All Questions
5 questions
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 ...
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 ...
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 ...
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 $M$, ...
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 ...