PAX

Penny Arcade Expo has been pretty fun! I've been competing in the PAX Pokemon League (I've beaten 9 gym leaders and 2 elite four members, and have a few challenges left for tomorrow), got a couple awesome T-shirts, had an impromptu hula hoop competition (which I lost), played Dragon Dice, an awesome game which I thought was lost to Time, and played some Four Calibur.... SOUL!

Overall it's been pretty fun. There's a lot going on, far too much to get to everything I'd like to. It's been keeping me busy and making it tough for me to take care of non-PAX things that I might like to be taking care of in the evenings. But overall it's been a great experience.

I'll be leaving for Germany Monday morning, pretty early, so I'm a little nervous about getting to the airport on time. I may just get a taxi so that I don't have to worry about it.

The Countdown

So I've been home in Santa Cruz hanging out for a while and that's been nice, but I'm about to get back on the road again on Wednesday, when I'll be driving up to Seattle for PAX. I'm excited to see my friends who are going, especially Eugene, Lucas, Lauren, Cambria, Gabe, and Puppo!

Today I made sure all my boxes were packed and moved them downstairs, as well as moving my bookshelves downstairs, and tonight I'll put my dresser downstairs and take apart my desk, because tomorrow morning I am moving all my stuff out into storage until I get back.

It's been a little bit frustrating having so much to do and not being able to do any of it because I don't have a car myself, but I'm glad that my friends are coming to help me out tomorrow and I'm confident we'll get it all done. And then it will be all fun times!

Back Home (mostly)

I'm back in the states, in San Francisco, waiting to say hi to some friends before I head down to Santa Cruz and settle back in (for all of ten days -_- ).

I'm pretty tired and ready to be at home for a while, where there is wireless all the time and plugs that fit my devices and a big bed and a substantially less public shower and a refrigerator and all that.

I'm a little weirded out by american money now, though. It's all... clothy? And dirty. And the denominations are weird. 2 euro coins are dreadfully convenient.

Travel Woes Update and excerpts from an awesome conversation

So I am now in Dusseldorf; it is 1:25am, and I am rather tired.

My flight yesterday from Rome to SFO started at 6:20 am. However the first bus to the Da Vinci airport, apparently, gets there at 6:00 am. Which is not soon enough. I probably should have taken a taxi. However, hindsight is 20/20 and it is far too late to change the past. All the flights for that day were overbooked, and the cheapest flights I could get on were over $2000. So I went on the internet at the price of .20 euro a minute to find a cheaper flight... and I found one for about half the price of the others, and snapped it up... before discovering that it left the NEXT EVENING then had a 30 hour+ layover in Dusseldorf.

So here I am in Dusseldorf. UGH.

Anyway that's depressing so here's something awesome from a conversation with my friend Cambria: (edited for clarity and conciseness)

9:13:16 PM Cambria Scalapino: what was the weirdest thing that has happened to you so far?

9:13:22 PM Mitchell Owen: Hmmm

9:13:57 PM Mitchell Owen: Candidate: being solicited by a prostitute

9:14:24 PM Mitchell Owen: Candidate: discovering a square filled with Harry potter posters

9:14:57 PM Mitchell Owen: Candidate: returning to a hostel and passing the David at sunrise

9:15:17 PM Mitchell Owen: Candidate: tossed my drink on someone

9:15:36 PM Cambria Scalapino: ...was it on purpose?

9:15:41 PM Mitchell Owen: Yes

9:15:36 PM Mitchell Owen: Candidate: had people debate whether or not I was gay

9:16:17 PM Mitchell Owen: Candidate: bought a glorious sparkly pink hat which is my best friend and we have followed each other everywhere

9:17:38 PM Cambria Scalapino: oh man HATS.

9:18:34 PM Mitchell Owen: Candidate: talked to a stranger about my Tegmark multiverse theory of self and applications

9:19:14 PM Mitchell Owen: Candidate: stayed up until all hours talking to Christians about my religious beliefs

9:19:21 PM Cambria Scalapino: XD

9:19:56 PM Mitchell Owen: Candidate: a philosophy student from Oxford got totally wasted and was hilarious

9:20:13 PM Mitchell Owen: Candidate: carried a German girl most of the way up a hill

Back in Rome

So I made my way back to Rome via Barcelona. I didn't have much time to spend in Barcelona, but I did have enough to meet two people at my hostel who were taking the same flight to Rome that I was! So hopefully I have some friends to visit the catacombs with this afternoon.

Then on Tuesday morning I'll be flying in to SFO. And then home!

Sato Tate for Picard Curves

So I gave my talk this morning and it went okay, though it was pretty short because I didn't really have all that much to say. I'm trying to do some more computations now but I have to do all kinds of software update crap so... bleh. Anyway here are my notes for my talk they are pretty thorough but I didn't follow them as closely as all that.

Sato Tate for Picard Curves

Benasque 2011

Mitchell Owen

Work conducted under Martin Weissman

As work for my undergraduate thesis, I did some computations for Sato-Tate over a certain extremely tame family of genus 3 curves known as Picard curves. A Picard curve is one isomorphic to a curve with projective model zy^3 = x^4 + ax^3z+bx^2z^2 +cxz^3 + dz^4. We can assume that the coefficient for cubes is zero by completing the quartic (replace x with x + a/4 or something) so long as our characteristic is not 2, but since I only dealt with equidistribution statements discarding a finite number of primes is fine; we will also want to discard primes dividing the discriminant which leaves us with primes of good reduction; however since this is a finite number of primes again it doesn't really matter. (i.e. we could count points over these fields and the information would be washed out.)

Picard curves have local L-factors that are sextic [insert here], and the work of Upton tells us that we should expect the image of Galois to be GU_3. I'm still not entirely sure what this means but you can see that the trace of Frobenius is equal to a_p, which allows us to make some very simple computations.

I wrote a program to use a naïve point count method to check the Sato-Tate distribution for these curves computationally. Of course as you all probably know counting points is very very slow, so this is definitely a computation worth revisiting using some methods with twists which Drew sent me an e-mail about but I haven't had a chance to read through in detail yet.

So what I did end up actually able to do was count points over F_p for p up to 30,000 for four curves, Then calculated the first couple moments and put together some fairly clunky histograms compared to the beautiful things Drew showed us last week.

So based on what I was told as an undergraduate I expected the traces to be distributed like random traces of matrices in U_3. One could use the Weyl integration formula with information about N_p^2 and N_p^3 but it's fairly messy and I don't know how.

What I could do was calculate the even moments and from Diaconis and Shashahani it is easy to see that the first two even moments are 2 and 12. My computations for the following three curves were fairly close, although a couple of the twelves looked a bit like 13s; I have been told that this is called Chebyshev's bias and that it isn't surprising.

Then I had another curve which obviously has some extra automorphisms and it's moments were totally different; 4 and 46.

And here are the histograms!

Chris' Third Talk

Consider the following elliptic curve:

C: y^2 = x^3 + 5^2

Which has additive reduction over Q5 but good reduction over Q5(5^1/3)

Now I_Q5 (the Inertia group, see previous posts) acts on VlE through C3 for all l other than 5. In our example, I_Q5 injects into VlE tensor Ql bar, and acts by a two by two diagonal matrix with entries phi and phi inverse, where phi is some character of order 3.

Define Xl:G_Qp -> Zl* as the action of G_Qp on the inverse limit of l^n roots of unity; then we call Xl the l-adic cyclotomic character.

Recall that TlE wedge TlE is isomorphic to Xl by the Weil paring.

I missed a bunch in the talk at this point.

Theorem("Tate Curve"): say E/Qp has split multiplicative reduction.

Then there is some q in Zp so that v(q)=v(disc(E))=-v(j)=n>0 (not sure what this v function is)

such that

E(Qp bar) ~ Qp bar cross modulo q^Z

here q^z represents a sort of spiral lattice; think of them as integer powers of a complex number; this allows us to "rotate" the p-adic space to within a certain sort of angle/absolute value pie chunk.

so E[l]= = (Z/lZ)^2, and E[l^n]=, and TlE=Zl^2.

Now G_Qp acts on TlE as an upper diagonal two by two matrix where the upper left entry is the l-adic cyclotomic character and the lower right entry is 1.

And I_Qp acts as an upper triangular matrix with 1s on the diagonal and v(q)*phi is in the upper right.

Then we define this phi to be the tame character; i.e. it is the part of Inertia that acts in a way we can observe like this.

So here I_Qp/G1 = the product of Zl for l not p, and we are projecting onto specific Zl.

Corollary: E/Qp has split multiplicative reduction -> Fp(T)=1-T. Yay local factors!

Now the case of potentially multiplicative reduction:

say E/K: y^2=x^3+ax+b

then the quadratic twist of E by d is

Ed/K: dy^2=x^3+ax+b

which is isogenous to E over K(d^1/2)

Let phi from the Galois group of that field extension be the nontrivial character.

Then Vl(Ed)=VlE tensor phi (this was an exercise to prove. I don't entirely know what it means)

Now say our E has pot. multiplicative reduction. Now consider E twisted by -6B, which has split multiplicative reduction (seeing why was maybe an exercise?)

now phi:Gal -> +-1 nontrivially means in the non-split multiplicative case, inertia acts trivially so Frobp |-> -1, in the additive inertia can act as -1 but this isn't too big a deal I guess? in the first case we can see that Fp(T)=1+T; in the additive case Fp(T)=1.

What this finally allows us to do is characterize Fp(T) for all cases. Yay!