We will be working over the projective line over a field. The basic question is that, since we know that there exist nontrivial groups between line bundles on , what explicit extensions do they correspond to?

For example, given that we we know , there should be a nontrivial extension

.

Recall that all vector bundles over splits into a direct sum of line bundles. So . By taking determinant bundles, we know , so . Furthermore, since it is a nontrivial extensions, we need both , or else we would have a section . Therefore, .

One way to produce an example of such a nontrivial extension is from the Koszul complex. Recall that the complex

is exact. Turning this into sheaves over , we get exactly a short exact sequence

.

Notice that this only works because is identified with the zero sheaf when we try to turn it from a -module to a sheaf over .

Repeating the argument above, we see that any extension

.

must have for and . We might ask if all such values of are possible. The answer is yes, and we can construct them, though I didn’t think of something as nice as the Koszul complex.

________________________________________________________________________________________________

Edit: Sometime after posting this, I realized there was a way to get these extensions in exactly the same way as above with the Koszul complex. Namely, we take the exact sequence

and turn it into sheaves over . This yields

.

However, it’s at least useful for myself to remember how the machinery below works in a basic example.

________________________________________________________________________________________________

To do so, we first recall the general recipe for how to get a extension from an element of the Ext group. Instead of just recalling the construction, we will derive part of it in a way that I think a beginner could come up with, as I don’t see this written down either. Those who aren’t interested can just skip the next section.

Since projectives are easier for me to think about than injectives, we’ll motivate this by constructing extensions of -modules for some (commutative) ring . A natural question is to ask, if we fix modules and , what modules can fit in the middle of the short exact sequence

?

I know I have had the experience of trying to ask this on mathstackexchange and getting a bunch of answers telling me to learn homological algebra and Ext, which is more effort than a poor undergraduate wants to spend to get at a concrete problem. However, I think there is a natural direct approach.

Instead of being in the middle of a short exact sequence, we would instead like to have a presentation for . For example, of and and are finitely generated (which is a lot of the time), then this allows us to compute explicitly. To do so, we want to find a surjection onto .

To do so, we can find a surjection onto from a free module, and lift it to a map . Then, we have a surjection . This gives us the diagram:

Here, is defined to be the kernel of . The top row is exact by the 9-lemma. Therefore, we have recovered the fact that there exists some map such that is the pushout of

.

Conversely, if we have such a map , then we can construct the first two rows of the commutative diagram above, and 9-lemma imples the exactness of the last row. Therefore, constructing all extensions if and are finitely generated abelian groups, say, is not a mysterious thing.

The main issue I see with this approach is that it’s not clear to me how to classify that two such maps give equivalent extensions precisely when it extends to a map directly from the large diagram.

Now, we want to use the section above to construct our extensions. One technical issue is that the quasicoherent sheaves don’t usually have enough projectives, so we would have to work with injective objects instead. The same argument in the previous section works exactly, so, we can fix an injective object containing . Let the cokernel of be . Then, each in the middle of a short exact sequence

.

is a pullback of the diagram

for some map and conversely, every map gives a pullback diagram, and the pullback fits into a short exact sequence .

Unfortunately, I don’t know of a good way of writing such an down. To work around this, we note that the argument we used to construct extensions didn’t use the full strength of injectivity. Namely, we only need to know that the injection extended to a map .

The long exact sequence applied to in yields . Therefore, it suffices for .

Therefore, instead of using an injective object for , we can use any object whose first cohomology vanishes. One source of such an object is the Cech complex. As an abuse of notation, if is a sheaf on , I’ll write for an open as , where is the inclusion.

The Cech complex gives us the exact sequence of sheaves

,

so we have and .

Recall that a basis of is represented by the sections that sends 1 to for . So it is natural to ask what extensions these sections represent.

Reading somebody else's computations isn't usually that enlightening, so I'll state the answer first. I got that the rank 2 bundle given by has the transition map given by

from to , where . To get this answer, we need to compute the pullback

Over , a section of the pullback is equivalent to choosing sections , over and respectively and a section of , such that .

Equivalently, . So a basis for over is given by , and is determined uniquely by and .

The situation over is exactly the same. To compute the transition map, we need to take a compatible choice over , express it in terms of the basis over , and then see what it should be in terms of the basis over . In the basis over , it is , and in the basis over , it is .

This means the transition matrix is given by , which is what we said above.

Finally, the identify , we need to diagonalize our transition matrix. Since we are allowed to change bases over and over , we are allowed to multiple on the right by an element in and on the left by an element of . Through row and column operations, we get

.

.

This means the section that sends 1 to represents an element in that corresponds to an extension

.

]]>I think in the future, I’ll add posts here either to 1) explain intuitively what happens in a paper of mine or 2) write the proof or idea of something that is hard for me to find in the literature. The audience for 1) is going to be really small, considering the lack of papers and people who read them (but I feel sorry for them), and 2) would be to help myself learn.

This time, I’ll try to outline intuitively what went on in my RSI project 5 years ago, the reason being that the main ideas (I think) are pretty well-disguised behind the technical details, which are really just a bunch of casework. The paper is here: http://arxiv.org/pdf/1107.1756.pdf.

As for motivation, we recall that there is the open problem of trying to find the largest subset of that contains no three term arithmetic progression. Most of the research has been focused on finding asymptotics for the size of given large . I won’t claim enough familiarity with the current literature to give the best current bounds, but I dimly recall there was a lower bound constructed by Elkin and an upper bound by Bourgain five years ago, and that they were the best then.

Disregarding the current literature, the most naive way to approach this problem is to use the greedy algorithm. Namely, you start with empty, and then repeatedly add the smallest number possible that does not violate the condition that there is no 3-term arithmetic progression.

It turns out that the greedy algorithm generates a sequence that contains precisely the integers that have only 0 or 1’s as digits when expanded in base 3. This extremely explicit solution is nice, but we can also easily tell that it is horrible asymptotically. For example, the greedy algorithm gives , while Elkin’s construction alluded to above gives .

The condition that the sequence has no three term arithmetic progression is equivalent to avoiding (nontrivial) solutions to , where nontrivial here means ruling out $x=y=z$. We would like to generalize this. For example, if you generate the greedy sequence that contains no solution to , where are all different, you get numbers of the form where is a number whose base 4 representation has only 0’s and 1’s, and . This is a result due to Layman, in a short paper in the journal of integer sequences.

What happens if we consider avoiding solutions to

where all the ‘s are different? (There is also the related (I think slightly easier but still hard) problem where you instead consider where not all the ‘s are the same, but let’s focus on this case for now.)

In general, I think we don’t understand this at all (or at least we didn’t understand it 5 years ago). From generating these sequences with a computer, it seems like, if you have a bunch of terms that are close together, then that forces the next terms in the sequence to be farther apart. Similarly, if a bunch of terms are far apart, then the next terms in the sequence are closed together. This makes complete sense heuristically.

I wish I still had the actual graphs to give an idea, but the graph of the sequence would typically be flat at the beginning, suddenly grow because of the terms bunched together. Then, since the terms are spread out, it would suddenly become flat again. This would repeat. Instead of simply cycling, the times where the graph is either flat or growing really fast increases with each iteration. I think they increase geometrically, but it was hard to compute enough terms to see.

In some very special cases, this pattern is so extreme that the part where the graph is growing really fast is instead a perfect jump. This is because the terms in the beginning are so bunched together that there are simply no gaps to stick any more terms. Consider, for example, the graph of the sequence that avoids :

My RSI paper was just to try to find these some of these sequences and their closed forms (for example the one above). The idea is the same as Layman’s paper, where the closed form is , where is some number, is some set of initial terms, and $M$ are the integers with 0’s and 1’s in base representation, where . Basically, the elements in prevent anything else less than $c$ from being in the sequence, and is so large that you can treat and independently when plugging into the equation . As you can tell, this only happens in extremely special cases, but, in particular, it encompasses the case where all the ‘s are 1.

In general, I think it’s interesting to try to find the growth rate of these sequences, but my guess is that it is difficult. In fact, I didn’t found anything in the literature that beats the silly counting argument (other than heuristic arguments) during RSI in terms of an upper bound. It’s messy, yet not random.

]]>So Dr. Khovanova described Nim-Chomp to me at RSI more than a year ago, and I thought I solved it. Then a few months ago I found a flaw, thought it was hopeless, and gave up. Then a few minutes ago I realized that the flaw was in fact fixable. The point of this paragraph is that I’m not sure I’m to be trusted regarding this problem, but I’ll try.

I won’t repeat the problem statement here. Too lazy. Just read her post. Because I find it easier to think about this way, I’ll make a slight modification to the Chomp perspective on Nim-Chomp: on a given turn, each player may only eat squares from a single *column* (rather than a single *row*).

First let’s pretend the bottom left square is not actually poisoned. We can transform this easy-Nim-Chomp into regular Nim as follows: for a given position in easy-Nim-Chomp, suppose the number of squares remaining in each column is a_{1}, a_{2}, …, a_{n} from left to right. Let b_{1} = a_{1} – a_{2}, b_{2} = a_{3} – a_{4}, …, b_{[n/2]} = (a_{n-1} – a_{n}) or (a_{n}) depending on whether n is even or odd. Then this position is equivalent to regular Nim with piles of b_{1}, b_{2}, …, b_{[n/2]}. Basically, we’re splitting the chocolate columns into pairs of adjacent columns and considering the differences between the members of each pair to be the piles of our regular game of Nim. Because the piles are in nonincreasing order, this is a well-defined transformation.

It works as follows: suppose the loser of the Nim-game (b_{1}, b_{2}, …, b_{[n/2]}) eats some squares from the kth column where k is odd. This decreases the value of a_{k}, thereby decreasing one of the Nim-piles as in a regular Nim-game, so the winner just makes the appropriate response. Instead, the loser might try to dodge by eating squares from the kth column where k is even, thus decreasing the value of a_{k} but increasing one of the Nim-piles rather than decreasing it, which can’t be done in regular Nim. But the winner can simply decrease a_{k-1} by the same amount and leave the loser in the same position as before. There are a finite number of squares, so the loser can’t keep doing this. Eventually they must go back to decreasing piles, and lose.

This is how far I got at RSI. I didn’t realize the poisoned lower-left square was a significant issue, but it is. Thankfully, all it really does (I think) is turn the game into misère Nim rather than normal Nim. We make the transformation to Nim-piles in the same way as before, and the winner uses nearly the same strategy as in the previous paragraph, but they modify it to ensure that the loser is eventually faced with “b_{k}”s which are all 0 or 1 with an odd number of 1s. (Maybe one day if I get around to writing a basic game theory post I’ll explain why this is possible. Or you can check Wikipedia. Or just think about it.) When the loser increases some b_{k}, the winner eats squares in the corresponding column to decrease it back to 0 or 1; when the loser decreases a 1 to a 0, the winner decreases another 1 to a 0.

Eventually, the loser is forced to hand the winner a chocolate bar consisting of pairs of adjacent equal columns. At this point the winner takes a single square from any column for which this is possible, leaving a bunch of 0s with a single 1—i.e. another misère 2^{nd}-player win. This continues until we run out of squares, at which point we conclude that the loser of the new game of misère Nim is indeed the player who consumes the poisoned square in the original game of Nim-Chomp.

Question I’m too lazy to think about right now: can we still do this or something like this if we poison not only the bottom left square of the chocolate bar, but some arbitrary section at the bottom left?

]]>I have yet to find a satisfactory closed-form expression for even . I obtain an ugly series in terms of Stirling numbers of the second kind. However, I suspect that is asymptotically linear:

Is this a well-known problem?

]]>So, first of all, Damien Jiang, Anirudha Balasubramanian, and I have each uploaded the papers resulting from our RSI projects to arXiv. I’ve been discussing the story of my project on representation theory and the mathematics around it on my personal blog (see in particular here and here). There are others from the program who have placed their papers on arXiv as well (but are not involved in this blog).

I’d like to congratulate my friend and fellow Rickoid Yale Fan for winning the Young Scientist award at the International Science and Engineering Fair for his project on quantum computation (which deservedly earned him the title “rock star”). I also congratulate his classmate and fellow rock star Kevin Ellis (who did not do RSI, but whom I know from STS) for winning the (again fully deserved) award for his work on parallel computation. There is a press release here.

RSI 2010 is starting in just a few more days. I’m not going to have any involvement in the program myself (other than potentially proofreading drafts of kids’ papers from several hundred miles away), nor do I know much about what kinds of projects (mathematical or otherwise) will be happening there. I think I’d be interested in being a mentor someday—maybe in six years time. I’m going to be doing a project probably on something geometric this summer, but it remains to be seen on what.

I don’t really know what’s going to become of this blog as we all now finish high school and enter college. It looks like most of us will be in Cambridge, MA next year; this is hardly surprising given the RSI program’s location there. Also, just to annoy Yale, I’m going to further spread the word that he is going to Harvard.

If anyone from RSI 2010 wants to join/revive this blog, feel free to send an email to deltaepsilons [at] gmail [dot] com.

]]>The first two judges I had were mathematics judges. The first one asked me what I would do if I were giving a talk about my project at a colloquium. He asked me to explain one of my results, which I initially did incorrectly (having not looked through the older proof in quite some time) but fixed along the way. He asked me how I had learned algebraic geometry (or, more precisely, that rather small subset I can claim to vaguely understand). Interestingly, he referred to a specific result in my paper by number (3.10; I didn’t remember what that was for sure)—one of the differences between Intel and ISEF is that the judges read the papers.

The second mathematician asked me to give an overview of my project in detail, so I went into my usual spiel. She asked me a few questions along the way about how the results were proved. Finally, she asked where I was going to go to college. I said that I didn’t know yet. This was a somewhat longer interview.

There were others who wanted a brief overview and then left. A computer scientist who had asked me earlier about certain algorithms and an engineer that asked about the law of atmospheres chatted with me about extensions of those problems.

The exhibits were then opened to the public. I met a few RSI 2009 alumni from the D.C. area. Most people were not mathematicians, which made explaining my project (on representation theory in complex rank) a somewhat difficult task, though there were some that knew, e.g. group theory. I wasn’t envious of my neighbor Joshua Pfeffer with mobs of people craning to hear about the super Kahler-Ricci flow though owing to me extreme hoarseness despite my consuming two bottles of fluids. Also, my parents stopped by to say hello and see the other projects.

I’m somewhat tired now, and there’s not that much more I really can say about it without going into technical details.

]]>I had my third judging interview today at about 11:40 am. The judging panel included a computer scientist who pointed to the seven wrapped chocolates on the desk and informed me that to each was assigned a number, and that I needed to discover which contained the median. I didn’t have the actual numbers, but I could compare any two. In the end, I said something about sorting algorithms (e.g. heapsort). He then asked me about counting paths from (0,0) to (m,n) where one can move either right or up on each move; I stated a recursive formula, but got the wrong closed form expression (it’s a binomial coefficient, and I said it was a power of 2). I was then asked by the other judge about what change I would make if I had to design the human body. I suggested eliminating cognitive biases and improving rationality but that wasn’t legal; I then suggested various ideas such as removing vestigial organs and improving our ability to type, but settled on increasing the efficiency with which energy can be extracted from food.

I then had lunch and went to the National Academy of Sciences, where we set up our projects. My poster had developed a slight tear from being sent through the mail, but I fixed it with construction paper.

]]>**(9:45)** So I’m at the 2010 Intel Science Talent Search in Washington, D.C. Presumably many of the folk that come across this blog will have heard of it; it’s a science competition for high school seniors. There are seven people from RSI here this year. I’ve also met many interesting people among the other finalists for the first time, all of whom seem to be rather beastly. Everybody arrived yesterday–I took the train–but nothing competitive actually happened. Today, we will be judged by a panel of ten or eleven scientists and mathematicians who are going to ask us general questions about science in general, and not our projects. My first interview is in about half an hour, so I’m basically procrastinating by writing this entry, if there was anything that I could do to prepare :-).

In any case, it was pretty cool to find that I’m in a room that has a TV in the bathroom. The hotel is ridiculously fancy.

After this, I’m going to go back to random Wikipedia surfing about diverse scientific topics. They told us this morning that the judges want to see the scientific process rather than technical knowledge–perhaps this is a license for me to babble? I’ve always enjoyed idle pontification. In any case, I promise more later after my judging interview.

**(11:31**) OK, phew. My judging interview (and recounts thereof to many fellow finalists) is now complete. I don’t know if I’m allowed to say who the judges were, but two were doctors and one a mathematician. The doctor went first and asked me to explain why when one flies (at 35000 ft) 80% of the atmosphere (by mass) is below you. Uhhh….I was confused. I’m not really all that solid on physics (took AP Physics B freshman year, haven’t thought about the subject since with any success). Anyway, I have to figure out the problem by Sunday, where he’s going to ask me it again. Next the mathematician said that she was going to ask me LOTS of questions about my project (hopefully a good thing). She then asked me to explain why representation theory was interesting to her two nonmathematical colleagues. So, I started babbling at full speed. I started with the remarks that “representation theory is how algebraic objects act on vector spaces” and mentioned how it reveals information about the objects themselves (mentioning Burnside’s theorem). She asked me to go back to what I said, and I said that the tools of linear algebra are much easier than studying complicated algebraic objects directly. I said that the classification of finite simple groups was something like 10000 pages in length. I then started to say that group representations were useful in chemistry and Lie groups and Lie algebras in theoretical physcis (but freely confessed my ignorance of said subjects). I said that the representation theory of Lie algebras was interesting itself, which is where I was cut off. I was posed some fairly simple combinatorial problem about The next person asked me about stem cells. I told her that I knew about stem cells but couldn’t tell her the difference between adult and embryonic stem cells. Anyway, I was next questioned about the ethical and scientific issues in transporting stem cells from her to me. I suggested that there were potential incompatibilities (like blood types). Our time soon ended and she concluded by saying that I was on the right track. Probably not horrible, for a math person at least.

I then had my picture taken. All the finalists have to do this today. My next interview’s after lunch.

(**3:11) **I just got back from my second judging interview. The interviewers included a distinguished astronomer, who went first; she opened with a question about what latitude and longitude was. I said that they were a system of coordinates on a sphere minus the two poles. I said something about their being an injective immersion, which may or may not have piqued their interest (none of them was a mathematician). She then asked why astronomers would use them, to which i replied something about “objects being spherical.” Next, she asked me to tell me what I knew about the universe—which has to be by far the broadest question I have received all day. I babbled mildly about the four main forces and that they could be described by mathematical laws, sand omething about the age of the universe and its expansion. This is all very rudimentary, but I don’t know anything about astronomy. The next interviewer asked me why France would launch satellites from New Guinea; I suggested it was closer to the equator, but didn’t realize that the explanation for that was simply that the earth’s rotation provided additional speed. (Whoops.) I was then asked how one might determine whether switchgrass was an efficient fuel (including environmental concerns); I suggested that would one have to compute the total expenditure (i.e. of the land deforested, inputs of irrigation, fertilizer, fermentation, etc.) and compare the obtained energy per unit cost to the analog for, say, coal. My questioner pointed out that I should have determined the energy inputs of the tractor and taken them into accouont. Finally, he asked me the following math problem that I embarrassingly failed at: given 248 arm-wrestlers, how many matches would it take to determine the best one? I said something like given that an elimination process was optimal. This is wildly wrong, of course, because there are roughly **levels, **each of which (except the last) contains omre than 1 match. So he said there was a trivial solution and asked me to work out how many matches would be necessary. After a short computation I arrived at…247. Which is utterly trivial, because there must be one loser per match, and 247 losers. The judges laughed and said they had enjoyed the time; I am not sure whether that was necessarily a good thing.

(**9:24) **I’ve now had my third judging interview and a very enjoyable dinner. The first person at the third judging interview started by immediately asking me the probability that a function where has cardinality has no fixed points. The answer is (rather trivially) , but for some reason it didn’t occur to me over several minutes as I babbled about inclusion-exclusion and inductive proofs. Next the same judge asked me the same for continuous, which is an easy application of the intermediate value theorem; it’s also a special (and relatively easy) proof of the Brouwer fixed point theorem. He asked me whether the same was true for the circle; I said no, a rotation would do the trick. He asked whether it was true for (unit disk in ); I replied that it was the Brouwer theorem, which he then asked me to prove (in that case). I gave the usual proof using singular homology (which he said was allowed); I actually sort of half-alternated between talking about the fundamental group and that. (If you haven’t seen it, cf. any book on algebraic topology.)

The next judge then told me that the chair he pointed to in the back of the room was magical and could take me 1 million years in the past. I could take three things wit h me, and I had to decide. I said that I’d take a weapon to defend myself, fertilizer to help grow food, and textbooks to entertain myself (I really couldn’t resist the last part).

The third judge in the panel asked me about gases. He inquired about what would happen if a box of gas had its volume changed (temperature kept constant); I remembered enough basic chemistry from sophomore year to explain the ideal gas law. He asked me about the deviations from the ideal gas law. I said that they were caused by van der Waals forces and other intermolecular attractions, though I couldn’t remember enough to say it well. The 15 minutes were up, but they said it was a good interview. Dropping algebraic topology references was probably a good idea, though the brain-freeze on a baby problem was embarrassing.

I next went to dinner, which included the other 39 finalists, the judges, various SSP people, and a couple of Intel alumni who had distinguished themselves. One of them was Roger Tsien, who said that his obsession with producing pretty colors in chemistry experiments was part of his motivation to study fluorescent proteins. Tsien gave an amusing recount of his life story and major discoveries. The other was an aerospace engineer from MIT who told us that each of the finalists get an asteroid named after him or her.

I’ve got one more (nonmathematical) interview tomorrow. On Sunday, we’re actually going to talk to the judges about our projects.

(**10:38) **A conversation with David Liu just made me realize that I omitted a part of the third judging interview. In particular, a doctor asked me to name a letter from A-Z (which, upon asking, was inclusive of the endpoints). I picked “A” and was asked what the adaptations of aardvarks are; I said something about their fur and warm-bloodedness.

In an unrelated note, if you haven’t already seen it, you ought to watch the MAA’s great debate. And that’s regardless what you think of the holiday “Pi Day”–I’m mostly in agreement with what John Armstrong has to say on this subject. Also, cf. this MO thread for some good alternatives to memorizing digits.

]]>Since we have been too lazy to post lately (and the so-not-lazy Akhil posts mostly elsewhere now), I’m going to post some problems that I probably should be able to solve, but haven’t.

First, an interesting (and apparently classic) problem given to me by Tim Chu, USA IMO alternate:

Find all such that f is multiplicative (for relatively prime positive integers), monotonically increasing, and f(2) = 2.

A second from an old IMO longlist:

Show that the union of all segments with endpoints in includes every point in .

EDIT: Well apparently this is wrong. However, part b of the problem asks if the convex hull of is equal to .

And a third from Russia 2005:

A quadrilateral ABCD without parallel sides is circumscribed around a circle with center O. Show that O is the barycenter of ABCD iff (OA)(OC) = (OB)(OD).

(Looks quite vector/complex-number friendly, but… I probably didn’t try hard enough.)

]]>