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!
Best Answer
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.