Recent MathOverflow Questions

Constructive proof of a rational version of Perron-Frobenius?

Math Overflow Recent Questions - Thu, 05/03/2018 - 07:20

In the following, we work with vectors and matrices whose entries are rational numbers. Inequalities between such vectors are understood to be coordinatewise: e.g., two vectors $a = \left(a_1,a_2,\ldots,a_n\right)^T$ and $b = \left(b_1,b_2,\ldots,b_n\right)^T$ are said to satisfy $a > b$ if and only if each $i \in \left\{1,2,\ldots,n\right\}$ satisfies $a_i > b_i$.

Theorem 1. Let $n$ be a positive integer. Let $A$ be an $n\times n$-matrix whose entries are nonnegative rational numbers. Then, there exists a nonzero vector $v \in \mathbb{Q}^n$ with nonnegative coordinates such that either $Av \geq v$ or $Av \leq v$.

The next theorem concerns a more restricted class of matrices. If $A$ is an $n\times n$-matrix whose entries are nonnegative rational numbers, then we define $D_A$ to be the directed graph whose vertices are the numbers $1,2,\ldots,n$, and which has an arc from vertex $i$ to vertex $j$ if and only if the $\left(i,j\right)$-th entry of $A$ is positive. We say that the matrix $A$ is irreducible if and only if this digraph $D_A$ is strongly connected. (Other equivalent definitions of irreducibility appear on the Wikipedia page for Perron-Frobenius.)

Theorem 2. Let $n$ be a positive integer. Let $A$ be an irreducible $n\times n$-matrix whose entries are nonnegative rational numbers. Then, there exists a nonzero vector $v \in \mathbb{Q}^n$ with positive coordinates such that either $Av > v$ or $Av = v$ or $Av < v$.

These two theorems can both be derived (with some nontrivial but not too hard work) from the Perron-Frobenius theorem. The latter constructs a Perron-Frobenius eigenvector of $A$ with real coordinates; one just needs to approximate its coordinates by rational numbers (unless it has eigenvalue $1$, in which case we have to make sure the approximation happens in the $1$-eigenspace).

Question. How can Theorem 1 and Theorem 2 be proven in constructive logic?

(Roughly speaking, this is probably tantamount to not using real numbers in the proof -- or building up the theory of real algebraics in constructive logic, which seems to have been done but I haven't seen it properly written up, and even then I don't know if Perron-Frobenius still applies and how you would prove it. Simon Henry's approximate version of Brouwer's fixed point theorem might help, but I don't know how it is proven.)

I'm asking partly because of recent uses of Theorems 1 and 2 in combinatorics, but this question has been in the back of my mind for many years, ever after I solved a restricted version of the $n = 3$ case (IMO Shortlist 2003 problem A1) (official solution) on an exam.

Diophantine equations and 'quasi-paucity'

Math Overflow Recent Questions - Thu, 05/03/2018 - 07:03

Let $X,Y \geq 1$. I am interested in the number of solutions of the following diophantine equations: $$S_1\colon \, \, x_1y_1^3 = x_2 y_2^3 $$ Let $N_1(X,Y) $ denote the number of solutions to $S_1$ with $1\leq x_i \leq X$ and $1\leq y_i \leq Y$. There are the obvious solutions $x_1=x_2$ and $y_1=y_2$ which contribute $XY$. The true order of magnitude should be $XY \log(XY)^A$ with some small power of log. I would like to get $A$ as small as possible.

Also consider the equation $S_2$ given by $$ x_1(y_1^3 - y_2^3) = x_2 (z_1^3 -z_2^3)$$ with $1\leq x_i\leq X$ and $1\leq y_i,z_i \leq Y$ with $y_1\neq y_2$ and $z_1\neq z_2$. Let $N_2(X,Y)$ denote the corresponding number of solutions. I expect something like $N_2(X,Y) \ll XY^2\log(XY)^B$, with $B$ again hopefully small.

Is there some clever way to get the exponent of the logarithm small?

Weak colimits in locally cartesian closed categories

Math Overflow Recent Questions - Thu, 05/03/2018 - 06:23

The general adjoint functor theorem implies that a complete locally small category has a weak colimit of a diagram if and only if it has a colimit of this diagram. It seems that this is also true for finite colimits in locally cartesian closed categories. For example, if a locally cartesian closed category has a weakly initial object $W$, then we can define the initial object using the internal language: $$\sum_{x : W} \prod_{f : W \to W} f(x) = x.$$ The proof that this object is initial is similar to the proof of this fact for complete categories. Does this fact about locally cartesian closed categories appears in the literature?

The second question is whether this fact is true for $\infty$-categories. It is true for locally small complete $\infty$-categories as was shown in this paper (Proposition 2.3.2), but I think that it does not hold for locally cartesian closed $\infty$-categories. So, what is an example of a locally cartesian closed $\infty$-category with a weakly initial object, but without the initial one?

Spaces without maximal homogeneous subspaces

Math Overflow Recent Questions - Thu, 05/03/2018 - 04:35

A homogeneous space $(X,\tau)$ is a topological space such that for all $x,y\in X$ there is a homeomorphism $\varphi:X\to X$ such that $\varphi(x)=y$. As a previous question implies, the union of an ascending chain of homogeneous spaces need not be homogeneous (example, see below).

What is an example of a Hausdorff space $(X,\tau)$ that does not contain a homogeneous subspace that is maximal with respect to $\subseteq$?

Note. The supremum (union) of an ascending chain of homogeneous spaces need not be homogeneous: Endow $\mathbb{N}$ with the discrete topology and consider the disjoint union $X = (\mathbb{N}\times\{0\}) \cup (\mathbb{Q}\times\{1\})$, where $\mathbb{Q}$ carries the Euclidean topology. Since $X$ is countable, it is the the union of an ascending chain of finite sets, all of which carry the discrete topology and are homogeneous, but $X$ is not.

The Modulus of a Polynomial are the Same is 1

Math Overflow Recent Questions - Thu, 05/03/2018 - 03:20

I had posted the following problem on stack exchange before.

Suppose the real number $\lambda \in \left( 0,1\right),$ and let $n$ be a positive integer. Prove that the modulus of all the roots of the polynomial $$f\left ( x \right )=\sum_{k=0}^{n}\binom{n}{k}\lambda^{k\left ( n-k \right )}x^{k}$$ are $1.$

I do not seem to know how to do it, to show if the roots of the polynomial have modulus one.Putnam 2014 B4Show that for each positive integer $n,$ all the roots of the polynomial $\sum_{k=0}^n 2^{k(n-k)}x^k$ are real numbers. This problem is very similar to the Putnam problem.

Regarding product of outer functions

Math Overflow Recent Questions - Thu, 05/03/2018 - 01:00

Please see the definition of Hardy spaces on the unit disc here. Let $0<p\leq\infty$. Let $f,g\in H^p$ be such that $f$ and $g$ are outer. Then is the product of $f$ and $g$ outer?

Linear optimization with one positive definite quadratic equality condition in P?

Math Overflow Recent Questions - Wed, 05/02/2018 - 19:16

I have the following minimization problem in $z \in \mathbb R^n$, which contains $x_1, \dots, x_t, y \in \mathbb R$.

$$\begin{array}{ll} \text{minimize} & y\\ \text{subject to} & xQx'= y\\ & 0 \leq x_i \leq1\\ & Az \leq b\end{array}$$

where $Q$ is diagonal and has positive diagonal integer values, $A \in \mathbb Z^{m \times n}$ and $b \in \mathbb Z^m$ are given.

Is this problem $NP$-hard or solvable in polynomial time?

name for monoids inducing bimonoids in Rel?

Math Overflow Recent Questions - Wed, 05/02/2018 - 12:38

Let Rel be the category of sets and relations, which is a (compact closed) symmetric monoidal category under the cartesian product of sets. We write $A \nrightarrow B$ to indicate a relation from $A$ to $B$, and given a function $f : A \to B$, we write $f^+ : A \nrightarrow B$ and $f^- : B \nrightarrow A$ for the corresponding functional/anti-functional relations.

Now, any ordinary monoid $(X,m,e)$ (i.e., monoid in Set, where $m : X \times X \to X$ is the multiplication and $e : 1 \to X$ is the unit) induces a pair of a "relational monoid" $(X,m^+,e^+)$ and a "relational comonoid" $(X,m^-,e^-)$, that is, an object $X$ which is both a monoid and a comonoid in Rel. I am interested in when $(X,m^+,e^+,m^-,e^-)$ forms a bimonoid in Rel.

The condition that $(X,m^+,e^+,m^-,e^-)$ is a relational bimonoid reduces to the following conditions on the original set-theoretic monoid (writing $m(x,y)$ as $x \cdot y$):

  1. $x\cdot y = x'\cdot y'$ if and only if there exist $t,u,v,w$ such that $x = t\cdot u$ and $y = v\cdot w$ and $x' = t\cdot v$ and $y' = u\cdot w$.
  2. If $x\cdot y = e$ then $x = e$ and $y = e$.

Question: Do monoids satisfying conditions (1) and (2) have a name?

Algebraic sum of relative commutants in a finite von Neumann algebra

Math Overflow Recent Questions - Wed, 05/02/2018 - 12:01

Let $(M,\tau)$ be a finite von Neumann algebra with faithful normal tracial state $\tau$. Suppose that $A \subset M$ is a finite-dimensional abelian unital subalgebra--say $A = W^*(p_1,\dots, p_n)$ where $p_1,\cdots, p_n \in M$ are mutually orthogonal projections with sum $1_M$--and $N \subset M$ is an arbitrary unital ($1_N = 1_M$) von Neumann subalgebra. Consider the relative commutants $A'\cap M$ and $N'\cap M$ and their respective $||\cdot||_2$-closures $\overline{A'\cap M}, \overline{N'\cap M} \subset L^2(M,\tau)$.

Under what conditions is the algebraic sum $\overline{A'\cap M} + \overline{N'\cap M}$ a closed linear subspace of $L^2(M,\tau)$?

Note: When $M$ is finite dimensional this sum is always closed. Also, recall that $A'\cap M = \sum_{i=1}^n p_iMp_i$.

Some references to consider are the following.

  1. Deutsch, Frank. The angle between subspaces of a Hilbert space. (English summary) Approximation theory, wavelets and applications (Maratea, 1994), 107–130, NATO Adv. Sci. Inst. Ser. C Math. Phys. Sci., 454, Kluwer Acad. Publ., Dordrecht, 1995. MathSciNet: MR1340886

  2. Luxemburg, W. A. J. A note on the sum of two closed linear subspaces. Nederl. Akad. Wetensch. Indag. Math. 47 (1985), no. 2, 235–242. MathSciNet: MR0799084

In [1], it is shown that two closed linear subspaces of a Hilbert space have a closed algebraic sum if and only if the cosine of the Friedrichs angle between them is strictly less than 1 (cf. Theorem 13). So an alternative approach to this question would be to compute the Friedrichs angle between these two relative commutant subspaces.

Reference [2] gives a number of functional analytic results on how to determine when the algebraic sum of two closed linear subspaces of a Banach space is closed.

Special case: A good start would be to resolve this question in the case where $N = uAu^*$ for a unitary $u \in M$.

Fast Algorithms for sum of independent random variables

Math Overflow Recent Questions - Wed, 05/02/2018 - 01:56

CLT implies the sum of n i.i.d random variables,after property normalized converge to a Normal distribution as n goes to infinity. Furthermore, Linderberg's condition points out not necessarily identically distributed ,the one sufficient condition for the sum converges to Normal distribution in the limit case. From a computational perspective, we want to know how numerically the sum converges to a Normal distribution, this is given by Berry-Esseen Theorem. Essentially, the theorem says the converges rate is determined by the 2,3 moments of those R.V . For convenience,we call this Berry-Esseen metric. B-E metric

With this theoretic results, a natural idea is when summing those R.V ,for some R.V we can well approximate their sums by a normal distribution ,and some 'exotic' r.v for simulation, hence reduce the computation complexity(simulation is time-consuming) .The question is how to build such 'cluster algorithms' Assuming we know 2 and 3rd moments of each R.V? --Noticed that ,unlike tradition Cluster algorithms in Machine learning, where they clustering is based on some 'distance',in this B-E metric, the expression is highly asymmetric for sigma and pho.

Approximating finite type algebras over a formal power series ring

Math Overflow Recent Questions - Tue, 05/01/2018 - 22:58

Let $k$ be a ring, let $A := k[x_{1},\dotsc,x_{d}]$ be the polynomial ring and let $A^{\wedge} := k[[x_{1},\dotsc,x_{d}]]$ be the formal power series ring. For a $d$-tuple $\mathbf{e} = (e_{1},\dotsc,e_{d}) \in \mathbb{Z}_{\ge 0}^{\oplus d}$ of nonnegative integers, let me denote $|\mathbf{e}| := e_{1} + \dotsb + e_{d}$ and $x^{\mathbf{e}} := x_{1}^{e_{1}} \dotsb x_{d}^{e_{d}}$. We may write any $a \in A^{\wedge}$ as \begin{align} \textstyle a = \sum_{\mathbf{e} \in \mathbb{Z}_{\ge 0}^{\oplus d}} c_{\mathbf{e}}x^{\mathbf{e}} \end{align} with $c_{\mathbf{e}} \in k$. For any nonnegative integer $\ell \ge 0$, let me denote \begin{align} \textstyle a^{\le \ell} := \sum_{\mathbf{e} \in \mathbb{Z}_{\ge 0}^{\oplus d} ; |\mathbf{e}| \le \ell} c_{\mathbf{e}}x^{\mathbf{e}} \end{align} the "truncation of $a$ at degree $\ell$".

To any polynomial $f \in A^{\wedge}[t_{1},\dotsc,t_{n}]$, we denote $f^{\le \ell} \in A[t_{1},\dotsc,t_{n}]$ the polynomial obtained by applying the truncation operation $(-)^{\le \ell}$ to the coefficients of $f$.

Let $f_{1},\dotsc,f_{m} \in A^{\wedge}[t_{1},\dotsc,t_{n}]$ be a collection of polynomials and set \begin{align} B &:= A^{\wedge}[t_{1},\dotsc,t_{n}]/(f_{1},\dotsc,f_{m}) \\ B^{\le \ell} &:= A^{\wedge}[t_{1},\dotsc,t_{n}]/(f_{1}^{\le \ell},\dotsc,f_{m}^{\le \ell}) \end{align} for all $\ell \ge 0$.

Is there a way to "approximate" $B$ (resp. $\operatorname{Spec}(B)$) by its "truncations" $B^{\le \ell}$ (resp. $\operatorname{Spec}(B^{\le \ell})$)? Is there a way to "approximate" the category of (e.g. finitely generated projective) modules $\mathrm{Mod}(B)$ by the $\mathrm{Mod}(B^{\le \ell})$?

How many rich directions a set in $\mathbb F_p^2$ determines?

Math Overflow Recent Questions - Tue, 05/01/2018 - 11:47

$\newcommand{\F}{\mathbb F}$ A subset $P$ of the affine plane $\F_p^2$ is said to determine a direction if there is a line in this direction containing at least two points of $P$.

A set of size $|P|>p$ determines all $p+1$ directions, a set of size $|P|\le p$ not contained in a single line determines at least $(|P|+3)/2$ directions; the former is an immediate consequence of the pigeonhole principle, the latter is a highly non-trivial result of Szonyi.

Let's say that a direction in $\F_p^2$ is rich if there is a line in this direction containing at least three points of $P$. If $P$ is trapped in a line, or a union of two lines, then there are just one or two rich directions.

What is the smallest possible number of rich directions for a set $P\subset\F_p^2$ of size $\frac53\,p<|P|\le 2p$ given that $P$ is not contained in a union of two lines?

It is easy to construct sets $P$ with about $|P|-p$ rich directions, but it is not clear to me how much better can one do.

The largest possible number of points in general position in $\F_p^2$ is $p+1$, which is quite easy to prove; thus, we are guaranteed to have at least one rich direction. Can the number of rich directions stay bounded as $p$ grows?

Understanding Quantum information theory

Math Overflow Recent Questions - Mon, 04/30/2018 - 13:15

The following theorem is well-known:

Let $(T(t))$ be a norm continuous semigroup on a Banach space, then its generator $A$ is bounded.

In Quantum mechanics one studies in the Heisenberg picture the following semigroup $T(t)(X)= e^{itH}X e^{-itH}$ on the Banach space of bounded operators.

The generator is found to be (modulo a sign maybe) $A=i[H,X]$ and so if $H$ is bounded, this generator is bounded and we know that the semigroup is norm continuous.

Quantum mechanists however are not only interested in positive dynamics, but rather completely positive dynamics. Therefore, they frequently use another norm, the so-called completely bounded norm

$$\left\lVert T(t) \right\rVert =\sup_{n \in \mathbb{N}} \left\lVert T(t) \otimes \operatorname{id}_{\mathbb C^n} \right\rVert.$$ Here is a reference for the predual norm on the trace-class operators click me.

Now this semigroup is only norm-continuous if the generator $A(X)=i[H,X]$ is completely bounded.

I ask: Under the assumption that $H$ is boundedIs it true that this generator is always completely bounded and the semigroup thus norm-continuous or are there counterexamples?

Is a local diffeomorphism with nice boundary values a diffeomorphism?

Math Overflow Recent Questions - Mon, 04/30/2018 - 13:01

Let $f:\mathbb{D}=\{z\in\mathbb{C}\mid |z|<1\}\rightarrow\mathbb{C}$ be a local diffeomorphism (i.e. an immersion) from an open disk in the plane to the plane.

The only situation I can image where $f$ is not injective is that $f$ sends $\mathbb{D}$ to a "self-overlapping'' region, in which case $f$ can not have continuous injective boundary values. But it seems non-trivial to proof that such boundary values guarantee injectivity:

Question. Assume that $f$ extends to a continuous map $\overline{\mathbb{D}}\rightarrow\mathbb{C}$ such that the boundary values $f|_{\partial\mathbb{D}}$ is injective and continuous, so that (by Jordan Curve Theorem) it maps $\partial\mathbb{D}$ homeomorphically to a Jordan curve which is the boundary of a simply connected domain $\Omega\subset\mathbb{C}$. Then it is true that $f$ is a homeomorphism from $\mathbb{D}$ to $\Omega$?

Eigenvalues of special sum of Hermitian matrices

Math Overflow Recent Questions - Mon, 04/30/2018 - 12:44

In my research on linear algebra and its applications, I have come across the following problem which has stumped me:

Let $ A $ be a positive definite matrix and let $ D $ be a positive diagonal matrix with entries on the main diagonal: $ d_1,d_2,...,d_n $, both $ A $ and $ D $ have the same dimension $ n \times n $. I was interested in understanding how the eigenvalues of the sum $ A + ADA $ qualitatively behave with respect to the eigenvalues of $ A $ and the entries $ d_1,...,d_n $.

I thought since this sum has a special form, one could hopefully say a bit more analytically than by exclusively using the fact that the two summands are Hermitian and commute. I thought about using some other techniques, perhaps expressing the product of $A$ and $ D $ as a polynomial of $ A $, but unfortunately, I am stuck. I certainly appreciate all help on this.

Fractional Sobolev spaces of order 0

Math Overflow Recent Questions - Mon, 04/30/2018 - 12:41

For $1\leq p <+\infty$, $0<s<1$ and $\Omega\subset R^n$ domain, the fractional Sobolev space $W^{s,p}$ is defined as

$$W^{s,p}(\Omega):=\big\{f \in L^p(\Omega)\colon \int_{\Omega} \int_{\Omega} \frac{|f(x)-f(y)|^p}{|x-y|^{s p + n}}dx dy<+\infty .\big\}$$

I wonder if this definition makes sense for $s=0$ for bounded $\Omega$, in particular, can one describe functions $f$ such that

$$\int_{K} \int_{K} \frac{|f(x)-f(y)|^p}{|x-y|^{n}}dx dy<+\infty$$

for any compact $K\subset R^n$?

Basis of coinvariant algebra on which reflection group acts as regular representation

Math Overflow Recent Questions - Mon, 04/30/2018 - 11:40

This question is almost a duplicate of a question of Christian Stump, except that Christian seems to ask about an isomorphism to irreducible representations rather than the regular representation: What does the regular representation of the coinvariant ring of a unitary reflection group look like?

If $W\subseteq GL(V)$ is a finite reflection group, then the Chevalley-Shephard-Todd theorem says that the invariant algebra $\mathbb{C}[V]^W$ is isomorphic to a polynomial algebra $\mathbb{C}[V]^W\simeq \mathbb{C}[e_1,e_2,\ldots,e_n]$ and that the coinvariant algebra $\mathbb{C}[V]^{\mathrm{co}W} := \mathbb{C}[V]/\mathbb{C}[V]^W_+$ (where $\mathbb{C}[V]^W_+$ is the set of invariant polynomials of positive degree) is isomorphic as a $W$-module to the left regular representation.

Question: can we write down some explicit basis of $\mathbb{C}[V]^{\mathrm{co}W}$ on which $W$ acts as the regular representation?

The only proofs I know of the fact that $\mathbb{C}[V]^{\mathrm{co}W}$ carries the regular representation use character computations which have a "non-constructive" flavor.

Really I am most interested just in the case of the symmetric group $W=S_n$.

Let me give a quick example of what this looks like for $W=S_2$. Then $\mathbb{C}[V] = \mathbb{C}[x_1,x_2]$ and we get $\mathbb{C}[V]^W = \mathbb{C}[x_1+x_2,x_1x_2]$ (these are the ``elementary symmetric polynomials''). (Maybe strictly speaking because $S_2$ acts on $\mathbb{R}^2/(1,1)$ I should write $\mathbb{C}[V] = \mathbb{C}[x_1,x_2]/\langle x_1+x_2 \rangle$ and $\mathbb{C}[V]^W =\mathbb{C}[x_1x_2]$ but I don't think this technicality matters.) At any rate we have that the coinvariant ring is $\mathbb{C}[V]^{\mathrm{co}W}=\mathbb{C}[x_1,x_2]/\langle x_1+x_2,x_1x_2 \rangle$. There are standard bases of the coinvariant ring for the symmetric group, like the staircase monomials or the Schubert polynomials. In this case both of those bases would be $\{x_1,1\}$ (note that those bases are homogeneous). But the symmetric group $S_2$ does not act on that basis as in the regular representation. Instead I would want a basis like $\{x_1+1,-x_1+1\}$.

Quotient of a motive by a finite group

Math Overflow Recent Questions - Mon, 04/30/2018 - 10:40

Given a smooth scheme $X$ over a field $k$, we can consider its motive $M(X)$. It is an object in Voevodsky's triangulated category of motives $DM(k,\Lambda)$ where $\Lambda$ is the ring of coefficients. It is constructed from the presheaf with transfers represented by $X$ by forcing $A^1$-invariances and invertibility of the Lefschetz motive. Now, if $X$ is acted on by a finite group in such a way that the categorical quotient $X/G$ exists in the category of smooth schemes, I can consider the motive $M(X/G)$. On the other hand, without any assumption on the action, I can construct a motive $M(X)_G$ by taking the homotopy quotient of $M(X)$ by $G$ in $DM(k,\Lambda)$. There is a slight abuse here as one needs to work at the point set level (using model categories or infinity-categories) in order to define this homotopy quotient. As far as I understand $M(X)_G$ is also the motive of the quotient stack $[X/G]$ Clearly there is a map $$M(X)_G\to M(X/G)$$ and I am wondering whether this map is an isomorphism. I think that it is the case when we work in $DM_{et}(k,\Lambda)$, the variant of $DM(k,\Lambda)$ that uses the étale topology instead of the Nisnevich topology because in that case the map $X\to X/G$ is an étale cover. Is it also true in $DM(k,\Lambda)$ ? And if it is not true in general are there conditions under which it is known to be true ?

Linear sections of Segre varieties and rational normal scrolls

Math Overflow Recent Questions - Mon, 04/30/2018 - 10:29

In a projective space $\mathbb{P}^{k+2}$ consider two complementary subspaces $\mathbb{P}^1,\mathbb{P}^k$, and let $C\subset\mathbb{P}^k$ be a degree $k$ rational normal curve. Fixed an isomorphism $\phi:\mathbb{P}^1\rightarrow C$ we consider the rational normal scroll

$$S_{(1,k)} = \bigcup_{p\in \mathbb{P}^1}\left\langle p, \phi(p)\right\rangle\subset \mathbb{P}^{k+2}$$

where $\left\langle p, \phi(p)\right\rangle$ is the line through $p$ and $\phi(p)$.

Now, let $\Sigma_{(1,k)}$ be the Segre embedding of $\mathbb{P}^1\times \mathbb{P}^k$ in $\mathbb{P}^{2k+1}$.

Can $S_{(1,k)}$ be recovered intersecting $\Sigma_{(1,k)}$ with a linear subspace of dimension $k+2$ of $\mathbb{P}^{2k+1}$? This is known to be true if $k = 2$. Indeed a general hyperplane section of $\Sigma_{(1,2)}\subset\mathbb{P}^5$ is a scroll surface of the form $S_{(1,2)}$.

Lower central series of nearly metabelian groups

Math Overflow Recent Questions - Mon, 04/30/2018 - 10:23

Let's say that $G \in \mathcal M_k$ if every $k$-generated subgroup of $G$ is metabelian. Obviously, $\mathcal M_{\geq 4} = \mathcal M_4 = \mathcal M$ is the variety of metabelian groups, but it's known that $\mathcal M \subsetneq \mathcal M_3 \subsetneq \mathcal M_2$. It's pretty easy to see that $\mathcal M_2$ is generated by single word $[[a, b], [a^{-1}, b]]$.

Associated lower central quotients of free metabelian groups are known to be free abelian of polynomial growth (exactly, for rank $r$ $n$-th lower central quotient has dimension $(n-1) {r+n-2 \choose n}$).

Question 1. Are there any results about lower central quotients of free groups in varieties $\mathcal M_k$, at least rationally?

Question 2. How well IA-automorphisms of free $\mathcal M_k$-groups lift to free groups? More generally, automorphisms inducing identity on $\gamma_k$?


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