Prove that the pullback $F^* g$ of a Riemannian metric $g$ is a Riemannian metric iff $F$ is a smooth immersion

Question

So I'm trying to prove that $F^*g$, where $F^*$ is the pullback of $F: M \rightarrow N$, and $g$ is a Riemannian metric on $N$ is a Riemannian metric on $F$ is and only if $F$ is a smooth immersion. Among trying to prove this exercise (Page 329 Lee Introduction to smooth manifolds second edition) I realized that my understanding of differentials and pullbacks is a bit flawed.

So, my understanding of $dF_p: T_pM \rightarrow T_{F(p)}N$ is that given $v \in T_pM$ and $f \in C^\infty(N)$ we have that $dF_p(v)(f)=v(f \circ F)$. In otherwords, it seems that the way $dF$ pushes forward the vectors $v \in T_pM$ is by composing it with the function that is defined on $N$.

However, with a Riemannian metric, the vectors that we use as arguements arn't really seen as differential opertors, we simply have two vectors that we plug into the metric to get a real number.

Okay, so let me try to define $F^*g$. Let $v_1,v_2 \in T_pM$.

Then $F^*g(v_1,v_2)=g(dF_p(v_1),dF_p(v_2))$.

I guess this is natural, since $dF_p$ is a linear map, so $(dF_p(v_1),dF_p(v_2))$ is a vector in $T_{F(p)}N$... But I guess i'm used to understanding them by viewing $dF_p(v_i)$ as a differential operator...

Anyway, I hope I'm not talking in circles. If anyone can lend me any insight what-so-ever into this situation or with the exercise I'm trying to work through (the first sentence of this post) I'd greatly appreciate it. Thanks!

Answer 1

Everything you wrote is correct (although you might want to write $$(F^*g)_p(v_1,v_2) = g_{F(p)}(dF_p(v_1),dF_p(v_2)),$$ in order to emphasise at which point the metric and its pullback are evaluated). Any vector is essentially a differential operator, so I guess I'm not really sure why the fact that $dF_p(v_i)$ is a differential operator is so problematic for you.

As for the exercise, you need to show that $F^*g$ is a metric, i.e., that it is symmetric, bilinear, and positive-definite. The first two properties follow easily from the corresponding properties of $g$. The last property is the one that requires $F$ to be an immersion, meaning that the derivative $dF$ is injective. Without injectivity of $dF$, $F^*g$ will still be positive, but will fail to be definite: if $v$ is a non-zero element of $\ker dF_p$, then $(F^*g)_p(v,v) = 0$ but $v\ne 0$. Injectivity of $dF$ eliminates this possibility.