In this question I previously asked how to think about the motivic complex $\mathbf{Z}(1)_{\mathcal{M}}$, whose Zariski hypercohomology should morally be the "singular cohomology" $H^*((-)\wedge S^{2n},\mathbf{Z})$, or the "singular cohomology of a pair" $H^*((-)\times\mathbf{P}^n_k,(-)\times\mathbf{P}^{n-1}_k)$.

Denis Nardin's answer explains a suggestive analogy, and here is a followup.

It happens that several motivic cohomology groups are infinitely generated, even if $X$ is a smooth projective variety over a field (even for $k = \mathbf{Q}$ and $X = \text{Spec}(k)$).

How to think about this? Is there an analogy with infinite generatedness of the singular cohomology of a pair, for instance?

In other words, what should be the "moral reason" why this infinite generatedness occurs? Any analogy coming from algebraic topology would be greatly clarifying.

  • 1
    $\begingroup$ For example, $H^{n,n}_\mathrm{mot}(k,\mathbf{Z}) = K^M_n(k)$ is Milnor $K$-theory of the field $k$, which is usually not finitely generated. $\endgroup$ – user19475 Feb 12 '18 at 11:56
  • $\begingroup$ Other examples are provided by the Beilinson-Lichtenbaum conjecture which relates motivic with étale cohomology. $\endgroup$ – user19475 Feb 12 '18 at 12:20

While waiting for someone more competent than me to answer, let me turn the question right back to you. Why should motivic cohomology be finitely generated?

The answer is, of course, that there's no reason for it. And it is not. Let us take a look at a special example

The Picard group

Let us fix the ground field to be $\mathbb{C}$. The Picard group of a variety is the group of isomorphism classes of line bundles with multiplication given by the tensor. It turns out to be a very "homotopical" invariant. We can consider various variants of it: topological Picard group, analytic Picard group, algebraic Picard group..., depending on what kind of line bundles we are interested in.

It is well known that the topological Picard group of a space $X$ is just $H^2(X;\mathbb{Z})$ (this comes from the identification of $\mathbb{CP}^\infty$ with $K(\mathbb{Z},2)$).

In the algebraic world, something similar happens. The Picard group is equivalent to $H^2_{mot}(X;\mathbb{Z}(1))$. But in this case we know the Picard group cannot be finitely generated: there is a map from the algebraic Picard group to the topological Picard group (called the first Chern class) but, for example, if $X$ is a curve its kernel is given by the points of an abelian variety, the Jacobian of $X$. This is a massively nonfinitely generated abelian group, in fact it is isomorphic to $(\mathbb{R}/\mathbb{Z})^{2g}$ where $g$ is the genus of $X$.

| cite | improve this answer | |

Your Answer

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