English

tmf Is Not a Ring Spectrum Quotient of String Bordism

Algebraic Topology 2018-07-03 v4

Abstract

This paper shows that tmf[1/6]\mathrm{tmf}[1/6] is not a ring spectrum quotient of MO8[1/6]\mathrm{MO}\langle8\rangle[1/6]. In fact, for any prime p>3p>3 and any sequence XX of homogeneous elements of πMO8\pi_*\mathrm{MO}\langle8\rangle, the πMO8\pi_*\mathrm{MO}\langle8\rangle-module π(MO8(p)/X)\pi_*\big(\mathrm{MO}\langle8\rangle_{(p)}/X\big) is not (even abstractly) isomorphic to πtmf(p)\pi_*\mathrm{tmf}_{(p)}. It does so by showing that, for any commutative ring spectrum RR and any sequence XX of homogeneous elements of π(R)\pi_*(R), there is an isomorphism of graded Q\mathbf{Q}-vector spaces π(R/X)QH(Tot(K(X)))Q,\pi_*(R/X)\otimes\mathbf{Q} \cong \mathrm{H}_*(\mathrm{Tot}(\mathrm{K}(X)))\otimes\mathbf{Q}, where the right-hand side is the rational homology of the (total) Koszul complex of XX, which is strictly bigger than π(R)/(X)Q\pi_*(R)/(X)\otimes\mathbf{Q} unless XX is a π(R)Q\pi_*(R)\otimes\mathbf{Q}-quasi-regular sequence. The result then follows from the fact that the kernel of the pp-local Witten genus cannot be generated by a πMO8Q\pi_*\mathrm{MO}\langle8\rangle\otimes\mathbf{Q}-quasi-regular sequence.

Keywords

Cite

@article{arxiv.1312.2440,
  title  = {tmf Is Not a Ring Spectrum Quotient of String Bordism},
  author = {Carl McTague},
  journal= {arXiv preprint arXiv:1312.2440},
  year   = {2018}
}

Comments

9 pages; statement of the new Theorem 3 made clearer