Here is a proof says that the differential of Gauss map is self-adjoint. But I seems there is an abuse of notation at (1) in it.

Since $dN_p$ is linear, it suffices to verify that $\langle dN_p(w_1), w_2 \rangle = \langle w_1, dN_p(w_2)\rangle$ for a basis ${w_1, w_2}$ of $T_p(S)$. Let $x(u, v)$ be a parametrization of $S$ at $p$ and ${x_u, x_v}$ the associated basis of $T_p(S)$. If $\alpha(t) = x(u(t), v(t))$ is a parametrized curve in $S$, with $\alpha(0) = p$, we have $$\begin{align} dN_p(\alpha'(0)) &= dN_p(x_uu'(0) + x_vv'(0)) \\ &= \frac d{dt}N(u(t),v(t))\mid_{t=0} & (1)\\ &= N_uu'(0) + N_vv'(0) \end{align}$$

I think it should rewrite as: $$\begin{align} dN_p(\alpha'(0)) &= dN_p(x_uu'(0) + x_vv'(0)) \\ &= dN_p(u(t),v(t))\mid_{t=0} & (2)\\ &= \frac d{dt} N(x(u(t),v(t)))\mid_{t=0} &(3)\\ &= N_uu'(0) + N_vv'(0) \end{align}$$

Reference : Differential Geometry of Curves and Surfaces Manfredo P. do carmo
Proposition 1.
I care this insomuch we have: $N:S\to S^2$ and $dN_p:T_p(S)\to T_p(S)$


Your second line $dN_p\bigl(u(t), v(t)\bigr)|_{t=0}$, should be $dN_p\bigl(x(u(t), v(t))\bigr)|_{t=0}$ (since as you note the Gauss map $N$ is defined on the surface $S$, not on the domain of the parametrization $x$), but your change to (1) looks right.

There still appear to be abuses of notation in the proposed calculation, particularly writing $N_u$ for the partial derivative $(N \circ x)_u$ (and similarly for $N_v$) in the last step. :)

| cite | improve this answer | |
  • $\begingroup$ $dN_p\bigl(u'(t), v'(t)\bigr)|_{t=0}$ Is right because $dN_p$ is defined over $T_p(s)$ $\endgroup$ – Hoseyn Heydari Jan 18 '14 at 13:47
  • $\begingroup$ Regarding (2), if I've understood your notation then $(u, v)$ takes values in the domain of the parametrization $x$ (not in $S$ itself), so $dN_p\bigl(u(t), v(t)\bigr)$ doesn't make sense (unless $N$ stands for $N \circ x$ in (2))...? $\endgroup$ – Andrew D. Hwang Jan 18 '14 at 16:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.