All Questions
Tagged with noncommutative-geometry homological-algebra
28 questions
2
votes
1
answer
306
views
Serre functors and global dimensions
Let $k$ be a field.
Let $\mathcal{C}$ be an abelian category (over $k$).
We say that $\mathcal{C}$ has a finite global dimension if there exists integer $n > 0$ such that
$$
\operatorname{Ext}^i(M,...
- 889
5
votes
1
answer
197
views
Examples of cyclic A-infinity algebra
I am wondering about (references to) examples of cyclic A-infinity algebras- especially including explicit descriptions of the structure maps and pairing.
Thanks a lot!
- 51
9
votes
1
answer
236
views
Formal smoothness of path algebras and connections
Let $k$ be a field of characteristic zero and $A = kQ$ the path algebra associated with a quiver $Q$. The algebra $A$ is said to be formally smooth over $k$ if
$$
\Omega^1_kA = \operatorname{Ker}(\...
- 985
7
votes
0
answers
573
views
Why is Hochschild homology interesting if its cohomology groups are infinite-dimensional?
I am trying to understand Hochschild homology, in particular the Hochschild–Kostant–Rosenberg theorem. As far as I understand this result gives an isomorphism between the algebraic (Kähler) ...
- 1,144
3
votes
1
answer
446
views
A set of objects classically generates the full subcategory of compact objects iff it generates the whole category
Sorry in advance if my question doesn't have the level of this community.
I am studying this paper of Bondal and Van Den Bergh and in particular section 2. Generators and resolutions in triangulated ...
- 224
12
votes
0
answers
688
views
Kontsevich's derived noncommutative geometry and Rosenberg's noncommutative 'spaces'
It appears to me (though I may be wrong) that the common opinion is that the main difference between derived noncommutative geometry and Rosenberg's noncommutative 'spaces' is that Rosenberg's version ...
- 439
6
votes
2
answers
329
views
Path algebras are formally smooth
In Ginzburg's notes Lectures on Noncommutative Geometry, he claim that the path algebra of a quiver is formally smooth (See Section 19.2).
I have two questions. First, how to show this claim and ...
- 661
5
votes
0
answers
183
views
Connes-Chern pairing, compatibility with periodicity operator in the odd case
Let $A$ be an algebra (say unital). For an odd (say $2n-1$) cyclic cocycle $\varphi$ and a class in $K_1(A)$ represented by invertible $u$ we define
$$\langle [\varphi],[u] \rangle:=\frac{2^{-(2n+1)}}...
- 9,330
5
votes
1
answer
354
views
Two approaches to periodic cyclic cohomology
Cyclic cohomology may be defined in several ways: the easiest way to define it is via a subcomplex $C^*_{\lambda}$of Hochschild complex consisting from cyclic cochains. There are also other ...
- 9,330
2
votes
1
answer
316
views
Normalization of cyclic cocycles
This question is a continuation of the discussion
Normalization of Hochschild cocycles
but this time in the cyclic context. I would like to ask whether the following is true:
The inclusion of ...
- 9,330
13
votes
1
answer
669
views
Is a "smooth" finite-dimensional algebra separable modulo its radical?
Let $k$ be a field, and let us write the "unadorned" tensor $\otimes$ in place of $\otimes_k$. For a unital finite-dimensional $k$-algebra $A$, let $A^e = A \otimes A^{op}$ denote the enveloping ...
- 5,407
3
votes
0
answers
98
views
Quasi isomorphism for bicomplexes both defining cyclic cohomology
Let $A$ be a complex unital algebra. Consider the cyclic (cohomological) bicomplex $\mathcal{C}(A)$. This is a bicomplex
where in the $p$-th row one has $C^p(A)$ (the space of all $p+1$-linear forms) ...
- 9,330
4
votes
0
answers
211
views
Inner automorphisms acts as identity on Hochschild homology
Let $A$ be a unital algebra and $u \in A$ be an invertible element. Let us consider $u_n(a_0 \otimes a_1 \otimes ... \otimes a_n):=ua_0u^{-1} \otimes ua_1u^{-1} \otimes ... \otimes ua_nu^{-1}$. Then $(...
- 9,330
8
votes
1
answer
601
views
Isomorphism in cyclic cohomology vs isomorphism in Hochschild cohomology
Let $A$ be a unital algebra over a field $K$, $C^n(A)$ a space of all $n+1$ linear maps into scalar field $k$ (I'm interested in case $k=\mathbb{C}$) and
$$(bf)(a_0,...,a_{n+1})=\sum_{i=0}^n(-1)^if(...
- 9,330
2
votes
1
answer
569
views
Why $k[x,y]$ is not a formally smooth algebra?
We could talk about the formal smoothness of an algebra. See for example Ginzburg's lecture notes For an associative algebra $A$ over a field $k$ we define
$$
D(A)=T(A+\bar{A})/(\bar{ab}=a\bar{b}+\bar{...
- 9,009
16
votes
0
answers
591
views
Lifting DG-categories to characteristic zero
The question of lifting (smooth projective) varieties from an algebraically closed field $k$ of characteristic $p$ to characteristic zero (i.e., to the Witt vectors $W(k)$) is a classical one. It's ...
- 25.6k
5
votes
0
answers
300
views
Can we obtain a derived category from an additive category? Like a category of Banach modules?
Let $A$ be a Banach algebra, let $A$-mod be the category of left Banach modules (as defined in Helemskii's "Banach and locally convex algebras"), $A$-mod is an additive category, but not abelian ...
- 211
1
vote
3
answers
450
views
Smooth affine algebras are Calabi-Yau
Are all smooth affine algebras over a field Calabi-Yau?
I'm thinking yes since they satisfy Van den Bergh duality with dualizing module themselves (have I made a mistake in this reasoning)/
- 27
5
votes
0
answers
490
views
Soft Question: What does periodic cyclic theory measure?
Ex1) The cyclic homology of $\mathbb{C}[X,Y]$ and that of the algebra of functions on the sphere $S^2$ have the same periodic cyclic homology. Clearly, however, these objects are topologically very ...
- 5,407
1
vote
1
answer
273
views
Jacobi-Zariski exact sequence question
Denote by $HC(A,M)$ the Hochschild homological complex of an algebra $A$ with coefficients in an $A$-bimodule $M$, and let $B\rightarrow A$ be an $R$-flat extension of $R$-algebras, for some $CRing$ $...
- 5,407
3
votes
0
answers
324
views
2 - Calabi Yau algebras and bimodule coherence
Let $\Pi:=\Pi(Q)$ be the preprojective algebra of a connected non-Dynkin quiver over an algebraically closed field of characteristic zero.
In H. Minamoto "Ampleness of two-sided tilting complexes", ...
9
votes
2
answers
2k
views
Global dimensions of non-commutative rings
This is related to my previous question: When is a quantum affine space $\mathbb{A}^{n}$ Calabi-Yau? I now would like to know the global dimension of the ring $R=\mathbb{C}\langle x_1,\dots,x_n\rangle/...
- 1,663
3
votes
1
answer
817
views
dg-lie structure on $HH^*$ and Koszul duality
This is shamelessly close to my other question: A Question on Koszul duality and $B(\infty)$ structures on $HH^*$. Maybe this one will get a better response. Rather than rewrite that one, I am going ...
- 3,758
4
votes
1
answer
535
views
A Question on Koszul duality and $B(\infty)$ structures on $HH^*$
The following theorem is known from a paper "Duality in Gerstenhaber Algebras" by Felix, Menichi, Thomas. Given a simply connected space X of finite type.
There is an equivalence of Gerstenhaber ...
- 3,758
7
votes
1
answer
611
views
Extension of the formality theorem?
The following question came up in a discussion the other day and I have been wondering whether something is known about it. Everything below takes place over $\mathbb{C}$. I don't have the expertise ...
- 3,758
3
votes
1
answer
408
views
Finite Homological Dimension of R/P for all P for module finite non-commutative rings
I have a reasonably precise question which I hope is clear enough to get a nice answer. Let R be a Noetherian non-commutative ring which is finite as a module (and flat/free if it helps) over it's ...
- 3,758
6
votes
1
answer
2k
views
Hochschild and cyclic homology of smooth varieties
Many of the standard sources which discuss the Hochschild Kostant Rosenberg theorem and cyclic homology for smooth varieties such as Loday and Weibel's paper "The Hodge Filtration and Cyclic Homology" ...
- 3,758
5
votes
0
answers
517
views
A smooth twisted tensor product of dg algebras?
I want to consider a Z/2Z dg algebra. As an algebra, it is generated over $\mathbb{Q}$ by two elements where x is even and e is odd with the relations $xe=ex$ and $e^2=1$(this makes it in particular ...
- 3,758