# Differential forms: The authors of a paper define $d(u\times du)$, but what is $u \times du$ supposed to mean?

I'm reading [1] recently and have another question about a remark in this paper. I tried to solve it myself (see below) but did not succeed. It could be just a notation problem.

The Setup:

• Let $u \in H^1(\Omega,\mathbb{C})$ where $\Omega = [-\pi,\pi]^3 \subset \mathbb{R}^3$.

• In [1, Rmk.4.2] the authors denote by $Ju$ the 2-form $$Ju \equiv \frac{1}{2} d(u \times du) = \sum_{1 \leq i < j \leq 3} (\partial_i u \times \partial_j u) dx_i \wedge dx_j.$$

• Furthermore they define $$\zeta_1(x)= - x_2 dx_1 \wedge dx_2 - x_3 dx_1 \wedge dx_3.$$

• From this definition it follows that $$\star \zeta_1 = -x_2 dx_3 + x_3 dx_2$$ which I checked. Here $\star$ denotes the Hodge-star-operator.

• I then calculated $$d(\star \zeta) = -2dx_2 \wedge dx_3 \tag{G1}$$ which I hope is correct.

• Now the authors state that $$(u \times du) \wedge d(\star \zeta_1) = 2 \langle i \partial_1 u, u \rangle dx_1 \wedge \ldots \wedge dx_3 \tag{G2}$$

My Question:

What is $u \times du$ supposed to be?

My Attempt:

I tried to find out myself, but I discovered the following difficulty:

• Let $\omega=\sum_{i=1}^3 \omega^i dx_i$ be the 1-Form $\omega=u \times du$. Then $$\frac{1}{2} d\omega = \frac{1}{2} \sum_{1 \leq i< j \leq 3} \omega^i_{x_j} dx_j \wedge dx_i\stackrel{!}{=} Ju = \sum_{1\leq i < j\leq 3} (\partial_i u \times \partial_j u) dx_i \wedge dx_j.$$ This suggests that $\omega_{x_j}^i = -2 (\partial_i u \times \partial_j u)$.

• On the other hand because of $(G1)$ equation $(G2)$ becomes $$\omega \wedge d(\star \zeta) = -2\omega^1dx_1 \wedge dx_2 \wedge dx_3 \stackrel{!}{=} 2 \langle i \partial_1 u, u \rangle dx_1 \wedge dx_2 \wedge dx_3,$$ which suggests $\omega^1=-\langle i \partial_1 u, u \rangle$.

So my question reduces to

How are these two equations compatible? What is the $\times$ supposed to mean?

[1] Béthuel, F., P. Gravejat und J. C. Saut: Travelling waves for the Gross- Pitaevskii equation. II. Comm. Math. Phys., 285(2):567–651, 2009.

• Isn't it simply $u\times du = u\, du$? Commented Jul 3, 2013 at 14:11
• Well if that was true, i.e. when $\times = \cdot$ then $d(udu)=d(\sum u u_{x_i} dx_i)=\sum u_{x_j} u_{x_i}+u u_{x_i x_j} dx_j \wedge dx_i \neq \sum u_{x_i} u_{x_j} dx_i \wedge dx_j$, right?
– mjb
Commented Jul 3, 2013 at 15:09
• $d(u\,du)=0$, so no help there. Commented Jul 3, 2013 at 15:20
• What does $H^1$ signify? Commented Jul 3, 2013 at 16:39
• @Muphrid The Sobolev space $W^{1,2}$
– mjb
Commented Jul 4, 2013 at 6:57

I believe this is all resolved if you have one typo. If $u$ maps $\Omega$ to $\mathbb R^3$, then you are taking the cross product of a vector in $\mathbb R^3$ with an $\mathbb R^3$-valued $1$-form. This is totally consistent with their formulas.

So, we both win.:) They are interpreting $\mathbb C \cong \mathbb R^2$ and defining the cross product of two vectors $u,v\in\mathbb R^2$ as the real number $(u\times v)\cdot e_3\in\mathbb R$. In particular, $$u\times du = \big(u_2(\partial_1u_1) - u_1(\partial_1 u_2)\big)dx_1 \pmod{dx_2,dx_3}\,.$$

All their formulas are consistent with this, recalling that they've defined $\langle u,v\rangle = \text{Re}(u\bar v)$, i.e., the real dot product of the vectors in $\mathbb C \cong \mathbb R^2$.

• That may be true but as you can see here www.math.polytechnique.fr/~gravejat/Recherch/BGS1-GP.pdf on page 44, this is not a typo. I need $u$ to be in $H^1(\Omega,\mathbb{C})$
– mjb
Commented Jul 3, 2013 at 15:01
• Thank you very much! I am still struggling to understand your post: What do you mean by $u \times v$ if $u,v \in \mathbb{R}^2$? and why is $(u \times v) \cdot e_3 \in \mathbb{R}$? And most important of all: what do you mean by $\mbox{mod } dx_2, dx_3$? Sorry for these questions, but I'm not very familiar with differential forms. Thanks in advance!
– mjb
Commented Jul 4, 2013 at 6:54
• Oh, I meant that you can take the cross product of two vectors in $\mathbb R^2$ by thinking if them as vectors in $\mathbb R^3$ with their last coordinate equal to $0$. the answer is a vector of the form $(0,0,\#)$; by taking just $\#$ we think of this as a real number. Next, since you are going to wedge your form with $dx_2\wedge dx_3$, any terms with $dx_2$ or $dx_3$ will then disappear, so by saying "mod" I meant we'll ignore those terms. Commented Jul 4, 2013 at 11:42
• Thanks a lot! That indeed explains equation (G2)! But how can I calculate $d(u \times du)$ from that? If $u \times du = (u_2 \partial_1 u_1 - u_1 \partial_1 u_2)dx_1 + \ldots$, then I would get mixed derivatives in $d(u \times du)$, right? For example $\partial_2 (u_2 \partial_1 u_1) = \partial_2 u_2 \partial_1 u_1 + u_2 \partial_2 \partial_1 u_1$. But in the definition of $d(u \times du)$ I don't see such mixed derivatives! Where is my mistake?
– mjb
Commented Jul 4, 2013 at 14:25
• Well, for that, we can't ignore the $dx_2$ and $dx_3$ terms. By the product rule, it is $du\times du$ and you get the authors' formula in your first displayed equation. $$du\times du = \big(\sum \partial_ku \,dx_k\big)\times \big(\sum \partial_ju \,dx_j\big)\,.$$ For example, the $dx_1\wedge dx_2$ coefficient will be $\partial_1(u_1+iu_2)\times\partial_2(u_1+iu_2)$. Work it out. Commented Jul 4, 2013 at 14:53