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In section 3-3 of do Carmo's Differential Geometry of Curves and Surfaces the author proves the following (formula (3) on page 155).

Let $\bf x$ be a parametrization of a surface embedded in $\mathbb R ^3$. Let $N=\frac{\mathbf x_u \wedge \mathbf x_v}{\|\mathbf x_u \wedge \mathbf x_v\|}$ and let $A$ be the matrix representing the shape operator endomorphism $\mathrm d_pN$ w.r.t the basis provided by $\bf x$. Furthermore, let $\mathrm I,\mathrm I\!\mathrm I$ respectively denote the matrices representing the first and second fundamental forms w.r.t the same basis. Then $-\mathrm I\!\mathrm I=\mathrm I A$.

The proof seems to just be about rearrangement, but I feel I'm missing geometric content.

Is there a conceptual explanation for this formula? Is there a conceptual proof for a general embedded oriented hypersurface in Euclidean space?

Added. Just to be clear, the rigorous conceptual definition of the Gauss map of an embedded oriented hypersurface $\iota:H\rightarrowtail V$ in an inner product space $V$ that I have in mind is the composite $$H\overset{\mathbf n}{\longrightarrow}(\mathrm TH)^\perp \longrightarrow\iota^\ast (\mathrm TV)\overset{\iota ^\ast \jmath}{\longrightarrow} H\times V\overset{\pi _2}{\longrightarrow}V$$ where $\bf n$ is the normal vector field and $\jmath$ is the canonical trivialization of the tangent bundle of $V$ itself.

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  • $\begingroup$ I'm sure I'm missing the intent of your question, but it's just the usual correspondence between a bilinear form and a linear map on an inner-product space. $\endgroup$ – Ted Shifrin Mar 11 '17 at 19:58
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Well...sort of.

First of all, the $ij$ entry of $II$ just tells you "as I move in the $x_i$ direction, how does $x_j$'s projection onto $N$ change?"

By contrast, $d_pN$ tells you how $N$ changes as you move in some direction $u$; dotting this change with some other tangent direction $v$ gives you a number. Picking $u$ and $v$ to be the $x_1$ and $x_2$ directions (or $x_u$ and $x_v$, as doCarmo would have it...) gives you the four entries of your matrix. The only problem is the sign: why is there that negative?

As I recall, it's roughly this: if, as you move along some path observing two vectors in a plane, vector $p$ rotates clockwise towards vector $q$, then if you observed things in a coordinate system consiting of $p$ and $p^\perp$, you'd see $q$ moving counterclockwise towards $p$.

And why is there that factor of $I$ in there? Well, if you move faster in the $x_1$ direction, the rate at which $N$ turns will be greater. So you need to scale down by the rate at which the position changes as a function of the coordinates...and that "rate" is precisely what the first fundamental form encodes.

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