0
$\begingroup$

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.

$\endgroup$
1
  • 1
    $\begingroup$ (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... $\endgroup$ Commented Apr 5, 2023 at 22:32

3 Answers 3

3
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ Thank you, but how do you arrive to the fact that quantity vanishes? Split the sum into two to get $\sum_{j<k}\frac{\partial f}{\partial x^j}\frac{\partial g}{\partial x^k}dx^j \wedge dx^k + \sum_{k<j}\frac{\partial f}{\partial x^j}\frac{\partial g}{\partial x^k}dx^j \wedge dx^k=\sum_{j<k}\frac{\partial f}{\partial x^j}\frac{\partial g}{\partial x^k}dx^j \wedge dx^k - \sum_{k<j}\frac{\partial f}{\partial x^j}\frac{\partial g}{\partial x^k}dx^k \wedge dx^j$? $\endgroup$ Commented Apr 5, 2023 at 22:42
  • $\begingroup$ Could do that way. In your second sum, you could rename $j\mapsto k$ and $k \mapsto j$... $\endgroup$
    – Kenny Wong
    Commented Apr 5, 2023 at 22:48
  • $\begingroup$ Thank you so much, makes sense now! $\endgroup$ Commented Apr 5, 2023 at 23:07
2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .