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Recall that complex $K$-theory is a cohomology theory on topological spaces, which can be described in several equivalent ways:

  • Given a finite complex $X$, $K^0(X)$ is the Grothendieck group of vector bundles on $X$. $K^*$ is even-periodic, and this determines the entire cohomology theory. Using the tensor product of vector bundles, $K$ becomes a multiplicative cohomology theory. There is a corresponding ring spectrum.
  • The classifying space $BU \times \mathbb{Z}$ for $K^0$ is, by a theorem of Atiyah, the space of Fredholm operators on a countably-dimensional Hilbert space. So we can think of classes in $K^0(X)$ as "families of Fredholm operators" parametrized by $X$: the "index" of such a family should be a virtual vector bundle, which connects to the previous definition.
  • $K$-theory is an even-periodic theory, so it is complex-orientable and corresponds to a formal group on $K^0(\ast) = \mathbb{Z}$. This formal group is the multiplicative one, which turns out to be Landweber-exact. Consequently, one can construct $K$-theory directly from the formal multiplicative group (once one has the spectrum $MU$) via $K_\bullet(X) = MU_\bullet(X) \otimes_{MU_\bullet} K_\bullet$.
  • The spectrum for $K$-theory can be obtained by taking the ring spectrum $\Sigma^\infty \mathbb{CP}^\infty_+$ (which is a ring spectrum as $\mathbb{CP}^\infty$ is a topological abelian monoid) and inverting the natural element in $\pi_2$. (This is a theorem of Snaith.)

It's sort of remarkable that $K$-theory can be described both geometrically (via vector bundles or operators) or algebraically (via formal groups or Snaith's theorem). The only explanation that I can think of for this is that the correspondence between (complex-orientable) ring spectra and formal groups is given more or less in terms of Chern classes of vector bundles, so a cohomology theory built directly from vector bundles would have a good chance of furnishing a fairly simple formal group law. (One can use this sort of argument to prove Snaith's theorem, for instance.)

A much less formal example of a formal group is that associated to an elliptic curve. If $E/\mathrm{Spec} R$ is an elliptic curve, then under appropriate hypothesis (Landweber exactness, or flatness of the map $\mathrm{Spec} R \to M_{1,1} \to M_{FG}$, or more concretely that $R$ is torsion-free and for each $p$, the Hasse invariant $v_1$ is a nonzerodivisor in $R/pR$) we can construct an "elliptic cohomology" theory $\mathrm{Ell}^*$ which is even-periodic and whose formal group is that of $E/R$. The associated formal group can have height up to $2$, so we get something much more complicated than $K$-theory.

It has been suggested that there should be a geometric interpretation of elliptic cohomology. This seems a lot more difficult, because the formal group law associated to an elliptic curve is less elementary than $\hat{\mathbb{G}_m}$. There are various programs (which start with Segal's survey, I believe), all of which I know nothing about, to interpret elliptic cohomology classes in terms of von Neumann algebras, loop group representations, conformal field theories, ...

I can understand why a geometric interpretation of elliptic cohomology would be desirable, but it's mystifying to me why researchers in this area are concentrating on these specific objects. Is there a "high-concept" explanation for this, and motivation (to someone without a background in geometry) for how one might "believe" in these visions? Is there a reason loop groups should be "height two" where the unitary group is "height one"?

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There is the relation between the group $String$ and tmf on one hand, and String as a 3-connected cover of $Spin$ and the canonical central extension (=2-connected cover) of $\Omega Spin$. String structures were first considered via loop groups, as there were no smooth constructions of String (we do as of about a year ago). Things like Witten genus should be mentioned. von Neumann algebras (and bimodules, and morphisms of bimodules) come up because they seem to be a good candidate for constructing categorified $L^2$ (see closing comments in arxiv.org/abs/0812.4969) for reps of 2-groups –  David Roberts Jul 6 '12 at 1:49
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...most smooth models of $String$ being 2-groups, including the the only finite-dimensional model so far, this is clearly something that needs further study. (the recent smooth construction of $String$ I mentioned earlier is a vanilla Frechet-Lie group, there were earlier Frechet-Lie 2-group models). Andre Henriques' research proposal, available from his website, has a lot of good material on looking at conformal nets and how they relate to tmf. –  David Roberts Jul 6 '12 at 2:20
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Very interesting. One thing that I've heard is that "chromatic level" is supposed to correspond to "categorical level" in some sense: i.e., if $K$-theory is supposed to come from vector bundles (a category), elliptic cohomology is supposed to come from a 2-category. (The place where I read that also said that ordinary cohomology was 0-categorical, which didn't make much sense to me: $H^1$, for instance, classifies torsors, which are a category and not a set.) –  Akhil Mathew Jul 6 '12 at 16:21
    
(Your mention of "categorified $L^2$" reminded me of this.) Anyway, it'll take me some time to digest this article, but I look forward to reading it! –  Akhil Mathew Jul 6 '12 at 16:23
    
@Akhil: the sense in which ordinary cohomology is 0-categorical is that $H^{\bullet}(X, \mathbb{Z})$ is at least morally speaking (derived) functions from $X$ to $\mathbb{Z}$ (which is a set), as opposed to K-theory, which at least morally speaking (derived) functions from $X$ to $\text{Vect}$ (which is a category). –  Qiaochu Yuan Dec 1 at 10:41

2 Answers 2

up vote 17 down vote accepted

David Roberts mentioned in his comments the relationship

K-theory : spin group
TMF : string groups

Let me recommend the first 6 pages of my unifinished article for a uniform construction of $SO(n)$, $Spin(n)$, and $String(n)$, which suggests the existence of a similar uniform construction of $H\mathbb R$, $KO$ (or $KU$), and $TMF$. You'll see that von Neumann algebras appear in the construction. More precisely, von Neumann algebras appear in the definition of conformal nets. The latter are functors from 1-manifolds to von Neumann algberas.

For a summary of the conjectural relationship between conformal nets and $TMF$, have a look at page 8 of this other paper of mine (joint with Chris Douglas).

Loop groups yield non-trivial examples of conformal nets. Those conformal nets are related (conjecturally) to equivariant $TMF$.

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By the way: in that note you obtain String as a topological 2-group. Have you, or can you, construct an equivalence to the topological 2-group underlying any one of the smooth 2-group models? (That would be good to have, since it would allow to associate fermionic net 2-bundles to smooth String-principal 2-bundles.) Last time that I looked into this with colleages we found a natural candidate homomorphism from the strict smooth 2-group version of String to that 2-group of net automorphisms in your writeup. But we didn't quite check a bunch of operator algebra things that one needs to check. –  Urs Schreiber Jul 6 '12 at 12:40
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Hi Urs. No, that remains to be done. That model of the string group doesn't look very smooth. Making sense of a smooth family of Hilbert spaces is already quite a tricky buisness: that's already something that I don't know how to do... So making sense of a smooth family of von Neumann algebras (or defects between conformal nets -- same issues) is much more tricky, and I also don't know how to do it. –  André Henriques Jul 6 '12 at 13:20
    
@André: On page 8 you mention that elements in K-theory can modeled by quasibundles (finite dimensional not necessarily locally trivial bundles) of Clifford modules. Do you have a reference for that? –  Dmitri Pavlov Jul 6 '12 at 16:45
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Awesome. Thank you! There's a lot to digest here, and maybe I'll comment after I've understood a bit more. But for starers I am curious if there is a "good" reason for the appearance of Clifford modules in $KO$-theory, especially the fact that Clifford modules can be used to produce $\pi_* KO$. (See this question mathoverflow.net/questions/85516/…). I take it general von Neumann algebras are supposed to be some sort of extension of Clifford algebras? –  Akhil Mathew Jul 6 '12 at 21:30
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Unfortunately, I don't know how to explain "why"... See also mathoverflow.net/questions/62654/… for more unanswered questions about the relationship between Clifford algebras and KO. –  André Henriques Jul 6 '12 at 22:31

There are various programs (which start with Segal's survey, I believe), all of which I know nothing about, to interpret elliptic cohomology classes in terms of von Neumann algebras, loop group representations, conformal field theories, ...

I can understand why a geometric interpretation of elliptic cohomology would be desirable, but it's mystifying to me why researchers in this area are concentrating on these specific objects.

You probably know most of this by now, but here are some thoughts. As usual I will be working at a very heuristic level throughout this answer.

Von Neumann Algebras. One place to start is the observation is that for $H$ an infinite-dimensional Hilbert space, the von Neumann algebra $B(H)$ has automorphism group the projective unitary group $PU(H)$. $PU(H)$ fits into a short exact sequence

$$1 \to U(1) \to U(H) \to PU(H) \to 1$$

and $U(H)$ is contractible by Kuiper's theorem; thus $PU(H)$ is a $B^2 \mathbb{Z}$, and $BPU(H)$ is a $B^3 \mathbb{Z}$. Hence $H^3(X, \mathbb{Z})$ is, in a suitable sense, a "Brauer group" of $X$ describing bundles of von Neumann algebras (isomorphic to $B(H)$) over $X$.

The significance of this observation is that $H^3(X, \mathbb{Z})$ is a natural cohomology group parameterizing twists of K-theory over $X$; equivalently, there is a natural map from $B^3 \mathbb{Z}$ to $BGL_1(KU)$. Given a bundle of von Neumann algebras over $X$, the corresponding twisted K-theory groups are something like the K-theory of module bundles of Hilbert modules over the bundle of von Neumann algebras, but don't trust me to have the specifics right here.

Now it's also known (see e.g. Ando-Blumberg-Gepner) that $H^4(X, \mathbb{Z})$ is a natural cohomology group parameterizing twists of tmf over $X$; equivalently, there is a natural map from $B^4 \mathbb{Z}$ to $BGL_1(tmf)$. If you could build a (higher) category which deloops von Neumann algebras in a suitable sense, you might hope to realize $BPU(H) \cong B^3 \mathbb{Z}$ as the automorphisms of an object in this category, and then families of those objects over $X$ could be a geometric avatar of these twists of tmf in the same way as above. I believe that conformal nets is explicitly intended to be such a delooping.

Loop group representations. This is the analogue of $G$-equivariant K-theory having something to do with the representation theory of $G$. One picture of tmf whose accuracy I can't comment on is that it should look heuristically like $K(ku)$, the cohomology theory presented by (Bass-Dundes-Rognes) 2-vector bundles. So $G$-equivariant tmf should look heuristically like $G$-equivariant 2-vector bundles, which over a point should look heuristically like representations of $G$ on (suitably dualizable) 2-vector spaces.

Whatever that means, such a thing ought to have a "character" which, rather than being a class function on $G$, or equivalently a function on the adjoint quotient $G/G$, is instead an equivariant vector bundle on $G/G$. Now $G/G$ looks heuristically like the classifying stack $BLG$ of the loop group $LG$, and hence an equivariant vector bundle on $G/G$ looks heuristically like a representation of $LG$. Freed-Hopkins-Teleman makes this precise. The non-equivariant version of this story is Witten's story about tmf having something to do with ($S^1$-equivariant?) K-theory of the free loop space $LX$.

Conformal field theories. This is the analogue of K-theory being presentable by vector bundles with connection. One way to think about a vector bundle with connection on a manifold $X$ is that it is a "$1$-dimensional topological field theory over $X$": that is, it assigns a vector space to a finite set of points equipped with signs in $X$ (the tensor product of either the fibers of the vector bundle or their dual depending on the orientation), and it assigns a map of vector spaces to every oriented cobordism between such points in $X$ (the tensor product of either the holonomies or the evaluation or coevaluation maps).

The historical motivation for generalizing this to $2$-dimensional conformal rather than topological field theories is, I think, to explain the modularity properties of the Witten genus. But again, don't trust me to have the specifics right here. (I guess it's $2$-dimensional topological rather than conformal field theories over $X$ that look more like $2$-vector bundles.)

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There's another set of hints involving $H^2(BG, \mathbb{C}^{\times})$ parameterizing both twists of $G$-equivariant K-theory and twists of 2d $G$-Dijkgraaf-Witten theory, and $H^3(BG, \mathbb{C}^{\times})$ parameterizing both twists of $G$-equivariant tmf (or something like that) and twists of 3d $G$-Chern-Simons theory. Chern-Simons is in turn also expected to be related to conformal nets, since it also has something to do with CFT; see, for example, André Henriques' answer at mathoverflow.net/questions/163301/…. –  Qiaochu Yuan Dec 4 at 7:14
    
What did you mean by this sentence: " One picture of tmf whose accuracy I can't comment on is that it should look at least a bit like K(ku)". It doesn't make sense to me. –  David Roberts Dec 4 at 7:15
    
@David: for example, Bass-Dundes-Rognes (math.uiuc.edu/K-theory/0629) showed that $K(ku)$ detects $v_2$-periodic phenomena. I'm not familiar with the details though. –  Qiaochu Yuan Dec 4 at 7:17
    
@DavidRoberts, it makes sense to me at least heuristically. It is natural to expect that tmf classes can be described (or at least are closely related) by K-theory of 2-vector bundles. A 2-vector space is a $\mathrm{Vect}$-module, and that looks a lot like $K(\mathrm{Vect})=K(ku)$-module. This extra group completion likely gives a different, but not too different cohomology theory. –  Anton Fetisov Dec 4 at 10:58
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I mean that the sentence doesn't seem to be grammatical, but maybe that's me. –  David Roberts Dec 4 at 19:48

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