From:  (Stephen A. Fenner)
Newsgroups:     sci.math,sci.physics
Subject:        The total curvature of a knot (long)
Summary:        It is > 4*pi
Keywords:       Knots, Curves, Curvature
Date:           29 May 90 02:43:10 GMT
Reply-To: (Stephen A. Fenner)
Organization:   Dept. of Comp. Sci., The University of Chicago

Ok, here is a the geometrical argument I promised on a previous post.
I'll show that any smooth, simple, closed curve in 3-space must have
total curvature at least 4*pi.  I'll try to keep the argument as
intuitive and geometrical as possible, although that's easier said than

First, I'll show that the total curvature of _any_ closed curve (not
necessarily knotted) is at least 2*pi.  The lower bound for knotted curves
uses the same idea.

Let c be a smooth, simple, closed curve in R^3.  Think of c as a smooth
map from (an open set containing) the closed interval [0,r] into R^3,
parameterized by arc length (that is, |c'(t)|=1 for all t in [0,r]).
Thus r is the length of the curve, and

     c(0) = c(r),
     c'(0) = c'(r),
     c''(0) = c''(r),

Let k(t) be the curvature of c at "time" t (0 < t < r).
If c is planar, then its total curvature must be at least 2*pi (equality
holding when c is convex) in order for its tangent vector to wind around
once.  Thus we can assume c cannot lie in any single plane.

For all t in [0,r], translate the unit tangent vector c'(t) back to the
origin.  As we go around the curve, this vector sweeps out its own curve
u on the unit sphere (i.e., u maps [0,r] into S^2 with u(t) = c'(t)).

Now note that the total curvature of c equals the total _length_ of u.
This is because for all t in [0,r]

     k(t) = |c''(t)| = |u'(t)|

     /r          /r
    |           |
    | k(t)dt  = | |u'(t)|dt  =  total length of u.
    |           |
   / 0         / 0

Thus it suffices to show that u has length at least 2*pi.  Now for t in
[0,r], define H_t to be the hemisphere of S^2 centered at u(t).  That

    H_t  =  { v in R^3 :  |v| = 1  and  u(t).v > 0 }.

I like to think of H_t as a cap on the sphere, moving around with the
vector u(t), and at each time t, H_t covers exactly half of the sphere.

The proof boils down to two main points: First, as H_t moves around, it must
eventually cover the entire sphere.  That is, every point on the sphere
must be covered by H_t for some t in [0,r].  Second, H_t can only cover
new points so fast (new area covered is proportional to |u'(t)|).  Thus,
for H_t to cover the entire sphere, u must have some minimum length, which
turns out to be 2*pi.

To be more precise, say that a point v in S^2 is _painted_ at time t_0 if
0 < t_0 < r, and v passes from being uncovered by H_t to being covered by
H_t at time t = t_0.  That is, immediately before t_0, v is not in H_t,
and at t_0, v is in H_t.

1.   I claim that for every v in S^2 there is a t_0 in (0,r] such that
     v is painted at t_0.

[Proof]: Suppose not.  Then there is a point v_0 on the unit sphere that
either always remains covered or always remains uncovered by H_t.  Assume
without loss of generality that v_0 is always covered (if v_0 is always
uncovered, look at its antipodal point).  But this means that v_0.u(t)
> 0 for all t in [0,r].  Therefore, c'(t) always has a nonnegative
component in the v_0 direction, so if c(t) is to loop back on itself
(c(0) = c(r)), c'(t) must always be perpendicular to v_0 (i.e.,
c'(t).v_0 vanishes everywhere).  Then c is planar, contrary to our
assumption.  Q.E.D.

2.   H_t paints new area on S^2 at a rate equal to 2*|u'(t)|.  That is,
     if A(t) is the area of that portion of the sphere painted before time
     t, then A(t) can increase at a rate of at most 2*|u'(t)|.

[Handwave proof]:  Consider a small interval [t,t+dt].  Within the
interval, u moves almost in a straight line (remember, u is smooth).
The points that are painted in this interval approximately constitute
a section of the sphere swept out by the advancing boundary of H_t.
This section is bounded by two great semicircles with the same endpoint,
with maximum width |u'(t)|*dt.  The area of this section is approximately
2*|u'(t)|*dt.  (For example, think of u taking a great circle route around
the sphere; its total length is 2*pi, and the total area swept out is 4*pi.)
Taking the limit as dt approaches 0 gives the result.  Q.E.D.

Now the main result for (unknotted) closed curves:
At time t=0, no points are painted (A(0) = 0).  At time t=r, all the
points on the sphere are painted (A(r) = 4*pi) by claim 1.  Thus u must
have total length at least 2*pi by claim 2.  Q.E.D.

Now suppose c is knotted.  I claim that each point on the sphere is
painted _twice_ (two coats :-).

Suppose there is a point v_0 on S^2 which is only painted once.  By
reparameterizing the curve c (actually a cyclic shift), we can assume
v_0 is painted at some point t_1, and v_0 is uncovered for all t in
(0,t_1), and is covered for all t in [t_1,r].  In this case, c has the
following behaviour with respect to the direction v_0: at t=0, c starts
moving with a component in the negative v_0 direction and keeps moving
in this direction until t_1, where it turns around and moves in the
positive (actually nonnegative) v_0 direction until t=r.  (Its motion in
the other two orthogonal directions may be arbitrary.)

The curve c therefore cannot be knotted: it resembles a circular
rubber band that is anchored at opposite ends and twisted.  It may twist
and turn arbitrarily in the middle, between the two ends, but it can
always be unwound into a circle again.

So if c is knotted, the sphere must be entirely painted twice, making
the total area to be covered 8*pi.  By claim (2) above, the spherical
curve u must have total length at least 4*pi, thus the total curvature
of c is at least 4*pi.  Q.E.D.

Comments?                      --Steve Fenner
or  ...ihnp4!gargoyle!steve
A sadist is someone who is kind to a masochist.