Correction to: Quadratic Presentations and Nilpotent Kähler Groups

James A. Carlson and Domingo Toledo

First correction: It has been pointed out to us by S. Chen that the inductive step (2.3) in the construction of 1-formal, 1-minimal models is not correct. By 1-formal, 1-minimal model we mean what is called formal 1-minimal model in the first parragraph of §2. We proceed to describe a correct inductive step.

Let V_{1} denote the vector space V of §2 over the field k, (one of ℚ, ℝ or ℂ). Let V_{2} denote the space C_{0} defined by (2.1), namely the kernel of the cup-product. Assume that i ≥ 2, that V_{1}, ⋯, V_{i} have been defined, let V^{i} = V_{1} ⊕ ⋯ ⊕ V_{i}, and let M_{i} = Λ(V^{i}). Assume that the differential d : M_{i} → M_{i} has also been defined and that it commutes with the action of the multiplicative group k^{∗} on Λ(V^{i}) by the algebra automorphisms defined by letting λ ∈ k^{∗} act on V_{j} as multiplication by λ^{j}. If W is a vector space with a k^{∗} action by semi-simple automorphisms, let W (j) denote the vector subspace where λ acts by multiplication by λ^{j}. Then define

V_{i+1} = Λ^{2}(V^{i})(i + 1) ∩ ker(d),

M_{i+1} = Λ(V^{i+1}), and the differential d on M_{i+1} to be the Hirsch extension of the differential algebra M_{i} obtained by defining d|V_{i+1} to be the inclusion of V_{i+1} in Λ^{2}(V^{i}).

This definition is clearly correct because, by the inductive assumption of a k^{∗} action, we have a decomposition

H^{2}(M_{i}) = ⊕_{j=2}^{2i}H^{2}(M_{i})(j).

We are inductively arranging H^{2}(M_{i})(j) = 0 for 2 < j ≤ i to achieve in the limit that H^{2}(M)(j) = 0 for j ≠ 2, which is equivalent to 1-formality. Now V_{i+1} is isomorphic to H^{2}(M_{i})(i+1), and d is defined precisely to then make H^{2}(M_{i+1})(j) = 0 for 2 < j ≤ i + 1.

The reason we choose this procedure is that the k^{∗} action is easily defined. Another correct procedure, closer to the definition of 1-formality, would have been to have a k^{∗} action on V_{i} with more weights, and to define V_{i+1} to be ⊕_{j>2}H^{2}(M_{i})(j).

With our chosen procedure, M_{i} is 1-formal if and only if H^{2}(M_{i})(j) = 0 for all j > 2. Note that this is in general a stronger condition than the vanishing of V_{i+1}.

The relation of the chosen procedure to §2 and the error there is now easily explained. The space C_{i} defined by (2.3) is the subspace V_{i+2} ∩ (V_{1} ⊗ V_{i+1}) of

Λ^{2}(V^{i+1}) = V_{1} ⊗ V_{i+1} ⊕ V_{2} ⊗ V_{i} ⊕ ⋯,

which in general is a proper subspace of V_{i+2}.

Fortunately, this error leaves all but one assertion in the paper unaffected. The reason is that for small i there is little difference between the correct and incorrect procedures. Namely, we have the following assertions:

Lemma. For any V_{1}, V_{2} and injective d : V_{2} → Λ^{2}V_{1}, if V_{i} and M_{i} are defined inductively as above, then 1. H^{2}(M_{2})(4) = 0.

2. H^{2}(M_{3})(5) = H^{2}(M_{3})(6) = 0.
3. H^{2}(M_{4})(7) = H^{2}(M_{4})(8) = 0.
4. V_{3} = C_{1}, where C_{1} is defined in (2.2).

5. If dim(V_{3}) = 1 and d(V_{2}) contains no decomposable element of Λ^{2}V_{1}, then V_{4} ≅ C_{2}, where C_{2} is defined as in (2.3). In fact, under these assumptions V_{4} = 0 unless V_{3} ⊂ V_{1} ⊗ V_{2} consists of decomposable elements, in which case dim(V_{4}) = dim(C_{2}) = 1. Proof: We prove typical assertions 2 and 5. To prove 2, note that

H^{2}(M_{3})(5) = ker(V_{2} ⊗ V_{3} → V_{1} ⊗ Λ^{2}V_{2} ⊕ Λ^{2}V_{1} ⊗ V_{3})

and the projection of this map to the second summand in the target is simply

d ⊗ 1 : V_{2} ⊗ V_{3} → Λ^{2}V_{1} ⊗ V_{3},

which is injective since d : V_{2} → Λ^{2}V_{1} is injective by definition. Thus H^{2}(M_{3})(5) = 0. Similarly

H^{2}(M_{3})(6) = ker(Λ^{2}V_{3} → V_{1} ⊗ V_{2} ⊗ V_{3} ⊕ Λ^{3}V_{2}),

which vanishes because the projection of this map to the first summand in the target is the composition of the inclusion Λ^{2}V_{3} → V_{3} ⊗V_{3} of skew-symmetric 2-tensors, followed by the inclusion V_{3} ⊗V_{3} → (V_{1} ⊗V_{2})⊗V_{3}.

To prove 5, observe that

V_{4} = ker(V_{1} ⊗ V_{3} ⊕ Λ^{2}V_{2} → Λ^{2}V_{1} ⊗ V_{2}).

If dim(V_{3}) = 1, then V_{3} ⊂ V_{1} ⊗ V_{2} has a basis element b of the form

b = ∑ x_{i} ⊗ c_{i},

where the c_{i} form a basis for V_{2} and the x_{i} ∈ V_{1}. The map in question is then

z = (v ⊗ b , a) → ∑(v ∧ x_{i}) ⊗ c_{i} + a,

where a ∈ Λ^{2}V_{2} ⊂ V_{2} ⊗ V_{2} is a skew-symmetric tensor. If z is in the kernel of this map, then

∑(v ∧ x_{i}) ⊗ c_{i} = −a ∈ V_{2} ⊗ V_{2}

is a skew-symmetric tensor. But this is only possible if

∑(v ∧ x_{i}) ⊗ c_{i} = 0 = a

because for each i, v ∧ x_{i} is decomposable, and c_{i} is by assumption not decomposable. Thus v ∧ x_{i} = 0 for all i. There are two possibilites: either v = 0, in which case C_{2} = V_{4} = 0, or v ≠ 0, in which case there is only one non-zero x_{i}, call it x, and v is a non-zero multiple of x, thus C_{2} and V_{4} are both one-dimensional with basis x ⊗ x ⊗ c, (x ⊗ x ⊗ c, 0) respectively. Thus we see that C_{2} ≅ V_{4} as asserted.

Corollary.

1. M_{2} is 1-formal if and only if V_{3} = 0.

2. M_{3} is 1-formal if and only if V_{4} = 0.

3. If dim(V_{3}) = 1, M_{3} is 1-formal if and only if C_{2} = 0.

¿From these assertions we see that the proofs of example 2.6, proposition 2.7 remain unchanged. In fact, the error affects only one assertion in the paper. Namely, the Lie algebra of example 3.5 is not 3step nilpotent, as asserted there. The reason is that the verification of assertion (3.2) of Lemma 3.4 is not sufficient for 3-step nilpotency (in this case dim(V_{3}) = 5, so part 5 of the Lemma does not apply). One needs to compute the kernel of the map

V_{1} ⊗ V_{3} ⊕ Λ^{2}(V_{2}) → Λ^{2}(V_{1}) ⊗ V_{2},

which turns out to be one-dimensional. Using the notation x_{1}, ⋯, x_{5}, c_{1}, c_{2}, c_{3} of Lemmas 3.3 and 3.4, a basis for the one-dimensional space V_{4} is the following. Let α ∈ V_{1} ⊗ V_{3} ⊂ V_{1} ⊗ V_{1} ⊗ V_{2} be

2x_{4}⊗(x_{1}⊗c_{1}+x_{2}⊗c_{2})+3x_{2}⊗(x_{4}⊗c_{2}−x_{2}⊗c_{3})+x_{5}⊗(x_{2}⊗c_{1}+x_{5}⊗c_{2}+x_{1}⊗c_{3})+x_{1}⊗(x_{4}⊗c_{1}+x_{3}⊗c_{2}+x_{5}⊗c_{3}),

and let β = c_{2} ∧ c_{1} = c_{2} ⊗ c_{1} − c_{1} ⊗ c_{2} ∈ Λ^{2}V_{2} ⊂ V_{2} ⊗ V_{2}. Then

(α, β) ∈ V_{1} ⊗ V_{3} ⊕ Λ^{2}V_{2}

is a basis for V_{4}. We believe that further computation shows that the corresponding M_{4} is 1-formal, but this would take us too far afield. Second correction: The statement near the end of §5 “This is the subgroup of smallest index in the integral Heisenberg group which we know to be a Kähler group.” is incorrect and misleading. The integral Heisenberg group itself is a Kähler group. This can be seen, for instance, by taking the first concrete example at the beginning of that paragraph, using a product of elliptic curves, with fundamental group the subgroup of the integral Heisenberg group (2.6) with x_{1}, ⋯, x_{4} even. Its subgroup of index two where z is also even is isomorphic to the integral Heisenberg group, and is of course a Kähler group since subgroups of finite index of Kähler groups are Kähler groups. This is a simple special case of the fact, pointed out to us by P. Deligne, that any two ℤ-forms of the same nilpotent graded ℚ-group can be embedded in each other as subgroups of finite index, therefore one is a Kähler group if and only if the other one is.