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Linear duality provides a correspondence between differentials on free tensor algebras and A-infinity algebras. Under this correspondence, what is the A-infinity counterpart of "stable, tame isomorphism" of dgas, a notion prevalent in Legendrian knot theory?

(I recall that "tame" refers to an automorphism which shifts a single generator by a combination of other generators -- or the composition thereof -- while "stable" means up to adding some pairs of generators a and b obeying da=b.)

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  • $\begingroup$ Aw, c'mon, I thought all those dg's and A-infinities would surely get the MO crowd out! As a further enticement for people to comment without full knowledge of the answer, let me confess that I had to edit my post several times before getting the definition of "tame" even reasonably accurate. $\endgroup$ Oct 30, 2014 at 13:57
  • $\begingroup$ Well, stabilizing corresponds to taking the direct sum with the $A_\infty$ (co)algebra which is the linear span of those two new generators and which has all (co)products zero except that differential. $\endgroup$ Oct 31, 2014 at 12:36
  • $\begingroup$ Thanks, Gabriel. Stabilization is clearer, but tameness? The actual definition of "stably tame isomorphic" is tamely isomorphic after possibly stabilizing one or both dgas. $\endgroup$ Oct 31, 2014 at 14:19
  • $\begingroup$ If I'm doing things right, a tame isomorphism is a "strict" isomorphism of $A_\infty$ (co)algebras, meaning a morphism $(f_1,f_2,\ldots)$ where $f_1$ is an isomorphism and $f_i$ is zero for $i>1$. It's not clear to me whether every strict isomorphism arises as a tame isomorphism, that's a question about linear algebra. What changes of basis of a linear space can you make with compositions of these automorphisms that shift a generator by a combination of other generators? $\endgroup$ Oct 31, 2014 at 15:08
  • $\begingroup$ I guess the best you could hope for is the whole special linear group? $\endgroup$ Oct 31, 2014 at 15:14

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