bio | website | math.ucdavis.edu/~greg |
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location | Davis, CA | |
age | 47 | |
visits | member for | 5 years, 9 months |
seen | 2 days ago | |
stats | profile views | 25,851 |
I am a professor at UC Davis.
Jun 1 |
awarded | Nice Answer |
May 27 |
awarded | Good Answer |
May 23 |
awarded | Enlightened |
May 23 |
awarded | Nice Answer |
May 8 |
comment |
Intuition behind the definition of quantum groups
See further comments. |
May 8 |
revised |
Intuition behind the definition of quantum groups
Answers in response to Semyon's further questions |
May 7 |
awarded | Nice Answer |
May 6 |
comment |
Intuition behind the definition of quantum groups
Okay, there is another asterisk to the formalism: By historical accident, the most standard $q$ in a quantum group is actually the square root of the more natural $q$ in $q$-analogues and Gaussian binomial coefficients. This discrepancy is controversial and could in theory still disappear one day. |
May 6 |
answered | Intuition behind the definition of quantum groups |
Apr 14 |
comment |
Open problems in Euclidean geometry?
Yes, uniformly random in that sense. |
Apr 10 |
answered | Dividing by two in the category of vector spaces |
Apr 10 |
comment |
Dividing by two in the category of vector spaces
(No, StackExchange, I would not like to move this discussion to chat.) Characteristic p is as far as I have gotten in "math by Google". |
Apr 10 |
comment |
Dividing by two in the category of vector spaces
Yeah, it looks like it just is a counterexample. |
Apr 10 |
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Dividing by two in the category of vector spaces
On the other hand, this conference proceedings, "Infinite Length Modules" by Krause and Ringel, says that Krull-Schmidt can fail for infinite-dimensional representations of a finite group, over an infinite field with positive characteristic. books.google.com/… |
Apr 10 |
comment |
Dividing by two in the category of vector spaces
Wikipedia says that this assertion, the "Krull-Schmidt theorem", holds for finite-length modules over a ring, but not in general for modules that are only Noetherian or only Artinian. en.wikipedia.org/wiki/… |
Apr 10 |
comment |
Dividing by two in the category of vector spaces
Note that one test of naturality in this case is whether it is true for infinite-dimensional vector spaces without the axiom of choice. |
Apr 6 |
awarded | Nice Answer |
Mar 27 |
awarded | Popular Question |
Mar 26 |
awarded | Good Question |
Mar 24 |
awarded | Nice Answer |