I am reading a little about cobordism and I have a basic question, which makes sense both in the topological and motivic setting. Let $\mathrm{Gr}_{n,\infty}$ denote the infinite $n$-Grassmanian and denote $\xi_n $ its universal bundle. The cobordism spectrum $\mathrm{MGL}$ is made out of the universal Thom spaces.That is to say, it is the spectrum $$ (S\, ,\, \mathrm{Th}(\xi_1)\, ,\, \mathrm{Th}(\xi_2)\, ,\, \ldots \, ,\, \mathrm{Th}(\xi_i)\, ,\, \ldots \ ) $$ with the natural bounding maps. My question is

Is $\mathrm{MGL}[2r](r)$ isomorphic, in $\mathbf{SH}$, to the spectrum $(\, \mathrm{Th}(\xi_r)\, ,\, \mathrm{Th}(\xi_{r+1})\, ,\, \ldots \, ,\, \mathrm{Th}(\xi_{r+i})\, ,\, \ldots \ )$? And if so, why?

It seems to me that this should be right and simple, but since $\mathrm{MGL}$ is not an $\Omega$-spectrum I fail to see a direct reason.


1 Answer 1


A simple way of seeing it is to explicitly spell out what we mean when we say that a spectrum is "presented" by a prespectrum. To say that that a spectrum $E$ is presented by $$(E_0,E_1,...)$$ means that, in whatever model for motivic spectra you are using, $$E\cong \mathrm{colim}_k \Sigma^{-2k,-k}\Sigma^{\infty}E_k$$ (here the colimit is a homotopy colimit). Since suspensions commute with colimits (being equivalences) we have that $$\Sigma^{2n,n}E\cong \mathrm{colim}_k \Sigma^{2(n-k),(n-k)}\Sigma^{\infty}E_k\cong \mathrm{colim}_k \Sigma^{-2k,-k}\Sigma^\infty E_{k+n}$$ where the last step is just reindexing the colimit. In particular, $\Sigma^{2n,n}E$ is presented by $$(E_n,E_{n+1},\dots)\,.$$

  • $\begingroup$ Thanks again for the answer, Denis. Could you encourage me a reference to learn these basics? thanks $\endgroup$
    – Tintin
    Apr 4, 2019 at 8:46
  • 1
    $\begingroup$ @Tintin Unfortunately references are known to be annoyingly spread out in homotopy theory. The trick I used in this answer is quite widespread whenever some version of spectra is used, I first learned it in the equivariant setting in the Hill-Hopkins-Ravenel paper (however it is useful even in the classical setting of spectra on topological spaces!). The best I can do is to encourage you to read some good introduction to spectra (possibly the best currently available is Schwede's book), although even there you won't find everything. $\endgroup$ Apr 4, 2019 at 8:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.