# Rationalised K-theory of number fields

Let $A$ be the ring of integers in a number field, and consider the rationalised algebraic $K$-theory groups $\mathbb{Q}\otimes K_*(A)$. A theorem of Borel calculates the ranks of these groups; the answer can be described as follows. The tensor product $\mathbb{R}\otimes A$ is isomorphic as a topological $\mathbb{R}$-algebra to $\mathbb{R}^s\times\mathbb{C}^t$ for some $s$ and $t$. It follows that $K_{\ast}(\mathbb{R}\otimes A)$ is isomorphic to $kO_{\ast}^s\oplus kU_{\ast}^t$ (if we use the version of $K$-theory that remembers the topology on $\mathbb{R}\otimes A$). Here $\mathbb{Q}\otimes kO_{\ast}=\mathbb{Q}[x]$ with $|x|=4$, and $\mathbb{Q}\otimes kU_{\ast}=\mathbb{Q}[y]$ with $|y|=2$. Borel's result says that $\mathbb{Q}\otimes K_{\ast}(A)$ is isomorphic, as a graded vector space, to $\Sigma\mathbb{Q}\otimes K_*(\mathbb{R}\otimes A)$ (where $\Sigma$ is the syspension, shifting degrees by one) except that we have to move one of the generators from degree one back down to degree zero. Of course this implies the corresponding relationship between $\mathbb{R}\otimes K_{\ast}(A)$ and $\mathbb{R}\otimes K_{\ast}(\mathbb{R}\otimes A)$.

My question: is there a natural isomorphism between these groups, either in the real case or the rational case? My feeling is that it should be possible to extract this from the work of Borel, if I understood it better. What is the conceptual explanation for the degree shift?

The answer is yes, but only after tensoring with $\mathbb R$.
The topological Chern character map $ch_n : kU_{i} \to H^{2n-i}(pt,\mathbb Q)$ is an isomorphism $kU_{2n} \otimes \mathbb Q \ \xrightarrow \cong \ \mathbb Q$ when $i=2n$.
The corresponding algebraic Chern character map with values on Deligne cohomology is $ch_n : K_{i}\mathbb C \to H^{2n-i}_{\cal D}(pt,\mathbb Q(n))$. Here $\mathbb Q(n)$ (or $\mathbb Z(n)$) denotes a certain cochain complex of sheaves for the analytic topology on a complex manifold $X$. It starts in degree 0 with $\mathbb Q$ (resp., $\mathbb Z$), in cohomological degree 1 it has $\mathbb C$, and the differential map $d^0 : \mathbb Q \to \mathbb C$ is multiplication by $(2 \pi i)^ n$. The term in degree $i+1$ is the sheaf of holomorphic differentials $\Omega^i$ if $i < n$ and is $0$ if $i \ge n$. The exponential map $\mathbb C \to \mathbb C ^ \times$ given by $z \mapsto e^z$ gives a quasi-isomorphism $\mathbb Z (1) \to \mathbb C ^ \times [-1]$; the degree shift there answers your second question, partially; another way of saying that is that there is a degree shift in the boundary map $c_1 : H^1(X,\mathbb C^\times) \to H^2(X,\mathbb Z)$. I say "partially", because one must know also that the regulator map involves no further degree shift; in degree 1 it's because the map $\mathbb C ^ \times \to \mathbb R$ given by $z \mapsto {\rm log} |z|$ involves no degree shift.
Now consider the projection $\mathbb C \to \mathbb R$ that sends $(2 \pi i)^n$ to $0$ and $i^{n-1}$ to $1$; perhaps there is a better normalization for this map, such as choosing to send $(2 \pi i)^{n-1}$ to $1$. It induces a map of cochain complexes $\mathbb Q(n) \to \mathbb R[-1]$; the map it induces on Deligne cohomology, composed with the Chern character map above, is the Beilinson regulator map $$ch_n : K_{i}\mathbb C \to H^{2n-i}(pt,\mathbb R[-1]) = H^{2n-i-1}(pt,\mathbb R),$$whose only nonzero possibility is the map $ch_{n} : K_{2n-1}\mathbb C \to H^{0}(pt,\mathbb R) = \mathbb R$. For the ring of integers $A$, we get a map $K_{2n-1} A \to K_{2n-1}( A \otimes \mathbb C ) \to H^{0}(Spec(A \otimes \mathbb C),\mathbb R) = \mathbb R^{s+2t}$. Borel's theorem is recast as saying that for $n > 1$ the map induces an isomorphism of $(K_{2n-1} A) \otimes \mathbb R$ with the appropriate eigenspace for the action of $G = Gal(\mathbb C/\mathbb R)$ on $$H^0(Spec(A \otimes \mathbb C), \mathbb R (n-1)) = \mathbb R^{s+2t},$$where now I use $\mathbb R (n-1)$ to remind us how $G$ acts on this real vector space of dimension 1.
(Added later: actually, it may be more natural to replace the $\Sigma$ in the question by $\Omega$ and to use the anti-invariants (or invariants, depending on the parity of n) under $G$ acting on $K^{top}(A \otimes \mathbb C) \otimes \mathbb C$ instead of the invariants. Thus the degree shift can be viewed as $-1 = 1 - 2$)