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Let $R=\mathbb{Q}[e^{2\pi i /3}]$. Does $H_3(GL(R))$ have $\mathbb{Z}$-rank $1$?

If so, what is the index of the map: $$ \mathbb{Z}\cong K_3(R)/{\rm Torsion} \to H_3(GL(R))/{\rm Torsion}\cong \mathbb{Z}? $$

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No, it doesn't have $Z$-rank 1. If you replaced $R$ with $Z[e2πi/3]$, it would. Or if you replaced $GL$ by $SL$. –  Ben Wieland Aug 9 '11 at 1:12
    
Let me add a proof to Ben's comment: The the multiplicative group $R^\times$ of a number field is isomorphic to the product of the unit group of its ring of integers and a free abelian group $A$ of countably infinite rank. Since $GL(R) \cong SL(R) \times R^\times$ it follows from Kunneth formula that $H^3(GL(R);\mathbb{Z})$ contains $H^3(A;\mathbb{Z})$ as a direct summand. But $H^3(A;\mathbb{Z})$ has a subgroup isomorphic to $\Lambda^3(A)$ (see [Brown, Cohomology of Groups, Theorem 6.4]), which has clearly infinite rank. –  Ralph Aug 9 '11 at 9:41

1 Answer 1

up vote 6 down vote accepted

Let $F = \mathbb{Q}(\zeta_3)$ and $R = \mathbb{Z}[\zeta_3]$ with a third root of unity $\zeta_3$. Then the Hurewicz homomorphism $$h_3: K_3(R) \to H_3(GL(R); \mathbb{Z})$$ induces an isomorphism on the torsion-quotients und the index of the map is $1$. This can be seen as follows:

1) rank $K_3(R) = r_2 = 1$, the number of pairs of complex embeddings of $F$ (see [W, Theorem 1.5])

2) Since $R$ is euclidian, $SL(R) = E(R)$, the subgroup generated by elementary matrices.

3) According to [A, after Cor. 5.1] the Hurewicz homomorphism $$h_n: K_n(R) = \pi_nBGL(R)^+ \cong \pi_nBE(R)^+ \to H_n(E(R);\mathbb{Z}) = H_n(SL_n(R); \mathbb{Z})$$ is surjective for $n=3$ with torsion kernel.

4) It follows that the ranks of $K_3(R)$ and $H_3(SL(R); \mathbb{Z})$ are equal and by 1) they are equal $1$.

5) $GL(R) \cong SL(R) \times R^\times = SL(R) \times \mathbb{Z} /6 \mathbb{Z}$.

6) From Kunneth formula it follows that $SL(R) \hookrightarrow GL(R)$ induces a monomorphisms $$H_3(SL(R),\mathbb{Z}) \hookrightarrow H_3(GL(R); \mathbb{Z})$$ with torsion cokernel.

7) It follows from 6) and 3) that the Hurewicz homomorphism $$\bar{h}_3: \pi_3(R)/(\text{torsion}) \to H_3GL(R); \mathbb{Z})/(\text{torsion})$$ is an isomorphism and the index that was asked for equals $1$.


Remarks:

a) An explicit description of $K_3(F)$ is given by [W, Theorem 1.7]

b) Note that $K_n(F) = K_n(\mathcal{O})$ for $n$ odd (see [W, Theorem 1.6])


References:

[A] Arlettaz, The exponent of the homotopy groups of Moore spectra and the stable Hurewicz homomorphism

[W] Weibel, Albebraic K-Theory of Rings of Integers in Local and Global Fields


Edit: To complete the story: The Hurewicz map $$h_3: K_3(F) \to H_3(SL(F);\mathbb{Z})$$ also induces an isomorphism on the torsion-quoients and is of index $1$ there.

For, from 1) and b) we know that the torsion free part of $K_3(F)$ has rank $1$ and 3) is also true with $F$ in place of $R$. Since $SL(F) = E(F)$ (this holds for every local ring) the assertin follows from 3).


That $H_3(GL(F);\mathbb{Z})$ is no good choice for approximating $K_3(F)$ was already discussed in the comments above.

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Thanks very much. Actually, just the fact that the Hurewicz homomorphism, $h_3: K_3(R) \to H_3(E(R);\mathbb{Z}$ is surjective is enough for me. That really helps. –  tkf Aug 9 '11 at 17:38

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