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The last paragraph in Two-Vector Bundles and Forms of Elliptic Cohomology remarks that neither the spectrum $K(ku)$ nor tmf is complex orientable. In the case of $K(ku)$: "...the unit map for $K(ku)$ cannot factor through the complex bordism spectrum $MU$, since $\pi_1(MU)=0$."

This confuses me, is it not the case that every elliptic cohomology theory represents a complex-orientable $E_\infty$-ring spectra and vice-versa?

On page 21 of Lurie's Survey, he mentions:

If we view $\mathbb{CP}^\infty$ as the classifying space for complex line bundles, then the group algebra $\Sigma[\mathbb{CP}^\infty]$ can be viewed as a universal cohomology theory in which it is possible to add line bundles. The above result can be viewed as saying that if we take this universal cohomology theory and invert the Bott element $\beta$ then we obtain a theory which classifies vector bundles. A very puzzling feature of the result is the apparent absence of any direct connection of the theory of vector bundles with the problem of orienting the multiplicative group.

Does this mean we have no functorial, multiplicative choice of Thom classes for complex vector bundles in tmf and $K(ku)$?

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Neither $K(ku)$ nor tmf are complex orientable, so neither $K(ku)$ nor tmf are elliptic cohomology theories in the strict sense. $K(ku)$ "is" an elliptic cohomology theory in the looser sense that it "detects $v_2$-periodic phenomena" (although I can't elaborate too much on what this means), and tmf "is" an elliptic cohomology theory in the looser sense that it is built out of all elliptic cohomology theories somehow.

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  • $\begingroup$ Why is Bott periodicity a $v_1$ periodic phenomena, (besides that K-theory is associated to the multiplicative formal group law which is of height 1)? I don't understand why the height of the formal group law associated to a complex orientable cohomology theory is relevant to the periodicity of the cohomology theory. $\endgroup$
    – Catherine Ray
    Jan 19, 2015 at 9:03
  • $\begingroup$ My understanding is that tmf is "built" out of elliptic cohomology theories in the following way: Take the taylor series expansion of an elliptic curve $E$ about the origin, you'll get a formal group law over a coefficient ring $C_E$. Varying the curve $E$ in the the Landweber-Stong construction, $MU^*(X)\otimes_{MU^*(pt)} C_E$, will give us a family of elliptic cohomology theories (assuming $C_E$ is Landweber-exact). Taking the limit of this family (over the category of elliptic curves), we get tmf. $\endgroup$
    – Catherine Ray
    Jan 20, 2015 at 5:09
  • $\begingroup$ @fractalcows: I believe that description of tmf is incorrect, roughly speaking because the structure of the category of elliptic curves does not know what it means for a family of elliptic curves to vary over a scheme. One needs something like the derived moduli stack of elliptic curves; this is Lurie's approach. I don't know anything about $v_n$-periodicity, so I'm not the one to ask about that; in particular I don't know what it has to do with Bott periodicity. $\endgroup$
    – Qiaochu Yuan
    Jan 20, 2015 at 6:38
  • $\begingroup$ @Qioachu I just read over Akhil Matthew's post. If I understand correctly, the description I gave of the Landweber-Stong construction was correct, but can be instead phrased as: An elliptic curve is a one-dimensional algebraic group (more generally, a group scheme) $C$ over a ring $R$, which we can write as $C \to \text{Spec}(R)$, where $\text{Spec}(R)$ is a ring spectrum (a commutative monoid in the stable homotopy category). (What is $R$ here? Is it what I called $C_E$ in my previous post, or is it like $y^2 = 4x^3 + ax + b; a, b \in R$) $\endgroup$
    – Catherine Ray
    Jan 20, 2015 at 16:46
  • $\begingroup$ Allow me to correct some of the errors in my past 2 comments: $C_E$ arises from the ring that parameterizes the coefficients in the family $C \to \text{Spec}(C_E)$ (e.g. $ y^2=4x^3+ax+b$; $a,b \in C_E$); $\text{Spec}(C_E)$ is emphatically not the a ring spectrum in the sense of a commutative monoid in the stable homotopy category, but instead the spectrum of a ring, in the sense of a functor from $\text{Ring} \to \text{Scheme}$. $\endgroup$
    – Catherine Ray
    Jan 22, 2015 at 23:09

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