This question is based on this lecture notes www.claymath.org/library/monographs/cmim02.pdf

I assume Beilinson-Soulé vanishing.

We can express motivic cohomology of a field via its algebraic Suslin homology $H^{i,p}(F,A)=H_{p-i}^{alg}(\mathbb G_m^{\wedge p},A)$. (Here $H_i(X,A)$ is $i$-th algebraic Suslin homology with value in some abelian group $A$). So Beilinson-Soulé vanishing $H^{0,p}(F,A)=0$ yields:

$$H_p^{alg}(\mathbb G_m^{\wedge p})=0$$

But why this does not give a contradiction? Intuitively $\mathbb G_m^{\wedge p}$ is p dimensional "sphere". It should have $p$-th homology! Formally we know that if $F=\mathbb C$, then $H_2^{alg}(\mathbb G_m^2,\mathbb Z/l\mathbb Z)=\mathbb Z/l\mathbb Z$ and $H_2^{alg}(\mathbb G_m,\mathbb Z/l\mathbb Z)=0$. (See https://citeseerx.ist.psu.edu/document?repid=rep1&type=pdf&doi=db86615a1a416252a5beaa0cfc9570da93fc1091)

But this of course implies that $H_2^{alg}(\mathbb G_m^{\wedge 2}, \mathbb Z/l\mathbb Z)=\mathbb Z/l\mathbb Z$ since

$$h(\mathbb G_m^{2})=h(\mathbb G_m^{\wedge 2})+h(\mathbb G_m^{\wedge 1})^{\bigoplus 2}+h(pt).$$

Any help is welcome.

  • $\begingroup$ The Beilinson-Soule conjecture says that $H^{i,p}(F,A)=0$ for $i<0$. $\endgroup$ Apr 8 at 12:53
  • 1
    $\begingroup$ Yes, but this is also called Belinson-Soule conjecture (If $p\ne 0$ and $i\leq 0$ then $H^{i,p}(F,A)=0$) For example, rationally, this statement is contained in Bloch, Kriz "Mixed Tate Motives". $\endgroup$ Apr 8 at 13:52


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