abstract algebra – Non-splitting short exact sequence given by nilpotent ideal

I’m trying to solve an exercise (Jon Carlson, Modules and Group Algebras, Exercise 1.1) which says

If $A$ is a finite dimensional $k$-algebra with $1$ over a field $k$ and $Nsubseteq A$ is a non-trivial nilpotent left ideal, then the short exact sequence
$$0longrightarrow Nlongrightarrow Alongrightarrow A/Nlongrightarrow 0$$
does not split.

Now, if there was an $A$-module morphism right inverse to the projection, let’s say $rho:A/Nlongrightarrow A$, it should map a class to one of its elements, namely
$$rho((a)_N)=a+n_a$$
for some $n_ain N$. My idea is to consider $a=1notin N$ and, by nilpotence $N^k=0$, for $k$ minimal, a nonzero product of $k-1$ elements in $N$, call it $m=m_1cdots m_{k-1}neq 0$. If we let this $m$ to act, we find

$$mrho((1)_N)=rho(m(1)_N)=rho((m)_N)=rho(0)=0$$

But therefore we get a contradiction since $$mrho((1)_N)=m(1+n_1)=m+0neq 0$$

My concern is about the correctness of this proof, because it seems odd to me that I never used finiteness condition on the dimension of $A$, neither that we are on a field. Maybe the result holds for a nil ideal, which is also nilpotent under the assumptions that I didn’t touch. Thanks in advance!

ra.rings and algebras – Special nilpotent elements

Let $R$ be a (noncommutative, associative) ring. Set $N_2:={xin R : x^2=0}$, the set of nilpotent elements of degree $2$ (also called the square-zero elements).

If $x,yin R$ satisfy $xy=0$, then $yxin N_2$, but not every element in $N_2$ arises in this way. (See the example below.)

Question: Has the set $F:={yx : xy=0}$ been studied before? Are there characterizations of these “special square-zero” elements?

Note that every element in $F$ is a commutator (if $xy=0$, then $yx=yx-xy=(y,x)$.) Thus, a precise question would be if $F=N_2cap(R,R)$?

Example: Consider $R=left{ begin{pmatrix} a & b \ 0 & aend{pmatrix} : a,binmathbb{R} right}$. Then $z=begin{pmatrix} 0 & 1 \ 0 & 0end{pmatrix}$ belongs to $N_2$ but not to $F$, since every commutator in $R$ is zero.

When is a nilpotent Lie algebra isomorphic to the associated graded of its lower central series?

All Lie algebras in this question are finite-dimensional and defined over a field $k$ of characteristic $0$ which I’m happy to take to be $mathbb{R}$ or $mathbb{C}$.

$DeclareMathOperatorgr{gr}$Let $L$ be a nilpotent Lie algebra. It is then filtered by its lower central series, and we have an associated graded nilpotent Lie algebra $gr L$. It is definitely not the case that $L$ and $gr L$ have to be isomorphic; see Malcev Lie algebra and associated graded Lie algebra for some examples.

Question: what kinds of conditions can I put on $L$ that ensure that it is isomorphic to $gr L$? E.g. if the field is $mathbb{R}$ are the there geometric/topological/algebraic conditions on the associated simply-connected nilpotent Lie group that ensure this?

reference request – Limit of a finite rank nilpotent

Let $H$ be a separable complex Hilbert space

Let $mathcal{F(H)} ={T in mathcal{B(H)}: text{rank}(T) < infty }$ ,

and $mathcal{N(H)} = { T in mathcal{F(H)}: T^k=0 text{ for some } k in mathbb{N} }$

Find the norm closure $overline{mathcal{N(H)}}$.

Ideas:

I know that $overline{mathcal{F(H)}} = mathcal{K(H)}$, where $mathcal{K(H)}$ is the set of compact operators.

So the closure will be compact operators.

And $mathcal{N(H)} subset mathcal{K(H)} $.

Now I have perform some sort of operation to on $mathcal{K(H)}$ to find the limit.

I also saw a somewhat similar problem here.

Quasi-nilpotent trace class operators as limits of nilpotents

Please let me know if you have any ideas or reference to this.

Any help will be appreciated.

Thank you in advance!

Dimension of kernel of nilpotent matrix

Let an $n times n$ real matrix $A$ such that $A^n=0$ and $A^{n-1} neq 0$. What can we say about the dimension of its kernel? If not a solution, a strong hint would be much appreciated. I am trying to apply the rank-nullity theorem but I don’t obtain something meaningful at the moment.

mg.metric geometry – Example of an invariant metric on a nilpotent group which is not asymptotically geodesic

Let $X$ be a metric space. We say that $X$ is asymptotically geodesic if for all $epsilon > 0$, there exists $R > 0$ such that, for all $x,y in X$, there exists some finite sequence of points $p_0=x,p_1,p_2,…,p_N = y in X$ such that each $d(p_{i-1},p_i) le R$ and
$$ sum_{i=1}^N d(p_{i-1},p_i) le (1 + epsilon)d(x,y). $$

I came across this definition while studying finitely generated nilpotent groups and their scaling limits. Pansu proved in “Croissance des boules et des geodesiques fermees dans les nilvarietes” that, if a finitely generated (virtually) nilpotent group $Gamma$ is endowed with a left-invariant metric $d$ such that $(Gamma,d)$ is asymptotically geodesic, then the sequence of metric spaces $(Gamma,frac{1}{n}d)$ has a Gromov-Hausdorff limit (which he describes).

However, I’ve been having a lot of difficulties coming up with examples of metrics on nilpotent groups which aren’t asymptotically geodesic, particularly under the extra assumption that the metric in question is bi-Lipschitz to a word metric on $Gamma$.

I believe that it follows pretty quickly from Fekete’s lemma that any invariant metric on $mathbb{Z}$ which is bi-Lipschitz to the standard metric is asymptotically geodesic, and, unless I made a mistake, the same is true for $mathbb{Z}^d$ by a more involved but fairly similar argument. But the property of being asymptotically geodesic is certainly not preserved by bi-Lipschitz maps a priori, and it seems implicit in the literature that there should be some metrics bi-Lipschitz to word metrics which are not asymptotically geodesic. (For instance, in Benjamini and Tessera, “First passage percolation in nilpotent Cayley graphs and beyond”, one of the key ingredients in a proof is showing that a certain invariant metric is asymptotically geodesic, even though that metric is assumed to be bi-Lipschitz to a word metric).

So my question is:

(1) What is an example of a left-invariant metric on a finitely generated nilpotent group which is bi-Lipschitz to a word metric but not asymptotically geodesic?

As it happens, I actually have had difficulty constructing such a metric even if I drop the assumption that it is bi-Lipschitz to a word metric, so I would also be interested in the weaker question:

(2) What is an example of a left-invariant metric on a finitely generated nilpotent group which is not asymptotically geodesic?

And even though I’m interested primarily in the case of nilpotent groups, I’m also generally trying to understand this property, so an answer to the even weaker question might help too:

(3) What is an example of a left-invariant metric on a (finitely generated?) discrete group which is not asymptotically geodesic?

Upper Triangular Invertible Matrices Group is Solvable but not Nilpotent

So I am suppose to show that the group of upper triangular matrices in $GL_n(mathbb{R})$ say S, is Solvable but not nilpotent.

If we define $G_1 = S$, $G_2=(G_1, G_1)$, and $G_{i+1} = (G_i, G_i)$, Then to show that $S$ is solvable if at some point we have $G_N = {id}$. I am unable to find $G_2$. I got some information about it. I think it should be a subset of $SL_n(mathbb{R})$ but I am unable to determine it precisely.

Also how should I determine $G_3$, $G_4 dots dots$ ?

nilpotent groups – Maximal subgroups of a subgroup

Let $G$ be a nilpotent group and $H$ a subgroup of $G$. It can easily be prove that if $M$ is a maximal subgroup of $G$ that is not contain $H$ then $H cap M$ is a maximal subgroup of $H$. I can prove the reverse(i.e. all maximal subgroups of $H$ are in this form) when $H$ is not a maximal subgroup of $G$ but not when $H$ is maximal. Is this true?

Can we relax the nilpotency condition?

ag.algebraic geometry – Deformation of p-divisible groups along nilpotent thickening

Let $S_0 rightarrow S$ be a nilpotent thickening of schemes (no divided power provided) where $p$ is nilpotent, let $G$ be a $p$-divisble group over $S_0$, how to describe all liftings of $G$ to $S$ as $p$-divisible groups (using linear algebra data)?

I am curious about the existence in good cases which is trivial for etale ones. In general the original Grothendieck-Messing theory is not available (as no divided power structure is introduced), unless we’re in some good cases like $mathbb Z/p^n rightarrow mathbb Z/p^m$ ($p>2$).

nilpotence – Jacobi identity, free nilpotent Lie algebra

Is there a general formula for the Jacobi identity on the free $nu$-nilpotent Lie algebra $mathfrak{FL}(nu,n)$ on $n$ generators?

In math overflow I found a list of the Hall bases. Looking at Hall’s paper, though, the proof only shows that the constructed generating system is a basis by showing every element has a unique representation as a linear combination.

Reutenauer’s “Free Lie Algebras” -which was also mentioned in the link above- seems to do what I’m asking for. However, it’s not accessible to me without first studying the entire theory on non-commutative polynomials.

Is there a general formula for, say, the Jacobi identity $(X_1,(X_2,(X_3,X_4)))$ and its cyclic permutations, and $((X_1,X_2),(X_3,X_4))$ and its cyclic permutations? Is there one for higher nilpotences?