Invariance of function under isometry of Riemannian manifolds

Question

Suppose that $(M,g)$ and $(N,g')$ are Riemannian manifolds and that $f: M \to N$ is an isometry. Now take smooth vector fields $X, Y, Z$ on $M$.

Is it true that $X\langle Y, Z\rangle_p = df_p(X)\langle(df_p(Y), df_p(Z)\rangle_{f(p)}$, where $p \in M$?

Answer 1

You may prove it in the following way:

(1) $X(f^*u)=df_p(X)(u)$ holds for all $u\in C^{\infty}(N)$.

(2) $\langle Y,Z\rangle_p=\langle df_p(Y),df_p(Z)\rangle_{f(p)}$ holds for all $p\in M$, since $f$ is an isometry.

(3) Let $w(p)=\langle Y, Z \rangle_p\in C^{\infty}(M)$, and let $q=w(p)$, define $u(q)=\langle df_{f^{-1}(q)}(Y),df_{f^{-1}(q)}(Z)\rangle_q$ be a $C^{\infty}$ function on $N$. Now we have that $$(f^*u)(p)=(u\circ f)(p)=u(f(p))=w(p)$$

Then apply this to (1), it gives you $X\langle Y,Z\rangle_{p}=df_p(X)\langle df_p(Y),df_p(Z)\rangle_{f(p)}$

Answer 2

If $\frac{d}{dt}p(t)=X,\ p(0)=p$ then $$ df_p X(df_p Y, df_p Z) =\frac{d}{dt} (df Y, df Z)_{f(p(t))} = \frac{d}{dt} (Y,Z)_{p(t)} $$ since $f$ is an isometry And $\frac{d}{dt} (Y,Z)_{p(t)}= X(Y,Z)$