All Questions
Tagged with noncommutative-geometry ra.rings-and-algebras
33 questions
5
votes
1
answer
883
views
Is this ring isomorphic to a quotient of a group algebra?
Consider the quotient of the free algebra $\mathbb{Q}\langle \alpha, \beta, \gamma, \delta, \varepsilon, \zeta \rangle$ by the two-sided ideal $I$ subject to the relations $$ \alpha\delta=\delta\alpha=...
- 1,093
5
votes
0
answers
226
views
Cohomology of representation varieties and the Hochschild cohomology
Let $k$ be a field, $A$ a $k$-algebra, and $V$ a $k$-vector space. Then we can consider the representation varieties of $A$ on $V$: $\mathrm{Hom}_{k\textrm{-alg}}(A, \mathfrak{gl}(V))$ and $\mathrm{...
- 985
2
votes
1
answer
306
views
Serre functors and global dimensions
Let $k$ be a field.
Let $\mathcal{C}$ be an abelian category (over $k$).
We say that $\mathcal{C}$ has a finite global dimension if there exists integer $n > 0$ such that
$$
\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
13
votes
0
answers
573
views
Classical (i.e. commutative) spaces with quantum symmetry but no classical symmetry
In a recent preprint (arXiv:2311.04889), my coauthors and I constructed a sequence of graphs with no classical symmetry which nevertheless have quantum symmetry.
For graphs this had been an open ...
- 2,339
5
votes
0
answers
125
views
Lie algebra cohomology of formal non-commutative vector fields
Let $k$ be a field of characteristic $0$ and $A=k\langle\langle x_1,\dotsc,x_n\rangle\rangle$ be a free completed associative algebra. The space of continuous derivations $\mathrm{Der}(A)$ is ...
- 985
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
291
views
Lie algebra cohomology of the space of vector fields
For a (closed and oriented) manifold $M$, the first Lie algebra cohomology $H^1(\mathrm{Vect}(M),C^\infty(M))$ of the space of vector fields with coefficients in smooth functions is isomorphic to $H^1(...
- 985
2
votes
0
answers
109
views
Automorphism group of the first Weyl field
A related question is this one (Automorphism group of the quantum Weyl field).
Let $A_1$ denote the rank 1 Weyl algebra (over the complex numbers), and $D_1$ its skew field of fractions, called the ...
- 3,318
5
votes
0
answers
302
views
Connections in non-commutative geometry
Let $K$ be a field, $A$ a unital associative $K$-algebra and $M$ a left $A$-module. A connection on $M$ is a $K$-linear map $\nabla:M\to \Omega^1A\otimes_AM$ which satisfies the Leibniz rule. ...
- 985
2
votes
1
answer
265
views
Gluing data for modules over a ring with idempotents
Let $A$ be a ring. If $e$ is an idempotent, then there is an abelian recollement involving the categories $A\text{-}\mathrm{Mod}$ and $eAe\text{-}\mathrm{Mod}$. This is Example 2.7 in Homological ...
- 1,071
2
votes
0
answers
88
views
Non-existence of idempotent via evaluation of higher order cocycle on a tuple of idempotents
The Kaplansky conjecture and Kaplanski-Kadison conjecture are classical conjectures about non existence of non-trivial idempotents in group algebra $\mathbb{C}\Gamma $ or the reduced group algebra $...
- 356
2
votes
0
answers
172
views
Simple modules of quantum planes
Let $k$ be an algebraically closed field.
Let $R := k\langle x,y \rangle/(yx-qxy) (q \in k^*)$.
We often call $R$ a quantum plane.
If $q$ is a primitive $n$-th root, then for any $(\zeta, \xi) \in k^* ...
- 889
4
votes
0
answers
91
views
Nullstellensatz for maximal left ideals of quantum plane
Let $R=\mathbb{C}\langle x,y\rangle/\langle xy=qyx\rangle$ be the quantum plane algebra. Does some sort of Nullstellensatz holds for the maximal left ideals of $R$? By this we mean all maximal left ...
- 91
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
6
votes
2
answers
449
views
Survey of recent developments of the Gelfand-Kirillov dimension
It is almost two decades since the now classical books by McConnell and Robinson's
[ Noncommutative Noetherian rings. With the cooperation of L. W. Small. Revised edition. Graduate Studies in ...
- 3,318
2
votes
1
answer
151
views
Automorphism of algebras with certain initial conditions on given idempotents
The First question
Let $A$ be a Banach or a $C^*$ algebra. Assume that $e,f$ are two idempotents or prjections in $A$ which satisfy $ef=fe=0$. Assume that there are two automorphisms $\phi, \psi: A \...
- 356
4
votes
0
answers
333
views
Other kinds of equivalence relations on the set of idempotents of a Banach or $C^*$-algebra or a ring (Can we get a new kind of K-theory?)
The standard equivalent relations on idempotents of a $C^*$ algebra or a Banach algebra are Murray von Neumann, similarity and homotopy equivalent. In this post we consider two other kinds of ...
- 356
2
votes
0
answers
60
views
Integrals in noncommutative graded algebras which are not necessarily Hopf
Let $\mathbf{k}$ be a field. Let $A$ be a finite dimensional $\mathbb{Z}_{\geq 0}$-graded $\mathbf{k}$-algebra such that $A^0=\mathbf{k}1$. Let $m$ be the maximal non-negative integer such that $A^m\...
- 1,066
8
votes
1
answer
739
views
A question regarding Kadison-Kaplansky idempotent conjecture (A nearest group element $g$ to a nontrivial self adjoint unitary element u )
Edit: According to answer and comments by Prof. Valette we edite the question.
The Kadison Kaplansky conjecture says:
Kadison-Kaplansky conjecture: If $G$ is a torsion-free discrete group then $C^*_{\...
- 356
5
votes
0
answers
219
views
Constructing a noncommutative algebra from a commutative algebra
I was told at a conference that one way to construct a noncommutative algebra from a commutative one is to "replace the product of finite spaces (which on the level of continuous functions corresponds ...
- 111
1
vote
0
answers
104
views
Hochschild coboundary on the space of alternative forms
Assume that $A$ is a complex algebra. By $C^{n}(A)$ we mean the space of all $n-$linear map $\phi:A^n \to \mathbb{C}$. An alternative $k-$ form is
an element $\phi \in C^{k}(A)$ ...
- 356
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
23
votes
0
answers
463
views
Topological loops vs. algebro-geometric suspension in Hochschild homology
Let $k$ be a base commutative ring, and let $A$ be a (unital but not necessarily commutative) $k$-algebra. The cone on $A$ is the ring $CA$ of infinite matrices $(a_{ij})_{i,j \geq 1}$ that are ...
- 6,069
3
votes
1
answer
181
views
Do representations of the universal enveloping algebra $\mathrm{U}\mathfrak{su}_2$ retain the Hopf algebra structure?
A Lie algebra $\mathfrak{g}$ generates its universal enveloping algebra $\mathrm{U}\mathfrak{g}$, which has the structure of a Hopf algebra. Modules of $\mathrm{U}\mathfrak{g}$ are exactly the of ...
- 183
6
votes
1
answer
936
views
Noncommutative HKR theorem
What is the analog of HKR theorem in the noncommutative world?
Recall that the well-known theorem by Hochschild-Kostant-Rosenberg says that for a smooth commutative algebra $A$ of finite type over a ...
- 1,276
0
votes
0
answers
208
views
A noncommutative analogy of the tube lemma
Assume that $A$ and $B$ are two unital commutative Banach algebras. Assume that $\phi \in \mathcal{M} (A)$ is an element of the maximal Ideal space. Define $\alpha: A\hat{\otimes} B \to \mathbb{C}\...
- 356
2
votes
1
answer
341
views
Weyl algebras $A_n(k)$ as tensor product of the first Weyl algebra
In afew threads I've read that the Weyl algebra $A_{n+1}(k)$ is isomorphic to the $k$-tensor product of $A_n(k)$ with $A_1(k)$, why is this true?
- 21
6
votes
1
answer
550
views
A generalized K- theory via generalized idempotents
Edit After the answer by Neil Strickland, I add the word "a ring" in this new version.
In the literature, there is a concept of generalized idempotent: an n- idempotent is an element $a$ of a Banach ...
- 356
1
vote
0
answers
134
views
Non commutative analogy of compact-open topology
Let $R$ be a ring, define a topology on $AUT(R)$(Or End(R)) with the following subbase:
For every two 2-sided Ideal $I$ and $J$, a subbase element is $B(I,J)=\{f\in AUT(R) \mid f(I)+J=R\}$.
We can ...
- 356
3
votes
1
answer
307
views
Global dimension of quantum $\mathbb{P}^{n}$
Let $k$ be a field. Given a (not necessarily commutative) $k$ graded ring $A$, M. Artin and J.J. Zhang introduced a notion of "noncommutative projective scheme" $Proj(A)$ in this paper. It is defined ...
- 65
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
13
votes
4
answers
2k
views
Geometric interpretation of Universal enveloping algebras
Given a complex Lie algebra $\mathfrak g$, we can form its universal enveloping algebra and interpret it as a noncommutative space.
Is this perspective useful? What does this space "look like"?
How ...
- 13.2k