Show that the gradient is normal everywhere on a regular level set

differential-geometrygeometryriemannian-geometrysemi-riemannian-geometry

I'm trying to solve Problem 2.9 in John Lee's book Introduction to Riemannian Manifolds. It says

Suppose that $(M,g)$ is a Riemannian manifold, $f\in C^{\infty}(M)$, and $\mathscr{R}\subseteq M$ is the set of regular points of $f$. Then for each $c\in \mathbb{R}$, the set $M_c=f^{-1}(c)\cap \mathscr{R}$ (if nonempty) is an embedded smooth hypersurface in $M$, and $\text{grad}(f)$ is everywhere normal to $M_c$.

I'm pretty sure that $M_c$ being an embedded hypersurface is from the Regular Level Set Theorem (Corollary A.26), but I have two questions at the moment.

  1. I'm confused on how to show that the gradient is everywhere normal to $M_c$. I guess the idea would be to show that given $p\in M_c$, we have $\langle \text{grad}(f), w \rangle_g=0$ for every $w\in T_pM_c$, but I'm a little lost on how to do this rigorously. How can you show this properly?

  2. Does this same result hold for pseudo-Riemannian metrics? I'm pretty sure it does, but I'm still unsure if this uses positive/negative definiteness somewhere along the way.

I'd appreciate thoughtful/detailed answers to these questions, thanks so much for the help!

Best Answer

First, ignore the Riemannian setting. One of the conclusions of the regular value theorem is that the tangent space to the submanifold equals the kernel of the derivative of the map whose level set you’re taking. Lee for sure talks about this (look in the sections where he talks about tangent spaces of submanifolds)

Regular value theorem.

Let $M,N$ be smooth manifolds of dimension $m,n$ respectively, $\Phi:M\to N$ a smooth map and $y\in N$ a regular value of $\Phi$. Then, the level set $M_{\Phi,y}:=\Phi^{-1}(\{y\})$ is a smooth properly embedded submanifold of $M$ of codimension $n$ (i.e of dimension $m-n$). Furthermore, for each point $p\in M_{\Phi,y}$, we have that $T_p(M_{\Phi,y})=\ker (T\Phi_p)$ where $T\Phi_p:T_pM\to T_{\Phi(p)}N$ is the tangent map.

(There is also a very useful generalization using transversality where you take the preimage of an entire submanifold, not just a single point). The first part I’m guessing you’re fine with (it’s the usual proof via the inverse/implicit function theorem). The second statement about tangent spaces is also the usual calculus argument, but here’s a reminder of that argument:

Let us first consider a tangent vector of the submanifold $v\in T_p(M_{\phi,y})$. One of the equivalent ways of defining tangent vectors is as equivalence classes of curves $v=[\gamma]$ (where this is the ‘formal way’ of trying to encode our usual idea of tangent vectors $\gamma’(0)$). Since $v$ is tangent to the submanifold, by definition this means the curve $\gamma$ has image lying entirely in $M_{\phi,y}$, i.e for each parameter $t$, we have $\gamma(t)\in M_{\phi,y}$, which thus implies $\Phi(\gamma(t))=y$, i.e $\Phi\circ\gamma$ is a constant curve. Thus, its tangent vector is the zero vector in $T_{\Phi(p)}N$. But now recall that (by definition or by chain rule, depending on how you develop the theory) we have $0=[\Phi\circ\gamma]=T\Phi_p([\gamma])=T\Phi_p(v)$, i.e $v$ belongs to the kernel (if this notation is confusing, perhaps the following is more reminiscent: $(\Phi\circ\gamma)’(0)=d\Phi_{\gamma(0)}(\gamma’(0))=d\Phi_p(\gamma’(0))$, and since the LHS vanishes, it follows $\gamma’(0)$ lies in the kernel of $d\Phi_p$, as claimed).

So, we have shown $T_p(M_{\Phi,y})\subset \ker(T\Phi_p)$. Next, on the one hand, we know that $M_{\phi,y}$ has dimension $m-n$. On the other hand, since $\Phi$ has full rank at $p$, the rank-nullity theorem tells us its kernel has dimension $m-n$, so these subspaces are actually equal.

Now how does this help us in the semi-Riemannian setting? Well, we consider a smooth function $f:M\to\Bbb{R}$ (i.e $N=\Bbb{R}$ and our $\Phi$ is now simply denoted $f$ because of tradition) which has $y\in\Bbb{R}$ as a regular value. In this case since we have a canonical isomorphism $T_{f(p)}\Bbb{R}\cong \Bbb{R}$, we can think of $Tf_p:T_pM\to T_{f(p)}\Bbb{R}$ as giving rise to a linear map $df_p:T_pM\to\Bbb{R}$ (this is one way of thinking about the exterior derivative of a function). Now suppose further that $g$ is a semi-Riemannian metric on $M$. Recall that the definition of the gradient vector is that it is the metrically-related vector to the covector $df_p$, i.e \begin{align} (\text{grad}_g f)(p):= (df_p)^{\sharp} \end{align} i.e $(\text{grad}_g f)(p)\in T_pM$ is the unique vector such that for all $v\in T_pM$, we have \begin{align} g\left((\text{grad}_g f)(p), v\right)&=df_p(v).\tag{$*$} \end{align} Note carefully that this only relies on non-degeneracy of the bilinear map $g$ (see here for a more detailed explanation of the musical isomorphisms). Positive-definiteness is not needed. In fact, you can also work with skew-symmetric forms (e.g symplectic forms) and consider the symplectic gradient and prove a similar result. So, putting everything together, we have \begin{align} T_p(M_{f,y})&=\ker(df_p)\\ &:=\{v\in T_pM\,:\, df_p(v)=0\}\\ &:=\{v\in T_pM\,:\, g\left((\text{grad}_g f)(p), v\right)=0\}\\ &:=\{v\in T_pM\,:\, \text{$v$ is $g$-orthogonal to $(\text{grad}_g f)(p) $}\} \end{align} Notice that the first equal sign is the only one with true content (and it’s a special case of the one we proved above). The second equal sign is definition of kernel, the third is by definition of gradient vector $(*)$, and the last is a rephrasing. You can now take the $g$-orthogonal complement of both sides to say that $(\text{grad}_gf)(p)$ spans the $g$-orthogonal complement of $T_p(M_{f,y})$.

So, in some sense, you can say that the definition of the gradient vector has been designed to make this claim true (note that the gradient vector is a different beast from the covector $df_p$).

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