Subject: Reply on cdga question Date: Sun, 10 Jun 2001 22:30:44 -0400 (GMT) From: Tornike Kadeishvili To: Don Davis Maybe so: if for a DGA (A,d) not only H(A) is commutative, but the induced A(\infty )-algebra (H(A),{m_i}) is commutative too, then SQB(H(A),{m_i}) (here B is your bar-tilda construction, Q is the functor of indecomposables with respect to shuffle product, so QB(H(A),{m_i}) is a DGLie coalgebra, and S is symmetric algebra) and (A,d) have same homology A(\infty )-algebras, thus they are weak equivalent: (A,d) \leftarrow \Omega B(H(A),{m_i})\rightarrow SQB(H(A),{m_i}). Of course over a field, but characteristic 0? Sq-s are not involved. Tornike Kadeishvili