Gauss's Theorema Egregium

Gauss's Theorema Egregium

Egregious originally meant “illustrious” or “remarkable”. e- for “out of” and greg- for “the flock”; it meant “standing out from the flock”. It only developed a negative connotation as an ironical usage. Thus, the Theorema Egregium meant the “Remarkable Theorem”.

Pull-backs, push-forwards, and isometry

Vectors

When a space is flat, we can think of vectors as lines between two points in the space. For instance, this is how we approach vectors in Newtonian and special-relativistic physics. However, when space curves, we no longer have an idea of how to “add” two lines, for example. For curved spaces embedded in $\mathbb{R}^n$ , we can revive the classical notion. However, many manifolds are not given as being embedded in $\mathbb{R}^n$ . Thus, we need to extend the notion of a vector.

We want to use the notion of vectors as directional derivatives. That is, we will essentially define a vector $\vec{v}$ to be
$\begin{displaymath} \vec{v} = \sum_i\, v^i\, \frac{\partial}{\partial x^i}\ . \end{displaymath}$
In fact, in this sense, we can show that the collection $\frac{\partial}{\partial x^i}$ is a basis for the vector space. Indeed, we often write these as vectors:
$\begin{displaymath} \vec{\frac{\partial}{\partial x^i}}\ . \end{displaymath}$

Given a manifold $M$ , write $\mathcal{F}$ as the space of all smooth functions from $M$ to $\mathbb{R}$ . We define a vector $\vec{v}$ at a point $p \in M$ as a map $\vec{v} : \mathcal{F} \longrightarrow \mathbb{R}$ that is (1) linear, and (2) obeys the Leibniz rule:

$\vec{v}(a f + b g) = a\vec{v}(f) + b\vec{v}(g)\ ,$
$\vec{v}(fg) = f(p)\, \vec{v}(g) + \vec{v}(f)\, g(p)\ ,$

for all $a,b \in \mathbb{R}$ and $f,g \in \mathcal{F}$ .

Pull-backs of maps and push-forwards of vectors

Let $M$ and $N$ be manifolds (not necessarily of the same dimension), and let $\phi : M \longrightarrow N$ be a $C^\infty$ map. In a natural manner, $\phi$ “pulls back” a function $f:N \longrightarrow \mathbb{R}$ to a function $f \circ \phi : M \longrightarrow \mathbb{R}$ . Similarly, in a natural manner, $\phi$ “pushes forward” vectors at $p \in M$ to vectors at $\phi(p) \in N$ . That is, it defines the map $\phi^\ast : V_p \longrightarrow V_{\phi(p)}$ as follows: For $\vec{v}\in V_p$ we define the vector $\phi^\ast \vec{v} \in V_{\phi(p)}$ by
$\begin{displaymath} \left(\phi^\ast \vec{v} \right)(f) = \vec{v}\left( f \circ \phi) \end{displaymath}$
for all smooth $f : N \longrightarrow \mathbb{R}$ .

Isometry

Intuitively, an isometry is any motion of a surface that leaves distances unchanged. Thus, if we think of an ant crawling along a surface as being completely unaware of the third dimension, an isometry would not be noticeable in any way to him. Alternatively, we might stretch the surface, for example. This would lengthen some lines, which the ant could notice. That would not be an isometry.

Now, suppose we have two manifolds $M$ and $N$ having the same dimension — each having a metric: $g_M$ and $g_N$ — and a diffeomorphism $\phi: M \longrightarrow N$ between them. Take any two vectors $\vec{X}, \vec{Y}$ on $M$ and push them forward to $N$ . We call $\phi$ an isometry if, for all vectors $\vec{X}, \vec{Y}$ at all points in $M$ ,
$\begin{displaymath}g_A(\vec{X},\vec{Y}) = g_B(\phi^*\vec{X},\phi^*\vec{Y})\ .\end{displaymath}$

Gaussian curvature

From now on, we’ll look at two-surfaces embedded in $\mathbb{R}^3$ . So, suppose we have coordinates $u,v \in \mathbb{R}^2$ giving a surface $\vec{x}(u,v) \in \mathbb{R}^3$ . Familiar examples are the plane
$\begin{displaymath} \bm{x}(u,v) = \left( \begin{array}{c} u \\ v \\ 0 \end{array} \right)\ ; \end{displaymath}$
the cylinder of radius $r$
$\begin{displaymath} \bm{x}(\phi,z) = \left( \begin{array}{c} r\, \cos\phi \\ r\, \sin\phi \\ z \end{array} \right)\ ; \end{displaymath}$
and the sphere of radius $r$
$\begin{displaymath} \bm{x}(\theta,\phi) = \left( \begin{array}{c} r\, \sin\theta\, \cos\phi \\ r\, \sin\theta\, \sin\phi \\ r\, \cos\theta \end{array} \right)\ . \end{displaymath}$

Now we want some notion of how much these surfaces curve in the surrounding geometry. From analytic plane geometry, we associate curvature with the second derivative of a function. Thus, we expect that $\vec{x}_{,\alpha \beta}$ should be associated with the curvature of the surface.

However, we can easily imagine that we could lay down different coordinates in the surface, without actually affecting the geometry. This means that derivatives along the direction of the surface don’t really mean anything; we can change them as much as we want simply by moving the coordinates a little or a lot. Thus, we take the normal component of $\vec{x}_{,\alpha \beta}$ to find its curvature. We define the extrinsic curvature as
$\begin{displaymath} K_{\alpha \beta} = \left\langle \vec{x}_{,\alpha\beta} , \vec{N} \right\rangle\ . \end{displaymath}$

Now, we have reduced the curvature to a symmetric, 2-by-2 matrix at each point of the surface. This matrix, then, has three independent components. Intuitively, however, we can see that there are just two curvatures: a curvature along one direction, and a curvature along the orthogonal direction. It turns out that the third degree of freedom is taken up by our ability to rotate the coordinates around the point of interest. Now, we can get rid of this useless freedom by looking at something more geometric: the eigenvectors and eigenvalues. [N.B., we must raise the first index of $K_{\alpha\beta}$ in order to think of this as a matrix and use our usual skills with matrices.]

The two eigenvectors correspond to the minimum and maximum values of the curvature along any direction in the surface, and are orthogonal (except in the case where the min and max are the same). We also define the two old-fashioned types of curvature:

Gaussian curvature, $K$ : $K = \kappa_1 \kappa_2 = \det K$
Mean curvature, $H$ : $H = \frac{\kappa_1 + \kappa_2}{2} = \text{tr} K$

Mention the surprising similarity between the ADM equations and the equation for the surface of a rock rolling down a hill.

Applications

Flat planes

Compute the Gaussian curvature of a flat plane. (This one’s easy.)
$\begin{displaymath} \bm{x}_{,\alpha\beta} = \left( \begin{array}{c} \left( \begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array} \right) \\ \left( \begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array} \right) \\ \left( \begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array} \right) \end{array} \right) \end{displaymath}$

$\begin{displaymath} K_{\alpha \beta} = \bm{x}_{,\alpha \beta} \cdot \bm{N} = 0 \end{displaymath}$
Thus, we also have $K^\alpha_{\phantom{\alpha}\beta} = 0$ , $\kappa_1 = \kappa_2 = 0$ , and $\det K = \kappa_1 \kappa_2 = 0$ .

Cylinders

$\begin{displaymath} \bm{x}_{,\alpha \beta} = \left( \begin{array}{c} \left( \begin{array}{cc} -r\, \cos\phi & 0 \\ 0 & 0 \end{array} \right) \\ \left( \begin{array}{cc} -r\, \sin\phi & 0 \\ 0 & 0 \end{array} \right) \\ \left( \begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array} \right) \\ \end{array} \right) \end{displaymath}$
The normal vector is
$\begin{displaymath} \bm{N} = \left( \begin{array}{c} \cos\phi \\ \sin\phi \\ 0 \end{array} \right) \end{displaymath}$
Finally, we take the dot product of these two:
$\begin{displaymath} K_{\alpha \beta} = \bm{x}_{,\alpha \beta} \cdot \bm{N} = \left( \begin{array}{cc} -r & 0 \\ 0 & 0 \end{array} \right) \end{displaymath}$
Raising the index, we have
$\begin{displaymath} K^\alpha_{\phantom{alpha} \beta} = \bm{x}_{,\alpha \beta} \cdot \bm{N} = \left( \begin{array}{cc} -1/r & 0 \\ 0 & 0 \end{array} \right) \end{displaymath}$

Spheres

Compute the Gaussian curvature of a sphere. (This one’s probably vaguely familiar.) We can compute
$\begin{displaymath} \bm{x}_{,\alpha\beta} = \left( \begin{array}{c} \left( \begin{array}{cc} -r\, \sin\theta\, \cos\phi & -r\, \cos\theta\, \sin\phi \\ -r\, \cos\theta\, \sin\phi & -r\, \sin\theta\, \cos\phi \end{array} \right) \\ \left( \begin{array}{cc} -r\, \sin\theta\, \sin\phi & r\, \cos\theta\, \cos\phi \\ r\, \cos\theta\, \cos\phi & -r\, \sin\theta\, \sin\phi \end{array} \right)\\ \left( \begin{array}{cc} -r\, \cos\theta & 0 \\ 0 & 0 \end{array} \right) \end{array} \right) \end{displaymath}$
The normal vector is
$\begin{displaymath} \bm{N} = \left( \begin{array}{c} \sin\theta\, \cos\phi \\ \sin\theta\, \sin\phi \\ \cos\theta \end{array} \right) \end{displaymath}$
Finally, we take the dot product of these two:
$\begin{eqnarray*} K_{\alpha \beta} = \bm{x}_{,\alpha \beta} \cdot \bm{N} &=& \left( \begin{array}{cc} -r\, \sin^2\theta\, \cos^2\phi & -r\, \sin\theta\, \cos\theta\, \sin\phi\, \cos\phi \\ -r\, \sin\theta\, \cos\theta\, \sin\phi\, \cos\phi & -r\, \sin^2\theta\, \cos^2\phi \end{array} \right) \\ && + \left( \begin{array}{cc} -r\, \sin^2\theta\, \sin^2\phi & r\, \sin\theta\, \cos\theta\, \sin\phi\, \cos\phi \\ r\, \sin\theta\, \cos\theta\, \sin\phi\, \cos\phi & -r\, \sin^2\theta\, \sin^2\phi \end{array} \right) \\ && + \left( \begin{array}{cc} -r\, \cos^2\theta & 0 \\ 0 & 0 \end{array} \right) \\ &=& \left( \begin{array}{cc} -r & 0 \\ 0 & -r\, \sin^2\theta \end{array} \right) \end{eqnarray*}$
Finally, we have:
$\begin{displaymath} K^\alpha_{\phantom{\alpha}\beta} = \left( \begin{array}{cc} -1/r & 0 \\ 0 & -1/r \end{array} \right) \end{displaymath}$

It’s not hard to see that the eigenvalues of this matrix are both $\kappa_1=\kappa_2=-1/r$ . Clearly, we also have $\det K = \kappa_1\,\kappa_2=1/r^2$ .

Note that we have proven the familiar result that you can’t isometrically represent the Earth with a flat map. All map projections change angles and distances.

The theorem

Suppose we have two manifolds $M$ and $N$ , and an isometry $\phi : M \longrightarrow N$ . Take coordinates $y^i$ in $N$ , and pull them back to define coordinates $x^i$ on $M$ :
$\begin{displaymath} x^i(p) \equiv y^i\left(\phi(p)\right)\ . \end{displaymath}$
Now we can push forward the coordinate basis $\vec{\frac{\partial}{\partial x^i}}$ , and find that
$\begin{displaymath} \phi^\ast \vec{\frac{\partial}{\partial x^i}} = \vec{\frac{\partial}{\partial y^i}} \end{displaymath}$
We can use this equation followed by the definition of an isometry to find that
$\begin{displaymath} g_{ij}^N(p) \equiv \left\langle \vec{\frac{\partial}{\partial y^i}} , \vec{\frac{\partial}{\partial y^j}} \right\rangle_N = \left\langle \phi^\ast \vec{\frac{\partial}{\partial x^i}} , \phi^\ast \vec{\frac{\partial}{\partial x^j}} \right\rangle_N = \left\langle \vec{\frac{\partial}{\partial x^i}} , \vec{\frac{\partial}{\partial x^j}} \right\rangle_M \equiv g_{ij}^M(p) \end{displaymath}$
That is, in these coordinates, the expressions for the metrics of the two surfaces are the same. Just switch $M$ and $N$ , and $x$ and $y$ .

Thus, any expression you can write down that just involves the metric and the coordinates will be an invariant. In particular, since we know how to write $R^{12}_{\phantom{12}12}$ in terms of the metric and its derivatives, this quantity is the same at corresponding points of the two surfaces.

Now, it is a moderately surprising fact (presented without proof) that, for a two-surface, the Riemannian has only one free component, which we can calculate to be $R^{12}_{\phantom{12}{12}} = K^1_{\phantom{1}{1}} K^2_{\phantom{2}{2}} - K^2_{\phantom{2}{1}} K^1_{\phantom{1}{2}} = \det K$ . Thus we find that the Gaussian curvature is invariant under isometry! Remarkable!

Discussion of the examples

Maps of the Earth

We’ve calculated that the Gaussian curvature of a sphere is $-1/r^2$ , whereas the curvature of a flat plane is $0$ . Thus, no flat map of the Earth can possibly be an isometric representation of Earth. That is, any flat map will introduce distortions.

Eating New York-style pizza

Compute the Gaussian curvature of a slice of pizza in two conditions. First, as a flat slice. Second, as a folded slice. Note that if the crust doesn’t stretch (a reasonable approximation), the two are isometric, so the Gaussian curvature is constant. Thus, the second curvature must also be zero.

So, take a slice of pizza, and bend it. We have two lines. The first line is along the bottom of the fold; the second is orthogonal to the fold. These represent the directions of the two eigenvectors. Now, clearly the eigenvalue along the second line is non-zero. But, in order for the Gaussian curvature to be zero — satisfying the theorem — the eigenvalue along the first line must be zero. That is, the fold must be a straight line.

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lunchtalks/theorema_egregium.txt · Last modified: 2007/08/23 11:46 by boyle