New AMS Blog Post: Looking At the Prism

I just finished my third blog post for the AMS Grad Student Blog. It’s called Looking At the Prism, and tries to discuss the issue of bias in academia. You should read it, and then tell me what you think!

Consider this awesome interactive chart created by Ben Schmidt that allows you to see how frequently various words are used in RateMyProfessor reviews by gender and field. So for example, if you search the word “genius” we see that in every field, but especially math, it is much more frequently used in the reviews of male instructors as apposed to those of female instructors.

Well this doesn't look good...

– See more at:


2015 AG Institute in Utah: Day 1

Roughly every ten year’s since 1954 the AMS and other organizations have sponsored a large meeting for algebraic geometers from all over the world. This year happens to be the ten years since the last one in Seattle, and it also happens that I am lucky enough to be in attendance for all three weeks!!!

The conference is being held on the campus of the University of Utah, and from what I’ve seen so far — that is what I could see in the taxi on the way to the dorms — Salt Lake City and the campus seem great. (See generic mountain shot below…)

That one bridge.

That one bridge.

As I said the conference is three weeks long with each week being denoted (roughly) to a different themes:

  • Week 1: Analytic Methods, Birational Geometry and Classification, Commutative Algebra and Computational Algebraic Geometry, Hodge Theory, Singularities and Characteristic p-Methods.
  • Week 2: Derived Algebraic Geometry, Derived Categories, Geometric Representation Theory, Gromov-Witten and Donaldson-Thomas Theories, Mirror Symmetry, Tropical Geometry.
  • Week 3: Algebraic Cycles, Cohomology Theories, p-adic Hodge Theory, Rational Points and Diophantine Problems, Topology of Algebraic Varieties.

The complete schedule can be found here. The days are broken into two sessions a morning session (9am – 11:30am MT) with plenary talks and an afternoon session for contributed talks. For those who were not able to make it the morning lectures will be live streamed here! While the afternoon lecture will be recorded and eventually posted.

Needless to say I am beyond excited for the next three weeks, and look forward to seeing some amazing talks by some of the worlds experts in algebraic geometry as well as get a chance to meet other geometers!!! As a treat for you — whoever happens to read this blog — I will be trying to blog at least weekly updates!! Stay tuned!

PS: This post is one day late. Yesterday (7/13) was the first day of the conference.

Some lecture notes on p-adic Lie groups

This post is short, and hopefully sweet.

For the past couple of months, I’ve been meaning to clean and upload these lecture notes from a talk I gave early this semester on p-adic Lie groups! The talk was given as part of a weekly seminar attended by faculty and grad students at GT and Emory to learn about Robert Coleman and Claude Chabauty’s remarkable attempts to provide an effective proof of the Mordell Conjecture (a.k.a. Falting’s Theorem) using methods from p-adic analysis. The seminar was organized by Matt Baker (whose own wonderful blog can be found here).

You’ll find very few proofs: the goal was to give the statements and intuition behind the main structure theorem. The target audience is someone with good general knowledge of (real) differential topology and of the p-adic numbers, but possibly little or no knowledge of (real) Lie group theory. (i.e., the audience is precisely myself prior to preparing these notes.) There are bits at the beginning and end which are intended to relate these ideas to Chabauty’s method and to Bill McCallum’s paper on Chabauty for Fermat curves. In the talk, they were intended as motivation, but for anyone seeking to learn about Lie theory, they’re basically beside the point.

Anyways, enough delay – the file can be downloaded by clicking here. I only ask that anyone using these notes let me know of any errors you may find!

P.S. Last semester (in the Yellow Pigs’ collective final semester at Michigan) I had a class with Bill Fulton on schemes and cohomology in which he presented a beautifully geometric proof of Serre duality for curves, purportedly due to George Kempf. At some point, I’d like to collect this proof from several badly organized notebooks and put it here as well, and in the meantime I’ll probably be posting a series of notes related to schemes, sheaves, and cohomology building towards that goal, so stay tuned!

What is group cohomology?

Of all the ubiquitous and mysterious algebraic tools that appear in number theory and arithmetic geometry – and surely, there are many – group cohomology remained an impenetrable black box to me for quite a while. Although I still consider myself a near absolute beginner in the topic, my goal in this post is to demystify it at least a little for those unfamiliar with it and hoping to get a window into its inner life.

The window through which we will peak is called Galois descent.

First, I’m going to to explain by example the type of glimpse I want to give. If you’ve ever studied any algebraic topology, you’re surely familiar with the singular cohomology of a topological space, and you also likely know that the cohomological cup product gives these groups (collectively) the structure of a graded-commutative ring. What you may not know, however, is that when the modern definition of cohomology was first formulated in the mid-1920’s, mathematicians like Schubert and Poincaré had already been computing these rings for more than half a century!

So how were they studying cohomology before cohomology existed? The answer is called the intersection ring. Consider a closed smooth manifold, and think about the free abelian group on the set of all closed submanifolds. We can define a natural equivalence relation by identifying submanifolds that can be wiggled onto eachother. (More precisely, we use cobordism: kill off a linear combination of submanifolds if their disjoint union is the boundary of a submanifold – not necessarily closed – in higher dimension.) Finally, we can define a product by wiggling pairs of submanifolds until they intersect transversally, and then taking intersections. This gives us a graded-commutative ring, where the grading is by codimension.

Now if we triangulate all the submanifolds in the picture, we can turn them into simplicial chains. In fact, the relation of cobordism turns out to be exactly equivalence-mod-boundaries, and the group we constructed is exactly the direct sum of singular homology groups! Better yet, since we’re on a closed manifold, we can use Poincaré duality to move the ring structure to cohomology, and there it miraculously agrees with the regular old cup product! This alternative intersection-ring description of the cohomology ring has a narrower definition (it doesn’t work on arbitrary topological spaces), but it has much richer geometric intuition, and it was the object of interest to the 19th century geometers mentioned above.

My goal here is less ambitious, but it has a similar flavor. Group cohomology is an invariant of a group acting on another group. In a very, very special case (in the case of a Galois group acting on the automorphism group of one of a few types of interesting object) I want to give an explicit and concrete description of the first cohomology of this action. We’ll finish up by using this description to give a wonderfully simple proof of Hilbert’s theorem 90, which I claim essentially just says that all one-dimensional vector spaces are isomorphic!

What is Galois descent?

Galois descent studies the following hilariously vague problem: let k be a field, probably not algebraically closed, and let X be a thing over k. It might be:

  1. An algebraic variety, maybe a plane conic, for instance.
  2. A vector space.
  3. An algebra.
  4. More generally, we could consider a vector space equipped with some other kind of extra linear structure, such as a linear or bilinear form. (Note that an algebra is a special case of this; it’s just a vector space with a bilinear form satisfying some identities.)

Pretty much anything goes, provided that it has one important property: we need to be able to “tensor” with field extensions, to move our object X to bigger fields. In cases 2-4, it’s obvious what we mean: a literal tensor product of vector spaces. In case 1, it’s also not too bad: tensor product just means regular old algebraic geometry base change. (Take the same equations and put them over a bigger field!)

In general, we’ll write X_L or X\otimes L for this base change when L/k is the field extension.

Galois descent attempts to answer the following question: given a field extension L/k and an L-thing X, which k-things become isomorphic to X when we tensor everything with L? In other words, when is Y_L=X? These are called the Galois twists or just twists of X.

When our extension is a separable closure, and X is the projective line, we are classifying plane conics up to isomorphism. When X is a matrix algebra, it is a (very nontrivial) fact that the objects Y are precisely the central simple k-algebras of the same dimension, and the theory presented here is the first step in describing the Brauer group of a field in terms of cohomology.

(In fact, with the right formalism, the examples of a matrix algebra and a projective space are basically the same!)

For a general extension, this is a hard problem, but when the extension L/k is Galois, we can exploit the Galois group in a fabulously cool way.

Actually doing descent

First, let’s set up some notation.

  1. G=\text{Gal}(L/k) is the Galois group of our field extension.
  2. A=\text{Aut}(X) is the group of automorphisms of our object X.

The surprising goal of the next couple paragraphs is to construct a map

\phi:\{Y\text{ with }Y_L\cong X\}\rightarrow\{\text{maps of sets } G\rightarrow A\}. 

Then, we’ll modify the target of this map to get a bijection, and find out that the set on the left can be naturally thought of as a cohomology set!

So how does this map work? Well we obviously need to start with some suitable Y, and then we need to choose an isomorphism f:X\rightarrow Y_L. (For now, our map will depend on this choice. We’ll fix that later.) Additionally, we need some \sigma\in G, and we need to explain where it goes in A. To do this, notice that an automorphism of the field extension induces an automorphism \sigma_X of X and one \sigma_Y of Y_L.

Beware: these automorphisms are weird! In particular, suppose Y is a real vector space and our field extension is \mathbb{C}/\mathbb{R}. Then complex conjugation acts on Y\otimes\mathbb{C}, but it is not a \mathbb{C}-linear map; it’s only \mathbb{R}-linear! This is something to be aware of, but it won’t be an issue, in the end.

Now we have maps \sigma_X:X\to X and \sigma_{Y}:Y_L\to Y_L, and together with our original isomorphism f, we get a square diagram of maps. But lo! This diagram does not commute! Of course, when we have a square diagram of isomorphisms, its failure to commute is measured by a particular automorphism of any one of the objects present – we’ll pick X. The point is that if we start at X and work around the square, composing as we go, we’ll get an automorphism – an element of A, just like we wanted! So our map is given by

\phi(Y,f)(\sigma) = \sigma_X^{-1}\circ\sigma_Y^f,

where exponentiation denotes conjugation, to simplify the notation: \sigma_Y^f=f^{-1}\circ\sigma_Y\circ f.

Note that we kept the isomorphism f as part of our notation, since the map isn’t quite well-defined on Y alone, yet. We’ll fix this soon. The rest of what we want to do will now be summarized as a series of exercises, some easy and some hard, and the interested reader is encouraged to try them all out!

From this point on, we’ll forget about case 1, where X is an algebraic variety. While this is an important and interesting case of descent, it is harder, and the statements are not as simple and explicit as what we’re about to prove. So now let’s take X to be a vector space, possibly with some additional linear structure.


  1. In your favorite case of our construction (variety, vector space, algebra, etc.), check that this the map \phi(Y,f) really lands in the automorphism group. That is, check that even though \sigma_X and \sigma_Y are only k-linear, the whole composite described above is actually L-linear.
  2. Verify the identity


    Such a map is called a twisted homomorphism, and we’ll denote the set of these maps by \text{THom}(G,A).

  3. Consider any twisted homomorphism \psi:G\to A. Produce some Y/k which becomes isomorphic to X over L (by some isomorphism f, for instance), and with the property that \phi(Y,f)=\psi. (Try to give an operation of G on X using your map, although it won’t quite be a group action. Now check that the subspace of invariants is a k-subspace – with whatever linear structure you imposed – and that the inclusion becomes an isomorphism when we pass back to L.)
    We now have a “surjection” (which is still not yet well-defined)

    \phi:\{Y\text{ with }Y_L\cong X\}\rightarrow\text{THom}(G,A).

  4. Next, consider another isomorphism g:X\to Y_L. The composite map \alpha=f^{-1}\circ g is an automorphism of X. Verify that the \phi(Y,f) and \phi(Y,g) are related by the formula


    Now if we let ~ denote the equivalence relation that we’ve just described, we have a well-defined surjection

    \phi:\{Y\text{ with }Y_L\cong X\}\rightarrow\text{THom}(G,A)/\sim.

  5. Finally, check that this map is injective by proving the converse to 4: if two twisted homomorphisms are related as in 4 by some automorphism \alpha\in A, then they in fact come from the same k-object Y, just with a different choice of isomorphism X\to Y_L.

So the moral is this: objects that become X after a base change along L/k (provided L/k is Galois) are the same thing as twisted homomorphisms G\to A up to some equivalence relation. Let’s call that latter set H^1(G,A).

While it may not be obvious whether this thing is any easier to compute than what we started with, it’s pretty easy to see that these sets, when A is abelian, have a natural group structure, and are really Ext-groups, which can be studied with all the machinery of homological algebra. When A is not abelian, the picture is not as nice, but often we can understand them by fitting A into short exact sequences with abelian groups, and then studying long exact sequences. (Details of all this can be found in any number of books, such as Serre’s Local Fields, the appendix to Silverman’s Arithmetic of Elliptic Curves, and Cassels and Froelich’s Algebraic Number Theory.)

We conclude with a final application, which will apply these ideas “in reverse,” computing a cohomology group by a trivial observation about base change. This theorem is commonly referred to as Hilbert’s theorem 90. If you’ve taken a basic graduate field theory or number theory class, you may have seen a form of this result, although you might be surprised by how easy the proof is using the observations here!

Theorem Let L/k be a Galois extension of fields with Galois group G. Then

H^1(G, L^{\times})=0.

Proof  The group L^\times is the automorphism group of a 1-dimensional L-vector space (with no extra linear structure). The k-vector spaces V with V\otimes L\cong L are really just the one-dimensional ones, and those are all isomorphic! Therefore, this cohomology set has only one element. \square

This fact, which may seem esoteric right now, can be combined with the long-exact group cohomology sequence to give an extremely easy proof of the main result of Kummer theory, which totally classifies cyclic extensions of fields with enough roots of unity. The results of Kummer theory are essentially the model and inspiration for class field theory, which is also typically proven using group cohomology.

Exercise  Note that L^\times = \text{GL}_1(L) . Generalize this proof to any \text{GL}_{n}(L) ! In fact, the same proof works for n an infinite cardinal.

I’ll end with a question about something that I suspect is probably well-studied, but that I certainly know nothing about. Essentially, the goal of representation is to relate arbitrary groups to linear groups like the ones whose cohomology we’ve just come to understand, and this begs the following question:

  • Under what circumstances does a subgroup of a linear group consist of the automorphisms of some linear structure?
  • Can we do this systematically, and find canonical linear structures associated to a wide class of groups, particularly linear structures whose twists are easy to understand?

Fields Medalists 2014!

The 2014 Fields medalists have been announced, and the winners are:

  • Artur Avila (IMPA – Dynamical Systems): “for his profound contributions to dynamical systems theory, which have changed the face of the field, using the powerful idea of renormalization as a unifying principle.”
  • Manjul Bhargava (Princeton – Number Theory): “for developing powerful new methods in the geometry of numbers, which he applied to count rings of small rank and to bound the average rank of elliptic curves.”
  • Martin Hairer (University of Warwick – Stochastic PDE’s): “for his outstanding contributions to the theory of stochastic partial differential equations, and in particular for the creation of a theory of regularity structures for such equations.”
  • Maryam Mirzakhani (Stanford – Teichmuller Theory & Ergodic Theory): “for her outstanding contributions to the dynamics and geometry of Riemann surfaces and their moduli spaces.”

The big news, aside from the announcement itself, is that for the first time in its almost 80 year history the Fields Medal is recognizing the outstanding work of a female mathematician! Hopefully this will continue in the future, and the award will continue to become more inclusive. Congratulations to these four outstanding mathematicians! (Hopefully in the coming days I’ll have time to track down/write a more complete biography/research sketches for the winners, but for right now you’ll just have to settle for their Wiki entry or homepage.)

Update #1 (8/12/14 – 1:50pm): I added the prized citations listed by the International Mathematical Union, which can be found here. The IMU also has brief sketches of each of their works.

Update #2 (8/12/14 – 7:43pm): It seems that traditional media sources have caught wind of this, and are beginning to release their stories on the announcement. Here are a couple of the more memorable quotes from some of these:

  • A Women Has Won the Fields Medal (Slate): “Which — and I apologize to mathematicians out there for whatever I do to butcher this description— roughly means that she considers abstract questions related to non-Euclidean entities such as, for example, the surface of a pretzel.”
  • Top Math Prize Has Its First Female Winner (NYT): “While women have reached parity in many academic fields, mathematics is still dominated by men, who earn about 70 percent of the doctoral degrees. The disparity is even more striking at the highest echelons. Since 2003, the Norwegian Academy of Science and Letters has awarded the Abel Prize, recognizing outstanding mathematicians with a monetary award of about $1 million; all 14 recipients so far are men. No woman has won the Wolf Prize in Mathematics, another prestigious award.”
  • Top Mathematics Prize Awarded to Women for First Time (Time): “Daubechies added: “At the IMU we believe that mathematical talent is spread randomly and uniformly over the Earth – it is just opportunity that is not. We hope very much that by making more opportunities available – for women, or people from developing countries – we will see more of them at the very top not just in the rank and file.”’
  • Stanford’s Maryam Mirzakhani wins Fields Medal (Stanford): ‘”I don’t have any particular recipe,” Mirzakhani said of her approach to developing new proofs. “It is the reason why doing research is challenging as well as attractive. It is like being lost in a jungle and trying to use all the knowledge that you can gather to come up with some new tricks, and with some luck you might find a way out.”‘

As you can see much of the focus is rightfully on the ground breaking recognition of Prof. Mirzakhani. That said most of the current articles are somewhat shallow, and I hope in the  coming days this inspires a more widespread and thoughtful discussion of the systemic barriers that have existed and continue to exist in the mathematical community, and how we can work to make mathematics more inclusive. (Remember this moment is remarkable in part because it took 70+ years and 52 medals before it happened…)

Some of the best coverage I’ve seen has come from the Simons Foundation and Quanta Magazine, which has not only great articles on each recipient and their work, but also lovely videos with each winner. (I do wonder how did the manage to get these done so fast. Did someone from the IMU tip them off to who would win (when are winners decided anyways? Or did they just go around to everyone they thought might win, and ask to do a video?)  Either way this is great!  

(PS: I am so excited about all of this that my dinner is now cold because I couldn’t stop… Shoot :/, but really :D)

Integrals, logarithms, and staircases

What noise does a drowning analytic number theorist make?

“log log log log” (Courtesy of Prof. Bob Vaughan, who must have made this joke at least a dozen times in a number theory class last semester. By the way, at least one source attributes this to number theorists Ram Murty and Barry Mazur.)

Alas, what I have to do say today has absolutely nothing to do with analytic number theory. Instead, I’d like to tell a story about the logarithm function, which will ultimately tie together topics in complex analysis, topology, and group theory. Before I do that though, I want to relate an interesting conversation I had with a couple of months ago with a first-year student, which led me to the first draft of this tale.

When I asked what he’d been working on lately, he mentioned two things: first of all, the second-semester introductory class for math majors that he’s taking was just starting to study groups, and this new manner of abstraction was difficult; and secondly, he had been reading some complex analysis on the side and was very fascinated by it.

Being a very algebraic sort of cat myself, I felt that this was an injustice that needed to be righted, and so proceeded our conversation in which I tried to motivate his study of groups via complex analysis. (I suspect that I did a pretty bad job. Fortunately, this was a rare occasion in pedagogy when one gets a second chance: he was at least interested enough to ask me about it again the next day, when I articulated the ideas a bit more clearly. Hopefully, this third chance will be even better!)

I’m going to start by saying a few words about a purely analytic object, the logarithm. We’ll attempt to formulate a definition for it as a function on the complex numbers, but quickly run into trouble. There are a number of ways to resolve the issue we’ll face, but (remarkably!) the most elegant solution requires tools from abstract algebra.

This post will be long, but hopefully I’ve organized it in a manageable way. Basically, I’m going to talk about the logarithm for a bit, and then a problem will arise. We’ll spend some time fixing it, and then we’ll solve a similar problem, and finally I’ll discuss the general situation. The last part is unfortunately going to require some group theory, but my hope is that the first few sections will be accessible to anyone with some calculus background, an understanding of complex numbers, and some time to spend pondering!

One last aside for readers not familiar with complex analysis: I’ll throw around the word “holomorphic” a lot. Don’t mind it! For a function defined on a subset of the complex plane, holomorphic means exactly the same thing as differentiable. For our discussion, the word is superfluous – it’s just such a part of my lexicon that I probably can’t avoid saying it!

The logarithm

Surely the reader, regardless of her familiarity with complex analysis, is more than familiar with the logarithm function. (If anyone asks, I’ll always use the word “logarithm” to refer to the natural logarithm, with base e. This’ll show up when I write down formulas, but it won’t be important at all.) The log is a beautiful object with many remarkable properties, but the one we’ll be most interested in here is that for any positive real number x, we have

\displaystyle\log(x) = \int_1^x \frac{dx}{x}  (1).

The logarithm is most important, or course, because it is inverse to the exponential function: for any real number x, e^x is positive, and in fact \log\left(e^x\right)=x. Unlike the log, the exponential function is easy to extend to the complex plane: if we look at its regular old Taylor series

\displaystyle e^x = \sum_{n=0}^\infty \frac{x^n}{n!}

from a calculus or analysis course, we’ll be relieved to know that this sum makes sense everywhere on the complex plane, so we can take it as our definition of the exponential. We now have enough background to really ask a question:

Question: Is there a holomorphic function f defined on \mathbb C \smallsetminus {0} so that f\left(e^x\right)=x?

(Why \mathbb C \smallsetminus {0}, you ask? Well the exponential function has no zeros, so there’s no need for its inverse function to be defined at zero! As we’ll eventually see, this is a necessary precaution.)

The first claim that I’ll make is that, if there is such a function, it definitely needs to satisfy something resembling equation (1). Why is this? Well, for any complex number x, we have

\displaystyle 1=\frac{d}{dx}x=\frac{d}{dx}f\left(e^x\right) = f'\left(e^x\right)\cdot e^x.

Now if we make the substitution y=e^x, this gives us that

\displaystyle f'(y) = \frac 1 y

for any nonzero complex number y. In other words, if there is a “complex logarithm,” whatever it is, we know what it’s derivative has to be.

So if we know it’s derivative, then it can’t be too hard to figure out what the function actually is: why don’t we just integrate? This works – sort of. Ultimately, we will just take equation (1) as our definition. The trouble is, equation (1) doesn’t actually make sense (yet)! The reason for this brings us briefly to the world of topology.

Integrals and paths

So what’s wrong with formula (1)? The issue here is that in the complex plane (unlike on the line) there many ways of getting from 1 to x, and hence of evaluating that integral.

Fortunately, this kind of thing isn’t usually an issue in complex analysis: if you have two different paths from a to b, and you integrate a function f which is holomorphic in the entire region between those paths, then you’ll get the same integral. (In other words, you can “slide integrals around,” as long as you stay inside the region where your function is holomorphic.) For this reason, we can still say that

\displaystyle e^x = \int_0^x e^x dx

without any issue. Since e^x is defined everywhere, it’s going to be defined in the region between any two paths we choose, so we always get the same integral. (Just to be clear, I’m only saying that the equation is at least a complete sentence – the fact that that it’s true isn’t quite obvious!)

Integrating e^x is no problem. 1/x, on the other hand, is problematic because it’s not defined at the origin. If we want to compute \log(-1), we can imagine choosing a path from 1 to -1 which goes above the origin, and one which goes below it, and we have no guarantee that these two paths have the same integral. In fact, they don’t.

In an extreme case, we could have one path which stays put (and so obviously has integral zero) and one path that branches out and then returns to where it began (which might have nonzero integral). It turns out that the question of closed loops integrating to zero is the only thing we need to worry about. Worded differently, if we can get rid of discrepancies coming from closed loops, we’ve gotten rid of all discrepancies!

(The interested reader might want to check this herself: given two paths from a to b whose integrals are different, can you come up with a closed loop that has nonzero integral?)

In the particular case of the logarithm, I can tell you exactly what all the integrals around closed loops are: if your loop circles the origin n times counterclockwise, then the integral is 2πni; if it does so clockwise, then the integral is –2πni. In fact, we might just stop here! A pessimist might remark that we have given a conclusive negative answer to our question: there is not any function holomorphic on \mathbb C \smallsetminus {0} which is inverse to the exponential.

(The details of this argument might be fun to work out. Suppose you had such a function, and compute its value in two different ways, using the integral formula. The two results you get have to be equal, since they’re the value of the same function at the same point! This’ll give us some silly formula like \log(x) = \log(x) + 2\pi i, which is obviously nonsense, so our original function must have been nonsense!)

But this is the wrong way to look at it; the trick here is to run away from our problems. Apparently, \mathbb C \smallsetminus {0} is the wrong world for our logarithm function to live in. If we want this to work, we need a world where there are no loops enclosing the origin.

Imagine a spiral staircase heading off to infinity above and below, and with steps that are infinitely long, extending away from the center. There’s no opening in the middle – this is the kind where the stairs all connect to a support beam in the center. Put that support beam perpendicular to the complex plane, sticking right through the origin. Believe it or not, I claim that this staircase is the place where the logarithm function lives. (The stairs themselves are a bit of a red herring – go ahead and imagine the stairs smoothed out into a ramp, it you prefer.)

So how does this work, exactly? Start by picking a spot on the stairs so that you’re directly above the number 1 on the complex plane. That point has logarithm zero. Now think about a path in the complex plane that runs from 1 to, say, 3+5i, and wraps around the hole in the origin – oh I don’t know – twice, counterclockwise. Start from where you’re standing on the staircase, and trace your way along that path.

Staircases like to change your altitude, so obviously as you circle the origin twice, you’ll be forced to move up two floors, and you’ll finish standing directly above 3+5i. Now consider the integral of the function 1/x along this path in the complex plane. Whatever number that gives, we say that this is the value of the logarithm at the spot you’re standing on the staircase, which we can think of as “one of the logarithms at a spot that’s above (or below) 3+5i.”

So what would happen if your path had wrapped around three times and ended up at the same spot? Well, you’d get a different integral, and hence a different value for your logarithm function, and this would be no problem at all, because you’d be at a different spot on the staircase: you’d circle around the support beam one extra time, and rise up one extra floor!

We had a rule for calculating a function, but our rule was ambiguous: there were many different ways to determine its value, and they didn’t always agree. We solved our problem by just using all the values, all at once, and creating a bigger space where there was no conflict. There were infinitely many different “logarithms” we could compute at each point, so our new “staircase space” had infinitely many points lying over each of our original points, each giving us one of the values of our functions.

A double-staircase

Now let’s play around with another example. Just like before, we’ll start with a function that’s defined globally by a simple formula, and try to define its antiderivative – just like before, we’ll run into issues when closed loops have nonzero integrals.

Problem: We’ll define an “antiderivative” for the function g on \mathbb{C}\smallsetminus \{0,1\} given by

g(x) = \frac{1}{x(x-1)}.

Just like I did before, I’ll just go ahead and tell you what the integrals are here. (If you believe the integrals that I gave you in the first example, there’s a simple trick for computing these ones from those. You might want to try thinking about it for a while.)

  1. If we take a loop encircling the origin once, counterclockwise, and not the point 1, then the integral will be –2πi.
  2. If we take a loop encircling the point 1 once, counterclockwise, and not the origin, then the integral will be 2πi.
  3. From this, we can see that a loop circling both points once has integral zero.
  4. In fact, if we take a loop which encircles the origin n times counterclockwise (taking negative if the loop circles it clockwise, for instance) and encircles 1 times, then its integral will be (n-m)2πi.

Now, as before, I’ll describe the space where this integral lives. In fact, it’s pretty simple; imagine a “building of infinitely many floors” as before, but this time with two stairwells positioned next to each other –  one going counterclockwise (over the origin) and the other running clockwise (over the point 1).

If you walk along a path that encircles both staircases, you’ll go halfway down one staircase, and then right back up the other one, and never change floors! This is consistent with bullet point 3. above. I’ll leave it to you to convince yourself that 1, 2, and 4, all make sense in this space as well – if you were comfortable with the logarithm example, then this one shouldn’t be much trouble at all.

Heading about groups

(Coming up with fun headings is hard.)

From now on, I’m going to assume that the reader knows a thing or two about groups. (The notions I’ll use: group, product group, homomorphism, kernel. Nothing very technical.)

In both of the examples we’ve studied so far, we solved our problem in essentially the same way: we “broke” the “bad” paths. We took the closed loops with nonzero integral (such as a loop around the origin) and created a new space where they were no longer closed.

I’ve claimed that there’s a group floating around somewhere, and that we can use it to understand this process. So what’s the group? The elements of this group are loops – or to be precise “sums of loops, with integer coefficients.” Let’s make this a bit more explicit:

Definition Let U be a subset of the complex plane. Then we define a group G to be the set of expressions of the form n1 α1 + … + nk αk where all the n‘s are integers and the α‘s are loops in U. The operation is the obvious thing: just collect terms and add up coefficients! (Note that this thing is Abelian.)

Well this is kind of silly – we have lots and lots of expressions, a lot of which basically represent the same loop.  For instance: we should be able slide loops around (within U) without changing which group element we’re talking about, and take a “pinched” loop (one that crosses over itself) and replace it by the sum of the two component loops, again without changing anything. Let’s do that.

(If you’re not sure why I think that we should be able to do these things, then consider this: we’re only really interested in integrals. Eventually we’l be “integrating along” these group elements. However, the two little tricks I just described will never change the value of the integral, so there’s no reason why we’d want to keep track of those little differences!)

Definition Let H denote the subgroup of G generated by elements of the following form:

  • If α and β are loops with the property that we can slide from one to the other without leaving U, then stick α-β in there. (Think about, for instance, two loops in \mathbb{C}\smallsetminus {0} which both run once clockwise around the origin.)
  • If α is a loop which is “pinched,” made up of two smaller loops β and γ , then let’s include α-β-γ. (Have in mind the image of a figure-eight loop, with two smaller loops that make it up.)
  • Finally, if α and β are the same loop but traversed in opposite directions, then take α+β too.

These are all things that we agree should be zero. Finally, define H1(U) to be the quotient group G/H.

G was way too big, but now magically we have a new group where lots of reasonable equations hold. For instance, if α and β are “reverses” of one another, then α = -β! This is the kind of group we want – unlike G, its structure actually reflects the geometry of U.

Also, note that we didn’t really use anything about being a “subset of the complex plane.” We could apply this construction, for instance, to our “staircase spaces” that we described before – in fact, we will do that.

(By the way, what we’ve just constructed is called the first integral singular homology group of U.)

Now let’s compute some H1‘s! My first claim is that if U is \mathbb{C}\smallsetminus {0} (the set from our logarithm example), then H1(U) is isomorphic to the group of integers (under addition). For a loop α in U, we’ll say that the corresponding integer is the number of times that α circles the origin counterclockwise. If it goes clockwise instead, it’ll be a negative integer. If it’s clockwise sometimes and counterclockwise other times, that’s not a problem either – more of those turns will “cancel,” and in the end we’ll still get just one integer.

It’s easy to see that this map H1(U) → Z is surjective – we can get any integer we want by circling the origin the right number of times. Seeing that it’s injective is harder, but fortunately it is true. We won’t worry about that.

Now how about our second example – what was the group there? I claim that, for a reason similar to the first example, this group is isomorphic to the product x Z. Can you see why? (We have two holes now – apply the construction from the first example to each one separately, and then stick them together to get the right map! Just like before, don’t worry about why it’s injective, unless you know some topology.)

There’s one last object we need in order to make use of these groups: a homomorphism. This part is easy, but we need one piece of extra structure. Up until now, we just had a set U. Now let’s fix a holomorphic function f on that set.

Definition We now define a map ∫ : H1(U) → \mathbb{C} taking a loop α to the complex number \int_\alpha f, the integral of f around that loop. As for sums of loops, just integrate the loops separately, and them add them all up with the right coefficients.

Of course to see that this is well-defined, we need to check that the integrals of all the things in the subgroup we killed are zero. This isn’t hard, and it’s also not terribly relevant, so I’ll forget about it.

Here’s the key observation about this homomorphism: we want it to be zero! If this is the zero homomorphism, then every loop integrates to zero, and we have no trouble, there’s no obstruction to defining an antiderivative, and we can just do it the naive way. On the other hand, every loop with nonzero integral presents a struggle. We need to break it, or else there’s no hope for our antiderivative.

Preferably, we don’t want to break anything extra. We only want to break loops with nonzero integrals, so that we can change the space as little as possible while making it a safe place for our antiderivative to live. In other words, we want to turn U into a new space \widetilde{U} with the property that H^1\left(\widetilde{U}\right) = \ker \int. We want a space whose loops are just the loops that already had zero integral.

(In particular, we don’t just want those groups to be isomorphic, we want them to be canonically isomorphic. Here’s what I mean: \widetilde{U} should come with a “projection map” to U, and that map will give us a homomorphism H^1\left(\widetilde{U}\right)\to H^1(U). This map should be injective, and its image should be the kernel. We really want even more than that, but you get the point – the isomorphism should arise from the geometry, not just the algebra.)

Finally, let’s look at the examples and check that this all makes sense. In the logarithm example, the integral map whs injective! Every loop that actually circled the origin had nonzero integral, so the kernel was zero. I’ll leave it to you to draw some pictures and convince yourself that H1 of the single staircase space was also zero. In other words, every loop in that space can be contracted to a point. (The idea is that you can circle the origin and still end up where you started: if you do, you’ve changed floors!)

In the second example, our map was not injective – some loops did integrate to zero, for instance those that looped around both of the holes the same number times, and in the same direction. In other words, if we think of the group as x Z, the kernel is the subgroup consisting of things like (n, n). This group is isomorphic to Z.

So is H1 of our double-staircase isomorphic to Z? It shouldn’t be very surprising that it is: any loop that doesn’t circle both wholes the same number of times lands on a different floor, so the only closed loops are the ones coming from (n, n)‘s anyways!

One last point

I’ve said a lot already and, this is getting pretty long, but I want to make one last point before moving on. We know now sort of what we need to do to get the right space for an antiderivative: compute H1 for our set, find the kernel of the integral map, and then try to cook up a space whose H1 is (canonically – see the note above) that kernel.

But where do we get that space? Well that’s unfortunately where I’m going to stop explaining. The fact that the right space exists is not totally obvious, and it requires a fair amount of topology and algebra to prove.

(The interested reader can find this theory in most books on algebraic topology, under the name “covering spaces.” Just so you’re not shocked, you won’t see any H1‘s there, since when the things we want aren’t subsets of the plane, that’s not really the right group to talk about.)

Anyways, I hope this lengthy piece can be worth someone’s time other than my own. Until next time, enjoy the following confused dogs.

Hi Everybody

It’s been a while since an introduction and so to get the monkey off of my back, here is my introduction.  My name is Solly Parenti.  Soon, I will be graduating U of M with a B.S. in Honors Mathematics, and I will be continuing my math studies at graduate school next year.

Like most of the rest of the Yellow Pigs, my interests lean are on the algebraic side.  Currently, I’m interested in Algebraic Geometry and Number Theory, but I still have a lot to learn and so these interests may change.  I’ll attempt to post some math stuff soon.

I’ll try to post mostly about math, but I don’t really know how this blogging thing works.  So if I get in a rut, I will probably post cat pictures.  I have one cat at home, her name is Lizzie and she is incredible.  I’ve got tons of pictures and I take more every chance I get, so I will not run out of material for that.  Here’s one of us taking a nap together: