Questions tagged [heisenberg-groups]

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Polynomials of growth for finite Heisenberg groups

Take a standard finite Heisenberg group with two standard generators and let's consider its growth polynomial - the polynomial which coefficients are equal to the sphere sizes. For example for $H_3(Z/...
Mikhail Evseev's user avatar
2 votes
0 answers
53 views

Distance distribution for Cayley graphs of the fintie Heisenberg groups H3(Z/nZ) approaches Gaussian for large "n"?

I wonder several questions about Cayley graphs of finite Heisenberg groups H3(Z/nZ). Question 1: do we know the diameter dependence on "n", at least for the standard choice of generators ? ...
Alexander Chervov's user avatar
1 vote
0 answers
134 views

Good references to understand sub-Riemannian geometry and Heisenberg groups

I'm looking for books and articles to understand a little about the Heisenberg group and sub-Riemannian geometry, specifically why the Heisenberg group is an example of a sub-Riemannian manifold, and ...
Ilovemath's user avatar
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1 answer
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Action of the Haar measure on the Heisenberg group

The Heisenberg group is $\mathbb{H}^N=\mathbb{R}^{2N+1} = \left\{ (x,y, \tau ) \right\}\in \mathbb{R}^N \times \mathbb{R}^N \times \mathbb{R}$ equipped with the group operation \begin{equation} (...
Ilovemath's user avatar
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1 vote
0 answers
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A question about semigroups in a Heisenberg group

I'm trying to understand if the regularity of solutions in Heisenberg groups works like in the Euclidean case. So far I haven't found any results, so I'm trying to check if the Regularity Theorems ...
Ilovemath's user avatar
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0 answers
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regularity theory of parabolic equations in Heisenberg group

I'm trying to understand if there are regularity results for mild solutions of partial differential equations in Heisenberg group. In this paper (Theorem 1.3 (iii) and proof of Theorem 1.1) the author ...
Ilovemath's user avatar
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0 answers
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Are there results of parabolic regularity in Heisenberg groups?

The $(2N +1)-$dimensional Heisenberg group $\mathbb{H}^N$ is the space $\mathbb{R}^{2N+1} = \left\{ (x,y, \tau ) \right\}\in \mathbb{R}^N \times \mathbb{R}^N \times \mathbb{R}$ equipped with the ...
Ilovemath's user avatar
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Sub Laplacian on the quaternion Heisenberg group $\mathbb{H}$

The sublaplacian is defined by $\mathcal{L}=-\left(X_{0}^{2}+X_{1}^{2}+X_{2}^{2}+X_{3}^{2}\right)$, which is independent of the choice of the orthonormal basis of $\mathbb{H}$. It is well known that ...
zoran  Vicovic's user avatar
0 votes
0 answers
149 views

Heisenberg group

Let $X_{j}=\frac{\partial}{\partial x_{j}}-\frac{1}{2}y_{j}\frac{\partial}{\partial t}$, $j=1,2,\dots,n$ $Y_{j}=\frac{\partial}{\partial y_{j}}+\frac{1}{2}x_{j}\frac{\partial}{\partial t},j=1,2,\dots,...
zoran  Vicovic's user avatar
2 votes
0 answers
190 views

Schrödinger representation of the Heisenberg group

Let $\Pi_{\lambda}$ be the the Schrödinger representations of the Heisenberg group $H^n=\Bbb C^n\times\Bbb R$. For $\phi\in L^2(\Bbb R^n)$, we have $$\Pi_{\lambda} (x,y,t)\phi(\xi)=e^{i\lambda t} e^{...
zoran  Vicovic's user avatar
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0 answers
118 views

Finite dimensional unitary representations of the discrete Heisenberg group

Let $H(\mathbb{Z})$ be the discrete Heisenberg group. What are the finite dimensional irreducible unitary representations of $H(\mathbb{Z})$? Do they all arise from the coordinate-wise quotient map to ...
Serge the Toaster's user avatar
1 vote
0 answers
289 views

Classification of the normal subgroups of the discrete Heisenberg group

Let $H$ be the discrete Heisenberg group, i.e., the set of matrices of the form $\begin{bmatrix} 1 & x & z \\ 0 & 1 & y \\ 0 & 0 & 1 \end{bmatrix}$ where $x,y,z \in \mathbb{Z}$...
Serge the Toaster's user avatar
1 vote
0 answers
438 views

Heat kernel on quaternion Heisenberg group

For the n Heisenberg($\Bbb C^n\times\Bbb R$) it is known that the heat kernel $q_s(z,t)=c_n\int_{\Bbb R} e^{-i\lambda t}\Big( \frac{\lambda}{\sinh(\lambda s)}\Big)^n e^{-\frac{\lambda|z|^2\coth(\...
user484672's user avatar
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80 views

Two pairings on the group $K(L)$ associated with a non-degenerate line bundle $L$ on an abelian variety

Let $A$ be an abelian variety over a field and let $L$ be a non-degenerate line bundle on $A$. Then $L$ gives rise to a morphism $\lambda:A\to A^*$ from $A$ to its dual. As usual, let $K(L):=\ker(\...
Damian Rössler's user avatar
7 votes
1 answer
324 views

What is the analogue of the Jacobi theta function in the Weyl representation?

It is known (see for example the associated Wikipedia entry) that the Jacobi theta function $$\vartheta(z; \tau) = \sum_{n\in\mathbb{Z}} \exp(\pi in^2\tau + 2\pi inz)$$ arises from a certain ...
Michael Barz's user avatar
1 vote
1 answer
165 views

Ratio limit results for restricted partition functions

This concerns difference/limit ratio results for special restricted partitions. Let $r,a, b$ be nonnegative integers; define $p(r,a,b)$ to be the number of partitions of the integer $r$ using at most $...
David Handelman's user avatar
1 vote
0 answers
316 views

Canonical commutation relations-bounded vs. unbounded picture

Suppose that $Q,P$ are self-adjoint operators which satisfy the relation $$(1) \ \ \ \ \ [Q,P]=iI$$ One can easily show that in this case $P,Q$ cannot be bounded. However one can find unbounded ...
truebaran's user avatar
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2 votes
1 answer
192 views

Semidirect product of metaplectic group and Heisenberg group

I know that Symplectic group has an action on Heisenberg group. I am wondering how to extend this to non-trivial two fold metaplectic covering? Thanks in advance!
Monty's user avatar
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20 votes
1 answer
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Carnot-Carathéodory metric

The metric in sub-Riemannian geometry is often called the Carnot-Carathéodory metric. Question 1. What is the origin of this name? Who was the first to introduce it? I believe that the "...
Piotr Hajlasz's user avatar
4 votes
1 answer
366 views

Heisenberg groups, Carnot groups and contact forms

The horizontal distribution in the Heisenberg group is the kernel of the standard contact form: $$ \alpha = dt + 2 \sum_{j=1}^n (x_j \, dy_j - y_j \, dx_j). $$ Question. Can one describe ...
Piotr Hajlasz's user avatar
1 vote
0 answers
55 views

Explicit Quasisymmetric embedding into Euclidean space

It is known that every doubling metric space admits quasisymmetric map into Euclidean space. My question is, is there a known explicit (closed-form) quasisymmetry from the Heisenberg group into a ...
ABIM's user avatar
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8 votes
1 answer
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The extension class of a finite Heisenberg group

Let $\mathbb{K}$ be a field of characteristic $\neq 2$ and let $(V, \omega)$ be a symplectic vector space. Then the Heisenberg group $\mathsf{Heis}(V, \, \omega)$ is the central extension of the ...
Francesco Polizzi's user avatar
5 votes
0 answers
196 views

Contact geometry: approximation of Legendrian mappings

Let $\alpha$ be a standard contact form on $\mathbb{R}^{2n+1}$. We say that a map $f:\mathbb{R}^k\to\mathbb{R}^{2n+1}$ contact if $f^*\alpha=0$. Question 1. Is it true that a $C^1$-contact ...
Piotr Hajlasz's user avatar
4 votes
1 answer
213 views

Legendre's symbol in Schrödinger model for the Weil representation

I have a question concerning the Schrödinger model for the Weil representation over a finite field $\mathbb{F}_q$. The way to present the action of the Weil representation $\omega$ of $Sp(2n,\...
Jesua Israel Epequin Chavez's user avatar
11 votes
3 answers
739 views

Non embedding of the Heisenberg group

It is well known that Heisenberg groups cannot be bi-Lipschitz embedded into Euclidean spaces. A standard proof uses the fact that a Lipschitz mapping from a Heisenberg group into a Euclidean space is ...
Piotr Hajlasz's user avatar
10 votes
0 answers
402 views

Lipschitz homotopy groups

There is an extensive literature on Lipschitz homotopies of Lipschitz maps. But I haven't seen anything about Lipschitz homotopy groups. We have introduced this notion in an article that you can find ...
Piotr Hajlasz's user avatar
5 votes
0 answers
130 views

Modules of algebras with idempotents and the Stone-von Neumann theorem

The Stone-von Neumann theorem tells us that all unitary irreducible representations of the integrated/exponentiated/Weyl form of the canonical commutation relations (CCR) algebra in finite dimensions ...
Carlos's user avatar
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1 vote
0 answers
95 views

Factoring in discrete Heisenberg group $H_3(\mathbb{Z})$

Let $H_3(\mathbb{Z})$ be the discrete Heisenberg group generated by $x=\begin{pmatrix} 1 & 1 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\\ \end{pmatrix},\ \ y=\begin{pmatrix} 1 & 0 &...
user avatar
3 votes
0 answers
127 views

About the purpose of introducing '"groups of Heisenberg type"

I would like to know, can we say that the "groups of Heisenberg type" where introduced by A. Kaplan in "Kaplan, A. (1980). Fundamental solutions for a class of hypoelliptic PDE generated by ...
Z. Alfata's user avatar
  • 640
1 vote
0 answers
182 views

Unitary dual of the Heisenberg group over non-archimedean local fields of characteristic two

What is the unitary dual of the Heisenberg group over non-archimedean local fields k of characteristic two? This is well-known for the real Heisenberg group and in fact, when local fields have ...
m07kl's user avatar
  • 1,672
10 votes
1 answer
552 views

Non-isomorphic Heisenberg groups over rings

Suppose $R_1,R_2$ are finite unital commutative rings. Consider Heisenberg groups $H_3(R_1)$ and $H_3(R_2)$ (upper unitriangular marticies $3 \times 3$). Proposition. If $R_1 \not\cong R_2$ (as ...
user35603's user avatar
  • 411
1 vote
0 answers
174 views

Why the sub-Laplacian $\Delta_{sub}$ on the Heisenberg group $H^3 $ is sub-elliptic but not elliptic?

I want to know why the sub-Laplacian $\Delta_{sub}= X^2 + Y^2$ on the Heisenberg group $H^3 = \mathbb C \times \mathbb R$ is sub-elliptic but not elliptic, where $X$ and $Y$ are the left-invariant ...
Z. Alfata's user avatar
  • 640
-1 votes
1 answer
311 views

Representations of the $3\times 3$ Heisenberg group [closed]

I am trying to understand how the Heisenberg group is defined because I would like to understand the (irreducible) representations. Following this article given a symplectic bilinear form $\langle, \...
Thomas's user avatar
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6 votes
1 answer
2k views

Difference between the Laplacian and the sub-Laplacian of a Lie group

Given a Lie group $G$, what is the difference between the Laplacian $\Delta$ and the sub-Laplacian $\Delta_{sub}$ of $G$. And what are the properties that we lose when going from sub-Laplace to ...
Z. Alfata's user avatar
  • 640
6 votes
0 answers
263 views

How to prove that the Laplace oparator of the Heisenberg group is sub-elliptic?

The Heisenberg group $H^3$ is the set $\mathbb C\times \mathbb R$ endowed with the group law $$ (z,t)\cdot(w,s) =\left (z+w, \,t+s+\tfrac{1}{2}\Im m(z \bar{w})\right). $$ For $z=x+ i y \in \mathbb C$ ...
Z. Alfata's user avatar
  • 640
5 votes
3 answers
531 views

Show that the Laplacian operator on the Heisenberg group is negative

The Heisenberg group $H^3$ is the set $\mathbb C\times \mathbb R$ endowed with the group law $$ (z,t)\cdot(w,s) =\left (z+w, \,t+s+\tfrac{1}{2}\Im m(z \bar{w})\right). $$ For $z=x+ i y \in \mathbb C$ ...
Z. Alfata's user avatar
4 votes
1 answer
217 views

dirichlet problem in the heisenberg group

Good morning everybody. I was looking just for a quick reference to know whether the Dirichlet problem has a solution in the Heisenberg group, that is $\mathbb R^3$ endowed with coordinates $(x,y,z)$ ...
guido giuliani's user avatar
4 votes
1 answer
352 views

Horizontal Sobolev space on Carnot group

This question is connected with my previous: Heisenberg group: function without vertical derivative. Here I am trying to look from another side: what is a difference between Sobolev space and ...
Nikita Evseev's user avatar
8 votes
0 answers
415 views

Heisenberg group: function without vertical derivative

Let $\mathbb H$ be Heisenberg group with vector fields $$ X=\partial_x - \frac12y\partial_t,\quad Y=\partial_y + \frac12x\partial_t,\quad T=\partial_t $$ and $U\subset\mathbb H$ is an open set. I am ...
Nikita Evseev's user avatar
4 votes
1 answer
1k views

Irreducible representation of Heisenberg group with characteristic 2?

As we all know that the irreducible representation for Heisenberg group can be classified easily when the group is over a finite field $\mathbb{F}_q$, where $q=p^n$ and $p$ is a prime greater than $2$....
Misaka01034's user avatar
2 votes
1 answer
249 views

totally geodesic submanifold of Heisenberg group

Let $G= \left\{ \begin{pmatrix} 1&a&c\\0&1&b\\0&0&0 \end{pmatrix} \mid a,b,c\in \mathbb{R} \right\}$ be the Heisenberg group. Is there a compact codimension one submanifold ...
Ali Taghavi's user avatar
6 votes
0 answers
164 views

Joint representation of the semi-direct product of the metaplectic group and Heisenberg group

Given a symplectic space $W$ over a local field $F$ and a additive character $\psi$ of $F$, we can construct the Weil representation $\omega_\psi$, which can be viewed as a representation of the semi-...
Math student's user avatar
3 votes
2 answers
402 views

Hesse pencil and Schrodinger representation of Heisenberg group

Let $E$ be a smooth elliptic curve over an algebraically closed field of characteristic zero. Let $\mathcal{L}$ be a line bundle of degree $3$. Heisenberg group $H_3$ acts on global sections of $\...
Sasha Pavlov's user avatar
  • 1,535
2 votes
0 answers
284 views

Finite Heisenberg groups action on cohomology of line bundles

Let $E$ be a smooth elliptic curve over algebraically closed field $k$ of characteristic zero, $\mathcal{L}$ is a line bundle over $E$, $\operatorname{deg}(\mathcal{L})=n \geq 1$. Then I define the ...
Sasha Pavlov's user avatar
  • 1,535
3 votes
0 answers
469 views

Harmonic analysis on the Heisenberg group

It is well known that: Theorem 1. For $f\in L^{2}(\mathbb H_{n}=\text{The Heisenberg group of dimesion } 2n+1)$ we have the expansion $$f(z, s)= (2\pi)^{-n} \sum_{k=0}^{\infty} \int_{0}^{\infty} f \...
Inquisitive's user avatar
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