Let $X$ be an algebraic variety over a field $k$.

Bloch defines the "algebraic singular complex" using the algebraic simplices

$$\Delta^n = \text{Spec}(k[x_0,\dots,x_n]/(x_0+x_1+\dots+x_n=1) \subset \mathbf{A}^{n+1}_k$$

that are easily arranged into a cosimplicial scheme $\Delta_k^{\bullet}$. Then one defines:

$$z^i(X, j)= \mathbf{Z}[C_{ij}]$$

the free abelian group on $C_{ij}$, where $C_{ij}$ is the set of integral closed subschemes in $X\times_k\Delta^j$ of codimension $i$ and in "good position", ie. intersecting every face of $\Delta_k^j$ in $X\times_k\Delta_k^j$ in codimension $\ge i$.

See https://www.uni-due.de/~bm0032/publ/CycleComplexes.pdf for a quick intro.

The cosimplicial structure on $\Delta_k^{\bullet}$ makes $z^i(X,\bullet)$ into a simplicial abelian group, and the Zariski hypercohomology of the associated complex (take the differential to be the alternating sum of the degeneracy maps) is denoted $H^{\bullet}(X, \mathbf{Z}(j))$ because if $X$ is smooth it agrees with motivic cohomology.

Let us construct a similar complex, using, instead of $\Delta_k^{\bullet}$, rather the cosimplicial scheme $\Gamma_k^{\bullet}$ where $\Gamma_k^n$ is the projective closure of $\Delta_k^n$ in $\mathbf{P}^{n+1}_k$, i.e.. the hyperplane $x_0 + \dots + x_n = x_{n+1}$ in $\mathbf{P}^{n+1}_k$. As before indeed $\Gamma_k^{\bullet}$ forms a cosimplicial scheme in a way that is completely analogous to $\Delta_k^{\bullet}$.

Then as before we define $Z^i(X, j)$ to be the free abelian group on integral closed subschemes in $X\times_k\Gamma_k^{j}$ intersecting every face in $\Gamma_k^j$ (included the one at $\infty$) in the product, in codimension $\ge i$.

It looks like the last condition still ensures that pullbacks between the abelian groups $Z^i(X,j)$ are defined, and hence we still get a simplicial abelian group, and then a complex as before.

There's an obvious map between complexes of abelian groups $z^i(X,\bullet)\to Z^i(X,\bullet)$.

My question is. Is the above map a quasi-isomorphism?

After all, if we analyze $Z^i(X,\bullet)$ in degree zero and one, we get the usual presentation for $\text{CH}^i(X)$ where rational equivalence is defined comparing cycle classes in $X\times_k\mathbf{P}_k^1$ and not in $X\times_k\mathbf{A}^1_k$.

Related question: What do higher Chow groups mean?


1 Answer 1


Good point. Can provide details as needed, but first let me suggest the answer in the form of an exercise.

Take $X$ to be a smooth projective curve over a number field $k$. Use de Franchis' Lemma to show your proposed complex has finite cohomology group in degree $3$ and weight $1$, while Bloch's complex typically won't.

It seems there are reasons from $K$-theory for why using $\Gamma_k^{\bullet}$ vs $\Delta_k^{\bullet}$ doesn't always work.

Another intuition should probably come from the Dold-Thom thm in topology, whose analogue should be seen as the quasi-isomorphism between the Bloch complex and the Suslin complex, this latter morally being "algebraic singular cohomology" of $\text{Sym}_k(X) = \varinjlim_{n\ge 0}\text{Sym}_k^n(X)$. Using $\Gamma_k^{\bullet}$, $k$-morphisms $\Gamma_k^i\to\text{Sym}^n_k(X)$ are "too few".


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