7
$\begingroup$

Apologies for any basic mistakes in this question; I'm a beginner to this theory and don't have anyone at my institution to consult for advice.

What I mean is, if you take an elliptic curve $E$ over $\mathbb{Q}$ say, you should get associated an cohomology theory by taking some deformation of it/putting it in a family and completing. (At least as far as I know, the deformation is necessary, since otherwise LEFT fails at supersingular primes. Is this correct?)

It seems to me that you could pretty much do this with any formal group over $\mathbb{Z}$; just find some deformation so that the heights are all generically $1$, then apply LEFT. (Or is this maybe harder to do than I think? Seems it wouldn't be, though, just from stacky considerations.)

My question is, is there some intrinsic interest in the cohomology theories actually arising from elliptic curves, as opposed to these ones? Or is the reason for the interest entirely because of the geometry of the universal tmf, and its speculative connections to loop groups, the Witten genus, etc.?

Or maybe is the principal interest in the direction of searching for a good equivariant theory which encodes all the elliptic curve's geometry, instead of just its formal Neron model?

Improve this question
$\endgroup$
15
  • 4
    $\begingroup$ I guess the "intrinsic" interest lies in the fact that these are explicit theories of chromatic level 2. That is, height never goes above 2, and some heights are 2. $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Sep 24, 2018 at 21:05
  • 1
    $\begingroup$ couldn't you write down an arbitrary fgl/Z that does that (using Hazewinkel's/Honda's Dirichlet series formalism for example), put it in a suitable family, and obtain much the same thing? is there any difference besides the elliptic curve one feeling more "god-given"? $\endgroup$
    – xir
    Commented Sep 24, 2018 at 21:20
  • 6
    $\begingroup$ I'm not an expert, but a great thing about elliptic curves is that they come in families, so you can naturally "deform" your cohomology theory. Ultimately this ends up giving you $TMF$ (that has roughly the same relationship with elliptic cohomology that $KO$ has with $KU$) $\endgroup$
    – Denis Nardin
    Commented Sep 24, 2018 at 22:01
  • 3
    $\begingroup$ Giving a geometric description of Tmf or TMF is a big open problem a lot of people are interested in. I agree that part of the reason TMF is not used that much in differential geometry is the lack of a geometric interpretation, however if you want an interesting intepretation look no further than the Witten genus (also known as the String orientation of Tmf) $\endgroup$
    – Denis Nardin
    Commented Sep 25, 2018 at 7:35
  • 1
    $\begingroup$ I'm confused by your claim that the Landweber exact functor theorem fails at supersingular primes; can you clarify? There are a few "intrinsic" reasons to find tmf and its variants interesting. For one, the Hurewicz image of tmf detects numerous elements in the stable homotopy groups of spheres (I think this fact featured in recent work of Zhouli Xu and Guozhen Wang). Another reason is that, using similar techniques, you can construct an E_oo-ring denoted tmf_1(3) --- which takes effort --- that is the only known E_oo-form of the spectrum BP<2>. I don't know of a geometric perspective on tmf. $\endgroup$
    – skd
    Commented Sep 27, 2018 at 18:21

1 Answer 1

Reset to default
4
$\begingroup$

Here's a partial answer which I find somewhat satisfying: elliptic genera have a natural geometric definition as genera which kill the projective total spaces of even-dimensional complex bundles over compact oriented manifolds, and are closely related to spin geometry and elliptic operators on free loop spaces. This gives a principled way to distinguish elliptic genera from others, coming from geometry. I would still be interested to see if this distinction is possible in purely homotopical terms, or if this is a case where the content really is purely geometrical (and "just happens" to link to the algebraizability of the associated formal group!).

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.