Should MathSciNet be a social network?

Jim Borger makes the following interesting suggestion in comments to the “Do you follow the arXiv?” thread:

What I think would be way better is if the MathSciNet sent out emails with abstracts of newly published papers. With some very basic filtering, based on which authors, subjects, key words, etc you like, they could probably keep the emails reasonably small. That would be much more useful than the archive digests. Then you could truly keep an eye on whatever fields you want without much effort.

This would presumably be easy to implement, given some kind of personal login to MathSciNet — but do you want MathSciNet keeping track of what abstracts you looked at, and which ones induced you to click through to the article?

The right way to do this, I guess, would involve allowing us to tag MathSciNet abstracts, so that overlaid on the AMS subject classification would be an emergent user-generated classification scheme which I’d expect to be much richer and more useful.  And it wouldn’t stop there — I imagine MathSciNet would keep track of everybody’s browsing in order to identify users with similar tastes, and make recommendations accordingly.  “People who looked up Deligne’s “Le Groupe Fondamental de la Droite Projective Moins Trois Points” also liked…”

Three questions:

1. Would you be into this?
2. Does Google Scholar already do this for people who use it while logged in to their Google accounts?  Is Google keeping track of my scholarly interests and ordering its Scholar search results accordingly, as it does for web search?
3. Why doesn’t the arXiv allow tagging?  Or does it, and I can’t find it?  There are already links on each article page for bookmarking at CiteULike, del.icio.us, and digg, all places where you can tag; so arXived articles are tagged, but the tags are scattered across different services used by different populations.  Why not tag where most people read?

By the way, I really like Jim’s new paper “Lambda-rings and the field with one element.” Yes, another definition of the category of schemes over F_1; but this approach smells particularly good to me.  Here’s one thing I like.  As always, you have some notion of which Spec Z – schemes descend to Spec F_1, and what should be meant by “descent data.”  In Jim’s story, you can do the same thing starting with S-schemes, where S is an algebraic curve over some finite field F_q.  But the target of this construction is not, as you might initially think, F_q-schemes — rather, there’s another category, “F_1^S-schemes”, which lies between S-schemes and F_q-schemes.  In this category you have, e.g., some rank 1 Drinfel’d modules.  I take this to be saying that you can’t take the slogan “Spec Z is like an algebraic curve over Spec F_1” too seriously; maybe there just isn’t anything which is to Spec Z as Spec F_q is to S.

F_1 and the braid group — a note of skepticism

Since I wrote this post, I’ve become less sure about this assertion that the braid group can be thought of as GL_n(F_1[t]). Here are three reasons to be doubtful:

• As Jim points out in commments, GL_n(F_q) embeds in GL_n(F_q[t]), but S_n doesn’t embed in the braid group. This has to be counted against the braid group, I think. Jim also says that in his version of F_1 geometry, which comes out of lambda-rings, GL_n(F_1[t]) is just S_n.
• Terry Tao observed that one might expect GL_n(F_1[t]) to embed into GL_n(F_q[t]), just as GL_n(F_1) embeds into GL_n(F_q). But this doesn’t appear to be the case, at least not for any obvious reason. Keep in mind, it was quite hard to prove that the braid group had any faithful linear representations at all! The n-dimensional linear representations developed by Lawrence and Krammer, and proven faithful by Bigelow (and then again by Krammer) have coefficients in Z[t,1/t,u,1/u]. So the idea that one might find the braid group inside GL(F_1[t,1/t,u,1/u]) remains, from this point of view, alive! But I wonder whether Jim thinks this latter group is also S_n…?
• Finally, the argument given by Kapranov and Smirnov looks like it’s making a case that the braid group admits a map to GL_n(F_1[[t]]), not so much that it should be thought of as GL_n(F_1[t]).

GL_n(F_1[[t]]), by the way, seems a little easier to get our hands on. Note that the order of the finite group GL_n(F_q[t]/t^k) is just a power of q times |GL_n(F_q)|. So, setting q = 1, one might expect

|GL_n(F_1[t]/t^k)| = |GL_n(F_1)| = n!

and in particular

GL_n(F_1[[t]]) = GL_n(F_1[t]/t^k) = S_n.

(Short version of this argument: “Pro-1 groups are trivial.”) In this case, the braid group certainly does map to GL_n(F_1[[t]])!

By the argument in the previous post, one would then want to say GL_n(F_1((t))) is the affine Weyl group Z^{n-1} semidirect S_n. Which means that the Hecke algebra

GL_n(F_1[[t]]) \ GL_n(F_1((t))) / GL_n(F_1[[t]])

is a pretty standard object — the double cosets above are in bijection with S_n-orbits on Z^{n-1}, which can be identified with the symmetrized monomials in n variables whose degree is a multiple of n, up to multiplication by x_1, … x_n. So the Hecke algebra should just be some version of the algebra of symmetric functions on n variables.

Believers that the braid group is GL_n(F_1[t]) are strongly encouraged to revivify my faith in comments.

F_1, buildings, the braid group, GL_n(F_1[t,1/t])

It used to be you had to talk about “the field with one element” very quietly, and only among people whose opinion of you was secure. The reason, of course, is that there is no field with one element. Which doesn’t stop people from saying “But if there _were_ a field with one element, what would it be like?”

Nowadays all kinds of people are musing about this odd question in the bright light of day, and no one finds it kooky. John Baez covered the basics in a 2007 issue of This Week’s Finds. And as of a few weeks ago the field with one element has its own blog, “Ceci N’est Pas Un Corps.”

From a recent post on CNPUC, I learned the interesting fact that the braid group on n strands can be thought of as GL_n(F_1[t]).

So here’s a question: what is GL_n(F_1[t,1/t])?

Proposed answer below the fold.