Extensions of Line Bundles on the Projective LineJanuary 26, 2015

Posted by Dennis in algebraic geometry.

The following is probably obvious to experts, as it is easy to find generalizations using google. However, this humble master’s student didn’t find this most basic case, so it’s probably worth writing down.

We will be working over the projective line $\mathbb{P}^1_k$ over a field. The basic question is that, since we know that there exist nontrivial ${\rm Ext}$ groups between line bundles on $\mathbb{P}^1$, what explicit extensions do they correspond to?

For example, given that we we know ${\rm Ext}^1(\mathscr{O},\mathscr{O}(-2))=H^1(\mathscr{O}(-2))=k$, there should be a nontrivial extension

$0\rightarrow \mathscr{O}(-2)\rightarrow E\rightarrow \mathscr{O}\rightarrow 0$.

Recall that all vector bundles over $\mathbb{P}^1_k$ splits into a direct sum of line bundles. So $E=\mathscr{O}(a)\oplus \mathscr{O}(b)$. By taking determinant bundles, we know $\mathscr{O}(a+b)=\mathscr{O}\otimes \mathscr{O}(-2)=\mathscr{O}(-2)$, so $a+b=-2$. Furthermore, since it is a nontrivial extensions, we need both $a,b<0$, or else we would have a section $\mathscr{O}\rightarrow E$. Therefore, $E=\mathscr{O}(-1)\oplus \mathscr{O}(-1)$.

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

$0\rightarrow k[x,y]\xrightarrow{\begin{pmatrix} -y\\x \end{pmatrix}}k[x,y]\oplus k[x,y]\xrightarrow{\begin{pmatrix} x & y\end{pmatrix}}k[x,y]\rightarrow k\rightarrow 0$

is exact. Turning this into sheaves over $\mathbb{P}^1_k$, we get exactly a short exact sequence

$0\rightarrow \mathscr{O}(-2)\rightarrow \mathscr{O}(-1)\oplus \mathscr{O}(-1)\rightarrow \mathscr{O}\rightarrow 0$.

Notice that this only works because $k$ is identified with the zero sheaf when we try to turn it from a $k[x,y]$-module to a sheaf over $\mathbb{P}^1_k$.

Repeating the argument above, we see that any extension

$0\rightarrow \mathscr{O}(-n)\rightarrow E\rightarrow \mathscr{O}\rightarrow 0$.

must have $E\cong \mathscr{O}(a)\oplus \mathscr{O}(b)$ for $a+b=-n$ and $a,b<0$. We might ask if all such values of $a,b$ are possible. The answer is yes, and we can construct them, though I didn’t think of something as nice as the Koszul complex.
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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

$0\rightarrow k[x,y]\xrightarrow{\begin{pmatrix} -y^j\\x^i \end{pmatrix}}k[x,y]\oplus k[x,y]$
$\xrightarrow{\begin{pmatrix} x^i & y^j\end{pmatrix}}k[x,y] \rightarrow k[x,y]/(x^i,y^j)\rightarrow 0$

and turn it into sheaves over $\mathbb{P}^1_k$. This yields

$0\rightarrow \mathscr{O}(-i-j)\rightarrow \mathscr{O}(-i)\oplus \mathscr{O}(-j)\rightarrow \mathscr{O}\rightarrow 0$.

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

Aside on correspondence between ${\rm Ext}^1$ and extensions

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

$0\rightarrow A\rightarrow E\rightarrow B\rightarrow 0$?

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 $E$ being in the middle of a short exact sequence, we would instead like to have a presentation for $E$. For example, of $R=\mathbb{Z}$ and $A$ and $B$ are finitely generated (which is a lot of the time), then this allows us to compute $E$ explicitly. To do so, we want to find a surjection onto $E$.

To do so, we can find a surjection $F\rightarrow B\rightarrow 0$ onto $B$ from a free module, and lift it to a map $F\rightarrow E$. Then, we have a surjection $F\oplus A\rightarrow E$. This gives us the diagram:

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

.

Conversely, if we have such a map $K\rightarrow A$, 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 $A$ and $B$ 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 $K\rightarrow A$ give equivalent extensions precisely when it extends to a map $F\rightarrow A$ directly from the large diagram.

Constructing the extensions

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 $I$ containing $\mathscr{O}(-n)$. Let the cokernel of $0\rightarrow \mathscr{O}(-n)\rightarrow I$ be $K$. Then, each $E$ in the middle of a short exact sequence

$0\rightarrow \mathscr{O}(-n)\rightarrow E\rightarrow \mathscr{O}\rightarrow 0$.

is a pullback of the diagram

for some map $\mathscr{O}\rightarrow K$ and conversely, every map $\mathscr{O}\rightarrow K$ gives a pullback diagram, and the pullback $E$ fits into a short exact sequence $0\rightarrow \mathscr{O}(-n)\rightarrow E\rightarrow \mathscr{O}\rightarrow 0$.

Unfortunately, I don’t know of a good way of writing such an $I$ 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 $\mathscr{O}(-n)\rightarrow I$ extended to a map $E\rightarrow I$.

The long exact sequence applied to $0\rightarrow \mathscr{O}(-n)\rightarrow E\rightarrow \mathscr{O}\rightarrow 0$ in ${\rm Ext}$ yields ${\rm Hom}(E, I)\rightarrow {\rm Hom}(\mathscr{O}(-n), I)\rightarrow {\rm Ext}^1(\mathscr{O},I)$. Therefore, it suffices for $H^1(I)=0$.

Therefore, instead of using an injective object for $I$, we can use any object whose first cohomology vanishes. One source of such an object $I$ is the Cech complex. As an abuse of notation, if $\mathscr{F}$ is a sheaf on $X$, I’ll write $\mathscr{F}|_U$ for an open $U\subset X$ as $\iota_{*}\iota^{*}\mathscr{F}$, where $\iota: U\rightarrow X$ is the inclusion.

The Cech complex gives us the exact sequence of sheaves

$0\rightarrow \mathscr{O}(-n)\rightarrow \mathscr{O}(-n)|_{D^+(x)}\oplus \mathscr{O}(-n)|_{D^+(y)}\rightarrow \mathscr{O}(-n)|_{D^+(xy)}\rightarrow 0$,

so we have $I=\mathscr{O}(-n)|_{D^+(x)}\oplus \mathscr{O}(-n)|_{D^+(y)}$ and $K=\mathscr{O}(-n)|_{D^+(xy)}$.

Recall that a basis of $H^1(\mathscr{O}(-n))$ is represented by the sections $s_i: \mathscr{O}\rightarrow \mathscr{O}(-n)|_{D^+(xy)}$ that sends 1 to $x^{-i}y^{i-n}$ for $1\leq i< n$. So it is natural to ask what extensions these sections $s_i$ 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 $E$ given by $s_i$ has the transition map given by

$\begin{pmatrix} t^n & - t^i\\ 0 & 1 \end{pmatrix}$

from $D^+(x)$ to $D^+(y)$, where $t:=\frac{y}{x}$. To get this answer, we need to compute the pullback

Over $D^+(x)$, a section of the pullback $E$ is equivalent to choosing sections $x^{-n}g$, $x_1^{-n} h$ over $\mathscr{O}(-2)(D^+(x))$ and $\mathscr{O}(-2)(D^+(xy))$ respectively and a section $f$ of $\mathscr{O}(D^+(x))$, such that $x^{-n}g-y^{-n}h=x^{-i}y^{i-n}f$.

Equivalently, $h = (\frac{y}{x})^n g -(\frac{y}{x})^i f$. So a basis for $E$ over $D^+(x)$ is given by $f,g\in \mathscr{O}(D^+(x))$, and $h$ is determined uniquely by $f$ and $g$.

The situation over $D^+(y)$ is exactly the same. To compute the transition map, we need to take a compatible choice $(x^{-n}g, y^{-n}h, f)$ over $D^+(xy)$, express it in terms of the basis over $D^+(x)$, and then see what it should be in terms of the basis over $D^+(y)$. In the basis over $D^+(x)$, it is $(g, (\frac{y}{x})^n g -(\frac{y}{x})^i f, f)$, and in the basis over $D^+(x)$, it is $((\frac{x}{y})^n h -(\frac{x}{y})^{n-i}, h, f)$.

This means the transition matrix is given by $\begin{pmatrix} (\frac{y}{x})^n & - (\frac{y}{x})^i \\ 0 & 1\end{pmatrix}$, which is what we said above.

Finally, the identify $E$, we need to diagonalize our transition matrix. Since we are allowed to change bases over $D^+(x)$ and over $D^+(y)$, we are allowed to multiple on the right by an element in $GL_2(k[t])$ and on the left by an element of $GL_2(k[t^{-1}])$. Through row and column operations, we get

$\begin{pmatrix} (\frac{y}{x})^n & - (\frac{y}{x})^i \\ 0 & 1\end{pmatrix}\rightarrow \begin{pmatrix} 0 & t^{-i} \\ t^{n-i} & 1\end{pmatrix} \rightarrow \begin{pmatrix} 0 & t^{-i} \\ t^{n-i} & 0\end{pmatrix} \rightarrow \begin{pmatrix} t^i & 0 \\ 0 & t^{n-i}\end{pmatrix}$.

.

This means the section $s_i: \mathscr{O}\rightarrow \mathscr{O}(-n)|_{D^+(xy)}$ that sends 1 to $x^{-i}y^{i-n}$ represents an element in $H^1(\mathscr{O}(-n))$ that corresponds to an extension

$0\rightarrow \mathscr{O}(-n)\rightarrow \mathscr{O}(-i)\oplus \mathscr{O}(i-n)\rightarrow \mathscr{O}\rightarrow 0$

.

Generalized Nonaveraging Integer SequencesJanuary 9, 2015

Posted by Dennis in Uncategorized.

Everybody who participated in RSI 2009 from this blog have now just finished their undergraduate degrees. I think it’d be interesting to have an update to see how everybody is doing.

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 $S$ of $\{0,1,\ldots,N\}$ that contains no three term arithmetic progression. Most of the research has been focused on finding asymptotics for the size of $S$ given large $N$. 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 $S$ 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 $S$ 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 $|S|=\Omega(n^{\log_3(2)})$, while Elkin’s construction alluded to above gives $|S| = \Omega(\frac{n}{2^{2\sqrt{2}\sqrt{\log_2(n)}}}\log^{1/4}(n))$.

The condition that the sequence has no three term arithmetic progression is equivalent to avoiding (nontrivial) solutions to $x+y=2z$, 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 $x+y+z=3w$, where $x,y,z,w$ are all different, you get numbers of the form $3M + R$ where $M$ is a number whose base 4 representation has only 0’s and 1’s, and $R\in \{0,1,2,3,4\}$. 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
\begin{aligned} a_0x_0+\cdots+a_nx_n = (a_0+\cdots+a_n)x_{n+1} \end{aligned}
where all the $x_i$‘s are different? (There is also the related (I think slightly easier but still hard) problem where you instead consider where not all the $x_i$‘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 $x_1+x_2+x_3+x_4=4x_5$:

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 $cM+R$, where $c$ is some number, $R$ is some set of initial terms, and $M$ are the integers with 0’s and 1’s in base $d$ representation, where $d=a_0+\cdots+a_n+1$. Basically, the elements in $R$ prevent anything else less than $c$ from being in the sequence, and $c$ is so large that you can treat $M$ and $R$ independently when plugging into the equation $a_0x_0+\cdots+a_nx_n=(a_0+\cdots+a_n)x_{n+1}$. As you can tell, this only happens in extremely special cases, but, in particular, it encompasses the case where all the $a_i$‘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.

Nim-ChompAugust 19, 2010

Posted by lumixedia in combinatorics, math, Problem-solving.
Tags: , ,

Wow, I’m really not cut out for helping to maintain a blog, am I? So what finally prompted me to post was Dr. Khovanova’s description of the game of Nim-Chomp at her blog, which she suggested I respond to.

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 a1, a2, …, an from left to right. Let b1 = a1 – a2, b2 = a3 – a4, …, b[n/2] = (an-1 – an) or (an) depending on whether n is even or odd. Then this position is equivalent to regular Nim with piles of b1, b2, …, 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 (b1, b2, …, b[n/2]) eats some squares from the kth column where k is odd. This decreases the value of ak, 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 ak 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 ak-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 “bk”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 bk, 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 2nd-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?

A PuzzleJune 25, 2010

Posted by genericme in Uncategorized.

Consider a set of $n$ objects from which $m$ are drawn randomly at a time, with replacement.  What is $E(n, m)$, the expected number of draws I have to make to have drawn all of the objects?

I have yet to find a satisfactory closed-form expression for even $E(n, 1)$.  I obtain an ugly series in terms of Stirling numbers of the second kind.   However, I suspect that $E(n, 1)$ is asymptotically linear:

Is this a well-known problem?

Posted by Akhil Mathew in Uncategorized.

It’s been another two months already since anyone last posted here, hasn’t it?

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.

Intel day 4March 14, 2010

Posted by Akhil Mathew in General.
Tags:

I had the last day of judging today at Intel. The 40 finalists first went to the Capitol to take a bunch of pictures, then to the Einstein statue at the NAS for another one.  We then went in the project exhibition hall.  I met with seven judges, two of who were mathematicians.  In order that future generations of Intelists may face the day of judgment without crushing uncertainty in the morning, I shall briefly describe my experience.

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.

Intel day 3March 13, 2010

Posted by Akhil Mathew in General.
Tags:

(Today was the third day of the Intel STS competition.)

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 $O(n \log n)$ 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.

Intel STS: Liveblogging day 2March 12, 2010

Posted by Akhil Mathew in General.
Tags:

I’m going to try liveblogging (insofar as possible) a science fair.  [9:24- It’s now pretty clear that what I’m doing is more like deadblogging.  Still, it’s better than nothing, I suppose.]

(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. (more…)

Is early specialization good?March 4, 2010

Posted by Akhil Mathew in General.

Thomas Sauvaget asked a question on MO on whether specializing early is a good thing.  It got into an interesting discussion, which continues on his blog.  I have placed some of my own thoughts there, so I won’t ramble here.

In an unrelated note, if you haven’t already seen it, you ought to watch the MAA’s great $\pi, e$ 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.

Some unsolved problemsJanuary 3, 2010

Posted by Damien Jiang in Problem-solving.
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