Compute $d\omega$ and $\int_{S^2}\omega$.

Question

I am wondering if my solution is correct? Thanks.

(a) On $\mathbb{R}^3$, let $\omega = y dx \wedge dz.$ Compute $d\omega$ and $\int_{S^2}\omega$, where $S^2$ is the unit sphere centered at the origin, oriented as the boundary of the ball.

Revised following Ted's advice: $$\int_{S^2}\omega = \int_{D^3}d\omega= -\int_{D^3}dx \wedge dy \wedge dz = -\operatorname{vol}(D^3) = -\frac{4}{3}\pi.$$

(b) Let $f: F \to S^2$ be a smooth map of degree $n$ from a closed, oriented surface of $F$. Compute $\int_F f^*\omega.$

Added according to Ted's hint.

Degree Formula. Let $f: X \to Y$ be an arbitrary smooth map of two compact, oriented manifolds of dimension $k$, and let $\omega$ be a $k$-form on $Y$. Then $$\int_X f^* \omega = \deg(f) \int_Y \omega.$$

Hence, $$\int_F f^*\omega = \deg (f) \int_{S^2} \omega = -\frac{4n}{3}\pi.$$

:-D It is very delightful to hear that! It has been a very difficult book for me, but the semester spent with it is worthy! Huh... LAST THREE DAYS!!!! — 1LiterTears, Aug 17 '13 at 00:09
But the interest developed in the field while taking this course is continuing, @JamesS.Cook. — 1LiterTears, Aug 17 '13 at 01:37

score 3 · Accepted Answer · answered Aug 16 '13 at 23:53

3

In (a), your final integral is over the unit ball, not the sphere. For (b), you need to apply the Degree Theorem. It's immediate.

answered Aug 16 '13 at 23:53

Ted Shifrin

115,160

Oh no.. That's a bad mistake.. So it should be $$\int_{S^2}\omega = \int_{Br^2}d\omega= -\int_{Br^2}dx \wedge dy \wedge dz = -\operatorname{vol}(Br^2) = -\frac{4}{3}\pi,$$ right? – 1LiterTears Aug 16 '13 at 23:58
2

I don't like your notation. I would write $D^3$ for the closed unit ball in $\mathbb R^3$. – Ted Shifrin Aug 17 '13 at 00:47

Compute $d\omega$ and $\int_{S^2}\omega$.

1 Answers1