# 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.

Exercises.
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)

or 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.

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

$dn_p:T_pS \rightarrow T_{p}S$

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 $n$.

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 $p$.

Examples.

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:

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

$II_p(\alpha')=e(u')^2+2fu'v'+g(v')^2$

where

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

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.