The manifold $S$ is seen as the image of the following embedding $\varphi:w\in W\mapsto\big(h(w),w\big)\in\mathbb{R}^{n_v +n_w}$ where $W$ is a domain of $\mathbb{R}^{n_w}$ and $h:\mathbb{R}^{n_w}\rightarrow\mathbb{R}^{n_v}$ is smooth. I set $H(w)\in\mathcal{M}_{n_v,n_w}(\mathbb{R})$ the jacobian matrix of $h$ evaluated on $w$.
How can I prove that $T(w)=I_{n_w}+(H(w)^T)H(w)$ is the riemannian metric of $S$ ?
Thank you.
You have to compute the (Euclidean) inner products between the partial derivatives of $\varphi$. Using total derivatives one can do this more easily. Since the total derivative is given by $D\varphi(w) = (Dh(w), {\rm Id})$. So, given $v_1,v_2 \in \Bbb R^{n_w}$, we'll have $$\langle D\varphi(w)v_1, D\varphi(w)v_2\rangle_{\Bbb R^{n_w+n_v}} = \langle Dh(w)v_1,Dh(w)v_2\rangle_{\Bbb R^{n_v}} + \langle v_1,v_2\rangle_{\Bbb R^{n_w}}.$$This is the definition of the Euclidean inner product in $\Bbb R^{n_v+n_w}$: it turns out to be the direct sum of the inner products in $\Bbb R^{n_v}$ and $\Bbb R^{n_w}$. In matrix form, regarding $v_1$ and $v_2$ as column vectors, the right side is $$(Dh(w)v_1)^\top(Dh(w)v_2) + v_1^\top v_2 = v_1^\top(Dh(w)^\top Dh(w) + {\rm Id}_{n_w})v_2,$$as required. (In your notation, $H(w) = Dh(w)$)