# Why can we think of the second fundamental form as a Hessian matrix?

Let $f: U \rightarrow \mathbb{R}^3$ be an immersion that parametrizes a piece of a surface, and let $(h_{ij})$ be the matrix for the second fundamental form of that surface.

According to pg. 70 of the text Differential Geometry by Wolfgang Kuhnel, we can think of the $(h_{ij})$ as "the Hessian of matrix of a function $h$, which represents the surface as a graph over its tangent plane".

I have a "heuristic" understanding of what's going on, but I'd like to be a bit more careful about this. What exactly is the function $h$? Can we write it down explicitly (perhaps in terms of the parametrization $f$, the unit normal $\nu$, and their derivatives), so that we can directly check that its Hessian is indeed the second fundamental form $(h_{ij})$?

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It can't be quite literally true -- consider for example $z=x^2-y^2$. Aside from normalization issues you've also got the question of how you interpret these objects -- the Hessian is a bilinear form, and the 2nd fundamental form is really a linear map $T_pM \to T_pM$. You can convert one into the other using the identification of a vector space with its dual coming from the inner product and I think after that identification the two become the negatives of each other -- but only at the point of tangency. –  Ryan Budney Feb 27 '11 at 21:13
I think the interpretation is that their matrices are the same, i.e. we think of the matrix element $(h_{ij})$ of the fundamental form as the second derivative $\partial^2 h / \partial u_i \partial u_j$ of some function $h$. Also, I'm not sure if there are different ways to define the second fundamental form, but in Kuhnel's text, it's defined as a bilinear form on the tangent space: $II(X, Y) = \langle LX, Y\rangle$ where $L: T_pM \rightarrow T_pM$ is the shape operator. –  Elliott Feb 27 '11 at 21:35
Ah, okay, I was referring to the "shape operator" rather than the "2nd fundamental form". In the example I give, the two matrices are additive inverses of each other, and a quick pencil-sketch argument tells me it's that way always. Just write down a surface that's the graph of a function and compute away. –  Ryan Budney Feb 27 '11 at 22:26
Oh thanks, that makes sense: when we have a parametrization of the form $f(u, v) = (u, v, h(u, v))$, the Hessian evaluated at h(0, 0) is precisely the second fundamental form at f(0, 0). I'm not sure how to generalize this to an arbitrary point of an arbitrary parametrization, but that's my own fault for not knowing the Implicit Function Theorem... –  Elliott Feb 28 '11 at 0:24
I don't think Kuhnel is talking about arbitrary parametrizations -- he says specifically in your quote that he's restricting to the situation as in your example $f(u,v)=(u,v,h(u,v))$ since up to a rotation of space, that's exactly what's going on when you express the surface as the graph of a function over the surface's tangent space (at a point). –  Ryan Budney Feb 28 '11 at 1:38

Thanks Ryan! So in conclusion, if we choose coordinates such that the surface is locally described by $f(u,v) = (u,v,h(u,v))$, then the second fundamental form at the point $f(0,0)$ is precisely the Hessian of $h$ at $(0,0)$.