why I am a nerd
I have a project I’m saving for when I get old and grey and unable to do real math anymore. Let me tell you about it.
My adviser, and many other folks besides, seem to have major beef with proof by contradiction. It works, but it doesn’t advance the field. Digging into a proof by contradiction, you’re not going to find a useful object to study.
But I find proofs by contradiction fascinating, and they feel quite natural to me somehow. So I want to write a book of them. More specifically, I want to see how many standard and/or famous proofs by contradiction can be interestingly tweaked into the ur-contradiction 0=1. I’ll call it Conjectures on the One-Element Field.
Most blow-up arguments (like Sesum’s that I just posted about) fall into this category, as do a lot of index-theory arguments (the Hairy Ball Theorem needs its own chapter, I think). The trick is to make the 0=1 part flow from the rest the proof, and not to be too obvious and artificial about it.
scalar curvature blowup at first singular time
I just figured out that Šešum’s proof that at a Ricci flow singularity in fact works, with a slight modification, to prove that
at these singularities as well.
To get the proof to work in this case, I note that an inequality like is how one establishes Glickenstein’s Lemma. If
, this inequality is immediate since
.
However, we can still get such an inequality even if isn’t bounded. First note that, for the rescaled flows
, we have a uniform Ricci bound
. So for each
we can choose
close enough to
that
Here ‘close enough’ is independent of . Then, using the definition of
, we see that
So if we want to prove, for any and
,
, we just have to write
, which can clearly be done. In fact we just have to find one such
. So we get a Glickenstein’s Lemma for a flow with merely
bounded.
The rest of the proof works just fine, because in the limit we still get , hence by the evolution equation
we see that .
Huisken and Sinestrari 2008 – II
I already gave the broad outline of Huisken and Sinestrari’s 2008 paper; I left out several big chunks that make the whole thing work.
Recall that we’re dealing with a family of 2-convex hypersurfaces evolving via mean curvature flow. We’ll assume uniform 2-convexity, i.e. for some
.
I won’t go into the analytic work, but we need to use it. There are two analytic results we need:
Cylindrical Estimate. For any
, there exists
also depending on
and the diameter of the initial hypersurface, so that along the flow
I call this a cylindrical estimate because the quantity vanishes on a cylinder, and when
is small it measures the sum of the squares of the differences of the higher eigenvalues.
Harnack Estimate. For any
, there exist
also depending on
, the initial diameter, so that along the flow
One first establishes these estimates for the smooth flow, and it’s rather easy to see that the surgery procedure doesn’t affect the constants, since surgery involves replacing approximate cylinders with approximate spheres.
Recall that the surgery procedure relies upon finding Hamilton necks–that is, regions of the hypersurface that are approximate round cylinders. In fact we want to do better than finding a region which looks like a cylinder at one time; we want to find regions that shrink like cylinders under the flow.
Definition. We call the spacetime region
an
shrinking neck if
- The final time-slice
is a Hamilton neck
-close to a round neck in
. Call the approximate radius of this neck
.
is
close to
, where
is the radius of a round cylinder at time
, which shrinks to radius
at time
.
, and each neck has length
.
All of the singularity analysis resides in the following theorem:
Neck Detection Lemma. Let
be constants as above. There exist constants
depending on
, and the initial diameter, so that if any
satisfies:
and
- The parabolic neighbourhood
doesn’t contain any points affect by a previous surgery.
then
is an
-shrinking neck.
To prove this theorem, we are going to argue by contradiction, using a blow-up argument.
Proof of the Neck Detection Lemma. If the theorem is not true, we may take a sequence of flows, each of which contains a point
so that
and each parabolic neighbourhood
is unaffected by surgery, and so that none of them has
as a shrinking neck.
Rescale and translate each of the flows: to get so that
. The parabolic neighbourhoods all rescale to the same
.
Tracing the Harnack estimate, we get a Harnack estimate for , so that the
are bounded uniformly on
. Then the cylindrical estimate says that
are uniformly bounded in
.
Then the Harnack estimate says that in fact all derivatives of $latex\tilde{A}_j$ are uniformly bounded in . So according to the Arzela-Ascoli theorem, there is a limit flow
, with convergence in
where
is the control we have on the derivatives during surgery.
Consider the properties of . We have
.
Now , and
satisfies a parabolic equation, so achieving its minimum at an interior point means in fact
. Then
. So in fact
has a zero eigenvalue and all other eigenvalues equal.
The evolution equation for the quantity (which we have just proved is identically zero) has a term involving
, so we see that
, i.e.
is covariant constant. A rigidity theorem of Lawson says that this characterises cylinders.
Thus is a shrinking cylinder on
. So the sequence
is approaching a shrinking cylinder, hence for large enough
,
is a shrinking neck. But
is just a rescaling of
. So
is a shrinking neck, in contradiction to our assumption.
Ricci blowup at first singular time
The major analytic task in geometric flows is to understand the structure of singularities. The first step toward understanding singularities and their development under the flow is a theorem like:
Characterisation of Ricci Flow Singularities. Suppose
is a Ricci flow, defined up to time
. Assume
is maximal. Then we have
, where
is the Riemann tensor.
That is to say, what goes wrong at a Ricci flow singularity is the curvature tensor. We could restate the theorem as its contrapositive:
Characterisation of When the Flow Can Be Extended. Suppose that up to time
, we have a bound
. Then the Ricci flow can be extended past
.
But of course the curvature tensor is a big nasty gadget, which is none too easy to understand. We want something easier to check than “Are my curvatures all bounded?”. Šešum proved that, in fact, one need only look at the trace of the Riemann tensor, which is the Ricci tensor:
Better Characterisation of Ricci Flow Singularities. As above, let
be the first singular time of a Ricci flow. Then
.
This is somewhat remarkable, because the Ricci tensor interacts weirdly with Ricci flow. For example, the condition is not preserved under the flow. Yet somehow the Ricci tensor carries the information about when singularities occur. Proof under the fold.
Huisken and Sinestrari 2008 – I
This is essentially the talk I prepared for my comprehensive exam. It’s a bit more detailed because, unlike in real life, there are no time limits on blogs.
The smooth mean curvature flow is a family of embeddings satisfying
, that is, moving by its mean curvature. For our purposes, we will consider
closed (compact without boundary). I’ll be fairly cavalier about identifying the embedding
with its image
.
Now , and we may take normal coordinates so that
and
is a normal vector. In these coordinates, then,
. So we may write
, and think of the mean curvature flow as the heat equation for hypersurfaces. In particular, the flow is parabolic, so we get maximum and comparison principles, as well as local existence and uniqueness.
As in the case of Ricci flow, mean curvature flow exhibits finite-time singularity development. To see this, let be any compact hypersurface. Then
is contained in the interior of some sphere
, of radius
. Now
collapses with radius
. By the comparison principle, the flow starting at
stays inside the region bounded by
; in particular
develops a singularity before time
. Sometimes, the singularity
achieves is collapse to a point, as in the case of the sphere. We have
Huisken 1984. Suppose
is uniformly convex, i.e.
where
are the principal curvatures. Then
is isotopic to a round sphere, with the isotopy given by a rescaled flow.
What happens when we relax the convexity assumption? We say a hypersurface is 2-convex if , where
are the principal curvatures. This is equivalent to there being at most one negative principal curvature, which is also the smallest.
Huisken-Sinestrari 2008. Suppose
is closed and 2-convex. Then
is diffeomorphic to
, i.e. the boundary of a handlebody. Moreover the flow detects the connect sums.
proving sup bounds using Lp bounds and iteration
Sorry for the break. Got a lot on my plate.
A trick that’s come up a lot in my recent readings is using an bound to get a bound on the supremum of some function. The trivial case is what happens when
where the constant
is independent of
; then we just use the fact that the sup norm is the limit of the
norms. But what if the constant
depends on
, and in fact blows up as
blows up? Turns out all is not lost.
I’ll use the main theorem from Huisken 1984 as an example, though the same trick is used in Hamilton 1982 and Huisken-Sinestrari 1999. The theorem is
Theorem. Suppose is a closed convex hypersurface. Then there exist constants
such that
along a mean curvature flow starting from
.
The way to prove this is to consider the function and try to prove that for some choice of
, it’s bounded. Clearly the function is bounded at each time, but we need to keep the bounds from blowing up as we approach the first singular time.
There is an important fact we’re going to use about the function . It can absorb powers of
; that is, for any integer
, we have
for
. Thus if we prove a
bound for
, we get
bounds for
as well.
Now with a lot of tinkering (which is the hard part of the paper, and is different for different geometric assumptions), you can show that
Lemma. There is a constant so that if
is large and
, we have
.
Now we want to consider the function . Notice that
and
differ only by a constant. Using the evolution equation for
, we can show without much fuss that
Now we use the Michael-Simon Sobolev inequality with
and the Holder inequality to get
where all integrals are taken over the support of , i.e. the region where
. Notice that the left-hand side occurs on the right as well.
Now on the support of
, so we have
, where
is the sup bound. So we have
Now integrate both sides in to get
and choose so large that
everywhere on
. In particular,
So we can estimate integrals of by integrals of
. Some more Holder and interpolation inequalities give
and using Holder again (stupidly this time–just pull a power of the volume out), this becomes
where is the space-time volume of the support of
. Now if
, we have
, so we get that
which is a sort of decay estimate on the function . If we choose
big enough that the exponent on the RHS is greater than 1, this estimate says that if
is small,
has to be much smaller. So the volume
is decaying pretty fast. More specifically,
Stampacchia’s Lemma. Suppose is a nonincreasing real function, which satisfies for all
,
where and
. Then there is a number
with
.
Thus we get a number for which
, that is, for which
on the entire flow.
for Megan especially
This article talks about the sort of math one can use to compare fossils (or living creatures), and the benefits that making said comparisons rigorous can have.
via Ars Mathematica
fuzzy sets and God
It’s been a while. Teaching does that. But back to the game.
Many times presuppositionalist apologists will bring up the nice crispness of logic and its unexplained utility for understanding the world as proof that God exists. There’s rationality out there, they say, which means there must be some intelligence, much like the rationality in our own heads is a product of intelligence.
I don’t find this convincing in the least, but I bring it up because today I came across an argument for the existence of God which takes the opposite tack, namely an appeal to fuzzy logic.
Huisken and Sinestrari 1999, pt. 4
So I think I figured out basically what’s going on with those cones . We needed them to be convex, so we could apply Hamilton’s maximum principle for systems.
Recall that the elementary symmetric polynomials on are defined by
If we let denote the polynomial formed by omitting from
without the terms involving
. It’s immediate that
Recall also the cones . We have
.
We’re going to prove
Theorem. are convex.
Proof. We’ll proceed by induction. The base case, , is clear, since
is a half-space.
Now suppose that is convex. Consider the functions on
,
,
. In fact,
, i.e.
is the epigraph of
.
Since we’ve assumed is convex, this means that convexity of
is equivalent to
, where
.
The relation implies
and
Dividing this last identity through by , we have
So that the function , whose concavity we’d like to establish, can be written in terms of the functions
.
Now , where the bar means setting the ith component to zero. It’s easy to check that
, and by our inductive hypothesis, every second derivative of
is nonpositive on this set. So we have
, i.e.
is concave on
.
The claim is now that the concavity of implies the concavity of
. This follows from a computation, but you can kind of see what’s going on by noting that
and
are degree-1 rational functions with nonpositive second derivatives.
Consider the 1-dimensional case. A degree-1 rational function is one which is asymptotic to a line; the nonpositive second derivative means the graph of
lies above the graph of the line. For example, let’s let
Then is asymptotic to the line
, but since
, we have that the second derivative is nonpositive everywhere. Here’s the graph of
:
Now consider what happens when we take the function , analogous to
. In our example, we have
. This function, you may notice, is also asymptotic to
, but it’s asymptotic to it from below, and its second derivative is nonnegative.
But the function we’re interested in is . Actually we’re interested in its second derivative, which means the linear term
doesn’t matter, so we can discard it, and consider
, which is analogous to
, whose graph looks like
The relevant fact about this graph is that it has the same concavity as the original !
So concavity of , which we have by the inductive hypothesis that
is convex, is enough to show that each
is concave, hence
is the sum of concave functions, hence itself concave. But this implies
is convex, which completes the induction.


