What does $df \wedge dg = 0$ imply?

Given $$f$$ and $$g$$ are smooth functions defined on some open set $$\Omega \subseteq \mathbb{R}^n,$$ what does $$df \wedge dg = 0$$ imply? Since $$df \wedge dg = \sum_{j=1}^n\frac{\partial f}{\partial x^j}dx^j \bigwedge \sum_{k=1}^n\frac{\partial g}{\partial x^k}dx^k=\sum_{j \neq k}\frac{\partial f}{\partial x^j}\frac{\partial g}{\partial x^k}dx^j \wedge dx^k =0,$$ does this imply $$\frac{\partial f}{\partial x^j}\frac{\partial g}{\partial x^k}=0$$ for $$j \neq k$$? That's what I thought at first, but now I am second guessing myself.

• (a) Do you mean $df \wedge dg = 0$ everywhere on $\Omega$, or just at some particular point? (b) You need to keep in mind that $dx^j \wedge dx^k = - dx^k \wedge dx^j$, so you can rewrite the last sum as $\sum_{j < k} \left( \frac{\partial f}{\partial x^j} \frac{\partial g}{\partial x^k} - \frac{\partial f}{\partial x^k} \frac{\partial g}{\partial x^j} \right) dx^j \wedge dx^k$, and then finally the $\{ dx^j \wedge dx^k \mid j < k \}$ are linearly independent. (c) You can also use that $df \wedge dg = 0$ if and only if $df$ and $dg$ are linearly dependent... Commented Apr 5, 2023 at 22:32

There's a small subtlety: remember that $$dx^j \wedge dx^k = -dx^k \wedge dx^j$$!

So in your expression $$\sum_{j \neq k}\frac{\partial f}{\partial x^j}\frac{\partial g}{\partial x^k}dx^j \wedge dx^k =0,$$ the coefficient in front of $$dx^j \wedge dx^k$$ is in fact $$\frac{\partial f}{\partial x^j}\frac{\partial g}{\partial x^k} - \frac{\partial f}{\partial x^k}\frac{\partial g}{\partial x^j}.$$ That's the quantity that vanishes.

• Thank you, but how do you arrive to the fact that quantity vanishes? Split the sum into two to get $\sum_{j<k}\frac{\partial f}{\partial x^j}\frac{\partial g}{\partial x^k}dx^j \wedge dx^k + \sum_{k<j}\frac{\partial f}{\partial x^j}\frac{\partial g}{\partial x^k}dx^j \wedge dx^k=\sum_{j<k}\frac{\partial f}{\partial x^j}\frac{\partial g}{\partial x^k}dx^j \wedge dx^k - \sum_{k<j}\frac{\partial f}{\partial x^j}\frac{\partial g}{\partial x^k}dx^k \wedge dx^j$? Commented Apr 5, 2023 at 22:42
• Could do that way. In your second sum, you could rename $j\mapsto k$ and $k \mapsto j$... Commented Apr 5, 2023 at 22:48
• Thank you so much, makes sense now! Commented Apr 5, 2023 at 23:07

As Kenny points out, in local coordinates this is just the condition that $$\frac{\partial f}{\partial x^j}\frac{\partial g}{\partial x^k} - \frac{\partial f}{\partial x^k}\frac{\partial g}{\partial x^j} = 0$$ for all pairs of indices $$j,k$$. In local coordinates, we can regard $$df$$ as being a smoothly varying field of covectors: $$df =\begin{bmatrix} \frac{\partial f}{\partial x^1} &\cdots& \frac{\partial f}{\partial x^n} \end{bmatrix}.$$ Arranging $$df$$ and $$dg$$ into a $$2\times n$$ matrix, $$\begin{bmatrix} \frac{\partial f}{\partial x^1} &\cdots& \frac{\partial f}{\partial x^n}\\ \frac{\partial g}{\partial x^1} &\cdots& \frac{\partial g}{\partial x^n} \end{bmatrix}$$ this is the condition of all the $$2\times 2$$ minors vanishing, which is equivalent to $$df$$ and $$dg$$ being scalar multiples of eachother.

The previous answers have shown that when $$df\wedge dg\equiv0$$ then there exists a function $$h$$ such that $$df=h\,dg\,.$$ If $$h$$ is nowhere vanishing then this has an interesting consequence in $$\mathbb R^2\,$$:

A vector field $$X$$ annihilates $$df$$ if and only if it annihilates $$dg\,.$$ In $$\mathbb R^2$$ a vector field $$X$$ that annihilates $$df$$ is tangent to the level sets of $$f$$ (same holds for $$g$$) and the integral curves of $$X$$ are the level sets. Therefore, $$f$$ and $$g$$ have the same level sets and it follows that

• $$f$$ depends only on $$g$$ and vice versa.

More formally: there exists a differentiable function $$F$$ on $$\mathbb R$$ such that $$f=F(g)\,.$$
From $$df=F'(g)\,dg$$ it follows that $$h=F'(g)\,.$$

• In $$\mathbb R^n$$ the same is true since the gradients of $$f$$ and $$g$$ are nowhere vanishing. This answer shows that then the level sets of $$f$$ and $$g$$ agree.