Why an immersion induces a Riemannian metric.

Question

I want to prove the following fact: “Let $f:M^n\rightarrow N^{n+k}$ be an immersion. If $N$ has a Riemannian structure, then $f$ induces a Riemannian structure on $M$ by $\langle u,v\rangle_p=\langle df_p(u),df_p(v)\rangle_{f(p)}$.”

I could check all conditions except that if $X$ and $Y$ are vector fields in a neighborhood $V$ of $M$, why $\langle X,Y\rangle$ will be a smooth map on $V$. I went as far as $\langle X,Y\rangle(p)=\langle X(p),Y(p)\rangle_p=\langle df_p(X(p)),df_p(Y(p))\rangle_{f(p)}$. Why is the last expression smooth?

Note: Notations are from DoCarmo’s Riemannian Geometry book.

Edit: I posted a sketchy solution. Is it correct?

Answer 1

You can check in coordinates, or for a coordinates-free solution, note that we have three varying pieces on the RHS, namely $$ (df)_p(X(p)),\quad (df)_p(Y(p)),\quad \langle\cdot,\cdot\rangle_{f(p)} $$ and they are smooth sections of $f^*TN$, $f^*TN$ and $f^*{\large\otimes}^2T^*N$ (or if you want, the subbundle $f^*\operatorname{Sym}^2T^*N$) because $f,X,Y,\langle,\rangle^N$ are smooth. So $$ p\mapsto((df)_p(X(p));(df)_p(Y(p));\langle\cdot,\cdot\rangle_{f(p)})\in (f^*TN\times f^*TN\times{\large\otimes}^2f^*T^*N)_p $$ is smooth. Contracting, we have $$ p\mapsto\langle(df)_p(X(p)),(df)_p(Y(p))\rangle_{f(p)} $$ is smooth.

Answer 2

I think I could solve it. Let $x$ be a parametrization at $p$ and $y$ a parametrization at $f(p)$. So the local representation of $df_p(X(p))$ in the associated basis {$\partial/\partial{y_i}$} is $[dy_i/dx_j]_{i,j}(x_i(p))$. The rest follows since the inner product is always bilinear.