Lines crossing a tetrahedron and the Bloch group

Let k be an arbitrary infinite field. Consider the projective spaces P n (k) with fixed sets of homogenous coordinates (t 0 : t 1 : • • • : t n ) ∈ P n (k). We call a subspace L ⊂ P n (k) of codimension r admissible if codim L ∩ {t i1 = • • • = t is = 0} = r + s for every s and distinct i 1 , . . . , i s . (Here codim(X) > n means X = ∅.) Let C r n = Z admissible L ⊂ P n (k) , codim(L) = r be the free abelian group generated by all admissible subspaces of P n (k) of codimension r. Then for every r we have a complex . . . (we assume that C r n = 0 when n < r) with the differential

where every {t i = 0} ⊂ P n (k) is naturally identified with P n-1 (k) by throwing away the coordinate t i . We are interested in the homology groups of these complexes

For example, one can easily see that H 1 1 ∼ = k * . Indeed, a hyperplane { α i t i = 0} is admissible whenever all the coefficients α i are nonzero, and if we identify

then the differential d :

(one can recognize Menelaus’ theorem from plane geometry behind this simple computation). Hence we have

Continuing the identifications of (2), C 1 • turns into the bar complex for the group k * (with the term of degree 0 thrown away) and therefore

Now we switch to r = 2 and try to compute H 2 3 . The four hyperplanes {t i = 0} form a tetrahedron ∆ in the 3-dimensional projective space P 3 (k) and the line ℓ is admissible if it 1) intersects every face of ∆ transversely, i.e. at one point P i = ℓ ∩ {t i = 0};

2. doesn’t intersect edges {t i1 = t i2 = 0} of ∆, i.e. all four points P 0 , . . . , P 3 ∈ ℓ are different .

Therefore it is natural to associate with ℓ a number, the cross-ratio of the four points P 0 , . . . , P 3 on ℓ. Namely, there is a unique way to identify ℓ with P 1 (k) so that P 0 , P 1 and P 2 become 0, ∞ and 1 respectively, and we denote the image of P 3 by λ(ℓ) ∈ P 1 (k) {0, ∞, 1} = k * {1}. We extend λ linearly to a map

(ii) Let L ⊂ P 4 (k) be an admissible plane and

(iii) The map induced by λ on homology

is surjective, where

is the Bloch group of k ( [5]).

(iv) We have

) and the kernel of (3)

fits into the exact sequence

where Tor(k * , k * ) ∼ is the unique nontrivial extension of Tor(k * , k * ) by Z/2, and T (k) is a 2-torsion abelian group (conjectured to be trivial).

We remark that Tor(k * , k * ) = Tor(µ(k), µ(k)) is a finite abelian group if k is a finitely-generated field. Furthermore, it is proved in [5] that B(k) has the following relation to K 3 (k): let K ind 3 (k) be the cokernel of the map from Milnor’s K-theory

In particular, if k is a number field then as a consequence of (5) and Borel’s theorem ( [1]) we have dim

where r 2 is the number of pairs of complex conjugate embeddings of k into C.

Proof of (i) and (ii). One can check that the diagram

is commutative, and therefore (i) follows. It is another tedious computation to check (ii).

In the next section we will prove the remaining claims (iii) and (iv) and also show that

H q (G 1 , Z) by Shapiro’s lemma. We have

) and E 2 0q = H q (k * , Z). This spectral sequence degenerates on the second term. Indeed, the embedding k * ֒→ G α → 1 0 0 α is split by determinant, and therefore all maps H q (k * , Z) -→ H q (G, Z) are injective. Consequently, E ∞ pq = E 2 pq and for every n ≥ 2 we have a short exact sequence

Let D n be the free abelian group generated by (n + 1)-tuples of distinct points in P 1 (k). Again we have the differential like (7) on D • and the augmented complex D • -→ Z -→ 0 is acyclic. We have a surjective map from C • to D • since a non-zero vector in k 2 defines a point in P 1 (k) and the group action agrees. The spectral sequence E 1 pq = H q (G, D p ) ⇒ H p+q (G, Z) was considered in [5]. In particular, E 1 p0 = (D p ) G is the free abelian group generated by (p-2)tuples of different points since G-orbit of every (p + 1)-tuple contains a unique element of the form (0, ∞, 1, x 1 , . . . , x p-2 ), and the differential

According to [5], terms E 2 pq with small indices are

where p(k) is the quotient of Z[k * {1}] by all 5-term relations as in right-hand side of (8), and the only non-trivial differential starting from p(k) is

Therefore E 4 30 = E ∞ 30 = B(k) and we have a commutative triangle

where both maps from H 3 (G) are surjective, hence the vertical arrow is also surjective. It remains to check that the vertical arrow coincides with λ * . A line ℓ in P 3 (k) is given by two linear equations and for an admissible line it is always possible to chose them in the form

This line corresponds to the tuple of vectors

which can be mapped to the points 0, ∞, 1, x1y2 y1x2 in P 1 (k), hence the vertical arrow maps it to [ x1y2 y1x2 ] (actually we need to consider a linear combination of lines which vanishes under d but for every line the result is given by this expression). On the other hand, four points of its intersection with the hyperplanes are P 0 = (0 : y 1 x 2y 2 x 1 : -x 2 : x 1 ) P 1 = (y 2 x 1y 1 x 2 : 0 : -y 2 : y 1 ) P 2 = (-x 2 : -y 2 : 0 : 1)

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