Imagine that there was a countably additive and translation invariant measure, , on a Hilbert Space such that every ball has positive measure, finite measure. Because is infinite dimensional, there exists such that . Note that the points and are far apart: , if . Now, the ball, , contains the countable collection of disjoint balls: (you can see where this is going). Thus, because all balls, have the same measure (by translation invariance), and is countable additive, it follows that . Thus, we’ve reached a contradiction.

In some cases, it’s desirable to have an analogue of a measure on a Hilbert space . The most common application is to make statements such as “Property X is true for a.e. function.” One way to do make these statements is through the concept of Prevalence. Another is the definition of a Gaussian like measure called the Abstract Wiener Space.

Note that I wrote this article before realizing that Wikipedia has a similar article.

## My First Two Years

Posted March 11, 2012

on:During the past one and a half years I’ve spent as a graduate student, I’ve developed a few principles. I have decided to post them here, and update them as I learn more.

1. Don’t be afraid to branch out and learn subjects in other disciplines. Broadening your interests is easy in graduate school, and you’re paid to do it. Further, you may discover that you’ve gone to grad school for the wrong subject. You may be studying number theory, topology, etc., because you’re already good at it, and you don’t know enough about other subjects. It’s easy to get lost in the achievement factor in math, and just because you can solve problems with ease, doesn’t mean you’re truly motivated by the subject’s goals. The best way to discover your true motivations is to think about the guiding problems in the field and decide if they’re interesting to you in their most isolated form (e.g. do you “care” about distinguishing smooth structures on manifolds?). If the subjects no longer seem interesting, then move on to something you really enjoy.

2. Talk to everybody: professors, graduate students, undergrads, etc. The best way to learn is from other people, but don’t depend on them. Everybody has a tool they’re fond of — learn it. Find out what professors are interested in and ask for a research project or a reading course in that topic. Professors generally want to speak with you after you pass your quals. It’s part of their job to graduate students. So don’t be discouraged by professors that turn you down, five no’s and a yes is still a yes.

3. Don’t be afraid to say you don’t understand. Once you realize you don’t understand, you’re in the best position to learn. Further, many students may give the impression that they truly understand a subject, but will be forced to reevaluate if you ask enough questions.

4. Don’t be afraid of the computer. There is a massive amount of data in the world, and you’ll be at a supreme advantage if you know how to compute with it.

5. Go to seminars and colloquiums in your field and other nearby fields. Learning what other people are working on is likely to give you research ideas. Sign up to give talks in front of these groups, even if you don’t know a thing about the topic. Put a lot of effort into explaining concepts clearly and accurately.

6. Think long term. I know that many PhD students are only thinking about the next five years of graduate school. Assuming that everything will “work out” will put you at a severe disadvantage. Before every major choice you make in grad school, ask yourself why you want a PhD, and how will this choice benefit your goal. No matter what field you plan to go into, ask yourself the following question: who are the experts, and am I good enough to take their job? There isn’t a lot of room at the top, so you’re likely going to have to push someone out of the way to make room for yourself.

7. Schedule enough down time so that you can be fresh every day. Make sure you attend social functions, go to the gym, or play a sport, etc.

8. Learn to manage your time effectively. How many projects are you working on? If you get side tracked on a particular problem, move on and come back to it. Try to skim a book before you do the problems. Then, go back and read carefully when you need a deep understanding.

9. Learn how to measure success. Set benchmarks for yourself. Did you achieve more today than you do on average? Did you master a difficult concept? Try to achieve more every day, but don’t be discouraged if you can’t. You can’t learn everything in finite time. Some things are long term projects and even reading a few pages can give you something to be proud of.

10. Apply for fellowships. Teaching eats up a lot of time, but it does give you valuable experience. If you can’t get an RA, having your own funding can make your research a lot more productive. Having your own funding also increases the chances of a professor working with you.

- In: Haskell | Knot Theory | Programming | Topology
- Leave a Comment

I’ve uploaded a Haskell implementation of a combinatorial algorithm to compute the “hat” version of Heegaard Floer knot homology. This program should be seen as a stepping stone to implementing the more general HF complexes for links, and ultimately for implementing HF^- for large integer surgeries on links. Generally, our approach is the same as the one found in the C++ implementation of Baldwin and Gillam, located here.

The major difference is how we compute generators in positive Alexander grading. Essentially, we noticed the number of possible solutions to a particular integer programming problem was very small. We used this fact to narrow the set of possible permutations with positive Alexander grading and used a recursive algorithm to generate the restricted set of permutations.

In general, Gillam and Baldwin’s program seems to be much faster. Though for some knots such as KT_{2,1} or obviously trivial, but huge, uknots, our program is comparable in speed, and for the uknot example, much faster.

To compile the code, you must install the Data.Vector Haskell and the Data.Array.Repa modules (via cabal install Vector and cabal install Repa) and run the terminal command:

`ghc --make -O2 HFKhat.hs -XBangPatterns -XTypeOperators -XTypeSynonymInstances`

If you’re having trouble installing Data.Repa, see the instructions here.

You can download the gzipped tar file here.

To Learn more about combinatorial Heegaard Floer homology, check out the references contained in thislink.

Note: If you’re a haskell beginner, I recommend downloading the Haskell platform for best results.

Final Note: More code, including a general module to compute homology over Z/2Z is available at my website.

I haven’t updated this blog in a while, due to qualifying exam preparation and research projects. I’ll write about the latter soon. While studying for the geometry/topology qual, I asked a basic question: Is path connectedness a homotopy invariant? Turns out the answer is yes, and I’ve written up a quick proof of the fact below. You can view a pdf of this entry here.

Proposition 1Let be a homotopy equivalence. If is path connected, then so is .

*Proof:* It’s clear that the image of is path connected. Thus, it is enough to show that any point of can be connected to a point of . Let be a map such that is homotopic to , via the homotopy . Let . Then, and is a path from to .

We can deduce the following result from the proof of proposition 1.

Corollary 2Let be a homotopy equivalence. Then is path connected if, and only if, is path connected.

*Proof:* If is path connected, then is path connected. Hence is path connected.

## A Note On Sheafification

Posted April 16, 2011

on:You can view a pdf of this entry here.

A presheaf on a topological space with values in a category (usually with values in **Grp**, **CommRing**, or **Set**), can be associated to a sheaf on in a natural way. In fact, this association is left adjoint to the inclusion functor from the category of sheaves on to the category of presheaves on .

The standard construction of is fairly straightforward: consider the topology on the disjoint union of stalks generated by the collection of sets

where is open, and is the image of in . Let be the natural projection. For every define to be the set of continuous sections of , i.e. . It follows that is a sheaf and there is a natural morphism , such that is a sheaf if, and only if, is an isomorphism, (this is a fairly easy exercise).

This shows that every sheaf arises as the sheaf of sections of some continuous map onto . This also justifies the terminology: an element is called a section. At first glance, it is hard to see how this sheaf relates to . Indeed, the canonical map fails to be either surjective or injective, in general. This says that we cannot “complete” by simply adding more elements, or by glueing together elements; we must do both. Thus, we can can’t view as a subpresheaf of or as a cover. However, there is another natural way to think of .

Let be a sheaf on with values in some category , as above. One of the advantages of having a sheaf on is that a section is uniquely determined by the collection of elements . This can fail for presheaves. Thus, we need to glue together elements such that for all . This is evident in the commutative diagram:

This says that and have the same image in for all . Thus, because is a sheaf, .

Aside from glueing sections together that “should” be equal, we also need to add more elements to our presheaf . Suppose we have an open cover of an open subset : . If we can find a collection , such that , then we should be able to find a “global” section on such that . Because our presheaf may be too “small” and may not contain this element, we need to ensure that our sheafification does. The set is compatible in the obvious sense, so we can glue them together to form a section . This map is continuous because . is open. Indeed, if for some , then there is an open subsets such that and in , so . It is also clear that .

To sum up our construction, we note the following: to construct we find the most “efficient” sheaf such that

- sections that have the same germ at every point are equal
- compatible sets of sections have a common extension.

This gives the following alternative description of :

where , if for all and , . Note that condition and the assumption that this extension is unique is just the sheaf axiom and, hence, it implies .

We claim that this definition of is equivalent to the definition given above. Indeed, we can view each collection as a section in the natural way: if . This is independent of the representative of by . Note that is continuous, as we showed earlier. Further, any section arises in this way. Indeed, let , then for a neighborhood of and an element . Thus, because is continuous there exists . Consider the collection . Because for all , it follows that corresponds to . This shows that the correspondence is surjective. We claim that this correspondence is also injective. Indeed, if and correspond to the same section , then for all . Thus, Finally we note that this correspondence is actually a morphism of presheaves: Restricting a collection to a subset , yields the element and this corresponds to the restricted section .

This new definition has some utility. It helps to illuminate the following claims:

Claim 1Let be the natural map: . Then is a sheaf if, and only if, is an isomorphism.

*Proof:* If is a sheaf then each collection can be written uniquely as for some . Conversely, suppose is an isomorphism, let be an open subset, let , and let be a collection such that . Then for all , so . Thus, there exists a unique element such that . Therefore, for each , for all . In particular, for each fixed , , whence, , as desired.

Now, suppose that is a sheaf on . Let be a morphism of presheaves.

Claim 2There exists a unique map , such that .

*Proof:* Let be an open subset and let . Observe that the commutative diagrams

ensure that for all . Thus, and so by the sheaf axiom for , there is a unique extension . For convenience, we can view as , with . Now, if is an open set, then

Thus, is a morphism . From the definition of , it is clear that and is unique.

A pdf of this post is available here.

**1. Introduction**

Burnside’s theorem, first proved in the early 20th century by William Burnside, shows that a group of order , where and are primes and , is solvable. The original proof of Burnside’s theorem utilized representation theory in an essential way. A proof using purely group theoretical tools was given by D. Goldschmidt around 1970. The aim of this note is to provide yet another proof of Burnside’s theorem, via representation theory. First we establish a fundamental definition: A group , with identity , is solvable, if there exists a chain of subgroups

such that is normal in and is abelian for . Such a chain is called an abelian tower. Thus, for instance, all abelian subgroups are solvable. We also get the easy result: If is a -group, then is solvable and has a non-trivial center. *Proof:* Note that acts on itself by conjugation, so by the class equation:

where the sum is over distinct conjugacy classes, and is the center of . Because each index is divisible and , it follows that . Now, note that a group of order is abelian, so it is solvable. Thus, by induction, is solvable. We can lift any abelian tower in to get an abelian tower in , as desired.

We fix some notation. If is a complex representation of , then we omit the homomorphism and simply write instead. We denote the character of the representation by . If , then , is the centralizer of in and is the conjugacy class of in . Note that , by the orbit-stabilizer formula. We write for the stabilizer of in .

**2. The Proof of Burnside’s Theorem**

The proof will proceed by a series of claims. We assume that is non-abelian and , otherwise is solvable by section one. We wish to derive a contradiction.

Claim 1We can, without loss of generality, assume that is simple.

*Proof:* If is a normal subgroup of , then and are solvable, by induction, therefore, is solvable.

Assuming this, we note that because is normal, and is non-abelian. This also implies that every non-trivial representation of is faithful.

Claim 2Let be an irreducible representation of . Then, for all

is an algebraic integer.

*Proof:* Let be the conjugacy classes of . Then

is in the center of because it is invariant under conjugation. Further, the center of is generated by these elements. Indeed, suppose is in the center of . Write . Suppose that , and choose such that . Then conjugation by fixes , so . It follows that is a class function and so . We have shown that the center of is finitely generated over , and hence it is integral over . Note that is a invariant map. Thus, by Schur’s lemma there exists such that . Because is integral over , and satisfies the same polynomial over that does, it follows that is integral over . Thus, and so is integral over .

Claim 3Let be an irreducible representation of . Then .

*Proof:* We maintain the notation from the previous claim. Note that by the second orthogonality relations:

If we divide both sides by , we get a sum of algebraic integers on the left, and a rational number on the right. Thus, .

Claim 4G has no non-trivial 1-dimensional representations.

*Proof:* If is a non-trivial one 1-dimensional representation, then , so is abelian. This is a contradiction.

Claim 5For every , , there exists an irreducible representation such that and for some .

*Proof:* Let be the irreducible representations of , where is the trivial representation. By claim 2.3, for all , so for some . If is the character of the regular representation, then is if , and otherwise. Now, suppose divides for . Then

because is the only 1-dimensional representation of . This is a contradiction: if we divide both sides by , we get an algebraic integer on the left, and a rational number on the right. In particular, it follows that there is an such that for some and .

The following claim holds for any finite group.

Claim 6If there exists and a representation such that , then .

*Proof:* Because is unitary, it is diagonalizable: . Note that with equality if, and only if, . Thus, because

it follows by induction that for all . Thus, , and so is in the center of the image of , but is simple so it is isomorphic to its image and hence .

By the Sylow theorems, has a -Sylow subgroup . By proposition 1.2, is non-trivial. Thus, we can choose . By taking an appropriate power of , we can assume that . Note that , so is not divisible by , i.e. for some . Let be a representation such that and . Note that the the map

is a 1-dimensional representation of , thus, it is trivial, so , for all . Let be the eigenvalues of listed with multiplicity. Note that is a -th root of unity for all , because but for all . Let and . By the formula

each conjugate of , appear with the constant multiplicity because the product is stable under the action of , in other words, if has multiplicity , then each has multiplicity for . Therefore, because each , has conjugates, we have . In particular, Recall that for any primitive -th root of unity , , thus,

By claim 2.2, it follows that is an algebraic integer, and hence an integer. However, , is relatively prime to , so , and for some , because . Now, because

it follows that and . By claim 2.6, it follows that , which is a contradiction. Thus, is solvable.

**3. Consequences**

There are several key steps in this argument. The most important assumptions are that is simple, and for any , there is a prime power dimensional representation such that . Retracing the steps of the proof yields the following consequence:

Corollary 1Suppose is a simple group and is an irreducible representation of of dimension , for some . Let be a -Sylow subgroup of . If of order , and , then is abelian.

You can view a pdf of this entry here.

Let be a finite group that acts on a finite set, . Given elements and , we introduce the cycle notation, to denote that , but for all . We say that is a -cycle in . Conceptually, this is a natural construction: the action of on induces a map, , of into the symmetric group of . This allows us to decompose the image of any element into disjoint cycles.

Let the expression denote the number of -cycles of . Let be the set of all -cycles contained in elements of . In particular, for , because -cycles are fixed points of elements of , and fixes every element of . Note that if is nonempty, then acts on by conjugation: if , then is a cycle in , so . Parker’s lemma is the following:

Lemma 1 (Parker’s Lemma)The number of orbits of on is

At first glance, this lemma appears to be a straightforward application of Burnside’s orbit counting lemma:

In fact, the case is Burnside’s lemma: the number, , is the number of -cycles of , i.e. . However, for , if is nonempty, then . Thus, we cannot simply equate terms: , in general. The difficulties described above suggest that Parker’s lemma is a stronger statement than Burnside’s lemma. In fact, I am unaware of any proof in the literature that deduces Parker’s lemma as a consequence of Burnside’s Lemma. The aim of this post is to establish Parker’s lemma as a simple consequence of Burnside’s lemma.

**Proof: **We assume that is nonempty, otherwise the result is trivial. We note that by Burnside’s lemma

Define subsets by

For each , choose such that . Let be the pointwise stabilizer of . Then is the set of elements of that contain , and is the stabilizer of . Note that for , with . Indeed, if , where , then is disjoint from , i.e. , which is a contradiction. Thus, because , we see that is a partition of . Therefore, , and hence

because . Now, we count the order of and by counting points along the fiber of the canonical projection

Indeed,

Thus,

as desired.