most recent 30 from http://mathoverflow.net 2013-06-18T06:52:35Z http://mathoverflow.net/feeds/question/119426 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/119426/ring-structure-for-the-motivic-spectrum-complex-that-represents-singular-cohomolo Mikhail Bondarko 2013-01-20T21:24:36Z 2013-01-20T23:38:12Z

<p>As the discussion here <a href="http://mathoverflow.net/questions/42693/is-singular-cohomology-representable-by-a-voevodskys-motivic-complex" rel="nofollow">http://mathoverflow.net/questions/42693/is-singular-cohomology-representable-by-a-voevodskys-motivic-complex</a> 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!</p>

http://mathoverflow.net/questions/119426/ring-structure-for-the-motivic-spectrum-complex-that-represents-singular-cohomolo/119434#119434 Marc Hoyois 2013-01-20T23:38:12Z 2013-01-20T23:38:12Z

<p>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: <a href="http://arxiv.org/pdf/0709.3905.pdf" rel="nofollow">http://arxiv.org/pdf/0709.3905.pdf</a>), $\mathbf{R}f$ preserves $E_\infty$-objects.</p> <p>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 $H\mathbb{Z}$.</p>