Line Bundle Twists for Unitary Bordism are Ghosts

Abstract

We prove that the canonical twist ζ : K(Z, 3) → BGL1(M Spin c ) does not extend to a twist for unitary bordism by showing that every continuous map f : K(Z, 3) → BGL1(M U ) loops to a null homotopic map. IntroductionTwisted cohomology, originally invented by Steenrod [Ste43], allows to classify geometric objects that cannot be classified by untwisted cohomology without further orientation conditions.While Steenrod defined twisted (co-)homology algebraically in terms of chain complexes with local coefficients, a modern approach to twisted (co)homology in terms of twisted spectra is provided by the work of [ABG + 14a], [ABG10], and [ABG + 14b]. Their construction goes roughly as follows: An A ∞ -ring spectrum R has a space of units GL 1 (R), which deloops to a classifying space BGL 1 (R). A twist is a map ξ : X → BGL 1 (R) from which we can construct a Thom spectrum X ξ , whose homotopy groups are the ξ-twisted homology groups of X.Our motivation for the work presented here arose from the results of [HJ20], see also [HS20]. The authors construct point set models for twisted Spin c bordism and twisted K-theory over K(Z, 3) as well as a model for a twisted Atiyah-Bott-Shapiro orientation α ABS : M Spin c K(Z,3) → K K(Z,3) . In [HJ20, App. C] it is shown that the two approaches agree, so that the point set model also arises from a twist map ζ : K(Z, 3) → BGL 1 (M Spin c ).One of its application concerns the bordism-determines-homology question. The classical results of [CF66] and [HH92] state that K-theory is completely described by the Conner-Floyd orientation CF : M U → K and the ABS orientation α ABS : M Spin c → K in the sense thatfor all spectra X. In [BJKS] the latter isomorphism is proved by geometric means in the twisted set-up using the K(Z, 3)-twisted ABS orientation.It is natural to ask whether one can also extend the first isomorphism to the K(Z, 3)twisted setup such that it reduces to the isomorphism of [BJKS] under the canonical map M U → M Spin c . The first step would be to construct a twisted unitary bordism spectrum over K(Z, 3) and a comparison map M U K(Z,3) → M Spin c K(Z,3) that extends the twisted Atiyah-Bott-Shapiro orientation 'to the left'. The first result of this article shows that this is impossible.Theorem A. The twist map ζ : K(Z, 3) → BGL 1 (M Spin c ) does not factor through unitary bordism. More precisely, there are no continuous maps S and T that make the following