3,562 reputation
11840
bio website ucs.louisiana.edu/~avm1260
location Lafayette, LA, USA
age 45
visits member for 5 years, 4 months
seen 8 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.


Jun
20
revised Representation of finite group
add tags
Jun
12
reviewed Reject Is regularity closed under products?
Jun
7
revised Min number of primes up to n
fix some latex
Jun
4
comment For a ring $k$ and a set $X$, what are the $k$-algebra homomorphisms $k^X \to k$?
@TomLeinster: Yes, you should keep LaTeX to a minimum in titles, and you definitely don't want to have an all-LaTeX title. On the other hand, ASCII art and pseudo-latex ( k^x )seem like a bad idea when you are talking about such a small part of the title. This is not a LaTeX-heavy title...
Jun
4
revised For a ring $k$ and a set $X$, what are the $k$-algebra homomorphisms $k^X \to k$?
latex title
Jun
1
reviewed Edit Quintic Equation
Jun
1
revised Quintic Equation
format edited
Jun
1
comment Quotients of finitely generated nilpotent groups
@YCor: I think yo umean "torsionfree" where you say "finite", but yes, I see there would be issues because the factors for $H/N_3$ (e.g., $(H/N_3)^{\rm ab}$) could be strictly proper factors of the corresponding ones for $H$, so we are not sure that $H$ itself has torsionfree factors. Thanks.
Jun
1
comment Quotients of finitely generated nilpotent groups
@YCor: Wouldn't we be able to do the same thing in general by induction on the class? First, go to a torsionfree subgroup of finite index $N$, so we may assume $N$ Is torsionfree. Then find a subgroup $H$ of finite index that contains $N_{k+1}$ such that $H/N_{k+1}$ has the desired property. But since $N_{k+1}\subseteq Z(N)\cap H\subseteq Z(H)$, then the commutator subgroup of $H/N_{k+1}$ is "essentially" the same as that of $H$ because the center is marginal in the commutator bracket, so the fact that $H/N_{k+1}$ has torsionfree quotients shoudl give $H$ does as well...
May
31
comment Quotients of finitely generated nilpotent groups
@YCor: Thanks! I confess to not knowing it before.
May
31
reviewed Edit Can one estimate the distribution of eigenvalues of a matrix by its Cauchy/Stieltje transform?
May
31
revised Can one estimate the distribution of eigenvalues of a matrix by its Cauchy/Stieltje transform?
Corrections
May
30
comment Quotients of finitely generated nilpotent groups
@YCor: How do we get to $N$ torsionfree to begin with?
May
30
revised Boundedness of solutions of a difference equation
As long as there was a recent bump, let's get rid of all those nasty typos, the annoying "how to proof", etc.
May
29
comment Quotients of finitely generated nilpotent groups
For arbitrary class, I would try to do some induction; assuming we can do it for class $c$, with $N$ of class $c+1$, we could try finding a subgroup of finite index $H$ such that $H/N_{c+1}$ has the desired property in $N/N_{c+1}$, so that the only possible problem lies in $H_{c+1}$, and then take a finite index subgroup of $H$ given by adequate powers of the generators so that we are in the torsionfree part of $H_{c+1}$ once we get down to it.
May
29
reviewed Reject A question on an set of 8 matrices related to the SU(3) generators
May
29
revised Quotients of finitely generated nilpotent groups
delete stray paragraph from draft version
May
29
answered Quotients of finitely generated nilpotent groups
May
29
comment Quotients of finitely generated nilpotent groups
@DaveWitteMorris: I'm not sure I follow what you are writing... are you reversing the roles of $N$ (original group) and $H$ (group we are looking for)? If we take the original group to be the integral Heisenberg group, then since $N/N_2\cong \mathbb{Z}^2$, $N_2/N_3\cong \mathbb{Z}$, and $N_3=\{e\}$, we can just take $H=N^1=N$. How did I get $H^2$ and $H^4$?
May
29
comment Quotients of finitely generated nilpotent groups
Is there some reason that you cannot just take $N^k$, where $k$ is a common multiple of the exponents of the torsion subgroups of all $N_i/N_{i+1}$ ?