How to reconcile two approaches to the Gaussian curvature

Question

Gaussian curvature can be found as the ratio of determinants of the second to the first fundamental forms (from here):

$$K =\frac{\det \mathrm {II}}{\det \mathrm I}=\det \left(\mathrm {I}^{-1}\rm {II}\right)=k_1({\rm I}^{-1} {\rm {II}})k_2(\rm {I}^{-1} \rm {II})\tag 1$$

where $k_1$ and $k_2$ are the eigenvalues of $\mathrm {II}^{-1}\rm I$ or principal curvatures.

In matrix notation

$$\rm I = \begin{bmatrix} E & F \\ F & G\end{bmatrix}=\begin{bmatrix}\langle r_u,r_u\rangle & \langle r_u,r_v\rangle \\ \langle r_u,r_v\rangle & \langle r_v,r_v\rangle\end{bmatrix}$$

and

$$\rm II = \begin{bmatrix} e & f \\ f & g\end{bmatrix}=\begin{bmatrix}\langle N,r_{uu}\rangle & \langle N, r_{uv}\rangle \\ \langle N, r_{uv}\rangle & \langle N,r_{vv}\rangle\end{bmatrix}$$

In tensor calculus the Gaussian curvature is

$$K= \det B^\alpha_\beta$$

with $B^\alpha_\beta$ being the covariant-contravariant form of the curvature tensor $B_{\alpha\beta}$

$$B_{\alpha\beta}=\vec r_\alpha \cdot \frac{\partial N}{\partial r_\beta}$$

and

$$B^\alpha_\beta=\rm I^\alpha_\gamma B_{\gamma\beta} \tag 2$$

where $I^\alpha_\gamma$ is the tensor notation for the first fundamental form.

How can I see that equations 1 and 2 are the same?

as explained by Pavel Grinfeld in here.

Answer 1

There are various mistakes here. $B_{\alpha\beta}$ is the negative of the usual second fundamental form. For example, differentiating $\langle N,r_u\rangle = 0$ yields $$\langle N_v,r_u\rangle = - \langle N,r_{uv} \rangle = -f.$$ $B^\alpha_\beta$ is the (negative of the) usual shape operator, the derivative of the normal (Gauss) mapping as a linear mapping from the tangent plane of the surface to itself.

We raise indices as is customary in tensor analysis by multiplying by the inverse of the first fundamental form. This is usually written classically as $(g^{\alpha\beta}) = (g_{\alpha\beta})^{-1}$. Indeed, to get the summation convention to work out, you need to have $$B^\alpha_\beta = I^{\alpha\gamma}B_{\gamma\beta},$$ and this is precisely the linear map $\text I^{-1}\text{II}$ given above, except it's off by a sign. The equation you've transcribed has indices making no sense whatsoever.

Answer 2

I believe the best answer to this question has been given Tristan Needham's Visual Differential Geometry

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Context:

Page-150

In the above, when we actually move an infinitesimal step $\epsilon \hat{v}$ in space, we generally fall off the surface (q). So, the naive idea of $ \lim_{\epsilon \to 0} \frac{n(q) - n(p)}{\epsilon}$ is senseless.

What we can do thought is to drop a perpendicular from $q$ to S, this gives us a new point $r$. Now, we can actually show that $pr$ is a geodesic segment of length $\epsilon$ and define $\delta n = n(p) - n(r)$. Furthermore, we can see this difference exists in $T_p M$, have a look at this picture:

If you find the above difficult to understand (how perpendicularity comes), think back of the velocity on a circle, and how the derivative of a radial position vector in circular motion gives a tangential velocity vector.

In 3D space, the tensor expression you wrote is mostly clear, but a question comes how are they related to the so called fundamental forms. The shape operator $S(v)$ is a linear transformation living in the tangent space telling us how displacement in certain direction are transformed to changes in the normal. The Fundamental forms, once the shape operator is defined have the following exps:

1. First Fundamental form $I(u,v) = u \cdot v$
2. Second Fundamental form $II(u,v) = S(u) \cdot v$
3. Third fundamental form $III(u,v)= S(u) \cdot S(v)$

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What the shape operator evaluated at a vector actually tells you:

The Geometric intuition is that, when we move in the direction of $v$, we can can think of running along a curve from $p$ to $r$, now this curve would have a curvature, and we are getting the curvature of this curve when we evaluate the shape operator in a certain direction. Btw curvature, here I mean tangent vector having magnitude of curvature.(Sec 15.6,page-159)

There is a theorem by Euler which shows that the curvature as said above is maximum in two specific directions(page-111). Now it turns out the Gaussian Curvature is product of these two max curvature. (page-134)

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The direct answer to the question using all the above:

Ok. Sorry I missed the point of the question in the beginning, sort of got distracted in my Love for Needham's book that I forgot I was answering a question. Anyways, with the above definition , let us actually find the matrix entries of first fundamental form and second with respect to an orthonormal basis $\{e_i, e_j \}$ pointing in the two tangent vectors be one in the principle direction. We have,

$$ II_{ij} = I(e_i, e_j) = S(e_i ) \cdot e_j$$

Now as I told before, the shape operator is diagonal in the basis I choose. So, this $II_{ij}$ is zero if $ i \neq j$, we have two terms $II_{11}$ and $II_{22}$, we find:

$$II_{11} = -\kappa_1$$

and, $$II_{22} = -\kappa_2$$

that is, the component of the second fundamental form are the principle curvatures. Now, let us see what happens with the first fundamental form / metric tensor. I have:

$$ g(e_i, e_j) = g_{ij}$$

Now, since principle direction are orthogonal , metric tensor evaluate two would give 0 unless $i = j$, but what if $i=j$? Well since I choose orthonormal basis, I have it as $g_{11}=g_{22}=1$.

Clearly, determinant of metric tensor/first fundamental form is one and that of second fundamental form is the gaussian curvature. The ratio is again gaussian curvature because the most famous rule that anything divided by 1 is the same thing.

Answer 3

Succinct answer:

We have ,

$$ K = \det ( \text{I}^{-1} \text{II})$$

What is entries of $\text{II}$? That is just your $B_{\alpha \beta}$(*), and what is entries of $\text{I}^{-1}$? that is just fancy notation for the contravariant metric tensor. In indical form, the matrix entry of product of inverse of first fundamental form times the second is given as:

$$ (\text{I}^{-1} \text{II})_{ij} =g^{i \alpha} B_{\alpha j} = B_j^{i} $$

Now, clearly

$$ K = \det(B_j^i)$$

And we are done.

*: Eqtn 11.16, Page-193, Pavel Grinfeld's Tensor Calculus book