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Dec
5
comment Compute adjugate matrix over commutative ring
@NicolasMalebranche: but the case of the generic matrix is a particular case of what you ask, if you consider general commutative rings (or fields), since one particular case is the function ring on $n^2$ indeterminates. If you have something else in mind, you should specify which kind of (computable) commutative rings you want to consider.
Dec
5
comment $\mathbb{Q}$-forms of $\mathrm{SL}_2\times \mathrm{SL}_2$
There are at least these ones: The non-$Q$-simple ones are the $G_1\times G_2$ where $G_1,G_2$ are $Q$-forms of $SL_2$. The $Q$-simple ones (or some of them?) are the Weil restrictions "$G(K)$" where $G$ is a form of $SL_2$ and $K$ is a quadratic extension of $Q$.
Dec
5
comment Compute adjugate matrix over commutative ring
If you take the generic matrix (the one with indeterminate entries $a_{11}\dots$), you can compute its inverse (inside the field of fractions), multiply by the determinant to eventually get the generic adjugate matrix. Anyway I'd expect the practical complexity to depend on the entries of the given matrix and not only on the size of the matrix...?
Dec
4
revised Quotients of the Higman Group
added 777 characters in body
Dec
4
comment Is the equational theory of commutative vN regular rings decidable?
@ThomasKlimpel I said that I used the definition of meadow given by Emil's link, which does not include $\phi(x)x\phi(x)=\phi(x)$. Alone, $x\phi(x)x=x$ does not put any constraint to $\phi(0)$.
Dec
2
comment Example of a polycyclic group which is not of polynomial growth?
Note that Derek's example contains RW's example with index two (because the square of the matrix $\begin{pmatrix}1 & 1 \\ 1 & 0\end{pmatrix}$ is $\begin{pmatrix}2 & 1 \\ 1 & 1\end{pmatrix}$)
Nov
28
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.
Nov
28
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.
Nov
27
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).
Nov
27
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.
Nov
27
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.
Nov
27
comment Does the associated Lie algebra determine a group?
Yes thanks Tobias I indeed mean "whose associated Carnot algebras (...) are isomorphic".
Nov
27
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 $(\mathfrak{g},c[\cdot,\cdot])$) ).