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Let $F$ be a number field and $r_{1}$ and $r_{2}$ the numbers of real and pairs of complex embeddings respectively of $F$. Then Borel computed that for $n\geq 2$

$$ K_{n}(F)_{\mathbb{Q}}\simeq \begin{cases} \mathbb{Q}^{r_{2}} & n=3 (\mod 4), \\ \mathbb{Q}^{r_{1}+r_{2}} & n=1 (\mod 4),\\ 0 & \text{else.} \end{cases} $$

Of course, $K_{0}(F)_{\mathbb{Q}}=\mathbb{Q}$ and $K_{1}(F)_{\mathbb{Q}}=F^{\times}\otimes\mathbb{Q}.$ I am now curious what the Adams graded parts are. More precisely, I have read that

$$ K_{2q-p}(F)_{\mathbb{Q}}^{(q)}\simeq \begin{cases} 0 & q<0,\\ 0 & q=0, p\neq 0,\\ \mathbb{Q} & q=p=0,\\ 0 & q>0, p\leq 0, \\ 0 & q>0,\text{even}, p=1,\\ F^{\times}\otimes \mathbb{Q} & q=p=1,\\ \mathbb{Q}^{r_{1}+r_{2}} & q>1, q=1 (\mod 4), p=1,\\ \mathbb{Q}^{r_{2}} & q>0, q=3 (\mod 4), p=1,\\ 0 & q>0, p>1. \end{cases} $$

Only the first three cases are really clear to me. So if someone could explain to me why we have the above in the other cases, that would be appreciated very much. Thank you.

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Fixed your LaTeX –  Yemon Choi Jan 21 '11 at 2:27
    
Thanks alot. Appreciate it. –  Devin Jan 21 '11 at 7:08
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1 Answer

One way of viewing Borel's theorem is that if $F$ is a number field the $p$-th Chern class with values in Deligne cohomology is a map from $K_{2p-1}(F)\to H^{1}_{\mathrm{Deligne}}(\mathrm{Spec}(F),\mathbb{R}(p))$ which embeds the $K$-group (mod torsion) as a lattice in a real vector space of the appropriate rank. (The point being that these Chern classes "are" the Borel regulator -- there is a nice book by Burgos explaining this; note that one must also take account of the action of complex conjugation.) But the p-th Chern class has the property that it converts the n-th Adams operation into multiplication by $n^p$. So the fact that a single Chern class detects $K_m(F)$ mod torsion implies that $K_m$ has pure weight.

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