text:
\begin{document}

\preprint{}

\title[Reconstruction of Hamiltonians from  given time evolutions]{Reconstruction of Hamiltonians from  given time evolutions}

\author{J. Bernatska}
\email{BernatskaJM@ukma.kiev.ua}
\affiliation{
National University of `Kiev-Mohyla Academy', 2 Skovorody Str., 04070 Kiev, Ukraine}
\altaffiliation[Also at ]{Bogolyubov Institute for Theoretical Physics.}

\author{A. Messina}
\email{messina@fisica.unipa.it}
\affiliation{Dipartimento di Fisica, Universit\`{a} di Palermo, Via Archirafi 36,
90123 Palermo, Italy}

\date{\today}

\begin{abstract}
In this paper we propose a systematic method to solve the inverse
dynamical problem for a finite-level quantum system governed by the von Neumann
equation: to find a class of Hamiltonians reproducing a
prescribed time evolution of a pure or mixed state of the system.
Our approach exploits the equivalence between the action of the
group of evolution operators over the state space and the adjoint
action of the unitary group over Hermitian matrices. The method is
illustrated by two examples involving a pure and a mixed state.
\end{abstract}

\pacs{02.20.Qs, 03.65.Aa}

\keywords{pure and mixed states, von Neuman equation, unitary group, adjoint action, stereographic parametrization}

\maketitle

\section{Introduction}
The problem of reconstructing a Hamiltonian or a Lagrangian from
observations on a physical system is a fundamental issue in physics.
Its history can be traced back to the time of Kepler and Newton when
the law of gravitation was discovered from observations of the
planetary motion. The reconstruction of an effective potential from
the data of particle scattering also belongs to this kind of problems.
In the context of quantum control theory
reconstructing Hamiltonians on demand represents a strategic intermediate step
toward the realization of noise-protected control
protocols governing the dynamics of a given, often
complex and multipartite, quantum system.

To drive the time evolution of a quantum system at will is at the
heart of the quantum control theory. Fruitful applications of this
theory in physical chemistry allow to achieve the optical control
over ultrafast molecular processes \cite{bonacic}, and to design
feedback control for quantum state preparations in quantum optics
\cite{handel}. A specific task of the quantum control theory is to
contrive schemes that allow to steer a quantum system to a desired
state starting from a given initial one. More ambitiously one could
even require a possibility to control an evolution of the system
under scrutiny at any instant of time. In other words, one could
require to design the control in such a way that the system evolves
along a prefixed path in its state space. Being
able to solve this problem, a dream since the early days of quantum
mechanics, would help to highlight dynamical
behaviors of microscopic systems, and would  provide additional tools and ideas applicable to quantum
computation. In a recently published paper \cite{Leggio} the
tomographic method has been used to show that a specially modeled
time-dependent state of two qutrits, violating Bell's inequalities
at $t\,{=}\,0$, loses this non-classical feature at successive
times. In this paper we introduce a systematic
method to solve the following inverse problem of the unitary quantum
dynamics: given the time evolution of a system is explicitly specified
by means of a density matrix $\rho$, to construct a class of generally time-dependent Hamiltonian
models whose associated von Neumann equations are satisfied by
this~$\rho$, describing a pure or a mixed quantum
state.

The paper is organized as follows. In Section 2 mathematical
notations are introduced. In Section 3 we explain our mathematical
procedure: how to reconstruct the
Hamiltonian reproducing the prescribed
time evolution, if this evolution obeys the von Neumann
equation. Every time evolution is realized by a class of Hamiltonian models.
We illustrate our method in
Section 4 where it is applied to time evolutions of both a pure
state and a mixed one.

\section{Mathematical notation}
To establish our notation we start with brief recalling the mathematical
grounds of quantum mechanics after J. von Neumann\cite{Neumann,Emch}.
A quantum system is generally described by three basic ingredients: states, observables,
and dynamics (the law of time evolution). Very often the state space is identified with a complex Hilbert space~$\Hilbert$.
In what follows we consider finite-level systems, so the Hilbert space is finite dimensional. After Dirac we denote its vectors  by $|\Phi\rangle$, $|\Psi\rangle$, \ldots, and an inner product by~$\langle \Phi | \Psi \rangle$.

But the Hilbert space  represents only pure states.
Every vector $|\Phi \rangle\,{\in}\,\Hilbert$ gives rise to the ray (the one-dimensional subspace) $\{\lambda |\Phi \rangle \,{\mid}\, \lambda\,{\in}\,\Complex\}$, associated with one pure state, say $\varphi$. In other words, a pure state is identified with an equivalence class of vectors of length~1 in~$\Hilbert$, and two vectors represent the same pure state if they differ only by a phase factor.

If we want to include mixed states in our pattern, it is better to represent  the pure state
$\varphi$ as a  projection operator $P_{\Phi}$ onto
the ray $\{\lambda |\Phi \rangle \,{\mid}\,\lambda\,{\in}\,\Complex\}$.  In~Dirac's notation
the operator $P_{\Phi}$ is expressed as $|\Phi
\rangle \langle \Phi|$. Every such  projector $P_{\Phi}$, and so every pure state $\varphi$,  is a self-adjoint linear transformation on the Hilbert space such that  $P_{\Phi}^2\,{=}\,P_{\Phi}$. The~collection of all pure states form a projective space.
A `statistical mix' of pure states \begin{equation*}
\rho = \sum_k p_k P_{\Phi_k},
\end{equation*}
where $0\,{\leqslant}\, p_k \,{<}\, 1$ and $\sum_k p_k
\,{=}\, 1$, represents a mixed state. The operator $\rho$  is not a projector, but a self-adjoint positive definite linear transformation. Therefore the state space of a quantum system can be identified with the subspace of positive definite matrices  in the space $\mathfrak{A}(\Hilbert)$ of self-adjoint linear transformations on~$\Hilbert$. In order to identify a quantum state with its density matrix we restrict the state space to matrices  of trace 1.

We introduce a scalar product in $\mathfrak{A}(\Hilbert)$, and the dual space to $\mathfrak{A}(\Hilbert)$ as follows:
\begin{equation*}
(\rho, A) \equiv \Tr \rho A,\qquad \rho \in \mathfrak{A}(\mathcal{H}),\  \  A\in \mathfrak{A}^{\ast}(\mathcal{H}).
\end{equation*}
The dual space $\mathfrak{A}^{\ast}(\mathcal{H})$ represents the associative  algebra of observables\footnote{Associating the space $\mathfrak{A}(\Hilbert)$ with quantum states and the dual space $\mathfrak{A}^{\ast}(\mathcal{H})$ with observables we are guided by their laws of time evolution. Namely, the evolution of a state $\rho$ is realized by the adjoint action $U_{t,t_0} \rho(t_0) U_{t,t_0}^{-1}$, and the evolution of an observable $A$ is realized by the coadjoint action $U_{t,t_0}^{-1} A(t_0) U_{t,t_0}$.}
of a quantum system.  The expectation value of an observable $A$
with respect to a state $\rho$ is defined by the formula
\begin{equation*}
\langle A \rangle_{\rho} = (\rho,A) \equiv \Tr \rho A.
\end{equation*}
Then for a pure state $\varphi\,{=}\,P_{\Phi}$ we have
\begin{equation*}
\langle A \rangle_{\varphi} = \Tr P_{\Phi} A =  \langle \Phi  \,{\mid}\, A \Phi \rangle.
\end{equation*}

The last ingredient of a quantum system is its dynamics, which is governed
by the well-known Schr\"{o}dinger equation
\begin{equation*}
i\hbar \frac{d |\Phi \rangle }{dt}= H |\Phi \rangle,\qquad  |\Phi \rangle \in \Hilbert,
\end{equation*}
if we deal with the pure state associated with the vector~$|\Phi\rangle$; or in general by
 the von Neumann equation
\begin{equation*}
i\hbar \frac{d \rho}{dt}  = [H,\rho],\qquad \rho \in  \mathfrak{A}(\mathcal{H}),
\end{equation*}
where $[\cdot,\cdot]$ denotes a commutator. The self-adjoint operator $H\,{\in}\,\mathfrak{A}^{\ast}(\mathcal{H})$
is called a \emph{Hamiltonian}.

According to Stone's theorem the time evolution generated by a time-independent self-adjoint operator $H$ is realized by
unitary operators $\{U_t\}$, forming a continuous one-parameter group in $t$, that is $U_{t}U_{s}\,{=}\,U_{t+s}$ for all $t,\,s\,{\in}\, \Real$ and $\lim_{t\to 0} U_t \,{\to}\, \Ibb$. Then the  Schr\"{o}dinger equation with an initial condition at $t_0$ is solved by the formula $|\Phi (t)\rangle \,{=}\, U_{t-t_0} |\Phi(t_0)\rangle$.  In the case of  a time-dependent $H$ time evolution is also realized by a continuous group of unitary operators\cite{Messiah,Woodhouse} $\{U_{t,t_0}\}$ such that $U_{t_2,t_1} U_{t_1,t_0}\,{=}\,U_{t_2,t_0}$, $U_{t_0,t_0}\,{=}\,\Ibb$; and the  Schr\"{o}dinger equation is solved by  $|\Phi (t)\rangle \,{=}\, U_{t,t_0} |\Phi(t_0)\rangle$.

Hereinafter we deal with a more general pattern, where the state space is identified with  the subspace $\mathcal{P}$ of positive definite matrices of trace 1 in the space $\mathfrak{A}(\Hilbert)$, and dynamics is governed by the von Neumann equation. Time evolution is realized by the group of unitary operators $\{U_{t,t_0}\}$ as follows
\begin{equation}\label{EvolDM}
\rho(t) = U_{t,t_0} \rho(t_0) U_{t,t_0}^{-1}.
\end{equation}
The Hamiltonian generating this evolution relates to the evolution operator $ U_{t,t_0}$
according to the formula
\begin{equation}\label{HamDef}
H(t)\,{=}\, i\hbar \frac{d U_{t,t_0}}{dt}\, U_{t,t_0}^{-1},
\end{equation}
which can be obtained by substituting the evolution of
$|\Phi\rangle$, or $\rho$, into the corresponding differential equation.

The mentioned well-known facts of quantum mechanics are naturally interpreted
from the position of the Lie group and algebra theory.
Evidently, time evolution of a quantum system is realized by the adjoint action \eqref{EvolDM}
of a unitary group. This gives us a hint how to treat the problem.
Suppose we deal with a $n$-level system, then the state space
can be identified with the positive definite subspace $\mathcal{P}$ of the affine linear space $i\mathfrak{su}(n)\,{\oplus}\,(1/n)\hat{\Ibb}_n$.  In our context the latter space can be replaced by the algebra $i\mathfrak{su}(n)$, what
 we show in the next section.
Therefore, from the mathematical point of view we deal with the adjoint action of a Lie group over its Lie algebra.

Moreover, it is easy to check that $\Tr
\rho^k$ remains invariant under time evolution for any
$k\,{\geqslant}\, 1$:
\begin{equation}\label{OrbitEqs}
\Tr \rho^k(t)\,{=}\,\Tr \rho^k\,(0) = \text{const},
\end{equation}
where only first $n\,{-}\,1$ equations are functionally independent for a $n$-level system.
In other words,
the adjoint action \eqref{EvolDM} divides the state space into non-intersecting orbits, and
every orbit is determined by the set of equations \eqref{OrbitEqs}.
This clarifies the geometry of  time evolution of a closed quantum system.
Every quantum state evolves within a single orbit, fixed by an initial state. It means:
only states within this orbit are reachable from the initial state.

\section{Scheme of reconstruction for $n$-level systems}
We aim at solving the problem of reconstructing a Hamiltonian
model from a given time evolution
satisfying the von Neumann equation.
Actually we reconstruct an evolution operator, which is
considered as an element\footnote{More precise, the proposed method of stereographic parametrization allows to
reconstruct an element of a coset space representing an adjoint orbit.} of a unitary group. For this
purpose we use  a generalized stereographic parametrization\cite{Skrypnyk, BernatskaGIQ}, applicable to
states with finite number of components, say $n$.  Therefore in
what follows we deal with matrices of size $n\,{\times}\, n$, and
mark them by the hat sign $\hat{\ }$.

\subsection{An orbit structure of the state space}\label{ss:OrbitStruct}
First of all, we show an equivalence between the described above physical and mathematical patterns.
Let us consider the algebra $i\mathfrak{su}(n)$ consisting of Hermitian traceless matrices. We denote
these matrices by $\hat{\mu}$, keeping $\hat{\rho}$ for matrices of trace~1. The
adjoint action of the unitary group $\mathrm{SU}(n)$ over  $i\mathfrak{su}(n)$ is naturally defined by
\begin{equation}\label{AdjActionUG}
\hat{\mu}\,{=}\,\hat{u} \hat{\mu}_{\rm in} \hat{u}^{-1}, \qquad
\hat{\mu}\,{\in}\,i\mathfrak{su}(n), \quad \hat{u}\,{\in}\,\mathrm{SU}(n).
\end{equation}
All $\mu$ obtained from the initial point $\hat{\mu}_{\rm in}$ when $\hat{u}$ runs the group  $\mathrm{SU}(n)$
form an orbit.  This orbit can be determined by the set of equations
\begin{equation}\label{OrbitEqsAlg}
\Tr \hat{\mu}^k\,{=}\, \Tr \hat{\mu}_{\text{in}}^k = \text{const},\quad  k=1,\,\dots\, n-1.
\end{equation}
The whole algebra $i\mathfrak{su}(n)$ is devided  into non-intersecting orbits by the adjoint action of $\mathrm{SU}(n)$.

It is easy to check that the adjoint action over $i\mathfrak{su}(n)$ in \eqref{AdjActionUG} and the time evolution over the state space~$\mathcal{P}$ in  \eqref{EvolDM} are realized by the same unitary group. It means that for a $n$-level quantum system the group of unitary operators $\{U_{t,t_0}\}$ is $\mathrm{SU}(n)$.
Also the algebra $i\mathfrak{su}(n)$ has the same orbit structure as the affine space $i\mathfrak{su}(n)\,{\oplus}\, (1/n)\hat{\Ibb}_n$ containig the state space of the system. Indeed, every orbit in $i\mathfrak{su}(n)$ determined by \eqref{OrbitEqsAlg} corresponds to a unique orbit \eqref{OrbitEqs} in $i\mathfrak{su}(n)\,{\oplus}\, (1/n)\hat{\Ibb}_n$. Therefore instead of the affine space we consider the Lie algebra $i\mathfrak{su}(n)$. Also we shift the state space~$\mathcal{P}$ into the corresponding subspace $\mathcal{M}\,{\subset}\,i\mathfrak{su}(n)$ by subtracting the matrix $(1/n)\hat{\Ibb}_n$.

Now we can use the knowledge of the orbit structure of a Lie algebra under the adjoint action of its Lie group in order to
describe the structure of the state space. Above we called this structure the geometry of time evolution.
Every orbit in $i\mathfrak{su}(n)$ is specified by its initial point taken from a fundamental Weyl chamber or its walls. The Weyl chamber is a subset of the Cartan subalgebra $\mathfrak{h}$ of $i\mathfrak{su}(n)$, which is defined as
the maximal Abelian subalgebra of a semisimple Lie algebra. For $i\mathfrak{su}(n)$ it consists of real diagonal matrices of zero trace. Thus a canonical form of $\hat{\mu}\,{\in}\,i\mathfrak{su}(n)$ can serve as an initial point for the orbit where $\hat{\mu}$ is located. But there is an ambiguity upon the order of diagonal entries. This  ambiguity is removed by introducing the Weyl group, which performs a permutation of the diagonal entries, and whose fundamental domain is called a fundamental Weyl chamber.

Here we choose the positive Weyl chamber as a fundamental one. Its interior is the open domain $$C = \{\hat{\mu} \in \mathfrak{h} \mid \langle \hat{\mu},\alpha \rangle >0, \, \forall \alpha \in \Delta^{+}\},$$ where $\Delta^{+}$ denotes the set of positive simple roots.
A wall of the positive Weyl chamber is the set  $$\Gamma_{\alpha} \,{=}\, \{\hat{\mu}\,{\in}\,
\mathfrak{h} \mid \langle \hat{\mu},\alpha \rangle = 0\}, \quad \alpha \in \Delta^{+}$$  or any intersection of such sets.

We omit the accurate definition of simple roots, instead we represent them in the standard for the algebra $i\mathfrak{su}(n)$ form. The simple roots belong to the dual subalgebra $\mathfrak{h}^{\ast}$ and form a basis in it. Because $\mathfrak{h}^{\ast}$ is equivalent to $\mathfrak{h}$ we can use the same representation. The set~$\Delta^{+}$ consists of $n\,{-}\,1$ diagonal matrices $\{\hat{\alpha}_{k}\,{=}\,\hat{E}_{kk}-\hat{E}_{k+1,k+1}\}$, where the matrix $\hat{E}_{jk}$ has the only nonzero entry in row~$j$ column~$k$ and this entry equals~1.
 It is easy to construct the dual basis $\{\hat{\gamma}^{(k)}\}\,{\subset}\,\mathfrak{h}$:
\begin{equation*}
\hat{\gamma}^{(k)} = \sum_{j=1}^k \hat{E}_{jj} - (k/n)\hat{\Ibb}_n.
\end{equation*}
Then every point of the positive Weyl chamber of  $i\mathfrak{su}(n)$ can be represented in the form
\begin{equation*}
\hat{\mu}_{{\rm in}}  = \sum_{k=1}^{n-1} b_k  \hat{\gamma}^{(k)}, \quad b_k \in \Real^{+}.
\end{equation*}
If we allow the coefficients $\{b_k\}$ to equal zero, we obtain the closed  positive Weyl chamber containig all its walls. Every wall spans a nontrivial subset of $\{\hat{\gamma}^{(k)}\}$, and  is represented by a hyperplane bounding the chamber.

Points from the interior of a fundamental Weyl chamber in $i\mathfrak{su}(n)$
give rise to generic orbits of dimension $n^2\,{-}\,n$. Every point of a wall of the Weyl chamber gives rise to a
degenerate orbit of lower dimension. In the case of a one-dimensional wall one obtains a maximal
degenerate orbit of dimension $2(n\,{-}\,1)$. The latter represents a set of pure states if being shifted back into the space $i\mathfrak{su}(n)\,{\oplus}\, (1/n)\hat{\Ibb}_n$ remains a projective space.

In particular,  the points  $b_1\hat{\gamma}^{(1)}$ and $b_{n-1}\hat{\gamma}^{(n-1)}$ give rise to maximal degenerate orbits in $\mathcal{M}$ if $b_1\,{\leqslant}\,1$, $b_{n-1}\,{\leqslant}\,1/(n\,{-}\,1)$. But only one of them, namely $\hat{\gamma}^{(1)}$, generates the set of pure states. Positive definiteness of the state space $\mathcal{P}$ implies that acceptable values of $\{b_k\}$ satisfy the inequalities
\begin{equation*}
n\sum_{k=j}^{n-1} b_k  -  \sum_{k=1}^{n-1} k b_k + 1  \geqslant 0,\quad  j=1,\dots, n.
\end{equation*}
The last inequality ($j\,{=}\,n$) appears to be sufficient  if one takes into account non-negativity of $\{b_k\}$.

Obviously, the elements of the basis $\{\hat{\gamma}^{(k)}\}$, except $\hat{\gamma}^{(1)}$, do not belong to the subspace~$\mathcal{M}$. We are able to construct a basis  in $\mathfrak{h}$ fulfiling this condition. This new basis consists of  traceless real diagonal matrices~$\{\hat{\pi}^{(k)}\}$ such that $(\hat{\pi}^{(k)})^2 \,{-}\, \hat{\pi}^{(k)}\,{=}\,0$.   It is easy to write these matrices explicitely:
\begin{equation*}
\hat{\pi}^{(k)} = \hat{E}_{kk} - (1/n)\hat{\Ibb}_n.
\end{equation*}
The basis $\{\hat{\pi}_k\}$ consists of projectors,
and the expansion
\begin{equation}\label{PrForm}
\hat{\mu}_{{\rm in}} =  \sum_{k=1}^{n-1} c_k \hat{\pi}^{(k)}
\end{equation}
 immediately displays a dimension of the orbit passing through $\hat{\mu}_{{\rm in}}$. The dimension is $(n\,{-}\,1)(m\,{+}\,1)$,
where $m$ denotes a number of nonzero summands.
Simply speaking the degeneracy of an orbit becomes evident from coincidence of diagonal entries of~$\hat{\mu}_{\text{in}}$. For example, if all but one entries are equal, then the orbit passing through $\hat{\mu}_{\text{in}}$ is maximal denerate.

Therefore, every evolution of a closed $n$-level quantum system is located within a certain orbit of the adjoint action of the group $\mathrm{SU}(n)$. The orbit is fixed by an initial state, say $\hat{\rho}(t_0)$. It means that only the states within the fixed orbit are reachable from~$\hat{\rho}(t_0)$ because orbits do not intersect.  All orbits together form the state space~$\mathcal{P}$.

Turning back to our problem, suppose we are given a sample of evolution $\hat{\rho}$ as a function in $t$, and we aim at constructing an operator $\hat{U}_{t,t_0}\,{\in}\,\mathrm{SU}(n)$  realizing the given evolution. First we should identify the orbit where $\hat{\rho}$ lives. We propose to identify it by the canonical form of $\hat{\rho}$, say $\hat{\rho}_{\rm in}$. The matrix $\hat{\rho}_{\rm in}$ is constant and characterizes the whole orbit.  In particular, it serves as a  canonical form for $\hat{\rho}(t_0)$. The constancy of $\hat{\rho}_{\text{in}}$ follows from the set of equations \eqref{OrbitEqs} determining an orbit. Indeed, fixed values of $\Tr \hat{\rho}^k$ guarentee that all poins of an orbit have the same characteristic polynomial, and so have the same eigenvalues. In order to detect a  degeneracy of the orbit, one should apply the expansion \eqref{PrForm} to the matrix $\hat{\rho}_{\rm in}\,{-}\,(1/n)\hat{\Ibb}_n$, that displays a dimension of the space where the given state $\hat{\rho}$ evolves. This determines a structure of the evolution operator $\hat{U}_{t,t_0}$ and some auxiliary matrices in the proposed method.

The next step of our scheme consists in constructing a generic form
of the unitary matrix $\hat{u}\,{\in}\,\mathrm{SU}(n)$
realizing motion within a chosen orbit. We call this matrix an `evolution' one,  indicating by quotes the difference between a motion and a time evolution realised by an evolution operator $\hat{U}_{t,t_0}$.

\subsection{Stereographic parametrization and `evolution' operators}\label{ss:StereogrProj}
As shown above, motion of a $n$-level quantum system  is realized within an orbit by elements of the unitary
group~$\mathrm{SU}(n)$. We call these elements
`evolution' matrices when they concerns motion within orbits, and evolution operators when they generate time evolution.
There is a connection between  an orbit and its `evolution' matrix.  An `evolution' matrix can be written
in terms of the complex coordinates parameterizing an orbit. In the paper we implement this  by the method of generalized stereographic parametrization.

First we briefly recall the idea of the generalized stereographic
parametrization\cite{Skrypnyk,BernatskaGIQ} for a compact Lie group $G$.
It consists in comparing  the
Iwasawa and Gauss decompositions of the complexified group
$G^{\Complex}$ whose maximal compact subgroup is $G$. Let $T$ be the
maximal torus of $G$, and $T^{\Complex}$ be the maximal torus of
$G^{\Complex}$. We also define the following subgroups of
$G^{\Complex}$: the maximal real Abelian subgroup $A$, a nilpotent
subgroup~$R$, and another nilpotent subgroup $Z\,{=}\,R^{\ast}$.
Note that $T^{\Complex}\,{=}\,TA$. The Iwasawa and Gauss
decompositions of $G^{\Complex}$, respectively, are
\begin{equation*}
G^{\Complex}=GAR,\qquad G^{\Complex}= Z T^{\Complex}R.
\end{equation*}
If we define a generic adjoint orbit by means of coset space we obtain the equality line
\begin{equation*}
\mathcal{O} \equiv \frac{G}{T} = \frac{G^{\Complex}}{P} \sim \frac{ZT^{\Complex}R}{T^{\Complex}R} \sim Z,
\end{equation*}
where $P\,{=}\,TAR$ is the minimal parabolic subgroup of
$G^{\Complex}$. Therefore, we can parameterize the orbit
$\mathcal{O}$ in terms of canonical coordinates of $Z$, we denote them $\{z_k\}$.

In general an orbit of the adjoint action of group $G$ is defined as the coset space $G/G_{\rho_{\rm in}}$, where  $G_{\rho_{\rm in}}$ denotes the stability subgroup of the point $\hat{\rho}_{\text{in}}$ identifying the orbit.
A~generic orbit arises when a stability subgroup coincides with the maximal torus $T$.
A larger stability subgroup gives rise to a degenerate orbit, then a nonminimal parabolic subgroup $P\,{=}\,G_{\rho_{\rm in}}AR$ should be taken. In this case the nilpotent subgroup $Z$ is generated by a smaller number of root vectors, and is parameterized by a smaller number of canonical coordinates.

Here we deal with the group $\mathrm{SU}(n)$. Its complexification
having $\mathrm{SU}(n)$ as a maximal compact subgroup is
$\mathrm{SL}(n,\Complex)$. All mentioned subgroups can be
represented in matrix form. The Abelian subgroup $A$ consists of
diagonal matrices $\hat{a}\,{=}\,\diag(d_1,\,d_2,\,\dots,\, d_n)$
with $\prod_{k} d_k\,{=}\,1$. Let $R$ be a subgroup of upper
triangular matrices $\hat{r}$, then $Z$ is a subgroup of lower
triangular matrices $\hat{z}$.

For a matrix $\hat{z}\,{\in}\,Z$, assigned to a point of an
orbit, we obtain an `evolution' matrix $\hat{u}\,{\in}\,
\mathrm{SU}(n)$ from the Iwasawa decomposition
of $\hat{z}$:
\begin{equation}\label{IwasawaDec}
\hat{z}\,{=}\,\hat{u}\hat{a}\hat{r}.
\end{equation}
At the same time we aim at expressing
the matrix $\hat{u}$  in terms of the canonical coordinates $\{z_k\}$ apperaing as entries of the matrix $\hat{z}$.
The easiest way to achieve this goal is to solve the equation
\begin{equation}\label{IwasawaCalc} \hat{z}^{\ast} \hat{z} =
\hat{r}^{\ast} \hat{a}^{\ast} \hat{u}^{\ast} \hat{u} \hat{a} \hat{r}
= \hat{r}^{\ast} \hat{a}^2 \hat{r}.
\end{equation}
for all entries of $\hat{a}$ and $\hat{r}$.
Having the matrices $\hat{a}$ and $\hat{r}$ expressed by means of $\{z_k\}$,
one easily calculates the matrix $\hat{u}$ from the equality \eqref{IwasawaDec}.

As a result, we obtain a generic form for the matrix $\hat{u}$,
parameterized by the complex canonical  coordinates~$\{z_k\}$. This matrix
realizes the `evolution' starting from zero-point $\{z_k\,{=}\,0\}$  within every orbit of the same type.

\begin{Remark}\label{r:1}
It should be noticed, that the obtained `evolution' matrix $\hat{u}$
is not a unique solution of the problem. One can multiply it by any matrix $\hat{v}\in G_{\rho_{\rm in}}$,
that gives another `evolution' within the same orbit.
\end{Remark}

\subsection{Pure states}
Here we consider how to apply the above approach to the
particular case of pure states. Usually such states are described
by vectors of the Hilbert space $\Hilbert$. Let $\{|k\rangle\}$
denote a set of basis quantum states, then a pure state can be
expressed in the following form:
\begin{equation*}
|\psi(t) \rangle = \sum_k c_k(t) |k\rangle,
\end{equation*}
where the coefficients $\{c_k\}$ satisfy the normalization condition
$\sum_k |c_k(t)|^2 \,{=}\, 1$. One can reduce this expression to the form of a projective vector by introducing the complex coordinates $z_k\,{=}\,c_k/c_1$ for $k\,{=}\,2$, \ldots, $n$ which parameterize the maximal
degenerate orbit where the state $|\psi\rangle$ lives, and $\hat{z}\,{\in}\,Z$ has the following form
\begin{equation*}
\hat{z} = \begin{pmatrix}
1 & 0 & 0 & \dots & 0 \\ z_2 & 1 & 0 &\dots & 0
\\ z_3 & 0 & 1 & \dots & 0  \\ \dots&\dots&\dots&\dots&\dots \\ z_n & 0 & 0 & \dots & 1
\end{pmatrix}.
\end{equation*}
Let $\hat{r}$ and $\hat{a}$ have the form
\begin{eqnarray*} \hat{r} =  \begin{pmatrix}
1 & r_{12} & r_{13} & \dots & r_{1n} \\ 0 & 1 & r_{23} &\dots & r_{2n}
\\ 0 & 0 & 1 & \dots & r_{3n}  \\ \dots&\dots&\dots&\dots&\dots \\ 0 & 0 & 0 & \dots & 1
\end{pmatrix},\\ \hat{a} =  \begin{pmatrix} a_2 & 0 & 0 & \dots & 0 \\
 0& a_3/a_2 & 0 & \dots & 0 \\ 0&0&a_4/a_3&\dots&0 \\ \dots&\dots&\dots&\dots&\dots \\
 0&0&0&\dots &1/a_{n} \end{pmatrix}
\end{eqnarray*}
After solving \eqref{IwasawaCalc} we obtain the expressions
\begin{eqnarray*}
a_k^2 = 1+\sum_{j=k}^n |z_j|^2,\quad r_{1k} = \bar{z}_k /a^2_{2},\ \ k\geqslant 2,
\\ r_{jk} = -z_j \bar{z}_k /a^2_{j+1},\ \ k\geqslant j+1,
\end{eqnarray*}
which provide a parameterization of the matrices
$\hat{a}$ and~$\hat{r}$.

The matrix $\hat{u}$ calculated by \eqref{IwasawaDec} reallizes evolution of the pure
state $|\psi\rangle$ starting from the initial state $|\psi(t_0)\rangle\,{=}\,|1\rangle$. In the case of other
initial state $|\psi_0 \rangle$, one should multiply $\hat{u}$ by the matrix $\hat{u}_0^{-1}$
such that $\hat{u}_0$ realizes the evolution form $|1\rangle$ to $|\psi_0 \rangle$.
Having the evolution operator $\hat{U}_{t,t_0}\,{=}\,\hat{u}\hat{u}_0^{-1}$,
one computes the corresponding Hamiltonian by the formula
\begin{equation*}
\hat{H}\,{=}\, i\hbar \frac{d \hat{U}_{t,t_0} }{dt} \hat{U}^{-1}_{t,t_0} =
i\hbar \frac{d \hat{u} }{dt} \hat{u}^{-1},
\end{equation*}
which does not depend on the
initial state.

The obtained evolution operator $\hat{U}_{t,t_0}$ realizes exactly the prescribed evolution of $|\psi\rangle$,
but is not a unique solution of the problem in question. According to Remark~\ref{r:1} any evolution operator
$\hat{U}_{t,t_0}\,{=}\,\hat{u}\hat{v}\hat{u}_0^{-1}$, where $\hat{v}\,{\in}\, \mathrm{U}(1)\,{\times}\,\mathrm{SU}(n\,{-}\,1)$, realizes the same evolution of $|\psi\rangle$. If $\hat{v}$ is a function in $t$, the corresponding Hamiltonian changes into
\begin{equation}\label{GaugeTransH}
\hat{H}_{\hat{v}}\,{=}\, \hat{H} +
i\hbar \hat{u} \frac{d \hat{v}}{dt}
\hat{v}^{-1} \hat{u}^{-1},
\end{equation}
which is a gauge transformation of $\hat{H}$.

\subsection{Mixed states}
Considering mixed states we represent them by density matrices, which are Hermitian positive definite matrices of unit trace. Because an orbit is specified by a diagonal matrix, one should diagonalize the given matrix~$\hat{\rho}$:
\begin{equation*}
\hat{\rho} = \hat{D} \hat{\rho}_{\text{in}} \hat{D}^{-1},
\end{equation*}
where $\hat{\rho}_{\text{in}}$ is the canonical form of $\hat{\rho}$.
As proven at the end of section~\ref{ss:OrbitStruct}, an orbit structure of the state space implies that the matrix $\hat{\rho}_{\text{in}}$ is constant and serves as the canonical form of $\hat{\rho}(t_0)$. Preserving the Hermitian character of $\hat{\rho}$, the matrix $\hat{D}$ should be a unitary operator, so it can serve as the desired evolution operator realizing the given evolution $\hat{\rho}$.
Here we show how to obtain such operator by means of the proposed method.

We start with constructing a generic element
the group $\mathrm{SU}(n)$ according to the scheme of
subsection~\ref{ss:StereogrProj} from the matrix
\begin{equation*}
\hat{z} = \begin{pmatrix}
1 & 0 & 0 & \dots & 0 \\ z_{12} & 1 & 0 &\dots & 0
\\ z_{13} & z_{23} & 1 & \dots & 0  \\ \dots&\dots&\dots&\dots&\dots \\
z_{1n} & z_{2n} & z_{3n} & \dots & 1 \end{pmatrix}.
\end{equation*}
In order to find  relations between the entries of $\hat{\rho}$ and the canonical coordinates
$\{z_{jk}\}$ we solve the equation
\begin{equation*}
\hat{\rho} = \hat{u} \hat{\rho}_{\text{in}}  \hat{u}^{-1}
\end{equation*}
for $\{z_{jk}\}$.
The relations allow to introduce the prescribed time evolution into the matrix $\hat{u}$.
The obtained time dependent matrix $\hat{u}$ realizes the evolution starting from $\hat{\rho}_{\text{in}}$, If $\hat{\rho}(t_0)$ does not coincide with its canonical form~$\hat{\rho}_{\text{in}}$,   the matrix $\hat{u}$ should be multiplied by the matrix $\hat{u}_0^{-1}$ such that $\hat{\rho}(t_0)\,{=}\,\hat{u}_0\hat{\rho}_{\text{in}}\hat{u}_0^{-1}$.
Finally, the given evolution from $\hat{\rho}(t_0)$ to $\hat{\rho}(t)$ is realized by the operator $\hat{U}_{t,t_0} \,{=}\, \hat{u} \hat{u}_0^{-1}$. Evidently, the Hamiltonian does not depend on $\hat{u}_0$:
\begin{equation*}
\hat{H}\,{=}\, i\hbar \frac{d \hat{U}_{t,t_0}}{dt} \hat{U}^{-1}_{t,t_0} =
i\hbar \frac{d \hat{u} }{dt} \hat{u}^{-1}.
\end{equation*}

We obtain the Hamiltonian model guaranteeing that $\hat{\rho}$ evolves according to the prescribed law. Again this model is not unique. Any evolution operator $\hat{U}_{t,t_0} \,{=}\, \hat{u}  \hat{v} \hat{u}_0^{-1}$ with the matrix $\hat{v}$ belonging to the stability subgroup of the orbit where the state $\hat{\rho}$ lives also realizes the prescribed evolution of $\hat{\rho}$. In this case $\hat{v}$ performs a gauge transformation of the model, where the Hamiltonian changes according to the law \eqref{GaugeTransH}.

\section{Illustration of the method}
To illustrate the method proposed in the previous section we
consider two examples: a pure state and a mixed one. In the both cases
we calculate evolution operators and the corresponding Hamiltonians,
and show how to construct other possible evolution operators.

\subsection{The case of a pure state}
As a time dependent normalized pure state we choose the following one\cite{Leggio}
\begin{eqnarray}\label{StateExmpl}
|\psi(t)\rangle = \frac{1}{\sqrt{2+\cos^2\omega t}}
\Bigl\{|1,-1\rangle +\cos\omega t
|0,0\rangle + \nonumber \\  e^{i\pi/6}|{-}1,1\rangle\Bigr\},
\end{eqnarray}
describing a time evolution of a pair
of interacting spins~1: $\bm{S}_1$ and
$\bm{S}_2$. The state $|m_1,m_2\rangle$ is a
simultaneous eigenstate of $S_{1z}$ and
$S_{2z}$, that is
$S_{iz}|m_1,m_2\rangle\,{=}\,m_i|m_1,m_2\rangle$, where
$m_i\,{=}\,{-}1,\,0,\,1$ for $i\,{=}\,1,\,2$.

It is worth noting that the evolution \eqref{StateExmpl} of $|\psi(0)\rangle$
exhibits the conservation of $m\,{=}\,(m_1\,{+}\,m_2)\,{=}\,0$, which is an eigenvalue
of the observable $S_z\,{=}\,S_{1z}\,{+}\,S_{2z}$.
As a result, $|\psi\rangle$, which is indeed a vector from the
nine-dimensional Hilbert space $\mathcal{H}_{12}$ of the two
qutrits, lives in a three-dimensional subspace
$\mathcal{H}_0\,{\subset}\,\mathcal{H}_{12}$ of
$S_z$ pertaining to $m\,{=}\,0$. Thus, we need not to take into consideration the whole
Hilbert space $\mathcal{H}_{12}$. It is enough to describe the evolution of $|\psi\rangle$
by a reduced Hamiltonian model living in $\mathcal{H}_0$.

Another interesting observation is the following.
Focusing on the mathematical nature of the problem we could replace
the specific evolution \eqref{StateExmpl} of two qutrits by the evolution of a single qutrit
$\bm{J}$ such that $J_z|M\rangle\,{=}\,M|M\rangle$ with
$M\,{=}\,0,\,{\pm}1$:
\begin{equation}\label{StateExmpl_red}
|\psi(t)\rangle = \frac{1}{\sqrt{2+\cos^2\omega t}} \left\{|1\rangle
+\cos\omega t |0\rangle + e^{i\pi/6}|{-}1\rangle\right\}.
\end{equation}
Both evolutions: \eqref{StateExmpl} and \eqref{StateExmpl_red}
can be described by the same Hamiltonian model constructed below.

For the sake of generality we consider a state of the form
\begin{equation}\label{State}
|\psi(t)\rangle = c_1(t)|1,{-}1\rangle + c_{2}(t) |0,0\rangle +
c_3(t) |{-}1,1\rangle
\end{equation}
with $|c_1(t)|^2+|c_2(t)|^2+|c_3(t)|^2=1$. While \eqref{State} is a
pure state, its evolution is located within a maximum degenerate orbit
$\mathrm{SU}(3)/(\mathrm{U}(1)\,{\times}\, \mathrm{SU}(2))$, which
is the projective space $\CP^2$. Reduction to the form of a
projective vector gives
\begin{equation*}
|\psi(t)\rangle = |1,{-}1\rangle + z_{2}(t) |0,0\rangle + z_3(t)
|{-}1,1\rangle,
\end{equation*}
where $z_2\,{=}\,c_2/c_1$, $z_3\,{=}\,c_3/c_1$. On the other hand,
this projective vector can be rearranged into
the matrix form:
\begin{equation*}
\hat{z} = \begin{pmatrix} 1&0&0\\
z_2&1&0 \\ z_3&0&1 \end{pmatrix}
\end{equation*}

From \eqref{IwasawaCalc} we obtain the following:
\begin{eqnarray*}
&a_2^2 = 1+|z_2|^2+|z_3|^2,\quad r_{12} = \frac{\bar{z}_2}{a_2^2},\quad r_{13} = \frac{\bar{z}_3}{a_2^2},\\
&a_3^2 = 1+ |z_3|^2,\quad r_{23} = -\frac{z_2 \bar{z}_3}{a_3^2}.
\end{eqnarray*}
Then an evolution matrix $\hat{u}$ looks like
\begin{equation}\label{UMatr}
\hat{u} = \hat{z} \hat{r}^{-1} \hat{a}^{-1} =
\begin{pmatrix} \frac{1}{a_2}&\frac{-\bar{z}_2}{a_2 a_3}& \frac{-\bar{z}_3}{a_3}\\
\frac{z_2}{a_2}&\frac{a_3}{a_2}&0 \\ \frac{z_3}{a_2}& \frac{-z_3
\bar{z}_2}{a_2 a_3}&\frac{1}{a_3} \end{pmatrix}.
\end{equation}
This matrix acting on the vector $\bm{e} \,{=}\,
(1,\,0,\,0)$ gives $\bm{z} \,{=}\,
\bigl(1/a_2,\,z_2/a_2,\,z_3/a_2\bigr)$, where $a_2$ serves as its
norm. For an arbitrary initial vector
$\bm{z}_0$ we need to find the matrix $\hat{u}_0$ such that $\hat{u}_0\bm{e}
\,{=}\, \bm{z}_0$. Then
\begin{equation*}
\bm{z} = \hat{u} \hat{u}_0^{-1} \bm{z}_0,
\end{equation*}
and $\hat{U}_t\,{=}\,\hat{u} \hat{u}_0^{-1}$ is the required evolution operator.

Now we come back to the state \eqref{StateExmpl}. In this case,
$z_2(t) \,{=}\, \cos \omega t$, $z_3(t) \,{=}\, e^{i\pi/6}$, and
$a_2^2 = 2+\cos^2 \omega t$, $a_3^2 \,{=}\, 2$. Then
\begin{eqnarray*}
&\hat{u} = \begin{pmatrix}\frac{1}{\sqrt{2+\cos^2 \omega t}} &
\frac{-\cos \omega t}{\sqrt{2(2+\cos^2 \omega t)}} & \frac{-e^{-i\pi/6}}{\sqrt{2}}
\\ \frac{\cos \omega t}{\sqrt{2+\cos^2 \omega t}} & \frac{\sqrt{2}}{\sqrt{2+\cos^2 \omega t}} &
0 \\ \frac{e^{i\pi/6}}{\sqrt{2+\cos^2 \omega t}} & \frac{-e^{i\pi/6}\cos\omega t}{\sqrt{2(2+\cos^2 \omega t)}}
& \frac{1}{\sqrt{2}} \end{pmatrix},\\
&\hat{u}_0 = \begin{pmatrix} \frac{1}{\sqrt{3}} &
\frac{-1}{\sqrt{6}} & \frac{-e^{-i\pi/6}}{\sqrt{2}}
\\ \frac{1}{\sqrt{3}} & \frac{\sqrt{2}}{\sqrt{3}} &
0 \\ \frac{e^{i\pi/6}}{\sqrt{3}} & \frac{-e^{i\pi/6}}{\sqrt{6}}
& \frac{1}{\sqrt{2}} \end{pmatrix},
\end{eqnarray*}
and the evolution matrix is
\begin{widetext}
\begin{equation*}
\small \hat{U}_t = \begin{pmatrix}
\frac{2+\cos \omega t + \sqrt{3(2+\cos^2 \omega t)}}{2\sqrt{3(2+\cos^2 \omega t)}} & \frac{1-\cos \omega t}{\sqrt{3(2+\cos^2 \omega t)}} &
\frac{2+\cos\omega t - \sqrt{3(2+\cos^2 \omega t)}}{2\sqrt{3(2+\cos^2 \omega t)}}\, e^{-i\pi/6} \\
\frac{\cos \omega t - 1}{2\sqrt{3(2+\cos^2 \omega t)}} & \frac{2+\cos \omega t}{\sqrt{3(2+\cos^2 \omega t)}} & \frac{\cos \omega t - 1}{2\sqrt{3(2+\cos^2 \omega t)}}\, e^{-i\pi/6} \\
\frac{2+\cos \omega t - \sqrt{3(2+\cos^2 \omega
t)}}{2\sqrt{3(2+\cos^2 \omega t)}}\,e^{i\pi/6} &
\frac{1-\cos \omega t} {\sqrt{3(2+\cos^2 \omega
t)}}\,e^{i\pi/6} & \frac{2+\cos \omega t + \sqrt{3(2+\cos^2 \omega
t)}}{2\sqrt{3(2+\cos^2 \omega t)}} \end{pmatrix}.
\end{equation*}
\end{widetext}
The Hamiltonian calculated by \eqref{HamDef} is the following
\begin{equation}\label{H}
\hat{H} = \frac{i \hbar \omega \sin \omega t}{2+\cos^2 \omega t}
\begin{pmatrix} 0& 1 & 0 \\ -1 & 0 & -e^{i \pi /6} \\
0&e^{-i\pi/6} & 0\end{pmatrix},
\end{equation}
and coincides with the result reported in \cite{Leggio}.

As mentioned in Remark~\ref{r:1}, the obtained evolution matrix $\hat{U}_t$ is not a unique solution of the problem. As well the problem is solved by any matrix $\hat{U}_t\,{=}\,\hat{u} \hat{v} \hat{u}_0^{-1}$ with $\hat{v}$ of the following form
\begin{equation*}
\hat{v} =
\begin{pmatrix} e^{i\gamma(t)/\hbar} & 0 & 0 \\ 0 & \alpha(t) & \beta(t) \\
0& -\beta^{\ast}(t) & \alpha^{\ast}(t)\end{pmatrix},
\end{equation*}
where $\gamma$ is an arbitrary real-valued function in $t$, and $\alpha$, $\beta$ are complex-valued functions in $t$ such that $|\alpha(t)|^2\,{+}\,|\beta(t)|^2\,{=}\,1$ for all $t$.

It is easy to check that the reduced to $\mathcal{H}_0$ operators $\hat{S}_{1z}\,{+}\,\hat{S}_{2z}$, $\hat{S}_{1z}\hat{S}_{2z}$,
$\hat{S}_1^+\hat{S}_2^-$, and $\hat{S}_1^-\hat{S}_2^+$, where $\hat{S}_i^{\pm}\,{=}\,\hat{S}_{ix}\pm i \hat{S}_{iy}$, $i\,{=}\,1,\,2$,
can be cast in the following matrix form:
\begin{eqnarray}
&\hat{S}_1^z+\hat{S}_2^z=
\begin{pmatrix}
0 & 0 & 0\\
0 & 0 & 0\\
0 & 0 & 0\\
\end{pmatrix},\quad
\hat{S}_1^z \hat{S}_2^z=
\begin{pmatrix}
-1 & 0 & 0\\
0 & 0 & 0\\
0 & 0 & -1\\
\end{pmatrix},  \nonumber \\
&\hat{S}_1^+ \hat{S}_2^-=
\begin{pmatrix}
0 & 2 & 0\\
0 & 0 & 2\\
0 & 0 & 0\\
\end{pmatrix},\quad
\hat{S}_1^-\hat{S}_2^+=
\begin{pmatrix}
0 & 0 & 0\\
2 & 0 & 0\\
0 & 2 & 0\\
\end{pmatrix}. \label{SS}
\end{eqnarray}
Exploiting the expressions \eqref{SS} a Hamiltonian model $\hat{H}_{\rm gen}$ acting in $\mathcal{H}_{12}$,
whose reduction to $\mathcal{H}_0$ is given by \eqref{H}, is written in the
following form
\begin{eqnarray}\label{H1}
&\hat{H}_{\rm gen}=\hbar\Omega_0(\hat{S}_1^z+\hat{S}_2^z)+ \nonumber \\ &i \hbar \lambda(t)\Bigl( (e^{-i\frac{\pi}{6}}\hat{S}_1^+\hat{S}_2^-+\hat{S}_1^-\hat{S}_2^+)\hat{S}_1^z \hat{S}_2^z - \nonumber\\
&\hat{S}_1^z \hat{S}_2^z(\hat{S}_1^+\hat{S}_2^-+e^{i\frac{\pi}{6}}\hat{S}_1^-\hat{S}_2^+)\Bigr)
\end{eqnarray}
where the coupling constant $\lambda$ has the form
\begin{equation}
\lambda(t)=-\frac{1}{2}\cdot \frac{\omega\sin\omega t}{2+\cos^2\omega t}
\end{equation}
and $\hbar\Omega_0$ is an arbitrary value of energy. Equation \eqref{H1} may
be interpreted as describing two interacting
qutrits $\bm{S}_1$ and $\bm{S}_2$ subjected to a
magnetic field. The Hamiltonian $\hat{H}_{\rm gen}$ commutes with $\hat{S}_z$ at any
$t$ and guarantees that the value of $m\,{=}\,(m_1\,{+}\,m_2)$ remains constant, thus the evolution of $|\psi(0)\rangle$
belongs to a maximal degenerate orbit of SU(3) over the state space.

\subsection{The case of mixed state}\label{Ex2}
To illustrate our method in the case of a mixed
state, let us consider a two-level atom coupled to a bosonic mode.
Atomic dynamical variables can be represented in terms of the
Pauli matrices $\sigma_x,\,\sigma_y,\,\sigma_z$ acting upon a
two-dimensional Hilbert space where a basis is
$\{|+\rangle,|-\rangle\}$ with $\sigma_z|\pm\rangle=\pm|\pm\rangle$.
The bosonic mode is represented, as usual, by an annihilation $a$ and
a creation $a^{\dagger}$ operators
such that $[a,a^{\dagger}]=1$, acting upon an
infinite-dimensional Hilbert space with the basis  $\{|n\rangle,
n=0,1,2,\ldots\}$ such that $a^{\dagger}a|n\rangle=n|n\rangle$. Suppose
that at $t=0$ the composed system is in the mixed state
\begin{equation}
\rho(0)=(3/4)|0,+\rangle\langle0,+| + (1/4)|1,-\rangle\langle1,-|,
\end{equation}
where $|n,\pm\rangle$ is a simultaneous eigenstate of
$a^{\dagger}a$ and $\hat{\sigma}_z$ pertaining to their eigenvalues $n$
and $\pm1$ respectively. Assume that the atom-mode interaction
 generates the following evolution:
\begin{eqnarray}\label{rho}
&\rho(t)=\Bigl(3/4-(1/2)\cos^2 gt \Bigr) |0,+\rangle\langle0,+|
 +  \nonumber \\ &\Bigl(1/4+(1/2)\cos^2 gt \Bigr)|1,-\rangle\langle1,-| +  \nonumber \\
&(1/4)\sin 2gt \Bigl(|0,+\rangle\langle1,-|+ |1,-\rangle\langle0,+|\Bigr).
\end{eqnarray}

It is suitable to represent $\rho$ in a matrix form as follow:
\begin{equation}\label{MStateExmpl}
\hat{\rho}(t)=\begin{pmatrix} 3/4 - (1/2) \cos^2 gt & (1/4) \sin 2 gt  \\
(1/4) \sin 2 gt  & 1/4 + (1/2) \cos^2 gt \end{pmatrix},
\end{equation}
which may be generalized by the following notation for entries of
$\hat{\rho}$:
\begin{equation*}
\hat{\rho}=\begin{pmatrix} \rho_{11}&\rho_{12} \\
\bar{\rho}_{12} & \rho_{22} \end{pmatrix},
\end{equation*}
Diagonalization of $\hat{\rho}$ leads to
\begin{eqnarray*}
&\hat{\rho}_{\text{in}}=\begin{pmatrix} \lambda_{1}&0 \\
0 & \lambda_{2} \end{pmatrix},\\ &\lambda_{1,2} = \frac{1}{2}\left(\rho_{11}+\rho_{22} \pm
\sqrt{(\rho_{11}-\rho_{22})^2+4|\rho_{12}|^2}\right).
\end{eqnarray*}
The group acting over the state \eqref{MStateExmpl} is $\mathrm{SU}(2)$,
and its generic element obtained by the explained above method of stereographic parametrization has the form
\begin{equation*}
\hat{u}=\frac{1}{\sqrt{1+|z|^2}} \begin{pmatrix} 1&-\bar{z} \\
z & 1 \end{pmatrix}.
\end{equation*}
Solving the equation $\hat{u}\hat{\rho}_{\text{in}}\hat{u}^{-1} \,{=}\, \hat{\rho}$
for $z$, we come to
\begin{equation*}
z = \frac{2\bar{\rho}_{12}}{\rho_{11}-\rho_{22} \pm
\sqrt{(\rho_{11}-\rho_{22})^2+4|\rho_{12}|^2}}.
\end{equation*}

In the case of state \eqref{MStateExmpl} we obtain
\begin{equation*}
\hat{\rho}_{\text{in}} = \begin{pmatrix} 1/4&0 \\
0 & 3/4 \end{pmatrix},
\end{equation*}
and $z \,{=}\, -\tan gt$. Then a possible evolution  operator
and its correspondent Hamiltonian acquire the forms
\begin{eqnarray}\label{U}
\hat{U}_t = \begin{pmatrix} \cos gt &  \sin gt \\
-\sin gt & \cos gt \end{pmatrix},\quad
\hat{H} =   i \hbar g \begin{pmatrix} 0 & 1 \\
-1 &0\end{pmatrix}.
\end{eqnarray}
The given evolution \eqref{MStateExmpl} is obtained by
\begin{equation*}
\hat{\rho}(t) = \hat{U}_t \hat{\rho}(0) \hat{U}_{t}^{-1}.
\end{equation*}

According to Remark~\ref{r:1} there exist other solutions of the problem.
Evolution operators $\hat{U}_t \,{=}\, \hat{u}\hat{v}$ such that
\begin{equation*}
\hat{v}=\begin{pmatrix} e^{i\alpha(t)/\hbar} & 0 \\
0 & e^{i \beta(t)/\hbar}\end{pmatrix},
\end{equation*}
where $\alpha$ and $\beta$ are arbitrary real-valued functions in $t$, also realizes
the same evolution~\eqref{MStateExmpl}.
The corresponding Hamiltonian $\hat{H}$ gets the form
\begin{eqnarray*}
\hat{H} = &  i \hbar g \begin{pmatrix} 0 & 1 \\
-1 &0\end{pmatrix} -  \frac{1}{2} \Bigl(\alpha'(t) + \beta'(t) \Bigr)
\begin{pmatrix} 1&0\\0&1 \end{pmatrix} + \\ & \frac{1}{2} \Bigl(\alpha'(t) - \beta'(t) \Bigr)
\begin{pmatrix} - \cos 2gt
 & \sin 2gt \\ \sin 2gt & \cos 2gt \end{pmatrix}.
\end{eqnarray*}

Once again we emphasize that although the Hilbert space of the
composite system is infinite-dimensional, the prescribed time
evolution belongs to the state space
$\mathcal{P}$ over the two-dimensional invariant subspace
$\Hilbert_1$ of the excitation number
operator $E\,{=}\,a^{\dagger}a\,{+}\,(1/2)\sigma_z + 1/2$, where only one
excitation is present with certainty. In this
physical context our method generates a Hamiltonian model reduced to
$\mathcal{P}$. It is easy to check that a
reduction of the time-independent Hamiltonian model
\begin{equation}\label{JC}
H_{\rm JC} = \hbar g(a^{\dagger}\sigma_-+a\sigma_+)
\end{equation}
has a matrix representation coincident with the model given by
(\ref{U}). The Hamiltonian $H_{\rm JC}$
represents the well known Jaynes-Cummings model of a two-level atom
resonantly coupled to a single electromagnetic field mode, taken in
the interaction picture with respect to
$H_0\,{=}\,\hbar\omega\left(a^{\dagger}a+\frac{1}{2}\sigma_z\right)$. Even
in this example, the same mathematical construction would be valid
if we substitute $|0,+\rangle\langle0,+|$ and
$|1,-\rangle\langle1,-|$ in (\ref{rho}) with
the two basis projectors $|+\rangle\langle+|$ and
$|-\rangle\langle-|$ of a single qubit. In
this case (\ref{U}) would describe the interaction of the qubit with
a particular magnetic field.

\section{Conclusion and discussion}
In the paper we have proposed  a method of reconstructing a Hamiltonian from a prescribed time evolution of a quantum state $\rho$. We~suppose that the quantum system is governed by the von Neumann equation, and so every evolution is represented by a unitary operator. This implies that the action of the group of evolution operators over the state space is equivalent to the adjoint action of the unitary group over Hermitian matrices. As a result the conditions $\Tr \rho^k(t)\,{=}\,{\rm const}$ for all  $k\,{\geqslant}\, 1$, determining an orbit of the adjoint action of the unitary group, are fulfilled. For every such state $\rho$ it is possible to construct
the class of evolution operators realizing the prescribed evolution, and this evolution is located within a single orbit.

In  the theory of open quantum systems
one deals with reduced density matrices. They are governed by more
complicate equations than the von Neumann one. These equations,
known as quantum master equations, are different for different
systems. Evidently, the dynamics of an open quantum
system, in general, can not be represented by a unitary operator.
Nevertheless, we hope that the inverse dynamical problem for reduced density matrices may be  solved within the
group-theoretical approach reported in this paper.

\begin{acknowledgments}
We would like to thank professor P.~Holod for fruitful discussion and pieces of sensible advice during preparation of the paper.
One of us (AM) thanks Elena Ferraro and Bruno Leggio for help
in the preparation of this manuscript and
scientific discussions.

This work is supported by the grant
MIUR-Cooperazione interuniversitaria internazionale --- capitolo 1706/6 --- A.F. 2007,  and a grant of the International Charitable Fund for Renaissance of Kiev-Mohyla Academy.
\end{acknowledgments}

\nocite{*}
\providecommand{\noopsort}[1]{}\providecommand{\singleletter}[1]{#1}


TITLE: summation notation for general sets
QUESTION [1 upvotes]: I'm working through an academic game theory paper and stumbled upon this summation notation in a proof and I'm not quite sure what it means:
$$\sum\limits_{j \in M \backslash\ \{i\}}$$
There is a set $M$ indexed by $j$. There are other terms in the expression indexed by $i$. I'm curious what the $\backslash\{i\}$ means. Does this mean "with the exception of $i$"?
This context is an actor ($i$)'s utility function that depends on what other actors in the set $M$ do. Other actors are indexed by $j$.

REPLY [0 votes]: Let X be a set and A be a subset of X, X\A = {x in X: x is not an element of A}.


text:
\chapter{Morita equivalence}\label{chap:morita}

In this chapter, we will will introduce and study various notions of Morita equivalence and the Picard group.
We will also review some strategies for computing Morita equivalences as well as take a look at a few examples.
Throughout this chapter, $\C$ is a good site (Definition~\ref{defn:goodsite}).

Recall that principal bibundles of $\C$-groupoids are precisely the isomorphisms in $\morpoid$.
Isomorphism in $\morpoid$ has a special name:
\begin{definition}
  Let $\G$ and $\H$ be $\C$-groupoids. A \emph{Morita equivalence} between $\G$ and $\H$ is a principal $(\G,\H)$-bibundle.
\end{definition}
This chapter will be organized as follows:
\begin{itemize}
  \item In Section~\ref{section:maplike} we explore the relationship between bibundles of $\C$-groupoids and $\C$-groupoid homomorphisms. In particular, we define a special class of $\C$-groupoid homomorphisms called \emph{weak equivalences}.
  \item In Section~\ref{section:moritaequiv} we prove the main results of this chapter. The main theorem essentially states that the equivalence relation generated by weak equivalences is the same as Morita equivalence.
  \item In Section~\ref{section:picard} we define the Picard group of a $\C$-groupoid $\G$, which turns out to be the group of automorphisms of $\B \G$.
  \item In Section~\ref{section:examples2} we look at a few examples of Morita equivalences and the Picard group for specific sites.
\end{itemize}

\section{Generalized morphisms and homomorphisms}\label{section:maplike}

In Chapter~\ref{chap:stacks} we saw that there is a functor $\B \colon \morpoid \to \mathbf{GStacks}$ from Groupoids with generalized morphisms to geometric stacks.
In this section we will note that there is a pseudofunctor (Definition~\ref{defn:pseudofunctor})
\[ \bP \colon \C\GRPD \to \morpoid \]
from the category of $\C$-groupoids (with homomorphisms as morphisms) to $\C$-groupoids with generalized morphisms.

\subsection{Homomorphisms of groupoids}

Recall that a $\C$-groupoid homomorphism from $\G \grpd M$ to $\H \grpd N$ covering $f \colon M \to N$, is a morphism $F \colon \G \to \H$ which is compatible with the groupoid structure maps.

\begin{definition}\label{defn:groupoidmaps}
  Let $F\colon \G \to \H$ be a $\C$-groupoid morphism covering $f \colon M \to N$. Then:
  \begin{itemize}
    \item $F$ is called \emph{essentially surjective} if $\t \circ \pr_1 \colon \H \times_N M \to N$ is a submersion.
    \item $F$ is called \emph{fully faithful} if $M \times_{f, \t} \H \times_{f, \s} M$ exists and:
    \[ (\t, F, \s ) \colon \G \to M \times_{f, \t} \H \times_{\s, f} M \]
    is an isomorphism.
    \item $F$ is called an \emph{weak equivalence} if it is both essentially surjective and fully faithful.
  \end{itemize}
  Observe that the essentially surjective condition depends only of the base map $f$. For this reason we say an arbitrary morphism $f \colon M \to N$ is essentially surjective if $1_M \to \H$ is essentially surjective.
\end{definition}

Unsurprisingly, one can make groupoids and groupoid homomorphisms into a bicategory.
To do this, we must give the analogue of a natural transformation internal to $\C$.
\begin{definition}\label{defn:cgroupoidnaturaltrasn}
  Let $F \colon \G \to \H$ and $G \colon \G \to \H$ be groupoid homomorphisms covering $f \colon M \to N$ and $g \colon M \to N$, respectively.
  A \emph{natural transformation} from $F$ to $G$ is a morphism $\eta \colon M \to \H$ such that:
  \begin{itemize}
  \item $ \s \circ \eta = f$ and $\t \circ \eta = g$;
  \item $\m(\eta \circ \t, F) = \m(G, \eta \circ s) \colon \G \to \H$
  \end{itemize}

  Given $F_i \colon \G \to \H$ covering $f_i \colon M \to N$ for $i=1,2,3$ and a pair of natural transformations $\eta_1 \colon F_1 \to F_2$ and $\eta_2 \colon F_2 \to F_3$ then the vertical composition $\eta_2 * \eta_1$ is defined to be:
  \[ \eta_2 * \eta_1  := \eta_2  \cdot \eta_1 \colon M \to \H \, . \]

  Now suppose we are given $\eta \colon F_1 \to F_2$ and $\zeta \colon G_1 \to G_2$ for $F_i \colon \G \to \H$ and $G_i \colon \H \to \K$. Then the horizontal product $\zeta \circ \eta$ is defined as below:
  \[
  \zeta \circ \eta (x) := (G_2 \circ \eta(x)) \cdot  (\zeta \circ f_1(x))
  \]
\end{definition}
These constructions give a natural bicategory structure on $\C$-groupoids which is analogous to the bicategory of standard groupoids.
The important caveat is that everything is performed internally to $\C$.
\newpage
\begin{definition}
Let $\C\GRPD$ be the bicategory defined as follows:
\begin{itemize}
  \item The objects of $\C\GRPD$ are $\C$-groupoids.
  \item The 1-morphisms of $\C\GRPD$ are $\C$-groupoid homomorphisms.
  \item The 2-morphisms of $\C\GRPD$ are natural transformations.
\end{itemize}
Composition of 1-morphisms is defined in the obvious manner. Horizontal and vertical composition of 2-morphisms is defined as above.
\end{definition}
It is not difficult to check that this satisfies the axioms of a bicategory. In fact, it is a \emph{strict} bicategory since the horizontal product is strictly associative.
We should offer a word of caution, however. Unlike the case of set-theoretic groupoids, a fully faithful and essentially surjective morphism of $\C$-groupoids does not necessarily admit a weak inverse.

\subsection{Pullback groupoids}

We would like to point out one construction which is particularly useful for computing Morita equivalences.
\begin{lemma}
  Let $\G \grpd M$ be a $\C$-groupoid and suppose $f \colon N \to M$ is essentially surjective. Then $N \times_M \G \times_M N$ exists and inherits a canonical groupoid structure such that projection to the middle component $N \times_M \G \times_M N \to \G$ is a groupoid homomorphism.
\end{lemma}
\begin{proof}
  Let $f^! \G$ be shorthand for $N \times_M \G \times_M N$. To give a groupoid structure on $f^! \G$ we should first define the source and target maps.

  Let the source $f^! \G \to N$ be projection to the last component and the target be projection to the first component. The source and target maps are submersions since they are the base changes of a submersion.

  Let $F \colon f^! \G \to \G$ be projection to the middle component. Then we define the product on $f^! \G$ as follows:
  \[ (z,g_2,y) \cdot (y,g_1, x):= (z, g_2 \cdot g_1 ,x). \]
  This clearly makes $F$ a $\C$-groupoid homomorphism.
  \end{proof}

\begin{definition}\label{defn:pullbackgroupoid}
  Let $G \grpd M$ be a $\C$-groupoid and $f \colon N \to M$ be essentially surjective. Then
  \[ f^! \G := N \times_M \G \times_M N \]
  is called the \emph{pullback groupoid} along $f$.
\end{definition}

The reason we do note use the notation $f^* \G$ to denote the pullback groupoid is due to possible confusion with the pullback of the trivial $\G$-bundle.
Recall that $f^* \G$ is defined to be $\G \times_M N$ while $f^! \G$ is $N \times_M \G \times_M N$.
Since the pullback of $\G$ as a bundle and as a groupoid do not coincide, it will spare us some possible confusion if we use distinct notation.

It is not necessary that $f \colon N \to M$ be essentially surjective for this construction to result in a $\C$-groupoid. It is clear however that it will not always work. Even if $f^! \G$ exists as an object in $\C$, it may not necessarily be the case that the source and target maps are submersions.
\begin{lemma}\label{lemma:pullbackequiv}
  Let $\G \grpd M$ be a $\C$-groupoid and suppose $f \colon N \to M$ is essentially surjective.
  The canonical homomorphism $F \colon f^! \G \to \G$ is a weak equivalence.
\end{lemma}
\begin{proof}
  The result is clear from the definition of fully faithful.
\end{proof}

\subsection{Map-like bibundles}

Given a $\C$-groupoid homomorphism $F \colon \G \to \H$ covering $f \colon M \to N$, there is a canonical construction of a generalized morphism.
Let
\[
\bP(F) := f^* \H  = \H \times_N M \, .
\]
Since this is just the trivial bundle associated to $f$, there is a canonical left principal $\H$-bundle structure over $M$.
To make $\bP(F)$ into a bibundle (and hence a generalized morphism) we must equip it with a right $\G$-action. Let:
\[
(h, \t(g)) \cdot  g := (h \cdot F(g), \s(g))
\]
\begin{definition}
  A 1-morphism $\G \to \H$ in $\morpoid$ is called \emph{map-like} if it is isomorphic to $\bP(F)$ for some homomorphism $F \colon \G \to \H$.
\end{definition}
The next lemma gives a geometric characterization of the map-like bibundles.
\begin{lemma}\label{lemma:maplike}
  A generalized morphism is map-like if and only if it admits a section (as a left $\G$-bundle).
\end{lemma}
\begin{proof}
  Let $F \colon \G \to \H$ be a $\C$-groupoid homomorphism and suppose $Q \isom \bP(F)$ is map-like. Since $\bP(F)$ clearly admits a section (it is trivial as a left $\H$-bundle) then $Q$ must admit a section.

  On the other hand, suppose $N \from Q \to M$ is a left principal $(\H, \G)$-bibundle and admits a section. We must show it is isomorphic to $\bP(F)$ for some $F \colon \G \to \H$. Let $\sigma \colon M \to Q$ be a section of $Q$. Now let $F$ be the unique morphism such that:
  \[
   \sigma(\t(g)) \cdot g = F(g) \cdot \sigma(\s(g))
  \]
  To see that this is a homomorphism, observe that:
  \begin{align*}
    F(g_1 \cdot g_2) \cdot \sigma(\s(g_1 \cdot g_2)) &= \sigma(\t(g_1 \cdot g_2)) \cdot g_1 \cdot g_2 \\
    &= \sigma(\t(g)) \cdot g_1 \cdot g_2 \\
    &= F(g_1) \sigma(\s(g_2)) \cdot g_2 \\
    &= F(g_1) \cdot F(g_2) \cdot  \sigma(\s(g_2)) \\
    &= F(g_1) \cdot F(g_2) \cdot \sigma(\s(g_1 \cdot g_2))
  \end{align*}
  Now we construct an isomorphism $\bP(F) = \H \times_N M \to Q$. Let
  \[ \phi(h,x) \mapsto h \cdot \sigma(x)  \]
  By Lemma~\ref{lemma:bundletrivialization} we know that $\phi$ is an isomorphism of principal $\H$-bundles. That $\phi$ is $\G$ equivariant follows from a simple computation:
  \[ \phi(h,x) \cdot g = h \cdot \sigma(x) \cdot g = h \cdot F(g) \cdot \sigma(\s(g)) = \phi(h F(g), \s(g)) \]
  \end{proof}
The assignment $F \mapsto \bP(F)$ is bifunctorial in the following sense.
\begin{lemma}
  Suppose $F \colon \G \to \H$ and $G \colon \H \to \K$ are $\C$-groupoid homomorphisms covering $f \colon M \to N$ and $g \colon N \to O$, respectively. Then there is a canonical isomorphism:
  \[ \mathbb{P}(G,F) \colon \bP(G) \otimes_\H \bP(F) \to \bP({G \circ F}) \]
\end{lemma}
\begin{proof}
  It suffices to construct an $\H$-invariant map $\Phi \colon \bP(G) \times_N \bP(F) \to \bP({F \circ G})$ which respects the left and right actions. Recall that
  \[ \bP(G) := g^* \K = \K \times_O N \quad \bP(F) := f^* \H = \H \times_O N \quad \bP({G \circ F}) := (g \circ f)^*\K = \K \times_O M \, . \]
  Hence we define
  \[
  \Phi((k,y),(h,x)) := ( k \cdot G(h), x) \, .
  \]
  It is a routine check that $\Phi$ is $\H$-invariant and respects the actions of $\K$ and $\G$. Therefore, it must descend to a bibundle isomorphism $\mathbb{P}(G,F)$.
\end{proof}

Isomorphisms of $\C$-groupoid homomorphisms can be related to bibundle isomorphisms by the following process.
Suppose $\eta \colon F \to G$ is a natural transformation of $\C$-groupoid homomorphisms from $\G$ to $\H$ covering $f \colon M \to N$ and $g \colon M \to N$, respectively.

Then let
\[ \bP(\eta)(h,x) := (h \cdot (\i \circ \eta)(x),x) \]
Note that this uses the (opposite) correspondence from Lemma~\ref{lemma:gbundlecycle}.
By that same lemma, this is $\H$ equivariant. That is is $\G$-equivariant follows from the natural transformation condition.
\begin{proposition}
  The mapping $\eta \mapsto \bP(\eta)$ gives a one-to-one correspondence between natural transformations $\eta \colon F \to G$ of $\C$-groupoid homomorphisms $\G \to \H$ and bibundle isomorphisms $\bP(F) \to \bP(G)$.

  Furthermore, this correspondence respects the vertical and horizontal products in the following senses:
  \begin{itemize}
    \item For all $\zeta \colon G_1 \to G_2$ and $\eta \colon F_1 \to F_2$ which are horizontally composable, the below diagram commutes:
    \begin{equation}\label{diagram:pcompatibility}
    \begin{tikzcd}
      \bP(G_1) \otimes \bP(F_1) \arrow[d, "{\mathbb{P}(G_1,F_1)}"]\arrow[rr, "{\bP(\zeta) \otimes \bP(\eta)}"] & & \bP(G_2) \otimes \bP(F_2)\arrow[d, "{\mathbb{P}(G_2,F_2)}"] \\
      \bP(G_1 \circ F_1) \arrow[rr, "{\bP(\zeta \circ \eta)}"] & & \bP(G_2 \circ F_2)
    \end{tikzcd}
  \end{equation}
    \item For all $\zeta$ and $\eta$ which are vertically composable $\bP(\zeta * \eta) = \bP(\zeta) \circ \bP(\eta)$.
  \end{itemize}
\end{proposition}
\begin{proof}
  The correspondence is one-to-one thanks to Lemma~\ref{lemma:gbundlecycle}.
  \begin{itemize}
    \item Suppose $\zeta$ and $\eta$ are horizontally composable such that $F_i \colon \G \to \H$ and $G_i \colon \H \to \K$. Then the bottom left path of Diagram~\ref{diagram:pcompatibility} is:
    \[
    \begin{tikzcd}
      (k , \t(h)) \otimes (h , x) \arrow[d, mapsto]  & \\
      (k \cdot G_1(h), x) \arrow[r] & (k \cdot G_1(h) \cdot (\i \circ \zeta)(f_1(x)) \cdot (\i \circ G_2 \circ \eta)(x), x)
      \end{tikzcd}
    \]
    Going the other direction, we get:
    \[
    \begin{tikzcd}
      (k , \t(h)) \otimes (h , x) \arrow[r, mapsto]
      & \left(k \cdot (\i \circ \zeta) (\t(h)), \t(h) \right) \otimes \left( h \cdot (\i \circ \eta)(x),x \right) \arrow[d, mapsto] \\
       & (k \cdot (\i \circ \zeta)(\t(h)) \cdot G_2(h) \cdot (\i \circ G_2 \circ \eta)(x),x )
      \end{tikzcd}
    \]
    But since $\zeta$ is a natural transformation, we have:
    \[
    (\i \circ \zeta)(\t(h)) \cdot G_2(h)  = G_1(h) \cdot (\i \circ \zeta)(\s(h)) = G_1(h)
    \]
    So the diagram commutes.
    \item Suppose $\zeta \colon F_2 \to F_3$ and $\eta \colon F_1 \to F_2$ are horizontally composable for $F_i \colon \G \to \H$ covering $f_i \colon M \to N$. Recall that $(\zeta * \eta)(x) := \zeta(x) \cdot \eta(x)$. Therefore:
    \[ \bP(\zeta * \eta)(h,x) := (h \cdot \i \circ (\zeta(x) \cdot \eta(x)), x) = (h \cdot (\i \circ \eta)(x) \cdot (\i \circ \zeta)(x), x) \]
    Which is the same as $\bP(\zeta) \circ \bP(\eta)$.
  \end{itemize}
\end{proof}
The following theorem summarizes our results so far:
\begin{theorem}
  $P \colon \C\GRPD \to \morpoid$ is a psuedofunctor from the bicategory of $\C$-groupoids to the bicategory of generalized morphisms.
  Furthermore, $P$ is a bijection on objects. The restriction of $P$ to the $\Hom$ categories
  \[ \Hom{(\G,\H)}_{\C\GRPD} \to \Hom{(\G,\H)}_{\morpoid} \]
  is a fully faithful embedding whose essential image consists of the the map-like generalized morphisms.
\end{theorem}
\begin{proof}
  We take $P$ to be the identity on objects. Except for coherence, the theorem follows from the propositions and lemmas we proved in this subsection. We will leave the proof of coherence for this functor to the bored reader.
\end{proof}

\section{Morita equivalence}\label{section:moritaequiv}
\subsection{Weak equivalences}
It is natural to ask whether the bifunctor defined in the previous section allows us to to study Morita equivalences purely in terms of $\C$-groupoid homomorphismse
To answer this question, we will begin with characterizing those morphisms which give rise to Morita equivalences.
\begin{lemma}
  Suppose $F \colon \G \to \H$ is a homomorphism of $\C$-groupoids. Then $\bP(F)$ is a Morita equivalence if and only if $F$ is a weak equivalence.
\end{lemma}
\begin{proof}
Suppose $F$ is a weak equivalence.
We only need to show that the right action of $\G$ on $\bP(F)$ is principal.
Since $F$ is a weak equivalence, we can assume without loss of generality that $\G \isom f^! \H$.
Recall that $\bP(F) := \H \times_N M$ and $f^!\H := M \times_N \H \times_N M$.
If $\C$ is made of sets and functions, the right action of $\bP(F)$ is of the form:
\[ (h, x) \cdot (x, h', y) = (h \cdot h', y) \, . \]
The total action is therefore:
\[ ((h,x), (x,h',y)) \mapsto ((h,x),(h \cdot h', y)) \]
This leads us to notice that
\[ ((h_1,x),(h_2,y)) \mapsto (x, h_1\inv h_2 , y) \]
is a division map for the right action.
Since this division map is unique, the action must be principal.
\end{proof}
Now that we have identified the $\C$-groupoid morphisms which give rise to Morita equivalences. It is reasonable to wonder if there are enough of these maps to generate the Morita equivalence relation. That is, given two Morita equivalent $\C$-groupoids, does there exist a chain of weak equivalences connecting them? The answer is yes and actually two weak equivalences suffice.
\begin{theorem}\label{thm:weakequivalences}
  The following are equivalent:
  \begin{itemize}
    \item$\G$ and $\H$ are Morita equivalent.
    \item There exists a $\C$-groupoid $\K$ and a pair of weak equivalences $\H \from \K \to \G$.
  \end{itemize}
\end{theorem}
\begin{proof}
  Going from the second to the first is clear.

  Let $P$ be a biprincipal $(\H,\G)$-bibundle for $\H \grpd N$ and $\G \grpd M$. By Lemma~\ref{lemma:pullbackequiv}, the mapping $(\s^P)^! \G \to \G$ is a weak equivalence. Let $\K := (\s^P)^! \G $. To conclude the proof, we must construct a weak equivalence $F \colon \K \to \H$. Let $F$ be the unique morphism satisfying:
  \[ p_1 \cdot g = F(p_1,g,p_2) \cdot p_2 \]
  The existence and uniqueness of $F$ follows from the existence and uniqueness of the division map for the left $\H$-action on $P$.
  A quick check verifies that this is indeed a homomorphism of $\C$-groupoids covering $\t^P \colon P \to N$.
  To prove that this is a weak equivalence it suffices to show that the induced map
  \[ \K = (\s^P)^! \G \to (\t^P)^! \H \]
  is an isomorphism.
  Since the right action is principal as well, by the same trick we can construct a homomorphism $F' \colon (\t^P)^! \H \to \K$ using the rule:
  \[ (p_1, h, p_2) \mapsto (p_1, g, p_2) \quad \Leftrightarrow \quad p_1 \cdot g = g \cdot p_2  \]
  Since $F'$ is clearly the inverse of $\K \to (\t^P)^! \H$ the result follows.
\end{proof}
The above proposition tells us that if we wish to show that something is Morita invariant, it suffices to show that it is invariant under weak equivalences.
\subsection{Examples of Morita equivalences}
\begin{example}[Manifolds]
    When $\C$ is the site of smooth manifolds. Then the $\C$-groupoids are precisely the Lie groupoids. Given a Lie groupoid $\G \grpd M$, the following are Morita invariants:
    \begin{itemize}
      \item The orbit space of $\G$, that is the topological space obtained by taking the quotient of $M$ by the equivalence relation on $M$ induced by the map $(\s,\t) \colon \G \to M \times M$.
      This is sometimes called the \emph{coarse moduli space} of $\G$.
      \item Suppose one is given an orbit $\O \subset M$ of $\G$ in $M$. In other words, a point in the coarse moduli space of $\G$. Then for any pair of points, $x,y \in \O$ the isotropy groups $\G_x := (\s,\t)\inv(x,x)$ and $\G_y$ are isomorphic.
      Hence, to each orbit we can associate an isomorphism class of an isotropy group, and these classes are preserved by Morita equivalences.
    \end{itemize}
    This is (not even close) to being an exhaustive list of the extensive number of Morita invariants which are known.
    However, these two Morita invariants fundamental in that they provide one with some justification for the intuition that the stack associated to a Lie groupoid should be thought of as a singular manifold equipped with additional symmetry groups.
\end{example}
\begin{example}[Dirac Structures]
  Suppose $\C = \DMan$ (see Chapter~\ref{chap:dman}). Then a $\C$-groupoid is a D-Lie groupoid. In the previous chapter, we say that a Morita equivalence of D-Lie groupoids always has an underlying Morita equivalence of Lie groupoids. Hence any Morita invariant of Lie groupoids is also a Morita invariant of D-Lie groupoids.

  A D-Lie groupoids often come with additional interesting invariants.
  \begin{itemize}
    \item Suppose $\G \grpd M$ is a symplectic groupoid. Then $M$ inherits a Poisson bivector. By choosing a density $\mu$ on $M$ one can construct what is called the \emph{modular vector field} $X_\mu$. It turns out that different choices of densities on $M$ give rise to modifications of $X_\mu$ by Hamiltonian vector fields. Hence, we obtain what is called the \emph{modular class}
    \[ [X_\mu] \in \X(M) / \X_{ham}(M) := H^1_\pi(M) \, .  \]
    Ginzburg and Lu showed~\cite{GinzburgPoissonCohomology} that symplectic Morita equivalences induced isomorphisms in $H^1_\pi(M)$.
    Later, it was shown by Crainic~\cite{crainicmod} that the modular class is preserved by such isomorphisms.
  \end{itemize}
\end{example}

\section{The Picard group}\label{section:picard}

Roughly, the Picard group is the group of Morita self equivalences.
\begin{definition}
  Suppose $\G \grpd M$ is a $\C$-groupoid. The Picard group of $\G$, denoted $\Pic(\G)$ is the set of principal $(\G,\G)$-bibundles, up to isomorphism.
\end{definition}
Under the correspondence developed in Chapter~\ref{chap:stacks}, we know that the Picard group is the automorphisms of the associated stack $\B\G$ modulo natural transformations. Of course, that point of view does not lend itself to calculations. In general, we should expect that the Picard group of a given groupoid is difficult, if not impossible, to compute.

\subsection{Outer automorphisms}
Let $P$ represent an element of the Picard group $\Pic(\G)$. Then we say that $[P] \in \Pic(\G)$ is \emph{map-like} if $P$ is map-like. Map-like elements of the Picard group do not form a subgroup in general since $[P]\inv$ need not be map-like.

\begin{definition}
  Consider the group homomorphism $\bP \colon \Aut(\G) \to \Pic(\G)$ which sends a $\C$-groupoid automorphism to the associated map-like element of $\Pic(\G)$. The kernel of this map is called the \emph{inner automorphisms} of $\G$, denoted $\InnAut(\G)$. The image of this map is canonically isomorphic to $\OutAut(\G) := \Aut(\G)/\InnAut(\G)$.
\end{definition}

We might wonder if one can characterize the image of $\bP$ geometrically. In order to do this, we need to introduce the notion of a bisection.

\begin{definition}
  Suppose $P$ is a principal $(\G,\H)$-bibundle for $\C$-groupoids $\G \grpd M$ and $\H \grpd N$. A \emph{bisection} of $P$ is a morphism $\sigma \colon N \to P$ such that $\t \circ \sigma \colon N \to M$ is an isomorphism.
\end{definition}

\begin{lemma}
  An element $[Q] \in \Pic(\G)$ comes from an automorphism $F \colon \G \to \G$ if and only if $Q$ admits a bisection.
\end{lemma}
\begin{proof}
  Suppose $Q$ admits a bisection. Then by Lemma~\ref{lemma:maplike} we can construct an $F \colon \G \to \G$ so that $Q \isom \bP(F)$. Since $P$ is principal we know that $F$ is a weak equivalence. Since $F$ is an isomorphism on objects it must be an isomorphism of $\C$-groupoids.

  The other direction is clear by taking the canonical section of $\bP(F)$ when $F$ is an isomorphism.
\end{proof}

We also have a characterization of inner automorphisms.

\begin{lemma}\label{lemma:innaut}
  A morphism $F \colon \G \to \G$ is an inner automorphism if and only if there exists a bisection $\sigma \colon M \to \G$ of $\G$ such that:
  \[ F(g) = (\sigma \circ \t)(g) \cdot g \cdot (\i \circ \sigma \circ \s(g))  \]
\end{lemma}
\begin{proof}
  By assumption there is an isomorphism $\Phi \bP(F) \to \G$ as $(\G,\G)$-bibundles. Let:
  \[
  \sigma(x) := \Phi(\u \circ f(x), x) \colon M \to \bP(F) \to \G
  \]
  Then we have that:
  \[
    \Phi(F(h),\s(h))  = \Phi(F(h)\cdot \u \circ f(\s(h)),\s(h)) = F(h) \cdot \sigma(\s(h))
  \]
  On the other hand:
  \[
  \Phi(F(h),x) = \Phi(F(h),x) = \Phi(\u \circ f(\t(h)),\t(h)) \cdot h =  \sigma(\t(h)) \cdot h
  \]
  Hence:
  \[ F(h) = \sigma(\t(h)) \cdot h \cdot (\i \circ \sigma)(\s(h)) \]
  On the other hand, suppose $F$ is given by such a formula. Then it is straightforward to check that:
  \[ \Phi \colon \P(F) \to \G \qquad \Phi(g,x) \mapsto g \cdot \sigma(x) \]
  yields the desired isomorphism.
\end{proof}
\section{Examples}\label{section:examples2}
\subsection{Manifolds}
Suppose $\C$ is the site of smooth manifolds.
\begin{definition}\label{defn:staticpicardgroup}
  Let $\G \grpd M$ be a Lie groupoid and let $M / \G$ denote the orbit space of $\G$ (as a purely topological object). The \emph{static Picard group} is the kernel of the homomorphism $\Pic(\G) \to \Aut(M / \G)$ denoted by $\Pic_Z (\G)$. In other words, it is the elements of the Picard group which act trivially on the orbit space.
\end{definition}
The static Picard group fits into an exact sequence:
\[  1 \to \Pic_Z (\G) \to \Pic(\G) \to \Aut(M / \G) \, . \]
Note that the last arrow is not surjective.
There is no guarantee that all topological symmetries of $M /\G$ can be witnessed by elements of the Picard group. We will treat a few instances for which the Picard group has been calculated.
\begin{example}[Trivial groupoids]
  Suppose $1_M \grpd M$ and $1_N \grpd N$ are trivial Lie groupoid. That is, every morphism is a unit. Then $1_M$ is Morita equivalent to $1_N$ if and only if they are diffeomorphic.
  A principal $(1_M,1_M)$-bibundle is the same as a diffeomorphism $f \colon M \to M$. Hence
  \[ \Pic(1_M) \isom \Diff(M) \, . \]
  Every element of the Picard group is map-like and the group of inner automorphisms of $1_M \grpd M$ is trivial. Lastly, the static Picard group is trivial as well since $\Pic(1_M) \to \Aut(M)$ is injective.
\end{example}
\begin{example}[Trivial line bundle]
  Let $ \G = \R_M \grpd M$ be the trivial line bundle. That is, the groupoid operation is just fiber-wise addition in $\R$. Suppose $P$ is a principal $(\R_M, \R_M)$-bibundle.
  The left action of $\G$ makes $P$ into an (affine) line bundle.
  Since every affine line bundle admits a global section, we know that $P$ comes from an automorphism of $\G$. Since $\G$ is abelian, the inner automorphisms of $\G$ are trivial and we conclude that
  \[ \Pic(\G) \isom \OutAut(\G) \isom \Diff(M). \]
  The static Picard group is trivial.
\end{example}
\begin{example}[Bundles of abelian groups]
  To the best of our knowledge, this case first appeared in a paper by Moerdijk~\cite{moerdijkgerbes}. It is a, more interesting, generalization of the previous example.

  Consider the case where $\G \grpd M$ is a Lie groupoid such that $\s = \t$ and is such that $\G_x := \t\inv(x)=\s\inv(x)$ is an abelian group for all $x \in M$. We can think of $\G$ as defining a sheaf whose local sections are bisections of $\G$. Any principal $(\G,\G)$-bundle $P$ defines a diffeomorphism $f_P \colon M \to M$ of the orbit space of $\G$.
  Furthermore, any $(\G,\G)$ bibundle is, in particular, a principal $\G$-bundle. Hence, there is an associated \v{C}ech cohomology class $c_P \in H^1(M,\G)$. This gives rise to a group isomorphism:
  \[ \Pic(\G) \to H^1(M,\G) \ltimes \Diff(M) \]
  where the action of $\Diff(M)$ on $H^1(M,\G)$ is by pullbacks.
\end{example}
\begin{example}[Lie groups]
  Suppose $\G \grpd \{ * \}$ is a Lie group. In other words, it is a Lie groupoid over a point. Then every $(\G,\G)$-bibundle admits a bisection and so $\Pic(\G) \isom \OutAut(\G)$. Note that, by Lemma~\ref{lemma:innaut}, the outer automorphisms of $\G$ as a groupoid are the same as the outer automorphisms of $\G$ as a Lie group.
\end{example}
\begin{example}[Transitive groupoids]
  Suppose $\G \grpd M$ is a Lie groupoid such that the orbit space $M /\ G$ is a point. For any $x \in M$, the inclusion of the isotropy group $\G_x \into M$ is a weak equivalence. Hence $\Pic(\G) \isom \Pic(\G_x) = \OutAut(\G_x)$.

  For example, if $\G = \Pi_1(M) \grpd M$ is the fundamental groupoid of $M$ and $M$ is connected, then we obtain $\Pic(\G) \isom \OutAut(\pi_1(M))$.
\end{example}
\subsection{Morita equivalence of Lie algebroids}

Recall that a Lie algebroid is a vector bundle $A \to M$ together with a Lie bracket
\[ [ \cdot , \cdot ] \colon \Gamma(A) \times \Gamma(A) \to \Gamma(A) \]
satisfying a Leibnitz type identity. When Pradines~\cite{Pradines} introduced the notion of Lie algebroid, he showed that any Lie groupoid can be differentiated to a Lie algebroid. If a Lie algebroid is isomorphic to one of this type, then we say it is integrable.
It is well known that not every Lie algebroid is integrable.

When a Lie algebroid is integrable, there is a unique (up to isomorphism) integration with simply connected source fibers called the \emph{canonical} integration.
For a reader interested in more details, we refer them to \cite{lecturesonint} and \cite{Cint}.
\begin{definition}
  Suppose $A \to M$ and $B \to N$ are integrable Lie algebroids, with canonical integrations $\G \grpd M$ and $\H \grpd N$, respectively. Then a Morita equivalence between $A$ and $B$ is a Morita equivalence of $\G$ and $\H$. Similarly, the Picard group of $A$ denoted $\Pic(A)$ is defined to be $\Pic(\G)$.
\end{definition}

\begin{example}[Tangent]
  The canonical integration of the tangent algebroid is the fundamental groupoid. Therefore, if $M$ and $N$ are connected, then $TM$ and $TN$ are Morita equivalent if and only if $\pi_1(M) \isom \pi_1(N)$. Furthermore, $\Pic(TM) = \OutAut(\pi_1(M))$.
\end{example}
\begin{example}
  If $A \to \{ * \}$ is a Lie algebroid over a point, then $A$ is the same as a Lie algebra. The canonical integration of $A$ is just the simply connected integration $G$ of the Lie algebra. Hence
  \[ \Pic(A) \isom  \OutAut(G) \isom \OutAut(\g). \]
  Two Lie algebras are Morita equivalent if and only if they are isomorphic.
\end{example}
\subsection{Symplectic Morita equivalence}
For $\C = \DMan$ (see Chapter~\ref{chap:dman}), there is a special class of $\C$-groupoids called \emph{symplectic groupoids}.
Morita equivalence of symplectic groupoids was originally developed by Xu~\cite{Morping} (without the use of Dirac structures). Picard groups of symplectic groupoids were first introduced and studied by Bursztyn and Weinstein\cite{BPic}.
We will recall the definition of these objects to make the following discussion more clear.
\begin{definition}
  A \emph{symplectic groupoid} is a Lie groupoid $\G$ together with a symplectic form $\Omega \in \Omega^2(\G)$ which is multiplicative:
  \[ \m^* \Omega = \pr_1^* \Omega + \pr_2^* \Omega \, . \]
  A \emph{Morita equivalence} of symplectic groupoids $\G$ and $\H$ is a principal $(\G,\H)$-bibundle $P$ together with a \emph{left} and \emph{right mulitplicative} 2-form $\omega$ on $P$.
  \[ \m_L^* \omega = \pr_1^* \Omega^\G + \pr_2^* \omega \qquad \m_R^* \omega = \pr_1^* \omega + \pr_2^* \Omega^\H \, . \]
  The \emph{Picard group} of a symplectic groupoid $\G$ is the set of isomorphism classes of Morita self equivalences.
\end{definition}

One important reason to be interested in symplectic groupoids is that they are the objects which integrate Poisson manifolds.
Given $(M, \pi)$ a Poisson manifold, then $T^*M$ inherits a Lie algebroid structure. When $T^*M$ is integrable, we say that $(M,\pi)$ is an integrable Poisson manifolds. The canonical integration of a Poisson manifold inherits a multiplicative symplectic form from the canonical symplectic form on $T^* M$.
\begin{definition}
  Suppose $(M, \pi_M)$ and $(N, \pi_N)$ are integrable Poisson manifolds. Then we say that $M$ and $N$ are Morita equivalent as Poisson manifolds if their canonical integrations are \emph{symplectically} Morita equivalent. Similarly, the Picard group of a Poisson manifold is defined to by the Picard group of its symplectic integration (as a $\DMan$-groupoid).
\end{definition}
\begin{example}
  Suppose $M$ and $N$ are any manifolds.
  Then $T^* M \grpd M$ and $T^* N \grpd N$ are symplectic groupoids via fiberwise addition.
  The Poisson structure that they integrate is given by the zero bivector.
  A principal $(T^* M, T^* N)$-bibundle $(P, \omega)$ always admits a bisection, $\sigma \colon N \to P$.
  The bisection gives rise to a diffeomorphism $f_P \colon N \to M$ and a 2-form $\beta := \sigma^* \omega$.
  Bursztyn and Weinstein showed that the resulting map:
  \[ \Pic(T^* M) \to H^2(M) \ltimes \Diff(M)  \]
  is a group isomorphism.
\end{example}
\begin{example}
  Suppose $(M, \omega\inv)$ is a connected Poisson manifold such that $\omega\inv \colon T^* M \to TM$ is the inverse of a symplectic form.
  It follows that $M$ is integrable and the canonical integration is the fundamental groupoid $\Pi_1(M)$.
  The symplectic structure on the fundamental groupoid is obtained by the pullbacks $\t^* \omega - \s^* \omega$.

  As we saw earlier, $\Pi_1(M)$ is Morita equivalent to the group $\pi_1(M,x)$ for any given base-point $x$.
  It turns out that this is Morita equivalence of symplectic groupoids.
  Hence, symplectic manifolds (interpreted as Poisson manifolds) are Morita equivalent if and only if their fundamental groups are isomorphic.

  This example tells us that we should not expect Morita equivalence to tell us very much about the \emph{leaf-wise} geometry of a Poisson manifold. Morita equivalences are fundamentally about the transverse geometry of the groupoid.
\end{example}
\begin{example}
  Suppose $\g$ is a Lie algebra. Then the dual $\g^*$ inherits a canonical Poisson structure.
  The canonical integration is $T^* G \isom G \times \g^*$ equipped with the (coadjoint) action groupoid structure, where $G$ is the simply connected integration of $\g$.

  For the compact and semi-simple case, Bursztyn and Fernandes~\cite{ruipic} proved the following theorem:
  \begin{theorem}
    Suppose $\g$ is a compact semi-simple Lie algebra. Then:
    \[ \Pic(\g^*) \isom \OutAut(\g) \]
  \end{theorem}
\end{example}

config: arxiv


TITLE: Complex Integrate $\int_{-\infty}^{\infty}e^{-|\lambda t|}e^{itx}dt$
QUESTION [1 upvotes]: I'm working through Big Rudin's (Real and Complex Analysis) Fourier Transform chapter, and the following complex integral is part of a discussion on the Inverse Transform that Rudin mentions briefly without resolving, and I'm very curious how this is done.
$\int_{-\infty}^{\infty}e^{-|\lambda t|}e^{itx}dt$
What I remember from my Complex Analysis days, is to take the limit of the integral evaluated around a contour from -R to R, and then around $C_R$, the semicircle connecting -R and R. Since this is closed I can use the Cauchy Residue Theorem, and by Jordan's Lemma hopefully the integral around $C_R$ will vanish, so I'm left with
$\int_{-\infty}^{\infty}e^{-|\lambda t|}e^{itx}dt=2\pi*\{\text{sum of residues}\}$
However, this integral has no residues, so simply taking the limit of this integral from -R to R should work, but my rusty integration techniques aren't helping and Mathematica is not giving me any results.
Any help in solving this is much appreciated!

REPLY [4 votes]: Let $F(x)$ be the function defined by the Fourier Integral
$$F(x)=\int_{-\infty}^\infty e^{-|\lambda t|}e^{itx}\,dt$$
Since the integral converges for $\lambda \ne 0$, we can write the integral as
$$F(x)=\lim_{R\to \infty}\int_{-R}^R e^{-|\lambda t|}e^{itx}\,dt$$
One might be tempted to evaluate this integral be closing the real line contour with a semi-circle, centered at the origin and with radius $R$, in the upper-half or lower-half plane and then using the Residue Theorem.  However, this approach is fatally flawed since $e^{-|\lambda||z|}$ is not an Analytic Function.  Therefore, we proceed along a different way forward.
First, we use Euler's Formula and exploit even and odd components of the integrand to reveal
$$F(x)=2\int_0^\infty e^{-|\lambda|t}\cos (xt)\,dt$$
Then, Integrating by Parts twice yields
$$\begin{align}
F(x)&=\lim_{R\to \infty}\left.\left(\frac{e^{-|\lambda|t}\left(-|\lambda|\cos (xt)+x\sin(x t)\right)}{x^2+|\lambda|^2}\right)\right|_{t=0}^{t=R}\\\\
&=\bbox[5px,border:2px solid #C0A000]{\frac{|\lambda|}{x^2+|\lambda|^2}}
\end{align}$$


TITLE: How to transform $z$ into $\hat z$
QUESTION [2 upvotes]: Assume that we have vector $a = \begin{bmatrix} a_1 \\ a_2 \\ \vdots \\ a_n \end{bmatrix} \in \mathbb R^n$, $b = \begin{bmatrix} b_1 \\ b_2 \\ \vdots \\ b_n \end{bmatrix} \in \mathbb R^n$, $c = \begin{bmatrix} c_1 \\ c_2 \\ \vdots \\ c_n \end{bmatrix} \in \mathbb R^n$.
Then, the vector $z = \begin{bmatrix} a \\ b \\ c \end{bmatrix}\in \mathbb R^{3n}$.
How to obtain the vector $\hat z = \begin{bmatrix} a \\ a  \\ b \\ b \\ c \\ c \end{bmatrix}\in \mathbb R^{2\times 3n}$ based on $z$?
The Kronecker product is not useful in this case, because it focuses on elements, instead of subvectors.

REPLY [2 votes]: I believe the Kronecker product can be pretty much useful. You could take.
Approach 1: Take:
$$z_{1} = \begin{pmatrix} 1 \\ 1 \\ 0 \\ 0 \\0  \\ 0\end{pmatrix} \quad z_{2} = \begin{pmatrix} 0 \\ 0 \\ 1 \\ 1 \\ 0 \\ 0 \end{pmatrix} \quad z_{3}=\begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 1 \\1 \end{pmatrix} $$
so that:
$$\hat{z} = z_{1}\otimes a + z_{2}\otimes b + z_{3}\otimes c$$
Approach 2: Take:
$$z_{1} = \begin{pmatrix} 1 & 0 & 0 \\ 1 & 0 & 0 \end{pmatrix} \quad z_{2} = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 1 & 0 \end{pmatrix} \quad z_{3} = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 1 \end{pmatrix}$$
so that:
$$z'=\begin{pmatrix}  a & b & c \\ a & b & c \end{pmatrix} = z_{1}\otimes a + z_{2}\otimes b + z_{3}\otimes c$$
and take $\hat{z}$ to be the vectorization of $z'$.


TITLE: Chebyshev's Inequality given mean and variance
QUESTION [0 upvotes]: Give examples of random variables Y with mean 0 and variance 1
such that
(a) $P(|Y | ≥ 2) = \frac{1}{4}$
(b) $P(|Y | ≥ 2) < \frac{1}{4}$
I know Chebyshev's equality must be used here but I am struggling with how to think of examples that fit the given restrictions.

REPLY [2 votes]: The first is the equality case in Chebyshev's inequality, and that's very tightly constrained. The variable must be supported on exactly three points: the mean, the point we're measuring at, and its reflection on the opposite side. That's $\{-2,0,2\}$ here.
So, what should the probabilities be to match the mean and variance? For the mean, we must have $P(-2)=P(2)$. For the variance, we must have $4\cdot P(-2)+0\cdot P(0)+4\cdot P(2)=1$, and $P(-2)+P(2)=2P(2)=\frac14$. Since the total probability is $1$, that's a probability mass function $P(-2)=\frac18$, $P(0)=\frac34$, $P(2)=\frac18$.
The second? Take absolutely any other distribution with the desired mean and variance.
Can't think of one? Whatever your favorite random variable $X$ is, if it has mean $\mu$ and variance $\sigma^2$, $Y=(X-\mu)/\sigma$ has mean zero and variance $1$.


TITLE: Deciding if a given $\mathsf{NC}^0$ circuit computes a permutation
QUESTION [27 upvotes]: What is the complexity of deciding whether an $\mathsf{NC}^0$ circuit
with $n$ input bits and $n$ output bits computes  a permutation
of $\{0,1\}^n$? in the other words, whether every bit strings in
$\{0,1\}^n$ is an output of the circuit for some input?
It looks like a problem which has been studied,
but I cannot find any references.

REPLY [29 votes]: Hardness
Following your comment on the question, we will call a circuit where each output bit depends on at most k input bits a “NC0k circuit.”  Using this term, your problem is coNP-complete in case of NC05 circuits.  That is, the following problem is coNP-complete.
Instance: A Boolean circuit C with n input bits and n output bits where each output bit depends on at most five input bits.
Question: Is the mapping from {0,1}n to itself computed by C bijective?
As Kaveh commented, it is clearly in coNP, even without the bound on the number of input bits on which each output bit depends.  To prove the coNP-hardness, we will reduce 3SAT to the complement of the current problem.  The key idea of the reduction is the same as the one used in the paper [Dur94] by Durand, which I mentioned in a comment on the question, but the whole reduction is much simpler in our case.
Given a 3CNF formula φ with n variables and m clauses, we construct a Boolean circuit C with (n+m) input bits and (n+m) output bits as follows.  We label the input bits as x1, …, xn, y1, …, ym, and the output bits as x′1, …, x′n, z1, …, zm.  We consider that the input bits x1, …, xn specify a truth assignment to the n variables in φ.

x′i = xi for 1≤i≤n.  That is, the first n bits of input are always copied to the first n bits of output.
For 1≤i≤m, if the ith clause of φ is satisfied, then zi = yi ⊕ yi+1, where the subscript is interpreted modulo m.  Otherwise, zi = yi.

Note that each output bit depends on at most five input bits.  I omit the proof of the correctness of the reduction, but the key idea (which I borrowed from [Dur94]) is that if φ is satisfiable and input bits x1, …, xn are set to a satisfying assignment of φ, then the m output bits z1, …, zm are constrained to have the even parity, and therefore the circuit cannot be a permutation.  On the other hand, if input bits x1, …, xn are set to a non-satisfying assignment of φ, then output bits z1, …, zm can be set to anything; because of this, if φ is unsatisfiable, then the circuit is a permutation.
Tractability
On the tractable side, your problem is in P in case of NC02 circuits.  This is shown as follows.  In general, each output bit in a Boolean circuit for a permutation is balanced; i.e., exactly half of the input strings set the output bit to 1.  However, every balanced Boolean function from {0,1}2 to {0,1} is affine; i.e., a copy of a single input bit, the XOR of the two input bits, or the negation of them.  Therefore, we can first check that each output bit is balanced, and then check the bijectivity by the Gaussian elimination.
I do not know the complexity in case of NC03 circuits or in case of NC04 circuits.
References
[Dur94] Bruno Durand.  Inversion of 2D cellular automata: some complexity results.  Theoretical Computer Science, 134(2):387–401, Nov. 1994.  DOI: 10.1016/0304-3975(94)90244-5.


text:
\begin{document}

\section{Conclusion}
In this work it was shown that a feed-forward properly regularized neural network, recast as a recurrent neural network and trained with the appropriate loss function, is able to qualitatively learn the global features of the phase space of dynamical systems, whose trajectories it was trained with. The examined characteristics were:
\begin{enumerate}
    \item Dissipative or conservative dynamics.
    \item Limit cycles.
    \item Unseen atractors.
    \item Frequency response.
    \item Parametric bifurcations.
\end{enumerate}

It was found that even very small neural networks with non-linear $max(0,x)$ activations can exhibit complex behaviors. This fact allows them to qualitatively fit into the mold of very different dynamics, also when the difference between the original and learnt system derivative, in terms of MSE, is considerable. This promotes the possibility of using simple networks with adjustable precision to replicate the parts of a system's dynamics that interest us, with a much lower computational cost.

With the Lorentz system experiment, it became apparent how a neural network could faithfully learn a parameterized system, with little or no added capacity with respect to the single configuration system learning problem.

Advancing in the work, we have been able to encode non-linear 1D PDEs derivative into a very compact set of hidden variables. This has enabled us to solve a system of ODEs with much fewer degrees of freedom.

It was demonstrated that the encoded space was capable of solving trajectories with a variable degree of proximity to the learned one. As well as encoding complex abstract characteristics of motion such as advection velocities, dissipation rates or transient times.

Finally, in this reduced representation, the requirements for temporal discretization, that ensure stability, were significantly relaxed compared to the spectral solver used.
\end{document}


TITLE: Prove $\frac{\ln(x)}{x} \le \log_{10} 2$
QUESTION [2 upvotes]: How would I go about proving the following inequality?
$\frac{\ln(x)}{x} \le \log_{10} 2$
Is there an algebraic solution and if not, why?

REPLY [2 votes]: $\log_{10}2 = \frac{\ln2}{\ln 10}$.
Then, for all positive $x$,
$$
x \leq 2^{\frac{x}{\ln 10}}
$$
Then you apply $\ln$ to both sides (given that it's monotonically increasing), and you have the inequality you want.


TITLE: Magdeburg Hemispheres
QUESTION [2 upvotes]: The Magdeburg Hemisphere experiment was the experiment that showed the effect of pressure differences on a vacuumed sphere.
We know that the Force caused by pressure is $\Delta p A$ and so you can calculate the force by using the area of the base of one of the hemispheres of the the vacuumed ball.
It is intuitive to me that the area we use is the area created by the diameter of the sphere, a line through the center of the sphere, but is there a clearer reason that we use this area? I'm not sure how to explain why this "important" area.

REPLY [2 votes]: At all points on the sphere the pressure points normal to the surface (because that it what pressure does...).
That said this system has a preferred direction: the line between the hoops where we hook the harness on for the horse to pull. And we have built the device such that the joint between the two halves lies at the equator perpendicular to that axis.
Now, the system therefore has axial symmetry, we are are only interested in the component of force parallel to the symmetry axis (because the components transverse to the axis cancel out when we add up the bits).
Measuring angle $\theta$ from the + end of the symmetry axis (either one, just pick your favorite), the axial part of the force from the pressure on a infintesimal area element $\mathrm{d}A$ (or very small element $\Delta A$ if you don't have calculus) is $P \sin (\theta ) \mathrm{d}A$ or the projection of that area onto a place normal to the axis.
Therefore the total force is that exerted on a circle of the same diameter.
This can be shown rigorously in a few lines of calculus which is what DumpsterDoofus set up (I've really just stated the same integral in a lot of words).


TITLE: How to know when to divide, factorise or move terms when making x the subject of a formula?
QUESTION [0 upvotes]: There are many ways of making something the subject of a formula, like making every term have the same denominator, factorisation, dividing every term by the same thing. However, I often get confused by which operation to use.
For example, if I have $\frac xa = \frac 2x - \frac xb$ and I want to make b the subject in terms of x and a, what should I do? The final answer is $\frac {(x^2)(a)}{2a-x^2}$. I got it using a really long-winded way where I tried everything I could but it wasn’t efficient. The suggested method is much simpler but I don’t see how they know when to use what operation.
This is the correct method:
$$\frac xa = \frac 2x - \frac xb$$
$$\frac 2x = \frac xa + \frac xb$$
$$\frac 2x = \frac {xb+xa}{ab}$$
$$2ab = (x^2) b+ (x^2)a$$
$$2ab-(x^2)b = (x^2) a$$
$$b(2a -x^2) = (x^2) a$$
$$b = \frac {(x^2)a}{2a-x^2}$$
OR it could be done this way (also the suggested method):
$$\frac xa = \frac 2x - \frac xb$$
$$\frac {x(xb)}{xab} = \frac {2(ab) - x(xa)}{xab}$$
$$x(xb) = 2(ab)-x(ab)$$
$$bx^2 = 2ab -(x^2)a$$
$$b(x^2 -2a) =-(x^2)a$$
$$b(2a-x^2) = (x^2)a$$
$$b = \frac {x^2 a}{2a-x^2}$$

REPLY [0 votes]: $$\frac{x}{b} = \frac{2}{x} - \frac{x}{a}$$
$$\frac{x}{b} = \frac{2a-x^2}{xa}$$
$$b = \frac{x^2a}{2a-x^2}$$


TITLE: Smoothness of real analytic space
QUESTION [0 upvotes]: Consider $\mathbb{R}^2$ with the polynomial $P(x,w):=w^2+xw+x$. The zero set $N\left(P\right)$ of $P$ is a real analytic submanifold of $\mathbb{R}^2$. That is, the stalks of $\mathcal{A}_D/J_{N(P)}$ are regular analytic $\mathbb{R}$-algebras, here $\mathcal{A}_D$ denotes the sheaf of real analytic functions and $J_{N(P)}$ the ideal of all functions vanishing on $N(P)$.
My question is: Is $J_{N(P)}= (P)?$
I.e. is the analytic space defined by the ideal $(P)$ generated by $P$ smooth or is it non reduced?
I know that in real algebraic geometry the ideal $J_{N(P)}$ would be the real radical of $(P)$, but I do not know of a similar result in the analytic category.
I have tried to find elements of $\mathcal{A}_{D}/(P)$ that are nilpotent but with no success. How does one approach this in the real analytic category?

REPLY [0 votes]: The answer is, no. After a change of coordinates around $0$, the polynomial may be assumed to be $P(x,w)=w^2+x$, as $w+1$ is invertible and positive around zero. Then by the Weierstraß division theorem one has
$\mathcal{A}_{D,0}/ (P)_0\cong \mathcal{A}_{D',0}^2.$
Here $D'$ is a domain in $\mathbb{R}$ containing $0$. Thus, the real analytic space defined by $P$ is not a real analytic manifold at $0$.


TITLE: How do I calculate this integral $\displaystyle\int_{0}^{+\infty} \exp(-ax^2)\sin(x) \,dx$?
QUESTION [0 upvotes]: So far I got: $\frac{1}{\sqrt{a}} \displaystyle\int_{0}^{\frac{1}{2\sqrt{a}}} \exp \left( x^2-\frac{1}{4a} \right) \,dx$. But, I ran into a snag. Any ideas how to pull out the value of the integral to within $O(a^3)$?

REPLY [3 votes]: It doesn't have a nice closed form. It can be stated in terms of the Dawson plus function
$$\int_0^\infty \exp(-xt^2)\sin(t)\mathrm{d}t=\frac{D_+\left(\frac{1}{2\sqrt{x}}\right)}{\sqrt{x}}$$
Near the origin it has the Taylor expansion
$$D_+(x)=\sum_{k=0}^\infty\frac{(-1)^k2^k}{(2k+1)!_2}x^{2k+1}$$
Whereas at infinity it has the asymptotic expansion
$$D_+(x)=\frac{1}{2x}+\frac{1}{4x^3}+\frac{3}{8x^5}+\cdots+\frac{(2k-1)!_2}{2^{k+1}x^{2k+1}}+O(x^{-2k-3})$$
edit: Where $!_2$ indicates the "double factorial". I dislike the "$!!$" notation.