Function and its exterior derivative

Question

Is there an example of a function $f:M\to \mathbb R$, where $M$ is a differentiable manifold, such that $f$ is constant on the hypersurface $\Sigma$ and its exterior derivative $df\neq 0$ on $\Sigma$?

My intuition of the exterior derivative is simply how $f$ changes in any arbitrary direction. Is this not right?

---

I ask this due to a proof in these notes on page 125 for the Lemma $K_{ab}=K_{ba}$. The first line of the proof suggests that such an example exists? Or perhaps I have misunderstood something?

Answer 1

No. (and Yes)

When first starting differentiable topology I suggest trying to use coordinates to try to answer questions like this. Sometimes the answer is less elegant, but it allows you to use your intuition from multivariable calc more effectively.

Now on to the question. There are two ways to interpret this. The first is that if $f|_\Sigma=$constant then $df|_{T\Sigma}$ can be nonzero. For this interpretation the answer is NO. The reasoning is below.

You should be able to locally view the hypersurfaces as a codimension 1 foliation. Then you can find a local coordinate system $(x_1, ..., x_n)$ where the coordinates $(x_1, ..., x_{n-1})$ are the coordinates of the hypersurface. If you write $df$ in these coordinates then it will just be the matrix of the partial derivatives. Since $f$ is constant on the first coordinates then $df=0$ on these coordinates (ie. the hypersurface).

The second is that if $f$ is constant on $\Sigma$ then can $df$ on all of $TM$ but $\textit{at a point in}$ $\Sigma$ be nonzero, and the point there is Yes. This is what happens in the link in the OP.

For example consider the function $\mathbb{R}^2\rightarrow\mathbb{R}$ given by $f(x, y)=x$. Letting the $y$-axis be $\Sigma$ we get exactly the situation above.

Answer 2

If $f$ is constant on the surface then $df$ is automatically zero on the surface. Because if $v\in T_p(\Sigma)$ them $df(v)=v(f)$. Now consider a curve $\gamma :(-1,1)\rightarrow \Sigma $ such that $\gamma(0)=p$ and $\gamma'(0)=v.$ Then $v(f)=\frac{d}{dt}f\circ \gamma(t)$ at $t=0.$ Now $f$ is conatant on $\Sigma$ hence $v(f)=\frac{d}{dt}f\circ \gamma(t)=0$ this is zero.