These are notes and problems for Rob Kusner's Spring 2020 honors
course on Differential Geometry of Curves & Surfaces (Math 563H at
UMassAmherst, Copyleft* 2010-20 by Rob Kusner), and are under
revision. (I may also prepare some .tex/.pdf Problem Sets, and revise
a .tex/.pdf version of notes that my 2013 grader, Dugan Hammock,
helped prepare). Homework is due weekly, but is retrospective by
a week; in other words, the problems marked ## are to be written up
and will generally be due in class the Wednesday 7-9 days after the
date noted - e.g. the first hw is due 29 Jan and consists of the ##
problems from the first day of class, while 27&29 Jan hw is due 05 Feb.
Intro Geometry.... Topology.... And the Universe!!!!!!!!
22 Jan Introduce ourselves. Review course logistics. Imagine how models
for the physical universe lead to the interplay the between local
geometry (curvature) and global topology (Euler number \chi=F-E+V,
where F=#faces, E=#edges, V=#vertices) of a (polyhedral) surface.
##Compute the Euler number of CUBE, TETRAHEDRON, SPHERE, TORUS and
PRETZEL surfaces with g>1 handles. See how subdviding may change F,
E or V, but not change \chi. Explore other ways to compute \chi,
like cutting out a pair of disjoint disks (reducing F by 2) and
gluing their boundary circles together (increasing g by 1); this
only gives surfaces with even \chi ≤ 2; what about odd \chi?!
Discuss surfaces that minimize "bending energy" (plastic panels and
metal plates) or some other geometric or physical quantity like area
or "stretching energy" (possibly with an enclosed-volume constraint
- soap films and bubbles minimize this).
Future: how might one define stretching or bending energy of a
curve in the plane or space (think of Hooke's law from mechanics)?
Part 1 Parametrized curves. Arc length and arc-length parametrization.
Unit tangent and normal vectors. Frenet and other adapted frames.
Curvature and torsion. Global questions for plane and space curves.
27 Jan ##Study the "twisted" cubic curve X:R -> R^3: t -> (at, bt^2, ct^3)
near t=0 for various values of (a, b, c). Note: a=0 is special.
(Draw pictures, projecting to the various coordinate planes. How
smooth is it, and how do you reconcile this with the pictures?)
##How smooth is f:R -> R where f(t)=e^{-1/t} for t>0 and 0 otherwise?
##Show that any interval I of the real numbers R is diffeomorphic to
either [0,1], R_+=[0,\infty) or R; here "diffeomorphic" simply means
there is a smooth map from I onto one of these (find it) with smooth
inverse; and show it's enough to find such a map with nonvanishing
derivative (implicit/inverse function theorem in dimension 1).
29 Jan ##If X(t) is smooth curve with X'(t) nonzero, show that the unit
tangent field T(t)=X'(t)/|X'(t)| is also smooth (or, if X is only
C^k, then T is C^{k-1}). What about the converse?
##Explicity parametrize the path of an "abrupt right turn" (the
union of the negative x2-axis and the positive x1-axis, with the
"turn" at the origin X=(x1,x2)=(0,0)) so that it's a C^k-smooth
parametrization. How about a C^\infty parametrization? (Compare
with the a=0 case – the "cusp" – of the "twisted" cubic.)
03 Feb Closed (periodic) curves X:R —> R^n with X(t)=X(t+1) can be viewed
as curves defined on the (abstract) circle S^1 = [0,1]/0~1.
Discuss: the length of a polygon (sum of edge lengths) & inscribed
polygons for a C^0 curve (index increases with parameter). Define
length of a C^0 curve as supremum of lengths of inscribed polygons.
If the supremum of the lengths over all such inscribed polygons
exists, we say X is "rectifiable" an define its length as this sup.
MR 1.6 & 1.10 show that a C^1 curve X is rectifiable and that its
sup length equals integral of its speed |X'| over the domain of X.
05 Feb Or... Define length of C^1 curve X as the (Riemann) integral of |X'|.
##Tedious exercise: Show these definitions agree for C^1 curves,
first by estimating how close the length of an inscribed polygon
approximates the corresponding Riemann sum for the integral of the
speed |X'|, and then by estimating how well Riemann sum approximates
this integral (your estimate may be easier and more explicit if you
assume X is C^2 with an explicit bound on the acceleration |X"|; see
MR sections 1.2 and 1.3, especially MR items 1.10, 1.11).
[Note: Jack, Matt and Nathan dropped by to ask about this exercise
above, which seems tricky and not-easy, as well as tedious! After
some discussion and false starts, we decided that a good strategy
would be to show: (A) Given any inscribed polygon \P, we can find a
polygon \P_N with vertices at X(k/N) (for k=0,1,...,N) such that
($) length(\P_N)≥length(\P)-\eps(N), where \eps(N)—>0 as N—>\infty
(the idea is that the vertices of \P are well-approximated by a
subset of vertices for \P_N, which works even for a C^0 curve X);
since (why?!) length(\P_N)≤sup(length(\P))=length(X([0,1])), and
assuming it's finite, ($) implies length(\P_N)—>length(X([0,1])).
(B) For any k, the intermediate value theoremm for each component
x_i' of X' yields a t_i in [k/N,(k+1)/N] with ($$) x_i'(t_i)/N=
x_i((k+1)/N)-x_i(k/N) (here we need X is C^1); then argue (again
use X is C^1, or for a more explicit estimate, assume X is C^2 so
that x'_i oscillates≤c/N over [k/N,(k+1)/N] where c=max_[0,1]|X"|)
if we move all the t_i to an endpoint of this interval, say t=k/N,
we can estimate ||X((k+1)/N)-X(k/N)|-|X'(k/N)|/N|≤c_n/N^2 (we use
($$) n times since X(t) in R^n, so c_n depends on c & n); now sum
from k=0 to N-1 to see that the Nth Riemann sum is within c_n/N of
length(\P_N).... (Aubrey made a good suggestion in class to use
the intermediate value th'm, but one must use care since it's NOT
TRUE that we can find t in [a,b] so that (b-a)X'(t)=X(b)-X(a).)
?!?!?!?MAYBE ONE OF YOU CAN FIND AN EVEN BETTER ARGUMENT?!?!?!?]
##Easy exercise: compute the length L of the cusp X:[0,1] —> R^2
where X(t)=(t^3,t^2); more generally, compute its length function
s(t) := length(X([0,t])) and its derivative s'(t)....
##Tricky exercise: ... you should see s'(t)>0 for t in (0,1]; and
since s=s(t) maps [0,1] onto [0,L], the inverse function theorem
shows there's an inverse function t=t(s); can you explicitly find
t(s) and the arclength reparametrization Y:[0,L]—>R^2, Y(s)=X(t(s))
for the cusp? (Extra credit: How smooth is t(s) at s=0?)
Discuss the Koch snowflake K. It's a limit of polygonal loops
K_n, where K_0 is an equilateral triangle of sidelegth 1, and
where K_{n+1} is obtained from K_n by replacing the middle third
of each edge of K_n by two more edges of an "outward" equilateral
triangle. Observe that K is not rectifiable - not even locally!
[For a Koch animation: en.wikipedia.org/wiki/Self-similarity]
Quick discussion of measuring length(K) by covering K with balls
of a given radius and counting how many are needed as this radius
tends to 0; this number grows like 1/radius if K is a rectifiable
curve, and otherwise leads to notion of Hausdorff measure & dim -
a possible project topic! (Try this for the Koch snowflake K.)
10 Feb Oriented curve is a parametrized curve modulo orientation-preserving
diffeomorphism (reparametrization) of the domain intervals.
Length of a C^1 curve is independent of C^1 reparametrization (chain
rule and change of variables formula for interval). Same is true for
a rectifiable C^0 curve and C^0 reparametrization (homeomorphism of
the domain intervals).
##Check the second statement above [it requires no calculation, just
the fact that the ordering of vertices of an inscribed polygon is
preserved (or reversed) if the homeomorphism is orientation-preserving
(or reversing) — and the definition of length as the sup...].
Proposition: Immersed C^1 curve X(t) has unit speed parametrization
Y(s)=X(t(s) with C^0 unit tangent vector T, unique up to orientation
and choice of starting point t(s=0)=0.
Proof sketch: "Solve" the differential equation s'(t)=|X'(t)| [just
integrate the given function |X'(t)|]. Apply inverse function th'm
[since s'(t)>0] to get t(s). Then compute Y'(s) [via chain rule].
[Discussion with Jarred suggested a simpler way to think about "up
to orientatation..." above: if S(t) is another arclength parameter
for the curve, then S'(t)=±s'(t); so writing S(t(s))=S(s) via the
inv fcn thm, the chain rule gives S'(s)=±1, and thus S=±s+S(0).]
Corollary: The "cusp" fails to be immersed at (0,0) with repsect to
any parametrization (because T changes sign there).
##Check the details of the above three things!
12 Feb Finish discussion of length (at long last ;-)!
##[Optional challenge question] Decide what extra conditions
on a C^k curve X(t) make the arclength function s(t) also C^k.
[It's true for an immersion, since |X'(t)|=s'(t)>0, and IFT
can be applied. Last class Aubrey warned about an s(t) with
s'(t)=|t| so that s(t) is C^1 but not C^2, and Shashank noted
the curve X(t)=(t^2/2,0,...,0) gives such an example....]
##Work out the u.s.p (unit speed parametrization) for the helix
X(t)=(r cost, r sint, qt).
The unit tangent vector defines a curve T:I -> S^{n-1} for curves
X:I -> R^n. Can recover X (up to translation) from T, again by
integration. If T is periodic, X may not close, but have a "period"
(the "translation" or "constant vector of integration").
##Check that helix above is an example. What's its period vector?
Give other examples of curves X in R^2 and in R^n with periodic T but
nonvanishing periods.
18 Feb [Tuesday is a Monday schedule, thanks to Washington and Lincoln! ;-]
Helices (and their limits, circles and lines) are important curves
in R^3 because they are orbits of Euclidean motions (1-parameter
subgroups of the Euclidean group E(3)). Discuss in class how E(3)
is a group generated by reflections in planes, and how to represent
via certain 4x4 matrices with [0 0 0 1] in the bottom row, with the
rotation/reflection part is 3\times3 block above the 3 zeros, and
translation part A=(a1, a2, a3) is first 3 rows of the 4th column.
If X(s) is a C^2 immersed curved in R^n, parametrized by arclength,
its velocity vector X'(s)=T(s) is the unit tangent vector, and its
curvature vector is its derivative K(s):=T'(s); thus K(s)=X"(s) is
the acceleration vector with respect to (any) arclength parameter.
Differentiating 1= shows K(s) is perpendicular to T(s).
##Use a u.s.p. for helix above to compute its curvature vector K(s)
and discuss various limits as r or q approach 0. [We saw in class
that |K| is a constant depending on r and q; if both approach 0,
their relative rate is important to see how |K| behaves.]
19 Feb ##For any C^2 curve X(u) in R^n, and any C^2 reparametrization u(t),
verify that the reparametrized curve Z(t)=X(u(t)) satisfies
Z"(t) = u"(t) X'(u) + (u'(t))^2 X"(u). Therefore, if u=s happens to
be an arclength parameter, the acceleration of X(s(t)) with respect
to the parameter t is a linear combination of T(s(t)) and K(s(t))
with coefficients (tangential acceleration) and (speed squared).
##In particular, to compute the curvature vector K at a point of an
immersed curve X(t) as the acceleration X"(t), it suffices that
|X'(t)|=1 and |X'(t)|'=0. Check that the second condition holds at
t=0 for the twisted cubic (at, bt^2, ct^3); use this idea to compute
K(0) in terms of a, b, c.
##Expres the curvature vector K for an arbitrarily parametrized curve
X(t) in R^n in terms of X' and X" (it involves projection of X" to the
hyperplane normal to X'). Try the case of curves in R^3 first, then
specialize to the R^2 case. (In R^3, the cross product can be used to
express projection of a vector W to the plane perpendicular to nonzero
vector V via proj_{V^\perp}(W) = (V \timesW \times V)/|V|^2.)
24 Feb All of the above means that the curvature is "geometric" - i.e. K is
independent of the parametrization. How does K change under rotation
or translation in space? Under scaling? Observe |K| has "dimension"
of 1/Length.
The osculating circle \O_p at a point p=X(0) on a C^2 curve X(s) makes
this "geometric" idea concrete: \O_p is the "best fitting" circle at p,
meaning it passes through p=X(0) with same unit tangent vector T=X'(0)
and curvature vector K=X"(0). Let \L_p=\{p+tT:t\in\R\} be the tangent
line at p, and \H_p=\{\L_p+uK:u≥0\} the halfspace containing \L_p & K;
then \O_p is the circle of radius 1/|K| centered at p+K/|K|^2 in \H_p
(note that K=0 iff \O_p=\L_p=\H_p).
##Work out the osculating circle at any point on the helix, and also
at the point (0,0,0) of the twisted cubic. [Later we noted these are
tilted, and that the \H_p slices the cylinder in an ellipse, whose
osculating circle in \H_ must agree with \O_p for the helix. Even
later we mentioned a century-old theorem of A. Kneser (the osculating
circles of a plane curve with monotone signed curvature "foliate" a
neighborhood of the plane — in particular, \O_p is disjoint from \O_q
if p≠q), and wondered if this could be used to prove something about
the collection of osculating circles on helix using Cici's nice view
(bird's-eye or worm's-eye, depending on your taste ;-) — yet another
great topic to explore if you're more "research-oriented"!]
If the curve is less smooth, osculating circles can be defined using
triples of distict points on the curve: any such triple defines a
circle or line in \R^n, and as triple of points converge to a point p
on the curve, any limit circle (or line) is an osculating circle at p.
FROM NOW ON (unless stated otherwise), WE ASSUME A CURVE IS IMMERSED!
##Describe the set of osculating circles at the point (0,0) of the
C^{1,1} curve X(t)=(t,t|t|)=(t,±t^2) if ±t≥0. [Hint: there are two
"extreme" osculating circles (that are limits of the unique osculating
circles at X(t) for t<0 or t>0, where X is C^2) and more "in between"!]
A C^1 curve X(s) is a C^1 graph over its tangent line \L_p at p=X(0)
for some small interval (s_-,s_+) around s=0 where T(s)•T(0)>0; if X
is C^2 (or C^{1,1}) with a curvature bound |K(s)=T'(s)=X"(s)|≤c, we
argued (using the geometry of the unit sphere) that T(s)•T(0)>0 is
true for at least |s|<π/2c.
##Compute the curvature vector when X(x) is expressed as a normal graph
over its tangent line \L_p at p=X(0). [Hint: without loss of generality
(by translation and rotation), assume X(x)=(x,x_2(x),...,x_n(x))=(x,y(x))
with y(0)=y'(0)=0 in \L_p^{perp}=\R^{n-1}, so p=X(0)=0, T=X'(0)=e_1, and
\L_p is the x_1-axis; x=x_1 is an arclength parameter along axis/tangent
line, but not necessarily a u.s.p. for the curve X(x); the unit tangent
vector T(x)=X'(x)/|X'(x)|=(1,y'(x))/\sqrt{1+|y'(x)|^2}, so (by the chain
rule and the fact that dx/ds=1/(ds/dx)=1/\sqrt{1+|y'(x)|^2})
K(x)=dT/ds=T'(x)dx/ds=((1,y'(x))/\sqrt{1+|y'(x)|^2})'/\sqrt{1+|y'(x)|^2})
=(0,y"(x))/{1+|y'(x)|^2}-(1,y'(x))y'(x)•y"(x)/{1+|y'(x)|^2}^2
which you should check, simplify, note K(x)•T(x)=0, and observe that two
terms in your more general formula (n>2) will cancel when n=2, giving a
formula you should recognize from calculus in case of a graphical curve
X(x)=(x,y(x)) in R^2 (as y(x) takes values in R^{n-1}=R^{2-1}=R^1=R).]
Schur's Lemma sharpens the normal graph estimate above: a curve
of a given length L≤π with |K|≤1 must have endpoints at distance at
least 2\sin(L/2) with equality iff the curve is a great circle arc.
[This might be part of a good topic for a project!]
26 Feb For a curve X(s) in R^2, there is a canonical extension of the unit
tangent vector field to an orthonormal frame field T, N=JT spanning
(the tangent space of) R^2 at any point along X. (Here J is just
the 2x2 matrix that rotates R^2 c.c. by a quarter-turn.) Thus we
express the curvature vector K=kN for a function k(s)=K(s)•N(s)
[some may call this \kappa, but I'll use lower case here], the
signed curvature of a plane curve.
Unlike the curvature vector K(s), the sign of k(s) flips when the
orientation of the parametrization flips (making the signs of T(s)
and N(s) flip)!
##Which plane curves have k=0? Which have constant k≠0? What can
be determined about X(s) from k(s) in general? [Hint: the ordinary
differential system T'(s)=k(s)N(s)=k(s)JT(s) can be solved by
integrating k(s)=\theta'(s) to get \theta(s) (up to a constant) and
T(s)=(\cos\theta(s),\sin\theta(s)) (up to rotation); and X(s) can be
recovered (up to translation) by integrating T(s)=X'(s).]
This can also be done using the moving frame for X(s) in R^2, namely
the 2\times2 matrix U(s)=[T(s)|N(s)] as follows:
##Verify the 2x2 matrix ODE for the frame U'(s)=U(s)Jk(s), where J
is the usual c.c. quarter-turn matrix. Solve this first order ODE.
[Hint: you should get U(s)=U(0)exp(J \int k(s) ds), integrating over
the interval [0,s]). Recall the exponential of any square (2\times2
here) matrix A is the usual power series exp(A)=I+A+...+A^k/k!+....]
Back to periods: we saw above that we have k(s)=\theta'(s) where
T(s)=\cos\theta(s)e_1+\sin\theta(s)e_2 is unit tangent vector; we
can recover T (up to initial angle \theta(0)) from integrating k(s),
and recover X (upto translation by initial position X(0)) from
integrating T.
##If X is periodic (i.e. a closed curve), T and k are also periodic;
what about the converses?
We saw that the total curvature \int k(s) of a closed curve in R^2
must be an integer multiple of its turning number w, the number of
times the unit tangent vector T winds around the circle S^1. Note
that the sign of w flips if the orientation of the curve flips.
For a curve X in R^2 with periodic curvature function, if the total
curvature (integral over a full period) is an integer multiple of
2\pi then T is periodic. What additional conditions guarantee X is
periodic, i.e. for X to be a closed curve?
##Draw pictures of (oriented) plane curves illustrating the various
values of w (among small integers, including w=0, which is the most
interesting case).
"Two wrongs don't make a right, but three lefts do!" [Remind me to
discuss this next week, in the context modern highways and turning
number w — which could also stand for Whitney, who proved a very nice
theorem about turning numbers and topology of immersed curves in R^2
— another great project topic!]
======Watch my "Coronavirus University" videos as they appear on YouTube======
[Some above material, and much remaining material, will be further revised as
we do our best to make this an on-line course: over the break I set up a home
studio in order to make short YouTube videos (so far about 10 URLs have been
shared "privately" with current students) under the "Coronavirus University"
(TM and copyleft 2020 by Rob Kusner ;-) and a 563hw.html page with links to
these videos is now under development (23 Mar 2020).]
02 Mar A k-frame in R^n is an ordered basis for k-dimensional subpace of R^n
(think of an n\times k matrix, whose columns are the basis vectors);
the k-frame defines an orientation of the subspace (in case k=n this
is the sign of the determinant, but for k R^2 has k≥0. For a
simple (i.e. injective) curve, try to argue the converse. Can you
find a non-simple (thus non-convex) plane curve with k>0. (Recall
that a curve is convex if it lies to one side of its tangent line;
the ideas in MR exercise 1(8) might be helpful.) [Hint: you might
do the converse by first showing there is a pair of points p1, p2
on X with a common tangent line, but where the arc of X between
p1, p2 is not a segment; then use the previous problem to show
(under the assumption of X being simple) the arc has zero total
curvature (show the "convexified" curve, gotten by substituting
the segment for the arc also has total curvature 2\pi); thus the
curvature changes sign on such an arc (contrapositive argument).]
You might also try the series of MR exercises 1(9-17) deal with
special properties of plane curves Y obtained by perturbing a plane
curve X. One nice result (not among these exercises) is a theorem
of Archimedes:
If X is a closed convex curve and Y is any closed curve
surrounding X then Length(Y)>=Length(X), with equality
iff Y = (a reparametrization of) X.
This generalizes the fact that straight lines are shortest paths
between points (think about the limiting case where X is a back-and-
forth line segement). [This topic might be part of a final project.]
##If X is closed curve in R^2, what are possible values for total
curvature \int k(s) ds? What about a simple (injective from S^1)
closed curve? (Try the case of a convex plane curve first, then
"dissect" the region bounded by a simple curve into convex pieces.)
[A possible way to do this is to approximate X with a circumscribed
simple polygon: First show, for fine enough approximation, the total
curvature of X between adjacent points of tangency is simply the
turning angle between adjacent endges of this polygon. Next, we
need a way to account for the sums of these turning angles. That
can be done by triangulating the interior of the polygon, then
clipping away triangles from the boundary, checking that the total
turning angle doesn't change. In the end we have a single triangle,
with total turning angle 2\pi (if oriented counter-clockwise).]
[NOTE: the hw above and below is due AFTER spring break — the first
batch (the 02-09 March ## problems) should be submitted as .pdf
DIRECTLY TO THE COURSE GRADER by 11:59PM Sunday 29 March VIA EMAIL –
I will email the submission details to you later!!!]
======Watch my "Coronavirus University" videos as they appear on YouTube======
04 Mar The unit tangent T(s) of u.s.p. curve X(s) in R^n (n>2) traces a
curve on unit sphere S^{n-1} (the tangent indicatrix). Note that
T(s) is not u.s.p (what's its speed?)! The curve X(s) has total
(absolute) curvature \int|K(s)|ds = \int|T'(s)|ds = length(T).
For n=2 we have the (signed) curvature k(s)=K(s)•N(s)=T'(s)•JT(s)
so |\int k(s)ds| ≤\int |k(s)|ds = \int|K(s)|ds = length(T) since
there will be length cancellations as T(s) sweeps back (clockwise,
k<0) and forth (counterclockwise, k>0) around the "direction" S^1:
this means \int k(s)ds measures the "signed" length for the tangent
indicatrix as T(s) wraps around S^1. Another way to think of this
involves signed-length(T), which we called "happiness" in class:
if T(s) points "east" (toward e1) and k(s)>0 ("smiling") then this
contributes positively to signed-length(T); if k(s)<0 ("frowning")
it contributes negatively to signed-length(T); and (by the wrapping-
around-S^1 picture, or the change of variable formula) we see that
length(T)=2π(average over e \in S^1 number of X(s) with T(s)=e) and
signed-length(T)=2π(signed-average of this number).
##Fill in the details of the above, and as a challenge, try to prove
the following result (for "pi day" we may do this in class next week):
Note that, if X is closed curve in R^3, then T closed on S^2; but not
all closed curves on S^2 are tangent indicatrices, whose center of
mass (weighted by arclength of X, not of T) must be at origin because
\int T(s)ds = \int X'(s)ds = 0. This can be used to prove a theorem
of W. Fenchel (1940):
The total curvature of a closed curve X in R^3 (or in R^n)
is at least 2\pi, with equality iff X lies in a plane as a
convex curve. (For knotted X, ... > 4\pi - Fary & Milnor.)
[This might make a good expository project. A good research project
might be to explore how the Frenet frame (viewed as a curve in SO(3)
or its double cover S^3) geometry compares with that of the original
curve in R^3 - we might say more about this, especially a formula
relating total torsion (or Twist) to the geometry and topology of
closed curve - cf. Writhe and Self-Linking Number (below).]
Frenet-Serret frame along C^3 curve X in R^3 with |K|=k>0 defined by:
T, N=K/|K|, B=TxN. For plane curve, B=normal to plane. In general,
B(s) normal to osculating plane span(T(s),N(s)). Assume k>0 below.
As usual, taking derivative of inner-products ** and **** using
product-rule gives B'(s)=t(s)N(s) for some function t, the torsion.
##Compute the torsion of a (r,q)-helix: t=-q/(r^2+q^2)=constant.
Discuss limiting cases. (Later, you will prove the converse: if
curvature k=constant and torsion t=constant then X is a helix;
see 1.37, 1(26&27).)
Note (unfortunate, but MR's and many other authors') sign convention:
left-handed helix has positive torsion!
[NOTE: the hw above and below is due AFTER spring break — the first
batch (the 02-09 March ## problems) should be submitted as .pdf
DIRECTLY TO THE COURSE GRADER by 11:59PM Sunday 29 March VIA EMAIL –
I will email the submission details to you later!!!]
======Watch my "Coronavirus University" videos as they appear on YouTube======
09 Mar ##Express torsion in terms of X', X" and X"' for an arbitrary C^3
parametrized curve X(u) in R^3 (hint: B = X'x X"/|X' x X"|; now
differentiate and find the normal part; the numerator is triple
product - explaining computationally why we need
C^3 curves for Frenet. I would have picked the opposite sign!)
##Compute the torsion of the twisted cubic at 0.
##Show that any two C^3 u.s.p curves X, X* in R^3 "busculate" (agree
through third derivatives at X(0)=X*(0)) iff they have the same
tangent T(0)=T*(0) and curvature K(0)=K*(0) vectors, and the same
torsion t(0)=t*(0) and first derivative of curvature k'(0)=k*'(0)
[thanks to M. Feller (2012) for pointing out this last necessary (and
sufficient) condition]. In particular, at least at critical point
for curvature (k'(0)=0) [and perhaps more generally?], there is a
busculating twisted cubic (or helix, generalizing the osculating
parabola or circle) at any point of a C^3 curve. [Careful: if only
the curvature functions and first derivatives agree (k(0)=k*(0)>0,
k'(0)=k*'(0)) then the two curves agree only after a rotation in
the normal plane about the common tangent direction; and what will
happen if k(0)=k*(0)=0 - do we need to exclude this case?!]
Frenet formula can be written as columns of 3x3 rotation matrix:
[T|N|B]'=[T|N|B]A where A is a transform (note signs) of the skew
symmetric matrix with curvature k and torsion t given in class:
|0 -k 0|
A = |k 0 t|
|0 -t 0|
[The weird sign convention would be fixed if a right-helix had t>0!]
The same form holds for any curve of orthonormal 3-frames [E1|E2|E3]
where = 1 or 0 iff i=j or not. This is from applying product
rule to these inner product equations: '=+ so
A_ji = = - = -A_ij.
Another way to see this uses the matrix transpose: if U=[E1|E2|E3]
is a path in SO(3) and U* is its transpose, then differentiating
the identity U*U=I and using U*'=U'* gives U'*U+U*U'=0, that is the
matrix U*U'=-U'*U=-(U*U')* is skew symmetric. [Recall (XY)*=Y*X*
and Z**=Z!]
##The above can also be expressed: there is a vector field W(s) along
the curve X so that [E1|E2|E3]'=W(s)\times[E1|E2|E3] - cf. 1(26); the
field W(s) is called the angular velocity; if [E1|E2|E3]=[T|N|B],
W(s)=?, and show W(s) constant iff X is helix - cf. 1(27) & below.
(The vector W(s) is often named after Darboux....)
More on helices' curvatures and torsions; limiting cases of circles
and lines. Pitch p=q/r dimensionless; sign of q gives handedness.
Helices are orbits of screw motions (the generic 1-parameter group of
Euclidean symmetries).
##Exercise 1.37 [Hint: We already computed k and t are constant for
a helix; for the converse, one looks at K(s):= X"(s) = kN(s), so
(by Frenet formulas) K'(s) = -k^2T-ktB => |K'|^2 = k^2(k^2+t^2);
K"(s) = -k(k^2+t^2)N => k_K = |K"(s)|/|K'(s)|^2 = 1/k; and also
K"'=(some multiple you can compute)K', so triple product
and torsion t_K must vanish; thus K is planar curve with curvature
k_K=1/k lying on sphere of radius k => K (great) circle of radius k.
Thus can write
| cos \sqrt{k^2+t^2}s |
X"(s) = K(s) = k| sin \sqrt{k^2+t^2}s |
| 0 |
which integrates (twice) to give a helix X(s) = (??? in k and t);
should get r = k/(k^2+t^2) and q^2=t^2/(k^2+t^2) in old notation.]
[NOTE: the hw above and below is due AFTER spring break — the first
batch (the 02-09 March ## problems) should be submitted as .pdf
DIRECTLY TO THE COURSE GRADER by 11:59PM Sunday 29 March VIA EMAIL –
I will email the submission details to you later!!!]
======Watch my "Coronavirus University" videos as they appear on YouTube======
11 Mar Parallel (Bishop) frames, holonomy, comparison with Frenet frame:
the latter is globally defined as long as k>0, but how to frame
a general curve in R^3? Go back to plane curves, but now thought
of as curves in R^3 with constant binormal B (so B'=0) and torsion
zero (N'=-kT); in particular, the derivatives of normal fields N,B
are multiples of T.
We call any normal field V (along u.s.p curve X(s) in R^3 with unit
tangent vector field T=X'(s)) _parallel_ provided V'(s)=f(s)T(s)
for some function f. Check (compute ') that a parallel field
has constant length. We often express parallel by V'(s)^\perp=0.
Let N1(0), N2(0) be an orthonormal basis for normal space of X
at X(0). By solving initial value problem V'(s)^\perp=0 we get
a parallel normal framing N1(s), N2(s) along X.
##Let N1, N2 be parallel orthonormal frame of normal fields along X
in R^3; the orthonormal frame M1(s)=cos\a(s)N1(s)+sin\a(S)N2(s),
M2(s)=-sin\a(s)N1(s)+cos\a(s)N2(s) is parallel iff \a(s)=constant.
Now for the bad news: if we follow a parallel field V around a closed
curve X in R^3, say with X(0)=X(1), we may not have V(0)=V(1), but
instead V(1) will be rotated in the normal plane from V(0). This
rotation (or the rotation angle \a) is the _holonomy_ (of parallel
translation along X).
##Find an explicit closed curve with an abritrary holonomy angle \a.
Can you characterize those curves with holonomy \a=0? [Research?]
Parallel (or Bishop) frame rotates about the binormal B - if we think
of osculating plane span{T,N} as the "equatorial" plane, then N1 and
N2 are tangent to "parallels" ("latitudes" or "poles") - in fact, we
know Ni'=f_iT, so these point in same direction as T (i.e. both Ni'
are normal to the same "longitude" as well). [This is my attempt to
say in words what I tried to draw on the board in class today! ;-]
Compare now with Frenet: write N(s)=cos\phi(s)N1(s)+sin\phi(s)N2(s)
for some twist angle \phi(s) for [T,N,B] versus [T,N1,N2]; and so
B(s)=-sin\phi(s)N1(s)+cos\phi(s)N2(s). Show torsion t(s)=-\phi'(s).
In other words, \int t(s) ds is the total (signed) twist of the Frenet
frame with respect to any parallel frame.
Wrap-up on curves and their global properties. We already mentioned
the Fenchel-Fary-Milnor theorem: a closed curve in R^3 has total
curvature >= 2\pi, with equality only for convex planar curves, and
knotted curves have total curvature > 4\pi (the latter result already
brings us to the "edge" of surface theory - sorry for the pun - since
one way to characterize unknotted closed curves is to say they are
the boundaries of embedded 2-disks. This result dates to the 1940s,
though as recently as 2 decades ago there was new work on this by
Eckholm-BWhite-Wienholz (soap films...) and Cantarella-Kuperberg-
Kusner[yours truly ;-]-Sullivan (nonempty second hulls...).
What about total torsion? This is the theorem of Calugareanu-JWhite-
Fuller: Self-link = Twist + Writhe. What do these mean? Again, most
are best understood using surfaces, but we can say a bit about each:
Twist = (1/2\pi)\int t ds is the total torsion (a single integral),
Writhe = (1/4\pi)\iint /|X(s)-X(r)|^3 ds dr
is the double integral of this (normalized) triple-product, and
Self-link = the linking number Lk(X,X+\epsN) where X is the original
curve and X+\epsN is a "push-off" of X by small \eps>0 along Frenet
normal N to a nearby, disjoint embedded closed curve (so this depends
on the choice of framing: we've made the standard assumption k>0 to
ensure the Frenet normal framing N,B is defined globally) and Lk(X,Y)
is defined as the number of times X crosses over Y in a right-handed
way minus the number of left-handed crossings (here we assume X and
Y are oriented simple closed curve, and when Y=X+\epsN we orient it
the same way we oriented X; we also assume a planar projection in
which we can depict the under- and over-crossings).
##Sketch some links X \union Y with Lk(X,Y) small.
##Check that Lk(X,Y) doesn't change as we deform X and Y so that they
remain disjoint and embedded. In particular, by flipping the plane
of projection, we get Lk(X,Y)=Lk(Y,X). (Gauss worked out a double-
integral formula for Lk(X,Y) which is essentially the same as that
for Writhe, but with Y(r), Y'(r) in place of X(r) and X'(r).)
======Watch my "Coronavirus University" videos as they appear on YouTube======
Part 2 Surfaces: parameterized and implicit; embedded (or immersed) in R^3
and R^n. The first and second fundamental forms. Length and area.
Mean and Gauss curvature.
23 Mar Just as we focused on immersed curves, the notion of a surface being
regular (or immersed) is very important, and I want you to explore it
in the following material:
A surface patch X:U->R^n for U\subset R^2 is a homeomoprhism to its
image and (if X is C^1) satisfies rank(DX)=2, i.e. X_u, X_v linearly
independent. For general n this means some 2\times2 minor in the
n\times2 matrix DX has nonzero determinant; for n=3, this means the
cross product X_u\times X_v is nonvanishing, so we can normalize and
define the (C^0) tangent map (Gauss map, unit normal) \nu:U->S^2 by
\nu = X_u\times X_v/|X_u\times X_v|; clearly \nu(u,v) is perpendicular
to the tangent plane span(X_u(u,v), X_v(u,v)) to the surface patch at
X(u,v).
##As we did for unit tangent vector for curves, check that if X
is C^k then \nu is C^{k-1}.
##Find a "cusplike surface patch" X which is C^1 but has rank(DX)=1
at a point (in other words, X_u and X_v are parallel but not both
zero); similarly, find a "conelike surface patch" where rank(DX)=0
at a point (i.e. both X_u and X_v must vanish). [Neither of these
is a (regular, immersed) surface patch, but sometimes it's good to
expand one's point of view. See the torus problem below. Hints:
The map X:R^2 -> R^2: (u,v) -> (u^2,v) is C^1 (in fact, smooth) but
rank(DX)=1 at (0,0). This isn't the example, but should give some
ideas. More directly: How would you parameterize a cone near its
vertex? How would you pametrize a surface of revolution made by
revolving a circle (other regular smooth curve) that's tangent to
the axis of revolution (that point of tangency becomes the cusp)?]
##Find patches covering a torus T^2=S^1\times S^1 of revolution in
R^3 which is a special case of a tube X^\eps around a closed curve:
if X*:S^1->R^3 is C^2 and 0<\eps<1/k_max(X*)=maximum curvature of X*
then X(u,v)=X*(u)+\eps(cos(v)N1+sin(v)N2), where [N1|N2] is a normal
framing (like [N|B] for the Frenet frame) of X*, is a doubly periodic
map R^2->R^3 (and thus a map from the torus). The special case
where X* is a circle of radius r>0 is worth checking explicitly:
X_u and X_v are linearly independent when r>\eps, and when r=\eps
there is one "cusp" point where rank[X_u|X_v]=1. What's the least
number of patches needed for a torus? [Hint: this will depend on
whether the patches are disk-like or annular!]
The sphere S^2 is covered by a "quilt" (atlas) of 6 surface patches
X^N, X^S, X^W, X^E, X^F, X^B which map an open unit disk onto the
various open hemispheres (X^N to the northern hemisphere, ... , X^B
to the back hemisphere). Each of these patches is graphical over
the tangent plane over the correponding "pole" (we have the north,
south, west, east, front & back poles, where the standard axes of
R^3 meet S^2).
##What's the fewest hemispherical patches needed to cover S^2?
(This is a discrete geometry problem: clearly neither 1 nor 2
are enough; are 3?)
If two coordinate patches overlap on the immersed surface, they
rise to a change-of-coordinates between domains in R^2. This is
the foundation for the idea of "intrinsic" surface (or manifold).
##Use stereographic projection to quilt S^2 with 2 patches X^N and
X^S (this is the minimum number, since S^2 is not region of R^2).
[Hint: recall that stereographic projection from a pole P takes the
punctured sphere S^2\setminus{P} to R^2, and its inverse gives a
patch X^P. (In class we point out that this works for S^n and
that everything follows from the n=1 picture. By using similar
triangles, we found the formula for the projection; for example,
we have X^N:R^1->S^1:t->(2t,t^2-1)/(t^2+1) - this rational map
parametrizes rational points on S^1 a.k.a. "Pythogorean triples"!)
For n>1, think of t as a vector, e.g. t=(u,v) for a surface....
What you need to check is that both X = X^N and X^S are immersions.]
In the lecture we began computing X_u\times X_v — you should finish
that and check that it is a multiple of X itself!]
[NOTE: I've successfully used Zoom for the first time to discuss
final projects (and some 563hw problems) with a small group of
students (24 March 2020), and hope a larger group will agree to
meet for a "Zoom party" some evening this week (I'll send an email
invitation if and when this is set). As far as 563hw, the next
batch (the 11-25 March ## problems) should be submitted as .pdf
DIRECTLY TO THE COURSE GRADER by 11:59PM Sunday 05 April VIA EMAIL –
I've already emailed the submission details to you!!!]
======Watch my "Coronavirus University" videos as they appear on YouTube======
25 Mar Here's why I want you to understand stereographic projection
better - it lets one go back and forth between R^n and S^n to
study the "bending" or Willmore energy W for surfaces:
Veronese embedding of S^2/{+,-} into R^6 (or R^5 or S^4 or R^4 via
stereographic projection) uses quadratic functions of linear coords
on S^2 (see !Newsflash! on my webpage for its importance). Also Boy's
surface in R^3 (see: http://en.wikipedia.org/wiki/Boy's_surface, or
http://www.mfo.de/general/boy/, or - for a more familiar face -
http://owpdb.mfo.de/detail?photo_id=12131)!
##Check that the composition of the patches {X^N,X^S}, {X^E,X^W}
and {X^F, X^B}: D\subset R^2 -> R^3 for S^2
(u,v) —> (x,y,z) [in my YouTube lecture (u1,u2) —> (x1,x2,x3)]
and the quadratic map R^3—>R^5
X=(x,y,z) —> Q=(q1=xy, q2=x^2-y^2, q3=yz, q4=y^2-z^2, q5=xz)
is an immersion, and if we consider as a map on antipodal pairs in
S^2, is actually 1-1. (We could also include a q6=z^2-x^2 in Q, so
QX would lie in the 5-dimensional hyperplane {q2+q4+q6=0} of R^6.)
[The above is one version of the "Veronese" surface, but clearly
we can adjust coefficients and change the shape a bit. I may say
more about these immersions later.]
[As you plan your final projects, please keep up with 563hw problems:
next batch (the 11-25 March ## problems) should be submitted as .pdf
DIRECTLY TO THE COURSE GRADER by 11:59PM Sunday 05 April VIA EMAIL –
I've already emailed the submission details to you!!!]
[Also, we had another "Zoom party" (hosted by Skandesh, to which you
were all invited) at 9PM Saturday 28 Mar to discuss final projects
(and a few hw/YouTube lecture questions). I urge *each of you* to
host (at least) one "Zoom party" before the semester ends, and I want
all of you to participate in many of these+!!!!!!!!–!!!!!!!!!!!!!!]
======Watch my "Coronavirus University" videos as they appear on YouTube======
29 Mar Finish discussing how to think of RP^2=S^2/{+,-} "intriniscally":
it's built from 3 patches, each corresponding to the pairs of patches
{X^N,X^S}, {X^E,X^W} and {X^F, X^B} for S^2.
##Fill in the details of the picture we made that RP^2\{N,S} is a
Moebius band build from the first two patches {X^E,X^W},{X^F, X^B},
and thus the third patch {X^N,X^S} "attaches" a disk along
the "border" of this band to yield all of RP^2.
Consider a C^1 surface patch X:U–>R^n. Its derivative matrix DX
matrix has 2\times 2 minors M_ij indexed by pairs of rows, giving
rise to (n choose 2) determinants µ_ij. The (ordered) set µ of
these generalizes the cross product X_u \times X_v, and when n=3
(µ_23, µ_31, µ_12)/|"| gives the unit normal \nu.
##Work out µ for the torus S^1\timesS^1 in R^4=C^2 defined by
X(u,v):= (cos u, sin u, cos v, sin v) = (e^2πiu, e^2πiv).
======Watch my "Coronavirus University" videos as they appear on YouTube======
01 Apr
=========================!!!being revised below!!!=========================
=========================!!!being revised below!!!=========================
[Extra remarks:]
Curvature and length.
#?Use Cauchy-Schwarz ≤ to show line segments minimize length. Try
giving alternate proof via (iterated) triangle inequality [MR 1.9].
(Generalize to Archimedes Theorem about curves surrounding a convex
curve?!)
Discuss the possible domains I for "complete" and "incomplete"
u.s.p. curves (this uses Cauchy sequences, i.e. {x_i} in I with
the property |x_i - x_j| is arbitrarily small for i,j arbitraily
large; completeness means {x_i} has a limit in I also).
Any closed interval of R is complete, as is R itself, but a bounded
(half) open interval is NOT complete. We agree I=S^1(L)=[0,L]/0~L
(circle of arclength L) also qualifies as a complete domain curve.
Are there any others (why or why not)?
This is, in effect, The Classification of (connected) 1-Manifolds
[see J. W. Milnor, Topology from the Differentiable Viewpoint]!
Recall the definitions of length. For a C^2 curve, the estimate in
MR 1.6 can be improved, as we hint in class: if |P|<\eps there is
a constant C (depending linearly on (|K|^2)_max Length(X) where K
is the curvature of X) such that |Length(P)-Length(X)| S^2/{+,-} = RP^2 is a covering projection
and that Veronese surface "factors through" this. Also we see
(with paper models) that an annular neighborhood of a great circle
projects to a M\"obius band. The antipodal map x -> -x is the deck
transformation, interchanging the two "decks" or "layers" which lie
over a given neighborhood of a point [x~-x] in RP^2.
We also introduce the notion of a fundamental domain ("patch")
for a surface. The upper hemisphere is one for the real projective
plane; a square in R^2 is one for the torus. But we saw there are
many more f.d.s (puzzle pieces assebled by deck transformations -
in case of the torus, translations by the subgroup Z^2 of R^2).
#Find fundamental domains in R^2 for other surfaces (later we will
see that most of these are more naturally viewed as being endowed
with hyperbolic rather than euclidean geometry) and see if you can
describe the deck transformations.
We introduce (mod 2) homology of closed curves on a surface S:
a and a' are homologous (a~a') if there is "subsurface" R of S with
boundary \dR = a+a'. This led to a Z/2 vector space H_1(S,Z/2) or
an abelian group (if we work over Z and keep track of orientations).
There is also a Z/2-quadratic form on this group, given by counting
intersection number (mod 2) of curves. (The subsurface R defining a
homology can be "singular" i.e. not immersed or embedded.)
#Check (by drawing pictures) that for a genus g surface S_g (connected
sum of g tori) we have dim(H_1(S_g,Z/2))=2g and that the intersection
form has matrix (with respect to a basis a_1,...,a_g,b_1,...,b_g of
curves where a_i meets b_i once and otherwise disjoint):
| 0_g I_g |
| I_g 0_g |
(and, over Z, one of the I_g blocks has a minus sign - use the disk
with bands picture of S_g \ disk introduced in class).
#Check (by drawing more pictures) that a nonorientable surface N_p
(connected sum of p RP^2s) has dim(H_1(N_p,Z/2))=p and that the
intersection form has matrix I_p (in a basis c_1,...,c_p of curves
whose neighborhoods are M\"obius bands - use the disk with half-
twisted bands picture of N_p \ disk to see c_i meets c_i once).
We introduce the fundamental (Poincare) group \pi_1(S) of a surface
S (with basepoint p): this the group generated by (oriented) loops
starting and ending at p with the relation that loops a and a' are
homotopic if the product (concatenation) a(a')* (in class used "bar"
instead of "*" to reverse orientation of the curve) is boundary of an
annulus A mapped into S (the mapping of A needn't be immersion...).
We sketch (last time, for case of the torus) why \pi_1(S) acts as
deck transformations on the universal covering space, translating one
fundamental domain to another. We also sketch why H_1(S,Z) is the
abelianization of \pi_1(S).
#Check that \pi_1(S^2)=\pi_1(R^2) is trivial group, \pi_1(RP^2)=Z/2,
and \pi_1(A)=\pi_1(M)=Z where A is annulus and M is M\"obius band.
#Check \pi_1(S_g)=group generated by a_1,...,a_g,b_1,...,b_g with
one relation [a_1,b_1]...[a_g,b_g]=1 (product of commutators, where
[a,b]=aba*b* - used "bar" instead of * in class, to denote inverse);
this is non-abelian if g>1.
#Can you figure out \pi_1(N_p) for p>1 (it's also non-abelian; the
case p=2 is Klein bottle, so may be useful to think of square with
edges identfied by deck transformations to work out \pi_1(N_2))?
[The preceding three problems could be a final project instead!!!]
Deck transformations of torus are translations z -> z + b of C=R^2
for b in a lattice Z^2=Z+iZ (or more general parallelogram) of C.
More general Euclidean motions are of the form z -> az + b, |a|=1.
We sketch a synthetic picture of hyperbolic plane (as unit disk
or upper-half-space \H in \C with "lines" as circles meeting boundary
orthogonally) and asserted that fractional linear maps
z->(az+b)/(cz+d) for a,b,c,d real, ad-bc=1
give hyperbolic motions of \H which take "lines" to "lines." In this
way, the octogon fundamental domain for S_2 is realized in \H and the
nonabelian group of deck transformations can be represented by 2x2
real matrices (SL(2,R) or PSL(2,R)=SL(2,R)/I~-I) as above. (More
general representations into PSL(2,C) treated in the book "Indra's
Pearls" by Mumford, Series and Wright lead to the study of surfaces
as complex curves or "Riemann surfaces" - also a good project!!!)
A surface S with boundary \dS has patches modeled on open sets of
the (closed) half-plane \R x [0,\infty), with points on \dS coming
from points on \d(\R x [0,\infty))=\R x {0}. Sketch pictures of
orientable surfaces with boundary for small genus g, ends e, and
boundary components b (may want to distinguish bounded components
of \dS from unbounded ones). Similarly for non-orientable ones.
We talk a bit about transversality: if surfaces S and S' meet
such that every point p in the intersection has the property that
their respective tangent spaces T_pS+T_pS'=R^3, then S intersects
S' in a regular curve.
This is the foundation for the "movies" we discussed a while back,
where S'=S_t is a family of level surfaces of a R-valued function
F on R^3 (with non-vanishing gradient), such as the height F(x)=
along the direction a:
The surfaces S_t above meet a fixed surface S transversally iff
t is a regular value of the restriction f of F to S. Typically,
the critical values t are where the level sets of this function f,
such as the height function, have a qualitative change. If there
are no critical values between a pair of regular values t1 and t2,
then the intersections of S_t are equivalent for all t in [t1,t2]
(this is the "fundamental lemma of Morse Theory" and is behind the
M\"obius classification of surfaces we sketched last week).
[End of a topological interlude....]
============================================================================
This appeared on arXiv awhile ago! Perhaps part of a project?!?
============================================================================
arXiv:1303.6448
Date: Tue, 26 Mar 2013 12:05:10 GMT (2195kb)
Title: Make your Boy surface
Authors: Eiji Ogasa
Categories: math.GT
Comments: 14pages, 15 figures
\\
This is an introductory article on the Boy surface. Boy found that RP2 can be
immersed into R3 and published it 1901. (The image of) the immersion is called
the Boy surface after Boy's discovery.
We have created a way to construct the Boy surface by using a pair of
scissors, a piece of paper, and a strip of scotch tape. In this article we
introduce the way.
\\ ( http://arxiv.org/abs/1303.6448 , 2195kb)\\
============================================================================
============================================================================
Mar [... Finally: back to geometry!]
We work out the derivative D\nu of the Gauss map \S -\nu-> S^2
for the case \S is the cylinder parametrized by (a fundamental domain
in) R^2 -X-> \S by the formula (u,v) -> (cos u, sin u, v) and show
how to think of this as a 2-by-2 matrix and as a quadratic form
(the second fundamental form) which records the normal curvature
in a given direction W in T_pS, i.e., the curvature of the (plane)
curve S intersected with the plane spanned by W and \nu(p).
##Do the same calculations in case \S is i) a plane, ii) a sphere of
radius r, iii) the graph (u,v, h(u,v)) of a function h, especially
for a quadratic function h(u,v) = au^2 + bv^2 at (0, 0, 0).
Show shape operator A is symmetric (with respect to the metric <,>)
since the associated 2nd fundamental form B = = -
is symmetric (i.e. B(v,w)=B(w,v) for any tangent vectors v,w), where
Hess(X) is vector of Hessians (second derivative forms) of the
coordinate components of X.
[Some of you already computed B for the sphere, and - thanks to the
confusion I may have caused above - conflated it with the shape
operator A, which on the sphere is 1/r times the identity, since the
sphere is totally umbilic (with principal curvature 1/r). The
clarification needed is discussed below!
Although in class, I pointed out that the second fundamental form
B(v,w) = can be computed by taking B = -, just a
little care needs to be taken when converting this into the shape
operator A. Indeed, if we let b denote the matrix
B(X_u,X_u) B(X_u,X_v)
B(X_v,X_u) B(X_v,X_v)
then the matrix for A (with respec to the ordered basis X_u, X_v for the
tangent plane) is a = g^{-1}b where g^{-1} is the inverse of the matrix
g for the "first fundamental form" in this ordered basis
Note that when X_u, X_v is an *orthonormal basis*, this confusion goes
away, since then g is just the identity matrix. In fact, the trace of
a quadratic form B (as we need for defining H, or rather 2H) is always
computed this way: either sum B(U,U) + B(V,V), evaluated on any
orthonormal basis {U,V}, or take the trace of the associated matrix
a=g^{-1}b. I hope this clears things up - it's so fundamental! ;-]
Review some linear algebra of linear maps and their eigenstuff. Symmetric
bilinear and quadratic forms. The main fact we need: if a 2x2 matrix
r s
A =
s t
is symmetric, then it has real eigenvalues and a corresponding basis
of orthonormal eigenvectors. Eigenvalues k1, k2 (the principal
curvatures in case A = a = b is the matrix for second fundamental form
taken with respect to an <,>-orthonormal basis, i.e. g=I_2)
are roots of the characteristic polynomial
c_b(t) = t^2 - (r+t)t + (rt - s^2)
whose disciminant (r+t)^2 - 4(rt - s^2) = (r-t)^2 + 4s^2 >=0
and so the roots are real. (When k1 \ne k2, corresponding eigenvectors
u1, u2 must be orthogonal [consider 0=-=(k2-k1)],
and when k1=k2 we have b is a multiple of I_2 and so any vector is an
eigenvector....)
Apr If we keep trying to clarify this subtle distinction between
A and B, when will we get to K? ;-) Another way to think of this:
if g is the matrix for <.,.> with respect to a basis, if a is the
matrix for A, and if the matrix b for B= is symmetric (b=b^t),
then a is "g-symmetric" meaning a^t=gag^{-1}. Of course, when
g commutes with a (as when we take an orthonormal basis with respect
to <.,.>) then a is symmetric in the usual sense (a=a^t).
Apr We work out (upward) normal for a graph, introducing tangential
gradient operator T(h)=\grad(h)/\sqrt{1+|\grad(h)|^2} where
\grad(h)=(h_u, h_v). Note that T(h) is the first two components
of \nu (up to sign).
#Compute the shape operator and 2nd fundamental form (again) for
a graph (u,v, h(u,v)) of h(u,v), expressing in terms of Hess(h)
Apr We continue the study of the second fundamental form and prove:
Theorem. A surface S is totally umbilic (k1=k2) iff S part of a
plane or sphere.
#Check that the following are equivalent ways to say S is umbilic
at p: 1) k1(p)=k2(p), 2) H^2(p)=K(p), 3) the shape operator A at p
is a multiple of the identity linear map from T_pS to itself.
#Find the umbilic points on an ellipsoid ax^2+by^2+cz^2=1. You
may first want to consider the case of an ellipsoid of revolution
(where, say a=b) using symmetry (the general case may be too
hard to do explicitly). Sketch the curves whose tangent vectors
are aligned with the principal directions associated with the larger
and smaller principal curvatures. [Hint: it may help to recall
that the unit normal for a level surface {f(x,y,z)=0} is given
by \nu = Df/|Df| viewed as a column vector, and then work out A
terms of Hess(f) - the formulas you get should specialize to the
earlier case of a graph {z=h(x,y)} by taking f(x,y,z)=z-h(x,y).)
#Does a torus of revolution have any umbilic points? Again, sketch
the curves whose tangent vectors are aligned with the principal
directions for the larger and smaller principal curvatures.
#One of the steps in proving the above Theorem is showing both
principal curvatures are constant fuctions on S. Can you find any
non-umbilc surfaces with this property? Can you characterize all
such S?
Apr Flat surfaces (k1=0) include cylinders and cones on, and tangent
ribbon of, space curves - are these all possibilities?
We discuss a bit about ribbon/ruled surfaces in general (more later).
Apr In the past we argued (using linear algebra):
If a surface S has a mirror symmetry plane P, then the
intersection curve S \cap P is a principal curve.
This time we argue (using synthetic geometry) a stronger result:
If a surface S is cut by a plane P, such that the surface
(*) normal along the intersection curve S \cap P lies in P,
then S \cap P is a principal curve.
Any surface of revolution is a union of such curves (congruent to
the "generating curve"), and the "orbits" of the revolution form the
orthogonal family of principal curves. This determines the shape
operator of S (the principal directions and curvatures): one
principal curvature is (up to sign) the curvature of the plane
curve S \cap P; the other (for a surface of revolution) is the
curvature of orbit projected to the surface normal (with the new
notation c" for the curvature vector of the orbit curve c, using
our sign convention: -=-(sin\phi)/r, where \phi is the
angle between the axis of revolution and \nu, and where r is the
distance to the axis).
#In fact, for any u.s.p. space curve c(s) along a surface S we get
the normal curvature of S in direction c'(0) at c(0) by computing
-.
Apr #A generalization of (*) is a version of Joachimsthal's Theorem:
If along S \cap P the unit normal of S make a constant angle
with P, then S \cap P is a principal curve on S. In fact,
this a special case of an even more general verion:
The intersection S \cap S' is a principal curve
on both S and S' iff they meet and constant angle
and (at least) S \cap S' is a principal curve on
at least one of S or S'.
[As I noted in class, one must assume the intersection curve
is principal on at least one of the surfaces, because there
are (counter-)examples of surfaces S and S' - take a pair of
helicoids with common axis but rotated about this axis - which
meet at a constant angle but where the intersection curve is
not principal - indeed, their axis is an "asymptotic" curve.
Perhaps there's a further generalization of J's theorem which
deals with that case?]
Apr Review some features of moving frames along a curve, clarify
angular velocity of such a frame. Compare holonomy of Bishop
(parallel) frame (in the normal bundle to a curve) from other
notions of holonomy (e.g. in the unit tangent bundle to a surface).
#And, now that we are dealing also with surfaces, we discuss the
Darboux frame [T|\eta|\nu] adapted to a curve c on a surface S: here
T is the unit tangent vector to c, and rotating T by 1/4-turn in the
tangent plane gives \eta, i.e. T \cross \eta = \nu (that's our sign
or orientation convention). Express the infinitesimal rotation
[T|\eta|\nu]'=[T|\eta|\nu]A=W\cross[T|\eta|\nu] of a Darboux frame
along c using the shape operator of S.
Apr We define an "asymptotic" direction for S as one with normal
curvature 0 and observe (by previous fact) that a ruled surface
has asymptotic direction along the rulings (the straight lines
which comprise the ruled surface).
For a "concrete" application of asymptotic directions and ruled
surfaces: one of the first steps in finishing concrete "flat" work
is to make it flat by "screeding" - rubbing a long straight board
back and forth across its surface. This will produce a ruled surface
in general, but not necessarily a planar surface (with shape operator
A=0 everywhere). Is it enough to have two independent rulings through
each point on a surface to get this?
#Check this for examples of ruled surfaces such as {z=xy} that
you can write down explictly.
#Check that the Gauss curvature K=det(A) is non-positive at p iff
there is an asymptotic direction. More generally, there are 0, 1
or 2 (independent) asymptotic directions if K>0 (resp. =0, <0).
#Check the the mean curvature H=trace(A)/2=0 at p iff the (pair of)
asymptotic directions at p are orthogonal (so the above "if" is not
"iff").
We also remark about where we expect umbilics to appear on an
ellipsoid and began to think globally: associate an index in Z[1/2]
to each umbilic, these add to the Euler number (2 for an ellipsoid
or any topological sphere), and so forth (how about higher genus?)!
Apr We discuss parallel surfaces S^t = S + t\nu (parametrized by
X^t(u,v) = X(u,v) + t\nu(u,v) if you like, and more nicely if we
take (u,v) to be coordinates along principal curves, possible
away from umbilics), and verify the "area form" dArea^t =
X^t_u x X^t_v dudv in these coordinates is (1+2Ht+Kt^2)dArea,
so that 2H is the "first variation" of Area while K is "second"!
#Redo the above calculation without assuming nice coordinates
and try to reach similar conclusions. What about the "first
fundamental form" g^t (given by the inner-products ,
etc. - see note on ?? Mar for more on this), as well as the second
fundamental form B^t and shape operator A^t for for the parallel
surfaces?
Apr In class we discuss the notion of the R^3-valued, normal
and tangent vector fields on a surface S in R^3, and begin to
discuss the divergence operators DIV and div.
#Fill in details of my claim that the divergence theorem for DIV
(in general, not just applied to the coordinate vector fields) is
equivalent to the statement that the (outer, unit) normal field
integrates to zero for any S=\d\Omega, where \Omega is a (smooth)
bounded domain in R^3 and S is its boundary surface.
Apr #Consider the shell domain between two surfaces S^t1, S^t2 in any
family of parallel surfaces S^t. The unit normals to S^t define
a vector field N on this shell domain. Compute DIV N along S^t in
terms of the mean curvature of S^t. [Hint: this can be done directly,
or by applying the divergence theorem to N and using the area formula
from last time.]
#Try proving the isoperimetric inequality via the divergence theorem
We define DIV V and div V as the trace and partial trace of
DV for any R^3-valued vector field V on R^3 or a surface S in R^3,
respectively.
#Show that our formulas for DIV V and div V are independent of the
choices of orthonrmal frames used to define the traces.
We give a couple arguments why div V = div V^T + 2H, and
looked at the consequences for a minimal (H=0) surface in R^3:
1) coordinate functions are harmonic [take V=e_i, so V^T=e_i^T=
grad x_i, for i=1,2,3] and 2) the conormal vector \eta integrates
to the zero vector for any closed curve c which bounds a domain
on a minimal surface (this goes back to Archimedes in the guise
of force-balancing, since surface tension in a soap film exerts a
constanr force per unit length on the boundary curve c in the
conormal direction).
#Suppose instead we have a region R of a *constant* mean curvature
surface bounded by a curve c. Now argue that the integral of \eta
around c must equal (perhaps with the constant factor -2H) the
integral of the unit normal \nu over R. This is a again force
balancing, except now there is both surface tension (the \eta
integral) force and pressure (the \nu integral) force to consider.
We show that force balancing in a cmc surface can be viewed as a
"conservation law": homologous curves C and C' have the same force,
defined as the integral of \eta over C (the tension force) minus 2H
times the integral of \nu over any cap K spanning C (the pressure
force). We also used this idea to get a first order ODE for cmc
surfaces of revolution.
#Analyze this first order ODE: check (by implicit differentiation)
that it is equivalent to the second order ODE for the radius from
the axis of a cmc surface of revolution; verify that for H=0 the
function r = cosh t solves the equation (its graph is a catenary,
revolving to the minimal catenoid); what happens as you vary the
magnitude m of the force parameter, in case H=0 and for general cmc
H?
Very Late April
Part 3: Differential forms and the intrinsic geometry of surfaces
Here are some problems about exterior algebra and differential
forms that were posed to the manifolds class a few years ago.
If you go to the bottom, you'll get to the problem I began to
outline in class:
-----------------------------------------------------------------------
Let V be an n-dimensional vector space over R. Compute the dimension of
the exterior algebra V. [Hint: if {e1, e2, ... , en} is a basis for V,
show that {1, e1, e2, ... , en, e1e2, ... , en-1en, ... , e1e2e3...en} is
a basis for the exterior algebra.
------------------------------------------------------------------------
From the above problem, you should have found the dimension of the degree
k vectors (or k-vectors, for short) in the exterior algebra of V to be
n \choose k = n!/k!(n-k)! Since n \choose k = n \choose n-k, there is
an isomorphism between k- and (n-k)-vectors. Here's a way to define
such an isomorphism, "Hodge duality": if {e1, e2, ... , en} is a basis
for V, then the n-vectors e1e2...en and -e1e2...en are ORIENTATIONS
for V; pick an orientation (say, the + one); then any k-vector is
paired with its Hodge dual (n-k)-vector according to this rule for
basis vectors:
*(ei1ei2...eik) := (+/-) e1...ei1^...eik^...en
where the v^ means v is omitted, and where the (+/-) is chosen to make
*(x)x a nonnegative multiple of the (+) orientation chosen, i.e.
*(ei1ei2...eik)ei1ei2...eik = e1e2...en.
Show that, on k-vectors, *^2 = (+/-)Id - how does the sign depend on k?
In case V = R^2 and k=1, *^2 = -Id, and * is "rotation by 90 degrees".
In case V = R^4 and k=2, *^2 = Id; compute the (+/-)1 eigenspaces for *,
i.e. the self- and anti-self-dual 2-vectors on R^4.
------------------------------------------------------------------------
If L:V -> V is a linear map, then L induces a linear map on the
exterior algebra. In particular, show that the induced map on
the 1-dimensional space of n-vectors is multiplication by det(L).
------------------------------------------------------------------------
If V* is the dual space of V, i.e. the space of all linear maps from V
to R, then we can define exterior algebra of V* as well. Show that the
k-vectors on V* can be regarded as skew-symmetric k-linear maps from V
to R (or k-forms, for short).
Show that L as above induces a linear map on k-forms - what is this map?
-----------------------------------------------------------------------
We have seen how to define Hodge duality for smooth forms: extend it
to commute with multiplication by smooth functions. For example,
let
\theta = (x dy - y dx)/(x^2 + y^2)
be a 1-form on R^2 \ {0}; then *dx = -dy and *dy = dx extend to
*\theta = (x dx + y dy)/(x^2 + y^2) = dr/r = d(ln r)
where r^2 = x^2 + y^2. Compute d\theta and d(*theta). Interpret
\theta in polar coordinates. Do these forms remind you of anything
from complex analysis?!
-----------------------------------------------------------------------
If we write dz = dx +i dy and d\barz = dx - i dy, then they span the
complexified (tensor with C) cotangent space of R^2. The complexified
tangent space is spanned by their duals \d_z and \dbar_z. Write \d_z
and \dbar_z in terms of \d_x and \d_y. Show that the Cauchy-Riemann
equations for f = f1 + i f2 can be expressed \dbar_z f = 0.
Also, express the exterior derivative operator d in terms of the complex
operators \d and \dbar defined by
\df = \d_z f dz and \dbar f = \dbar_z d\barz
and conversely.
-----------------------------------------------------------------------
The differential operators grad, curl and div from advanced calculus
have a nice interpretation in terms of exterior derivative d and Hodge
duality *, as follows.
Let I, J and K be the standard constant vector fields on R^3. Then any
vectorfield
W = aI + bJ + cK
on R^3 is determined by smooth functions a, b and c. We may identify
W with a 1-form \omega = a dx + b dy + c dz via
W = #\omega or \omega = &W
(here # or "sharp" is what Marcel Berger calls a "musical isomorphism"
- its inverse is called "flat": we'll use & for that so &# = #& = Id).
Considering the spaces of forms on R^3, the operator d, and Hodge *:
<---*--->
\Omega^0 -d-> \Omega^1 -d-> \Omega^2 -d-> \Omega^3
<---------------*----------------->
please check the following (up to sign):
grad f = #df
curl W = #(*d(&W))
div W = *(d*(&W))
and in particular, since d^2 = 0, we have the calculus facts
curl(grad f) = 0
and
div(curl W) = 0.
-----------------------------------------------------------------------
Note that the operator div above is what I've been calling DIV
in earlier problems here - it's the R^3 divergence. If you can
do this last problem, you've understood much of what one needs
to know about differential forms - so please try it....
Tangential projection of D as covariant differentiation (connection)
on a surface S in R^3; notion of parallel transport. Connection
1-form (so(2)-valued) \om for an orthonormal (co)frame e1,e2
(\th1, \th2) on S.
First structure equations d\th1=\om\wedge\th2 and d\th2=-\om\wedge\th1
|\th1| | 0 \om| |\th1|
(or the matrix equation d| | = | |\wedge| |)
|\th2| |-\om 0 | |\th2|
depend only on intrinsic information (the metric g=\th1^2+\th2^2)
about S.
Compute d\om for basic examples:
sphere (extrinsic) S^2 in R^3 (\th1=cosv du, \th2=du in usual
longitude, latitude coord's) we get d\om=-\th1\wedge\th2;
and
hyperbolic plane (upper-half plane with metric g=(dx^2+dy^2)/y^2
where we get d\om=\th1\wedge\th2).
Second structure equation gives curvature K as ratio of 2-forms:
d\om = -K \th1\wedge\th2 = -K da
where \th1\wedge\th2 can also be viewed the area form. (Note: even if
S is non-orientable, this gives a well-defined K!)
Check that this K agrees with determinant of shape operator for S
in R^3. This is Gauss's *Theorem Egregium*!
Also check that if the o.n.frame is adtaped to a curve c in S (say
e1=T is tangent along c) then \om(T)=k_g the geodesic curvature of
c in S, which extrinsically is just component of acceleration for
(pbal) c tangent to S (see the Darboux frame problem above).
A metric on S can always be given as g = e^2h (du^2 + dv^2)
where e^2h is the "conforrmal factor" for some function h=h(u,v).
Check that g is a flat metric (K=0) iff h is harmonic (h_uu+h_vv=0).
More generally, find K in terms of h.
What happens if we integrate the second structure equation (using
Stokes's Theorem)? We get an even more remarkable theorem (really
the same as the Theorem Egregium, rather than "more egregium" as
MR suggest, or - a double entendre - *more* of the same ;-):
\int_\dR k_g ds \int_\dR \om = \iint_R d\om = -\iint_R K da
the Gauss-Bonnet Theorem! (Here we've adapted the (co)frame to
the boundary curve \dR.)
But we need to be careful, since this doesn't work for a disk in R^2
(k_g=1 and K=0) or hemisphere in S^2 (k_g=0 and K=1)! We've made the
same mistake Chern made when he first discovered this (intrinsic)
proof: we forgot that the adapted (co)frame may have a singularity
in R! When we account for this, we see that for a disk-like region
R on S, we have instead:
\int_\dR k_g ds + \iint_R Kda = 2\pi
which agrees with the disk examples above. More careful accounting
(breaking R into geodesic polygons and counting turning angles at
vertices as atomic contributions to k_g - this gives both the edge
the vertex counts E and V) replaces r.h.s. with 2\pi(F-E+V) where
F is the number of faces, i.e. with the Euler number 2\pi\chi(R).
For a closed surface S (compact, with empty boundary \dS) in R^3,
this says the degree of the Gauss map is \chi(S)/2 in case S is
orientable. In case S is nonorientable, the Gauss map goes to
RP^2 and we need to be a bit more careful about degree.
There's also an nice (extrinsic) proof using Morse Theory of height
functions and the change-of-variable formula for the Gauss map to
express (1/2\pi times) the total curvature as the (signed) number
of critical points averaged over the sphere of directions.
=====================================================================
May Geometry Research Project Presentations
[10AM till 5PM, in LGRT ???? - liquid refreshment & pizza possible,
and if the weather is nice, afterward we can all hike over to my
place and build a roaring fire in the outdoor hearth to celebrate
the end of the semester!]
=====================================================================
Here's the list of final Research Project topics (where I took some
liberties) and a tentative schedule for the presentations:
???day ???day
10:00
10:30
11:00
11:30
12:00
12:30
13:00
[Pizza delivery!?!]
14:00
14:30
15:00
15:30
16:00
16:30
[Possibly over two days?!]
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NOTA BENE: Please bring a printed draft of your presentation/paper for our
???&???day sessions, and send the final paper (preferably in .tex or .doc
format, but ALSO in .pdf) as an email attachment to my usual address:
profkusner@gmail.com
BEFORE 11:59PM ?????????????? (and please email there with any questions).
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**