Tag Archives: open question

## Open problem in number theory

3 Dec

A number theorist was trying to convince me that while you can do things easily for finite sets of prime numbers, it’s really hard to make a leap to infinitely many primes.  She gave me a sketch of an example, and I thought I’d share it/some other interesting number theory things.

We say that an integer is squarefree if for every prime number p, $p^2$ does not divide a.  Remember, the primes are the numbers that are only divisible by themselves and 1, like 2, 3, 5, 7, etc.  Every integer can be written as a product of prime numbers, so squarefree means what we think it does: no perfect square factors.  Here are the first squarefree numbers which aren’t primes: 6, 10, 14, 22, 26, 30,…

So you can ask a few questions as soon as you get this definition.  How many squarefree numbers are there?  A: infinitely many, because any product of two primes is squarefree, and there are infinitely many primes.  But how big is that infinity?  Remember, not all infinities are equal.  Maybe a better question is, what’s the ratio of squarefree numbers vs. not-squarefree numbers?  This seems a little nuts because we’ll have infinity divided by infinity, but it’s not nuts to say that half the numbers are even.  So we’re not nuts!  Huzzah!

This is from wolfram alpha and shows 10000 numbers: the white pixels are squarefree. 1-100 are the bottom horizontal line of pixels, you can see 1, 4, 8,9,12,16,18,20 all blacked out in the lower left.

A: about $6/\pi^2$ much of the numbers are squarefree, and this has been known since at least 1951.  Just from the first 20, we have 60% squarefree, and $6/\pi^2$ means that as you look at bigger and bigger numbers, the proportion of squarefree numbers approaches 60.97….%  But this is crazy, right?  Where did that $\pi$ come from?

Well, you can go prime by prime.  For instance, about 1/9 of numbers are divisible by 9, so 8/9 of numbers are 9-free.  And 3/4 are 4-free.  So 3/4*8/9=2/3 are 4-and-9-free.  You could do this for any finite set of primes, and say that the chance that is $latex p^2$-free for all of your finite set is $\prod_{p\in S} (1- \frac{1}{p^2})$.  Then we’d conjecture that you could make this an infinite product $\prod_{primes} (1-\frac{1}{p^2})$.

By this crazy formula from 1737 (oh that Euler!), $\sum_1^{\infty} n^{-s} = \prod_{primes} \frac{1}{1-p^{-s}}$.  And then that sum on the left is the definition of the Riemann zeta function, and $\zeta(2) = \frac{\pi^2}{6}$, which was proven by Euler 91 years after Basel posed the question.

But that step where we conjectured that you could jump to an infinite product?  It works in this case, but through a lot of hard work!  Which I do not understand nor will explain.

Here’s a related question.  Suppose you have a polynomial (a function in the form $f(x) = a_nx^n + a_{n-1}x^{n-1}+\ldots+ a_1x + a_0$), and you put numbers into it.  How many of those resulting function values are squarefree?  If you have something silly like $f(x) = 9x+9$, all values will be divisible by a square.  So let’s cut out polynomials like that.  Do squarefree values happen infinitely often?  And if so, how often [in the infinite sense that half of numbers are even]?

Open question: Does $f(x) = x^4+2$ take infinitely many square-free values?  If so, what is the probability that f(x) is squarefree, if you randomly choose an x?

You can ask a question with just putting in primes instead of putting in all numbers for too.  A polynomial is irreducible if you can’t write it as a product of other polynomials.  For instance, $x^3-1$ is not irreducible, since $x^3-1 = (x-1)(x^2+x+1)$, but $x^2+x+1$ is irreducible.  The degree of a polynomial is the biggest number that appears as an exponent of x, so the degree of $x^3-1$ is 3.

Open question (with conjecture): If f(x) is an irreducible polynomial of degree 3 or more, how many squarefree values does it take?

Dude, number theory is full of unsolved problems that are easy to state!  When reading up for this post, I ran into this magic squares problem.  Remember a magic square is one where the sum of all the numbers in each column, in each row, and along the diagonals is all the same number.

Sum of rows, columns, and diagonals is all 15.  I’m sorry I didn’t bother to make you a magic square; these values are from wikipedia.

Open question: Build a magic 3×3 square of square numbers.

I hope I’ve managed to convey some interesting number theory to you today!  Personally I’m not that into number theory, but I tried to not let that color this post.

Look at this bibimbap I made for dinner the other day!  My first time making bibimbap.

A photo posted by Yen Duong (@yenergyyyy) on Dec 1, 2015 at 7:18am PST

//platform.instagram.com/en_US/embeds.js

## Current research: lifting geodesics to embedded loops (and quantification)

19 Nov

Last week we learned about covering spaces, and I made a promise about what we’d talk about in this post.  For those who are more advanced, this all has to do with Scott’s separability criterion, so you can take a look back at that post for a schematic.  I’ll put the picture in right here so this post isn’t all words:

Left side is an infinite cover, the real numbers covering the circle. Middle is a happy finite cover, three circles triple covering the circle. Right is a happy finite cover, boundary of the Mobius strip double covering the circle.

In my friend Priyam Patel’s thesis, she has this main theorem:

Theorem (Patel): For any closed geodesic g on a compact hyperbolic surface $\Sigma$ of finite type with no cusps, there exists a cover $\tilde{\Sigma}\to\Sigma$ such that g lifts to a simple closed geodesic, and the degree of this cover is less than $C_{\rho}\ell_{\rho}(g)$, where $C_{\rho}$ is a constant depending on the hyperbolic structure $\rho$.

We know what geodesics are, and we say they’re closed if the beginning and end are the same point (so it’s some sort of loop, which might intersect itself a bunch).  But wait, Yen, I thought that geodesics were the shortest line between two points!  The shortest path from a point to itself is not leaving that point, so how could you have a closed geodesic?  Nice catch, rhetorical device!  A closed geodesic is still going to be a loop, but it won’t be the shortest path between endpoints because there are no endpoints.  Instead, just think locally: if a closed geodesic has length l, then if you look at any two points x and y less than l/2 apart from each other, the closed geodesic will describe an actual geodesic segment between x and y.  It’s locally geodesic.

What about hyperbolic surfaces of finite type with no cusps?  Well, we say a surface $\Sigma$ is of type (g, b, n) if it has genus (that’s the number of holes like a donut), boundary components, and punctures or cusps.

Pink: (4,0,0)
Orange: (3,0,2)
Green: (1,2,1)
Ignore the eyes they’re just for decoration

Boundary components are sort of like the horizontal x-axis for the half plane: you’re living your life, totally happy up in your two-dimensional looking space, and then suddenly it stops.  This is also what a boundary of a manifold is: where the manifold locally looks like a half-space instead of all of $\mathbb{R}^n$.  Surfaces are 2-manifolds.

Finally, I drew punctures or cusps suggestively- these are points where you head toward them but you never get there, no matter how long you walk.  These points are infinitely far from the rest of the surface.

I think we know all the rest of the word’s in Priyam’s theorem *(hyperbolic structure is a hyperbolic metric).  The important thing to take from it is that she bounds the degree of the cover above by a constant times the length of the curve.  This means that she finds a cover with degree smaller than her bound (you can always take covers with higher degree in which the curve still embeds, but the one she builds has this bound on it).

Just looking at this old picture again so you can have a sort of idea of what we’re thinking about

She’s looking for a minimum degree cover and finds an upper bound for it in terms of length of the curve.  Let’s write that as a function, and say $f_{\rho}(L)$ gives you the minimum degree of a cover in which curves of length embed (using the hyperbolic structure $\rho$).   What about a lower bound?

Here’s where a theorem (C in that paper) by another friend of mine, Neha Gupta, and her advisor come in:

Theorem (Gupta, Kapovitch): If $\Sigma$ is a connected surface of genus at least 2, there exists a $c=c(\rho, \Sigma)>0$ such that for every $L\geq sys(\Sigma), f_{\rho}(L)\geq c(\log(L))^{1/3}$.

So they came up with a lower bound, which uses a constant that depends on both the surface and the structure.  But it looks like it only works on curves that are long enough (longer than the systole length, which we’ve seen before in Fanoni and Parlier’s research: the length of the shortest closed geodesic on the surface).  Aha!  If you’re a closed geodesic, you’d better be longer than or equal to the shortest closed geodesic.  So there isn’t really a restriction in this theorem.  Also, that paper is almost exactly 1 year old (put up on arxiv on 11/20/2014).

Now we have $c_{\rho,\Sigma}(\log(L))^{1/3}\leq f_{\rho}(L) \leq C_{\rho}L$.

This is where it gets exciting.  We know from Scott in 1978 that this all can be done, and then Patel kickstarts the conversation in 2012 about quantification, and then two years later Gupta and Kapovich do the other bound, and boom! in January 2015, just three months after Gupta-Kapovich is uploaded to the internet, my buddy Jonah Gaster  improves their bound to get $\frac{1}{c}L\leq f_{\rho}(L)$, where his constant doesn’t even depend on $\rho$.  He does this in a very short paper, where he uses specific curves that are super hard to lift and says hey, you need at least this much space for them to not run into each other in the cover.

Here’s a schematic of the curves that are hard to lift (which another mathematician used to prove another thing [this whole post should show you that the mathematical community is tight]):

This curve in the surface goes around one part of the surface 4 times, and then heads over to a different part and circles that. This schematic is a flattened pair of pants, which we’ve seen before (so the surface keeps going, attached to this thing at three different boundary components).  I did not make this picture it is clearly from Jonah’s paper, page 4.

So that’s the story… for now!  From Liverpool (Peter Scott) to Rutgers in New Jersey (Priyam) to Urbana/Champaign in Illinois (Gupta and Kapovitch) to Boston (Jonah), with some quick nods to a ton of other places (see all of their references in their papers).  And the story keeps going.  For instance, if you have a lower bound in terms of length of a curve, you automatically get a lower bound in terms of the number of times it intersects itself ($K\sqrt{i(g,g)}\leq \ell(g)$, same mathematician who came up with the curves).  So an open question is: can you get an upper bound in terms of self-intersection number, not length?