Why is $\phi^* g = g$ a PDE for a pseudo-Riemannian metric $g$ on a manifold?

Question

Given a (locally trivial) bundle $\pi: E \to M$ a PDE of order $k$ is usually defined to be a submanifold of the jet-bundel $J^k(E)$.

Now assume $E = M \times M$ and $\pi$ is the projection on the first factor. Thus $J^k(E) = J^k(M,M)$ and local sections of $\pi$ are nothing else than smooth maps $\phi: U \to M$ defined on an open subset of $U \subset M$. Furthermore assume, we have a given metric $g$ on $M$. Now we are interested in all local diffeomorphisms (that is all diffeomorphisms between open subsets of $M$), that act isometrically. In other words: We are looking for local diffeomorphisms $\phi$ for which $\phi^*g = g$ holds. Now everybody (except me!) seems to be sure, that this relation definies a PDE of order 1.

Hence my question is: Why is the set $S= \lbrace j^1_p f \in J^1(M,M), (f^*g -g)(p)=0 \rbrace$ a submanifold of $J^1(M,M)$?

This cleary links to my former question: Why is a differential equation a submanifold of a jet bundle? (this one beeing an example of the other one): If $\rho: H \to M$ is the bundle of (2,0)-tensors over $M$, we can define a map $D_f: \Gamma_{loc}(\pi) \to \Gamma_{loc}(\rho)$, which takes a local section $\phi$ of $\pi$ to the section $p \to (\phi^*g -g)(p)$ of $\rho$. It is not hard to see, that there is a bundle morphism $f: E=M\times M \to H$ for which $D_f (\phi)(p)=f(j^1_p \phi) $ holds for everey local section $\phi \in \Gamma_{loc}(\pi)$, namly $f(j^k_p \phi)(X,Y)=g_{\phi(p)}(d\phi_p X, d\phi_p Y)-g_p(X,Y)$. Obviously $S$ ist the preimage of the zero-section in $\rho$ under $f$. But as Anthony Carapetis pointed out in his answer, we need to require something about the rank of the map $f$ for the preimage of a section of $\rho$ to be a submanifold. (Anthony Carapetis asked surjectivity of $Df$ restricted to the vertical bunde, but actually I think constant rank should be enough). Anyhow, I can't figure out, why such a condittion should hold.

All of the manifolds and maps in this post are $C^{\infty}$.

I'm gratefull for any hint.

Answer 1

You're right that the condition you need to check is that $Df$ has constant rank when restricted to the vertical bundle. I'll show that it is in fact surjective when we restrict the codomain to be the bundle $\mathfrak S = \operatorname{Sym}^2(T^*M)$ of symmetric 2-tensors.

Given an arbitrary $j^1 \phi_0 \in S_p$, $h \in \mathfrak S_p$, it suffices to find a family $\phi(t)$ of local diffeomorphisms such that $\phi(0) = \phi_0$ and $$\frac d{dt}\Big|_{t=0} f(j^1 \phi(t)_p)=h.$$ Substituting in the expression for $f$, the LHS becomes $$\frac d{dt}\Big|_{t=0}\phi^* g_p.$$

When $\phi_0$ is the identity and $\phi(t)$ is the flow of a vector field $X$, this is $\mathcal L_X g_p$. A standard computation shows that this Lie derivative of the metric is twice the symmetrization of $\nabla X$. If we choose $X$ to be a linear function in normal coordinates about $p$, then $\nabla X_p$ has the same matrix, and thus choosing this to be the symmetric matrix $h$ we get the desired result.

When $\phi_0$ is some other isometry, the same argument applies after conjugating by $\phi_0$.