$\mathrm{grad}\ f|_p$ is normal to the level set of $f$ through $p$ if $p$ is a regular point of a function $f$.

Question

Please allow me to write with reference to John Lee's Introduction to Smooth Manifolds. Thank you.

Let $(M,g)$ be a Riemannian manifold and let $f$ be a smooth function on $M$ with a regular point $p$. I'd like to show that $\mathrm{grad}\ f|_p$ is normal to the level set of $f$ through $p$. Below is what I have tried so far:

Let $f(p)=c$. If I understand correctly, $f^{-1}(c)$ is the level set of $f$ through $p$. Thus, I will have to show that for each $x\in f^{-1}(c)$ and for each $u\in T_x M$, $$\langle u,\mathrm{grad}\ f|_p\rangle_g=0.\tag{1}$$ From now on, I will always denote local coordinates by $(x^i)$, meaning that I will write $$u=u^k\frac{\partial}{\partial x^k}|_x\tag{2}$$ and $$\mathrm{grad}\ f|_p=g^{ij}(p)\frac{\partial\hat{f}}{\partial x^i}(p)\frac{\partial}{\partial x^j}|_p\tag{3}$$ with $\hat{f}$ denoting the coordinate representation of $f$ in $(x^i)$. Plugging (2) and (3) into (1), I get $$\langle u,\mathrm{grad}\ f|_p\rangle_g=u^k g^{ij}(p)\frac{\partial\hat{f}}{\partial x^i}(p)\langle\frac{\partial}{\partial x^k}|_x,\frac{\partial}{\partial x^j}|_p\rangle_g,$$ which seems to lead me nowhere. I'm aware that I haven't appealed to the definition of a regular point, but it seems to make no difference after I try it. What I will get is merely a surjective map $df_p$, which can be locally expressed as $$df_p=\frac{\partial\hat{f}}{\partial x^i}(p)dx^i|_p.\tag{4}$$ What should I do next? Thank you.

Answer 1

The question asks you to show

If $p$ is a regular point of $f$, then $\mathrm{grad}f|_p$ is orthogonal to $\Sigma$, the level set of $f$ through $p$.

Watch out, "$\mathrm{grad}f|_p$ is orthogonal to $\Sigma$, the level set through $p$" does not mean, as you seem to believe, the following:

$\forall x \in \Sigma, \forall u \in T_x\Sigma,\quad \langle \mathrm{grad}f|_p,u\rangle = 0$.

There are two reasons why this does not even make sense:

• First, while $p$ is by assumption a regular point of $f$ and therefore $\Sigma$ is a submanifold in a neighbourhood of $p$, there is no reason for $\Sigma$ to be regular away from $p$, and $T_x\Sigma$ does not even make sense there.
• Second, $\mathrm{grad}f|_p$ is a vector tangent at $p$, so it can only be compared to other vectors tangent at $p$. If $u$ is tangent at any other point than $p$, its scalar product with $\mathrm{grad}f|_p$ has no meaning. For instance, you wrote $$ \left\langle\left.\frac{\partial}{\partial x^k}\right|_x,\left.\frac{\partial}{\partial x^j}\right|_p\right\rangle_g $$ which makes absolutely no sense at all. It is a good and important exercise to convince yourself about that precise point.

What it really means is:

$\forall u \in T_p\Sigma, \quad \langle \mathrm{grad}f|_p,u\rangle = 0$.

Below is a proof of this correct interpretation of the question.

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You do not need to use any computation, only the definition of the gradient of a function and the characterisation of tangent vectors as derivatives of curves.

Let $p\in M$ be a regular point for $f\colon (M,g)\to \Bbb R$, and let $\Sigma = f^{-1}(f(p))$ be the level set through $p$. Since $p\in \Sigma$ is a regular point, $\Sigma$ is a smooth hypersurface in a neighbourhood of $p$. Let $u\in T_p\Sigma$ be a tangent vector, and $\gamma\colon (-\varepsilon,\varepsilon)\to \Sigma$ a smooth curve such that $\gamma(0) = p$, $\gamma'(0) = u$. Then $$ \DeclareMathOperator{\grad}{grad} \newcommand{\d}{\mathrm{d}} \langle \grad\! f|_p,u\rangle = \d_pf(u) = \left.\frac{\d \left(f\circ\gamma(t)\right)} {\d t}\right|_{t=0}. $$ Since $\gamma$ has range in $\Sigma$ which is by definition the level set through $p$, $f\circ \gamma\colon (-\varepsilon,\varepsilon) \to \Bbb R$ is constant, equal to $f(p) \in \Bbb R$. Consequently, one has $\langle \grad\! f|_p,u\rangle = 0$. This being true for any $u\in T_p\Sigma$, one concludes that $\grad f|_p \perp T_p\Sigma$.