3,152 reputation
11637
bio website ucs.louisiana.edu/~avm1260
location Lafayette, LA, USA
age 45
visits member for 4 years, 10 months
seen 3 hours ago

With the move of MathOverflow into the SE network, this account is now associated with dormant accounts in math.SE and other sites in the network. While I plan to continue my (generally low-level) participation in MO, my current plans do not include restarting my participation in those other sites. Therefore, I will be ignoring any comments or pings that reach me from those sites, unless and until I resume my active participation there.

I remain "gone for the foreseeable future" from math.SE, tex.SE, and meta.SE.

Please do not send me private e-mail to call my attention to comments, questions, or other matters related to those sites. Thank you. Also, as I no longer participate in those sites, I do not wish to be sent, by private e-mail, questions that you can just as well ask on those sites. I would have thought it was obvious, but apparently I need to say so explicitly.


Dec
17
comment Schur covering group
@Taj: And then you re-posted your full question in Group Pub Forum. But I for one don't plan to think about it, because I don't want to have to check all the places where you may have asked the question to see if my efforts are wasted or not. You should have posted in one place, and then waited. If no answer was forthcoming after a reasonable time, then go ahead and post elsewhere, mentioning that you had already posted the question at the other place.
Dec
16
comment Schur covering group
@StefanKohl: It was asked there, and the first question was answered (whether any finite group is a Schur cover); the second question was not answered there yet, last I checked. I'm more bothered by the fact that the OP decided to ask the same question in several fora rather than pose the question in one and then wait to see if he would get a satisfactory answer there. It invites wasted effort by others.
Dec
16
comment Schur covering group
You asked this question in the Group Pub Forum; it was pointed out there that $(2,2)$ is not a covering group, and that many finite abelian groups are not covering groups.
Dec
15
reviewed Approve A question about intuitionistic analysis
Dec
9
reviewed Approve Local fractional Sobolev inequality
Dec
1
reviewed Approve perturbation of Invariant subspaces
Nov
28
comment Is it provable in $\mathsf{ZF}$ that there is a group structure on any set $X$?
And in any case, you need to assume $X$ is nonempty!
Nov
28
awarded  Pundit
Nov
16
reviewed Approve Can you efficiently solve a system of quadratic multivariate polynomials?
Nov
6
reviewed Approve Is this a definition of equivariant derived category?
Oct
23
reviewed Approve Second Hardy-Littlewood Conjecture theme
Oct
5
reviewed Approve The Convergence of Jacobi and Gauss-Seidel Iteration
Sep
30
awarded  Explainer
Sep
8
reviewed Approve Amenability as a geometric property
Sep
7
reviewed Approve Dimension of totally reflexive modules
Sep
3
reviewed Approve Irreducibility of $x^m-g(y)$
Sep
1
revised Group structure on an arbitrary completely regular topological space that makes $(x,y)\mapsto xy^{-1}$ continuous at $(1,1)$
more accurate title
Aug
23
comment Who defined and who coined “module”?
I would say, in answer to your edit, that if that is your definition of "define" then Dedekind both coined and defined the term (just like he coined the term "ideal", as opposed to "ideal number"). Noether generalized, just like she generalized a lot of the theory of rings and ideals, but according to Stillwell in the work quoted by anon, she was fond of saying "Es steht schon bei Dedekind" ("It is already in Dedekind") when talking about ideals/modules.
Aug
11
comment Self-duality of the subgroup lattice of $G\times H$
I realize the observation above doesn't answer the question; but it now looks more like: if the product of two lattices is self-dual, is one of the lattices self-dual? And that does indeed look suspect.
Aug
10
comment Self-duality of the subgroup lattice of $G\times H$
The condition on the orders of $G$ and $H$ imply that every subgroup of $G\times H$ is of the form $A\times B$ with $A\leq G$ and $B\leq H$. Hence, the lattice of subgroups of $G\times H$ is equal to the product of the lattices of the subgroups of $G$ and of $H$.