MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

As the discussion here Is singular cohomology representable by a (Voevodsky's) motivic complex? shows, the singular cohomology of (smooth) complex varieties is represented by a motivic complex (and also by a motivic spectrum). My question is: what can be proved about the ring structure for this complex/spectrum? Is it known that this a 'weak' ring spectrum? an $A_{\infty}$-spectrum? a highly structured ring spectrum? Any hints would be very welcome!

share|cite|improve this question
You should take a look to this article by Joshua Roy (Theorem 1.1) The Rational and Integral motivic complex. In the rational case it is represented by commutative DGA over Q (since zoo can always strictify over rationals) and the second is represented by $E_{\infty}$-differential graded $\mathbf{Z}$-algebra. $ – Ilias Amrani Aug 11 '14 at 21:36
Here is an "easy" argument in the case of topological space (that you can perform in the stable motivic category). Suppose that $R$ is an $E_{\infty}$ ring spectrum and $X$ is a space, use the internal Hom $F(-.-)$ in the category of spectra, then $F(\Sigma^{\infty} X_{+}, R)$ is an $E_{\infty}$ ring spectrum, since $\Sigma^{\infty} X_{+}$ is an $E_{\infty}$ coalgebra. – Ilias Amrani Aug 11 '14 at 21:56
up vote 6 down vote accepted

Singular cohomology is represented by an $E_\infty$ motivic ring spectrum. That spectrum is $\mathbf{R}f(H\mathbb{Z})$ where $f$ is right adjoint to the stable topological realization functor and $H\mathbb{Z}$ is the topological Eilenberg-Mac Lane spectrum. Since this is a symmetric monoidal Quillen adjunction (see Theorem A.45 here:, $\mathbf{R}f$ preserves $E_\infty$-objects.

The same argument should show that the motivic complex is also $E_\infty$, but I don't know a reference for the required symmetric monoidal adjunction. At least if you work with $(\infty,1)$-categories this adjunction comes for free from the fact that motivic complexes are the same thing as modules over the motivic Eilenberg-Mac Lane spectrum, whose topological realization is $H\mathbb{Z}$.

share|cite|improve this answer
Thank you!!!!!! – Mikhail Bondarko Jan 21 '13 at 4:59

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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