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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!

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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. $people.math.osu.edu/joshua.1/mdga_short.pdf –  Fedotov Aug 11 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. –  Fedotov Aug 11 at 21:56

1 Answer 1

up vote 5 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: http://arxiv.org/pdf/0709.3905.pdf), $\mathbf{R}f$ 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 $H\mathbb{Z}$.

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Thank you!!!!!! –  Mikhail Bondarko Jan 21 '13 at 4:59

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