Gaussian Curvature and the Second Fundamental Form

In the last post we saw that the first fundamental form was an inner product on the the tangent space T_pS, and we used this to study the intrinsic geometry of surfaces. The goal of this post will be to understand the differential of the Gauss map, which we will use to define Gaussian curvature and the second fundamental form – a quadratic form on the tangent space T_pS. We will start by defining the Gauss map, which we have already seen in class. I will give a few examples/exercises in which we actually compute some of these. We will see that the differential of the Gauss map is a self-adjoint linear map, and the second fundamental form is the associated quadratic form. The Gauss map very much depends on how the surface is sitting in \mathbb{R}^3, i.e., on its parametrisation \phi. The remarkable fact is that the Gaussian curvature, which we will define in terms of this Gauss map, is an intrinsic property of S, that is, it doesn’t depend on the specific embedding \phi. This is the statement of Gauss’ Theorma Ergergium.

The Gauss map and its differential

We want to consider parametrised surfaces in \mathbb{R}^3 such that the parametrisation \phi is differentiable. Given such a parametrisation \phi: U\subseteq \mathbb{R}^2 \rightarrow \mathbb{R}^3 of  S, we can choose a unit normal vector at each point p in \phi(U) by the rule

n(p) = \displaystyle\frac{\phi_u \wedge \phi_v}{|\phi_u \wedge \phi_v|}

Indeed, \{\phi_u, \phi_v\} is basis for T_pS, and the cross product (see last post) gives us a vector perpendicular to both. The map n:\phi(U) \rightarrow \mathbb{R}^3 is differentiable, and we say we have a differentiable field of unit normal vectors (on the image \phi(U)). There has been some confusion in class over the definition of orientable. For us, an orientable surface is one that admits a differentiable field of normal vectors on the whole surface S, and we call a choice of such a field n an orientation of S.

Remark. Given an orientation of S, we get an induced orientation on each tangent space T_pS as follows: call a basis \{a, b\} of T_pS positive if \langle a \wedge b, n\rangle is positive. Now the set of all positive bases for T_pS is an orientation of T_pS.

Now, given an orientable surface we can define the Gauss map,

Definition. Let S \subseteq \mathbb{R}^3 be a surface with an orientation n. The Gauss map is the map

n:S \rightarrow \mathbb{R}^3

taking a point p \in S to the normal vector n(p) \in S^2 \subseteq \mathbb{R}^3.

The Gauss map is a differentiable map, and its differential at p is a linear map,

dn_p:T_pS \rightarrow T_{n(p)}S^2

Exercise. Compare T_pS with T_{n(p)}S^2. Aren’t they similar? This is an important idea and we will come back to it.

In your prelim you were asked to compute a Gauss map. Let’s look at a few examples. Each time we can think about the above exercise.

1. An easy example is the plane \pi. The unit normal vector is constant here, so dn_p=0 for all p \in \pi. Notice that T_{n(p)}S^2 is parallel to T_p\pi.

2. Let \mathcal{C} = \{ (x, y, z) : x^2+y^2=1\} be a cylinder. Choosing the inward pointing normal, we see by inspection that the Gauss map is

n(x, y, z) = (-x, -y, 0)

cylinder normalor if you prefer, let \alpha(t) = (x(t), y(t), z(t)) be a parametrized curve on \mathcal{C} and observe that 2xx'+2yy'=0 and thus xx'+yy'=0. This shows that either (x, y, 0) or (-x, -y, 0) are normal to (x, y, z) on \mathcal{C} (outward and inward pointing normals, respectively).

Now n(\alpha(t))=(-x(t), -y(t), 0) and therefore we have

dn_{\alpha(t)}=(-x'(t), -y'(t), 0)

Exercise. Describe dn_p in the two cases
(a) Our curve \alpha_1 is parallel to the z-axis (consider a tangent vector v_1 parallel to z-axis)
(b) Our curve \alpha_2 is parallel to the xy-plane (consider a tangent vector v_2 parallel to xy-plane)

v_1 and v_2 are eigenvectors of dn_p, and you have just computed their eigenvalues.

3. Exercise. Compute the Gauss map for the unit sphere \mathcal{S} using the same method as above.

Self Adjoint Linear Maps

You should have seen the following in 2230, so this is just a quick recap. Let V be a vector space with an inner product \langle-,-\rangle. We will be thinking about T_pS with the first fundamental form.

Definition. A linear map A:V \rightarrow V is self-adjoint if \langle Av, w\rangle = \langle v, Aw\rangle.

Exercise. Fix an orthonormal basis for V. Show that the corresponding matrix for A is symmetric.

We associate to A a symmetric bilinear form

B:V \times V \rightarrow \mathbb{R}        by        B(v, w) = \langle Av, w\rangle

Conversely, if B is a symmetric bilinear form then one can define a linear map A:V \rightarrow V by \langle Av, w\rangle = B(v, w), and it will be self-adjoint.

There is an associated quadratic form Q(v) = B(v, v) which totally determines B since

B(u, v) = \frac{1}{2}(Q(u+v)-Q(u)-Q(v))

and thus there is a one-to-one correspondence between self-adjoint linear maps of V and quadratic forms in V. The main result we will use here is the following,

Theorem. Given a self-adjoint linear map A:V \rightarrow V, there exists an orthonormal basis for V such that the corresponding matrix for A is diagonal. Moreover the eigenvalues of this diagonal matrix are precisely the maximum and minimum values of the quadratic form Q(v)=\langle Av, v\rangle on the unit circle of V.
(Remark. If this wasn’t covered in 2230 I can supply a proof)

The Second Fundamental Form

The key observation to make, that I hinted at above, is that T_pS and T_{n(p)}S^2 are parallel planes (e.g., see picture below), and so

dn_p:T_pS \rightarrow T_{n(p)}S^2

can be thought of as a linear map

dn_p:T_pS \rightarrow T_{p}S

Gauss Map

Lemma. The differential dn_p:T_pS \rightarrow T_pS is a self-adjoint linear map

Proof. Let \phi be a parametrization of S at p, and \{\phi_u, \phi_v\} be the associated basis for T_pS. Let \alpha(t) = \phi(u(t), v(t)) be a parametrized curve in S with \alpha(0)=p then

dn_p(\alpha'(0)) = dn_p(\phi_uu'(0)+\phi_vv'(0))
=\displaystyle\frac{d}{dt}n(u(t), v(t))\vert_{t=0}
= n_uu'(0)+n_vv'(0)

In particular dn_p(\phi_u)=n_u and dn_p(\phi_v)=n_v. To show self adjoint need to show

\langle dn_p \phi_u, \phi_v \rangle = \langle \phi_u, dn_p \phi_v \rangle

(and vice versa) which is equivalent to showing

\langle n_u, \phi_v \rangle = \langle \phi_u, n_v \rangle

Exercise. finish the proof. Hint: take derivative of \langle n, \phi_u \rangle=0 with respect to v and of \langle n, \phi_v \rangle=0 with respect to u.

We thus have an associated quadratic form Q. It turns out to be more convenient to use -Q.

Definition. The second fundamental form II_p is defined on T_pS by II_p(v)=-\langle dn_p(v),v \rangle. It is the quadratic form associated to self-adjoint linear map dn_p.

Normal, Principle, Gaussian and Mean Curvature

I first describe the normal curvature of a regular curve \alpha in S. Recall (or perhaps this is a new definition) that if \alpha:I \rightarrow \mathbb{R}^3 is a curve (parametrized by arc length t \in I) then k(t):=|\alpha''(t)| is the curvature of \alpha at t.

Exercise. If this is a new definition then
1. Show that the the curvature of a straight line is 0.
2. Show that in general \alpha''(t) is normal to \alpha'(t). Hint: consider the expression \alpha'(t)\cdot\alpha'(t)=1.

Let p be a point on \alpha, k the curvature of \alpha at p, and let N_\alpha denote the (unit) normal vector to \alpha at p (\frac{\alpha''(t)}{|\alpha''(t)|}). Then we have that \cos\theta=\langle N_\alpha, n\rangle, where n is the normal to the surface S at p.

Definition. The normal curvature of \alpha at p is k_\alpha:=k\cos\theta. It is simply the projection of k to n.

Normal Curvature!

We can think of the second fundamental form in terms of normal curvature by making a simple calculation: We have that \langle n(s), \alpha'(s)\rangle=0. Therefore

\langle n(s), \alpha''(s)\rangle + \langle n'(s), \alpha'(s)\rangle =0

and we get

II_p(\alpha'(0)) = -\langle dn_p(\alpha'(0)), \alpha'(0)\rangle =-\langle n'(0), \alpha'(0) \rangle=\langle n(0), \alpha''(0)\rangle=k_\alpha(p)

So the second fundamental form II_p of a vector v \in T_pS is the normal curvature of a(ny) curve \alpha passing through p and tangent to v. In particular this tells us that any two curves passing though a given point with the same tangent have the same normal curvature at that point (this is known as Meusnier’s theorem).

We have shown that dn_p is a self-adjoint linear map – and so we can apply the theorem stated above to conclude that there exists an orthonormal basis \{e_1, e_2\} of T_pS such that dn_p is diagonal, that is dn_p(e_1)=-k_1e_1 and dn_p(e_2)=-k_2e_2, and moreover, we know that (WLOG) k_1 is the maximum value of II_p restricted to the unit circle in T_pS and k_2 is the minimum. That is, k_1 and k_2 are the extreme values of the normal curvature at the point p. We can picture this as follows: fix a point p \in S, and consider all possible unit tangent vectors at that point (a circle’s worth of them). For each choice of vector we find the normal curvature of a(ny) associated tangent curve, we pick out the maximum and minimum values we find around the circle, and call them k_1 and k_2.

Definition. k_1 and k_2 are called the principle curvatures at p, and the directions of their corresponding eigenvectors e_1 and e_2 are called the principle directions at p.


  1. For the plane or the sphere all directions at any point are extremal, and so all directions are principle directions. In particular k_1=k_2. Thought. Can you think of another surface with this property?
  2. For the cylinder we showed that the principle curvatures are 0 and 1. In the specific example we covered, the principle directions at any given point are parallel to the z-axis (corresponding to the minimum normal curvature, 0) and parallel to the xy-plane (corresponding to the maximum, 1).

We know that the determinant and the trace of linear map are independent of the choice of basis, and therefore we have that \det(dn_p)=k_1k_2 and tr(dn_p)=-(k_1+k_2).

Exercise. What would happen to the determinant and to the trace of dn_p if we were to change to orientation of S? Do you think this would be the same if we were considering 3-manifolds? What about n-manifolds?

Definition. Let dn_p:T_pS \rightarrow T_pS be the differential of the Gauss map at p\in S. The Gaussian curvature K is the determinant of dn_p and the mean curvature H is the negative of half of the trace. In terms of principle curvatures that is,

K=k_1k_2        and        H = \displaystyle\frac{k_1+k_2}{2}

The Gauss map takes a point p \in S and gives you its unit normal vector. If we think intuitively about the differential dn_p, it should tell us how that normal vector is changing near to the point p. For example, if the Gaussian curvature is positive, that is telling us that both the principle curvatures have the same sign, and we can picture the surface ‘falling away’ from the tangent plane at p in all directions:
elliptical point
Here are some terms you may have come across before that relate to this point of view.

Definition. A point on a surface S is called

  1. Elliptic if K>0
  2. Hyperbolic if K<0
  3. Parabolic if K=0 but dn_p \neq0
  4. Planar if dn_p=0

Hopefully it is clear that this classification does not depend on the choice of orientation of S.

Exercise. Give examples of a points on surfaces satisfying each of 4 definitions above.

Gauss’ Remarkable Theorem

Let’s recap what we just saw. The fundamental construction was the differential of the Gauss map dn_p which was defined in terms of the Gauss map n = \displaystyle\frac{\phi_u\wedge\phi_v}{|\phi_u\wedge\phi_v|} which itself we defined in terms of the parametrisation \phi. This is an important observation – the second fundamental form very much depends on the specific embedding \phi of S \subseteq \mathbb{R}^3. We also can see that the principle curvatures depend on the embedding. So unlike the first fundamental form, which was used to describe the intrinsic geometry of S, the second fundamental form is used to understand its extrinsic properties. It is in this setting that Gauss’ theorem really does seem remarkable. Indeed, Gauss’ theorem tells us that k_1k_2 is an intrinsic property of S, that is, does not depend on the choice of \phi. Indeed, k_1k_2=det(dn_p)=K the Gaussian curvature, and Gauss’ theorem is precisely that K is an intrinsic property.

A Word on Local Expressions

You may be wondering why we even bothered to define the second fundamental form? We didn’t use it directly to compute the anything as we did with the first fundamental form. Is it so fundamental? In fact its power can be seen via its use in making local computations. Given a parametrised surface we can describe the local situation very naturally, i.e., on a so called chart \phi: U\subseteq \mathbb{R}^2 \rightarrow S, we have a family of linear maps

dn=\left(\begin{array}{cc}a & b \\c & d\end{array}\right)

Fixing a basis \{\phi_u, \phi_v\} for T_pS one computes



e=\langle n, \phi_{uu} \rangle, f=\langle n, \phi_{uv} \rangle, g=\langle n, \phi_{vv} \rangle

One obtains useful local expressions that are easy to compute in terms of e, f, g, for example, it can be shown that the Gaussian curvature is given by,

K = \det\left(\begin{array}{cc}a & b \\c & d\end{array}\right)=\displaystyle\frac{eg-f^2}{EG-F^2}

where E, F, G are the coefficients of the first fundamental form. Similar expressions exist for the mean curvature and principle curvatures for example.

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