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3 votes
0 answers
261 views

On the Hochschild cohomology of the minimal model of an $A_\infty$ algebra

Suppose $(A, (\mu_k))$ is a (curved) $A_\infty$ algebra, and let $(\tilde A, (\tilde\mu_k))$ be its minimal model. Now, we have two Hochschild cohomology rings $HH^*(A)$ and $HH^*(\tilde A)$. (It may ...
Hang's user avatar
  • 2,789
10 votes
0 answers
268 views

Isomorphisms between minimal $A_\infty$-algebras having identical $k$-truncations

Let $A_m =(A,0,m_2,m_3,\dots)$ and $A_n=(A,0,n_2,n_3,\dots)$ be two $A_\infty$-structures on a vector space $A$. Assume that i) $A_m$ and $A_n$ are isomorphic, and ii) $A_m$ and $A_n$ have the same ...
Bashar Saleh's user avatar
8 votes
1 answer
674 views

Frobenius $A_{\infty}$-bialgebras?

Recall that a finite dimensional associative algebra $A$ over a field $k$ is called a symmetric Frobenius algebra (sometimes called "closed" Frobenius algebra) if its equipped with a symmetric non ...
Manuel Rivera's user avatar
8 votes
1 answer
353 views

Does there exist a model of chains on oriented manifolds with both a strict intersection pairing and strict functoriality for closed embeddings?

Let $M$ be a smooth oriented $n$-dimensional manifold. My favorite model of $\operatorname{Chains}_\bullet(M) \otimes \mathbb R$ is the space of smooth compactly-supported de Rham forms on $M$, ...
Theo Johnson-Freyd's user avatar
5 votes
0 answers
677 views

Is a certain A-infinity algebra (homologically) smooth?

An A-infinity algebra $A$ is smooth a'la Kontsevich if it is perfect as an $A$-$A$-bimodule. I am wondering about the standard tricks to show smoothness of given algebras. A relatively basic example ...
Daniel Pomerleano's user avatar