# How to reconcile two approaches to the Gaussian curvature

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.

• So your question is how this two derivations connects to each other? May 23 at 20:11
• @BinxuWang王彬旭 Yes, I would like to see either in conceptual terms or mathematically how the last expression is the same as the first.
– JAP
May 23 at 20:12
• Sure thing this is very interesting! May 23 at 20:17
• I think you get a typo here $B_{\alpha\beta}=\vec r_\alpha \cdot \frac{\partial N}{\partial r_\alpha}$, you missed $\beta$ May 23 at 20:18
• Also I think $\det \left(\mathrm {II}^{-1}\rm I\right)$ is a typo... the 2nd foundamental form on top May 23 at 20:26

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.

• Much obliged! Perfectly expressed. I see the indices are a mess, and your corrections in the answer are a reminder of the live and contracted indices as they should be used.
– JAP
May 23 at 21:46
• One silly follow-up question: I thought the sign depended on which one of two normal vectors you chose.
– JAP
May 23 at 21:49
• @JAP yes/ no, but with the same normal vector, the sign Ted Shifrin points out just comes from the rule of differentiation. May 23 at 21:52
• You are misusing terminology when you refer to the curvature tensor. That terminology is usually reserved for the Riemann curvature tensor, not for (up to sign) the shape operator or Weingarten map. The shape operator is ordinarily defined with the negative sign so that its matrix representation is precisely $\text I^{-1}\text{II}$. May 23 at 22:01
• It is pretty immediate from the hint. Write the system of linear equations in matrix form. May 23 at 22:56

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

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 $$III(u,v)= S(u) \cdot S(v)$$

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)

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.

• I am pretty sure you yourself will be able to calculate out the fundamental forms once you see the picture I have shown. May 23 at 22:10
• This actually has almost nothing to do with the question. It nicely explains the meaning of the shape operator. May 23 at 22:16
• Cool! Thanks +1
– JAP
May 23 at 22:18
• I have done it now @TedShifrin May 23 at 22:43
• You can’t choose an orthonormal basis when your computation comes specifically from an arbitrary parametrization. This, unfortunately, is not the method of moving frames. May 23 at 22:45

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

• Thank you for summarizing it on this answer.
– JAP
May 24 at 14:41
• Hi I just realized one of the point I mentioned was non obvious and I have linked the proper page in pavel's book discussing it. Take the equation I linked and dot both side with $N$. @JAP May 24 at 15:38