book:content:torus - Geometry of Differential Forms

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%/*
\begin{figure}
\centering
%\epsfysize=2in\epsffile{\TOP figures/torus.eps}
\includegraphics[height=2in]{\TOP figures/torus}
\caption{A torus in $\RR^3$.}
\label{Torus}
\end{figure}
%*/
%*{{ book:figures:torus.png?200 }}
%* **Figure 1:**
%* //A torus in $\RR^3$.//

%/*
\begin{figure}
\centering
%\epsfysize=2in\epsffile{\TOP figures/torus2d.eps}
\includegraphics[height=2in]{\TOP figures/torus2d}
\caption[Parameterizing the torus as a surface of revolution.]
{Parameterizing the torus as a surface of revolution about the $z$-axis.}
\label{Torus2d}
\end{figure}
%*/
%*[{{ book:figures:torus2d.png?200|
%* **Figure 2:**
%* //Parameterizing the torus as a surface of revolution about the
%* $z$-axis.//}}]

The torus $\TTT$, shown in
%/*
Figure~\ref{Torus},
%*/
%* Figure~1,
can be parameterized as a surface of revolution about the $z$-axis,
as shown in
%/*
Figure~\ref{Torus2d},
%*/
%* Figure~2,
resulting in
\begin{align}
x &= (R+r\,\cos\theta)\,\cos\phi \\
y &= (R+r\,\cos\theta)\,\sin\phi \\
z &= r\,\sin\theta
\end{align}
from which it follows by direct computation that
\begin{equation}
ds^2 = r^2\,d\theta^2 + (R+r\,\cos\theta)^2\,d\phi^2
\label{dstorus}
\end{equation}
Alternatively, (\ref{dstorus}) follows directly from the Pythagorean theorem,
using arclength to compute small distances in the $\theta$ and $\phi$
directions.

Our orthonormal basis of 1-forms is therefore
\begin{align}
\sigma^\theta &= r\,d\theta \\
\sigma^\phi &= (R+r\,\cos\theta)\,d\phi
\end{align}
The structure equations are now
\begin{align}
-\omega^\theta{}_\phi\wedge\sigma^\phi &= d\sigma^\theta = 0 \\
-\omega^\phi{}_\theta\wedge\sigma^\theta &= d\sigma^\phi
 = -r\,\sin\theta\,d\theta\wedge d\phi
\end{align}
from which it follows that
\begin{equation}
\omega^\theta{}_\phi
 = \sin\theta\,d\phi
\end{equation}
and therefore that
\begin{equation}
d\omega^\theta{}_\phi
 = \cos\theta\,d\theta\wedge d\phi
 = \frac{\cos\theta}{r(R+r\,\cos\theta)}\>\omega
\end{equation}
since $\omega^\phi{}_\theta=-\omega^\theta{}_\phi$, and where
$\omega=\sigma^\theta\wedge\sigma^\phi$ is a choice of orientation.  Thus, the
Gaussian curvature of the torus is given by
\begin{equation}
K = \frac{\cos\theta}{r(R+r\,\cos\theta)}
\end{equation}

This result makes sense: the curvature in the $\theta$ direction is clearly
just $1/r$, and the curvature in the $\phi$ direction should go like the
inverse radius, or $1/(R+r\,\cos\theta)$.  But why is there a factor of
$\cos\theta$?

Recall that the principal curvatures tell you how the normal vector is
changing.  Read this sentence carefully: the principal curvatures tell you how
the normal vector \textit{to the surface} is changing along (principal)
directions \textit{in the surface}.  When computing the curvature of a circle,
the relevant normal vector is the outward-pointing radial vector, which lies
in the plane of the circle.  However, when computing the curvature of a
surface \textit{along} a circle, the relevant normal vector is that of
the \textit{surface}.  For circles on the torus in the $\theta$ direction,
there is no difference; these two normal vectors are the same.  However, for
circles in the $\phi$ direction, this is not the case.

Consider the circles on the top or bottom of the torus.  The \textit{surface}
normal vector is  always straight up (or down); it does \textit{not}
change in the $\phi$ direction.  Thus, one principal curvature vanishes at the
top and bottom of the torus, and hence so does the Gaussian curvature.

Furthermore, on the outer half of the torus, the change in the normal vector
in the $\phi$ direction has a positive component in that direction, whereas on
the inner half of the torus, the opposite is true.  This property is most
easily visualized in the equatorial plane; since the surface normal vectors
have opposite orientations on the inner and outer rings, the principal
curvature in the $\phi$ direction has opposite signs there.  The factor of
$\cos\theta$ accounts precisely for the difference in angle between the normal
vector to the surface, and the (horizontal) normal vector to the circles of
constant ``latitude''.
%/*
\footnote{
%*/
%*((
A similar argument holds on a sphere, where at most latitudes the normal
vector to the sphere is \textit{not} in the plane of a circle of constant
latitude.
%*))
%/*
}
%*/

Thus, the principal curvature in the $\phi$ direction, and hence the Gaussian
curvature, must be positive on the outer half, and negative on the inner half
(and zero in between).

This should not be surprising.  The Gauss-Bonnet Theorem tells us that
\begin{equation}
\int_T K\,\omega = 2\pi \chi(\TTT)
\end{equation}
and it is easy to check by example that the Euler characteristic of a torus is
zero, that is, $\chi(\TTT)=0$.  Thus, the \textit{total} curvature of any torus
must be zero, so that regions of positive curvature must be counterbalenced by
regions of negative curvature.  This is a \textit{topological} statement; no
matter how you twist a torus, its total curvature must be zero.

Can a torus be flat, that is, can its curvature be zero \textit{everywhere}?
Surprisingly, the answer is yes --- and if you have ever played a video game,
it is likely that you have already encountered a flat torus.  Take a
rectangular piece of paper, and identify opposite edges; this is just the
``wraparound'' feature of many video games, in which leaving the screen on the
right causes you to reenter the screen on the left, etc.  The curvature is
clearly zero; locally, it's just a piece of paper.  And the topology is indeed
that of a torus, as can be shown by experimentally determining the Euler
characteristic.

Can you actually make such a torus?  Not in three dimensions!  This torus
naturally lives in \textit{four} dimensions, as can be seen by the following
parameterization in $\RR^4$
\begin{align}
x = r\,\cos\theta \\
y = r\,\sin\theta \\
z = s\,\cos\phi \\
w = s\,\sin\phi
\end{align}
which also shows why we often write the torus $\TTT$ as $\SSS^1\times\SSS^1$,
the product of two circles.
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