pdf but this is an exercise given by my teacher and it is not in the notes, and I copied the exercise correctly – Leafar Jun 8 '15 at 23:28
• I retract my previous comments. Your question is reasonable. I'm working on it now. – James S. Cook Jun 9 '15 at 2:27
• Btw, it seems these questions are not well-understood by the MSE, I see math.stackexchange.com/questions/1302681/… is still unanswered at this time. – James S. Cook Jun 9 '15 at 3:19

Assuming Euclidean metric on $\mathbb R^2$ and given a weighted "volume form" $\omega=e^{xy} dx \wedge dy$ which is not the natural volume form for this metric, we're asked to find its induced "volume form" $i^{*}(\iota_\nu \omega)$ on $S^1$, where we first take the interior product with the unit outer normal $\nu=x\partial_x+y\partial_y$ and then pull back to $S^1$ along the inclusion map.
Using the linearity of the interior product we arrive directly at @James's result: $\iota_\nu \omega=(-ydx+xdy)e^{xy}$. One could transform it further to either $\frac{e^{xy}}{y}dx$ or $-\frac{e^{xy}}{x}dy$ by differentiating and using again the $x^2+y^2=1$ and then for some reason changing the sign as @Ted noted.
I guess what I write here is wrong. But, I'm not sure exactly why, so I post it here in the hope someone points out the error of my ways. The two-form $\omega = e^{xy}dx \wedge dy$ may be written as $\omega = e^{r^2 \sin \theta \cos \theta} r \, dr \wedge d\theta$. Then, in polar coordinates, the unit circle is $r=1$ and $\partial / \partial r$ is an outward-pointing vector field on the circle. If I understand the induced orientation of $\omega$ correctly, then all I need to do is to take the inner product (sometimes called "hook") of $\omega$ with $\partial/\partial r$ and pull that back to the circle. Calculate $$\omega(\partial/\partial r, \cdot) = e^{r^2 \sin \theta \cos \theta} r dr(\partial/r)d\theta = e^{r^2 \sin \theta \cos \theta} r d\theta$$ but, this is to be pulled-back to the circle. In Cartesians, $$d\theta = \frac{-ydx+xdy}{x^2+y^2}$$ thus, using a hybrid notation momentarily, $$\omega(\partial/\partial r, \cdot) = e^{r^2 \sin \theta \cos \theta} r \frac{-ydx+xdy}{x^2+y^2} = \frac{-ydx+xdy}{\sqrt{x^2+y^2}}e^{xy}$$ But, I would think the pull-back of this form to the curve $x^2+y^2=1$ simply gives us: $$\phi = (-ydx+xdy)e^{xy}$$ Given all this, I'd think $\phi$ is consistent with $dx$ when $y<0$ but, opposite $dx$ when $y>0$. I'm not sure how to get the quoted answers in the stated question. Roughly I follow Lee's Smooth Manifolds, which is also discussed in Inducing orientations on boundary manifolds .
• James and Leafar: The $e^{xy}$ is a complete red herring, inasmuch as it is always positive. The $2$-form $dx\wedge dy$ gives the "usual" orientation on $\Bbb R^2$, and the boundary orientation on $S^1=\partial D^2$ is, indeed, given by the $1$-form obtain by contracting against the outward normal ($\partial/\partial r$), so the induced orientation is given by $d\theta = -y\,dx+x\,dy$. – Ted Shifrin Jun 9 '15 at 14:32
• @TedShifrin Thanks for the comment, do you have any idea where they get $(e^{xy}/y) dx$ or $-(e^{xy}/x) dy$ as the induced orientations? – James S. Cook Jun 9 '15 at 15:00