# (Algebraic) cobordism and the rank function

I write the question for algebraic cobordism but I have the analogue question for classic cobordism.

The spectrum representing algebraic cobordism $$\mathbf{MGL}=(*, \mathrm{Th}(1) , \ldots , \mathrm{Th}(n), \ldots)$$ is given by the Thom spaces $$\mathrm{Th}(n)$$ of the universal bundles of the Grassmanians $$\mathrm{Gr}_n$$. Let $$\mathbf{E}=(E_n)_{n\in\mathbb{N}}$$ be the spectrum of a cohomology theory with Chern classes (or oriented), it is very easy to write the universal map $$\varphi^E\colon \mathbf{MGL}\to \mathbf{E}$$ since it is the morphism of spectra having at the $$n$$-th level the Thom class $$\mathit{th}^\mathbf{E}(\mathrm{Th}(n)) :\mathrm{Th}(n)\to E_n$$.

Now, for the concrete case of motivic cohomology (or singular cohomology in the topological case) one can check that at the (0,0)-level the map $$\mathbf{MGL}^{0,0}(X)\to H^{0}(X,\mathbb{Z}(0))$$ induced by $$\varphi^\mathbf{H}$$ "is" the rank function. More concretely, the above arrow is the composition of the following: $$\mathbf{MGL}^{0,0}(X)\xrightarrow{\varphi^K} K_{0}(X)\xrightarrow{\mathrm{rank}}H^{0}(X,\mathbb{Z}(0))$$ One can check this explicitly, for example, whenever you know that $$\mathbf{MGL}^{0,0}(X)$$ is generated by morphisms $$f\colon Y\to X$$ where $$\mathrm{dim} Y=\mathrm{dim}X$$. This suggests that the unstable map $$(\varphi^H)_0: \Omega^\infty \mathbf{MGL}\to K(0,0)$$ defined as the adjoint of the composition in $$\mathbf{SH}$$ of $$\Sigma^\infty \Omega^\infty \mathbf{MGL}\to \mathbf{MGL}\xrightarrow {\varphi^H} \mathbf{H}$$, should equal the composition in the unstable homotopy category of $$\Omega^\infty \mathbf{MGL} \xrightarrow{(\varphi^K)_0}\mathbb{Z}\times \mathrm{Gr}\xrightarrow{\mathrm{rank}} K(0,0),$$ where $$(\varphi^K)_0$$ is the defined analogously to $$(\varphi^H)_0$$ and $$K(0,0)$$ is the space at level zero of the spectrum $$\mathbf{H}$$ representing motivic cohomology. To sum up, my question is:

Does $$(\varphi^H)_0$$ equal to $$\mathrm{rank}\circ (\varphi^K)_0$$?

On top of the motivation I gave above and to rephrase the question. It is already known that, over a field, $$\Omega^\infty \mathbf{MGL}=\mathbb{Z}\times\mathrm{Hilb}_\infty^{\mathrm{lci}}(\mathbb{A}^\infty)^+$$ (see [1]) and the rank function on $$K$$-theory is given by the projection towards the first factor of $$\mathbb{Z}\times \mathrm{Gr}$$, so it is very likely that $$(\varphi^H)_0$$ is also the projection towards the first factor composed with its natural map towards $$K(0,0)$$.

If you know any reference of a computation similar to this for classic cobordism I would also thank that.

• I am not sure I understand your question, but does section 9 of this paper help (in particular corollary 9.2)? We wrote it to give a neat picture of the current state of affairs of geometric models for motivic cohomology theories May 27 at 15:04

Let me first write what happens for classical cobordism. You are basically asking whether the map $$\operatorname{MU}\to H\mathbb{Z}$$ factors through the projection $$\operatorname{ku}\to H\mathbb{Z}$$. But this is clear, since all spectra in sight are connective and we have an equivalence $$\operatorname{Map}(E,H\mathbb{Z})\cong \operatorname{Map}(\pi_0E,\mathbb{Z})$$ for every connective spectrum $$E$$ given by the fact that $$\pi_0$$ is the left adjoint of the inclusion of discrete spectra (i.e. abelian groups) into connective spectra. This adjunction by the way works both in spectra and abelian groups and $$E_\infty$$-ring spectra and commutative rings, in which case it's clear that the right hand side is just a point both for $$\operatorname{MU}$$ and $$\operatorname{ku}$$.
Something similar happens in motivic cohomology. Here the statement is that the map $$\operatorname{MGL}\to H\mathbb{Z}$$ factors uniquely through the orientation $$\operatorname{MGL}\to\operatorname{kgl}$$ (here with $$\operatorname{kgl}$$ we mean the very effective cover of the motivic spectrum $$\operatorname{KGL}$$). Instead of taking homotopy groups the correct thing to do is taking slices. Indeed the both map $$\operatorname{MGL}\to H\mathbb{Z}$$ and $$\operatorname{kgl}\to H\mathbb{Z}$$ exhibit $$H\mathbb{Z}$$ as the zero slice of the respective spectrum and $$\operatorname{MGL}\to\operatorname{kgl}$$ induces an equivalence on 0-slices.
• Thank you very much, Denis. That indeed is a nice solution. One more thing, do you know a reference where I can read about the adjunction between groups/rings and spectra/ring spectra? The references I see only do the construction of $H$ and I am interested in reading in detail the adjoint map $1\to H\pi_0$, I don't see how to do it. Thanks in advance Jun 3 at 13:37
• @Tintin For spectra this is just the standard axioms of a t-structure (the functor is the 0-truncation functor from connective spectra to the heart). For ring spectra, a reference is Higher Algebra, 7.1.3.15, but it's probably simpler to prove by identifying connective $E_\infty$-ring spectra with models for a certain Lawvere theory, where you can just apply the corresponding statement for spaces. Jun 3 at 20:24