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Recall that the higher Chow groups $CH^k(X,m)$ are defined as the homology of the complex $Z^k(X,\bullet)$, where $Z^k(X,m)$ is the subgroup of codimension $k$ cycles of $X\times \Delta^m$ which meet all faces $X\times \Delta^n$ (with $n<m$) in codimension $k$, and the boundary maps are the obvious ones. Here, $\Delta^n = Spec (\mathbb{C}[t_0,\ldots,t_n]/(1-\sum_jt_j)$ is the algebraic $n$-simplex.

In the expository paper "Indecomposable Higher Chow Cycles", by Gordon and Lewis, they mention in 1.3.4 that elements in the higher Chow groups $CH^k(X,1)$ can be represented by "higher Chow cycles": $\mathcal{Z} = \sum_i Z_i \otimes f_i$, where $Z_i$ are irreducible subvarieties of codimension $k-1$ and $f_i$ is a non-zero rational function on $Z_i$, such that $\sum_i div(f_i)=0$.

The proof goes by $K$-theory, which I don't know anything about. Is there a more direct (and geometric) way of describing $CH^k(X,1)$ in terms of higher Chow cycles? What happens for $CH^k(X,m)$, for $m>1$?

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There is a rather concrete construction in Landsburg's 1991 paper "Relative Chow groups" which gives an explicit isomorphism from $CH^k(X, 1)$ to the degree 1 homology of the Gersten complex, which is precisely the quotient of the group of higher Chow cycles by some equivalence relation.

(It would be quite difficult to remove K-theory from the picture entirely, since the equivalence relation by which one has to quotient the higher Chow cycles in order to get $CH^k(X, 1)$ is given by $K_2$ of function fields of codimension $k-2$ subvarieties, via the tame symbol map.)

Landsburg's paper also constructs some analogous maps for $CH^k(X, m)$ for $m > 1$, but he doesn't claim that they're isomorphisms.

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  • $\begingroup$ Although probably one can always use Milnor K-theory, which is significantly more elementary that Quillen's higher algebraic K-theory (not that algebraic K-theory is not worth learning ;)). $\endgroup$ Commented Jul 19, 2022 at 6:56
  • $\begingroup$ Indeed. This is part of the reason that Landsburg's map isn't an isomorphism for $m > 1$: because it factors through the natural map from Milnor to Quillen K-theory (applied to residue fields) $\endgroup$ Commented Jul 19, 2022 at 7:03

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