text:
\begin{document}

\author{Constantin C\u alin and Mircea Crasmareanu}
\title{From the Eisenhart problem to Ricci solitons in $f$-Kenmotsu manifolds}
\date{Dedicated to the memory of Neculai Papaghiuc 1947-2008}
\maketitle

\begin{abstract}
The Eisenhart problem of finding parallel tensors is solved for the symmetric case in the regular $f$-Kenmotsu framework. On this way, the Olszack-Rosca example of Einstein manifolds provided by $f$-Kenmotsu manifolds via locally symmetric Ricci tensors is recovered as well as a case of Killing vector fields. Some other classes of Einstein-Kenmotsu manifolds are presented. Our result is interpreted in terms of Ricci solitons and special quadratic first integrals.
\end{abstract}

\noindent {\bf 2000 Math. Subject Classification}: 53C40; 53C55; 53C12; 53C42.

\noindent {\bf Key words}: $f$-Kenmotsu manifold; parallel second order covariant tensor field; irreducible metric; Einstein space; Ricci soliton.

\medskip

\section*{Introduction}

In 1923, Eisenhart \cite{l:e} proved that if a positive definite Riemannian ma\-ni\-fold $(M,g)$ admits a second order parallel symmetric covariant tensor other that a constant multiple of the metric tensor, then it is reducible. In 1926, Levy \cite{h:l} proved that a second order parallel symmetric non-degenerated tensor $\alpha $ in a space form is proportional to the metric tensor. Let us point out that this question can be considered as dual to the the problem of finding linear connections making parallel a given tensor field, problem which was considered by Wong in \cite{y:w}. Also, the former question implies topological restrictions namely if the (pseudo) Riemannian manifold $M$ admits a parallel symmetric $(0,2)$ tensor field then $M$ is locally the direct product of a number of (pseudo) Riemannian manifolds, \cite{h:w} (cited by \cite{g:z}). Another situation where the parallelism of $\alpha $ is involved appears in the theory of {\it totally geodesic maps}, namely, as is point out in \cite[p. 114]{o:c}, $\nabla \alpha =0$ is equivalent with the fact that $1:(M,g)\rightarrow (M,\alpha )$ is a totally geodesic map.

While both Eisenhart and Levy work locally, Ramesh Sharma gives in \cite{r:s1} a global approach based on Ricci identities. In addition to space-forms, Sharma considered this {\it Eisenhart problem} in contact geometry \cite{r:s2}-\cite{r:s4}, for example for $K$-contact manifolds in \cite{r:s3}. Since then, several other studies appear in various contact manifolds: nearly Sasakian \cite{t:m}, (para) $P$-Sasakian \cite{t:d1}, \cite{d:e} and \cite{z:l}, $\alpha $-Sasakian \cite{l:d2}. Another framework was that of quasi-constant curvature in \cite{x:j}. Also, contact metrics with nonvanishing $\xi $-sectional curvature are studied in \cite{g:s}.

Returning to contact geometry, an important class of manifolds are introduced by Kenmotsu in \cite{k:k} and generalized by Olszack and Rosca in \cite{o:r}. In the last time, there is an increasing flow of papers in this direction e.g. that of our professor N. Papaghiuc \cite{n:p1}-\cite{n:p2} to which we dedicate this short note. Motivated by this fact we studied the case of $f$-Kenmotsu manifolds satisfying a special condition called by us {\it regular} and show that a symmetric parallel tensor field of second order must be a constant multiple of the Riemannian metric. There are three remarks regarding our result:\\
i) it is in agreement with what happens in all previously recalled contact \\ geometries for the symmetric case, \\
ii) it is obtained in the same manner {\bf as} originated in Sharma's paper \cite{r:s1}, \\
iii) yields a class of Einstein manifolds already indicated by Olszack and Rosca but with a more complicated proof. \\
Let us point out also that the anti-symmetric case appears without proof in \cite{v:m}.

Our main result is connected with the recent theory of Ricci solitons, a subject included in the Hamilton-Perelman approach (and proof)  of Poincar\'e Conjecture. Ricci solitons in contact geometry were first studied by Ramesh Sharma in \cite{g:sc} and \cite{r:s5}; also the preprint \cite{mm:t} is available to arxiv. In these papers the $K$-contact and $(k, \mu )$-contact (including Sasakian) cases are treated; then our treatment for the Kenmotsu variant of almost contact geometry seems to be new.

Our work is structured as follows. The first section is a very brief review of Kenmotsu geometry and Ricci solitons. The next section is devoted to the (symmetric case of) Eisenhart problem in a $f$-Kenmotsu manifold and several situations yielding Einstein manifolds are derived. Also, the relationship with the Ricci solitons is pointed out. The last section offers a dynamical picture of the subject via Killing vector fields and quadratic first integrals of a special type.

{\bf Acknowledgement} Special thanks are offered to Gheorghe Pitis for some useful remarks as well as sending us his book \cite{gh:p}, a source of several references. Also, we are very indebted to Marian-Ioan Munteanu and the referees who pointed out major improvements.

\section{$f$-Kenmotsu manifolds. Ricci solitons}

Let $M$ be a real $2n+1$-dimensional differentiable manifold endowed with an almost
contact metric structure $(\varphi, \xi, \eta, g)$:
$$
\begin{array}{l}
(a) \quad \varphi ^{2} = -I + \eta \otimes \xi,\ \ (b) \quad \eta(\xi) = 1,\ \ (c) \quad \eta \circ \varphi = 0,\\ (d) \quad \varphi (\xi) = 0, \ \
(e) \quad \eta(X) = g(X, \xi),\\  (f) \quad g(\varphi X, \varphi Y)=g(X, Y)-\eta (X)\eta(Y),
\end{array} \eqno(1.1)
$$
for any vector fields $X, Y\in {\cal X}(M)$ where $I$ is the identity of the tangent
bundle $TM$, $\varphi $ is a tensor field of $(1,1)$-type, $\eta $ is a 1-form,
$\xi $ is a vector field and $g$ is a metric tensor field.
Throughout the paper all objects are differentiable of class $C^{\infty }$.

\medskip

We say that $(M,\varphi ,\xi ,\eta ,g)$ is an $f$-{\it Kenmotsu manifold} if the Levi-Civita connection of $g$ satisfy \cite{v:m}:
$$
(\nabla _X\varphi )(Y)=f(g(\varphi X,Y)\xi -\varphi (X)\eta (Y)), \eqno(1.2)
$$
where $f\in C^{\infty }(M)$ is strictly positive and $df\wedge \eta =0$ holds. A $f=constant\equiv \beta >0$ is called $\beta $-{\it Kenmotsu manifold} with the particular case $f\equiv 1$-Kenmotsu manifold which is a usual {\it Kenmotsu manifold} \cite{k:k}.

\medskip

In a general $f$-Kenmotsu manifold we have, \cite{o:r}:
$$
\nabla _X\xi =f(X-\eta (X)\xi), \eqno(1.3)
$$
and the curvature tensor field:
$$
R(X,Y)\xi =f^2(\eta (X)Y-\eta (Y)X)+Y(f)\varphi ^2X-X(f)\varphi ^2Y \eqno(1.4)
$$
while the Ricci curvature and Ricci tensor are, \cite{k:pt}:
$$
S(\xi ,\xi ) = -2n(f^2+\xi (f)) \eqno(1.5)
$$
$$
Q(\xi )=-2nf^2\xi -\xi (f)\xi-(2n-1)grad f. \eqno(1.6)
$$

In the last part of this section we recall the notion of Ricci solitons according to \cite[p. 139]{r:s5}. On the manifold $M$, a {\it Ricci soliton} is a triple $(g,V,\lambda )$ with $g$ a Riemannian metric, $V$ a vector field and $\lambda $ a real scalar such that:
$$
{\cal L}_Vg+2S+2\lambda g=0. \eqno(1.7)
$$
The Ricci soliton is said to be {\it shrinking, steady} or {\it expanding} according as $\lambda $ is negative, zero or positive.

\section{Parallel symmetric second order tensors and \\ Ricci solitons in $f$-Kenmotsu manifolds}

Fix $\alpha $ a symmetric tensor field of $(0,2)$-type which we suppose to be parallel with respect to $\nabla $ i.e. $\nabla \alpha =0$. Applying the Ricci identity \\ $\nabla ^2\alpha (X,Y;Z,W)-\nabla ^2(X,Y;W,Z)=0$ we get the relation $(1.1)$ of \cite[p. 787]{r:s1}:
$$
\alpha (R(X,Y)Z,W) +\alpha (Z,R(X,Y)W)=0, \eqno(2.1)
$$
which is fundamental in all papers treating this subject. Replacing $Z=W=\xi $ and using $(1.4)$ it results:
$$
f^2[\eta (X)\alpha (Y,\xi )-\eta (Y)\alpha (X,\xi)]+Y(f)\alpha
(\varphi ^2X, \xi)-X(f)\alpha (\varphi ^2Y, \xi)=0, \eqno(2.2)
$$
by the symmetry of $\alpha $. With $X=\xi $ we derive:
$$
[f^2+\xi (f)][\alpha (Y,\xi )-\eta (Y)\alpha (\xi , \xi)]=0
$$
and supposing $f^2+\xi (f)\neq 0$ it results:
$$
\alpha (Y,\xi )=\eta (Y)\alpha (\xi , \xi). \eqno(2.3)
$$
Let us call {\it regular} $f$-{\it Kenmotsu manifold} a $f$-Kenmotsu manifold with $f^2+\xi (f)\neq 0$ and remark that $\beta $-Kenmotsu manifolds are regular.

\smallskip

Differentiating the last equation covariantly with respect to $X$ we have:
$$
\alpha (\nabla _XY, \xi )+f[\alpha (X,Y)-\eta (X)\eta (Y)\alpha (\xi ,\xi )]=X(\eta (Y))\alpha (\xi , \xi), \eqno(2.4)
$$
which means via $(2.3)$ with $Y \rightarrow \nabla _XY$:
$$
f[\alpha (X,Y)-\eta (X)\eta (Y)\alpha (\xi ,\xi )]=[X(g(Y,\xi ))-g(\nabla _XY, \xi )]\alpha (\xi ,\xi )=
$$
$$
=g(Y,\nabla _X\xi )\alpha (\xi ,\xi )=
f[g(X,Y)-\eta (X)\eta (Y)]\alpha (\xi ,\xi ). \eqno(2.5)
$$
From the positiveness of $f$ we deduce that:
$$
\alpha (X,Y)=\alpha (\xi ,\xi )g(X,Y) \eqno(2.6)
$$
which together with the standard fact that the parallelism of $\alpha $ implies the constance of $\alpha (\xi ,\xi )$ via $(2.3)$ yields:

\medskip

{\bf Theorem} {\it A symmetric parallel second order covariant tensor in a regular $f$-Kenmotsu manifold is a constant multiple of the metric tensor. In other words, a regular $f$-Kenmotsu metric is irreducible which means that the tangent bundle does not admits a decomposition $TM=E_1\oplus E_2$ parallel with respect of the Levi-Civita connection of $g$}.

\medskip

{\bf Corollary 1} {\it A locally Ricci symmetric} $(\nabla S\equiv 0)$ {\it regular $f$-Kenmotsu manifold is an Einstein manifold}.

\medskip

{\bf Remarks} 1) The particular case of dimension three and $\beta$-Kenmotsu of our theorem appears in Theorem 3.1 from \cite[p. 2689]{d:m}. The above corollary has been proved by Olszack and Rosca in another way. \\
2) In \cite{b:tdt} it is shown the equivalence of the following statements for an Kenmotsu manifold:\\
i) is Einstein, \\
ii) is locally Ricci symmetric, \\
iii) is Ricci semi-symmetric i.e. $R\cdot S=0$ where:
$$
(R(X, Y)\cdot S)(X_1, X_2) = -S(R(X, Y)X_1, X_2)-S(X_1, R(X, Y)X_2).
$$
The same implication iii) $\rightarrow $ i) for Kenmotsu manifolds is Theorem 1 from \cite[p. 438]{j:dp}. But we have the implication iii) $\rightarrow $ i) in the more general framework of regular $f$-Kenmotsu manifols since $R\cdot S=0$ means exactly $(2.1)$ with $\alpha $ replaced by $S$. Every semisymmetric manifold, i. e. $R\cdot R=0$, is Ricci-semisymmetric but the converse statement is not true. In conclusion:

\medskip

{\bf Proposition 1} {\it A Ricci-semisymmetric, particularly semisymmetric, re\-gu\-lar} $f$-{\it Kenmotsu manifold is Einstein.}

\medskip

Another class of spaces related to the Ricci tensor was introduced in \cite{h:q}; namely a Riemannian manifold is a {\it special weakly Ricci symmetric space} if there exists a 1-form $\rho $ such that:
$$
(\nabla _XS)(Y,Z)=2\rho (X)S(Y,Z)+\rho (Y)S(Z,X)+\rho (Z)S(X,Y). \eqno(2.7)
$$
The same condition was sometimes called {\it generalized pseudo-Ricci sym\-metric manifold} (\cite{j:s}) or simply {\it pseudo-Ricci symmetric manifold} (\cite{c:k}). Making $X=Y=Z=\xi $ it results:
$$
\xi (S(\xi ,\xi ))=4\rho (\xi )S(\xi ,\xi) \eqno(2.8)
$$
and then for a $\beta $-Kenmotsu manifold we get $\rho (\xi )=0$. Returning to $(2.7)$ with $Y=Z=\xi $ will results $\rho (X)=0$ for every vector field $X$ and then we have a generalization of Theorem 3.3. from \cite[p. 96]{a:go}:

\medskip

{\bf Proposition 2} {\it A} $\beta $-{\it Kenmotsu manifold which is special weakly Ricci symmetric is an Einstein space.}

\medskip

We close this section with applications of our Theorem to Ricci solitons:

\medskip

{\bf Corollary 2} {\it Suppose that on a regular $f$-Kenmotsu manifold the} $(0,2)$-{\it type field} ${\cal L}_Vg+2S$ {\it is parallel where} $V$ {\it is a given vector field. Then} $(g, V)$ {\it yield a Ricci soliton. In particular, if the given regular} $f$-{\it Kenmotsu manifold is Ricci-semisymmetric or semisymmetric with} ${\cal L}_Vg$ {\it parallel, we have the same conclusion.}

\medskip

Naturally, two situations appear regarding the vector field $V$: $V\in span{\xi }$ and $V\bot \xi$ but the second class seems far too complex to analyse in practice. For this reason it is appropriate to investigate only the case $V=\xi $.

We are interested in expressions for ${\cal L}_{\xi }g+2S$. A straightforward computation gives:
$$
{\cal L}_{\xi }g(X,Y)=2f(g(X,Y)-\eta (X)\eta (Y))=2fg(\varphi X, \varphi Y). \eqno(2.9)
$$

A general expression of $S$ is known by us only for the the 3-dimensional case and $\eta $-Einstein Kenmotsu manifolds. Let us treat these situations in the following:

\medskip

I) \cite[p. 251]{d:t}:
$$
S(X,Y)=\left(\frac{r}{2}+\xi (f)+f^2\right)g(X,Y)-
$$
$$
-\left(\frac{r}{2}+\xi (f)+3f^2\right)\eta (X)\eta (Y)-Y(f)\eta (X)-X(f)\eta (Y) \eqno(2.10)
$$
where $r$ is the scalar curvature. Then, for a 3-dimensional $f$-Kenmotsu manifold we get:
$$
\alpha:=({\cal L}_{\xi }g+2S)(X,Y)=(r+2\xi (f)+2f+2f^2)g(X,Y)-
$$
$$
-(r+2\xi (f)+2f+6f^2)\eta (X)\eta (Y)-2Y(f)\eta (X)-2X(f)\eta (Y) \eqno(2.11)
$$
while, for $\beta $-Kenmotsu:
$$
\alpha (X,Y)=(r+2\beta +2\beta ^2)g(\varphi X, \varphi Y)-4\beta ^2\eta (X)\eta (Y), \eqno(2.12)
$$
$$
(\nabla _Z\alpha )(X,Y)=Z(r)g(\varphi X, \varphi Y)-
$$
$$
-\beta (r+2\beta +6\beta ^2)[\eta (X)g(\varphi Y, \varphi Z)+\eta (Y)g(\varphi X, \varphi Z)]. \eqno(2.13)
$$
{\bf Substituting} $Z=\xi , X=Y\in (span{\xi })^{\bot }$ respectively $X=Y=Z\in (span{\xi })^{\bot }$ in $(2.13)$ we derive that $r$ is a constant, provided $\alpha $ is parallel. Thus, we can state the following:

\medskip

{\bf Proposition 3} {\it A 3-dimensional} ${\beta }$-{\it Kenmotsu Ricci soliton} $(g, \xi , \lambda )$ {\it is expanding and with constant scalar curvature}.

\medskip

{\bf Proof} $\lambda =-\frac{1}{2}\alpha (\xi, \xi)=2\beta ^2$. \quad $\Box $

\medskip

At this point we remark that the Ricci solitons of almost contact geometry studied in \cite{r:s5} and \cite{mm:t} in relationship with the Sasakian case are shrinking and this observation is in accordance with the diagram of Chinea from \cite{d:c} that Sasakian and Kenmotsu are opposite sides of the trans-Sasakian moon. Also, the expanding character may be considered as a manifestation of the fact that a $\beta $-Kenmotsu manifold can not be compact.

\medskip

II) Recall that the metric $g$ is called $\eta $-{\it Einstein} if there exists two real functions $a, b$ such that the Ricci tensor of $g$ is:
$$
S=ag+b\eta \otimes \eta .
$$
For an $\eta $-Einstein Kenmotsu manifold we have, \cite[p. 441]{j:dp}:
$$
S(X,Y)=\left(\frac{r}{2n}+1\right)g(X,Y)-\left(\frac{r}{2n}+2n+1\right)\eta (X)\eta (Y) \eqno(2.14)
$$
and then:
$$
\alpha (X,Y)=\left(\frac{r}{n}+4\right)g(X,Y)-\left(\frac{r}{n}+4+4n\right)\eta (X)\eta (Y) \eqno(2.15)
$$
$$
(\nabla _Z\alpha )(X,Y)=\frac{1}{n}Z(r)g(\varphi X, \varphi Y)-
$$
$$
-\left(\frac{r}{n}+4n+4\right)[\eta (Y)g(\varphi X, \varphi Z)+\eta (X)g(\varphi Y, \varphi Z)]. \eqno(2.16)
$$

\medskip

{\bf Proposition 4} {\it An} ${\eta }$-{\it Einstein Kenmotsu Ricci soliton} $(g, \xi , \lambda )$ {\it is ex\-pan\-ding and with constant scalar curvature, thus Einstein}.

\medskip

{\bf Proof} $\lambda =-\frac{1}{2}\alpha (\xi, \xi)=2n$. The same computation as in Proposition 3 implies constant scalar curvature. \quad $\Box $

\section{The dynamical point of view}

We begin this section with a straightforward consequence of the main Theorem, which also appears in the Olzack-Rosca paper, and is related to the last part of Corollary 2:

\medskip

{\bf Corollary 3} {\it An affine Killing vector field in a $\beta $-Kenmotsu manifold is Killing. As consequence, that scalar provided by the Ricci soliton} $(g, V)$ {\it of a Ricci-semisymmetric} $\beta $-{\it Kenmotsu manifold is} $\lambda =-S(V,V)$.

\medskip

{\bf Proof} (inspired by \cite[p. 504]{g:s}) Fix $X\in {\cal X}(M)$ an affine Killing vector field: $\nabla {\cal L}_Xg=0$. From Theorem it results that $X$ is {\it conformal Killing} i.e. ${\cal L}_Xg = cg$; more precisely $X$ is {\it homothetic} since $c$ is a constant. Lie differentiating the identity
$(1.5)$ along $X$ and using ${\cal L}_XS = 0$ (since $X$ is homothetic) and equation $(1.6)$
we get $g({\cal L}_X\xi ,\xi ) = 0$. Hence $c =({\cal L}_Xg)(\xi ,\xi )=-2g({\cal L}_X\xi ,\xi )= 0$. Thus $X$ is Killing. \quad $\Box $

\medskip

Let us present another dynamical picture of our results. Let $(M, \nabla)$ be a $m$-dimensional manifold endowed with a symmetric linear connection. A {\it quadratic first integral} (QFI on short) for the geodesics of $\nabla $ is defined by ${\cal F}=a_{ij}\frac{dx^i}{dt}\frac{dx^j}{dt}$ with a symmetric 2-tensor field $a=(a_{ij})$ satisfying the {\it Killing-type equations}:
$$
a_{ij:k}+a_{jk:i}+a_{ki:j}=0, \eqno(3.1)
$$
where, as usual, the double dot means the covariant derivative with respect to $\nabla $.

\medskip

The QFI defined by $a$ is called {\it special} (SQFI) if
$a_{ij:k}=0$ and the maximum number of linearly independent SQFI a pair $(M, \nabla)$ can admit
is $\frac{m(m+1)}{2}$; a flat space will admit this number.
In \cite[p. 117]{l:k} it is shown that a non-flat Riemannian
manifold may admit as many as $M_S(m)=1+\frac{(m-2)(m-1)}{2}$
linearly independent SQFI. Therefore, for an almost contact manifold
($m=2n+1$) the maximum number of SQFI is $M_S(2n+1)=1+n(2n-1)>1$.

\medskip

Our main result implies that for a regular $f$-Kenmotsu manifold the number of SQFI is exactly 1 and the only SQFI is the {\it kinetic energy} ${\cal F}=g_{ij}\frac{dx^i}{dt}\frac{dx^j}{dt}$. So:

\medskip

{\bf Proposition 5} {\it There exist almost contact manifolds which does not admit $M_S(2n+1)$ SQFI}.

\medskip

It remains as an open problem to find examples of almost contact metrics with exactly $M_S(2n+1)$ SQFI.


TITLE: Bounded linear operator property and its spectral radius
QUESTION [0 upvotes]: Let $A$ be a closed linear operator densely defined in a Banach space $X$. Furthermore, we suppose $\sigma(A)$ as its spectrum. We define $r_{A} := \sup\limits_{\lambda \in \sigma(A)} |\lambda|$ as the spectral radius of $A$. Assume $A$ is bounded linear operator of $X$ and bounded, then it is known that $r_{A} = \max\limits_{\lambda\in\sigma(A)}|\lambda| = \lim\limits_{n\to\infty}||A^{n}||^{\frac{1}{n}}$ and $r_{A} \leq ||A||$. Now, we let $r \in \mathbb{R}$ such that $r_{A} < r < \infty$.
My problem is I want to show that $\lim\limits_{n\to\infty}\frac{||A^{n}||}{r^{n}}=0$ but I don't know where to start. Any help is much appreciated!
Thank you very much!

REPLY [1 votes]: Let $0<\epsilon <r-r_A$.  Then there exists $k$ such that $\|A^{n}\| ^{1/n} <r_A+\epsilon$ for $n \geq k$. Hence $\frac {\|A^{n}\|} {r^{n}} <(\frac {r_A+\epsilon} r)^{n}\to 0$ since $r_A+\epsilon <r$.


TITLE: RSA Cryptography math problem
QUESTION [1 upvotes]: I have this math problem I'm kind of stuck on.

You intercept the message 27284682555982882069237 which was encrypted
   using a public modulus of 124137798108168664109413 and an encryption
   exponent 257. The modulus is now too large to be factored by testing
   successive candidate divisors (although it can be done using more
   sophisticated algorithms). However Bob, who is the intended recipient
   of the message, runs several online businesses. and the public key
   information for his other business is modulus 283967477199546905990801
   and encryption exponent 257.
You have heard a rumor that Bob has a ‘lucky prime’, which he uses as
   one of the prime factors for all the moduli he generates. Verify that
   this rumor is true (how do you check if the two moduli have a common
   prime factor?) and then use this information to decrypt the message.

I know that this is an RSA problem... so I believe that $p = 124137798108168664109413$ and $e = 257$, if that's correct... However, I'm sot sure how to continue. Thanks.

REPLY [0 votes]: Call the moduli $m_1$ and $m_2$, to avoid my having to refer to those big long numbers.  The rumor that Bob uses a "lucky prime" $p$ is equivalent to the claim that there exist two other primes $q_1, q_2$ such that $m_1 = pq_1$ and $m_2 = pq_2$.
In that case, the totient values of the two moduli are
$$
\phi(m_1) = (p-1)(q_1-1)
$$
$$
\phi(m_2) = (p-1)(q_2-1)
$$
We know that for both keys, the encryption exponent is $e = 257$.  So then the decryption exponents $d_1, d_2$ are characterized by
$$
ed_1 \equiv 1 \pmod{\phi(m_1)}
$$
$$
ed_2 \equiv 1 \pmod{\phi(m_2)}
$$
Hopefully, you can take it from here.


TITLE: Does this property hold for conditional expectation.
QUESTION [1 upvotes]: Assume you have a probability space $(\Omega,\mathcal{A},P)$. Assume you have a sub sigma-algebra and two random variables $X,Y$ such that $E(|X|), E(|Y|), E(|XY|)<\infty$. Do we then have
$$E(X|\mathcal{G})\cdot E[Y|\mathcal{G}]=E[XY|\mathcal{G}]?$$
Both sides are $\mathcal{G}$-measurable, so we only need to check the integrability requirement. So let $A\in \mathcal{G}$, what we need to show is that
$$E[E(X|\mathcal{G})\cdot E[Y|\mathcal{G}]\cdot1_A]=E[XY\cdot 1_A].$$
Do you see if this holds?

REPLY [2 votes]: No. Let $B_t$ be Brownian motion and $\mathcal{F}_t$ be its natural filtration. Set $X = Y = B_t$. Observe that $B_t$ and $B_t^2 - t$ are martingales. Then, for $s \leq t$,
$$E(X | \mathcal{F}_s)E(Y | \mathcal{F}_s) = B_s^2$$
But,
$$E(XY |\mathcal{F}_s) = E(B_t^2 \, | \, \mathcal{F}_s) = E(B_t^2 - t\, | \, \mathcal{F}_s) + t = B_s^2 -s + t$$


TITLE: Integer $N$, in base $b$, is $6789$. If $N$ is a multiple of $b-1$, and $b<16$, then what is the successor of $b$?
QUESTION [1 upvotes]: An integer $N$, expressed in base $b$, is $6789$. If $N$ is a multiple of $b-1$, and $b$ is less than $16$, then what is the successor of $b$?

I couldn't develop anything more than
$$N_b = \left(\;n\cdot (b-1)\;\right)_b = 6789_b$$
(with subscripts indicating "in base $b$").

REPLY [2 votes]: If $N$ is a multiple of $b-1$ then by the rule of nines[1] the sum of the digits is a multiple of $b-1$ as well.
So $b-1| 6+ 7 + 8 + 9 = 30$.  So $ b-1 = 1,2,3,5,6,10,15,30$
Bu $9$ is a digit so $b\ge 10$.  And $b < 16$ so $b-1 < 15$.
... so $b-1 < 15$ and so $b-1 =10$ and $b =11$ and then successor of $b$ is $12$.
[1]  Rule of nines:  $\sum_{k=0}^n a_k b^k \equiv \sum_{k=0}^n a_k \pmod {(b-1)}$.
Pf:  $\sum_{k=0}^n a_k b^k - \sum_{k=0}^n a_k = \sum_{k= 0}^n a_k (b^k-1) $
Now each $b^0 - 1 = 0$ and $b^1 - 1 =b-1$ and for $k > 1$ then $b^k -1 = (b-1)(b^{k-1} + b^{k-1} + ...... + 1)$ so $b-1|b^k-1$.  So $b-1|\sum_{k= 0}^n a_k (b^k-1)$
And so $\sum_{k=0}^n a_k b^k - \sum_{k=0}^n a_k\equiv 0 \pmod {(b-1)}$.
In other words, $N =\sum_{k=0}^n a_k b^k$ and the sum of the digits of $N$ (in base $b$) have the same remainder when divided by $b-1$.
And therefor $b-1|N$ if and only if $b-1|$ the sum of the digits if $N$ (in base $b$)


TITLE: Existence of two natural number satisfying a given condition in a given set
QUESTION [2 upvotes]: Suppose $A=\{1,2,\dots,112\}$, $B \subset A$ and the number of elements in $B$ is greater or equal to 37. Then, is it true that there always exist two elements $x,y \in B$ such that $x-y\in \{9,10,19\}$? Thanks in advance.

REPLY [1 votes]: Let's prove the following ...
Claim: Let $S$ be a subset of $X=\{1,2,\dots , 28\}$ for which $a-b\not\in\{9,10,19\}$ for all $a,b\in S$.  Then $|S|\le 9$.
By way of contradiction, assume that $|S|\ge 10$.  Now arrange the elements of $X$ as follows.
\begin{array}{rrrrrrrrrr}
1&2&3&4&5&6&7&8&9&\\
10&11&12&13&14&15&16&17&18&19\\
20&21&22&23&24&25&26&27&28&
\end{array}
Note that $S$ can contain at most one element from each column.  However, as $|S|\ge 10$, $S$ must contain at least one element from every column.  In particular, $19\in S$ is forced.
Now consider the following arrangement of the elements of $X$.
\begin{array}{rrrrrrrrrr}
1&2&3&4&5&6&7&8&9&10\\
11&12&13&14&15&16&17&18&19\\
20&21&22&23&24&25&26&27&28&
\end{array}
By using the exact same reasoning as above, we see that $10\in S$ is forced.  But this is a contradiction because $19-10=9$.  Thus $|S|\le 9$ as claimed.
We may repeat this argument on the three remaining subintervals $\{29,\dots,56\}$, $\{57,\dots,84\}$, and $\{85,\dots,112\}$.  This shows there are at most $9+9+9+9=36$ elements of $\{1,2,\dots,112\}$ with the desired avoidance property.  Hence $|A|\ge 37$ will force the difference of some pair of elements of $A$ to be $9$, $10$ or $19$.


TITLE: Subspace $X \in Mat_{2 \times 2}(\mathbb{R}):tr(YX)=0$, where $Y =\begin{bmatrix} 2 & 5 \\ 3 & 4 \end{bmatrix}$
QUESTION [0 upvotes]: I want to prove that set of $X \in Mat_{2 \times 2}(\mathbb{R}):tr(YX)=0$, where $Y =\begin{bmatrix}
2 & 5 \\
3 & 4
\end{bmatrix}$
is subspace of space $Mat_{2 \times 2}(\mathbb{R})$. But how can i find $\textbf{1}:\textbf{1} A = A\textbf{1} = A$, where A is an element of our subspace. And i also have problems with finding basis of this subspace. How can i do it?

REPLY [1 votes]: The map $X\mapsto\operatorname{Tr}(YX)$ is linear because it is a composition of linear maps: $X\mapsto YX$ and the trace. Your set is the kernel, so it is a subspace.
In order to find a basis of the kernel, you need to compute; if
$$
X=\begin{bmatrix} x_{11} & x_{12} \\ x_{21} & x_{22} \end{bmatrix}
$$
then
$$
YX=\begin{bmatrix}
2x_{11}+5x_{21} & 2x_{12}+5x_{22} \\
3x_{11}+4x_{21} & 3x_{12}+4x_{22}
\end{bmatrix}
$$
so the trace is zero when
$$
2x_{11}+3x_{12}+5x_{21}+4x_{22}=0
$$
and finding a basis should be easy.


TITLE: Dark energy stars disproven by Event Horizon images?
QUESTION [4 upvotes]: I was reading about the hypothesis advanced by George Chapline and Robert Laughlin that black holes might actually be 'dark energy stars', regions where spacetime has undergone a phase transition (e.g. this lay article and this paper).
Does the black hole image produced by the Event Horizon Telescope Collaboration falsify this hypothesis for good?
Or the hypothesis may still be compatible with the data?
Disclaimer: I am not a physicist, so please take it into account in your answers. I am just curious about the topic and would love to hear some expert opinions on this idea.

REPLY [2 votes]: No. The image obtained by EHT does not falsify this hypothesis since this image is mainly an evidence of a photon sphere, rather than the event horizon of the black hole.
But the “dark energy star” (one of black hole proposed alternatives) is expected to have a normal photon sphere up to rather small distance from the would-be horizon. And so the image from the EHT is compatible with black hole and dark energy star and also with other proposed black hole alternatives that are characterized by clear photon sphere.
For an overview of the multitude of exotic compact objects (ECO, the general name for black hole alternatives) and observational prospects for understanding their precise properties see the following (Open Access) review:

Cardoso, V., & Pani, P. (2019). Testing the nature of dark compact objects: a status report. Living Reviews in Relativity, 22(1), 4,
doi:10.1007/s41114-019-0020-4

Regarding the EHT images specifically, in section 5.3 the authors state:

In particular, the Event Horizon Telescope Collaboration has very recently obtained a radio image of the supermassive BH candidate in M87 (Akiyama et al. 2019) and similar results for Sgr A* are expected soon. The Event Horizon Telescope images of Sgr A* and M87* in the millimeter wavelength so far are consistent with a point source of radius $r_0=(2−4)M$
  (Doeleman et al. 2008; Doeleman 2012; Johannsen et al. 2016; Akiyama et al. 2019), or
  $$ ϵ∼1. \tag{102}$$
  This corresponds to the size of the photon sphere, which as we described in Sect. 2 will be the dominant relevant strong-field region for these observations. The absence of an horizon will influence the observed shadows, since some photons are now able to directly cross the object, or be reflected by it. There are substantial differences between the shadows of BHs and some horizonless objects [most notably boson stars (Cunha et al. 2015, 2017b; Cunha and Herdeiro 2018)]. Nevertheless, because of large astrophysical uncertainties and the focusing effect for photons when $ϵ→0$
  [Eq. (9)], all studies done so far indicate that it is extremely challenging to use such an effect to place a constraint much stronger than Eq. (102) (Vincent et al. 2016; Cunha et al. 2018; Cardenas-Avendano et al. 2019).

The parameter $ϵ$ used in the quote is a measure of how close a given ECO is to a ordinary black hole of the same mass: $ϵ=1-2M/R$, where $M$ is the mass of the ECO and $R$ is its radius (or more generally the radius where ECO becomes significantly different than the black hole), so that $ ϵ \to 0$ corresponds to an unmodified black hole of general relativity.


TITLE: Proof of the inequality $\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b} \geq \frac{3}{2}$
QUESTION [12 upvotes]: I am currently attempting to prove the following inequality
$\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b} \geq \dfrac{3}{2}$ for all $ a,b,c>0$
My instinctive plan of attack is to use the AM/GM inequality with $x_1=\dfrac{a}{b+c}$ etc.
Using that I get this
$\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b} \geq 3\times \sqrt[3]{\dfrac{abc}{(a+b)(b+c)(a+c)}}$
From here, I used the fact that $(a+b)(b+c)(a+c)\geq 8abc$, which can be easily proven by considering that $a+b\geq 2\sqrt{ab}$
But by using this, I get the following...
$\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b} \geq 3\times \sqrt[3]{\dfrac{abc}{(a+b)(b+c)(a+c)}} \leq 3 \times \sqrt[3]{\dfrac{abc}{8abc}} = \dfrac{3}{2}$
Everything seems so perfect because I get the value $\dfrac{3}{2}$ as required, but this method isn't valid due to the change in direction! What is going on?
Is there a way of proving this inequality otherwise then?

REPLY [2 votes]: \begin{eqnarray*}
  % \nonumber to remove numbering (before each equation)
   && \frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}-\frac{3}{2}\\
 &=& \frac{2(x^3+y^3+z^3)-(x^2y+xy^2+y^2z+yz^2+z^2x+zx^2)}{2 (x + y) (x + z) (y + z)}\\
     &=& \frac{(x+y)(x-y)^2+(y+z)(y-z)^2+(z+x)(z-x)^2}{2 (x + y) (x + z) (y + z)}\geq0
  \end{eqnarray*}
with equality holds if and only if $x=y=z$.


TITLE: criteria for equivalence to the trivial category $\ast$
QUESTION [1 upvotes]: I'm trying to determine when a category is equivalent to the category $\ast$ - the category with one object and one morphism. Such category clearly cannot have a morphism which is not an isomorphism, and I think that the condition $|Hom_C (X,Y)|=1$ for every two objects $X,Y$,  is also needed. Am I right? how can I prove a necessary and sufficient condition?

REPLY [0 votes]: Remark first that a category $\mathbf C$ as a terminal object if and only if the functor $!:\mathbf C \to \mathbf 1$ has a right adjoint.
Then, $!:\mathbf C \to 1$ is an equivalence if and only if it has a right adjoint $R$ such that $R \circ {!} \simeq \mathrm{id}_{\mathbf C}$. By the previous remark, the latter condition is equivalent to $\mathbf C$ having a terminal object which is isomorphic to every object in $\mathbf C$, which in turn is equivalent to $\mathbf C$ being non empty and having exactly one morphism between each pair of objects.


TITLE: Haldane model, why the nearest hopping is not changed?
QUESTION [1 upvotes]: I am leaning the Haldane model :
https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.61.2015
Haldane imaged threading magnetic flux though a graphene sheet, and the net flux of a unit cell is zero.
He argued that since the loop integral $exp[ie/ℏ∮A⋅dr]$ along a path of nearest bonds vanishes, the nearest hopping is not changed.
However, I cannot see the connect between the vanishing loop integral and the unchanged nearest hopping, can anybody help?

REPLY [1 votes]: That is because, in a closed loop, the states can only change by a phase factor due to Aharonov-Bohm effect. Since the magnetic flux is
$$\Phi_B = \oint\limits_{\partial S} \mathbf{A} \cdot d\boldsymbol{\ell}$$
Its clear that you aquire a zero phase in the n.n.n. hopping, and a non-zero phase $\varphi$ in the s.n.n. hopping. Ok, the issue is, you take this phase factor as a part of the hopping terms: a loop is represented by changing the hopping terms in the form
$$t_1 \rightarrow t_1 \qquad t_2 \rightarrow t_2 e^{i\varphi}$$
i.e. you are taking the effect of the magnetic flux always by shifting the hopping terms, but the $t_1$ is not changing because the flux on the unit cell is zero, so $e^0 = 1$.
You can see that introduction to topological insulators from Carpentier and Fruchart: https://arxiv.org/abs/1310.0255 .Though it has some minor errors (like the vectors in the $d$ functions of the Hamiltonian), I think this part is well explained.


TITLE: All unique combinations of n numbers
QUESTION [12 upvotes]: How is the algorithm or process called which calculates all unique combinations of n numbers? By unique I mean that this 1234 is the same as this 1243.
Example:
Take this 4 numbers and list all unique combinations:
1
2
3
4

Output:
1
2
3
4
12
13
14
23
24
34
123
124
134
234
1234

REPLY [14 votes]: There are $2^n-1$ non-empty subsets of the set $\{1,...,n\}$. Given a number $k=1,...,2^n-1$, we can write the $k$ as an $n$-digit binary number, and then put $i$ in the set $A_k$ if the $i$th binary digit of $k$ is $1$.
For example, $n=3$ yields:
$$\begin{array}{cc}\text k & \text {binary} & \text{set}\\
1 & 001 & \{3\}\\
2 & 010 & \{2\}\\
3 & 011 & \{2,3\}\\
4 & 100 & \{1\}\\
5 & 101 & \{1,3\}\\
6 & 110 & \{1,2\}\\
7 & 111 & \{1,2,3\}
\end{array}$$


TITLE: Non analytic, non constant , continuous complex valued function
QUESTION [1 upvotes]: I am trying to construct a non analytic , non constant, continuous function $f $ such that $f$ :$ \mathbb C\setminus\{0\}\to  \mathbb C$ with $f(z)= f\left(\frac{z}{|z|}\right)$.
I got stuck how to start this problem

REPLY [0 votes]: Note that the condition of non-constant is trivial for any non-analytic $f$.
Using polar coordinates $z=r.e^{i\theta}$
$f(r.e^{i\theta})=f(\frac{r.e^{i\theta}}{|r.e^{i\theta}|})$
$f(r.e^{i\theta})=f(e^{i\theta})$
A convenient choice will be $f(r.e^{i\theta})=\theta^2\implies f(z)=(Arg(z))^2$ which is defined on $\mathbb C/{0}$.


TITLE: What exactly is the CW complex structure on a geometric realisation?
QUESTION [6 upvotes]: This is likely a silly question.
Definitions:
$\bullet$ $\Delta_n = \{ (t_0, \dots, t_n) \: | \: 0 \leq t_i \leq 1, \sum_i t_i = 1 \}$
$\bullet$ Given $f: \underline{m} \to \underline{n}$ in $\Delta$, we have a map $\tilde f: \Delta_m \to \Delta_n$ given by $(s_0, \dots, s_m) \mapsto (t_0, \dots, t_n)$, where $t_i = \sum_{f(j) = i} s_j$.
$\bullet$ Given a simplicial set $F: \Delta^{\text{op}} \to \underline{\text{Sets}}$, we have its geometric realisation

$|F| := \coprod_{n \geq 0} F(n) \times \Delta_n/ \sim$

where $(x, \tilde f (s)) \in F(n) \times \Delta_n$ is equivalent to $(F(f)(x), s) \in F(m) \times \Delta_m$ for $f: \underline{m} \to \underline{n}$.
$|F|$ is a CW complex with a $n$-cell for each nondegenerate $n$-simplex.

May you please explicitly describe how $|F|$ is given the structure of a CW complex?

REPLY [2 votes]: To see why the geometric realization of a simplicial set is a CW complex, let us describe the geometric realization inductively. The zero skeleton of $|F|$ will be $\coprod_{\Delta^0\to F}|\Delta^0|$, which is just a bunch of points. Now suppose that we have described the $n-1$ skeleton, $sk_{n-1}|F|$ at this point, we attaching a copy of $|\Delta^n|$ for each non-degenerate n-simplex, $\sigma:\Delta^n\to F$ along the map $|\partial\sigma|:|\partial\Delta^n|\to |F|$  (we know that this maps is defined by the induction hypothesis). In categorical language, we are forming the pushout of the following diagram to obtain the n-skeleton.
$$sk_{n-1}|F|\leftarrow \coprod_{\sigma\in N_nF}|\partial\Delta^n|\rightarrow \coprod_{\sigma\in D_nF}|\Delta^n|$$ where $N_nF$ is the set of non-degenerate n-simplices. For a reference to this see page 8 of Goerss and Jardine.


TITLE: Confusion related to the concatenation of two grammars
QUESTION [3 upvotes]: I have this confusion. Lets say I have two languages produced by type 3 grammar such that
L(G1) = <Vn1,Vt,P1,S1>
L(G2) = <Vn2,Vt,P2,S2>

I need to find a type3 grammar G3 such that
L(G3) = L(G1)L(G2)
I can't use $S3 \rightarrow S1S2$  to get the concatenaion, because the production is not type 3 as well. So what should I do?

REPLY [2 votes]: First change one of the grammars, if necessary, to make sure that they have disjoint sets of non-terminal symbols.
If you’re allowing only productions of the forms $X\to a$ and $X\to Ya$, make the new grammar $G$ generate a word of $L(G_2)$ first and then a word of $L(G_1)$: replace every production of the form $X\to a$ in $G_2$ by the production $X\to S_1a$.
If you’re allowing only productions of the forms $X\to a$ and $X\to aY$, replace every production of the form $X\to a$ in $G_1$, by the production $X\to aS_2$.
If you’re allowing productions of both forms, you’re not talking about type $3$ grammars.


TITLE: Arriving at odd possible solutions for functions
QUESTION [0 upvotes]: Say $f(x)=\dfrac{3x+1}{2}$
I want to find out for which values of $x$ is the value of $f(x)$ an odd number.
So I reframe $f(x)$ to
$\dfrac{3x+1}{2}=2k_1+1$
on simplifying further...
$\dfrac{3x-1}{4}=k_1$ or $\dfrac{4k_1+1}{3}=x$
For $x=3,7,11,15,19,...$ we find that $k_1$ is a natural number and hence we get the solution.
Now I am attempting to find out for which values of $x$ is the value of $f^2(x) $or $f(f(x))$ is an odd number.
The answer is $x=7,15,23,31...$
How do I frame the equation?
I tried putting
$\dfrac{3(2k_1+1)+1}{2}=2k_2+1$ and
$f(f(x))=f^2(x)= \dfrac{3(\frac{3x+1}{2})+1}{2}=2k_2+1$
and both of them lead to incorrect answers.

REPLY [1 votes]: Continuing what you have written,
$f^2(x)=\frac{9x+5}{4}=2k+1$
$k=\frac{9x+1}{8}$, where k is any positive integer.
Note that $x=7$ satisfies this.
Further, let $x=7+y$ also gives integral $k$.
Then, $$k=\frac{9(7+y)+1}{8}$$ $$k=8+\frac{9y}{8}$$ $$\implies 8|y $$.
It gives $x=7,15,23,31...$


TITLE: Find the average velocity of gas atoms
QUESTION [1 upvotes]: The following problem arose from trying to solve a textbook problem and I don't know if I made some miscalculation or I should assume something that I don't know. I'm given helium in a balloon and I know its pressure $p$ and its density $d$; I'm asked to find the average velocity $v$ of the atoms.
What I know is that in general the average kinetic energy $\langle E_{k}\rangle$ for a monoatomic gas is given by $\frac{3}{2}k_{b}T$ where $k_{b}$ is the boltzmann constant and $T$ the temperature. From this formula I got that my average velocity should satisfy $$v=\sqrt{\frac{3k_{b}T}{m}}.$$ Using the ideal gas law and the relation $R=N_{A}k_{b}$ I got
$$v=\sqrt{\frac{3k_{b}pV}{mnR}}.$$ Now I used the fact that $d=\frac{m}{V}$
to conclude that $$v=\sqrt{\frac{3p}{dnN_{a}}}.$$
It's probably a dumb question but if this is correct I should know how to find the number of moles in the balloon, something that I'm not quite sure how to do. Am I missing something trivial? Thanks in advance for any hint.

REPLY [0 votes]: As stated by @Wolphram jonny  my mistake was in using blindly the formula without questioning the nature of what I called $m$.


TITLE: Find $f \in \mathbb{Q}[X]$ such that $f(a) = a^{-1}$, where $a := \sqrt{3} + \sqrt{5}$
QUESTION [0 upvotes]: I would like to get a hint fo the following problem:

Find $f \in \mathbb{Q}[X]$ such that $f(a) = a^{-1}$, where $a :=\sqrt{3} + \sqrt{5}$.

I know that $a^{-1} = \frac{1}{\sqrt{3} + \sqrt{5}} = \frac{1}{2}\left( \sqrt{5} - \sqrt{3}\right)$ and the minimal polynomial of $a$ is $g = x^4 - 16x^2 + 4$ and the minimal polynomial of $a^{-1}$ is $4x^4 - 16x^2 + 1$.

REPLY [0 votes]: Hint: $a^4 - 16a^2 = -4$. Try factoring the LHS.