TITLE: Symmetric linear least squares solution
QUESTION [2 upvotes]: Given an overdetermined linear system $AX=Y$ with known $A$ and $Y$, how would I go about finding the least squares solution $X$ under the constraint that it is symmetric ($X=X^T$)?

REPLY [4 votes]: Let $P=A^TA$ and $S=A^TY+Y^TA$. When $X$ is symmetric,
$$
\|AX-Y\|_F^2=\operatorname{tr}(XPX-SX+Y^TY).\tag{1}
$$
Since the set of all symmetric matrices form a closed linear subspace, there is always a global minimiser and any such global minimiser must be a critical point of $(1)$. However, as $P$ is positive semidefinite, the converse is also true, i.e. each critical point of $(1)$ is a global minimum.
At each critical point $X$, we have
$$
0=\operatorname{tr}\left((\Delta X)PX+XP(\Delta X)-S(\Delta X)\right)
=2\operatorname{tr}\left((PX+XP-S)(\Delta X)\right)
$$
for any symmetric matrix $\Delta X$. In particular, the trace is zero if we put $\Delta X=PX+XP-S$. But then $\operatorname{tr}((PX+XP-S)^2)=0$, meaning that
$$
PX+XP-S=0.\tag{2}
$$
Thus the global minimisers of $(1)$ are the symmetric solutions to $(2)$.
When $A$ has full column rank, $P=A^TA$ is positive definite. Therefore the above equation (known as Lyapunov equation) has a unique solution, which can be expressed as
$$
\operatorname{vec}(X)=(I\otimes P+P\otimes I)^{-1}\operatorname{vec}(S)
$$
or
$$
X=\int_0^\infty e^{-tP}Se^{-tP}dt.
$$
When $A$ has deficient column rank, we may solve $(2)$ without the symmetry constraint first, and then obtain a global minimiser by symmetrising the solution to $(2)$. More specifically, let $M = I\otimes P+P\otimes I$. Then the general solution to $(2)$ is given by
$$
\operatorname{vec}(X_0)=M^+\operatorname{vec}(S) + (I-M^+M)\operatorname{vec}(T)
$$
where $T$ is any matrix with the same size as $S$. The general symmetric solution to $(2)$ is therefore given by $X=\frac12(X_0+X_0^T)$.


TITLE: In the case of the Dirac Equation,what forbids the free electron to absorb a photon? (electron magnetic moment)
QUESTION [2 upvotes]: It is straightforward to show from relativistic kinematics that a free electron cannot absorb a photon, as shown in this previous thread.
However, It is also known that using the Dirac equation, you can derive a magnetic moment associated with including an electromagnetic field (for example, in the derivation on page 11 in these lecture notes or page 2 of these notes).
Are the above derivations consistent/coherent only because the electromagnetic field is not quantized there, or is there some other feature (such as spin) that allows photon absorption to take place?
As a separate but related question, why is the Feynman diagram corresponding to the single loop correction of the anomalous magnetic moment even allowed? It seems that energy-momentum conservation as shown in the first link above forbids this process. What am I missing?

REPLY [2 votes]: It's true that a free electron can't absorb a single photon because of kinematics, but that certainly doesn't mean electrons can't interact with the electromagnetic field! It is kinematically allowed to, e.g. absorb multiple photons, or absorb one photon and reemit it.
We measure the magnetic moment from precession experiments in a classical field, and a classical electromagnetic field arises from the coherent superposition of macroscopically many photons. Interaction with such a field has very little in common with absorption or emission of single photons. If you wanted to describe it with a Feynman diagram, it roughly involves the electron absorbing and emitting very many photons, going somewhat off-shell in the intermediate stages.


TITLE: Expositions of Stallings's fibration theorem
QUESTION [7 upvotes]: In his famous paper
Stallings, John,
On fibering certain 3-manifolds. 1962 Topology of 3-manifolds and related topics (Proc. The Univ. of Georgia Institute, 1961) pp. 95–100 Prentice-Hall, Englewood Cliffs, N.J.
Stallings proves a theorem that (roughly stated, I'm ignoring some hypotheses) says that if $M$ is a 3-manifold, then every short exact sequence
$$1 \longrightarrow G \longrightarrow \pi_1(M) \longrightarrow \mathbb{Z} \longrightarrow 1$$
with $G$ finitely generated comes from a fiber bundle $M \rightarrow S^1$.  In particular, $G$ is a surface group.
This is an oft-quoted theorem, but I have trouble reading the paper.  Are there any expository accounts of it anywhere?

REPLY [1 votes]: In the book knots by Gerhard Burde, Heiner Zieschang, Michael Heusener they give a proof for the special case of a knot exterior (Theorem 5.1). It becomes a bit easier, because if M is not closed, then G is a free-group.
Be careful and don't stumble on illegal copies of the book e.g. https://www.maths.ed.ac.uk/~v1ranick/papers/burdzies.pdf
It is not directly an exposition but perhaps it clarifies a missing part for you.


TITLE: Checking if a linear code exists - singleton , hamming and gilbert-varshamov bounds do not help.
QUESTION [1 upvotes]: Suppose I want to check if a (11, 6, 4) code exists.
I cannot prove non-existence using the singleton and the hamming bound. I also cannot prove existence using the gilbert-varshamov bound. I'm not sure what is the best way to go then -  shall I just try to create a parity-check matrix?

REPLY [1 votes]: I tried to simply create a parity check matrix for the code and it turned out to be quite easier than I thought.
Assuming $r = n - k = 5$
$$
H=\left(\begin{array}{l}
11100\\
01110\\
00111\\
10011\\
11001\\
11111\\
10000\\
01000\\
00100\\
00010\\
00001
\end{array}\right)
$$
Any 3 $(d-1)$ rows are linearly independent and there is at least one set of of 4 $(d)$ linearly dependent rows.


TITLE: Calculating the expected value of free tickets in a paytable?
QUESTION [0 upvotes]: Lets says I have a lottery game where the ticket costs $1 and has the following probability/prize distribution:

0.3 -> \$1
0.2 -> X
0.5 -> \$0

If X = \$1, then the expected value is: 0.3(\$1) + 0.2(\$1) = \$0.50
If X = FREE_TICKET, then I've calculated (and confirmed) via sampling that the EV is either:

0.3(\$1) / (0.3+0.5) = 0.375
0.3(\$1) + 0.2 * \$1 * 0.3/(0.3+0.5) = 0.375

Obviously, a FREE_TICKET isn't the same as \$1, the price of a ticket.
However, I'm stumped when calculating the EV of the table for the following two cases:

X = 2 FREE_TICKETS
X = \$5 + FREE_TICKET

How do I calculate them?

REPLY [1 votes]: My guess:
EV = 0.3 + 0.2 * (2 * EV) in the first case
EV = 0.3 + 0.2 * (5 + EV) in the second


TITLE: Acceleration in a rotating frame
QUESTION [2 upvotes]: From "Centrifugal force", Wikipedia:

Acceleration
[...]
By applying the transformation above from the stationary to the rotating frame three times the absolute acceleration of the particle can be written as:
\begin{aligned}{\boldsymbol {a}}&={\frac {\operatorname {d} ^{2}{\boldsymbol {r}}}{\operatorname {d} t^{2}}}={\frac {\operatorname {d} }{\operatorname {d} t}}{\frac {\operatorname {d} {\boldsymbol {r}}}{\operatorname {d} t}}={\frac {\operatorname {d} }{\operatorname {d} t}}\left(\left[{\frac {\operatorname {d} {\boldsymbol {r}}}{\operatorname {d} t}}\right]+{\boldsymbol {\omega }}\times {\boldsymbol {r}}\ \right)\\&=\left[{\frac {\operatorname {d} ^{2}{\boldsymbol {r}}}{\operatorname {d} t^{2}}}\right]+{\boldsymbol {\omega }}\times \left[{\frac {\operatorname {d} {\boldsymbol {r}}}{\operatorname {d} t}}\right]+{\frac {\operatorname {d} {\boldsymbol {\omega }}}{\operatorname {d} t}}\times {\boldsymbol {r}}+{\boldsymbol {\omega }}\times {\frac {\operatorname {d} {\boldsymbol {r}}}{\operatorname {d} t}}\\&=\left[{\frac {\operatorname {d} ^{2}{\boldsymbol {r}}}{\operatorname {d} t^{2}}}\right]+{\boldsymbol {\omega }}\times \left[{\frac {\operatorname {d} {\boldsymbol {r}}}{\operatorname {d} t}}\right]+{\frac {\operatorname {d} {\boldsymbol {\omega }}}{\operatorname {d} t}}\times {\boldsymbol {r}}+{\boldsymbol {\omega }}\times \left(\left[{\frac {\operatorname {d} {\boldsymbol {r}}}{\operatorname {d} t}}\right]+{\boldsymbol {\omega }}\times {\boldsymbol {r}}\ \right)\\&=\left[{\frac {\operatorname {d} ^{2}{\boldsymbol {r}}}{\operatorname {d} t^{2}}}\right]+{\frac {\operatorname {d} {\boldsymbol {\omega }}}{\operatorname {d} t}}\times {\boldsymbol {r}}+2{\boldsymbol {\omega }}\times \left[{\frac {\operatorname {d} {\boldsymbol {r}}}{\operatorname {d} t}}\right]+{\boldsymbol {\omega }}\times ({\boldsymbol {\omega }}\times {\boldsymbol {r}})\ .\end{aligned}

Here I have seen that the last term, generally $\mathbf{\omega} {\times}\left(\mathbf{\omega} {\times} \mathbf{r} \right)$. The other terms nullify out possibly.
Question:  Why do the rotating frame problems require usage of non-inertial frames instead of ground frames for general calculations?

REPLY [0 votes]: Chiming in with the answer provided by David Hammen:
Depending on the thing you want to compute you make an assessment as to which is more practical.
Some examples:
Let's say you want to calculate ballistic trajectories, in the simplified case of an Earth with no atmosphere.
You have a computer, and you have the skill to program calculations.
Then it is practical to use the non-rotating coordinate system that is co-moving with the Earth's center. You find the starting velocity and direction of the projectile, and using that you find numerically how long its flight lasts. During the flight the Earth has turned underneath the flying projectile, so the last step of that calculation is to factor in how much the Earth has turned during the flight.
Advantage of using the inertial coordinate system: the trajectory of the flight is a Kepler orbit; very straightforward to compute.
Now let's say you want a more accurate simulation, and you do want to take air friction into account. The Earth's air mass is co-moving with the Earth, so it has a velocity relative to the non-rotating coordinate system that you have to factor in.
Another example of a case where all the input data are given relative to the coordinate system that is co-rotating with the Earth: Meteorology. By using the coordinate system that is co-rotating with the Earth all of the input data go straight in. So that would be a case where it is more practical to use the co-rotating coordinate system.

General remark:
As emphasized by John Rennie: the basis of understanding motion is to think in terms of the motion with respect to the inertial coordinate system.
To use laws of motion there is only one choice of reference frame: the equivalence class of inertial coordinate systems. There are no exceptions to that; the 'omega' in the expression of the transformation is the angular velocity of the rotating coordinate system with respect to the equivalence class of inertial coordinate systems.
That is: the equations for the rotating coordinate system work because they reference the inertial coordinate system.


TITLE: modification of Doob inequality
QUESTION [1 upvotes]: Hi everyone, please consider the following problem:
Let $(M_t)_{t\geq 0}$ be a continuous and positive submartingale and $S_t=\sup_{0\leq s\leq t}M_s$. Please prove that for any $\lambda>0$ we have
$$\lambda P(S_t>2\lambda)\leq E[M_t1_{\{M_t>\lambda\}}]$$
This inequality makes me remember the Doob inequality
$$\lambda P(S_t>\lambda)\leq E[M_t1_{\{S_t>\lambda\}}]$$
So it is enough to show that
$$E[M_t1_{\{S_t>\lambda\}}]\leq 2E[M_t1_{\{M_t>\lambda\}}]$$
But I have no idea to deal about the term $1_{\{M_t>\lambda\}}$, even by introducing a stopping time $T_{\lambda}$. Could someone help me prove this inequality or give some idea? Many thanks!

REPLY [1 votes]: Nice little exercise, it should go to math.stackexchange.com, though. I'll give you a hint: Let $T$ be the hitting time of $2\lambda$. Then, $E[M_t 1_{M_t\ge \lambda}] \ge E[M_t 1_{M_t\ge \lambda,\ T\le t}] \ge E[M_t1_{T\le t}]- E[M_t1_{M_t\le \lambda}1_{T\le t}].$


text:
\begin{document}
\begin{abstract} For a  bornology $\mathcal B$ on a cardinal
$\kappa$, we  prove that
 the $\mathcal B$-macrocube is normal if and only if $\mathcal B$  has a  linearly ordered  base.
As a corollary, we  get   that the   hyperballean of bounded subsets of an ultradiscrete ballean is not normal. These  answer Question 1 from \cite{b2}  and Question 14.4 from \cite{b1}.

\end{abstract}
\maketitle

MSC: 54E05, 54E15, 05D10

Keywords: coarse structure, ballean, macrocube, hyperballeans, Ramsey ultrafilter.

\section{ Introduction  and preliminaries}

Given a set $X$, a family $\mathcal{E}$  of subsets of $X\times X$ is called a
{\it  coarse structure} on $X$ if

\begin{itemize}
\item{} each $E \in \mathcal{E}$  contains the diagonal $\bigtriangleup _{X}:=\{(x,x): x\in X\}$ of $X$;
\vspace{3 mm}

\item{}  if  $E$, $E^{\prime} \in \mathcal{E}$  then  $E \circ E^{\prime} \in \mathcal{E}$  and
$ E^{-1} \in \mathcal{E}$,    where  $E \circ E^{\prime} = \{  (x,y): \exists z\;\; ((x,z) \in E,  \ (z, y)\in E^{\prime})\}$,    $ E^{-1} = \{ (y,x):  (x,y) \in E \}$;
\vspace{3 mm}

\item{} if $E \in \mathcal{E}$ and  $\bigtriangleup_{X}\subseteq E^{\prime}\subseteq E$  then  $E^{\prime} \in \mathcal{E}$.
\end{itemize}

Elements $E\in\mathcal E$ of the coarse structure are called {\em entourages} on $X$

For $x\in X$  and $E\in \mathcal{E}$ the set $E[x]:= \{ y \in X: (x,y)\in\mathcal{E}\}$ is called the {\it ball of radius  $E$  centered at $x$}.
Since $E=\bigcup_{x\in X}\{x\}\times E[x]$, the entourage $E$ is uniquely determined by  the family of balls $\{ E[x]: x\in X\}$.
A subfamily ${\mathcal E} ^\prime \subseteq\mathcal E$ is called a {\em base} of the coarse structure $\mathcal E$ if each set $E\in\mathcal E$ is contained in some $E^\prime \in\mathcal E$.

The pair $(X, \mathcal{E})$  is called a {\it coarse space}  \cite{b10} or  a {\em ballean} \cite{b7}, \cite{b9}.

In this paper, all balleans under consideration are supposed to be
 {\it connected}: for any $x, y \in X$, there is $E\in \mathcal{E}$ such $y\in E[x]$.
A subset  $Y\subseteq  X$  is called {\it bounded} if $Y= E[x]$ for some $E\in \mathcal{E}$,
  and $x\in X$.
The family $\mathcal{B}_{X}$ of all bounded subsets of $X$  is a bornology on $X$.
We recall that a family $\mathcal{B}$  of subsets of a set $X$ is a {\it bornology}
if $\mathcal{B}$ contains the family $[X] ^{<\omega} $  of all finite subsets of $X$
 and $\mathcal{B}$  is closed   under finite unions and taking subsets. A bornology $\mathcal B$ on a set $X$ is called {\em unbounded} if $X\notin\mathcal B$.
A subfamily  $\mathcal B^{\prime}$ of $\mathcal B$ is called a base for $\mathcal B$ if, for each $B \in \mathcal B$, there exists $B^{\prime} \in \mathcal B^{\prime}$ such that $B\subseteq B^{\prime}$.

Each subset $Y\subseteq X$ defines a {\it subbalean}  $(Y, \mathcal{E}|_{Y})$  of $(X, \mathcal{E})$,
 where $\mathcal{E}|_{Y}= \{ E \cap (Y\times Y): E \in \mathcal{E}\}$.
A  subbalean $(Y, \mathcal{E}|_{Y})$  is called  {\it large} if there exists $E\in \mathcal{E}$
 such that $X= E[Y]$, where $E[Y]=\bigcup _{y\in Y} E[y]$.

Let $(X, \mathcal{E})$, $(X^{\prime}, \mathcal{E}^{\prime})$
 be  balleans. A mapping $f: X \to X^{\prime}$ is called
  {\it  macrouniform }  if for every $E\in \mathcal{E}$ there
  exists $E^{\prime}\in \mathcal{E}$  such that $f(E(x))\subseteq  E^{\prime}(f(x))$
    for each $x\in X$.
If $f$ is a bijection such that $f$  and $f ^{-1 }$ are macrouniform, then   $f  $  is called an {\it asymorphism}.
If  $(X, \mathcal{E})$ and  $(X^{\prime}, \mathcal{E}^{\prime})$  contains large  asymorphic  subballeans, then they are called {\it coarsely equivalent.}

Every  metric $d$ on a set  $X$ defines the coarse  structure
$\mathcal{E} _{d}$  on $X$  with  the base $\{\{(x,y): d(x, y)< n\} : n\in \mathbb{N}\}$.
A ballean $(X, \mathcal{E})$ is called {\it metrizable}  if there is a metric $d$ on such that $\mathcal{E} = \mathcal{E} _{d}$.

\vspace{5 mm}

{\bf Theorem 1.1. } ([9, Theorem 2.1.1])
{\it A ballean $(X,\mathcal{E})$ is metrizable  if and only if $\mathcal{E}$  has a countable base.}

\vspace{5 mm}

Let $(X,\mathcal{E})$  be a ballean. A subset $U \subseteq X$  is called an
 {\it asymptotic neighbourhood  } of a subset $Y\subseteq X$ if for every $E\in \mathcal{E}$ the set $E [Y] \setminus U$  is bounded.

Two subset $Y, Z$ of $X$  are called  {\it asymptotically disjoint  (separated)}  if for every $E\in \mathcal{E}$ the intersection
  $E[Y] \cap E[Z] $  is bounded ($Y$ and $Z$ have disjoint asymptotic neighbourhoods).
We say that $Y,Z$  are {\it  linked} if $Y,Z$  are not asymptotically disjoint.

A ballean $(X, \mathcal{E})$  is called {\it normal}  \cite{b6} if any
two asymptotically disjoint subsets of $X$  are asymptotically separated. Every ballean $(X, \mathcal{E})$
 with linearly ordered base of $\mathcal{E}$  is  normal.
In particular,   every  metrizable ballean is normal, see [6, Proposition 1.1].

A function $f: X\to \mathbb{R}$  is called   {\it slowly oscillating} if
 for any $E\in \mathcal{E}$  and $\varepsilon > 0$,  there exists a bounded  subset $B$ of $X$  such that $ diam \   f (E[x])<  \varepsilon $ for each $x\in X \setminus B$.   $ \ \  \Box $

\vspace{5 mm}

{\bf Theorem 1.2. } ([6, Theorem 2.1.])
{\it A ballean  $(X, \mathcal{E})$ is normal if and only if
 for any two disjoint  asymptotically disjoint
subsets $Y, Z$ of $X$ there exists a slowly oscillating function  $f: X\to [0,1]$ such that
$f| _Y \equiv   0$  and $f| _Z  \equiv 1$.}
\vspace{5 mm}

{\bf Theorem 1.3. } ([1, Theorem 1.4 .])
{\it If the product $X \times Y$  of  two unbounded balleans $X, Y$ is normal then the  bornology  $B _ {X\times Y}$ has  a linearly  ordered base.  }

\section{ Macrocubes and hyperballeans}

Let     $\mathcal{B}$  be a bornology on an infinite cardinal  $\kappa$  such that $\kappa \notin \mathcal{B}$,
$\{ 0,1 \}^{\mathcal{B}}=$
  $ \{ (x_{\alpha})_ {\alpha  < \kappa})\in  \{ 0,1 \}^\kappa :$
$\{ \alpha: x_\alpha =1\} \in \mathcal{B}. $
We take the family
$$\{ \{(( x_\alpha)_ {\alpha  < \kappa},  \    (y_\alpha)_ {\alpha  < \kappa}) \in  \{ 0,1 \}^{\mathcal{B}  } \times
  \{ 0,1 \}^{\mathcal{B}}:
x_{\alpha} = y_{\alpha}, \   \alpha\in \kappa \setminus B \} : B\in \mathcal{B}\}  $$
as a base  of the coarse structure on  $\{ 0,1 \}^{\mathcal{B}}$.
The obtained ballean is called the $\mathcal{B}$-macrocube \cite{b2}.
If $\kappa = \omega$  and  $\mathcal{B}= [\omega]^{<\omega}$
then we get the well-known  Cantor macrocube whose coarse characterization was given in \cite{b3}.

Given a ballean $(X, \mathcal{E})$   the {\it hyperballean}   $(X, \mathcal{E})^\flat$   is a ballean on   $\mathcal{B}_{(X, \mathcal{E})}$ endowed  with  a coarse  structure  with the  base  $\{ E^\flat : E \in  \mathcal{E}\}, $    $E^\flat [Y]= \{Z: Y \subseteq E[Z]$,  $Z \subseteq E[Y]\}$, see \cite{b4},  \cite{b8}.

Every bornology $\mathcal{B}$  on a cardinal $\kappa$  defines the {\it discrete ballean}
$(\kappa,  \mathcal{E}_\mathcal{B} )$,  $ \mathcal{E}_\mathcal{B}$
has the base
$\{  E_\mathcal{B}: B \in \mathcal{B} \}$, $E_\mathcal{B} [x]= B$
if
 $x\in B$  and
$E_\mathcal{B} [x]=\{x\}$
if
$x\in \kappa \setminus B$.
If the set
$\{\kappa \setminus B: B\in   \mathcal{B} \}$
 is an ultrafilter, then
$(\kappa,  \mathcal{E}_\mathcal{B} )$
 is called {\it ultradiscrete}.
We denote by
$[\kappa]^\mathcal{B}$
 the  hyperballean
$(\kappa,  \mathcal{E}_\mathcal{B} )^\flat$.

\vspace{5 mm}

{\bf Proposition 2.1. }
{\it  Let $\kappa$  be a cardinal, $\mathcal{B} )$
be a  bornology on $\kappa$,  $K_0 = \{ A \in [\kappa]^\mathcal{B}  :  0\in A$.
Then the characteristic  function  $f: [\kappa]^\mathcal{B}  \longrightarrow \{0,1\}^\mathcal{B}$
is macro-uniform and the restriction
of $f$  to $K_0$ is an asymorphism between  $K_0$  and   $f(K_0)$. }

\vspace{5 mm}

{\bf Proposition 2.2. }
{\it   For any  cardinal  $\kappa$ and bornology  $\mathcal{B}$ on $\kappa$,
the balleans
$\{0,1\}^\mathcal{B}$ and $[\kappa]^\mathcal{B}$  are not coarsely equivalent.
\vspace{5 mm}

Proof.}
Following  \cite{b3}, we say that a ballean $(X, \mathcal{E}$ has asymptotically isolated balls if, for any bounded subset $B$ and  $E\in \mathcal{E}$  there exists $x\in X\setminus B$ such that  $E^\prime [x]\setminus E[x]= \emptyset$.
If  $(X^\prime , \mathcal{E} ^\prime)$ is  coarsely equivalent  to   $(X, \mathcal{E})$    then  $(X^\prime , \mathcal{E} ^\prime)$
has asymptotically isolated balls.

We observe
$\{0,1\}^\mathcal{B}$
 does not have asymptotically isolated balls,
but the subballean
$[\kappa]_1 ^\mathcal{B} = \{ A\subset \kappa : |A|=1\}$
 of
$[\kappa] ^\mathcal{B} $
 has asymptotically isolated balls. Since
$[\kappa]_1 ^\mathcal{B} $
 and
$[\kappa] ^\mathcal{B}\setminus  [\kappa]_1 ^\mathcal{B}$
  are asymptotically disjoint, we see that
$[\kappa] ^\mathcal{B}$
 has asymptotically isolated balls.
$ \ \  \Box $

\vspace{5 mm}

{\bf Proposition 2.3. }{\it
Let
$\mathcal{B} $
 be a bornology on a cardinal   $\kappa$  such that
$\{ \kappa \setminus B: B\in  \mathcal{B}\} $  is not an ultrafilter. Then
$[\kappa] ^\mathcal{B} $
 is normal if and only if
$\mathcal{B} $
  has a linearly ordered base.

\vspace{5 mm}

Proof.}
If  $\mathcal{B} $ has a linearly ordered base then
$[\kappa] ^\mathcal{B} $
 is normal by  [6, Proposition 1.1].

We assume that $[\kappa] ^\mathcal{B} $ is normal, partition $\kappa$  in two unbounded subsets $K_1$, $K_2$ and put
$\mathcal{B}_1=\mathcal{B} | _{K_1}$,
$\mathcal{B}_2=\mathcal{B} | _{K_2}$.
Then
$\{0,1\}^\mathcal{B}$
 is asymorphic to
$\{0,1\}^{\mathcal{B}_1}\times   \{0,1\}^{\mathcal{B}_2}$.
By Theorem 1.3,  $\mathcal{B} $ has a linearly ordered base.
$ \ \  \Box $

\section{Some algebra on ultrafilters }

Let $G$ be an Abelian group.  We endow $G$ with the discrete topology and identify the Stone-$\check{C}$ech compactification $\beta G$ of  $G$ with the family of all ultrafilters on $G$.

For $A\subseteq G$, we denote $\bar{A} =
 \{\mathcal{ U} \in \beta G: A\in \mathcal{ U}\}$.

Then the family $\{\bar{ A}: A\subseteq G \}$ forms a base for the topology on $\beta G$.
Every mapping $f: G\longrightarrow [0,1]$ can be extended to the continuous  mapping
$ \  f^\beta : \beta G \longrightarrow [0,1]$.

Following \cite{b5}, we extend the addition  $+ $  on $G$ onto $\beta G$ by the following rule: for
$\mathcal{ U}$,  $\mathcal{ V}\in  \beta G$
 we take
$U\in \mathcal{ U}$
  and, for each $x\in U$, pick $V_x \in \mathcal{ U}$.
Then $\bigcup_{x\in U} (x+V_x)\in \mathcal{ U}+\mathcal{V}$, and the family of all these subsets  forms a base of the ultrafilter
$\mathcal{ U}+\mathcal{V}$.

\section{Results }

{\bf Theorem 4.1. }
{\it Let $ \mathcal{B}$ be a  bornology on a cardinal
$\kappa$.
The  $\mathcal B$-macrocube $X$ is normal if and only if $\mathcal B$ has a  linearly ordered base.
\vspace{5 mm}

Poof.}
In light of Proposition 2.3,  it suffices to assume that $ \{ \kappa  \setminus B  : B\in \mathcal{B}\}$ is an ultrafilter   and  show that $X$  is not normal.

We consider   $X$  as a group with pointwise addition $mod \ 2$.
For $A \subseteq \kappa$ and  $\alpha  < \kappa$,  $\chi_A$  denotes the characteristic function of $A$,  $[0.\alpha]= \{\gamma < \kappa: \gamma \leq \alpha \}$.
Replacing $\kappa$ to some cardinal    $\kappa^\prime\leq  \kappa$,  we may suppose that $[0, \alpha]\in \mathcal B $ for  each
$\alpha < \kappa$.

We consider two subsets $Y, Z$ of  $X$ defined by
$$Y=\{ y_\alpha : \alpha < \kappa\},  \  y_\alpha  = \chi _{\{\alpha \}}  , $$
$$Z=\{ z_\alpha : \alpha < \kappa\},  \  z_\alpha  = \chi _{[0, \alpha]}  , $$
and show that $Y, Z$ are asymptotically disjoint.
We take an arbitrary $B\in \mathcal B$,  denote by $H$ the subgroup
of $X$ generated by
$\{ y_\alpha : \alpha  \in  B \} $
 and take the minimal
$\gamma < \kappa$
such that $\gamma \in \kappa\setminus B$.
Then
$(y_\beta  + H)\cap Z = \emptyset $
  as  soon as $\beta \in \kappa\setminus B$, $\beta > \gamma$.

We suppose that $X$ is normal and use Theorem 1.2  to  choose a slowly oscillating function $f: X\longrightarrow [0,1]$
such that
$f|_{Z}  \equiv  1$,
$f|_{Y}  \equiv  0$,
and denote by
$\mathcal{U} $  and $\mathcal{V} $
ultrafilters on $X$ with the bases
$$\{\{  y_\alpha : \alpha\in  \kappa \setminus B\} : B\in \mathcal B\}, \  \{\{  z_\alpha : \alpha\in  \kappa \setminus B\} : B\in \mathcal B\}. $$

Let $f^\beta (\mathcal{V}  + \mathcal{U}) = r  $,   $r\in [0.1]. $
We take an arbitrary
$W\in \mathcal{V}  + \mathcal{U} $
and pick  $ C \in \mathcal B$
and
 $ D _\alpha \in  \mathcal B  $,
$\alpha \in  C  $
such that
$$\bigcup \{ z_\alpha +y_\beta:  \alpha \in \kappa\setminus    C,  \
 \beta \in \kappa\setminus  D_\alpha \}\subseteq W. $$
Then we construct inductively two mappings
$\psi: \kappa \setminus  C\longrightarrow \kappa,  $
$\phi: \kappa \setminus  C\longrightarrow \kappa $
such that
$\psi(\alpha) \in \kappa \setminus  D_\alpha $, $\phi(\alpha)\in \kappa\setminus  D_\alpha  $
and
$\psi (\kappa \setminus  C) \cap \phi(\kappa\setminus  C) =\emptyset$.
Since
$\{  \kappa \setminus B : B\in \mathcal B\}$
is an ultrafilter, either
$\psi(\kappa \setminus  C)\in \mathcal B$ or
$\phi(\kappa \setminus  C)\in \mathcal B$.
We assume that
$\psi(\kappa \setminus   C)\in \mathcal B$
 and denote by $H$ the subgroup of $X$ generated by
$\{ y_\alpha: \alpha \in \psi(\kappa \setminus  C)\}.$

Then
$z_\alpha \in  W+H$
 for each
$\alpha \in  \kappa \setminus  C$.
Hence, $W$ and $Z$ are linked.
Since $f$ is slowly oscillating, we conclude that $r=1$.

On the other hand, for
$\alpha\in \kappa \setminus  C$,
 $ z_\alpha +  \{y_\beta:
 \beta \in \kappa\setminus  D_\alpha \}\subseteq W. $
It follows that $W$ and $Y$ are  linked.
Since $f$ is slowly oscillating, we get $r=0$ contradicting above paragraph. $\ \Box$

\vspace{5 mm}

{\bf Corollary 4.2. }
{\it  Let $\mathcal B$  be a bornology on a cardinal $\kappa$ such that
$\{  \kappa \setminus B : B\in \mathcal B\}$
 is an ultrafilter.
Then the hyperballean
$[\kappa ]^{\mathcal B}$
  is not normal.

\vspace{5 mm}

Poof.}  Since a subballean of a normal ballean is normal, to apply Theorem 4.1, we use Proposition 2.1.  $ \ \Box$
\vspace{5 mm}

Theorem 4.1 answers Question 1 from \cite{b2},
Corollary 4.2 answers  Question 14.4 from \cite{b1}.
\vspace{5 mm}

For  a bornology  $\mathcal B$ on a cardinal  $\kappa$, the subballean of  all characteristic functions of finite subsets of
$\kappa$ of the
ballean $\{ 0,1\}^\mathcal B$
is called the {\it finitary $\mathcal B$-macrocube. }
It follows from the proof of  Theorem 4.1 that the finitary $\mathcal B$-macrocube on $\omega$ is not normal provided that $\{ \omega\setminus B : B\in \mathcal B\}$ is ultrafilter.

\vspace{5 mm}

{\bf Question 4.3. }
{\it  Let $\mathcal B$  be a bornology on a cardinal $\kappa$ such that
$\{  \kappa \setminus B : B\in \mathcal B\}$
 is an ultrafilter.
Is it true that the finitary $\mathcal B$-macrocube is not normal?}

\vspace{5 mm}

For $n\in \omega$, $n>0$ and a bornology   $\mathcal{B}$  on $\kappa$, we denote
 $[\kappa]^\mathcal{B}_n  = \{ A\in [\kappa]^\mathcal{B} : |A|\leq n\}$,
 $\{ 0,1 \}^\mathcal{B}_n  = \{  (x_\alpha)_{\alpha<\kappa}: | \{ \alpha : x_  \alpha  =1\} |  \leq n \}$.
If  $\{ \kappa  \setminus B: B \in \mathcal{B} \} $ is an ultrafilter then
 $[\kappa]^\mathcal{B}_n $ is normal  [1, Theorem 1.13].

A ballean $(X, \mathcal{E})$ is called {\it ultranormal} if any two unbounded subsets of $X$ are linked.

\vspace{3 mm}

{\bf Theorem 4.4. }
{\it Let $ \mathcal{B}$ be  a  bornology on $\kappa$  such that
$\{ \kappa \setminus B: B \in \mathcal{B}\}$  is an ultrafilter.
Then the ballean
 $\{ 0,1 \}^\mathcal{B}_n  $ is ultranormal.
\vspace{5 mm}

Proof.}  We consider  $\{ 0,1 \}^\mathcal{B}_1  $ as a subballean of    $\{ 0,1 \}^\mathcal{B}_n  $.
Since
 $\{ \kappa  \setminus B: B \in \mathcal{B} \} $  is an ultrafilter, it suffices to show that $\mathcal{A}$,
$\{ 0,1 \}^\mathcal{B}_1  $ are linked for any unbounded subset $\mathcal{A}$ of $\{ 0,1 \}^\mathcal{B}_n  $.

Applying $n$-times Lemma 7.1 from \cite{b1}, we pick $P\subseteq \kappa$ such that $ \kappa \setminus P\in \mathcal{B}$
and  $ \mathcal{A}^\prime = \{ A \in  \mathcal{A}:  |A\cap P|=1\}$ is unbounded.
 We put
$H= \{ (x_\alpha) _ {\alpha <\kappa} : x_\alpha =0$ for each $\alpha \in P\}$.
Then $(x+ H) \cap  \mathcal{A}^\prime \neq \emptyset$ for each $x\in P$.
Hence, $ \mathcal{A}^\prime$  and
 $\{ 0,1 \}^\mathcal{B} _1  $ are linked.  $ \ \  \Box $

config: arxiv


TITLE: Two Fundamental Polygons for the Double Torus?
QUESTION [1 upvotes]: The answer to this question shows two different fundamental polygons to represent a double torus, i.e.

Let's call $I=A,II=B,III=C$ and $IV=D$, start from left and go counter-clockwise, then their identifications of their sides are:

left: $\,\,\,A\phantom{^{-1}}D^{-1}C^{-1}B^{-1}A^{-1}D\phantom{^{-1}}C\phantom{^{-1}}B\phantom{^{-1}}$
right: $A^{-1}B\phantom{^{-1}}A\phantom{^{-1}}B^{-1}C^{-1}D\phantom{^{-1}}C\phantom{^{-1}}D^{-1}$

What's difference between them?
  Are there more $8$-sided fundamental polygons for the double torus?

UPDATE: I just found the two above listed here:

flipped and relabelled versions of the above...

REPLY [3 votes]: There are four octagon gluing patterns (up to rotation and relabelling) which give a double torus. Starting with the two that you have depicted, here they all are:

a b c d A B C D
a b A B c d C D
a b A c B d C D
a b c A B d C D

To convert these into pictures, start from one corner of the octagon and walk around it in the counterclockwise direction. When you see lower case, write that letter and a counterclockwise arrow. When you see upper case, write the lower case letter and a clockwise arrow.
What distinguishes these patterns is that all 8 corners are identified to give $V=1$ point. The 8 sides are identified in pairs to give $E=4$ edges, and of course there is $F=1$ face. The Euler characteristic formula gives
$$\chi = V-E=F = 1 - 4 + 1 = -2
$$
For an oriented surface of genus $g$ --- the double torus has $g=2$ --- one has $\chi = 2-2g$. So all together we get $-2=2-2g \implies g=2$, as expected.
Other (oriented surface) octagon gluing patterns will identify the 8 corners into either 3 or 5 points, as it turns out, with (respective) Euler characteristics $\chi = 0$ or $2$, giving the surface of genus $1$ which is the torus), or of genus $0$  which is the sphere).
Rigorous proofs of all these relations between Euler characteristic and genus can be found in any treatment of the Classification of Surfaces. The other statements here, regarding octagon gluing patterns, are proved by simply enumerating combinatorial possibilities.


TITLE: Expected determinant of a random symmetric matrix
QUESTION [7 upvotes]: The three distinct entries of a $2 \times 2$ symmetric matrix are drawn from the uniform distribution over $[-60, 60]$. What is the expected determinant of the matrix?
I assume it is $0$ but I am not good at proving it efficiently. Thanks.

REPLY [9 votes]: Let the matrix be in the form of
$$\begin{bmatrix}a & b \\ b & d \end{bmatrix}$$
The determinant is $ad-b^2$.
Assuming the $3$ entries to be independent.
\begin{align}\mathbb{E}[ad-b^2]&=\mathbb{E}[ad]-\mathbb{E}[b^2]\\&=\mathbb{E}[a]\mathbb{E}[d]-\mathbb{E}[b^2] \\
&=-(\mathbb E[b^2]-\mathbb{E}[b]^2)\\
&=-Var(b)\end{align}