On the page 4 of these notes it is stated that an $A_{\infty}$-algebra $A$ is necessarily is quasi-isomorphic to an $A_{\infty}$-algebra $HA$ with trivial differential. Moreover, $HA$ is unique up to a non-unique $A_{\infty}$-isomorphism. My question, is the morphism between the underlying vector spaces induced by this non-unique isomorphism unique?
2 Answers
A minimal model of an $A_\infty$-algebra $A$ is an $A_\infty$-algebra $B$ with trivial differential together with an $A_\infty$-quasi-isomorphism $B\to A$. This gives an isomorphism $H^*A=H^*B$, and we have $H^*B=B$ since $B$ has trivial differential. Together these show that the underlying vector space of $B$ is indeed canonically isomorphic to the homology of $A$.
(Instead of $B$, you wrote $HA$ for the minimal model. That's of course correct in a sense, as the underlying vector space of the minimal model is indeed canonically the homology of $A$, but it is sure to cause confusion when explaining the above.)
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$\begingroup$ What is the canonical isomorphism between $H^*(S^{2n+1})$ and it minimal model $\Lambda(x)$? Could you write it down? Thanks. $\endgroup$ Nov 12, 2018 at 19:12
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$\begingroup$ @NajibIdrissi We are speaking of different definitions of the word "minimal model", that is all. $\endgroup$ Nov 12, 2018 at 20:55
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$\begingroup$ Right, I took commutative algebras, my bad. Let $A = \mathbb{R}[x]$ (viewed as an algebra with no higher operations) and let $HA = \mathbb{R}[y]$ be its minimal model in the sense given by OP. Let $HA' = \mathbb{R}[z]$ be another minimal model. What is the canonical isomorphism $HA \cong HA'$? I still don't get your answer. $\endgroup$ Nov 13, 2018 at 8:04
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$\begingroup$ @NajibIdrissi: What is the canonical isomorphism between the cokernel of the diagonal map $\mathbb Z/3\to(\mathbb Z/3)^2$ and $\mathbb Z/3$? Answer: there is none, but that has no bearing on whether or not the cokernel of a map of abelian groups is well defined up to canonical isomorphism. $\endgroup$ Sep 2, 2019 at 23:07
You're essentially asking whether an $A_\infty$-algebra with zero differential has nontrivial auto-isotopies (where an isotopy is an $\infty$-morphism whose linear part is an isomorphism). Indeed, if there is a nontrivial one, then you can compose your isomorphism with it to get a different one; conversely if you have two isomorphisms, then you take $fg^{-1}$ to get a nontrivial automorphism.
Nontrivial automorphisms exist in general, so the answer to your question is no, there is no unicity. For a simple example, consider $A = HA = \mathbb{R}[x]/(x^2)$ (for some $x$ of degree $n > 0$) with the obvious algebra structure. Then there is a nontrivial automorphism given by $f: A \to A$, $f(x) = 2x$.
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$\begingroup$ You're taking the definition of "minimal model of $A$" to be "an $A_\infty$-algebra $B$ with trivial differential such that there exists an $A_\infty$-quasi-isomorphism $B\to A$" (compare the definition in my answer). While the question of which definition is "correct" is meta-mathematical (so we may disagree), here is an analogy which I find convincing: $\endgroup$ Nov 13, 2018 at 16:43
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$\begingroup$ Let $G$ be a group and let $N\leq G$ be a subgroup. We could define a quotient of $G$ by $N$ to be a group $Q$ such that there exists a surjection $G\to Q$ with kernel $N$. This is clearly a bad definition, for one because the quotient is now only unique up to non-unique isomorphism. Instead, we should define the quotient to be a group $Q$ together with a surjection $G\to Q$ with kernel $N$. Now it's unique up to unique isomorphism. Much better! (Yes, I didn't say anything about $N$ being normal, since that's only relevant for existence.) $\endgroup$ Nov 13, 2018 at 16:46