Amalgamated Free Products, Unstable Homotopy Invariance, and the Homology of SL_{2}(Z[t])

Kevin P. Knudson^{⋆}

Department of Mathematics, Northwestern University, Evanston, IL 60208

e-mail: knudson@math.nwu.edu

Received: / Revised version:

Abstract We prove that if R is a domain with many units, then the natural inclusion E_{2}(R) → E_{2}(R[t]) induces an isomorphism in integral homology. This is a consequence of the existence of an amalgamated free product decomposition of E_{2}(R[t]). We also use this decomposition to study the homology of E_{2}(Z[t]) and show that a great deal of the homology of E_{2}(Z[t]) maps nontrivially into the homology of SL_{2}(Z[t]). As a consequence, we show that the latter is not finitely generated in all positive degrees.

Key words amalgamated free product – group homology – homotopy invariant presheaf – elementary matrices

Introduction

The fundamental theorem of algebraic K-theory asserts that if A is a regular ring, then there is a natural isomorphism K_{i}(A[t]) ≅ K_{i}(A) for all i ≥ 0. More generally, a presheaf F on the category of schemes over a base S is homotopy invariant if F(X × A^{1}) ≅ F(X) for all schemes X. The utility of homotopy invariant presheaves is well-documented (see e.g., [10–12]).

The existence of homotopy invariance in K-theory suggests that one search for unstable analogues. In a previous work [4], the author showed that such a statement is true for infinite fields k: the natural map G(k) → G(k[t]) induces an isomorphism in integral homology for G = SL_{n}, GL_{n}, P GL_{n}. Since homotopy invariance in K-theory holds for all regular rings, one might hope that the above isomorphism holds for rings other than infinite fields.

⋆ Supported by an NSF Postdoctoral Fellowship, grant no. DMS-9627503

For a ring A, denote by E_{n}(A) the subgroup of GL_{n}(A) generated by the elementary matrices. In many cases, the group E_{n}(A) coincides with SL_{n}(A) (e.g., A local or Euclidean), but this need not always be so. In this paper, we study the homology of the group E_{2}(R[t]). Our first result is the following.

Theorem A If R is an integral domain with many units, then the inclusion E_{2}(R) → E_{2}(R[t]) induces an isomorphism

H_{•}(E_{2}(R), Z) ⟶^{≅} H_{•}(E_{2}(R[t]), Z).

The definition of a ring with many units will be recalled below. The principal example of such a ring is a local ring with infinite residue field (or an algebra over such a ring).

Let k be an infinite field and denote by Sch/k the category of smooth affine schemes over k. In this setting, Theorem A admits the following geometric interpretation.

Corollary The presheaves H_{i} = H_{i}(E_{2}(−), Z) are homotopy invariant on Sch/k.

One would hope that a similar statement holds for H_{i}(E_{n}(−), Z) for n > 2, but so far we have been unable to prove it. What is clear, as the following example of Weibel [13] shows, is that if Theorem A is to admit a generalization to the case n > 2, then the hypotheses on the ring R will have to be strengthened.

Example Consider the local ring of the node at the singularity— R = (k[x, y]/(y^{2} −x^{3} −x^{2}))_{(x,y)}. This is a one-dimensional noetherian local domain, and, if k is infinite, is a ring with many units. However, if H_{•}(E_{n}(R), Z) ≅ H_{•}(E_{n}(R[t]), Z) for all n, then in particular H_{2}(E(R), Z) ≅ H_{2}(E(R[t]), Z); i.e., K_{2}(R) ≅ K_{2}(R[t]). But in [13] it is shown that K_{2}(R) ≠ K_{2}(R[t]).

Thus, it seems that one must assume that R is at least regular, and it is not at all clear how to use this property.

There is another reason to study the homology presheaves H_{i} = H_{i}(E_{n}(−), Z). If one could show that the H_{i} are homotopy invariant and admit transfer maps (i.e., for a finite flat morphism R → S there is a map H_{i}(S) → H_{i}(R) satisfying the usual properties), then one could prove the Friedlander–Milnor conjecture [2] for SL_{n}: If G is a reductive group scheme over an algebraically closed field k and if p ≠ char k, then

H_{et}^{•}(BG_{k}, Z/p) ≅ H^{•}(BG(k), Z/p).

Here, BG_{k} is the simplicial classifying scheme of G_{k} and BG(k) is the classifying space (= simplicial set) of the discrete group G(k) of k-rational points. Unfortunately, the H_{i} do not appear to admit transfers. However, it would still be interesting to know if the H_{i} are homotopy invariant.

Theorem A is a consequence of the existence of an amalgamated free product decomposition

E_{2}(R[t]) ≅ E_{2}(R) ∗_{B(R)} B(R[t])

where, for a ring A, B(A) denotes the upper triangular subgroup. This follows from Nagao’s decomposition of SL_{2}(k[t]) (k a field) and some straight-forward group-theoretic considerations. In particular, for R = Z we have E_{2}(Z[t]) ≅ SL_{2}(Z) ∗_{B(Z)} B(Z[t]) (note that E_{2}(Z) = SL_{2}(Z)). This allows us to make partial computations of the homology of SL_{2}(Z[t]). Grunewald, Mennicke, and Vaserstein [3] have shown that SL_{2}(Z[t]) has free quotients F of countable rank and that the projection SL_{2}(Z[t]) → F may be taken so that all unipotent elements (and hence all elements of E_{2}(Z[t])) lie in the kernel. This was generalized and improved by Krstić and McCool [5], who showed that the same is true for any domain R which is not a field. This implies that H_{•}(SL_{2}(Z[t]), Z) contains the homology of a countably generated free group as a direct summand. Note, however, that this only shows that H_{1}(SL_{2}(Z[t]), Z) has infinite rank.

The amalgamated free product decomposition for E_{2}(Z[t]) allows us to compute its homology. We show that H_{i}(E_{2}(Z[t]), Z) contains a countably generated free summand for i ≥ 1. It is not clear, a priori, that any homology classes map nontrivially into H_{•}(SL_{2}(Z[t]), Z). We show that this is indeed the case.

Theorem B For each i ≥ 0, the map

H_{i}(E_{2}(Z[t]), Z) ⟶ H_{i}(SL_{2}(Z[t]), Z)

is nontrivial.

Corollary For each i ≥ 1, the group H_{i}(SL_{2}(Z[t]), Z) is not finitely generated.

In the case i = 1 we see that H_{1}(E_{2}(Z[t])) maps nontrivially into H_{1}(SL_{2}(Z[t])). Since all elementary matrices map to 1 in the free quotient F, the image of H_{1}(E_{2}(Z[t])) intersects the copy of H_{1}(F) ⊂ H_{1}(SL_{2}(Z[t])) trivially. Hence we have found a nontrivial, nonfinitely generated summand orthogonal to H_{1}(F).

Theorem B is proved by considering the (surjective!) homomorphisms φ_{p} : E_{2}(Z[t]) → SL_{2}(F_{p}[t]) obtained by reducing polynomials modulo p. We show that classes in H_{i}(E_{2}(Z[t]), Z) map nontrivially into H_{i}(SL_{2}(F_{p}[t]), Z) under φ_{p∗}. The result follows since the map φ_{p} factors through SL_{2}(Z[t]).

The paper is organized as follows. In Section 1 we deduce the amalgamated free product decomposition for E_{2}(R[t]). In Section 2 we prove Theorem A. In Section 3 we discuss the homology of E_{2}(Z[t]). In Section 4 we study the homology of SL_{2}(F_{p}[t]). Section 5 contains the proof of Theorem B. Finally, Section 6 deals with other homology classes in H_{•}(SL_{2}(Z[t]), Z).

Notation If A is a ring (always assumed commutative with unit), then A^{×} denotes the multiplicative group of units. If G is an abelian group, then nG denotes the subgroup of G annihilated by n. If no coefficient module is specified, homology groups are to be interpreted as integral homology.

Acknowledgements I would like to thank Dick Hain and Andrei Suslin for many valuable conversations. I am indebted to Chuck Weibel for the example above. Finally, I thank Mark Walker for listening to me talk about homotopy invariance.

1 Amalgamated Free Products

Recall the following theorem of Nagao [6]. Let k be a field. Then we have an amalgamated free product decomposition

SL_{2}(k[t]) ≅ SL_{2}(k) ∗_{B(k)} B(k[t])

where for a ring A, B(A) denotes the subgroup of upper triangular matrices. We will make use of the following result of Serre ([8], Prop. 3, p. 6).

Proposition 1.1 Suppose that G is an amalgamated free product, G = G_{1} ∗_{A} G_{2}. Let H_{i} ⊆ G_{i} be subgroups such that H_{1} ∩ A = B = H_{2} ∩ A. Denote by H the subgroup of G generated by the H_{i}. Then the evident homomorphism

H_{1} ∗_{B} H_{2} ⟶ H

is an isomorphism. ⊓ ⊔

Let R be an integral domain with field of fractions Q and consider the subgroups E_{2}(R) ⊆ E_{2}(Q) (= SL_{2}(Q)) and B(R[t]) ⊆ B(Q[t]). Observe that E_{2}(R) ∩ B(Q) = B(R) = B(R[t]) ∩ B(Q).

Corollary 1.2 There is a decomposition

E_{2}(R[t]) ≅ E_{2}(R) ∗_{B(R)} B(R[t]).

Proof It is clear that the group H generated by E_{2}(R) and B(R[t]) is contained in E_{2}(R[t]). The reverse inclusion follows since E_{2}(R)

contains the permutation matrix ( \begin{matrix}0 & −1 \\ 1 & 0 \end{matrix} ) and hence H contains all

upper and lower triangular matrices. ⊓ ⊔

2 Rings with Many Units

In this section, we shall prove Theorem A.

Definition 2.1 A ring A is an S(n)-ring if there exist a_{1}, …, a_{n} ∈ A^{×} such that the sum of any nonempty subfamily of the {a_{i}} is a unit. If A is an S(n)-ring for every n, then we say that A has many units.

Examples of rings with many units include local rings with infinite residue fields and algebras over such rings. However, a ring having infinitely many units need not be a ring with many units. An example is the local ring Z_{(p)}. In fact, Z_{(p)} is an S(p−1)-ring, but not an S(p)- ring.

Rings with many units satisfy the following property (see [7], Thm. 1.10).

Proposition 2.2 If R has many units, then the inclusion B(R) → B(R[t]) induces an isomorphism in integral homology. ⊓ ⊔

Theorem 2.3 Suppose R is an integral domain with many units. Then the inclusion E_{2}(R) → E_{2}(R[t]) induces an isomorphism

H_{•}(E_{2}(R), Z) ⟶^{≅} H_{•}(E_{2}(R[t]), Z).

Proof The amalgamated free product decomposition yields a Mayer– Vietoris sequence for computing H_{•}(E_{2}(R[t]), Z). Since the evident map H_{•}(B(R), Z) → H_{•}(B(R[t]), Z) is split injective for any domain R, the long exact sequence breaks up into short exact sequences

0 → H_{i}(B(R)) → H_{i}(B(R[t])) ⊕ H_{i}(E_{2}(R)) → H_{i}(E_{2}(R[t])) → 0 (this is valid for any domain R). By Proposition 2.2, H_{i}(B(R)) ≅ H_{i}(B(R[t])) if R has many units so that the map

H_{i}(E_{2}(R), Z) ⟶ H_{i}(E_{2}(R[t]), Z)

is an isomorphism. ⊓ ⊔

Remark 2.4 Evidently, this approach will not work for computing H_{•}(E_{n}(R[t])) for n ≥ 3 since no such amalgamated free product decomposition exists.

3 The Homology of E_{2}(Z[t])

We now shift gears and study the homology of E_{2}(Z[t]). Theorem 2.3 does not apply in this case since Z does not have many units. It is possible to calculate H_{•}(E_{2}(Z[t]), Z) explicitly, but the final answer is a bit complicated. For this reason, we will compute only a part of the integral homology in detail. We also compute the homology of E_{2}(Z[t]) with F_{p}-coefficients.

For each i we have a short exact sequence

0 → H_{i}(B(Z)) → H_{i}(B(Z[t])) ⊕ H_{i}(SL_{2}(Z)) → H_{i}(E_{2}(Z[t])) → 0.

Observe that B(Z[t]) = B(Z) × tZ[t] so that for any principal ideal domain k we have the Künneth exact sequence

0 → ⨁_{l+m=i} H_{l}(B(Z), k) ⊗ H_{m}(tZ[t], k) → H_{i}(B(Z[t]), k) →

⨁_{l+m=i−1} Tor_{1}^{k}(H_{l}(B(Z), k), H_{m}(tZ[t], k)) → 0.

In particular, if k is a field we have

H_{•}(B(Z[t]), k) ≅ H_{•}(B(Z), k) ⊗ H_{•}(tZ[t], k).

Suppose k is the finite field F_{p}. Then we have the following result.

Proposition 3.1 If p ≥ 3, then for R = Z, Z[t],

H_{•}(B(R), F_{p}) = H_{•}(R, F_{p}) = ⋀_{F_{p}}^{•}(R ⊗ F_{p}).

If p = 2, then

H_{•}(B(Z[t]), F_{2}) = (H_{•}(Z/2, F_{2}) ⊗ H_{•}(Z, F_{2})) ⊗ ⋀_{F_{2}}^{•}tF_{2}[t].

Proof Recall that if A is an abelian group then there is an isomorphism

⋀_{F_{p}}^{•}(A ⊗ F_{p}) ⊗ Γ_{F_{p}}(pA) ⟶^{≅} H_{•}(A, F_{p})

where Γ_{F_{p}} is a divided power algebra [1], p. 126. The isomorphism is natural if p ≠ 2. The first result follows from the fact that B(R) = Z/2 × R (R = Z, Z[t]) and that H_{i}(Z/2, F_{p}) vanishes for i > 0 if p ≥ 3. Note that in this case pR = 0. The second assertion is just a restatement of the Künneth formula. ⊓ ⊔

Corollary 3.2 If p ≥ 3, the homology of E_{2}(Z[t]) is

H_{i}(E_{2}(Z[t]), F_{p}) = \{ \begin{matrix}F_{p} & i = 0 \\ tF_{p}[t] ⊕ (Z/12 ⊗ F_{p}) & i = 1 \\ ⋀_{F_{p}}^{i} F_{p}[t] ⊕ H_{i}(SL_{2}(Z), F_{p}) & i ≥ 2. \end{matrix}

In particular, if p ≥ 5, then for i ≥ 2

H_{i}(E_{2}(Z[t]), F_{p}) = ⋀_{F_{p}}^{i}F_{p}[t].

Proof Consider the short exact sequence

0 → H_{i}(B(Z), F_{p}) → H_{i}(B(Z[t]), F_{p}) ⊕ H_{i}(SL_{2}(Z), F_{p}) → H_{i}(E_{2}(Z[t]), F_{p}) → 0.

The result for i ≥ 2 follows easily since

H_{i}(B(Z), F_{p}) = H_{i}(Z, F_{p}) = 0.

For i = 1, we have H_{1}(B(Z), F_{p}) = F_{p} and H_{1}(B(Z[t]), F_{p}) = F_{p}[t]. The map F_{p} → F_{p}[t] is the usual inclusion and the calculation of H_{1}(E_{2}(Z[t]), F_{p}) follows easily. The final assertion is a consequence of the fact that the abelianization map SL_{2}(Z) → Z/12 induces an isomorphism on integral homology. ⊓ ⊔

The calculation for p = 2 is more complicated, and instead of writing down the complete answer we make the following observation.

Proposition 3.3 For each i, H_{i}(E_{2}(Z[t]), F_{2}) contains ⋀_{F_{2}}^{i} tF_{2}[t] as a direct summand.

Proof Since B(Z[t]) = B(Z) × tZ[t], it follows that the quotient H_{i}(B(Z[t]), F_{2})/H_{i}(B(Z), F_{2}) consists of

⨁_{p+q=i,p<i} H_{p}(B(Z), F_{2}) ⊗ H_{q}(tZ[t], F_{2}),

and this contains H_{0}(B(Z), F_{2}) ⊗ H_{i}(tZ[t], F_{2}) = ⋀_{F_{2}}^{i} tF_{2}[t]. ⊓ ⊔

The integral homology is similarly complicated and we leave it to the interested reader to write it down. What is important for our purposes is the following.

Proposition 3.4 For each i, H_{i}(E_{2}(Z[t]), Z) contains ⋀_{Z}^{i} tZ[t] as a direct summand. Moreover, the element t^{l_{1}} ∧ ⋯ ∧ t^{l_{i}} maps to the element t^{l_{1}} ∧ ⋯ ∧ t^{l_{i}} in ⋀_{F_{p}}^{i} tF_{p}[t] ⊆ H_{i}(E_{2}(Z[t]), F_{p}) under the map induced by reducing coefficients modulo p.

Proof Arguing as in the proof of Proposition 3.3 we see that the group H_{i}(B(Z[t]), Z)/H_{i}(B(Z), Z) contains a copy of the tensor product H_{0}(B(Z), Z)⊗H_{i}(tZ[t], Z). Since tZ[t] is a torsion-free abelian group, its integral homology is simply an exterior algebra; i.e., H_{i}(tZ[t], Z) = ⋀_{Z}^{i} tZ[t]. The second assertion follows by considering the homomorphism of Mayer–Vietoris sequences induced by reducing coefficients modulo p. ⊓ ⊔

4 The Homology of SL_{2}(F_{p}[t])

In this section we compute the homology of the group SL_{2}(F_{p}[t]). As before, the amalgamated free product decomposition of SL_{2}(F_{p}[t]) yields a collection of short exact sequences

0 → H_{i}(B(F_{p})) → H_{i}(B(F_{p}[t])) ⊕ H_{i}(SL_{2}(F_{p})) → H_{i}(SL_{2}(F_{p}[t])) → 0.

Observe that if ℓ is a prime distinct from p, then the natural map

H_{i}(B(F_{p}), F_{ℓ}) ⟶ H_{i}(B(F_{p}[t]), F_{ℓ})

is an isomorphism for all i. This follows by considering the Hochschild-Serre spectral sequence associated to the extension

0 ⟶ R ⟶ B(R) ⟶ F_{p}^{×} ⟶ 1  (4.1)

for R = F_{p}, F_{p}[t]. Since H_{q}(R, F_{ℓ}) = 0 for q > 0, the claimed isomorphism follows. As a consequence, we see that for all i,

H_{i}(SL_{2}(F_{p}[t]), F_{ℓ}) = H_{i}(SL_{2}(F_{p}), F_{ℓ})

for ℓ ≠ p. (This was proved, in greater generality, by C. Soulé [9].)

If we consider integral or F_{p} coefficients, however, the situation is much different.

Lemma 4.1 For R = F_{p}, F_{p}[t], we have

H_{i}(B(R), F_{p}) = H_{0}(F_{p}^{×}, H_{i}(R, F_{p})).

Proof Consider the Hochschild–Serre spectral sequence associated to the extension (4.1); it has E^{2}-term

E_{r,s}^{2} = H_{r}(F_{p}^{×}, H_{s}(R, F_{p})).

Since |F_{p}^{×}| = p − 1 is invertible in F_{p}, we see that E_{r,s}^{2} = 0 for r > 0 (see [1], p. 84). ⊓ ⊔

The action of α ∈ F_{p}^{×} on R is α : x ↦ α^{2}x. Thus, if p = 2 or 3, we have the following.

Corollary 4.2 If p = 2 or 3, then for all i ≥ 0,

H_{i}(B(R), F_{p}) = H_{i}(R, F_{p}).

Proof In this case, (F_{2}^{×})^{2} = {1} = (F_{3}^{×})^{2} so that F_{p}^{×} acts trivially on H_{i}(R, F_{p}). ⊓ ⊔

Corollary 4.3 If p = 2 or 3, then for all i ≥ 0,

H_{i}(SL_{2}(F_{p}[t]), F_{p}) = H_{i}(SL_{2}(F_{p}), F_{p})
⊕H_{i}(B(F_{p}[t]), F_{p})/H_{i}(B(F_{p}), F_{p}).

In particular, H_{i}(SL_{2}(F_{p}[t]), F_{p}) contains a copy of ⋀_{F_{p}}^{i} tF_{p}[t] as a direct summand.

Proof If p = 2 or 3, then B(F_{p}[t]) = B(F_{p}) × tF_{p}[t]. Hence the group H_{i}(B(F_{p}[t]), F_{p})/H_{i}(B(F_{p}), F_{p}) consists of

⨁_{l+m=i,l<i} H_{l}(F_{p}, F_{p}) ⊗ H_{m}(tF_{p}[t], F_{p}).

This contains H_{i}(tF_{p}[t], F_{p}) as a summand and this, in turn, contains ⋀_{F_{p}}^{i} tF_{p}[t] as a summand. ⊓ ⊔

The calculation of the integral homology is obviously more complicated, even for p = 2, 3. However, we make the following observation.

Proposition 4.4 If p = 2 or 3, then for all i ≥ 0, H_{i}(SL_{2}(F_{p}[t]), Z) contains a copy of ⋀_{Z}^{i} tF_{p}[t]. Furthermore, the element t^{l_{1}} ∧⋯∧t^{l_{i}} ∈ ⋀_{Z}^{i} tF_{p}[t] maps to t^{l_{1}} ∧ ⋯ ∧ t^{l_{i}} ∈ ⋀_{F_{p}}^{i} tF_{p}[t] ⊆ H_{i}(SL_{2}(F_{p}[t]), F_{p}) under the map induced by reducing coefficients modulo p.

Proof For any abelian group A, the map ⋀_{Z}^{•} A → H_{•}(A, Z) is injective [1], p. 123. It follows that H_{i}(B(F_{p}[t]), Z) contains a copy of H_{0}(B(F_{p}), Z) ⊗ H_{i}(tF_{p}[t], Z), which, in turn, contains a copy of ⋀_{Z}^{i} tF_{p}[t]. Thus, the group H_{i}(SL_{2}(F_{p}[t]), Z) does also. The second assertion follows from the naturality of the map induced in homology by the coefficient homomorphism Z → F_{p}. ⊓ ⊔

The situation for primes greater than 3 is made more difficult by the fact that (F_{p}^{×})^{2} ≠ {1} for p ≥ 5. Thus, the calculation of H_{•}(B(F_{p})) and H_{•}(B(F_{p}[t])) is more involved. Since considering the primes 2 and 3 is sufficient for our purposes here, we leave the case p ≥ 5 to the interested reader.

5 The Homology of SL_{2}(Z[t])

In this section we prove Theorem B (which was stated in the introduction). We first demonstrate the following result.

Proposition 5.1 If p = 2 or 3, then the natural map

H_{i}(E_{2}(Z[t]), Z) → H_{i}(SL_{2}(F_{p}[t]), Z)

induced by φ_{p} : E_{2}(Z[t]) → SL_{2}(F_{p}[t]) maps the element t^{l_{1}}∧⋯∧t^{l_{i}} ∈ ⋀_{Z}^{i} tZ[t] to t^{l_{1}} ∧ ⋯ ∧ t^{l_{i}} ∈ ⋀_{Z}^{i} tF_{p}[t]. The same is true with F_{p}- coefficients.

Proof Consider the commutative diagram

H_{i}(B(Z)) → H_{i}(B(Z[t])) ⊕ H_{i}(SL_{2}(Z)) → H_{i}(E_{2}(Z[t]))
↓ ↓ ↓
H_{i}(B(F_{p})) → H_{i}(B(F_{p}[t])) ⊕ H_{i}(SL_{2}(F_{p})) → H_{i}(SL_{2}(F_{p}[t])).

Since t^{j} ∈ tZ[t] maps to t^{j} ∈ tF_{p}[t], the claim is clear. The second assertion follows similarly. ⊓ ⊔

Consider the inclusion j : E_{2}(Z[t]) → SL_{2}(Z[t]) and the induced map j_{∗} on homology. For each i-tuple \underline{ℓ} = (l_{1}, …, l_{i}), l_{1} < ⋯ < l_{i}, denote by x\underline{ℓ} the homology class j_{∗}(t^{l_{1}} ∧ ⋯ ∧ t^{l_{i}}) ∈ H_{i}(SL_{2}(Z[t]), Z).

Theorem 5.2 For each i-tuple \underline{ℓ}, the class x\underline{ℓ} is nontrivial. Moreover, if \underline{m} is another i-tuple distinct from \underline{ℓ}, then x\underline{ℓ} and x\underline{m} are distinct. Thus, the group H_{i}(SL_{2}(Z[t]), Z) is not finitely generated for all i ≥ 1.

Proof Let p = 2 or 3. Consider the commutative diagram

E_{2}(Z[t]) ⟶^{j} SL_{2}(Z[t])
φ_{p} ↘ ↓ π_{p}
SL_{2}(F_{p}[t])

and the induced commutative diagram

H_{i}(E_{2}(Z[t]), Z) ⟶^{j_{∗}} H_{i}(SL_{2}(Z[t]), Z)
φ_{p∗} ↘  ↓ π_{p∗}
H_{i}(SL_{2}(F_{p}[t]), Z)

in homology. Since φ_{p∗}(t^{l_{1}} ∧ ⋯ ∧ t^{l_{i}}) = t^{l_{1}} ∧ ⋯ ∧ t^{l_{i}}, we see that π_{p∗}j_{∗}(t^{l_{1}} ∧ ⋯ ∧ t^{l_{i}}) ≠ 0 and hence x\underline{ℓ} is nontrivial. Moreover, since φ_{p∗}(t^{l_{1}} ∧⋯∧t^{l_{i}}) ≠ φ_{p∗}(t^{m_{1}} ∧⋯∧t^{m_{i}}), we see that x\underline{ℓ} ≠ x\underline{m} provided \underline{ℓ} ≠ \underline{m}. ⊓ ⊔

Corollary 5.3 For each i-tuple \underline{ℓ}, the element x\underline{ℓ} has either infinite order or order divisible by 6.

Proof Note that φ_{p∗}(t^{l_{1}} ∧ ⋯ ∧ t^{l_{i}}) is an element of order p in the group H_{i}(SL_{2}(F_{p}[t]), Z) (since this group is all p-torsion). It follows that x\underline{ℓ} has order divisible by the least common multiple of 2 and 3 (if it is not infinite). ⊓ ⊔

Remark 5.4 We can glean further information by reducing modulo primes p ≥ 5. By Lemma 4.1,

H_{i}(B(F_{p}[t]), F_{p}) = H_{0}(F_{p}^{×}, H_{i}(F_{p}[t], F_{p})),

where α ∈ F_{p}^{×} acts on F_{p}[t] via α : x ↦ α^{2}x. Now, H_{i}(F_{p}[t], F_{p}) contains ⋀_{F_{p}}^{i} tF_{p}[t] as a direct summand and if i is a multiple of (p − 1)/2, then, by Fermat’s Little Theorem, F_{p}^{×} acts trivially on ⋀_{F_{p}}^{i} tF_{p}[t]. This implies that for i = n(p − 1)/2, H_{i}(SL_{2}(F_{p}[t]), F_{p}) contains a copy of ⋀_{F_{p}}^{i} tF_{p}[t]. Hence, if \underline{ℓ} is an i-tuple, i = n(p−1)/2, then x\underline{ℓ} has order divisible by 2·3·p. For example, the x\underline{ℓ} in H_{2n} have order divisible by 2 · 3 · 5 = 30, those in H_{3n} have order divisible by 2 · 3 · 7 = 42, etc. Given this, it seems reasonable to conjecture that the x\underline{ℓ} have infinite order.

6 Other Classes in H_{•}(SL_{2}(Z[t]), Z)

In [5], Krstić and McCool exhibit a basis of a free retract F of the quotient SL_{2}(Z[t])/U_{2}(Z[t]) (here, U_{2}(Z[t]) denotes the subgroup generated by unipotent matrices). It consists of elements h_{p,k} where p ≥ 2 and k ≥ 1. The elements h_{p,k} are defined as

h_{p,k} = ( \begin{matrix}1 + pt^{k} & t^{3k} \\ p^{3} & 1 − pt^{k} + p^{2}t^{2k} \end{matrix} ) .

Thus, the elements \overline{h}_{p,k} ∈ H_{1}(SL_{2}(Z[t]), Z) form the basis of a free direct summand; denote this summand by Z\{\overline{h}_{p,k}\}. Then we may write

H_{1}(SL_{2}(Z[t]), Z) = H_{1}(SL_{2}(Z), Z) ⊕ Z\{\overline{h}_{p,k}\} ⊕ X,

where X is some abelian group.

Since E_{2}(Z[t]) ⊂ U_{2}(Z[t]), the elements

x_{k} = ( \begin{matrix}1 & t^{k} \\ 0 & 1 \end{matrix} )

map to 1 in F. Thus, the elements \overline{x}_{k} map into the summand X ⊂ H_{1}(SL_{2}(Z[t]), Z). Hence we see that X is also not finitely generated.

Consider the homomorphism π_{p} : SL_{2}(Z[t]) → SL_{2}(F_{p}[t]). Then we have the following.

Proposition 6.1 The kernel of the map

π_{p∗} : H_{1}(SL_{2}(Z[t]), Z) → H_{1}(SL_{2}(F_{p}[t]), Z)

is not finitely generated.

Proof If p ≥ 5, then H_{1}(SL_{2}(F_{p}[t]), Z) = H_{1}(SL_{2}(F_{p}), Z) = 0 so that π_{p∗} = 0. If p = 2 or 3 then it is not clear that any summand lies in the kernel. But note that π_{p}(x_{k}) = π_{p}(h_{p,k}) so that π_{p∗}(\overline{x}_{k} −\overline{h}_{p,k}) = 0 for all k ≥ 1. ⊓ ⊔

Consider the elements g_{p,k} defined by

g_{p,k} = ( \begin{matrix}1 & −t^{k} \\ −p & 1 + pt^{k} \end{matrix} ) .

Note that each g_{p,k} is not unipotent; i.e., the matrix

n_{p,k} = ( \begin{matrix}0 & −t^{k} \\ −p & pt^{k} \end{matrix} )

has infinite order. However, we do have the following.

Lemma 6.2 For each k, l, g_{p,k} ≡ g_{p,l} modulo U_{2}(Z[t]).

Proof An easy calculation shows that

g_{p,k}^{−1}g_{p,l} = ( \begin{matrix}1 & t^{k} − t^{l} \\ 0 & 1 \end{matrix} ) .

The result follows. ⊓ ⊔

Thus, the set of all g_{p,k} (for a fixed p) lies in a single coset of U_{2}(Z[t]). Denote by \overline{g}_{p} the induced (nontrivial!) element of Z\{\overline{h}_{p,k}\}.

Proposition 6.3 If p = 2 or 3, then \overline{g}_{p} + \overline{x}_{k} lies in the kernel of π_{p∗} for all k ≥ 1.

Proof Note that π_{p}(g_{p,k}) = x_{k}^{−1}. It follows then that π_{p∗}(\overline{g}_{p}) = −t^{k} ∈ ⋀_{Z}^{1} tF_{p}[t] ⊂ H_{1}(SL_{2}(F_{p}[t]), Z). Since π_{p∗}(\overline{x}_{k}) = t^{k}, the result follows.

⊓ ⊔

Note that we also have \overline{g}_{p} + \overline{h}_{p,k} ∈ ker(π_{p∗}) since \overline{g}_{p} + \overline{x}_{k} and \overline{h}_{p,k} − \overline{x}_{k} lie in the kernel.

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