These are notes and problems from Rob Kusner's Spring 2012 honors course
on Differential Geometry of Curves & Surfaces (Math 563H at UMassAmherst,
Copyleft* 2010-12 by Rob Kusner). The notation "MR" refers to items in the
Montiel & Ros text. Please write up the problems marked # (generally due
in class the Wednesday 12-16 days after the date noted - e.g. the first hw
is due 08 Feb and consists of those from the first week):
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.
23 Jan We talked about how geometry and physics interact, and we examined
plaster models of surfaces that minimize their "bending energy" or
some other physical quantity (e.g. soap films and bubbles minimize
their area or "stretching energy" possibly with an enclosed-volume
constraint). How might you define the stretching or bending energy
of a curve (think of Hooke's law from mechanics)?
#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.
#How smooth is f:R -> R where f(t)=e^{-1/t} for t>0 and 0 otherwise?
25 Jan #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).
#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?
27 Jan 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.
Proposition: Immersed C^1 curves X admit unit speed parametrization
with C^0 unit tangent vector T, unique up to "orientation" and s=0.
Proof: "Solve" the differential equation s'(t)=|X'(t)| (i.e. just
integrate the given function |X'(t)|.
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!
#Work out the u.s.p for the helix X(t)=(r cost, r sint, qt).
30 Jan We discussed possible domains I for "complete" and "incomplete"
u.s.p. curves (with a little "fugue" on Cauchy sequences in, and
completenes of, metric spaces ;-). We agreed I=S^1(L)=[0,L]/0~L
(the circle of arclenght L) also qualifies as a complete domain,
as do [0,L], [0, \infty[ and R. #Any others (why or why not)?
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 peroidic, X may not close, but have a "period"
(the "translation" or "constant vector of integration").
#Give example of curves X with periodic T but nonvanishing periods.
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).
01 Feb #For any C^2 curve X(s) in R^n, and any C^2 reparametrization s(t),
verify that the reparametrized curve Y(t)=X(s(t)) satisfies
Y"(t) = s"(t) X'(s) + (s'(t))^2 X"(s). Therefore, if 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, and use this idea to compute K(0) in terms
of a, b, c.
#More generally, work out an expression for the curvature vector K
for an arbitrarily parametrized curve X(t) in terms of X' and X".
(Try the case of curves in R^2 and R^3 first; in R^3, the cross
product can be used, then specialized to R^2; the special case
where X is a graph over its tangent line is also worth trying.)
All of the above means that the curvature is "geometric" - i.e. K is
indepenent of the parametrization. How does K change under rotation
or translation in space? Under scaling? (Show |K| has "dimension"
of 1/Length.)
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 by a quarter-turn.) Thus we can
express the curvature vector K=kN for a function k(s) [In the past I
called this \kappa, but will use lower case here]. What curves have
k=0? Which have constant k? 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) and
exponentiating the resulting 2x2 matrix [more next time in class].)
For a curve in R^2n=C^n one can also define a 2n x 2n version of J
which corresponds to multiplication by i.
03 Feb We briefly discussed polygonal approximation of a curve, and we
promised to say more in these notes. Even for a continuous (C^0)
X one can make a partition (see MR 1.6) and associated inscibed
polygon; if the supremum of the length of such inscribed polygons
exists, we say X is "rectifiable"; MR 1.6 & 1.10 show that a C^1
curve X is rectifiable and that this supremum equals the length of
X defined by integrating |X'|. For a C^2 curve, the estimate in
MR 1.6 can be improved, as hinted in class: if |P|<\eps then 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)|=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).]
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?
10 Feb More on helices and their curvatures; limiting cases of circles
and lines. Pitch p=q/r dimensionless; sign of q gives handedness.
Orbits of screw motions.
Frenet-Serret frame (for points on a C^3 curve X in R^3 with |K|=k>0:
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.
13 Feb #Work out the osculating circle of a helix at any point. (Second
derivatives agree).
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 and many other authors) sign convention:
left-handed helix has positive torsion!
#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). In particular, there is a busculating twisted
cubic (or helix, generalizing the osculating circle) at any point
of a C^3 curve. [Careful: if only the curvature functions agree
(k(0)=k*(0)>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?!]
The unit tangent vector T(s) of a u.s.p curve X(s) in R^3 traces out
a curve on unit sphere S^2 called 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|K(s)|ds = \int|T'(s)|ds =
Length(T). [Fill in the details, and as a challenge, try to prove
the following result:]
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 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.]
15 Feb 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 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)x[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.
#Exercise 1.37 [Hint: We already computed k and t are constant for
a helix; for the converse, one looks at A(s):= X"(s) = kN(s), so
(by Frenet formulas) A'(s) = -k^2T-ktB -> |A'|^2 = k^2(k^2+t^2);
A"(s) = -k(k^2+t^2)N -> k_A = |A"(s)|/|A'(s)|^2 = 1/k; and also
A"'= (some multiple you can compute)A', so triple product
and torsion t_A must vanish; thus A is planar curve with curvature
k_A=1/k lying on sphere of radius k -> A (great) circle of radius k.
Thus can write
| cos \sqrt{k^2+t^2}s |
X"(s) = A(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.]
Back to plane curves: 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 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.]
17 Feb 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.
To compare 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)| is the total twisting of the Frenet
frame with respect to a parallel frame.
#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?]
20 Feb [No class on account of Lincoln and Washington ;-}]
22 Feb 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 10 years 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).)
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.
24 Feb #A surface patch X:U->R^n for U\subsetR^2 has X homeomoprhism to its
image and (if X is C^1) satisfies rank(DX)=2, i.e. X_u, X_v linearly
independent; for n=3, this means X_u x X_v nonvanishing, so we can
normalize and define the (C^0) Gauss map (unit normal) \nu:U->S^2 by
\nu = X_u x X_v/|X_u x 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)?]
27 Feb #Find patches covering a torus T^2=S^1xS^1, say, 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).
29 Feb 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?)
#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 pointed 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 projection, but got a bit stuck
figuring out the inverse. The missing factor was 2. 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"!)]
Here's why I wanted 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:
!!!!!!NEWSFLASH!!!!!! for you W-rated differential geometers: the
Willmore conjecture has been resolved! This is one of what the late
Bob Osserman referred to (in his early 1980s Berkeley Colloquium
Lecture) as "Three Obdurate Conjectures in Differential Geometry" and
concerns the total bending energy W of closed surface in 3-space. The
round sphere achieves the absolute minimum W=4\pi and it's been known
since work of Leon Simon from the 1980s that for higher genus surfaces
the absolute minimum is strictly greater; indeed Willmore conjectured
in the 1960s that W is at least the area 2\pi^2 of the Clifford
minimal torus in the 3-sphere, and this is part of what Fernando Coda
Marques and Andre Neves have just proven. [I've done some work on this
problem myself over the years and believe their methods might help
resolve a related conjecture of mine for closed surfaces in n-space: W
exceeds 6\pi for any surface not diffeomorphic to the 2-sphere, with
equality if and only if the surface is a Veronese minimal (see below)
projective-plane in the 4-sphere.]
02 Mar 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 above !!!Newsflash!!! 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)!
05 Mar Classifying surfaces of finite or infinite topology (notion of
genus for orientable surfaces, number of ends, nonorientability,
bands and Mobius bands).
#Sketch "movies" of slices of surfaces in space such as the sphere
(in various embedded configurations), the torus (begun in class) and
Klein bottle, the real projective plane (see websites above) immersed
as a Boy's surface, etc.
#Make a series of sketches to indicate why the various examples of
1-ended surfaces with infinite genus (drawn after one class !?) are
diffeomorphic to each other.
07 Mar We indicated how to form the connected sum S#S' for two surfaces
(or surfaces with boundary) S and S': remove an open disk from the
interior of S, an open disk from the interior of S', and attach these
(smoothly) along the common boundary circle. If S and S' are both
orientable, orientations can be matched to make the connected sum
S#S' orientable. If S or S' is non-orientable, so is S#S'.
We can also define (with caveats!) a boundary sum operation in case
S and S' each have nonempty boundary by smoothly attaching an arc of
\dS to an arc of \dS': the resulting surface is denoted S #_\d S'
though some care needs to be taken in case \dS or \dS' have several
components (of circle or line (unbounded) type).
Both of these operations (sum and boundary sum) are commutative up
to diffeomorphism, giving the set of surfaces (up to diffeomorphism)
the structure of an abelian semi-group under # (or under #_\d with
the above caveats!) whose neutral element represented by S^2
(or the by disk D^2 under #_\d). This leads to a decomposition into
"prime" pieces. For closed bounded surfaces (without boundary) this
semi-group is generated by the torus T=S^1xS^1 (orientable case) and
projective plane P=\RP^2 (non-oruentable case) subject to the
sole relation 3P=P#P#P=P#T. And if we also include the operations
of deleting a point (puncturing, possibly on the boundary) and of
deleting an open disk (or a point from boundary), then all finite
topology surfaces (with boundary) can be generated in this way.
#Check how genus g, number of ends e, number of boundary compenents
b=(b_bounded,b_unbounded) behave under each connected sum operation
for orientable surfaces. Note that in this case, the genus g is
simply the number of T's in the prime decomposition (no P's). For
non-orientable surfaces, the number of T's can be taken as 0, and
the resulting number p of P's is sometimes referred to as the non-
orientable genus (though one must be careful when saying "genus p"
out loud ;-). How do g and p behave under # and #_\d?
#Trousers means a surface diffeomorphic to a 3-punctured sphere.
Show that (except for the sphere S^2, plane \R^2, cylinder S^1 x \R
and torus T^2) any orientable, finite topology surface (without
boundary) can be decomposed into (how many?) trousers by deleting a
finite number (how many?) of simple closed curves.
[Thanks to Dugan Hammock for this week's live illustrations! :-]
09 Mar We discussed (using Stanford-red-sweatshirt double-cover) covering
projections and spaces, as well as "deck transformations" on the
universal cover of a surface. We also explained what "smooth" maps
between surfaces means, in particular, the notion of diffeomorphic
surfaces (we'll come back to smooth maps when we study the "Gauss"
or normal map in more detail next month).
We checked that S^2 -> S^2/{+,-} = RP^2 is a covering projection
and that Veronese surface "factors through" this. Also we saw
(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 introduced 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.
12 Mar We introduced (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).
14 Mar We introduced 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 sketched (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 sketched 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 sketched 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!!!)
========================================================================
[Something I did for 563H back in 2010 to help celebrate Pi Day:]
We introduced ideas from integral geometry (Buffon needle problem),
especially the "Crofton principle" that the Length(c) of a curve c
is proportional to the average number of intersections with standard
geometric objects (such as lines in R^2 or great circles in S^2).
For curves on S^2 we computed the Crofton Formula proportionality
constant by looking at the "test" case where c is a great circle
and discovered (in honor of the upcoming Pi Day?!) that it's \pi.
We then applied this to complete a key lemma in a proof of Fenchel's
theorem about total absolute curvature (from 2 Feb - like Groundhog
Day, we keep repeating this ;-), namely that if a curve on S^2 is not
contained in any hemisphere it has length at least 2\pi.
Use these ideas to work out Crofton's Formula for a curve c in R^2,
equating Length(c) = (univeral constant)(average number of times c
intersects a straight line in R^2); in other words, find (univeral
constant) for this case. [Hint: it also involves \pi. Bigger hint:
use a unit length line segment as your "test" curve.]
For fun, check out my friend Bob Palais' article "Pi is Wrong":
http://www.math.utah.edu/~palais/pi.html
========================================================================
16 Mar Wrap-up of topology of surfaces (review things above)!
#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 talked 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....]
19 Mar Spring Break (please finish the problems from 05 Mar through now -
to be turned in 04 Apr - and work on your research project! :-)
26 Mar Week after Spring Break (stop procrastinating on the project! ;-)
[I'll be at a meeting in Germany this week, so we'll next meet:]
02 Apr [... Finally: back to geometry!]
We worked 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 showed
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 intresected 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).
We worked 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).
Showed shape operator A is symmetric since the associated 2nd
fundamental form B = = - is symmetric, 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 A = g^{-1}b where g^{-1} is the inverse of the matrix g for the "first
fundamental form"
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! ;-]
#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)
Reviewed some features of moving frames along a curve, clarified
angular velocity of such a frame. Compared 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 discussed 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.
04 Apr We continued the study of the second fundamental form and proved:
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 was 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?
06 Apr We argued: if a surface S has a mirror symmetry plane P, then
the intersection curve S \cap P is a principal curve. As any
surface of revolution is a union of such curves (congruent to
the "generating curve"), 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 pbal space curve c(s) along a surface S we get
the normal curvature of S in direction c'(0) at c(0) by computing
-.
We defined an "asymptotic" direction for S as one with normal
curvature 0 and observed (by previous fact) that a ruled surface
has asymptotic direction along the rulings (the straight lines
which comprise the ruled surface).
#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 remarked 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?)!
We continued clarifying some issue raised in the 02 Apr problems
about A and B (when will we get to K? ;-)! We (re)discovered that if
g is the matrix for <.,.> with respect to a basis, and if a is the
matrix for A in this basis, and b for B= is symmetric then
a is "g-symmetric" in the sense that 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.
09 Apr We discussed 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 verifed 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 09 Mar for more on this), as well as the second
fundamental form B^t and shape operator A^t for for the parallel
surfaces?
11 Apr In class we have discussed the notion of R^3-valued, normal
and tangent vector fields on a surface S in R^3, and began 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.
13 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
17 Apr We defined 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 formula for DIV V and div V are independent of the
choices of orthonrmal frames used to define the traces.
18 Apr We gave 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.
20 Apr We showed 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?
Part 3: Differential forms and the intrinsic geometry of surfaces
23 Apr 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?
-----------------------------------------------------------------------
25 Apr
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.
-----------------------------------------------------------------------
27 Apr
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....
30 Apr 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 depend only on intrinsic information (the
metric g=\th1^2+\th2^2) about S. Compute d\om for basic examples:
(extrinsic) S^2 in R^3 (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 give 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.
==============================================================
Here's the list of final Research Project topics (I took some
liberties) and the tentative schedule for the presentations:
==============================================================
09 May Research Project Presentations (pizza provided)
3:45 Course Evaluations
4:00 Dan "Plateau's problem"
4:15 Ian "Minimal surfaces based on the catenoid"
4:30 Harry "Veronese minimal surface in S^4"
4:45 John "Bending energy and total curvature of surfaces"
5:00 Andrew "Variations of Length for space curves"
5:15 Mark "Linking number integral formulas"
5:30 Andrey "Total curvature of knots"
5:45 Matt "The Whitney Umbrella and branch points"
6:00 Melissa "Koch snowflake and its Hausdorff dimension"
6:15 Nate "Hyperbolic metrics on surfaces"
6:30 Kevin "Surfaces of revolution"
11 May (Please finish your take-home final exam, which is basically a
careful - and carefully referenced! - expanded write-up of your
presentations, sent as .pdf file to me at profkusner@gmail.com -
it is due 11:59:59PM 11 May! After that, have a great summer!!! :-)
-----------------------------------------------------------
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