On the realisability problem of groups for Sullivan algebras

In this paper, we prove that any group occurs as the group of homotopy classes of self-equivalences of a none elliptic Sullivan algebra.

where the latter is the group of the self-equivalences dga-maps of ( V, ∂) modulo the relation of homotopy. We write E( V ) = aut(( V, ∂))/ for this group.

Strongly connected digraphs and theorem of de Groot
A digraph (i.e., a directed graphs) G = (V (G), E(G)), where V (G) denotes the set of the vertices of G and E(G) the set of its edges, is strongly connected if for every v, u ∈ V (G), there exists an integer m ∈ N and vertices v = v 0 , v 1 , . . . , v m = u such that (v i , v i+1 ) ∈ E(G), i = 1 = 0, 1, . . . , m − 1.
The following theorem, due to J. de Groot [8, p. 96], plays a central role in this paper Theorem 2.1 Any group G is isomorphic to the automorphism group of a some strongly connected digraph G.

The main result
3.1 Construction of the Sullivan algebra M G associated to a digraph G Definition 3.1 Let G be a strongly connected digraph with more than one vertex. Define the Sullivan algebra where the degrees of the generators are given by and where the differential is given by

Lemma 3.2 The Sullivan algebra M G is not elliptic.
Proof It is easy to see from the definition of the differential ∂ that for every k ∈ N, the cocycles x k 1 and x k 119 is spanned by where almost every coefficients A 7,v , A 8,v and A 9,v is zero, then So the space of 119-cocycles is spanned by But we have Consequently, any cocycle is a coboundary.
Next we use Lemma 3.3.
where almost every coefficients λ (v,u),(r,s) , a (v,u) is zero. Note that as α is an equivalence of homotopy, it induces an isomorphism on the indecomposables, therefore β 1 , β 2 , p 1 , p 2 , p 3 = 0 and at least one of the coefficients a (v,u) is not zero . Likewise, as ξ is an isomorphism at least one of the coefficients λ (v,u),(r,s) is not zero.
As, for i = 1, 2, 3, we have α(∂(y i )) = ∂(α(y i )), it follows that Implying that α(X ) = β 9 1 β 5 2 X, where X is given in (1). Notice that, by the definition of the group D 119 40 and the commutativity of diagram (A), there exists ϕ (v,u) Proof It worth mentioning that the strong connectivity of the graph implies that for every v ∈ V (G), v is the starting vertex of an edge (v, u) ∈ E(G). Therefore the coefficients a (v,u) , γ 1,v , γ 2,v involved in α(w v ) (see (4)) can be entirely determined by using (6). Indeed, we have Expanding the expression leads to the following monomials But none of them appears in the expression (8), giving ∂(ξ(z (v,u) )). Moreover, any linear combination of the monomials in (9), which has degree 120, is not coboundary. Indeed, according to (2) if φ is an element of degree 119, then, by (3), ∂(φ) is a linear combination of the monomials x 10 1 x 4 2 , . Therefore, all the coefficients in (9) are nil. As at least one of the coefficients a (v,u) is not zero, it follows that γ 1,v = γ 2,v = 0 and only one coefficient among a (v,u) ∈ V (G) is not zero. Let us denote it by a (v,t) . Thus, by going back to the formula (4), we proved that for every v ∈ V (G) there is a unique vertex t ∈ V (G) such that α(w v ) = a (v,t) w t .