bio | website | math.berkeley.edu/~mbtucker |
---|---|---|
location | Berkeley | |
age | 32 | |
visits | member for | 4 years, 6 months |
seen | yesterday | |
stats | profile views | 2,674 |
I like quantum groups, representation theory, noncommutative algebra, and homological algebra, mainly in the context of noncommutative geometry.
Apr 6 |
comment |
How to recognize a Hopf algebra?
For the sake of those reading this question (including me!), could you specify the sense in which you mean "regular"? It's a pretty overloaded term... |
Feb 26 |
accepted | Software for noncommutative Groebner bases over rational function fields |
Feb 25 |
comment |
Software for noncommutative Groebner bases over rational function fields
I've seen that other question, but it didn't address the question of which base fields are allowed. |
Feb 25 |
comment |
Software for noncommutative Groebner bases over rational function fields
@NeilHoffman, that seems to be for commutative rings. I did find this: magma.maths.usyd.edu.au/magma/handbook/text/900#9876 which indicates that Magma can do Groebner basis calculations for noncommutative algebras, but it doesn't say much about what coefficient fields are possible. |
Feb 25 |
asked | Software for noncommutative Groebner bases over rational function fields |
Feb 25 |
comment |
Basis of a Finite Dimensional Algebra with a Finitely Generated Relation Set By Computer
@Leandro, do you know if there is a way to do this sort of thing in GAP when the algebra is defined over a field of rational functions? |
Nov 14 |
awarded | Necromancer |
Oct 19 |
awarded | Yearling |
Sep 30 |
awarded | Caucus |
Sep 28 |
revised |
A matrix algebra has no deformations?
Fixed small mistake since question came back to front page anyway. |
Sep 10 |
awarded | Nice Answer |
Aug 23 |
comment |
Versions of the spectral theorem
Convergence in the weak topology does not imply convergence in norm. |
Aug 7 |
comment |
What is a complex inner product space “really”?
I think it's important to note what translating the concept of "normal operator" to the complex setting yields. The point is that any operator can be decomposed into real and imaginary parts, which are self-adjoint and hence diagonalizable. It's easy to check that an operator is normal if and only if its real and imaginary parts commute, so that they are simultaneously diagonalizable. |
Aug 7 |
awarded | Enlightened |
Aug 7 |
awarded | Nice Answer |
Aug 4 |
revised |
Conjugate linear maps between $*$-algebra modules
Fixed a small mistake |
Aug 4 |
answered | Conjugate linear maps between $*$-algebra modules |
Aug 2 |
answered | What are good ways to present proofs of theorems requiring auxiliary lemmas? |
Aug 1 |
comment |
About generator and isomorphism problems for free groups operator algebras
To the downvoters: it is customary and polite to explain your reasons for downvoting a question. Perhaps you might like to leave suggestions for how the question could be improved. |
Jul 26 |
comment |
Name for a Specific Type of Non-Symmetric Bilinear Form
Do you mean $B(i)$ rather than $I(j)$ in the superscript? And is $I=\lbrace 1, \dots, N \rbrace$? |