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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 $dN_p:T_p(S)\to T_p(S)$

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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. :)

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  • $\begingroup$ $dN_p\bigl(u'(t), v'(t)\bigr)|_{t=0}$ 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

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