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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 ...
Aaron Mazel-Gee's user avatar
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 ...
Jan Weidner's user avatar
  • 13.2k
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 ...
Manny Reyes's user avatar
  • 5,407
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 ...
David Roberson's user avatar
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/...
user2013's user avatar
  • 1,663
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}(\...
Qwert Otto's user avatar
8 votes
1 answer
740 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^*_{\...
Ali Taghavi's user avatar
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(...
Qwert Otto's user avatar
7 votes
0 answers
575 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) ...
Jake Wetlock's user avatar
  • 1,144
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 ...
Ali Taghavi's user avatar
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 ...
jg1896's user avatar
  • 3,318
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 ...
Sasha Patotski's user avatar
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=...
Bumblebee's user avatar
  • 1,093
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!
Jak's user avatar
  • 51
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{...
Qwert Otto's user avatar
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 ...
Qwert Otto's user avatar
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. ...
Qwert Otto's user avatar
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 ...
TerryL's user avatar
  • 111
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 ...
user498029's user avatar
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 ...
Ali Taghavi's user avatar
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 ...
josh's user avatar
  • 183
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 ...
Michel's user avatar
  • 65
2 votes
1 answer
307 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,...
YkMz's user avatar
  • 889
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 ...
Sergey Guminov's user avatar
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?
WEylmaster's user avatar
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 \...
Ali Taghavi's user avatar
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 ...
jg1896's user avatar
  • 3,318
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 $...
Ali Taghavi's user avatar
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^* ...
YkMz's user avatar
  • 889
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\...
Christoph Mark's user avatar
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 $\phi \in C^{k}(A)$ ...
Ali Taghavi's user avatar
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 ...
Ali Taghavi's user avatar
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}\...
Ali Taghavi's user avatar