My question is simple, I just want to compute the differential of the map
$$
N(p) = \frac{Bp + b}{|Bp + b|}
$$
where $B$ is a non-null symmetric matrix of order three, $b \in \Bbb{R}^3$ and $p \in S$ where $S$ is the following surface (a quadric):
$$
S = \{p \in \Bbb{R}^3 \ | \ \langle Bp, p \rangle + 2\langle b, p \rangle + c = 0\},
$$
with $c \in \Bbb{R}^3$.
The context is the following: the question asks to show that $N(p)$ is a Gauss map on $S$ and to compute the corresponding second fundamental form. I am stuck in this second part. The solution is $$ \sigma_p = - \frac{1}{|Bp + b|} \langle Bv, v \rangle \quad p \in S, v \in T_pS $$
EDIT:
Following the hint by Jean Marie:
The jacobian matrix of $q \mapsto \frac{q}{|q|}$ is given by $$1/|q|^3 \begin{bmatrix} q_2^2+q_3^2 & - q_1q_2 & - q_1q_3 \\ & q_1^2 + q_3^2 & -q_2q_3 \\ & & q_1^2+q_2^2 \end{bmatrix} $$ Then, if $N = f \circ g$ we have $(dN)_p(v) = df(g(p)) dg(p) (v) = \frac{1}{|Bv+b|^3} * [matrix] * Bv$. Is this correct? If so, how do I arrive at $$ \sigma_p(v, v) = - \frac{1}{|Bp + b|}\langle Bv, v \rangle $$ ?
Thanks in advance.
I managed this with the help of Renan Lima.
We use the definition to compute the differential of $N$ at $p \in S$. Let $\alpha: (- \varepsilon, \varepsilon) \longrightarrow S$ be a curve such that $\alpha(0) = p$ and $\alpha'(0) = v$, for $v \in T_pS$. Then $$ (dN)_p(v) = \frac{d}{dt}|_{t = 0} N(\alpha(t)) = \frac{Bv}{|Bp + b|} + \frac{\langle Bv, Bp + b \rangle}{|Bp + b|}(Bp + b). $$
Since we are interested only on the tangential component, that is $(dN)_p|_{T_{N(p)}\Bbb{S}^2}$, is suffices to take only the first term. This gives the expression we are seeking.
For more details about the computations, see this link.