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The problem is: Consider the differential form $a=p_1dq_1+p_2dq_2-p_1p_2dt$ in the space of $R^5$ with coordinates $(p_1,p_2,q_1,q_2,t)$.

(a) compute $da$ and $da\wedge da$

(b) Evaluate the integral $\int_S t da\wedge da$ where $S$ is the 4-dim surface(with boundary) in $R^5$ defined by the relations: $p_1^2+p_2^2\le1, q_1^2+q_2^2\le2$ and $t=3$.

I just learn differential forms, and I have difficulty in this problem. I know it is a problem on Stokes theorem but I do not know how to use it.

For part (a), I don't know my calculation is right or wrong, since this is the first time I do such a long differentiation. My work is

$$da=d(p_1dq_1+p_2dq_2-p_1p_2dt)$$ $$=dp_1\wedge dq_1+dp_2\wedge dq_2-d(p_1p_2)\wedge dt$$ $$=dp_1\wedge dq_1+dp_2\wedge dq_2+p_2dp_1\wedge dt-p_1dp_2\wedge dt$$

$$da\wedge da=-2dp_1\wedge dp_2\wedge dq_1\wedge dq_2-2p_1\wedge dp_1\wedge dp_2\wedge dq_1\wedge dt+2p_2\wedge dp_1\wedge dp_2\wedge dq_2\wedge dt$$

My question is

(1) Is there any quick way(or some known result) to calculate $da\wedge da$, or the only way is to expand it one by one, just like expanding the product of polynomials? Since this is the first time for me to do this kind of calculation, I am not sure I am doing it in the right or usual way.

(2) For part (b), I know I need to use the $\int_{\partial\Omega}\Phi=\int_{\Omega}\partial\Phi$ kind of formula. But I am kind of confuse of the higher dimension "surface". Should we set $S$ in the problem to be $\Omega$ or $\partial\Omega$? And if it is $\Omega$, what is $\Phi$ and what is $\partial\Omega$?

I saw most of the books only gave a abstract form of stokes theorem and then it is over. Could you please help me with this concrete example? Is there any book that contains this kind of worked problems?

Thank you very much!

breezeintopl
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2 Answers2

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You should set $S$ to be $\Omega$, which is (as they say) a "surface" with boundary.

One reason to see this: $a$ is a 1-form: it only involves, in each addend, a single $dx$ term. $da$ is then a 2-form, and so $da \wedge da$ is a $2 + 2 = 4$-form. Thus you can only integrate it over a 4-dimensional space, which $\Omega$ is. So the expression $$ \int_\Omega da \wedge da $$ makes sense, whereas $$ \int_{\partial\Omega} da \wedge da $$ would necessarily be zero, since $\partial\Omega$ is 3-dimensional.

In this case, what you need is a concrete description of the surface and its boundary. The surface is given as (essentially) the set $D_1 \times D_2$ where $D_r$ is a disk of radius $r$. The boundary of a cartesian product is given by $$ \partial (A \times B) = \partial A \times B \sqcup A \times \partial B $$ which should allow you to more readily compute the required integrals. I suppose that the only thing remaining is to either a) write $da \wedge da = d(\omega)$ for some 3-form $\omega$ (and then to use Stoke's theorem, as you noted), or to just perform the computation by hand...

Simon Rose
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For part (a), I believe you missed a sign in the third term on the third line.

For part (b), part (a) is almost a red herring, write $i: S \rightarrow \mathbb{R}^5$ for the inclusion. Since wedging and applying $d$ commute with pulling back, and the pullback $i^*a$ is just $p_1 dq_1 + p_2 dq_2$ ($t$ is constant), we have that $i^* (t da \wedge da) = -6 dp_1 \wedge dp_2 \wedge dq_1 \wedge dq_2$, i.e. $-6$ times the volume form on $\mathbb{R}^4 \times \{3\} \subseteq \mathbb{R}^5$. So now we're asking for the volume of a product of disks of radius squared 1 and 2, which is $2\pi^2$. So the integral is $-12 \pi^2$.

uncookedfalcon
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  • Thank you so much for your answer! Now I almost understand it. I can understand the integrand is -6 times the volume. But when I do this integral, I did not realize it corresponds to a volume. So I just do it by parameterization and I got $-2\pi^2$. I know I must be wrong, but I cannot find the mistake. I think the problem must be in the parameter matrix. Does the parameters order matters in the Jacobian Matrix? I mean, if I do the parameter form of the two disks with $r_1,\theta_1, r_2,\theta_2$, does that matter if I take the columns in the order $r_1,r_2,\theta_1,\theta_2$? Thank you! – breezeintopl Apr 11 '14 at 02:59
  • You're welcome, and you are absolutely correct - to integrate a differential form you need to consider 'orientation', and the swapping coordinates map $(r_1, r_2, \theta_1, \theta_2) \mapsto (r_1, \theta_1, r_2, \theta_2)$ has determinant -1, i.e. reverses orientation. Plainly, it's why you add a sign when swapping $dr_1 d\theta_1 dr_2 d\theta_2 = - 1 dr_1 dr_2 d\theta_1 d\theta_2$. – uncookedfalcon Apr 11 '14 at 06:27
  • By the way, I should have been more careful about this earlier, but I don't see how $S$ comes with a preferred orientation, i.e. $\pm 12 \pi^2$ both look equally nice to me. I could be wrong though, perhaps ask someone in your class. – uncookedfalcon Apr 11 '14 at 06:32
  • Thank you so much for your help! It's really helpful. – breezeintopl Apr 12 '14 at 00:30
  • Glad to help - I also found differential forms challenging when learning (and relearning, and relearning...) them, and had the same experience re: few concrete calculations. Best wishes – uncookedfalcon Apr 12 '14 at 10:03