TITLE: Finding a matrix given eigenvalues and eigenvectors.
QUESTION [4 upvotes]: I am asked to construct a $4 \times 4$ symmetric matrix, with given eigenvalues and eigenvectors. I understand how to actually get $A$ as a product of $P^T, D$ and $P$, when $D$ is the diagonal matrix, and $P$ is a matrix with the eigenvectors as columns.
The problem is that there is only three given eigenvectors, along with three eigenvalues (one is repeated), so my question is, how do you construct a $4 \times 4$ matrix with three eigenvectors?
For more information here is the actual question:
Let $A$ be a symmetric $4 \times 4$ matrix with real entries whose eigenvalues
are $−1$ and $2$. If $(1, 0, 0, −1)$, $(0, 1, 1, 0)$ is a basis for the eigenspace of eigenvalue $-1$ and $(1, 0, 0, 1)$ is an eigenvector of $A$ with eigenvalue $2$, find the matrix $A$.
Thank you.

REPLY [1 votes]: Consider the usual dot product
$$
(x_{1}, x_{2}, x_{3}, x_{4})
\cdot
(y_{1}, y_{2}, y_{3}, y_{4})
=
x_{1} y_{1} + x_{2} y_{2} + x_{3} y_{3} + x_{4} y_{4}.
$$
Then it is easy to see that $A$ is symmetric iff for all $x, y$
$$
x A \cdot y = x \cdot y A.
$$
(just take the $x = e_{i}, y = e_{j}$, where $e_{i}$ is the vector which is all zero except for a $1$ in the $i$-th position).
Now prove that if $x$ is an eigenvector with respect to $\lambda$, $y$ is an eigenvector with respect to $\mu$, and $\lambda \ne \mu$, then $x, y$ are orthogonal:
$$
\lambda (x \cdot y)
=
(\lambda x) \cdot y
=
x A \cdot y
=
x \cdot y A
=
x \cdot (\mu y)
=
\mu (x \cdot y)
$$
and $\lambda \ne \mu$ implies $x \cdot y = 0$.
Therefore in your case the missing eigenvector $v$ relative to the eigenvalue $2$ must be orthogonal to both $(1, 0, 0, −1)$ and $(0, 1, 1, 0)$. It is easy to check that this means
$$
v = (a, b, -b, a)
=
a (1, 0, 0, 1) + b (0, 1, -1, 0),
$$
for some $a, b$.


TITLE: When are there infinite stationary points?
QUESTION [0 upvotes]: Say we are given:
$$f(x_1,x_2) = \alpha (x_1^2+x_2^2) + \beta x_1x_2 + x_1 + x_2$$
then $$\nabla f = \langle f_{x_1}(x_1,x_2), f_{x_2}(x_1,x_2) \rangle = 0$$
for stationary points.
which will give us: $$\nabla f = \langle (2x_1\alpha + \beta x_2 + 1), (2x_2 \alpha + \beta x_1 + 1) \rangle = 0 $$
I need to find
a) $\alpha, \beta \in \mathbb{R}$ such that there is no stationary points.
b) "..." exactly $1$ SP, and its a global strict minimum
c) "..." infinite SP, and all of them global minimizers.
For b) and c) this requires that $\nabla ^2 f(x) $ is positive definite, and positive semidefinite for all SP, respectively.
However for what conditions imposed on $\alpha ,\space \beta$ will there be infinite Stationary points?
Also am I correct for $a) $ in saying $\alpha, \beta = 0$ since then $\nabla f \neq 0$? Or is there something I am overlooking?

REPLY [0 votes]: Critical points solve the linear system
$$
\begin{pmatrix}
2 \alpha & \beta \\
\beta &2 \alpha
\end{pmatrix}
\begin{pmatrix}
x_1 \\
x_2
\end{pmatrix}
=
\begin{pmatrix}
-1 \\
-1
\end{pmatrix}.
$$
A necessary condition to have infinitely many critical points is that the matrix of this system be singular, i.e. $4\alpha^2=\beta^2$. Can you go on with this suggestion?


TITLE: What is the derivative of $c\mapsto F(cx)$ for fixed $x$
QUESTION [0 upvotes]: Let $F:\mathcal B\to \mathbb R$ be a nice functional on Banach space $\mathcal B$. Define for fixed $x\in \mathcal B$ and for $c\in \mathbb R$ the map $\phi_x:\mathbb R\to \mathbb R$ by
$$\phi_x(c)=F(cx).$$
Is there a nice expression for $$\frac{d}{dc}\phi_x(c)?$$
The naive way to do it is $F'(cx)\cdot x$. However the derivative and the multiplication are both unclear.

REPLY [0 votes]: Choice of derivative for $G:X \rightarrow Y$ ($X, Y$ Banach spaces)  in $x$ should be a linear map $D_xG:X \rightarrow Y$ such that
$$
\lim_{h \rightarrow 0} \frac{\lVert G(x+h) -G(x)-D_xG(h)\rVert_Y}{\lVert h \rVert_X} = 0.
$$
It's called Fréchet derivative. Chain rule also applies to Fréchet-derivatives. If some $f$ is Fréchet-differentiable in $x$ and $g$ is Fréchet-differentiable in $f(x)$, then
$$
(D_x(g\circ f)(h)) = (D_{f(x)}g\circ D_xf)(h).
$$
Proof works exactly as it does in $\mathbb{R}^n$.
So if $j:\mathbb{R} \rightarrow \mathcal{B}$ is $j(c) := cx$, then you can easily calculate its Fréchet-derivative by $D_cjh =hx$. And so, for $h \in \mathbb{R}$:
$$
D_c(\phi_x)(h) = D_c(F\circ j)(h) = (D_{j(c)}F\circ D_cj)(h) = (D_{cx}F(hx)) \underset{\text{linearity}}{=} (D_{j(c)}F(x))h
$$
Fréchet-derivative and normal derivative are the same in $\mathbb{R}$, so you would normally write $\phi_x'(c) = (D_{cx}F(x))$.
Example: If $\mathcal{B} = L^2(\mathbb{R})$, $x:\mathbb{R} \rightarrow \mathbb{R}$, $x(t):= \exp(-t^2)$ and $\displaystyle F(y) := \int_{\mathbb{R}} f(y)^2~\mathrm{d}y$, you can easily compute that
$$
D_yF(h) = \int_{\mathbb{R}} 2y(t)h(t)~\mathrm{d}t
$$
for all $h \in \mathcal{B}$. Then
$$
\phi_x'(c) = \int_{\mathbb{R}} 2c\exp(-t^2)\exp(-t^2)~\mathrm{d}x
$$
This example is easy enough for you to also compute it the convential way.


TITLE: Prove an integral inequation
QUESTION [1 upvotes]: Prove that if a function is descending on [0; 1], then
$\forall\theta\in(0;1)\quad \theta\int\limits_0^1f(x)dx \leq \int\limits_0^\theta f(x)dx$.
If a see $\leq$, then there is an idea to use limits. I've tried to play with integral sums but all I've got is that $\int\limits_0^\theta f(x)dx \gt f(\theta)\theta$ (but it's obvious without integral sums).

REPLY [1 votes]: Without assuming that $f$ is differentiable, we can show that $F:[0,1] \to \mathbb{R}$ where $F(\theta) =\int_0^\theta f(x) \, dx$ is concave.
Assuming that $f$ is bounded we have that $F$ is continuous. Since $f$ is nonincreasing we have for $\theta_1 < \theta_2$
$$F(\theta_2) - F\left(\frac{\theta_1 + \theta_2}{2}\right) \leqslant F\left(\frac{\theta_1 + \theta_2}{2}\right) - F(\theta_1)$$
Hence,
$$F\left(\frac{\theta_1 + \theta_2}{2}\right) \geqslant \frac{1}{2} ( F(\theta_1) + F(\theta_2))$$
Since $F$ is continuous and midpoint concave it must be concave and for all $\theta \in (0,1)$,
$$\frac{F(\theta)}{\theta} =\frac{F(\theta)-F(0)}{\theta-0}\geqslant \frac{F(1)-F(0)}{1-0} = F(1)$$
Thus,
$$\frac{1}{\theta}\int_0^\theta f(x) \, dx \geqslant \int_0^1 f(x) \, dx$$


TITLE: limit superior question
QUESTION [0 upvotes]: Suppose that you can apply the Ratio Test to $\Sigma a_{n}$. Let $r$ be the limit of $|a_{n+1}|/|a_{n}|$. Show that $\lim\sup|a_{n}|^{1/n}=r$ as $n\rightarrow\infty$.
I know by definition of lim sup that $\forall\epsilon>0$,  $\exists N_{\epsilon}$  s.t. $x_{n}<r+\epsilon$  $\forall n>N_{\epsilon}$. Not sure how to apply that here though.

REPLY [1 votes]: Let $\epsilon>0$. There is an $N$ so that for all $n\ge N$, $|a_{n+1}|\le (r+\epsilon)|a_n|$
From this it follos that
$\ \ \ |a_{N+1}|\le (r+\epsilon)|a_N|$
$\ \ \ |a_{N+2}|\le (r+\epsilon)^2|a_N|$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \vdots$
$\ \ \ |a_{N+k}|\le (r+\epsilon)^k|a_N|$
Using the above, if $n>N$:
$$
|a_n|=|a_{N+(n-N)}|\le (r+\epsilon)^{n-N}|a_N|
$$
Let $A= |a_N|/(r+\epsilon)^N$.
Then
$$
\root n\of {|a_n| }\le A^{1/n}(r+\epsilon)
$$
for all $n>N$.
Now, $A^{1/n}\rightarrow1$ as $n\rightarrow\infty$; so,
$$
\limsup_{n\rightarrow\infty} \root n\of{|a_n|}\le r+\epsilon.
$$
Since $\epsilon$ was arbitrary, we have
$$
\limsup_{n\rightarrow\infty} \root n\of{|a_n|}\le r.
$$
Now show that $\liminf\limits_{n\rightarrow\infty} \root n\of{|a_n|}\ge r$. I'll leave that for you.

Note that the  above can be modified slightly to show that $\limsup\limits_{n\rightarrow\infty}\root n\of {|a_n|}\le\limsup\limits_{n\rightarrow\infty}\root n\of {\Bigl|{a_{n+1}\over a_n}\Bigr|}$.  One can also show that $\liminf\limits_{n\rightarrow\infty}\root n\of {|a_n|}\ge\liminf\limits_{n\rightarrow\infty}\root n\of {\Bigl|{a_{n+1}\over a_n}\Bigr|}$.
From this, your result easily follows.
Note also please, that the result has little to do with series...

REPLY [0 votes]: We will prove the following:
If $ \lim|\frac{a_{n+1}}{a_n}|=l$ then it is $\lim |a_n|^{1/n} =l \quad (\spadesuit)$
To prove this we need the following two inequalities:
(a) $\limsup |a_n|^{1/n} \leq \limsup|\frac{a_{n+1}}{a_n}|$
(b) $\liminf|\frac{a_{n+1}}{a_n}| \leq \liminf |a_n|^{1/n}$
Proof of $(\spadesuit)$: From (a) and (b) we have that $\liminf|\frac{a_{n+1}}{a_n}| \leq \liminf |a_n|^{1/n} \leq\limsup |a_n|^{1/n} \leq \limsup|\frac{a_{n+1}}{a_n}| \quad (\diamond)$
Since $ \lim|\frac{a_{n+1}}{a_n}|=l$ exists we have that $ \liminf|\frac{a_{n+1}}{a_n}|= \limsup|\frac{a_{n+1}}{a_n}|= \lim|\frac{a_{n+1}}{a_n}|=l$ and then $(\diamond)$ gives:
$\liminf |a_n|^{1/n}=\limsup |a_n|^{1/n}=\lim |a_n|^{1/n}=l$. Q.E.D
The converse does not hold Here it is a counterexample:
Consider the sequance $(a_n), \quad n\in\mathbb{N}$ with $a_{2n-1}=\frac{1}{2^n}$ and $a_{2n}=\frac{1}{2^{n-1}}$


TITLE: Definition : The Distribution of laplacian of a function?
QUESTION [0 upvotes]: Let $u\in H_0^1(\Omega)$(but i think we can be way more general than than).
How is $\langle\Delta u,\phi\rangle$ (distribution) defined for $\phi \in D(\Omega)$ a test function ?

REPLY [1 votes]: As with most definitions in distribution theory, it boils down to "formal integration by parts":
$$\langle \Delta u,\phi \rangle = -\langle \nabla u,\nabla \phi \rangle = \langle u,\Delta \phi \rangle.$$
In general when doing this you need to be careful to be sure that the boundary terms in each of these integrations actually vanish in your setting. In the case described in the question this is trivial since $\phi \in D(\Omega)$.


text:
\section{Introduction}\vspace{-.4em}
In linear programming, the goal is to make a optimal decision under a linear objective and subject to linear constraints. Traditionally, a linear program is designed using knowledge of relevant costs and constraints. More recently, methodologies that are data-driven have emerged. For example, in the ``predict-then-optimize'' paradigm \citep{Elmachtoub19}, linear programs are learned from direct observations of previous costs or constraints.

Inverse optimization~(IO) \citep{Burton92,Troutt95,Ahuja01}, in contrast, learns linear programs from observations of optimal decisions rather than of the costs or constraints themselves. The IO approach is particularly important when observations come from optimizing agents (e.g., experts \citep{Chan14,Barmann17} or customers \citep{Dong18}) who make near-optimal decisions with respect to their internal (unobserved) optimization models.

From a machine learning perspective, the IO setup is as follows: we are given feature vectors $\{\bu_{1}, \bu_{2}, \dots, \bu_{N}\}$ representing conditions (e.g., time, prices, weather) and we observe the corresponding decision targets $\{\xt_{1}, \xt_{2}, \dots, \xt_{N}\}$ (e.g., quantities, actions) determined by an unknown optimization process, which in our case is assumed linear. We view IO as the problem of inferring a constrained optimization model that gives identical (or equivalent) decisions, and which generalizes to novel conditions $\bu$. The family of candidate models is assumed parametrized by some vector~$\bw$.

Learning a constrained optimizer that makes the observations both feasible \emph{and} optimal poses multiple challenges that have not been explicitly addressed. For instance, parameter setting $\bw_1$ in Figure~\ref{fig:example1} makes the observed decision $\xt_1$ optimal but not feasible, $\bw_2$ produces exactly the opposite result, and some $\bw$ values (black-hatched region in Figure~\ref{fig:example1}) are not even admissible because they will result in empty feasible regions. Finding a parameter such as $\bw_3$ that is consistent with the observations can be difficult. We formulate the learning problem in a novel way, and tackle it with gradient-based methods despite the inherent bi-level nature of learning. Using gradients from backpropagation or implicit differentiation, we successfully learn linear program instances of various sizes as well as learning the costs and right-hand coefficients of a minimum-cost multi-commodity flow problem.

\begin{figure}[t]

\hspace{-3.95em}\includegraphics[width=1.104\textwidth]{Figure/fig1_final.pdf}\vspace{-.35em}
\caption{A depiction of our constrained learning formulation. We learn a parametric linear program (PLP), here parametrized by a feature $u$ and weights $\bw \!=\! (w_1, w_2)$ and using a single training observation $(u_1, \xt_1)$. The PLP corresponding to three parameter settings $\bw_1, \bw_2, \bw_3$ are shown, with the cost vector and feasible region corresponding to $u_1$ emphasized. The goal of learning is to find solutions such as $\bw^* = \bw_3$. (See Appendix for the specific PLP used in this example.) \label{fig:example1} \vspace{-1em}}
\end{figure}

{\bf Parametric Linear Programs \;} In a linear program (LP), decision variables $\bx \in \bbR^D$ may vary, and the cost coefficients ${\bc \in \bbR^D}$, inequality constraint coefficients ${\bA \in \bbR^{M_1 \times D}}$, ${\bb \in \bbR^{M_1}}$, and equality constraint coefficients ${\bG \in \bbR^{M_2 \times D}}$, ${\bh \in \bbR^{M_2}}$ are all constant. In a {\em parametric linear program} (PLP), the coefficients (and therefore the optimal decisions) may depend on features~$\bu$. In order to infer a PLP from data, one may define a suitable hypothesis space parametrized by $\bw$. We refer to this hypothesis space as the form of our {\em forward optimization problem} (FOP).\\[-0.4em]

\noindent\begin{minipage}{.25\linewidth}

\begin{equation} \label{eq:LP} \tag{LP}
\begin{aligned}
\textstyle \min_{\bx} \;\; & \bc^T \bx \\
\text{s.t.}           \;\; & \bA \bx \leq \bb \\
                           & \bG \bx =    \bh
\end{aligned}
\end{equation}

\end{minipage}

\begin{minipage}{.34\linewidth}

\begin{equation} \label{eq:PLP} \tag{PLP}
\begin{aligned}
\textstyle \min_{\bx} \;\; & \bc(\bu)^T \bx \\
\text{s.t.}           \;\; & \bA(\bu)\bx \leq \bb(\bu)\\
                           & \bG(\bu)\bx =    \bh(\bu)
\end{aligned}
\end{equation}
\end{minipage}
\begin{minipage}{.41\linewidth}

\begin{equation} \label{eq:FOP} \tag{FOP}
\begin{aligned}
\textstyle \min_{\bx} \;\; & \bc(\bu, \bw)^T \bx \\
\text{s.t.}           \;\; & \bA(\bu, \bw)\bx \leq \bb(\bu, \bw)\\
                           & \bG(\bu, \bw)\bx =    \bh(\bu, \bw)
\end{aligned}
\end{equation}
\end{minipage}\\[.4em]

A choice of hypothesis $\bw$ in \eqref{eq:FOP} identifies a PLP, and a subsequent choice of conditions $\bu$ identifies an LP. The LP can then be solved to yield an optimal decision $\bx^*$ under the model. These predictions of optimal decisions can be compared to observations at training time, or can be used to anticipate optimal decisions under novel conditions $\bu$ at test time.

config: arxiv


TITLE: Disjunctive Normal Form
QUESTION [1 upvotes]: I need some help understanding how to convert a formula into disjunctive normal form.
Can anybody explain how one would write φ = ((p ∨ q ∨ r) → (¬p ∧ r)) in disjunctive normal form?
Is it possible to use truth tables to help converting to DNF form?

REPLY [1 votes]: I think a truth table could work for simpler propositions, but this one in question is relatively complicated - using logical equivalences might be easier.
Consider a conditional proposition $P \implies Q$, this is equivalent to $(\neg P \vee Q)$, which can be verified using a truth table.
Thus considering, $\varphi = ((P \vee Q \vee R) \implies (\neg P \land R))$,
\begin{align}
&\phantom{{}\equiv{}} ((P \vee Q \vee R) \implies (\neg P \land R))\\
&\equiv \neg (P \vee Q \vee R) \vee (\neg P \land R) \\
&\equiv ((\neg P) \land (\neg Q) \land (\neg R)) \vee (\neg P \land R) \\
&\equiv ((\neg P \land R) \vee \neg P) \land ((\neg P \land R) \vee \neg Q) \land ((\neg P \land R) \vee \neg R) \\
& \phantom{((P \vee Q \vee R) \implies} \vdots
\end{align}
which can be simplified using logical equivalences.
Hope this helps.


text:
\begin{document}

\title{Nonlinear Connections and Exact Solutions\\
in Einstein and Extra Dimension Gravity}
\author{Sergiu I. Vacaru \\

Departamento de Fisica, Instituto Superior Tecnico, \\
Av. Rovisco Pais 1, Lisbon 1049-001, Portugal,\\
and\\
Departamento de Fisica Aplicada, \\
Facultad de Ciencias Experimentales,\\
Universidad de Huelva, 21071 Huelva, Spain }
\maketitle

\begin{abstract}
We outline a new geometric method of constructing exact solutions of
gravitational field equations parametrized by generic off--diagonal metrics,
anholonomic frames and possessing, in general, nontrivial torsion and
nonmetricity. The formalism of nonlinear connections is elaborated for
(pseudo) Riemannian and Einstein--Cartan--Weyl spaces.
\end{abstract}

\section{ Introduction:\ Nonlinear Connections}

We consider a $(n+m)$--dimensional manifold $V^{n+m},$ provided with general
metric and linear connection structure and of necessary smooth class. It is
supposed that in any point $u\in V^{n+m}$ there is a local splitting into $
n- $ and $m$--dimensional subspaces, $V_{u}^{n+m}=V_{u}^{n}\oplus
V_{u}^{mn}. $ The local/abstract coordinates are denoted $u=(x,y),$ or $
u^{\alpha }=\left( x^{i},y^{a}\right) ,$ where $i,j,k,...=1,2,...,n$ and $
a,b,c,...=n+1,n+2,...,n+m.$ The metric is parametrized in the form
\begin{equation}
\mathbf{g}=g_{ij}(u)\mathbf{e}^{i}\otimes \mathbf{e}^{j}+h_{ab}(u)\mathbf{e}
^{a}\otimes \mathbf{e}^{b}  \label{metr}
\end{equation}
where
\begin{equation}
\mathbf{\vartheta }^{\mu }=[\vartheta ^{i}=dx^{i},\vartheta
^{a}=dy^{a}+N_{i}^{a}(u)dx^{i}]  \label{ddif}
\end{equation}
is the dual frame to
\begin{equation}
\mathbf{e}_{\nu }=[e_{i}=\frac{\partial }{\partial x^{i}}-N_{i}^{a}(u)\frac{
\partial }{\partial y^{a}},e_{a}=\frac{\partial }{\partial y^{a}}].
\label{dder}
\end{equation}

Let us denote by $\pi ^{\top }:TV^{n+m}\rightarrow TV^{n}$ the differential
of a map $\pi :V^{n+m}\rightarrow V^{n}$ defined by fiber preserving
morphisms of the tangent bundles $TV^{n+m}$ and $TV^{n}.$ The kernel of $\pi
^{\top }$is just the vertical subspace $vV^{n+m}$ with a related inclusion
mapping $i:vV^{n+m}\rightarrow TV^{n+m}.$

\begin{definition}
A nonlinear connection (N--connection) $\mathbf{N}$ on space $V^{n+m}$ is
defined by the splitting on the left of an exact sequence
\begin{equation*}
0\rightarrow vV^{n+m}\rightarrow TV^{n+m}\rightarrow
TV^{n+m}/vV^{n+m}\rightarrow 0,
\end{equation*}
i. e. by a morphism of submanifolds $\mathbf{N:\ \ }TV^{n+m}\rightarrow
vV^{n+m}$ such that $\mathbf{N\circ i}$ is the unity in $vV^{n+m}.$
\end{definition}

Equivalently, a N--connection is defined by a Whitney sum of horizontal (h)
subspace $\left( hV^{n+m}\right) $ and vertical (v) subspaces,
\begin{equation}
TV^{n+m}=hV^{n+m}\oplus vV^{n+m}.  \label{whitney}
\end{equation}
A space provided with N--connection structure will be denoted $V_{N}^{n+m}.$
We shall use boldfaced indices for the geometric objects adapted to
N--connections. The well known class of linear connections consists a
particular subclass with the coefficients being linear on $y^{a},$ i. e. $
N_{i}^{a}(u)=\Gamma _{bj}^{a}(x)y^{b}.$

To any sets $N_{i}^{a}(u),$ we can associate certain anholonomic frames (\ref
{ddif}) and (\ref{dder}), with associated N--connection structure,
satisfying the anholonomy relations
\begin{equation*}
\lbrack \mathbf{\vartheta }_{\alpha },\mathbf{\vartheta }_{\beta }]=\mathbf{
\vartheta }_{\alpha }\mathbf{\vartheta }_{\beta }-\mathbf{\vartheta }_{\beta
}\mathbf{\vartheta }_{\alpha }=W_{\alpha \beta }^{\gamma }\mathbf{\vartheta }
_{\gamma }
\end{equation*}
with (antisymmetric) nontrivial anholonomy coefficients $W_{ia}^{b}=\partial
_{a}N_{i}^{b}$ and $W_{ji}^{a}=\Omega _{ij}^{a},$ where $\Omega
_{ij}^{a}=e_{[i}N_{j]}^{a}$ are the coefficients of the N--connection
curvature.

Essentially, the method to be presented in this work is based on the notion
of N--connection and considers a Whitney-like splitting of the tangent
bundle to a manifold into a horizontal (see discussion and bibliography in
Refs. \cite{v1,v2,v3}). Here we emphasize that the geometrical aspects of
the N--connection formalism has been studied since the first papers of E.
Cartan \cite{car1}\ and A. Kawaguchi \cite{ak1,ak2}\ (who used it in
component form for Finsler geometry), then one should be mentioned the so
called Ehressman connection \cite{eh}) and the work of W. Barthel \cite{wb}
where the global definition of N--connection was given. The monograph \cite
{mhss} consider the N--connection formalism elaborated and applied to
geometry of generalized Finsler--Lagrange and Cartan--Hamilton spaces. There
is a set of contributions by Spanish authors, see, for instance, \cite
{ml1,fe,af}.

We considered N--connections for Clifford and spinor structures \cite
{vsp1,vsp2}, on superbundles and (super) string theory \cite{vstr1} as well
in noncommutative geometry and gravity \cite{vnc}. The idea to apply the
N--connections formalism as a new geometric method of constructins exact
solutions in gravity theories was suggested in Refs. \cite{vex1,vex2} and
developed and applied in a number of works, see for instance, Ref. \cite
{vt,vs1,dv}). This contribution outlines the author's and co--authors'
results.

\section{\protect\bigskip N--distinguished Torsions and Curvatures}

The geometric constructions can be adapted to the N--connection structure:

\begin{definition}
A distinguished connection (d--connection) $\mathbf{D}=\{\mathbf{\Gamma }
_{\beta \gamma }^{\alpha }\}$ on $V_{N}^{n+m}$ is a linear connection
conserving under parallelism the Whitney sum (\ref{whitney}).
\end{definition}

Any d--connection $\mathbf{D}$ is represented by irreducible h- \
v--components $\mathbf{\Gamma }_{\beta \gamma }^{\alpha }=\left( L_{jk}^{i},
\widetilde{L}_{bk}^{a},C_{jc}^{i},\widetilde{C}_{bc}^{a}\right) $ stated
with respect to N--elongated frames (\ref{ddif}) and (\ref{dder}). This
defines a N--adapted splitting into h-- and v--covariant derivatives, $
\mathbf{D}=D^{[h]}+D^{[v]},$ where $D^{[h]}=(L,\widetilde{L})$ and $
D^{[v]}=(C,\widetilde{C}).$ A d--tensor (distinguished tensor, for instance,
a d--metric like (\ref{metr})) formalism and d--covariant differential and
integral calculus can be elaborated \cite{v1} for spaces provided with
general N--connection, d--connection and d--metric structure and nontrivial
nonmetricity
\begin{equation*}
\mathcal{Q}_{\alpha \beta }\circeq -\mathbf{Dg}_{\alpha \beta }.
\end{equation*}
The simplest way is to use N--adapted differential forms like $\mathbf{
\Gamma }_{\beta }^{\alpha }=\mathbf{\Gamma }_{\beta \gamma }^{\alpha }
\mathbf{\vartheta }^{\gamma }$ with the coefficients defined with respect to
(\ref{ddif}) and (\ref{dder}).

\begin{theorem}
The torsion $\mathcal{T}^{\alpha }\doteqdot \mathbf{D\vartheta }^{\alpha }=d
\mathbf{\vartheta }^{\alpha }+\Gamma _{\beta }^{\alpha }\wedge \mathbf{
\vartheta }^{\beta }$ of a d--connection has the irreducible h- v--
components (d--torsions),
\begin{eqnarray}
T_{\ jk}^{i} &=&L_{\ [jk]}^{i},\ T_{\ ja}^{i}=-T_{\ aj}^{i}=C_{\ ja}^{i},\
T_{\ ji}^{a}=\Omega _{\ ji}^{a},\   \notag \\
T_{\ bi}^{a} &=&T_{\ ib}^{a}=\frac{\partial N_{i}^{a}}{\partial y^{b}}-L_{\
bi}^{a},\ T_{\ bc}^{a}=C_{\ [bc]}^{a}.  \label{dtors}
\end{eqnarray}

\begin{proof}
By a straightforward calculation we can verify the formulas.
\end{proof}
\end{theorem}

The Levi--Civita linear connection $\nabla =\{^{\nabla }\mathbf{\Gamma }
_{\beta \gamma }^{\alpha }\},$ with vanishing torsion and nonmetricity, is
not adapted to the global splitting (\ref{whitney}). One holds:

\begin{proposition}
There is a preferred, canonical d--connection structure,$\ \widehat{\mathbf{
\Gamma }}\mathbf{,}$ $\ $on $V_{N}^{n+m},$ constructed only from the metric
coefficients $[g_{ij},h_{ab},N_{i}^{a}]$ and satisfying the conditions $
\widehat{\mathbf{Q}}_{\alpha \beta }=0$ and $\widehat{T}_{\ jk}^{i}=0$ and $
\widehat{T}_{\ bc}^{a}=0.$

\begin{proof}
By straightforward calculations with respect to the N--adapted bases (\ref
{ddif}) and (\ref{dder}), we can verify that the connection
\begin{equation}
\widehat{\mathbf{\Gamma }}_{\beta \gamma }^{\alpha }=\ ^{\nabla }\mathbf{
\Gamma }_{\beta \gamma }^{\alpha }+\ \widehat{\mathbf{P}}_{\beta \gamma
}^{\alpha }  \label{cdc}
\end{equation}
with the deformation d--tensor
\begin{equation*}
\widehat{\mathbf{P}}_{\beta \gamma }^{\alpha }=(P_{jk}^{i}=0,P_{bk}^{a}=
\frac{\partial N_{k}^{a}}{\partial y^{b}},P_{jc}^{i}=-\frac{1}{2}
g^{ik}\Omega _{\ kj}^{a}h_{ca},P_{bc}^{a}=0)
\end{equation*}
satisfies the conditions of this Proposition. It should be noted that, in
general, the components $\widehat{T}_{\ ja}^{i},\ \widehat{T}_{\ ji}^{a}$
and $\widehat{T}_{\ bi}^{a}$ are not zero. This is an anholonomic frame (or,
equivalently, off--diagonal metric) frame effect.
\end{proof}
\end{proposition}

The torsion of the connection (\ref{cdc}) is denoted $\widehat{\mathbf{T}}
_{\beta \gamma }^{\alpha }.$ In a similar form we can prove:

\begin{theorem}
The curvature $\mathcal{R}_{\ \beta }^{\alpha }\doteqdot \mathbf{D\Gamma }
_{\beta }^{\alpha }=d\mathbf{\Gamma }_{\beta }^{\alpha }-\mathbf{\Gamma }
_{\beta }^{\gamma }\wedge \mathbf{\Gamma }_{\gamma }^{\alpha }$ of a
d--connection $\mathbf{\Gamma }_{\gamma }^{\alpha }$ has the irreducible h-
v-- components (d--curvatures),
\begin{eqnarray}
R_{\ hjk}^{i} &=&e_{k}L_{\ hj}^{i}-e_{j}L_{\ hk}^{i}+L_{\ hj}^{m}L_{\
mk}^{i}-L_{\ hk}^{m}L_{\ mj}^{i}-C_{\ ha}^{i}\Omega _{\ kj}^{a},  \notag \\
R_{\ bjk}^{a} &=&e_{k}L_{\ bj}^{a}-e_{j}L_{\ bk}^{a}+L_{\ bj}^{c}L_{\
ck}^{a}-L_{\ bk}^{c}L_{\ cj}^{a}-C_{\ bc}^{a}\Omega _{\ kj}^{c},  \notag \\
R_{\ jka}^{i} &=&e_{a}L_{\ jk}^{i}-D_{k}C_{\ ja}^{i}+C_{\ jb}^{i}T_{\
ka}^{b},  \label{dcurv} \\
R_{\ bka}^{c} &=&e_{a}L_{\ bk}^{c}-D_{k}C_{\ ba}^{c}+C_{\ bd}^{c}T_{\
ka}^{c},  \notag \\
R_{\ jbc}^{i} &=&e_{c}C_{\ jb}^{i}-e_{b}C_{\ jc}^{i}+C_{\ jb}^{h}C_{\
hc}^{i}-C_{\ jc}^{h}C_{\ hb}^{i},  \notag \\
R_{\ bcd}^{a} &=&e_{d}C_{\ bc}^{a}-e_{c}C_{\ bd}^{a}+C_{\ bc}^{e}C_{\
ed}^{a}-C_{\ bd}^{e}C_{\ ec}^{a}.  \notag
\end{eqnarray}
\end{theorem}

Contracting the components of (\ref{dcurv}) we prove:

\begin{corollary}
a) The Ricci d--tensor $\mathbf{R}_{\alpha \beta }\doteqdot \mathbf{R}_{\
\alpha \beta \tau }^{\tau }$ has the irreducible h- v--components
\begin{equation}
R_{ij}\doteqdot R_{\ ijk}^{k},\ \ R_{ia}\doteqdot -R_{\ ika}^{k},\
R_{ai}\doteqdot R_{\ aib}^{b},\ R_{ab}\doteqdot R_{\ abc}^{c}.
\label{dricci}
\end{equation}
b) The scalar curvature of a d--connection is
\begin{equation*}
\overleftarrow{\mathbf{R}}\doteqdot \mathbf{g}^{\alpha \beta }\mathbf{R}
_{\alpha \beta }=g^{ij}R_{ij}+h^{ab}R_{ab}.
\end{equation*}
c) The Einstein d--densor is computed $\mathbf{G}_{\alpha \beta }=\mathbf{R}
_{\alpha \beta }-\frac{1}{2}\mathbf{g}_{\alpha \beta }\overleftarrow{\mathbf{
R}}.$
\end{corollary}

In modern gravity theories one considers more general linear connections
generated by deformations of type $\mathbf{\Gamma }_{\beta \gamma }^{\alpha
}=\widehat{\mathbf{\Gamma }}_{\beta \gamma }^{\alpha }+\mathbf{P}_{\beta
\gamma }^{\alpha }.$We can split all geometric objects into canonical and
post-canonical pieces which results in N--adapted geometric constructions.
For instance,
\begin{equation}
\mathcal{R}_{\ \beta }^{\alpha }=\widehat{\mathcal{R}}_{\ \beta }^{\alpha }+
\mathbf{D}\mathcal{P}_{\ \beta }^{\alpha }+\mathcal{P}_{\ \gamma }^{\alpha
}\wedge \mathcal{P}_{\ \beta }^{\gamma }  \label{deformcurv}
\end{equation}
for $\mathcal{P}_{\beta }^{\alpha }=\mathbf{P}_{\beta \gamma }^{\alpha }
\mathbf{\vartheta }^{\gamma }.$

\section{Anholonomic Frames and Nonmetricity\newline
in String Gravity}

For simplicity, we investigate here a class of spacetimes when the
nonmetricity and torsion have nontrivial components of type
\begin{equation}
\mathcal{T}\doteqdot \mathbf{e}_{\alpha }\rfloor \mathcal{T}^{\alpha
}=\kappa _{0}\phi ,\ \mathcal{Q}\doteqdot \frac{1}{4}\mathbf{g}^{\alpha
\beta }\mathcal{Q}_{\alpha \beta }=\kappa _{1}\phi ,\ \mathbf{\Lambda }
\doteqdot \mathbf{\vartheta }^{\alpha }\mathbf{e}^{\beta }\rfloor (\mathcal{Q
}_{\alpha \beta }-\mathcal{Q}\mathbf{g}_{\alpha \beta })=\kappa _{2}\phi
\label{aux01}
\end{equation}
where $\kappa _{0},\kappa _{1},\kappa _{2}=const$ and $\phi =\phi _{\alpha }
\mathbf{\vartheta }^{\alpha }.$ The abstract indices in (\ref{aux01}) are
''upped'' and ''lowed'' by using $\eta _{\alpha

{\acute{}}

\beta

{\acute{}}

}$ and its inverse defined from the vielbein decompositions of d--metric, $
\mathbf{g}_{\alpha \beta }=\mathbf{e}_{\alpha }^{\alpha ^{\prime }}\mathbf{e}
_{\beta }^{\beta ^{\prime }}\eta _{\alpha ^{\prime }\beta ^{\prime }}.$

Let us consider the strengths $\mathbf{H}_{\nu \mu }\doteqdot \widehat{
\mathbf{D}}_{\nu }\phi _{\mu }-\widehat{\mathbf{D}}_{\mu }\phi _{\nu
}+W_{\mu \nu }^{\gamma }\phi _{\gamma }$ (intensity of $\phi _{\gamma })$
and $\widehat{\mathbf{H}}_{\nu \lambda \rho }\doteqdot \mathbf{e}_{\nu }
\mathbf{B}_{\lambda \rho }+\mathbf{e}_{\rho }\mathbf{B}_{\nu \lambda }+
\mathbf{e}_{\lambda }\mathbf{B}_{\rho \nu }$ (antysimmetric torsion of the $
\mathbf{B}_{\rho \nu }=-\mathbf{B}_{\nu \rho }$ from the bosonic model of
string theory with dilaton field $\Phi $) and introduce
\begin{eqnarray*}
\mathbf{H}_{\nu \lambda \rho } &\doteqdot &\widehat{\mathbf{Z}}_{\nu \lambda
\rho }+\widehat{\mathbf{H}}_{\nu \lambda \rho }, \\
\widehat{\mathbf{Z}}_{\nu \lambda } &\doteqdot &\widehat{\mathbf{Z}}_{\nu
\lambda \rho }\mathbf{\vartheta }^{\rho }=\mathbf{e}_{\lambda }\rfloor
\widehat{\mathbf{T}}_{\nu }-\mathbf{e}_{\nu }\rfloor \widehat{\mathbf{T}}
_{\lambda }+\frac{1}{2}(\mathbf{e}_{\nu }\rfloor \mathbf{e}_{\lambda
}\rfloor \widehat{\mathbf{T}}_{\lambda })\mathbf{\vartheta }^{\gamma }.
\end{eqnarray*}
We denote the energy--momentums of fields:
\begin{equation*}
\mathbf{\Sigma }_{\alpha \beta }^{[\phi ]}\doteqdot \mathbf{H}_{\alpha }^{\
\mu }\mathbf{H}_{\beta \mu }-\frac{1}{4}\mathbf{g}_{\alpha \beta }\mathbf{H}
^{\nu \mu }\mathbf{H}_{\nu \mu }+\mu ^{2}(\phi _{\alpha }\phi _{\beta }-
\frac{1}{2}\mathbf{g}_{\alpha \beta }\phi _{\nu }\phi ^{\nu }),
\end{equation*}
$\mu ^{2}=const,\ \mathbf{\Sigma }_{\alpha \beta }^{[mat]}$ is the source
from any possible matter fields and $\mathbf{\Sigma }_{\alpha \beta }^{[T]}(
\widehat{\mathbf{T}}_{\nu },\Phi )$ contains contributions of torsion and
dilatonic fields.

\begin{theorem}
\label{theorfields}The dynamics of sigma model of bosonic string gravity
with generic off--diagonal metrics, effective matter and torsion and
nonmetricity (\ref{aux01}) is defined by the system of field equations
\begin{eqnarray}
\widehat{\mathbf{R}}_{\alpha \beta }-\frac{1}{2}\mathbf{g}_{\alpha \beta }
\overleftarrow{\mathbf{\hat{R}}} &=&k(\mathbf{\Sigma }_{\alpha \beta
}^{[\phi ]}+\mathbf{\Sigma }_{\alpha \beta }^{[mat]}+\mathbf{\Sigma }
_{\alpha \beta }^{[T]}),  \label{eecdc1} \\
\widehat{\mathbf{D}}_{\nu }\mathbf{H}^{\nu \mu } &=&\mu ^{2}\phi ^{\mu },\
\widehat{\mathbf{D}}^{\nu }(\mathbf{H}_{\nu \lambda \rho })=0,  \notag
\end{eqnarray}
where $k=const,$ and the Euler--Lagrange equations for the matter fields are
considered on background $V_{N}^{n+m}$).

\begin{proof}
See details in Ref. \cite{v2}.
\end{proof}
\end{theorem}

In terms of differential forms the eqs. (\ref{eecdc1}) are written
\begin{equation}
\eta _{\alpha \beta \gamma }\wedge \widehat{\mathcal{R}}_{\ }^{\beta \gamma
}=\widehat{\Upsilon }_{\alpha },  \label{eecdc2}
\end{equation}
where, for the volume 4--form $\eta \doteqdot \ast 1$ with the Hodje
operator ''$\ast $'', $\eta _{\alpha }\doteqdot \mathbf{e}_{\alpha }\rfloor
\eta ,$ $\eta _{\alpha \beta }\doteqdot \mathbf{e}_{\beta }\rfloor \eta
_{\alpha },$ $\eta _{\alpha \beta \gamma }\doteqdot \mathbf{e}_{\gamma
}\rfloor \eta _{\alpha \beta },...,\widehat{\mathcal{R}}_{\ }^{\beta \gamma
} $ is the curvature 2--form and $\Upsilon _{\alpha }$ denote all possible
sources defined by using the canonical d--connection. The deformation of
connection (\ref{cdc}) defines a deformation of the curvature tensor of type
(\ref{deformcurv}) but with respect to the curvature of the Levi--Civita
connection, $\ ^{\nabla }\mathcal{R}_{\ }^{\beta \gamma }.$ The
gravitational field equations (\ref{eecdc2}) transforms into
\begin{equation}
\eta _{\alpha \beta \gamma }\wedge \ ^{\nabla }\mathcal{R}_{\ }^{\beta
\gamma }+\eta _{\alpha \beta \gamma }\wedge \ ^{\nabla }\mathcal{Z}_{\
}^{\beta \gamma }=\widehat{\Upsilon }_{\alpha },  \label{eecdc3}
\end{equation}
where $^{\nabla }\mathcal{Z}_{\ \ \gamma }^{\beta }=\nabla \mathcal{P}_{\ \
\gamma }^{\beta }+\mathcal{P}_{\ \ \alpha }^{\beta }\wedge \mathcal{P}_{\ \
\gamma }^{\alpha }.$

\begin{corollary}
A subclass of solutions of the gravitational field equations for the
canonical d--connection defines also solutions of the Einstein equations for
the Levi--Civita connection if and only if $\eta _{\alpha \beta \gamma
}\wedge \ ^{\nabla }\mathcal{Z}_{\ }^{\beta \gamma }=0$ and $\widehat{
\Upsilon }_{\alpha }=\ ^{\nabla }\Upsilon _{\alpha },$ (i. e. the effective
source is the same for both type of connections).
\end{corollary}

\begin{proof}
It follows from the Theorem \ref{theorfields}.
\end{proof}

This property is very important for constructing exact solutions in Einstein
and string gravity, parametrized by generic off--diagonal metrics and
ahnolonomic frames with associated N--connection structure (see Refs. in
\cite{v1,v2} and \cite{v3}) and equations (\ref{eecdc3})).

Let us consider a five dimensional ansatz for the metric (\ref{metr}) and
frame (\ref{ddif}) when $u^{\alpha }=(x^{i},y^{4}=v,y^{5});i=1,2,3$ and the
coefficients
\begin{eqnarray}
g_{ij} &=&diag[g_{1}=\pm
1,g_{2}(x^{2},x^{3}),g_{3}(x^{2},x^{3})],h_{ab}=diag[h_{4}(x^{k},v),h_{5}(x^{k},v)],
\notag \\
N_{i}^{4} &=&w_{i}(x^{k},v),N_{i}^{5}=n_{i}(x^{k},v)  \label{ansatz1}
\end{eqnarray}
are some functions of necessary smooth class. The partial derivative are
briefly denoted $a^{\bullet }=\partial a/\partial x^{2},a^{^{\prime
}}=\partial a/\partial x^{3},a^{\ast }=\partial a/\partial v.$

\section{Main results:}

\begin{theorem}
\label{ricci}The nontrivial components of the Ricci d--tensors (\ref{dricci}
) for the ca\-no\-nical d--connection (\ref{cdc}) are
\begin{eqnarray}
R_{2}^{2} &=&R_{3}^{3}=-\frac{1}{2g_{2}g_{3}}[g_{3}^{\bullet \bullet }-\frac{
g_{2}^{\bullet }g_{3}^{\bullet }}{2g_{2}}-\frac{(g_{3}^{\bullet })^{2}}{
2g_{3}}+g_{2}^{^{\prime \prime }}-\frac{g_{2}^{\prime }g_{3}^{\prime }}{
2g_{3}}-\frac{(g_{2}^{^{\prime }})^{2}}{2g_{2}}],  \notag \\
R_{4}^{4} &=&R_{5}^{5}=-\frac{1}{2h_{4}h_{5}}[h_{5}^{\ast \ast }-h_{5}^{\ast
}(\ln \left| \sqrt{\left| h_{4}h_{5}\right| }\right| )^{\ast }],
\label{riccia} \\
R_{4i} &=&-w_{i}\frac{\beta }{2h_{5}}-\frac{\alpha _{i}}{2h_{5}},\ R_{5i}=-
\frac{h_{5}}{2h_{4}}[n_{i}^{\ast \ast }+\gamma n_{i}^{\ast }],  \notag
\end{eqnarray}
\begin{equation*}
\alpha _{i}=\partial _{i}h_{5}^{\ast }-h_{5}^{\ast }\partial _{i}\ln \left|
\sqrt{\left| h_{4}h_{5}\right| }\right| ,\ \beta =h_{5}^{\ast \ast
}-h_{5}^{\ast }[\ln \left| \sqrt{\left| h_{4}h_{5}\right| }\right| ]^{\ast
},\ \gamma =3h_{5}^{\ast }/2h_{5}-h_{4}^{\ast }/h_{4}
\end{equation*}
$h_{4}^{\ast }\neq 0$ and $h_{5}^{\ast }\neq 0.$
\end{theorem}

\begin{proof}
It is provided in Ref. \cite{v2}.
\end{proof}

\begin{corollary}
The Einstein equations (\ref{eecdc2}) for the ansatz (\ref{ansatz1}) are
compatible for vanishing sources and if and only if the nontricial
components of the source, with respect to the frames (\ref{dder}) and (\ref
{ddif}), are any functions of type
\begin{equation*}
\widehat{\Upsilon }_{2}^{2}=\widehat{\Upsilon }_{3}^{3}=\Upsilon
_{2}(x^{2},x^{3},v),\ \widehat{\Upsilon }_{4}^{4}=\widehat{\Upsilon }
_{5}^{5}=\Upsilon _{4}(x^{2},x^{3})\text{\mbox{ and }}\widehat{\Upsilon }
_{1}^{1}=\Upsilon _{2}+\Upsilon _{4}.
\end{equation*}

\begin{proof}
The proof, see details in \cite{v2}, follows from the Theorem \ref{ricci}
with the nontrivial components of the Einstein d-tensor, $\widehat{\mathbf{G}
}_{\ \beta }^{\alpha }=\widehat{\mathbf{R}}_{\ \beta }^{\alpha }-\frac{1}{2}
\delta _{\ \beta }^{\alpha }\overleftarrow{\mathbf{\hat{R}}},$ computed to
satisfy the conditions
\begin{equation*}
G_{1}^{1}=-(R_{2}^{2}+R_{4}^{4}),G_{2}^{2}=G_{3}^{3}=-R_{4}^{4}(x^{2},x^{3},v),G_{4}^{4}=G_{5}^{5}=-R_{2}^{2}(x^{2},x^{3}).
\end{equation*}
\end{proof}
\end{corollary}

Having the values (\ref{riccia}), we can proove \cite{v2} the

\begin{theorem}
The system of gravitational field equations (\ref{eecdc1}) (equivalently, of
(\ref{eecdc2})) for the ansatz (\ref{ansatz1}) can be solved in general form
if there are given certain values of functions $g_{2}(x^{2},x^{3})$ (or,
inversely, $g_{3}(x^{2},x^{3})$), $h_{4}(x^{i},v)$ (or, inversely, $
h_{5}(x^{i},v)$) and of sources $\Upsilon _{2}(x^{2},x^{3},v)$ and $\Upsilon
_{4}(x^{2},x^{3}).$
\end{theorem}

Finally, we note that we have elaborated a new geometric method of
constructing exact solutions in extra dimension gravity and general
relativity theories. The classes of solutions define very general integral
varieties of the vacuum and nonvacuum Einstein equations, in general, with
torsion and nonmetricity and corrections from string theory, and/or
noncommutative/quantum variables. For instance, in five dimensions, the
metrics are generic off--diagonal and depend on four coordinates. So, we
have proved in explicit form how it is possible to solve the Einstein
equations on nonholonomic manifolds (see mathematical problems analized in
Refs. \cite{leit1,leit2}), in our case, provided with nonlinear connection
structure.

{\vskip 4pt}

\textbf{Acknowledgement: }The author is grateful to the Organizes for
supporting his participation at the Conference ''Geometr\'{\i}a de Lorentz,
Murcia 2003, 12-14 de Noviembre'', for partial support from ''Consejeria de
Educacion de la Junta de Andalucia'' and the ''Grupo de Estructura de la
Materia'' of the University of Huelva for his kind hospitality. He also
thanks the referee for valuable remarks and D. Leites, M. de Leon and F.
Etayo for important discussions on the geometry of nonholonomic manifolds
and nonlinear connection formalism.


TITLE: Do Carmo Problem 3.1
QUESTION [0 upvotes]: So I was going through the exercises in Do Carmo's Riemannian Geometry, but I had an issue on problem 1 in chapter 3. I've done 1a, 1b but I am stuck at 1c.
We have a geodesic;
\begin{equation}
\gamma(t)= (f(v(t))\cos u(t), f(v(t))\sin u(t), g(v(t))).
\end{equation}
We can compute,
\begin{equation}
 |\gamma'(t)|^2= f^2\left(\frac{\text{d}u}{\text{d}t}\right)^2+ [(f')^2+ (g')^2]\left(\frac{\text{d}v}{\text{d}t}\right)^2,
\end{equation}
and,
\begin{equation}
\frac{\text{d}}{\text{d}t}|\gamma'(t)|^2= 2ff'\frac{\text{d}v}{\text{d}t}\left(\frac{\text{d}u}{\text{d}t}\right)^2+ 2f^2\frac{\text{d}u}{\text{d}t}\frac{\text{d}^2u}{\text{d}t^2}+ 2[f'f''+g'g'']\left(\frac{\text{d}v}{\text{d}t}\right)^3+2[(f')^2+(g')^2]\frac{\text{d}v}{\text{d}t}\frac{\text{d}^2v}{\text{d}t^2}.
\end{equation}
I do not see how the vanishing of this is equivalent to the second equation... That is,
\begin{equation}
 \frac{\text{d}^2v}{\text{d}t^2}- \frac{ff'}{(f')^2+(g')^2}\left(\frac{\text{d}u}{\text{d}t}\right)^2+\frac{f'f''+g'g''}{(f')^2+(g')^2}\left(\frac{\text{d}v}{\text{d}t}\right)^2=0.
\end{equation}
Moreover, any help on the second part of this question would also help.

REPLY [2 votes]: The first part of part c) is solved by substituting in the equations from part b). These are:
\begin{equation*}
    \frac{\text{d}^2u}{\text{d}t^2} + \frac{2ff'}{f^2}\frac{\text{d}u}{\text{d}t} = 0\\
    \frac{\text{d}^2v}{\text{d}t^2} - \frac{ff'}{(f')^2+(g')^2}\left(\frac{\text{d}u}{\text{d}t}\right)^2+\frac{f'f''+g'g''}{(f')^2+(g')^2}\left(\frac{\text{d}v}{\text{d}t}\right)^2 = 0
\end{equation*}
Note the first equation can be rewritten as:
\begin{equation*}
        f^2\frac{\text{d}^2u}{\text{d}t^2} = -2ff'\frac{\text{d}u}{\text{d}t}\frac{\text{d}v}{\text{d}t}
\end{equation*}
For clarity I will enclose simplifications from the equations derived in b) in square brackets:
\begin{align*}
    \frac{\text{d}}{\text{d}t}|\gamma'(t)|^2 &= 2ff' \frac{\text{d}v}{\text{d}t}\left(\frac{\text{d}u}{\text{d}t}\right)^2 + 2f^2 \frac{\text{d}u}{\text{d}t} \frac{\text{d}^2u}{\text{d}t^2} + 2(f'f'' + g'g'')\left(\frac{\text{d}v}{\text{d}t}\right)^3 + 2((f')^2 + (g')^2)\frac{\text{d}v}{\text{d}t}\frac{\text{d}^2v}{\text{d}t^2}\\
    &=f^2\frac{\text{d}u}{\text{d}t}\left(\left[\frac{2ff'}{f^2}\frac{\text{d}v}{\text{d}t}\frac{\text{d}u}{\text{d}t} + \frac{\text{d}^2u}{\text{d}t^2}\right]+ \frac{\text{d}^2u}{\text{d}t^2}\right) + 2\frac{\text{d}v}{\text{d}t}\left((f'f'' + g'g'')\left(\frac{\text{d}v}{\text{d}t}\right)^2 + ((f')^2 + (g')^2)\frac{\text{d}^2v}{\text{d}t^2}\right)\\
    &= \left[f^2\frac{\text{d}^2u}{\text{d}t^2}\right]\frac{\text{d}u}{\text{d}t} + 2\frac{\text{d}v}{\text{d}t}((f')^2+(g')^2)\left(\frac{f'f''+g'g''}{(f')^2+(g')^2}\left(\frac{\text{d}v}{\text{d}t}\right)^2+\frac{\text{d}^2v}{\text{d}t^2}\right)\\
    &=\left(-2ff'\frac{\text{d}u}{\text{d}t}\frac{\text{d}v}{\text{d}t}\right)\frac{\text{d}u}{\text{d}t} + 2\frac{\text{d}v}{\text{d}t}((f')^2+(g')^2)\left(\frac{f'f''+g'g''}{(f')^2+(g')^2}\left(\frac{\text{d}v}{\text{d}t}\right)^2+\frac{\text{d}^2v}{\text{d}t^2}\right)\\
    &=2\frac{\text{d}v}{\text{d}t}((f')^2+(g')^2)\left[\frac{-ff'}{(f')^2+(g')^2}\left(\frac{\text{d}u}{\text{d}t}\right)^2+\frac{f'f''+g'g''}{(f')^2+(g')^2}\left(\frac{\text{d}v}{\text{d}t}\right)^2+\frac{\text{d}^2v}{\text{d}t^2}\right]\\
    &=2\frac{\text{d}v}{\text{d}t}((f')^2+(g')^2)0 = 0
\end{align*}
For the second half of c) I will lay out an approach that I think you can fill in the gaps for. Attached is the image on page 79 of the textbook.

Consider a geodesic path $\gamma$ and for any $t$ its angle of incidence $\beta(t)$. For any given $t_0$, $\beta(t_0)$ would be determined by the angle between $\gamma'(t_0)$ and $p'(s_0)$ where $p$ is the (unique up to reparametrization) parallel of $S$ passing through $\gamma(t_0)$. We may write these as:
\begin{align*}
    p(s)&=\big(f(v_p)\cos(u_p(s)),f(v_p)\sin(u_p(s)),g(v_p)\big)\\
    \gamma(t) &= \big(f(v_\gamma(t))\cos(u_\gamma(t)),f(v_\gamma(t))\sin(u_\gamma(t)),g(v_\gamma(t))\big)
\end{align*}
After choice of $p$ of constant speed and some calculation you will see:
\begin{align*}
    \langle \gamma'(t),p'(s) \rangle_{\gamma(t_0)} &= f(v_\gamma(t_0))f(v_p)\frac{\text{d}u_\gamma}{\text{d}t}\\
    &= f(v_\gamma(t_0))^2\frac{\text{d}u_\gamma}{\text{d}t}
\end{align*}
Where the simplification comes from that $p$ and $\gamma$ intersect at $\gamma(t_0)$. Now these vectors are in $T_{\gamma(t_0)}S$ and are also in $T_{\gamma(t_0)}\mathbb{R}^3$. Using our knowledge of the usual inner product in $\mathbb{R}^3$:
\begin{align*}
    \langle\gamma'(t),p'(s)\rangle_{\gamma(t_0)} &= |p'(s_0)|\,|\gamma'(t_0)|\cos(\beta(t_0))
\end{align*}
Where $|\gamma'(t_0)|$ and $f(v_\gamma(t_0))^2\frac{\text{d}u_\gamma}{\text{d}t}$ are constant by calculations of $\frac{\text{d}}{\text{d}t}|\gamma'(t)|^2$ and by manipulation of $\frac{\text{d}^2 u}{\text{d}t^2} + \frac{2ff'}{f^2}\frac{\text{d}u}{\text{d}t}\frac{\text{d}v}{\text{d}t}=0$ to $\frac{\text{d}}{\text{d}t}\left(f^2\frac{\text{d}u}{\text{d}t}\right) = 0$, respectively.


TITLE: A single electron (and positron) annihilation, is my face melted?
QUESTION [2 upvotes]: To help clarify my understanding of scale in regard to microscopic particle energy levels (specifically, an electron), I came up with this thought exercise:
What would happen if an electron and a positron annihilated one meter in front of my face?  Would my face melt?  Would I even notice?
I am ignoring the details/difficulties in creating the annihilation scenario.  Assume it it possible.
I usually learn the most by making an attempt myself, so here is my crack at an answer:
From Wikipedia, an electron (and a positron) is 0.510998928 MeV/c^2.  So their annihilation would yield their sum, conveniently also directly noted in Wikipedia along with a bit of rounding and setting c=1: 1.022 MeV.
Now we need something to compare against.  Once again, Wikipedia provides...  A flying mosquito is about 1 TeV of kinetic energy.  It's definitely looking like I'm not noticing anything.  The annihilation expressed in TeV is: 0.000001022 TeV, therefore it would take 978,474 total particles (489,237 electrons, 489,237 positrons) to reach that level of the mosquito.  We are at an energy level that under the right conditions could be 'felt'.  However, at a meter from my face and the energy being dissipated in an omnidirectional manner, I'm not noticing anything.
Now let's go a little further and (very) roughly estimate how many would be needed to notice.  Taking another comparison from this Boom Table, a firecracker is about 100 J, equivalent to about 624,150,965 TeV.  We are probably roughly in the realm of something I would feel, but wouldn't hurt me (a true firecracker may due to shrapnel, but no shrapnel with particle annihilation).  To reach that level, we need about 3.053577457x10^14 electrons and the same number of positrons.
Are my assumptions and ballpark numbers reasonable?  Did I leave out anything significant in my logic?
The number of particles to reach the firecracker energy level is larger than I would have expected.  Then again, it's difficult to fathom just how small electrons really are.

REPLY [7 votes]: It turns out there is a good way to estimate this.
In PET (positron emission tomography), the patient is injected with a radioactive tracer that emits positrons (hence, the name). These positrons annihilate with electrons, and emit two 511 keV photons. Detecting these photons, one can estimate the location of the radiation. Doing this many times over, you can create an image of the tracer uptake - used in a wide variety of medical diagnostics.
So no, one annihilation won't kill you - even when it happens inside your body. Typical injections are on the order of 10 mCi (370 MBq) - so even 370 million annihilations per second won't kill you.
Long before the "energy" will kill (by cooking you), it is the ionization properties that will get to you. The typical lethal dose is considered to be 20 Gy - J/kg or about 1 kJ spread evenly over your body. Much lower levels would make you quite sick. The amount of damage done by radiation to individual organs changes quite a bit - some organs are much more sensitive than others. In radiation therapy, they try to deliver a local dose of about 70 Gy to tumors, while sparing the surrounding tissue; but they do this in "fractions" - giving just a little bit of dose every day for several weeks in order to give the tissue a chance to recover. And that is usually enough to make people feel pretty sick.
You should expect that you will really notice the effects of 1 J of radiation being delivered "locally" (say, to your hand). That would correspond to 6 x 1012 electrons. Not a crazy number.


TITLE: Easy way (just from definition) to prove that $H(A,B) \leq H(A)+H(B)$
QUESTION [0 upvotes]: I am looking for a way to prove that :
$$ H(A,B) \leq H(A)+H(B) $$
from the definition of the shannon entropy :
$$H(X)=-\sum_l P(X=x_l) log(P(X=x_l))$$
Is there a very simple way to prove it with the definition only ?
I ended up with the fact that the problem is equivalent to prove that :
$$ \sum_b P(b) ln(P(b)) - \sum_a P(a) \sum_b P(b/a) ln(P(b/a)) \leq 0$$  but I am stuck...
I am looking for a way using definition if it exists because I don't really know anything about information theory, I am mainly focusing on thermodynamic (where as you know entropy is a very important notion).

REPLY [2 votes]: The chain rule for entropy
states $H(A,B) = H(A) + H(B \mid A)$, so it suffices to prove $H(B \mid A) \le H(B)$, which I believe is the last inequality that you have written.
The quantity $H(B) - H(B \mid A)$ is known as the mutual information of $A$ and $B$, denoted $I(A ; B)$. One can prove that mutual information (in fact, any KL-divergence) is nonnegative, using Jensen's inequality.


TITLE: Adjacent transposition - explain in layman terms
QUESTION [1 upvotes]: I've red this article many-many times, as well as some more supplementary reading on groups, permutations, generating sets etc. and I give up. :( I'm not a mathematician (I study combinatorics, but I'm not there yet, so I can handle a moderate amount of terminology).
My problem: I want to write a function that generates successive unique permutations of a sequence of an arbitrary length by swapping two elements of the sequence at a time.
What I don't understand in the article: what are the numbers inside the parenthesis. Are they the offset into the sequence or are they members of the sequence? How long is the sequence being discussed? (two elements long, or five?) What do the dots between parenthesis mean? It is not / can't be a dot product or multiplication or function composition - but I can't find any more meanings / searching for dot is clearly unproductive...
Ideally, I would be most happy if you took a sequence of 3 (better 4, but it might be a lot of work - so I'm not asking to do it all, if it's 4) and do something like this:

ABCD - swap A and B because
BACD - swap B and C because
CABD - ...

In case you are bored / have no time to write one - below is an example I've found empirically - but I can't formulate the algorithm for producing permutations for the sequence of a given length.
1   1 2 3 4    #(0 1 2) 1 |
    - -                   |
2   2 1 3 4 *  #(0 1 2) 2 |
    -   -                 |
3   3 1 2 4    #(0 0 2) 2 |
    - -                   |
4   1 3 2 4    #(0 1 2) 2 |
    -   -                 |
5   2 3 1 4    #(0 0 2) 2 |
    - -                   |
6   3 2 1 4 ** #(0 1 2) 3 |
    -     -               |
7   4 2 1 3    #(0 1 1) 3 |
      -   -               |
8   4 3 1 2    #(0 1 0) 3 |
      - -                 |
9   4 1 3 2    #(0 0 2) 3 |
      -   -               |
10  4 2 3 1    #(0 0 1) 3 |
      - -                 |
11  4 3 2 1    #(0 0 0) 3 |
      -   -               |
12  4 1 2 3 +  #(0 1 2) 3 |
    -   -                 |
13  2 1 4 3    #(0 1 1) 3 |
    -     -               |
14  3 1 4 2    #(0 1 0) 3 |
    - -                   |
15  1 3 4 2    #(0 0 2) 3 |
    -     -               |
16  2 3 4 1    #(0 0 1) 3 |
    - -                   |
17  3 2 4 1    #(0 0 0) 3 |
    -     -               |
18  1 2 4 3 +  #(0 1 2) 3 |
      - -                 |
19  1 4 2 3    #(0 1 1) 3 |
    -     -               |
20  3 4 2 1    #(0 1 0) 3 |
    -   -                 |
21  2 4 3 1    #(0 0 2) 3 |
    -     -               |
22  1 4 3 2    #(0 0 1) 3 |
    -   -                 |
23  3 4 1 2    #(0 0 0) 3 |
    -     -               |
24  2 4 1 3 ***#(0 1 2) 3 |

Asterisks and + signs are marks for myself which mean I know how to find these stages. (it first permutes the sub-sequence of length of 2, then of length of 3 and so on).
Minus signs stand between the elements being swapped:
ABCD
- -
CBAD

for example.
Kind thanks in advance!
EDIT:
It looks like my problem has to do with Levi-Civata symbol
EDIT2:
If you were looking for implementation, here's something I could come up with: http://pastebin.com/jszrQV1V

REPLY [2 votes]: According to http://en.wikipedia.org/wiki/Steinhaus%E2%80%93Johnson%E2%80%93Trotter_algorithm, "The Steinhaus–Johnson–Trotter algorithm or Johnson–Trotter algorithm, also called plain changes, is an algorithm named after Hugo Steinhaus, Selmer M. Johnson and Hale F. Trotter that generates all of the permutations of n elements. Each permutation in the sequence that it generates differs from the previous permutation by swapping two adjacent elements of the sequence." I'm not going to copy out the details --- it's all there at the Wikipedia page.


TITLE: Motivation for the definition of $L^p$ norm for quadratic and Beltrami differentials
QUESTION [1 upvotes]: According to Riemann surfaces, dynamics and
geometry
 by C. McMullen (Course notes), the definition for a quadratic differential $\phi$ on a Riemann surface $X$ is given by
$$
\|\phi\|_p = \left(\int_X \rho^{2-2p} |\phi|^p\right)^{1/p}, \|\phi\|_\infty = \sup_X \frac{|\phi|}{\rho^2},
$$
where $\rho |dz|$ is the hyperbolic metric.
For a Beltrami coefficient $\mu,$
$$
\|\mu\|_p = \left(\int_X \rho^{2} |\mu|^p\right)^{1/p}, \|\mu\|_\infty =\sup_X |\mu|.
$$
I am confused about the mysterious exponents like $2-2p$ and I am not sure how the definition of $L^p$ can pass "smoothly" to $L^\infty.$ I also do not know why we are dividing $\rho^2$ in the first expression but not in the second one.
Why must we put that certain power of $\rho$ into the expression?

REPLY [3 votes]: The powers of $\rho$ are necessary to make the integrals well-defined. In probably excessive detail:

Because $X$ is a Riemann surface, the integrands here need to be $(1,1)$-forms. This is just the familiar fact (change of variables theorem) that integrals over (oriented) manifolds are only well-defined on objects $\omega$ which are given in local coordinates by functions $\omega(x)$ satisfying the transformation rule
$$
\omega(x) = \det(D\psi_x) \omega(y)
$$
whenever $y = \psi(x)$ a smooth change of coordinates. If $M = X$ is a Riemann surface, the coordinate changes $w = \psi(z)$ are holomorphic, so that $\det(D\psi) = |\psi'|^2$ (Cauchy-Riemann), and thus the above rule can be rewritten as
$$
\phi(z) = |\psi'(z)|^2 \phi(w) = \psi'(z) \overline{\psi'(z)} \ \phi(w).
$$
This condition can be summarized in the language of Riemann surface $(n, m)$-tensors (see Lyubich's book, p. 99) by saying that $\phi$ can be integrated over $X$ iff it is a $(1,1)$-form, i.e. an object with local form
$$
\phi = \phi(z) \ dz \ d\bar{z} = \phi(z) \ |dz|^2.
$$

Quadratic differentials are $(2, 0)$-forms. Beltrami forms are $(-1, 1)$-forms. Take $\phi$ a quadratic differential. Then $|\phi|^p$ is a $(p, p)$-form. In local coordinates:
$$
\phi = \phi(z) \ dz^2 \implies
|\phi|^p = |\phi(z) dz^2|^p = |\phi(z)|^p |dz|^{2p} = |\phi(z)|^p dz^p d\bar{z}^p.
$$
In particular, $|\phi|$ is a $(1, 1)$-form, so can be integrated over $X$. This shows that the $L^1$-norm on quadratic differentials is canonically defined, even when $X$ is not hyperbolic (and thus has no preferred conformal metric), in the way you'd expect:
$$
||\phi||_1 = \int_X |\phi|.
$$
When $p \neq 1$, however, $|\phi|^p$ is not a $(1,1)$-form, so its integral over $X$ is not defined, and thus the naive definition of the $L^p$-norm for quadratic differentials will not work! Fortunately, this issue can be remedied in the presence of a conformal metric $\rho = \rho(z) |dz|$, because we can adjust the tensor type of our candidate integrand using powers of $\rho$. In particular, multiplying by $\rho^{2-2p}$ gives
$$
\rho^{2-2p} |\phi|^p = \rho(z)^{2-2p} |dz|^{2-2p} |\phi(z)|^p |dz|^{2p} = \rho(z)^{2-2p} |\phi(z)|^p |dz|^{2},
$$
which is a $(1,1)$-form, and so can be integrated over $X$. Therefore we can define an $L^p$-norm on quadratic differentials by the expression you gave above:
$$
||\phi||_p = \left( \int_X \rho^{2-2p} |\phi|^p \right)^{1/p}.
$$
Moreover, though this definition appears to depend on an arbitrary choice of $\rho$, when $X$ is a hyperbolic Riemann surface there is a canonical choice, namely the unique conformal metric of constant curvature -1. So, in the hyperbolic case with $\rho$ taken to be this distinguished metric, it makes sense to refer to the above definition as the $L^p$-norm on quadratic differentials. The Beltrami case works similarly.

The construction here is exactly analogous to the problem of defining $L^p$-norms for spaces of functions on a smooth manifold $M$. The $L^\infty$-norm of a function is, of course, defined as usual by
$$
||f||_\infty = \displaystyle\text{ess sup}_M |f|.
$$
But without further structure, the $L^p$-norms for $p < \infty$ are not well-defined, precisely because functions cannot be integrated over $M$. After fixing a Riemannian metric on $M$, however, the $L^p$-norm of a function can be defined in the obvious way by integrating against the associated volume form:
$$
||f||_p = \left( \int_M |f|^p \ \text{dVol} \right)^{1/p}
$$
Taking this approach in our context leads to the same result as above. That is, we could have alternatively begun by modifying $\phi$ by $\rho$ to obtain a function on $X$, then applied the Riemannian definition given above to our special case. In detail, the local computation
$$
\dfrac{|\phi|}{\rho^2} = \dfrac{|\phi(z)||dz|^2}{\rho(z)^2 |dz|^2} = \dfrac{|\phi(z)|}{\rho(z)^2},
$$
shows that $\dfrac{|\phi|}{\rho^2}$ is a function on $X$. Then we can define
$$
||\phi||_\infty = \displaystyle\text{ess sup}_X \dfrac{|\phi|}{\rho^2}$$
and, using the volume form $\rho^2 = \rho(z)^2 |dz|^2$ associated to the conformal metric $\rho$, define
$$||\phi||_p = \left( \int_X \left( \dfrac{|\phi|}{\rho^2} \right)^p \ \rho^2 \right)^{1/p} = \left( \int_X \rho^{2-2p} |\phi|^p\right)^{1/p}
$$
for $p < \infty$. Voila!


TITLE: Contraction of Bellman Operator under general $L_p$ norms
QUESTION [2 upvotes]: We know that the Bellman Operator
$$
TV(s) = \max_a r(s,a) + \sum_{s' \in S}p(s'|s,a)V(s')
$$
is a contraction under $L_\infty$ norm.For reference one can see the following link Proof that Bellman update is a contraction
A couple of definitions are warranted:
$p(s'|s,a)$ is the probability of hitting state $s'$ under action $a$ at state $s$. $S$ is the set of all possible states. $r(s,a)$ is the reward, incurred at state $s$ when playing action $a$. $V(s')$ is the reward incuured at state $s'$ over an entire horizon when optimal actions are chosen at each subsequent state.
I was wondering what was stopping it from being a contraction in any $L_p$ norm. Ofcourse in finite dimensions it should not be a problem because all norms are equivalent in finite dimensions, so I guess the only interesting thing is in infinite dimensions.
My idea is that
$$
\begin{split}\|TV - T\bar{V}\|_p& = \bigg(\sum_s\big(\sum_{s'}p(s'|s,a^*)(V(s')-\bar{V}(s'))\big)^p\bigg)^{1/p}\\
&\leq \sum_{s'}p(s'|s,a^*)\bigg(\sum_s\bigg(V(s')-\bar{V}(s')\bigg)^p\bigg)^{1/p}
\end{split}$$ by Jensen Inequality
and then $\sum_{s'}p(s'|s,a) = 1$
to get
$$
\|TV - T\bar{V}\|_p \leq \|V - \bar{V}\|_p.
$$
What is the mistake?

REPLY [1 votes]: So I found my error. My definition of Jensen's was wrong.
The correct inequality would be in terms of $L_2$ norm
$$\|TV - T\bar{V}\|_2 \leq \sqrt{|S|}\|V-\bar{V}\|_2$$
and hence obviously cannot be a contraction even in finite dimensions