All Questions
Tagged with noncommutative-algebra differential-graded-algebras
7 questions
3
votes
2
answers
255
views
Is being graded commutative a necessary condition on $A$ such that $H^*(A)$ is commutative?
If we consider any differential graded algebra $A^\bullet$, then its homology is a graded algebra, since the tensor product interacts well with homology.
A sufficient condidtion for the homology to be ...
- 4,418
2
votes
0
answers
111
views
Correct notion of "connected" for dga of bundle-valued forms
Consider a vector bundle $E$ over a manifold $M$ with flat connection, $\nabla$. From this data I can form the associative/unital differential graded algebra $\mathcal{A} = \left(\Omega^{\bullet}(M, ...
- 1,611
1
vote
0
answers
85
views
Exponential of a sum in a non-commutative graded algebra
Let $a,b$ be two elements of a graded algebra $A$ such that $\deg(a)=1$, $\deg(b)=0$ and $[a,b]\neq 0$.
I would like to know whether there exits an explicit expression for the degree 1 component
$$\...
- 789
1
vote
0
answers
60
views
Universal bimodule for homotopy biderivations
Recall the commutative story: for a commutative algebra $A$, its module of differentials $\Omega (A)$ is characterized by the universal property that any derivation $\delta \colon A \to M$ is in a ...
- 765
8
votes
2
answers
2k
views
Reason to apply the Koszul sign rule everywhere in graded contexts
The Koszul sign rule is a sign rule that arises from graded-commutative algebras. For instance, let $\bigwedge(x_1,\dots, x_n)$ be the free graded-commutative algebra generated by $n$ elements of ...
- 499
2
votes
1
answer
515
views
dg-resolution of the polynomial algebra
I am intersted in constructing a cofibrant resolution of the commutative polynomial algebra in some number of variables in the category of dg-algebras(not necceserily commutative).
The resolutions ...
- 165
2
votes
0
answers
99
views
Turning left modules into right modules over a homotopy Gerstenhaber algebra
For simplicity's sake, let $A$ be a dg-algebra over $\mathbb{Z}/2\mathbb{Z}$.
In the case when $A$ is a commutative algebra, we can turn a left $A$ module into a right $A$ module trivially. Of course ...
