The purpose of this post is to pick up where the last one left off: the story of curvature from a, like, “geometer’s geometer” standpoint.

## Table of Contents

## 1. Surfaces in \(\mathbb{R}^3\).

Just as we can *parametrize* curves \(\gamma\) in \(\mathbb{R}^2\),
there is a notion of a *parametrized surface* in \(\mathbb{R}^3\).
This is given by *three* functions of *two* variables,
i.e. \((x(s,t), y(s,t), z(s,t))\).
One common example is the graph of a function of two variables,
i.e. \((x, y, f(x,y))\),
like the function \(z = x^2 + y^2\) that came up in an earlier post.
Just as a parametric curve is a slight generalization
of the graph of a function in \(\mathbb{R}^2\),
so too should you think of the notion of a parametric surface.

There are also *implicit* surfaces,
like the surface of a sphere when defined as
the set of points \((x,y,z)\) satisfying the equation
\(x^2 + y^2 + z^2 = 1\).
We’ll use both freely and not attempt to worry about converting between them.
There are, of course, plenty of pathological examples,
but you should imagine the things one attempts to mean with this definition,
like the surface of a donut,
or the graph of \(z = x^2 + y^2\).

## 2. Curvatures at a point

For a smooth surface,
i.e. similar to smooth curves like \(y = x^2\)
or parametrized curves like \((\cos t, \sin t)\)
and implicit curves like \(y^2 - x^2 = 1\),
there is, rather than a tangent *line,*
a whole *plane* tangent to the surface at a point \(p\).
Now, for every curve on the surface through \(p\),
we have a tangent line, and one can think of
the tangent plane as the plane containing,
for every such curve,
the tangent line at \(p\).
The fact that there is a unique such plane
is baked into the idea for what it means for the surface to be smooth,
essentially.

### 2.1. Normal lines

Now, given a plane containing a point \(p\) in three dimensional space,
there is a unique line *normal* to that plane through \(p\).
Like, if the plane is the $xy$-plane,
the normal line is the $z$-axis:
it’s perpendicular to every line contained in the plane.
Mathematicians typically use a convention called the “right hand rule”
for determining a *direction* for this line,
but it depends on a choice of *orientation* of the plane,
that is, of two ordered, oriented lines in the plane that cross at \(p\).
Going back to our previous example, maybe our lines are the $x$-axis
and the $y$-axis.
Anwyay, you’re supposed to imagine orienting your closed right hand
with your thumb sticking up
such that your fingers curl (counterclockwise!)
from the direction of the first oriented line to the second.
The direction that your thumb points is the orientation for the normal line.

That is, in our situation we go $x$-axis, then $y$-axis and the $z$-axis is directed upwards. It’s just a convention, though—any choice will do, as long as it’s consistent.

### 2.2. Principal curvatures

Anyway, back to curvature.
So, given our tangent plane,
we can choose any straight line in it through \(p\)
and the normal line.
This gives us a plane that *slices through* the surface—and
a plane curve, namely the intersection of this plane
with the surface itself.
Since any such slicing plane contains the normal vector,
we can compute the signed curvature of the curve
with respect to *that* normal vector.

In general, as we change the tangent line through \(p\),
the curvature changes.
The *principal* curvatures at \(p\)
are the *maximum* and *minimum* signed curvatures at \(p\).

We can compute them for some simple surfaces.
For the sphere \(x^2 + y^2 + z^2 = 1\), the paraboloid \(z = x^2 + y^2\)
and the hyperboloid \(z^2 - x^2 - y^2 = 1\),
for the special point where \(x = y = 0\),
since \(z\) is really a function of \(x^2 + y^2\),
that is, because we have *rotational symmetry*
all the curvatures will be equal; we computed them in the previous post.

For something like the saddle \(z = x^2 - y^2\), we see different behavior at the origin: for the line \(y = 0\), we end up looking at the function \(z = x^2\), whose curvature we computed to be \(-2\) (negative, right, because we are curving towards the positive \(z\) axis) last time. For the line \(x = y\), the function itself is a line, with curvature \(0\). For the line \(x = 0\), we’re looking at the function \(z = -y^2\), whose curvature is \(+2\). It turns out the curvatures in the \(x = 0\) and \(y = 0\) lines are principal.

### 2.3. Gaussian curvature

The *Gaussian curvature* of a surface at the point \(p\)
is the *product* of the principal curvatures at the point.

Gauss’s *Theorema egregium* states that the Gaussian curvature of a surface,
although it is *defined* in terms of an embedding of that surface
into three-dimensional space,
is actually an *intrinsic* property of the surface
that does not change if the surface is not stretched.
Moreover, even if the surface is stretched,
the *sign* of the Gaussian curvature should not change.
Right? replacing \(z = x^2 - y^2\) with \(z = ax^2 - by^2\)
will still have negative curvature at \((0,0)\)
provided \(a\) and \(b\) have the same sign.

## 3. Manifolds in \(\mathbb{R}^n\) in general

In the general case, how would one go about defining curvature?
As it turns out, there are several ways.
For (smooth) manifolds embedded in \(\mathbb{R}^n\),
one way is to define the *sectional curvatures*
of a point \(p \in M\)
to be the *Gaussian curvatures*
of any *tangent $2$-plane* at that point.
For a manifold of dimension \(m > 2\),
there are infinitely many tangent $2$-planes through any point,
and hence many many sectional curvatures.
If we believe Gauss's theorem,
each of these Gaussian curvatures should be an intrinsic property
of \(M\), and hence should not change if we embed \(M\) in \(\mathbb{R}^n\)
differently without distorting the geometry of the embedding.

There’s a way to make this precise with reference only to the manifold \(M\) and not to the space it is embedded into. As a kid (that is, when I was 21 and taking Riemannian geometry) I thought this was the neatest part of the subject—that you can do all of this fancy computation with just the manifold and not any space it is embedded into.

Now, don’t get me wrong, I still think that’s extremely neat.
But as a mathematician working geometrically with other spaces,
I actually rarely meet a manifold that doesn’t come equipped
with a way of embedding it (or its universal cover)
into \(\mathbb{R}^n\),
and the manifolds which I *do* think of sort of “by themselves”
tend to have dimension at most three,
so one can actually tend to reason about these things “by hand” a little more.