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17h
comment Does the associated Lie algebra determine a group?
@Zuriel: OK but "Carnot Lie algebra" is a terminology, while "$E_0^*$" is a notation, these are two different things.
1d
comment Does the associated Lie algebra determine a group?
@Zuriel: several distinct communities use this notion of Lie algebras and use distinct terminologies with no attempt of unification. I tried to gather as many as possible of these terminologies... where did you find $E^*_0(G)$? never seen... and I don't know how to find it with Google.
1d
comment Does the associated Lie algebra determine a group?
@Zuriel: there's no point in posting twice the same comment... for your second question maybe you could specify the kind of properties you'd like. If $G$ is a finite nilpotent group then the associated Carnot algebra retains the cardinal and the nilpotency length (as well as the cardinal of all terms in the central series).
1d
comment Does the associated Lie algebra determine a group?
It's the word the subriemannian geometers use. I think I heard it from Stefan Wenger. Google "Carnot Lie algebra". "graded" is possibly more common but it's hopelessly ambiguous.
1d
comment Does the associated Lie algebra determine a group?
For a 2-nilpotent 2-group, the Carnot algebra also captures the structure of the abelian 2-groups $G/[G,G]$ and $[G,G]$, which are not always 2-torsion (think of the Heisenberg group modulo $2^n$). Also even in the case these are 2-elementary (or in the case of complex unipotent groups), when the center is more than 1-dimensional, the classification is much more complicated than just one symplectic form: it corresponds to $k$-tuples of alternating forms up to simultaneous conjugation and these have a quite wild classification.
1d
comment Does the associated Lie algebra determine a group?
Yes thanks Tobias I indeed mean "whose associated Carnot algebras (...) are isomorphic".
2d
comment Does the associated Lie algebra determine a group?
There are 5-dimensional complex unipotent groups (or real/rational/discrete analogues) that are not isomorphic but whose associated Carnot algebras (i.e. the graded Lie algebra you construct) are not isomorphic. So even when $G$ is nilpotent, $\mathcal{L}_G$ is only some kind of 1st-order approximation of $G$.
Nov
23
comment Unitary representations of Tarski Monsters and other beasts
There is no universal definition of Tarski monster, could you give one? Do you mean an infinite f.g. quasi-finite group (quasi-finite = in which every proper subgroup is finite)? or something else such as "every proper subgroup is cyclic"?
Nov
22
comment Parallel topologies on a Prüfer group with the trivial group topology as the only group topology contained in both
Let me prove only one direction (the one you need), in a wide generality (one group $C$, two topological groups $G_1,G_2$). Assume that the diagonal map $g=f_1\times f_2:C\to G_1\times G_2$ has a dense image, let $B_i$ be a symmetric neighborhood of $1$ in $G_i$ and $A_i=f_i^{-1}(B_i)$. Let's show $A_1A_2=C$. Let $x$ be an element of $C$. Then by density, we can approximate $(f_1(x),1)$ by the image of $g$. This implies that there exists $y\in C$ such that $(f_1(x)^{-1}f_1(y),f_2(y))\in B_1\times B_2$. Thus $x^{-1}y\in A_1$ and $y\in A_2$. Thus $x=y(x^{-1}y)^{-1}$ is in $A_1A_2$.
Nov
22
comment Parallel topologies on a Prüfer group with the trivial group topology as the only group topology contained in both
The definition in wikipedia is enough to figure out how to the existence of such a homomorphism.
Nov
22
comment Groups with a unique composition series
Ah OK sure, you're right.
Nov
22
comment Parallel topologies on a Prüfer group with the trivial group topology as the only group topology contained in both
(I write $C_{p^\infty}$ for the Prufer group, to avoid confusion with $p$-adics) Well, tranversal means that $(f,g):C_{p^\infty}\to\mathbb{T}\times\mathbb{T}$ has a dense image. Yes this exists and this is easy: indeed by Pontryagin duality it amounts to proving that there is an injective homomorphism of $\mathbb{Z}^2$ into the $p$-adics $\mathbb{Z}_p$.
Nov
21
comment Groups with a unique composition series
In other words, you ask which finite groups $G$ have $N(G)$ totally ordered, where $N(G)$ is the set of normal subgroups. Some discussion is here: people.bath.ac.uk/masgcs/problem/commentary10.html
Nov
21
comment Rigid nilpotent Lie algebras
Yes you quotient by isomorphism (obviously $(\mathfrak{g},[\cdot,\cdot])$ is always isomorphic to $(\mathfrak{g},c[\cdot,\cdot])$) ).
Nov
18
comment torsion free for the 2nd cohomology group?
Where do you use Property T? For instance, the vanishing of $H^1(G,\mathbf{Z}X)$ for every $G$-set $X$ holds for many more groups.
Nov
18
comment torsion free for the 2nd cohomology group?
Do you know in the case $SL_3(\mathbf{F}_p[t])$? For this group possibly there is a geometric way of understanding $H^2(G,\mathbf{Z}G)$, and this group is not f.p. (Behr) and has Property T.
Nov
18
comment amenable + without $BS(m,n)$+finite $K(G,1)$implies virtually cyclic?
Another comment: the assumption "does not contain any Baumslag-Solitar is equivalent to "does not contain any solvable Baumslag-Solitar", i.e., does not contain $BS(1,n)$ for any $n\ge 1$ (including $BS(1,1)\simeq\mathbf{Z}^2$). The reason is that $BS(m,n)$ for $\min(|m|,|n|)\ge 2$ contains a copy of $\mathbf{Z}^2$. This is not a hard fact, but it can sound surprising (at least two experts told me that $BS(2,3)$ does not contain $\mathbf{Z}^2$).
Nov
18
comment amenable + without $BS(m,n)$+finite $K(G,1)$implies virtually cyclic?
I guess the main expectation for the "open problem 1" is that no, there are examples, they are just hard to find. If so, your question is probably even harder. Or else you expect a positive answer for amenable groups, but it would sound difficult as well. Very few structural results for groups are known even under a strong assumption such as finite $K(G,1)$.
Nov
16
comment Are profinite groups of cardinality $|\mathbb{R}|$ determined by their finite quotients?
This hypothesis "have at most countably many finite Hausdorff quotients" for a profinite group, is the same as being metrizable, and is also equivalent to be either finite or homeomorphic to a Cantor set.
Nov
14
comment Are homotopy braid groups residually nilpotent?
Braid groups are not residually nilpotent (for $n\ge 5$, $[B_5,B_5]$ is perfect as far as I remember); you probably mean pure braid groups (which are indeed residually [torsion-free nilpotent]).