@Drew Excellent. Thanks.

Let ML_n denote the group of pairs (A, z) with A in GL_n(C) and z in GL_1(C) such that det(A) = z^2. Then BML_n classifies complex vector bundles with a chosen square root (I think). What is known about BML_n, BML, MML?

what's the difference between vector bundles with a chosen square root vs any old vector bundle, thought of as the square root of its square? do equivalences not directly respect the square root data?

@EricPeterson I think they are vector bundles equipped with a square root of the determinant, not a square root of the vector bundle itself

@Tom I suspect that MSL = MML, because SL-oriented spectra are automatically ML-oriented. The main point is that the J-homomorphism Vect(X) → pointed spaces/X is C_2-invariant for the action of C_2 on Vect(X) by duality, and in particular a line bundle L and its inverse L^{-1} have the same Thom spaces. Both L+L-1-L^2 and L+L^{-1}-2 have canonically trivialized determinant, so from the perspective of an SL-oriented theory the Thom spectrum of L^2-1 is trivial.

@EricPeterson what denis said. sorry

@MarcHoyois what's the isomorphism between Th(V) and Th(DV)?

o

@Tom Let H(V) ⊂ V ⊕ DV consist of pairs (v,f) where f(v)=1. Then both projections H(V) → V-0 and H(V) → DV-0 are affine bundles, so V-0 ≃ DV-0. I learned this trick from Ananyevskiy's paper arxiv.org/abs/1406.2894

very cool! Thanks

I suppose it's conceivable that an SL-oriented spectrum can be ML-oriented in more than one way, in which case one would only expect MSL to be a summand of MML...