TITLE: Reference request - Jacquet module and asymptotic of matrix coefficients
QUESTION [6 upvotes]: Hello,
I would like to know some nice references about the relation between asymptotics of matrix coefficients of representations of reductive groups over local fields, and the pairing between the Jacquet module of the representation and the Jacquet module of its dual.
I would like to know reference for the p-adic case, as well as for the real case (where one uses Jacquet-Casselmann functor instead of Jacquet functor).
As I understand, both cases are due to Casselmann.
Thank you,
Sasha

REPLY [1 votes]: For real groups, You  find a precise  answer to your question in the work of Hecht Henryk and Wilfried Schmid, best regards jorge


TITLE: How to see that $\Delta(X)\le nw(X)$ for any regular spce?
QUESTION [0 upvotes]: How to see that $\Delta(X)\le nw(X)$ for any regular spce?
$\Delta(X)$ and $ nw(X)$ are two  cardinal function on $X$.
Thanks for any help.

REPLY [2 votes]: $\newcommand{\cl}{\operatorname{cl}}$Let $\mathscr{N}$ be a network for $X$ of cardinality $nw(X)$. Suppose that $U$ is an open nbhd of some $x\in X$. $X$ is regular, so there is an open $V$ such that $x\in V\subseteq\cl V\subseteq U$. Then there is an $N\in\mathscr{N}$ such that $x\in N\subseteq V$, so $x\in\cl N\subseteq\cl V\subseteq U$. Thus, $\{\cl N:N\in\mathscr{N}\}$ is a network for $X$, and we may therefore assume without loss of generality that each $N\in\mathscr{N}$ is closed.
Let $$\mathscr{P}=\left\{\{M,N\}\in[\mathscr{N}]^2:M\cap N=\varnothing\right\}\;,$$
where as usual $[\mathscr{N}]^2$ is the family of two-element subsets of $\mathscr{N}$. For each $P=\{M,N\}\in\mathscr{P}$ let $\mathscr{U}(P)=\{X\setminus M,X\setminus N\}$, an open cover of $X$.
Fix $x\in X$. If $y\in X\setminus\{x\}$, there is a $P=\{M,N\}\in\mathscr{P}$ such that $x\in M$ and $y\in N$. Clearly, then, $y\notin\operatorname{st}(x,\mathscr{U}(P))=X\setminus N$. Thus,
$$\bigcap_{P\in\mathscr{P}}\operatorname{st}(x,\mathscr{U}(P))=\{x\}\;.$$
And $x$ was arbitrary, so $\Delta(X)\le|\mathscr{P}|\le|\mathscr{N}|=nw(X)$.


TITLE: Proving modular statements
QUESTION [4 upvotes]: Show that for any even number $k$,
$$2^{nk} \equiv 1 \pmod {2^n + 1} $$
I tried using induction with base case $k = 0$, but I am unsure how to proceed.

REPLY [1 votes]: Let $$2^n \equiv −1\, {(\mod2^n+1)}\implies {(2^n)}^k \equiv -1^k \pmod {2^n + 1}\implies$$
$$2^{nk} \equiv (-1)^k \pmod{2^n + 1} $$
Therefore
$$ 2^{nk} \equiv 1 \pmod {2^n + 1} \iff k := \{2x\,|\, x\in \mathbb{Z}\}$$


TITLE: Evaluating limit making it $\frac{\infty}{\infty}$ and using L'Hopital Rule
QUESTION [2 upvotes]: Let $P(x)=x^n+\displaystyle\sum\limits_{k=0}^{n-1}a_kx^k$. Find $$ \lim_{x \to +\infty} ([P(x)]^{1/n}-x) $$
I know that in order to solve this problem I need to multiply it by something that will make it $\frac{\infty}{\infty}$ and then use L'Hopital Rule. I also know that the answer should be $1/n$ if I am not mistaken. I have tried multiplying the expression by $\frac{e^x}{e^x}$ and then using L'H. Rule but with not much success. Any suggestions on how I should proceed? Thank you for the help.

REPLY [2 votes]: Here is a way of answering this question without appealing to l'Hospital's rule.  In principle, any limit involving radicals and rational functions can be resolved by algebraic manipulations.  In fact, it's the same idea that went into the answers provided by DonAntonio and Cronus.
We wish to find the limit: $$\lim_{x\to\infty} (P(x)^{1/n} - x)$$ where $P(x) = x^n + \sum_{k=0}^{n-1} a_k x^k$ is a polynomial of degree $n$.
First we remember the polynomial equation: $$(y-x)(x^{n-1}y^0 + x^{n-2}y^1 + \cdots + x^0y^{n-1}) = y^n - x^n$$
Now take $y = P(x)^{1/n}$.  This enables us to rewrite the limit as:
$$\begin{array}{rcl}\lim_{x\to\infty} (P(x)^{1/n} - x) &=& \lim_{x\to\infty} \frac{P(x) - x^n}{(x^{n-1}(P(x)^{1/n})^0 + x^{n-2}(P(x)^{1/n})^1 + \cdots + x^0(P(x)^{1/n})^{n-1})}\\
&=& \lim_{x\to\infty} \frac{\sum_{k=0}^{n-1} a_k x^k}{(x^{n-1}(P(x)^{1/n})^0 + x^{n-2}(P(x)^{1/n})^1 + \cdots + x^0(P(x)^{1/n})^{n-1})}\end{array}$$
Now the key observation is that we can view every term in the denominator as being $x^{n-1}$ in a limiting sense:
$$\begin{array}{rcl}x^{n-k}(P(x)^{1/n})^{k-1}&=&x^{n-k}((x^n + a_{n-1} x^{n-1} + \cdots)^{1/n})^{k-1}\\
&=&x^{n-k}(x (1 + a_{n-1} x^{-1} \cdots )^{1/n})^{k-1}\\
&=&x^{n-1}(1 + a_{n-1} x^{-1} \cdots )^{(k-1)/n}\end{array}$$
Note that $\lim_{x\to\infty} (1 + a_{n-1} x^{-1} \cdots )^{(k-1)/n} = 1$.  For convenience, note that we can write these terms as: $$x^{n-1}(P(x)/x^n)^{(k-1)/n}$$
Now let us return the the limit we are concerned with:
$$\begin{array}{rcl}\lim_{x\to\infty} (P(x)^{1/n} - x) &=& \lim_{x\to\infty} \frac{\sum_{k=0}^{n-1} a_k x^k}{(x^{n-1}(P(x)^{1/n})^0 + x^{n-2}(P(x)^{1/n})^1 + \cdots + x^0(P(x)^{1/n})^{n-1})}\\
&=&\lim_{x\to\infty} \frac{\sum_{k=0}^{n-1} a_k x^k}{x^{n-1}(((P(x)/x^n)^{1/n})^0 + ((P(x)/x^n)^{1/n})^1 + \cdots + ((P(x)/x^n)^{1/n})^{n-1})}\\
&=&\lim_{x\to\infty} \frac{a_{n-1} + a_{n-2} x^{-1} + \cdots + a_0 x^{-(n-1)}}{((P(x)/x^n)^{1/n})^0 + ((P(x)/x^n)^{1/n})^1 + \cdots + ((P(x)/x^n)^{1/n})^{n-1}}\end{array}$$
Finally notice this is simply a combination of limits we already know.  The numerator tends to $a_{n-1}$ as $x \to \infty$, and each of the terms in the denominator (of which there are $n$ terms) tends to 1 as we already stated.
Thus,
$$\lim_{x\to\infty} (P(x)^{1/n} - x) = \frac{a_{n-1}}{n}.$$


TITLE: Understanding part of a proof for Chebyshev inequality
QUESTION [1 upvotes]: I'm trying to wrap my head around this proof I'm reading for the discrete case of the Chebyshev inequality:
$$P(|X-EX|<\epsilon)\ge 1-\frac{\operatorname{Var}(X)}{\epsilon^2}$$
Proof:
Without loss of generality we assume that there exists $\epsilon > 0$ for which the first $k$ values of the random variable $|X-EX|$ are less than that $\epsilon$, and the rest are not less than that $\epsilon$. Then
$$P(|X-EX|<\epsilon)=1-\sum_{i=k+1}^\infty p_i,$$
where $p_i=P(X=x_i)$. To find $\sum_{i=k+1}^\infty p_i$ lets look at the variance of the random variable $|X-EX|$(I think there might be a mistake here since in the next line we are looking at $Var(X)$):
$$\operatorname{Var}(X)=\sum_{i=1}^k p_i(x_i-EX)^2+\sum_{i=k+1}^\infty p_i(x_i-EX)^2 \ge \sum_{i=k+1}^\infty p_i(x_i-EX)^2,$$
from where
$$\sum_{i=k+1}^\infty p_i \ge \frac{\operatorname{Var}(X)}{\epsilon^2}.$$
Finaly we get, that
$$P(|X-EX| < \epsilon) \ge 1-\frac{Var(X)}{\epsilon^2}$$
I don't get that last part, i.e. why $\sum_{i=k+1}^\infty p_i \ge \frac{\operatorname{Var}(X)}{\epsilon^2}$?

REPLY [0 votes]: Since every value of $(x_i-EX)^2$ is greater than $\epsilon^2$ we have
$$\operatorname{Var}(X)\ge \sum_{i=k+1}^\infty p_i(x_i-EX)^2 \ge \sum_{i=k+1}^\infty p_i\epsilon^2.$$
From there by dividing by $\epsilon^2$ we get $\sum_{i=k+1}^\infty p_i \le \frac{Var(X)}{\epsilon^2}$ and not the other way around.
From $\sum_{i=k+1}^\infty p_i \le \frac{\operatorname{Var}(X)}{\epsilon^2}$ it follows that $1- \sum_{i=k+1}^\infty p_i \ge 1-\frac{\operatorname{Var}(X)}{\epsilon^2}$ and from there we get
$P(|X-EX|<\epsilon) \ge 1-\frac{\operatorname{Var}(X)}{\epsilon^2}$.


TITLE: Undecidability in Conway's Game of Life
QUESTION [11 upvotes]: I strongly believe that - given the rules of Conway's Game of Life and an initial configuration - it is not decidable by a Turing Machine whether a given pattern will emerge, let alone as a stable pattern, be it static, moving, and/or rotating.

How can this be proven?

I guess, this kind of uncomputability would go far beyond the "simple" unpredictability of non-linear systems.

REPLY [5 votes]: This is proved (or, really, a sketch of the proof is given) in the second volume of the extraordinary [Elwyn R. Berlekamp, John H. Conway, and Richard K. Guy, Winning Ways for your Mathematical Plays. Academic Press, 1982]


TITLE: Is there a countable cover of $\mathbb{R}^2$ by balls $B(x_n, n^{-1/2})$?
QUESTION [15 upvotes]: Is it possible to cover all of $\mathbb{R}^2$ using balls $\{ B(x_n,n^{-1/2})\}_{n=1}^\infty$ of decreasing radius $n^{-1/2}$? I know that if we chose e.g. radius $n^{-1}$ it could never work because $\sum \pi (n^{-1})^2 < \infty$. But in this case the balls cover an infinite amount of area, so it seems that it may be possible to construct.
Edit: It can definitely be done with balls of radius $n^{-1/2+\epsilon}$ for any $\epsilon > 0$.

REPLY [1 votes]: It's true for any $r_n$ with $\sum r_n^2=\infty$.
Say $S(x,r)$ is the square with center $x$ and side length $r$.
It's enough to show that if $\sum r_n^2>4$ then there exist $x_n$ such
that $S(x_n,r_n)$ covers $[0,1]^2$.
For every $n$ there exists $j$ with $2^{-j}\le r_n<2^{-j+1}$. Replace $r_n$ with $2^{-j}$:
It's enough to show that if each $r_n$ is of the form $2^{-j}$ and $\sum r_n>1$ then there exist $x_n$ such that $S(x_n,r_n)$ cover the unit square.
A dyadic interval is an interval $[j2^{-n},(j+1)2^{-n}]$. A dyadic square is the product of two dyadic intervals with the same side length.
Start: If there exists $n$ with $r_n\ge1$ we're done. Suppose not. Divide $[0,1]^2$ into four dyadic squares of side length $1/2$. Call them $Q$'s.
If there are four $n$ such that $r_n=1/2$ we're done. Suppose not. Cover as many $Q$'s as possible with $S(x_n,r_n)$ with $r_n=1/2$.
Discard the $r_n$'s we just used. Note we used all the $r_n$ that equal $1/2$. Discard the $Q$'s we just covered. Divide each remaining $Q$ into four subsquares of side length $1/4$.
If there are enough $n$ with $r_n=1/4$ to cover all the reminaing $Q$'s we're done. Suppose not. Cover as many $Q$'s as possible with $S(x_n,r_n)$ where $r_n=1/4$. Note we've used all the $r_n$ that equal $1/4$. Discard those $r_n$, discard the $Q$'s we just covered. Divide each remaining $Q$ into four suubsquares of side length $1/8$.
Etc.
If at some point we're done then we're done. Suppose not. Then we've used up all the $r_n$ covering essentially disjoint subsquares of $[0,1]^2$; hence $\sum r_n^2\le 1$.


TITLE: Is the Apophis asteroid a concern?
QUESTION [7 upvotes]: I was watching the TED lecture How to defend Earth from asteroids , and the speaker Phil Plait spoke of the possibility of this asteroid hitting Earth. He also said something about a Mayan legend. He says it's not a concern, is there a consensus on that? And what is the source for this Mayan legend?

REPLY [13 votes]: Apophis is of no real concern to us as far as we can tell.  That said, it doesn't mean another asteroid doesn't have our name on it...
According to the most reliable data we have regarding this particular asteroid it only has a 1 chance in about 250,000 of hitting the earth.  University of Hawaii states:

“Our new orbit solution shows that Apophis will miss Earth’s surface in 2036 by a scant 20,270 miles, give or take 125 miles,” Tholen said. “That's slightly closer to Earth than most of our communications and weather satellites.” He credits the large telescopes and superb atmospheric conditions on Mauna Kea for being able to make these determinations.

Considering how big space is, this is pretty much a hit in cosmic terms.  Keep in mind that this information has been readily available since 2009 at the NASA website as well.
As for the bat guano crazy 2012 stuff...  Suffice it to say that anyone that honestly buys into that is mentally feeble, and should seek psychiatric help...


TITLE: Finite self-maps exist on rigid CY3s
QUESTION [3 upvotes]: Let $X$ be a smooth projective rigid Calabi-Yau threefold.

Question. Does there exist a finite map $X\to X$ of degree $>1$?

REPLY [3 votes]: One can show that such a map in the question doesn't exist (no need to assume simply-connectedness).
As abx pointed out, any finite map between smooth projective varieties with trivial canonical bundle must be étale.
Let $\chi(X)$ be the topological Euler characteristic of $X$ and $d$ be the degree of the map. Since $X$ is rigid,
$$\chi(X) = 2 \left( h^{1,1}(X)- h^{1,2}(X) \right)  =2  h^{1,1}(X) > 0.$$
Noting $d \cdot \chi(X) = \chi(X)$, one has $d=1$.


text:
\section{Star shape of the rotation set}\label{s.agujas}

We show that for $f\in \mathcal{A}_0(\Sigma)$ the set $\rho(f)\subset H_1(\Sigma_g;\mathbb{R})$ is a union of convex sets all containing the origin.

The proof will be based on our characterization of the rotation set as $$\rho(f) = \bigcup_{\mathcal{C}}\rho_{\mathcal{C}}(f)$$ where $\mathcal{C}$ ranges over the maximal chains of basic pieces of $f$, given by Theorem \ref{t.rhorhochain}. This is a union of finitely many convex sets, but there may be chains $\mathcal{C}$ for which $\rho_{\mathcal{C}}(f)$ does not contain $0$. We shall show that the rotation sets of such chains are {\em radial} segments (i.e. contained in a $1$-dimensional subspace), and that $\rho(f)$ contains the continuation to $0$ of these segments.

\begin{lemma}\label{l.agujas}
Let $f\in \mathcal{A}_0(\Sigma)$ and assume $\Lambda_1\prec \Lambda_2$ are basic pieces such that there is no $\Lambda$ with $\Lambda_1\prec \Lambda \prec \Lambda_2$. Then one of the following holds:
\begin{itemize}
\item[a)] $\Lambda_1\cup\Lambda_2$ contains a zero-rotation periodic point of $f$, or
\item[b)] there is a closed curve $\alpha$ such that $\rho_{\Lambda_i}(f) \subset \langle [\alpha]\rangle $ for $i=1,2$.
\end{itemize}
\end{lemma}

We remark that $\alpha$ in statement (b) might be inessential, in which case $\rho_{\Lambda_i}(f) = \{0\}$.

\begin{proof}
Take $k\in\N$ so that the basic pieces of $f^k$ are topologically mixing. By Remark \ref{r.piecespower}, there exist $\Lambda_1'\subseteq\Lambda_1$ and $\Lambda_2'\subseteq\Lambda_2$ basic pieces of $f^k$ satisfying the same hypothesis on the heteroclinical relation. Notice that it is enough to prove the lemma for $f^k$, $\Lambda'_1$ and $\Lambda'_2$. Thus we shall assume that $\Lambda_1$ and $\Lambda_2$ are topologically mixing for $f$.

Recall that by Proposition \ref{p.propfaxa}, the set $\Lambda^u  =\textrm{cl}\left[\bigcup_{x\in\Lambda}W^u(x,f)\right]$ is an attractor for any basic piece $\Lambda$ of $f$. So we can find a subsurface $U$ containing $\Lambda_1^u$ such that
\begin{itemize}
\item $\textrm{cl}[f(U)]\subseteq U$
\item $\Omega(f)\cap U= \Lambda_1^u\cap\Omega(f)=\cup_{\Lambda\succeq\Lambda_1}\Lambda$
\end{itemize}

Moreover, we can assume that $U$ is W-filled and that all its connected components intersect $\Omega(f)$.
Define $$\mathcal{F}=\{\Lambda:\Lambda_1\prec\Lambda,\text{ }\Lambda\neq\Lambda_2\}.$$
The basic pieces of $f$ that are contained in $U$ are exactly $\Lambda_1$, $\Lambda_2$ and those in $\mathcal{F}$. On the other hand, the pieces $\Lambda\in\mathcal{F}$ do not precede $\Lambda_1$ nor $\Lambda_2$, so the closed set $$\bigcup_{\Lambda\in\mathcal{F}}\Lambda^u  $$ does not meet $\Lambda_1\cup\Lambda_2$. We shall construct a curve $K$ joining $\Lambda_1$ and $\Lambda_2$ within $U$ that does not meet $\cup_{\Lambda\in\mathcal{F}}\Lambda^u $.

Let $V_1$ and $V_2$ be neighbourhoods of $\Lambda_1$ and $\Lambda_2$ respectively, that are contained in $U$ and disjoint from $\cup_{\Lambda\in\mathcal{F}}\Lambda^u $.  Since $\Lambda_1\prec\Lambda_2$ there are $x\in\Lambda_1$ and $y\in \Lambda_2$ such that $W^u(x,f)$ meets $W^s(y,f)$ at some point $z$. Let $\beta$ be the arc of $W^u(x,f)$ from $x$ to $z$ and $\gamma$ the arc of $W^s(y,f)$ from $z$ to $y$. We can assume that $\gamma\subset V_2$, otherwise we apply a high enough power of $f$ and redefine $x$, $y$ and $z$. This yields that $K:=\beta\cup\gamma$ is contained in $U$, since $\beta\subset \Lambda_1^u \subset U$. Next we prove that $K$ is disjoint from $\cup_{\Lambda\in\mathcal{F}}\Lambda^u$. That is clear for $\gamma$, since it lies in $V_2$. For each $p\in \beta$ there is some neighbourhood $V_p$ and some $k>0$ with $f^{-k}(V_p)\subset V_1$.  Now if $\Lambda^u$ met $\beta$ for some $\Lambda\in\mathcal{F}$, we consider some point of intersection $p$ and see that $W^u(\Lambda,f)$ must meet $V_p$. Since $W^u(\Lambda,f)$ is $f^{-1}$ invariant, this would mean that $W^u(\Lambda,f)$ meets $V_1$, which contradicts the choice of $V_1$.

We apply Lemma \ref{l.construquasi} to the subsurface $U$, the basic pieces in $\mathcal{F}$, and the continuum $K$. This gives a quasi-invariant subsurface $S$ such that:
\begin{itemize}
\item $S\cap\Omega(f)=\Lambda_1\cup\Lambda_2$
\item $K\subset S$
\end{itemize}

Let $S_0$ be the connected component of $S$ that contains $K$, which is also quasi-invariant by Remark \ref{r.quasiconexas}. By construction of $K$ we get that both $\Lambda_1$ and $\Lambda_2$ meet $S_0$. Therefore Lemma \ref{l.qisurface} item (1) lets us deduce that $\Lambda_1\cup\Lambda_2$ is contained in $S_0$. If $S_0$ is contained in a disk, we can apply Lemma \ref{l.qisurface} point (3) to show that $\rho_{\Lambda_1}(f) = \rho_{\Lambda_2}(f)$ and this is a point $v\in \langle[\alpha]\rangle$ where $\alpha$ is some closed curve in $\Sigma$. Thus statement (b) holds.

If $S_0$ is contained in an essential annulus $A$ (but not in a disk) we take an essential simple closed curve $\alpha$ with $\langle[\alpha]\rangle = H_1(A) = H_1(S_0)$. From Lemma \ref{l.qisurface} point (4) we get statement (b) for this $\alpha$.

In the remaining case, $S_0$ is not contained in a disk or an annulus , so Lemma \ref{l.filling} point (1) gives statement (a).
\end{proof}

We obtained a little more from the proof of Lemma \ref{l.agujas}, that we state below as a remark.

\begin{rem} \label{r.agujas} If statement (b) in Lemma \ref{l.agujas} holds with a curve $\alpha$ that is not essential or not simple, then we can say more: $\rho_{\Lambda_1}(f)=\rho_{\Lambda_2}(f)=\{v\}$ for $v\in \langle [\alpha]\rangle$.

\end{rem}

\begin{lemma} \label{l.cadenas}
Let $f\in \mathcal{A}_0(\Sigma)$ and $\mathcal{C}$ be a maximal chain of basic pieces of $f$. Then exactly one of the following holds.
\begin{enumerate}
\item[A)] $0\in\rho_{\mathcal{C}}(f)$.
\item[B)] There is an essential closed curve $\alpha$ in $\Sigma$ with $\rho_{\mathcal{C}}(f)\subset \langle[\alpha]\rangle - \{0\}$.
\end{enumerate}

\end{lemma}

We observe that in case (B) the curve $\alpha$ must be not only essential, but non-trivial in homology.

\begin{proof}

Write $\mathcal{C} = \{\Lambda_1 \prec \cdots \prec \Lambda_k \}$. If some $\Lambda_i$ has $0\in\rho_{\Lambda_i}(f)$ then $\mathcal{C}$ satisfies (A). We assume that is not the case, in particular that no $\Lambda_i$ contains a zero-rotation periodic point of $f$. By the maximality of $\mathcal{C}$ we can apply Lemma \ref{l.agujas} to $\Lambda_i\prec\Lambda_{i+1}$ for $i=1,\ldots,k-1$, to conclude that there are closed curves $\alpha_i$ in $\Sigma$ such that $\rho_{\Lambda_i}(f)$ and $\rho_{\Lambda_{i+1}}(f)$ are contained in $\langle [\alpha_i]\rangle-\{0\}$. Under our assumptions it is clear that $[\alpha_i]\neq 0$, thus $\alpha_i$ are essential and $\langle [\alpha_i]\rangle$ are one dimensional. For $i=1,\ldots,k-2$ we have that $\left(\langle [\alpha_i]\rangle \cap \langle [\alpha_{i+1}]\rangle\right) -\{0\}$ contains $\rho_{\Lambda_{i+1}}(f)$, thus it is non empty. So the subspaces $\langle [\alpha_i]\rangle$ are all the same, and we can choose $\alpha_1=\cdots=\alpha_{k-1}$ and call this curve $\alpha$. Then we have  $\rho_{\mathcal{C}}(f)\subset \langle[\alpha]\rangle$. Depending on whether $\rho_{\mathcal{C}}(f)$ contains $0$ or not we are in case (A) or (B).

\end{proof}

\begin{rem} If $\mathcal{C}$ is in case (B) of Lemma \ref{l.cadenas} with a curve $\alpha$ that is not simple, then $\rho_{\mathcal{C}}(f)$ is a single point. This follows from the proof of Lemma \ref{l.cadenas} and Remark \ref{r.agujas}.
\end{rem}

\begin{theorem} \label{t.starshape}
If $f\in\mathcal{A}_0(\Sigma_g)$ then $\rho(f)$ is star-shaped about $0$.
\end{theorem}

\begin{proof}
By Theorem \ref{t.rhorhochain} we write $\rho(f) = \bigcup_{\mathcal{C}}\rho_{\mathcal{C}}(f)$ where $\mathcal{C}$ ranges over the maximal chains of basic pieces of $f$. We take an arbitrary $v\in\rho(f)$, aiming to prove that $[0,v]\subset \rho(f)$, and let $\mathcal{C}$ be a maximal chain so that $v\in\rho_{\mathcal{C}}(f)$. By Lefschetz's theorem $f$ has a zero-rotation fixed point, so we can find a maximal chain $\mathcal{C}'$ that has a basic piece containing said zero-rotation fixed point. In particular $\rho_{\mathcal{C}'}(f)$ is a convex set containing $0$.

Recall that the equivalence relation generated by $\prec$ on the set of basic pieces of $f$ has only one class ($\mathcal{G}_f$ is connected). So we can find maximal chains $\mathcal{C}=\mathcal{C}_1,\ldots,\mathcal{C}_k=\mathcal{C}'$ so that $\mathcal{C}_i$ intersects $\mathcal{C}_{i+1}$ in at least one basic piece for all $i=1,\ldots,k-1$. We may assume that for $i<k$ the $\mathcal{C}_i$ are in case (B) of Lemma \ref{l.cadenas}, otherwise we redefine $\mathcal{C}'$ to be the first $\mathcal{C}_i$ in case (A).

If $k=1$ then $\rho_{\mathcal{C}}(f)$ is a convex set containing $0$ and $v$, so we are done. Otherwise, for $i<k$ we have that $\rho_{\mathcal{C}_i}(f)\subset\langle[\alpha_i]\rangle-\{0\}$ for some essential closed curve $\alpha_i$. Since $\mathcal{C}_i$ and $\mathcal{C}_{i+1}$ have a basic piece in common, $\rho_{\mathcal{C}_i}(f)$ intersects $\rho_{\mathcal{C}_{i+1}}(f)$ for all $i=1,\ldots,k-1$. This implies that the $\langle[\alpha_i]\rangle$ for $i<k$ are all the same subspace of dimension $1$. The sets $\rho_{\mathcal{C}_i}(f)$ for $i<k$ are segments in the line $\langle[\alpha_1]\rangle$, each intersecting the next, thus $I=\bigcup_{i=1}^{k-1} \rho_{\mathcal{C}_i}(f)$ is a segment in $\langle[\alpha_1]\rangle$. Now $\rho_{\mathcal{C}_k}(f)$ is a convex set that contains $0$ and intersects $I$, thus $\bigcup_{i=1}^{k} \rho_{\mathcal{C}_i}(f) = I\cup \rho_{\mathcal{C}_k}(f) $ contains $\text{conv}(I\cup\{0\})\subset \langle[\alpha_1]\rangle$. In particular $\bigcup_{i=1}^{k} \rho_{\mathcal{C}_i}(f)$ contains $[0,v]$.
\end{proof}

\section{Bounding the complexity of the rotation set: trivial pieces}\label{s.triviales}

Given $f\in \mathcal{A}_0(\Sigma)$ we consider a Conley decomposition of $\Sigma$ as in section \ref{s.conley}, and for each basic piece $\Lambda$ of $f$ we denote by $S_\Lambda$ the Conley surface associated to $\Lambda$ in this decomposition. Recall that a basic piece $\Lambda$ is called trivial if every component of $S_{\Lambda}$ is trivial. Equivalently, if $\text{Fill}(S_\Lambda)$ is a disjoint union of finitely many disks.

The aim of this section is to prove the following result, that is the first step towards achieving a topological bound for the number of convex sets needed to express $\rho(f)$ for $f\in\mathcal{A}_0(\Sigma)$.

\begin{proposition} \label{sacar.triviales} Let $\mathcal{C}$ be a maximal chain for $f\in\mathcal{A}_0(\Sigma)$. Then there is a chain $\mathcal{C}^*$ that contains no trivial pieces such that $\rho_{\mathcal{C}}(f)\subset \rho_{\mathcal{C}^*}(f)$.
\end{proposition}

Proposition \ref{sacar.triviales} will allow us to write $$\rho(f) = \bigcup_{\mathcal{C}}\rho_{\mathcal{C}}(f)$$ with $\mathcal{C}$ ranging over the chains of basic pieces of $f$ that contain no trivial pieces. Of course, we can restrict this union by asking for $\mathcal{C}$ to be maximal among the chains with no trivial pieces, as a way of avoiding the obvious repetition. In short, this says we can forget about trivial pieces when computing $\rho(f)$.

We recall that if $\Lambda$ is a trivial piece, then $\mathcal{D} = \text{Fill}(S_\Lambda) $ is a disjoint union of disks satisfying that $f(\mathcal{D})$  either contains or is contained in $\mathcal{D}$. Moreover, if we write $\mathcal{D}=D_1\cup\cdots\cup D_k$ as a disjoint union of disks, then $f(D_j)$ either contains or is contained in $D_{\sigma(j)}$, for $\sigma$ a cyclic permutation of $\{1,\ldots,k\}$.

\begin{definition} Let $\mathcal{C}$ be a chain of basic pieces of $f\in\mathcal{A}_0(\Sigma)$. We say that $\mathcal{C}$ is {\em completely trivial} if $$\mathcal{D}:=\text{Fill}\left(\bigcup_{\Lambda\in\mathcal{C}}S_\Lambda\right)$$ is a disjoint union of finitely many disks, so that $f(\mathcal{D})$ either contains or is contained in $\mathcal{D}$.
\end{definition}
If $\mathcal{C}$ is a completely trivial chain, then
every basic piece of $\mathcal{C}$ meets every connected component of $\mathcal{D}$:
Otherwise, we can decompose $\mathcal{D}$ as a union of two disjoint non-empty open sets $U$, $V$ that are both either forward or backward invariant (according to whether $\mathcal{D}$ is forward or backward invariant). Then each basic piece in $\mathcal{C}$ is contained in either $U$ or $V$. On the other hand, by definition every connected component of $\mathcal{D}$ meets some basic piece of $\mathcal{C}$. But this is absurd, for basic pieces contained in $U$ cannot precede basic pieces in $V$ and the other way around, which contradicts the existence of the chain $\mathcal{C}$.

Thus Lemma \ref{l.qisurface} applied to a suitable power of $f$ gives that
$\rho_{\mathcal{C}}(f)$ is a single point whenever $\mathcal{C}$ is completely trivial.
This fact will be relevant in what follows.

\begin{lemma} \label{l.ctrivial} Let $\mathcal{C}$ be a completely trivial chain of $f\in\mathcal{A}_0(\Sigma)$. Then there exists $\Lambda$ a non-trivial basic piece with $\rho_{\mathcal{C}}(f)\subset \rho_{\Lambda}(f)$.

\end{lemma}

\begin{proof} Let $v\in H_1(\Sigma)$ be the point with $\rho_{\mathcal{C}}(f)=\{v\}$. Denote $\mathcal{D}=\text{Fill}(\bigcup_{\Lambda\in\mathcal{C}}S_\Lambda)$, which by hypothesis is a disjoint union of disks. Recalling Remark \ref{r.discos} for all the non-trivial surfaces in the Conley decomposition, we see that there exists some non-trivial Conley surface $S$ so that $\mathcal{D}\subseteq \text{cl}[\text{Fill}(S)\setminus S]$. Let $\Lambda$ be the (non-trivial) basic piece associated to $S$. By item (1) of Lemma \ref{l.conleyesquasi} there is  $k\in\N$ so that $S$ is quasi-invariant under $f^k$. Let $D_0$ be a connected component of $\text{cl}[\text{Fill}(S)\setminus S]$ that meets $\mathcal{D}$.
Applying Lemma \ref{l.filling} we get $x_0\in S$ a fixed point of $f^k$ so that $\rho(f^k,x_0)$ equals $\rho_{\Lambda'}(f^k)$ for every basic piece $\Lambda'$ of $f^k$ that is contained in $D_0$. Observe that for every such piece $\Lambda'$ we have $\rho_{\Lambda'}(f^k)=\{k\cdot v\}$, thus $\rho(f^k,x_0)=k\cdot v$, and also note that $x_0\in S\cap\Omega(f)=\Lambda$. Therefore $x_0\in \Lambda$ is a periodic point of $f$ whose rotation set equals $\rho_\mathcal{C}(f)$, so $\Lambda$ is the basic piece that satisfies the statement.

\end{proof}

Clearly Lemma \ref{l.ctrivial} proves the simplest case of Proposition \ref{sacar.triviales}, that is when the maximal chain $\mathcal{C}$ is completely trivial. Next we shall study the structure of a chain in terms of the trivial pieces it contains.

\begin{lemma} \label{l.trivialsegment} Let $\mathcal{C} =\{ \Lambda_0\prec \cdots \prec \Lambda_k \}$ be a chain of $f\in\mathcal{A}_0(\Sigma)$. If $\Lambda_j$ is a trivial piece then either the initial segment $\Lambda_0\prec \cdots \prec \Lambda_j$ or the final segment $\Lambda_j\prec \cdots \prec \Lambda_k$ is completely trivial (or both).
\end{lemma}

\begin{proof}
Let $\mathcal{D}=\text{Fill}(S_{\Lambda_j})$, which is a union of disks since $\Lambda_j$ is a trivial piece. If $f(\mathcal{D})\subset \mathcal{D}$ then $\Lambda_j^u$ is contained in $\mathcal{D}$. Recalling that $\Lambda_j^u$ contains every piece that follows $\Lambda_j$, we obtain that the final segment of $\mathcal{C}$ starting at $\Lambda_j$ is completely trivial. On the other hand, if $\mathcal{D}\subset f(\mathcal{D})$ we can give the same argument with $\Lambda_j^s$, and obtain that the initial segment is completely trivial.

\end{proof}

Lemma \ref{l.trivialsegment} allows us to write any chain $\mathcal{C}$ of $f\in\mathcal{A}_0(\Sigma)$ as a union of three segments, some of them possibly empty, in the following way: First we have an initial segment $\mathcal{C}_0$ that is completely trivial, all of its pieces contained in a disjoint union of disks $\mathcal{D}_0$ with $f(\mathcal{D}_0) \supset \mathcal{D}_0$. Next we have the middle segment $\mathcal{C}_{NT}$ containing the non-trivial pieces, and finally the completely trivial segment $\mathcal{C}_1$ with its pieces contained in a disjoint union of disks $\mathcal{D}_1$ such that $f(\mathcal{D}_1)\subset \mathcal{D}_1$.

When $\mathcal{C}$ contains non-trivial pieces this decomposition is unique. On the other hand, $\mathcal{C}$ is completely trivial if it can be written as either $\mathcal{C}=\mathcal{C}_0$ or $\mathcal{C}=\mathcal{C}_1$. We remark that a chain might contain only trivial pieces without being completely trivial. The next step towards Proposition \ref{sacar.triviales} is to exclude that case.

\begin{lemma} \label{l.nctrivial} Let $\mathcal{C}$ be a chain of $f\in\mathcal{A}_0(\Sigma)$. Then there is a chain $\mathcal{C}'$ of $f$ that contains a non-trivial piece and such that $\rho_{\mathcal{C}}(f)\subset \rho_{\mathcal{C}'}(f)$.

\end{lemma}

\begin{proof} By Lemma \ref{l.ctrivial} the only case we need to consider is when $\mathcal{C}$ has trivial pieces only, but is not completely trivial. We take a decomposition $\mathcal{C}=\mathcal{C}_0\cup\mathcal{C}_1$ as in the discussion above. Since $\mathcal{C}_0$ and $\mathcal{C}_1$ are completely trivial chains, there must exist $\mathcal{D}_0$ and $\mathcal{D}_1$ disjoint unions of disks such that:
\begin{itemize}
\item $\mathcal{D}_j$ contains all the basic pieces of $\mathcal{C}_j$ for $j=0,1$,
\item There exist non-trivial Conley surfaces $S_j$ such that $\mathcal{D}_j$ consists of a disjoint union of connected components of $\text{cl}[\text{Fill}(S_j)\setminus S_j]$ for $j=0,1$,
\item $f(\mathcal{D}_1)\subseteq \mathcal{D}_1$ and $f(\mathcal{D}_0)\supseteq \mathcal{D}_0$.
\end{itemize}
Notice that by this construction, the trivial basic pieces that are not contained in $\mathcal{D}_0\cup \mathcal{D}_1$ have associated Conley surfaces that are disjoint from $\mathcal{D}_0\cup \mathcal{D}_1$. This is the key for the following claim:

\vspace{.1cm}
{\bf Claim:} { If we have a chain $\Lambda_0\prec\Lambda\prec\Lambda_1$ with $\Lambda_j\subset \mathcal{D}_j$ for $j=0,1$ and $\Lambda$ disjoint from $\mathcal{D}_0\cup \mathcal{D}_1$, then $\Lambda$ must be non-trivial.}

\vspace{.1cm}
To prove the claim by contradiction, assume that $\Lambda$ is trivial. Then it is contained in a disjoint union of disks $\mathcal{D}$ that is disjoint from $\mathcal{D}_0\cup \mathcal{D}_1$. But this is absurd: If $\mathcal{D}$ is attracting, every piece that follows $\Lambda$ is contained in $\mathcal{D}$ and we cannot have $\Lambda\prec\Lambda_1\subset \mathcal{D}_1$. If $\mathcal{D}$ is repelling, we argue similarly with $\Lambda_0$.

\vspace{.1cm}
Taking a suitable power of $f$ and applying Lemma \ref{l.qisurface} point (3) to it, we see that there exist $v_j\in H_1(\Sigma)$ such that $\rho_{\Lambda_j}(f)=\{v_j\}$ for every piece $\Lambda_j\subset \mathcal{D}_j$ and $j=0,1$. In particular $\rho_{\mathcal{C}}(f)=\text{conv}(v_0,v_1)$. Together with the Claim, this fact means we can prove the lemma by finding a chain of the form $$\mathcal{C}'=\{\Lambda_0\prec\Lambda\prec\Lambda_1\}$$ where $\Lambda_j\subset \mathcal{D}_j$ for $j=0,1$, and $\Lambda$ is disjoint from $\mathcal{D}_0\cup \mathcal{D}_1$.

Define $U_0 := \mathcal{D}_1\cup\bigcup_{\Lambda_0\subset \mathcal{D}_0} \Lambda_0^u$. Since it is an attractor, taking a neighbourhood of it we an find a subsurface $U$ such that: \begin{itemize}
\item $\text{cl}[f(U)]\subseteq\text{int}(U)$
\item $U\cap \Omega(f)=U_0\cap\Omega(f)$, which is the union of the following set of basic pieces: $$\{\Lambda : \Lambda\subset\mathcal{D}_1 \}\cup\{\Lambda:
\Lambda\succeq\Lambda_0\mbox{ for some }\Lambda_0\subset\mathcal{D}_0\}$$
\end{itemize}
Moreover, we can assume that every connected component of $U$ meets $\Omega(f)$ and that $U$ is W-filled. Notice that this surface contains $\mathcal{D}_0$, since every $x\in \mathcal{D}_0$ has $\alpha$-limit in $\mathcal{D}_0$ and therefore $x$ must belong to $\Lambda_0^u$ for some basic piece $\Lambda_0\subset \mathcal{D}_0$. The existence of the chain $\mathcal{C}$ implies that we can find a basic piece $\Lambda'\subseteq \mathcal{D}_0$, an arc $\gamma$ contained in $W^u(\Lambda',f)$ and disks $D'_0$, $D'_1$ which are connected components of $\mathcal{D}_0$ and $\mathcal{D}_1$ respectively, so that $\gamma$ joins $D'_0$ with $D'_1$. Denote by $K$ the union of $\gamma$ and these disks, i.e. $K:=\gamma\cup D'_0\cup D'_1$, and notice that $K$ is contained in $U$.

Let $\mathcal{F}$ be the family of basic pieces $\Lambda$ that are not contained in $\mathcal{D}_0$, that follow some piece in $\mathcal{D}_0$, but precede no piece in $\mathcal{D}_1$. In other words, recalling the definition of $U$ we can write $$\mathcal{F} = \{\Lambda: \Lambda\subset U\setminus\mathcal{D}_0,  \Lambda^u\cap \mathcal{D}_1 = \emptyset \}.$$ If $\Lambda\in\mathcal{F}$ then $\Lambda^u$ must be disjoint from $\mathcal{D}_0\cup\mathcal{D}_1$, and arguing as in Lemma \ref{l.agujas} we can show that $\Lambda^u$ is disjoint from $\gamma$ as well. Therefore $K$ does not meet $\bigcup_{\Lambda\in \mathcal{F}}\Lambda^u$.

Consider $k\in\N$ so that the basic pieces of $f^k$ are topologically mixing, and recall Remark \ref{r.piecespower}, stating that each basic piece of $f$ decomposes as a finite and disjoint union of basic pieces of $f^k$. We are in conditions to apply Lemma \ref{l.construquasi} to $f^k$, with the subsurface $U$, the family of basic pieces of $f^k$ that arises from $\mathcal{F}$, and the continuum $K$.  Thus we obtain a subsurface $S$ with the following properties:
\begin{itemize}
\item $S$ is quasi-invariant under $f^k$,
\item $S\cap\Omega(f^k)=(\Omega(f)\cap U)\setminus \bigcup_{\Lambda\in\mathcal{F}}\Lambda^u$
\item $K\subset S$
\end{itemize}

Denote by $S'$ the connected component of $S$ containing $K$, which is quasi-invariant under $f^k$ by Remark \ref{r.quasiconexas}. We consider  $\mathcal{D}''_0:=f^{-k}(\mathcal{D}_0)\subset \textrm{int}(\mathcal{D}_0)$ and $\mathcal{D}''_1:=f^{k}(\mathcal{D}_1)\subset \textrm{int}(\mathcal{D}_1)$, whose components are either contained in $\textrm{int}(S')$ or disjoint from $S'$. This allows us to ensure that
 $$S'' = \textrm{cl}[S'-(\mathcal{D}''_0\cup \mathcal{D}''_1)]$$  is a subsurface, that is quasi-invariant for $f^k$. Also notice that $D''_0:=f^{-k}(D'_0)$ and $D''_1:=f^{k}(D'_1)$ are components of $\mathcal{D}''_0$ and $\mathcal{D}''_1$ that do meet $S'$, thus give $S''$ two boundary components. Then $S''$ cannot be an annulus, otherwise $\Sigma$ would be $S^2$, so $f^k$ must  have some fixed point in $S''$ by Remark \ref{r.filling}. This fixed point of $f^k$ must belong to some basic piece $\Lambda$ of $f$. We see  that $\Lambda\subset S$, since $\Omega(f)\cap S$ is $f$-invariant, and that $\Lambda$ must be disjoint from $\mathcal{D}_0\cup\mathcal{D}_1$, for otherwise it would be contained in $\mathcal{D}''_0\cup\mathcal{D}''_1$ and could not meet $S''$.

This gives the chain we were looking for: Since $\Lambda\subset S\setminus(\mathcal{D}_0\cup \mathcal{D}_1)$ we get, on one hand, that $\Lambda\subset U\setminus\mathcal{D}_1$ so there is a basic piece $\Lambda_0\subset\mathcal{D}_0$ with $\Lambda_0\prec\Lambda$. On the other hand, $\Lambda\notin\mathcal{F}$ while following $\Lambda_0\subset\mathcal{D}_0$, so there must be a basic piece $\Lambda_1\subset\mathcal{D}_1$ with $\Lambda\prec\Lambda_1$. Then the chain $\mathcal{C}'= \{\Lambda_0\prec\Lambda\prec\Lambda_1\} $ verifies the lemma, recalling the Claim and the fact that $\Lambda$ is disjoint from $\mathcal{D}_0\cup \mathcal{D}_1$.

\end{proof}

We are ready to provide a proof or Proposition \ref{sacar.triviales}. The proof will be
supported on the previous lemmas, using some of the constructions introduced
in their proofs.

\smallskip

{\bf Proof of Proposition \ref{sacar.triviales}:}

Let $\mathcal{C}$ be a chain for $f\in\mathcal{A}_0(\Sigma)$. Write $\mathcal{C} = \mathcal{C}_0\cup \mathcal{C}_{NT} \cup \mathcal{C}_1$ where $\mathcal{C}_{NT}$ consists of all the non-trivial pieces of $\mathcal{C}$. By Lemma \ref{l.nctrivial} we can assume $\mathcal{C}_{NT}$ is not empty, and write $\mathcal{C}_{NT} = \{ \Lambda_1\prec\cdots\prec\Lambda_n \}$. As we observed before, Lemma \ref{l.qisurface} gives that for $j=0,1$ there are $v_j\in H_1(\Sigma)$ such that $\rho_{\mathcal{C}_j}(f)=\{v_j\}$. We finish the proof by the following claim.

\vspace{.1cm}
{\bf Claim:} {\em There are non-trivial pieces $\Lambda'_j$, for $j=0,1$, such that $\Lambda'_0\preceq \Lambda_1$, $\Lambda_n\preceq \Lambda'_1$, and $v_j\in\rho_{\Lambda'_j}(f)$.}
\vspace{.1cm}

This claim completes the proof, taking $\mathcal{C}^*=\{\Lambda'_0\preceq \Lambda_1\prec\cdots\prec\Lambda_n\preceq \Lambda'_1 \}$. What follows is the proof of the claim.

\smallskip

We prove the existence of $\Lambda'_0$, the argument for $\Lambda'_1$ being analogous. As in the proof of Lemma \ref{l.nctrivial}, we can find $\mathcal{D}_0$ a disjoint union of disks so that:
\begin{itemize}
\item $\mathcal{D}_0\subseteq f(\mathcal{D}_0)$,
\item $\mathcal{D}_0$ contains all the basic pieces of $\mathcal{C}_0$
\item $\mathcal{D}_0$ is a union of connected components of $\text{cl}[\text{Fill}(S_0)\setminus S_0]$ for some non-trivial Conley surface $S_0$.
\end{itemize}

Therefore, every trivial piece not contained in $\mathcal{D}_0$ is contained in some attracting or repelling union of disks that is disjoint from $\mathcal{D}_0$.

Let $\mathcal{F}$ be the family of basic pieces $\Lambda$ that satisfy the following:
\begin{itemize}
\item $\Lambda\preceq\Lambda_1$,
\item $\Lambda$ is not contained in $\mathcal{D}_0$,
\item there is some basic piece $\Lambda_0\subset \mathcal{D}_0$ with $\Lambda_0\prec\Lambda$.
\end{itemize}

Notice that $\Lambda_1\in\mathcal{F}$. We also see that the pieces in $\mathcal{F}$ are non-trivial: Take a trivial piece $\Lambda$ that is not contained in $\mathcal{D}_0$. Then it is contained in a quasi-invariant union of disks $\mathcal{D}$ that is disjoint from $\mathcal{D}_0$. If $\mathcal{D}$ is repelling then there is no $\Lambda_0\subset \mathcal{D}_0$ that precedes $\Lambda$. If $\mathcal{D}$ is attracting, then $\Lambda_1$ cannot follow $\Lambda$.

Let $\Lambda'$ be a minimal element of $\mathcal{F}$ with respect to the partial order $\prec$. We will show that $v_0\in\rho_{\Lambda'}(f)$, thus $\Lambda'$ works as the basic piece $\Lambda'_0$ in the statement of this claim.

Define $R=\cup_{\Lambda\subseteq \mathcal{D}_0}\Lambda^u$. Since $R$ is an attractor we can find a subsurface $U$ containing $R$ so that:
\begin{itemize}
\item $U\cap\Omega(f)=R\cap\Omega(f)$
\item $\textrm{cl}[f(U)]\subseteq\text{int}(U)$
\end{itemize}
We can also assume that every connected component of $U$ meets $\Omega(f)$. Notice that $U$ contains $\mathcal{D}_0$ by definition of $R$ and the fact that $\mathcal{D}_0$ is a repellor (we gave the explicit argument in the proof of Lemma \ref{l.nctrivial}).

Recall that $\Lambda'$ is the minimal element of $\mathcal{F}$, so there is some basic piece $\Lambda_0 \subset \mathcal{D}_0$ with $\Lambda_0\prec\Lambda'$. Then there is an arc $\gamma$ joining $\Lambda_0$ and $\Lambda'$ that is a concatenation of $\gamma^u\subset W^u(\Lambda_0,f)$ and $\gamma^s\subset W^s(\Lambda',f)$. Moreover, we can assume $\gamma^s$ is within a small neighbourhood of $\Lambda'$, which gives us that $\gamma\subset U$, since $\gamma^u\subset \Lambda_0^u\subset U$ and $\Lambda'\subset \textrm{int}(U)$. Let $D'_0$ be the component of $\mathcal{D}_0$ with the starting point of $\gamma$, and let $K:=D'_0\cup\gamma$, observing that it is contained in $U$.

Let $\mathcal{U}$ be the family of basic pieces $\Lambda$ of $f$ that are contained in $U
\setminus \mathcal{D}_0$ but are different from $\Lambda'$, i.e.: $$\mathcal{U} = \{\Lambda \subset U\setminus \mathcal{D}_0:\Lambda\neq\Lambda'\}. $$ Notice that the pieces of $\mathcal{U}$ cannot precede $\Lambda'$, for if there was $\Lambda\in \mathcal{U}$ with $\Lambda\prec\Lambda'$ then $\Lambda'$ would not be minimal in $\mathcal{F}$. Thus

\begin{equation}\label{eqloca}(U\cap\Omega(f))\setminus \left(\bigcup_{\Lambda\in\mathcal{U}}\Lambda^u\right)=\Lambda'\cup \left( \mathcal{D}_0\cap\Omega(f) \right) \end{equation}

Using equation \ref{eqloca} we can argue as in Lemmas \ref{l.agujas} and \ref{l.nctrivial} to show that $K$ is disjoint from $\cup_{\Lambda\in\mathcal{U}}\Lambda^u$. Take $k\in\N $ so that the basic pieces of $f^k$ are topologically mixing, and apply Lemma \ref{l.construquasi} to $f^k$, the subsurface $U$, the family of basic pieces of $f^k$ that arises from $\mathcal{U}$, and the continuum $K$. This gives a surface $S$ that is quasi-invariant under $f^k$ and satisfies the following:

\begin{itemize}
\item $S\cap\Omega(f^k)=\Lambda'\cup (\mathcal{D}_0\cap\Omega(f))$, by equation \ref{eqloca},
\item $K\subset S$
\end{itemize}

Let $S'$ be the connected component of $S$ containing $K$, which is quasi-invariant under $f^k$ by Remark \ref{r.quasiconexas}. We take $\mathcal{D}''_0:=f^{-k}(\mathcal{D}_0)$, so we have that the components of $\mathcal{D}''_0$ that meet $S'$ are contained in $\text{int}(S')$, and $$S''=S'\setminus \mathcal{D}''_0 $$ is a surface that is quasi-invariant under $f^k$. This construction gives that $\Omega(f^k)\cap S''\subset \Lambda'$. We see that this set is non empty: The endpoint of $\gamma^u$ is in $K\setminus \mathcal{D}''_0$, so it belongs to $S''\cap \Lambda'$. Thus there exists $\Lambda''\subset\Lambda'$ a basic piece of $f^k$ that is contained in $S''$. On the other hand, notice that $D''_0 := f^{-k}(D'_0)$ is a component of $\text{Fill}(S'')\setminus S''$, that it is a repellor for $f^k$ and that every basic piece of $f^k$ in $D''_0$ has rotation vector $k\cdot v_0$.

If $S''$ is trivial, Lemma \ref{l.qisurface} implies that every basic piece in $\text{Fill}(S'')$ has the same rotation vector. In particular, since $D_0''\subseteq \text{Fill}(S'')$, we have that $\rho_{\Lambda''}(f^k)=k\cdot v_0$ and therefore $v_0\in\rho_{\Lambda'}(f)$ as desired.

In case that $S''$ is non-trivial we use Lemma \ref{l.filling} point (2) to deduce that there exists $p$ a fixed point of $f^k$ in $S''$ with rotation $k\cdot v_0$. Since $p\in \Omega(f)\cap S'' \subset \Lambda'$, this again shows that $v_0\in\rho_{\Lambda'}(f)$.

config: arxiv


TITLE: Recurrence with varying coefficient
QUESTION [6 upvotes]: Problem 1
$$
{\rm f}\left(n\right)
=
\frac{1}{n}\,
\left[{\rm f}\left(n - 1\right)k_{0} + {\rm f}\left(n-2\right)k_{1}\right]\tag{1}
$$
( This can also be written as
${\rm Q}\left(n\right)
=
k_{0}\frac{{\rm Q}\left(n - 1\right)}{n - 1}
+k_{1}\frac{{\rm Q}\left(n - 2\right)}{n - 2}$ provided
${\rm Q}\left(n\right) = n{\rm f}\left(n\right) $ )
where $k_{0},\,k_{1}$ are constants.
${\rm f}\left(0\right)=3,\,{\rm f}\left(1\right)=5$. The variable coefficient $\frac{1}{n}$  is making lot of issues here. It blocks the proper expansion.
Question
Please help me to find the solution of the recurrence  in terms of $n$
( implies ${\rm f}\left(n\right)$ )  and also the summation of the recurrence up to infinity ( $ sum = \sum_{n = 0}^{\infty}{\rm f}\left(n\right)$ ).
Note::

Partial answers are also welcome. We can discuss that here. This recursion forms a structure of Fibonacci look like tree. I have attached a pic here. You can see Red nodes   to indicate $k_0$ multiplication and green to indicate $k_1$ multiplication. You can see the index of the tree expands like a Fibonacci sequence(starts from 5)
Some of the generating function smart manipulations can be looked in to this famous e book for reference link:-generatingfunctionology!
I know a method of using ODE. But I am trying to solve it with out ODE so that I can extent this to higher dimension like matrices in similar structure questions. Please avoid ODE solution or any matrix exponential. A method using series would be more desirable

Problem 2( extended Matrix version of problem 1)
We have a given matrix recurrence ,
$  (\curlyvee_i,\curlyvee_{i-1})_{1\times2}= (\curlyvee_{i-2},\curlyvee_{i-3})_{1\times2}{\begin{bmatrix}A_{i-1}A_i+B_i & A_{i-1} \\B_{i-1}A_i & B_{i-1} \end{bmatrix}}_{2\times2} \tag 1$
$A_i= r\frac{\Im1}{i},B_i= r^2\frac{\Im2}{i} \tag 2$
$   \Im1=\left( \begin{array}{ccc}
   0 & -n_0 & m_0 \\
   n_0 & 0 & -l_0 \\
  -m_0 & l_0 & 0 \\
   \end{array} \right),
   \Im2=\left( \begin{array}{ccc}
   0 & -(n_1-n_0) & (m_1-m_0) \\
   (n_1-n_0) & 0 & -(l_1-l_0) \\
   -(m_1-m_0) & (l_1-l_0) & 0 \\
   \end{array} \right).\tag3$
$l_1,m_1,n_1,l_0,m_0,n_0,r$ all are constants can't alter at all
Given Data about the problem

Dimensions of $A_i,B_i,\curlyvee_i$ are $3\times3$ and $A_i,B_i$ are skew symmetric matrices with diagonal zero
Our initial condition is given as $(\curlyvee_1,\curlyvee_{0})_{1\times3} $
Properties of  $\Im1,\Im2$ useful for simplification are given below
For each matrices  $\Im1 (  a=l_0,b=m_0,c=n_0),  \Im2  (  a=l_1-l_0,b=m_1-m_0,c=n_1-n_0)$,let $p = \sqrt{a^2+b^2+c^2}$. It is true that $\wp^3 = -(a^2+b^2+c^2)\wp = -p^2\wp$ ($\wp$ represents matrices $\Im1,   \Im1$ ,select a,b,c accordingly).
 Hence, $\wp^{2m+1} = (-1)^mp^{2m}\wp \tag 4$ and $\wp^{2m} = (-1)^{m-1}p^{2m-2}\wp^2 \tag 5$.  In essence it says both matrices  $\Im1,  \Im2$ satisfies this property. courtesy @JimmyK4542

Question
 1. How do we solve this recurrence in terms of n. Is there any way to take out  n from the $3\times3$ matrices in recursion so that we can go for easy simplification. ?
NB :: I would like to have a solution with out using ODE or matrix exponential as I can extent to tensors etc. Please avoid such solution. Those were discussed earlier. Another version of the same question I posted as graph problem is given here. I didnt get any response to that,so I made it simplified as matrix recursion here. I haven been trying lot of methods on it none is successful coz of the properties of $A_i$
Note :: Bounty will be awarded to the helpful suggestions which leads to solution.
Efforts done so far

Tried to evaluate using power series simple method. Result:: failed ,Reason ::   varying n blocks the procedure
Tried to make another equivalent series by substitution. Result :: failed,Reason :: it still went back to old issues.  (not sure any nice substitution exists such that we can get form that is solvable..trying)
Made it as graph problem .I see a finite structure in Tree and node coloring.. If we go with graph theory method to get an expression for general case n and later add all as series up to infinity , we can find a sequence some thing like GP. We may end up in a finite expression. Result :: Failed ,Reason:: It was tough to get that common expression for a graph based on node contribution
Made it as matrix recursion, Result :: failed, could not decompose the matrix recursion component to separate n
Tried to compare or extract some properties from Fibonacci tree as it has same structure. Result :: Failed,Reason:: we will still end up in varying coefficient ..

REPLY [5 votes]: So I do realize you wanted to avoid a matrix exponential answer, now I would like to post one here just so other people running into this question can see what that looks like:
$$ \textbf{MATRIX SOLUTION}$$
Let:
$$L(f(n)) = f(n) - f(n-1)$$
Notice therefore:
$$L^2(f(n)) = f(n) - 2f(n-1) + f(n-2)$$
Therefore we can re-write as
$$f(n) = \frac{k_1}{n}(L^2 (f(n)) - 2L(f(n)) + f(n)) + \frac{k_0}{n}(f(n) - L(f(n)))$$
use $f = f(n)$ for cleaner appearance and use standard operator notation (no parenthesis between L and f)
$$f = \frac{k_1}{n}(L^2 f - 2Lf + f) + \frac{k_0}{n}(f - Lf)$$
Which simplifies to
$$\frac{k_1}{n}L^2f + \left(-\frac{2k_1+k_0}{n}\right)Lf + \left(\frac{k_0 + k_1 - n}{n} \right)f = 0 $$
We now multiply through by $n$ and divide by $k_1$
$$L^2f - \left(\frac{2k_1+k_0}{k_1}\right)Lf + \left(\frac{k_0 + k_1 - n}{k_1} \right)f = 0 $$
And move all terms besides $L^2f$ to the other side
$$L^2f  =  \left(\frac{2k_1+k_0}{k_1}\right)Lf - \left(\frac{k_0 + k_1 - n}{k_1} \right)f $$
Now generate a vector $G = \begin{bmatrix} f \\ Lf  \end{bmatrix} $
At this point it becomes evident that if we define $L$ to act component by component on matrices then:
$$LG = \begin{bmatrix} - \frac{k_0 + k_1 - n}{k_1} && 0 \\ 0 && \frac{2k_1 + k_0}{k_1} \end{bmatrix}G$$
Traditionally to solve equations of the type
$$Lf = bf$$
exploits the fact that
$$L(2^{Q(n)}) = (2^{LQ(n)}-1)2^{Q(n)}$$
In other words we are seeking a matrix $H$ such that
$$ 2^{LH(n)} - I_2  = \begin{bmatrix} - \frac{k_0 + k_1 - n}{k_1} && 0 \\ 0 && \frac{2k_1 + k_0}{k_1} \end{bmatrix}$$
Meaning
$$2^{LH(n)} =  \begin{bmatrix} - \frac{k_0  - n}{k_1} && 0 \\ 0 && \frac{3k_1 + k_0}{k_1} \end{bmatrix} $$
At this point we must find the components of:
$$\log_2 \left( \begin{bmatrix} - \frac{k_0  - n}{k_1} && 0 \\ 0 && \frac{3k_1 + k_0}{k_1} \end{bmatrix} \right) $$
This is a tricky calculation which requires we replace the 0 components with a variable z. Then compute the logarithm treating, z as a real variable. Then for each component $c_i$ solve
$$Lu = c_i$$
to assemble the components of $H(n)$
from here
$$2^{H(n)}$$
will be the solution of the system
And therefore the dot product of the top row of $2^{H(n)}$ with the vector $\begin{bmatrix} c_0 \\ c_1 \end{bmatrix}$ is the general solution to our problem
At this point take the limit as $z \rightarrow 0$ to recover the solution
$$ \textbf{SERIES SOLUTION} $$
We note that given the linear operator described above
$$Lf = f(n) - f(n-1)$$
We can generate a Taylor-like series by first observing the existence of a product rule
$$L(f*g) = f(n)g(n) - f(n-1)g(n-1) = g*Lf + f*Lg - Lg*Lf $$
And second observing that
$$L \left( \frac{1}{n!} \prod_{i=1}^{n}{(x +i-1)}\right) =\frac{1}{n!} \prod_{i=1}^{n-1}{(x+i-1)}  $$
Thereby allowing us to generate a "series"
$$f(x) = f(a) + Lf(a)x + \frac{L^2f(a)}{2!}x(x+1) + \frac{L^3f(a)}{3!}x(x+1)(x+2)... $$
We now recall the original problem:
$$L^2f  =  \left(\frac{2k_1+k_0}{k_1}\right)Lf - \left(\frac{k_0 + k_1 - n}{k_1} \right)f $$
We can consider the values of
$$ \begin{matrix} L^2f && L^3f && L^4f && ... && L^rf \end{matrix}$$
At which point it becomes evident
$$ L^rf = \left(2 + \frac{k_0}{k_1}\right)L^{r-1}f + \frac{r-2}{k_1}L^{r-3}f + \left(\frac{n}{k_1} - 1 - k_0+ \frac{r-2}{k_1}  \right)L^{r-2}f$$
This can be re-written by noting $L^rf = X(r)$ and therefore
$$X(r) = aX(r-1) + (r-2)bX(r-3)+ (c + b(r-2))X(r-2)$$
Whereas $X(0) = 3, \ X(1) = (5-3) = 2$ and
$$a = \left(2 + \frac{k_0}{k_1}\right), \ b = \frac{1}{k_1}, \ c = \left(\frac{n}{k_1} - 1 - k_0 \right) $$
Now the solution to the problem is
$$f(n) =  \sum_{i=0}^{\infty} \left[ \frac{X(i)}{i!} \prod_{j=1}^{i} (n + j - 1) \right]$$
$$ f(n) = X(0) + X(1)n + \frac{X(2)}{2}n(n+1) + \frac{X(3)}{3!}n(n+1)(n+2) ... $$
Of course finding a closed form for $X$ seems to be a still intriguing problem but not impossible since it is of the exact same form as the original problem: we can create a matrix solution or series solution for $X$ (depending on personal preference...) but to address the concerns of dividing by $k_0$ and $k_1$ our solution terms can instead be written as:
$$ k_1*L^rf = \left(2k_1 + k_0\right)L^{r-1}f + (r-2)L^{r-3}f + \left(n - k_1 - k_0*k_1+ (r-2)  \right)L^{r-2}f$$
Thus we can still attempt to solve for each individual $L^{r}f$ and generate series terms... I am curious how this will play out and whether any inconsistent systems will be formed (if they are formed: does that mean there cannot exist a solution to said equation, or does it require the creating of a new flavor of functions outside of all others, much like how we can create an imaginary constant to deal with the unsolvable $\sqrt{-1}$ problem)


TITLE: Is $(t,x)\mapsto T(t)x$ continuous for a semigroup of operators $T(t)$
QUESTION [0 upvotes]: Is there a strongly continuous one-parameter semigroup of operators  $(T(t))_{t\geq 0}$ such that the map $M:[0,\infty)\times X\to X$ defined by $M:(t,x)\mapsto T(t)x$ is not continuous? where $X$ is a Banach space.
I notice that such a map $M$ is continuous if the semigroup $T(t)$ is continuous in the operator norm.

REPLY [1 votes]: No, there isn't; the mapping $M$ is continuous for any strongly continuous one-parameter semigroup.
Hint: Fix $(t,x) \in [0,\infty) \times X$, then
$$|T(t)x-T(s)y| \leq |T(t)x-T(s)x| + |T(s)x-T(s)y| \leq  c |T(|t-s|)x-x| + c|x-y| $$
for any $y \in X$, $s \in [0,2t]$ where $$c := \sup_{r \in [0,2t]} \|T(r)\|<\infty.$$


TITLE: Is there a mathematical connection between Zipf’s Law and the Birthday Problem?
QUESTION [0 upvotes]: According the the Wikipedia article on Zipf’s Law, a mere 135 words suffices to cover 50% of the Brown Corpus, and according to the Wikipedia article on the Birthday Problem, a mere 23 randomly-chosen people suffices to have a probability > 50% that two of them will have the same birthday (month and day, ignoring leap years). Is this merely a coincidence? Or is there some underlying mathematical reason for the Pareto-Principle-like behavior of these two phenomena? Does the Pareto Principle in fact explain them both?

REPLY [1 votes]: There may be psychological similarities in that surprisingly few things make something likely, but that belongs on another site.  The English word case just comes about because some words are much more common than other words, so if you add up the frequencies of the most common words you get to $50\%$ rather quickly.  The birthday paradox comes about because each pair of birthdays has $\frac 1{365}$ chance to match.  As you add more people into the mix, the number of pairs goes up as the square of the number of people.  The number to get a match is then about $\sqrt {365} \approx 19$.  Here the birthdays are equally probable, there are just more chances for a match than people recognize.


TITLE: Transformation of Uniform Random Variable
QUESTION [1 upvotes]: Let $X\sim \mathcal{U}(0,1)$  and $Y = \ln\left((1-X)^{-c}\right)$ for some $c>0$.
We need to show that $Y \sim {\mathrm{Exp}}(1/c)$.
We got to here but couldn't go further:
$$
   \mathbb{P}(Y \leqslant y) = \mathbb{P}( X \leqslant 1- \mathrm{e}^{-c y})
$$
Thanks in advance.

REPLY [1 votes]: Here is a different approach. Given that $X\sim \mathcal{U}(0,1)$, $f_X(x)=1$.
From $y = \ln\left((1-x)^{-c}\right)$ we get $x=1-\mathrm{e}^{-\frac{y}{c}}$ and thus, $$\frac{\text{d}x}{\text{d}y}=\frac{1}{c}\mathrm{e}^{-\frac{y}{c}}~~~.$$ Thus
$$\begin{align*}
f_Y(y) & = f_X\left(1-\mathrm{e}^{-\frac{y}{c}}\right)\left|\frac{\text{d}x}{\text{d}y}\right|\\
& = \frac{1}{c}\mathrm{e}^{-\frac{y}{c}}~~,
\end{align*} $$  which is what we wanted.


TITLE: System of equation doesn't want to get solved
QUESTION [2 upvotes]: I've been trying to solve the following system of equations for hours, to no avail:
$$(1)\;\frac{x}{y}+\frac{y}{x}=\frac{10}{3}\\\\$$
$$(2)\;x^2+y^2=8$$
Can anyone give me a hint?

REPLY [4 votes]: Hint:
The first equation is equivalent to $x^2+y^2=\dfrac{10}3xy$.
Given the second equation, we have $\dfrac{10}3xy=8, $ so $\color{blue}{xy=\dfrac{12}5}$.
Also, $(x+y)^2=x^2+y^2+2xy=8+\dfrac{24}5=\dfrac{64}5, $ so $\color{blue}{x+y=\pm\dfrac8{\sqrt5}}$.
Can you take it from here?

REPLY [1 votes]: As Peter Foreman already mentioned, given this equation system
$$\begin{equation*}\ x/y+y/x=10/3  \\ x^2+y^2=8\end{equation*}$$
we can multiply the first equation by $xy$ and get the following equivalent equation system
$$\begin{equation*}\ x^2+y^2=(10/3)xy  \\ x^2+y^2=8\end{equation*}$$
From this we see that $$x^2+y^2=(10/3)xy=8$$ and $$(10/3)xy=8$$ or $$\color{red}{y = (12/5)/x}$$
Pluging $y = (12/5)/x$  into  $(x/y)+(y/x)=(10/3)$ we get $$x/((12/5)/x)+((12/5)/x)/x=10/3$$ or this: $$(5/12)x^2+(12/5)/x^2=10/3$$
Multiplying by $x^2$, we get: $$(5/12)x^4+(12/5)=(10/3)x^2$$ or this: $$(5/12)x^4-(10/3)x^2+(12/5)=0$$
Multiplying by $(12/5)$, we get: $$x^4-8x^2+(12/5)^2=0$$ Substituting $u = x^2$, we get: $$u^2-8u+(12/5)^2=0$$
Completing the square, we get: $$u^2-8u+(12/5)^2=(u+(-4))^2-4^2+(12/5)^2=0$$
This leads to $$x_{1,2,3,4}=\pm\sqrt{u_{1,2}} = \pm\sqrt{4\pm\sqrt{4^2-(12/5)^2}}=\begin{cases}6/\sqrt{5} &\rightarrow \color{red}{y_1 = (12/5)/(6/\sqrt{5})} = 2/\sqrt{5}  \\ -6/\sqrt{5} &\rightarrow \color{red}{y_2 = (12/5)/(-6/\sqrt{5})} = -2/\sqrt{5}\\2/\sqrt{5} &\rightarrow \color{red}{y_3 = (12/5)/(2/\sqrt{5})}=6/\sqrt{5}\\ -2/\sqrt{5}&\rightarrow \color{red}{y_4 = (12/5)/(-2/\sqrt{5})}=-6/\sqrt{5}\end{cases}$$
Another way was suggested by rsadhvika can be done this way:
Start with the first equation of the original equation system $$x/y+y/x=10/3$$ Set $y = \alpha x$, then we get $$x/(\alpha x)+(\alpha x)/x=10/3$$
or
$$1/\alpha + \alpha =10/3$$
Multiply by $\alpha$ and it follows
$$1+ \alpha^2 =(10/3) \alpha$$
or
$$ \alpha^2-(10/3) \alpha+1 =0$$
Completing the square:
$$ \alpha^2-(10/3) \alpha+1 =(\alpha+(-5/3))^2-(5/3)^2+1=0$$
This leads to
$$ \alpha_{1,2} = 5/3\pm\sqrt{(5/3)^2-1} = \begin{cases}\alpha_1 = 3 \\ \alpha_2 = 1/3 \end{cases}$$
The equations $\color{blue}{y = \alpha_1 x= 3x}$ and $\color{blue}{y= \alpha_2 x= x/3}$ can be substituted seperately for $y$ in the original equation system and they will lead to the same results!
A third method was suggested by J. W. Tanner
Because we have $xy=12/5$ and $x^2+y^2=8$, we use these to evaluate $$(x+y)^2=x^2+2 \cdot xy+y^2=x^2+y^2+2\cdot xy=8+2\cdot (12/5)=8+24/5=40/5+24/5=64/5$$
This means that $(x+y)^2=64/5$ and it follows that $x+y=\pm 8/\sqrt{5}$. From that we get $\color{orange}{y= 8/\sqrt{5}-x} $ or $\color{orange}{y= -8/\sqrt{5}  -x}$ that we can again use to eliminate $y$ in the original equation system and to solve for $x$. With these equations for $y$ we will also get the same results at the end!
In all methods we try to express $y$ in terms of $x$. We try to find an equation $y = f(x)$ with $y$ on the left and $x$ on the right. Then we go with our minds back to the original equation system and plug the $f(x)$ expression for the $y$ each time we encounter an $y$ in the original equation system. This way of doing things gives us equations that only contain $x$. We gain equations with one unknown variable $x$ instead of equations with two variables, instead of equations with $x$ and $y$. Equations with just one variable are usually easier to solve.


TITLE: Turning $\mathbb R^n$ into field
QUESTION [4 upvotes]: I am reading Apostol's fascinating text Mathematical Analysis. In a footnote on P117, he writes:

If it were possible to define multiplication in $\mathbb R^3$ so as to make $\mathbb R^3$ a field including $\mathbb C,$ we could argue as follows: for every $\bf x$ in $\mathbb R^3$, the vectors $1,\bf x,\bf x^2,\bf x^3$ would be linearly dependent. Hence for each $\bf x$ in $\mathbb R^3,$ a relation of the form $a_0+a_1{\bf x}+a_2{\bf x^2}+a_3{\bf x^3}=0$ would hold, where $a_0,a_1,a_2,a_3$ are real numbers.But every polynomial of degree three with real coefficients is a product of a linear polynomial and a quadratic polynomial with real coefficients. The only roots such polynomials can have are either real numbers or complex numbers.

I have these couple of questions:

Does above argument show that $\mathbb R^3$ can't be made a field?  Or just that $\mathbb R^3$ can't be field such that $\mathbb C$ is its subfield?
How are we so sure that there are no other roots than complex numbers? Maybe we have not explored enough!

REPLY [4 votes]: We can show that $\Bbb R^3$ cannot be assigned a multiplication operation which turns it into an extension field of $\Bbb R$ without assuming such a field contains a subfield isomorphic to $\Bbb C$ as follows:
If $\Bbb R^3$ were such a field, we would have
$[\Bbb R^3:\Bbb R] = 3; \tag 1$
being an extension field of $\Bbb R$, $\Bbb R^3$ contains a multiplicative identity $1$ and a subfield $1\Bbb R = \Bbb R1$ isomorphic to $\Bbb R$ in the usual manner, that is
$\Bbb R \ni r \leftrightarrow r1 \in 1\Bbb R \subsetneq \Bbb R^3; \tag 2$
by virtue of (1), there exists
$\mathbf v \in \Bbb R^3 \setminus \Bbb R1 \tag 3$
such that $1, \mathbf v, \mathbf v^2, \mathbf v^3$ are linearly dependent over $\Bbb R1 \cong \Bbb R$; that is
$\exists c_i \in \Bbb R, \; 0 \le i \le 3, \tag 4$
not all $c_i$ zero, with
$c_3 \mathbf v^3 +c_2 \mathbf v^2 +  c_1 \mathbf v + c_0 = 0; \tag 4$
let us first consider the case
$c_3 = 0; \tag 5$
then
$c_2 \mathbf v^2 +  c_1 \mathbf v + c_0 = 0; \tag 6$
now if
$c_2 = 0, \tag 7$
then if
$c_1 = 0 \tag 8$
as well, we find
$c_0 = 0, \tag 9$
contradicting our hypothesis that not all the $c_i = 0$; and if
$c_1 \ne 0 \tag{10}$
we may write
$\mathbf v = -\dfrac{c_0}{c_1} \in \Bbb R 1 \cong \Bbb R, \tag{11}$
which contradicts (3); thus we have that
$c_2 \ne 0, \tag{12}$
and we may write (6) as
$\mathbf v^2 + b_1 \mathbf v + b_0 = 0, \tag{13}$
where
$b_i = \dfrac{c_i}{c_2} \in \Bbb R; \tag{14}$
we write (13) as
$\mathbf v^2 + b_1 \mathbf v = -b_0, \tag{15}$
and complete the square:
$\left (\mathbf v + \dfrac{b_1}{2} \right )^2 = \mathbf v^2 + b_1 \mathbf v + \dfrac{b_1^2}{4}  = \dfrac{b_1^2}{4} - b_0 = d; \tag{16}$
if
$d \ge 0, \tag{17}$
(16) yields
$\mathbf v = -\dfrac{b_1}{2} \pm \sqrt d \in \Bbb R, \tag{18}$
in contradiction to (3); thus,
$d < 0, \tag{19}$
and (16) becomes
$\dfrac{1}{{\sqrt{-d}}^2} \left (\mathbf v + \dfrac{b_1}{2} \right )^2 = -1, \tag{20}$
which shows the existence of an element
$\mathbf i \in \Bbb R^3 \tag{21}$
with
$\mathbf i^2 = -1, \tag {22}$
and in the usual manner we see that the subalgebra
$\Bbb R + \Bbb R \mathbf i = \{ s + t \mathbf i \mid s, t \in \Bbb R \} \cong \Bbb C \tag{23}$
is a subfield of $\Bbb R^3$ with
$[\Bbb C: \Bbb R] = 2; \tag{24}$
but this is impossible since it implies
$3 = [\Bbb R^3:\Bbb R] =[\Bbb R^3:\Bbb C] [\Bbb C: \Bbb R] = 2[\Bbb R^3:\Bbb C]; \tag{25}$
but $2 \not \mid 3$; we conclude then that no such $\mathbf v$ satisfying (6), (13) can exist in $\Bbb R^3$.
Now if
$c_3 \ne 0, \tag{26}$
then $\mathbf v$ satisfies the full cubic (4), and as above setting
$b_i = \dfrac{c_i}{c_3}, \; 0 \le i \le 2, \tag{27}$
we obtain the real monic cubic
$p(\mathbf v) = \mathbf v^3 +b_2 \mathbf v^2 +  b_1 \mathbf v + b_0 = 0, \tag{28}$
which as is well-known always has a root
$r \in \Bbb R, \tag{29}$
whence
$p(\mathbf v) = (\mathbf v - r)q(\mathbf v) \tag{30}$
for some monic real quadratic polynomial $q(\mathbf v)$; thus,
$(\mathbf v - r)q(\mathbf v) =p(\mathbf v) = 0; \tag{31}$
but
$\mathbf v - r \ne 0 \tag{32}$
since
$\mathbf v \notin \Bbb R; \tag{33}$
it follows that
$q(\mathbf v) = 0, \tag{34}$
and we have reduced the cubic to the previous (quadratic) case, which we have reduced to the absurd; we thus conclude that $\Bbb R^3$ admits no multiplication operation compatible with the field axioms, and we are done.
We close with the observation that our argument requires no assumption that $\Bbb R^3$ contains a subfield isomorphic to $\Bbb C$; indeed, we have shown that the existence of such a subfield follows from the assertion that $\Bbb R^3$ is an extension field of $\Bbb R$, from which a contradiction is deduced.
Finally, as for our OP Silent's two closing questions, Apostol's proof indeed makes use of the assumption that $\Bbb R^3$ has a subfield isomorphic to $\Bbb C$ to show that $\Bbb R^3$ can't be made into a field; and the issue that there are "other" roots of the polynomial in $\mathbf x$ than the usual complex numbers falls once we have $\Bbb C \subset \Bbb R^3$, for then the familiar factorizations in $\Bbb C[x]$ hold, and since a polynomial of degree $n$ over any field has at most $n$ zeroes, we see that all the roots of a real polynomial in $\mathbf x$ must lie in $\Bbb C$; we need look no further.


TITLE: Show a rational function is transcendental over a field.
QUESTION [7 upvotes]: Let $u=\frac{x^3}{x+1}\in F(x)$, where $F(x)$ is the field of quotients of $F[x]$ ($F$ some field, $x$ an indeterminate over it). Show that $u$ is transcendental over $F$.

This is an exercise in Hungerford.
I'm having some trouble even grasping the concepts involved.  For instance, I know that if $v$ is transc. over $F$, then $F[v]\cong F[x]$. Or that if $v$ is transc. over $F$, then $F[v]\subsetneq F(v)$.  But I have no idea how to use this to my advantage.
I'm also confused about what it even means for $u$ as above to be transc. over $F$.  Am I going to have to consider "polynomials of polynomials"?

REPLY [2 votes]: The following way of saying it is a bit more concrete, in case anyone prefers it that way.
If the coefficients $c_k$ in the sum
$$
\sum_{k=0}^n c_k u^k
$$
are in $F$, then the point is to show that that sum cannot be $0$ unless all of the coefficients are $0$.  The sum is
$$
\sum_{k=0}^n c_k \left( \frac{x^3}{x+1} \right)^k.
$$
If that $=0$ then multiplying both sides by $(x+1)^n$, one concludes
$$
\sum_{k=0}^n c_k x^{3k} (x+1)^{n-k} = 0.
$$
So a polynomial in $x$ evaluates to $0$.  Since $x$ is transcendental, that can't happen unless all of the coefficients are $0$.


TITLE: Conditional Probability Question for A Continuos Random Variable
QUESTION [2 upvotes]: Suppose $X$ is a continuous random variable with density $p$ with respect to the Lebesgue measure. According to Radon-Nikodym theorem $p$ is measurable itself so it can be seen as a random variable. Question is what is the following expression equal to
$$p(X=x_0|p(X=x_0)=3)\overset{?}{=}3$$
Or in a more general but vaguer sense
$$p\Big(X=x_0\Big|p(X=x)\ {\rm is\ given\ and\ is\ }f(x)\Big)\overset{?}{=}f(x_0)$$
I have guessed what they need to be equal to --after the question marks-- but I do not have any rigour argument to justify them. So I would really appreciate if you could please guide me how to do it (or prove me wrong).

REPLY [1 votes]: Let $p_{\small X}(x)$ be the probability density for the continuously distributed random variable $X$.   Then $p_{\small X}(X)$ is a random variable, and you seem to want to talk about the condtional measure $p_{\small X\mid p_{\small X}(X)=3}(x)$.
The nature of the conditional measure, and whether it even exists, entirely depends on the distribution for the random variable.

If $X$ has a probability density of $3$ in an integrateable interval of uncountable many points, say $A$, then the conditional probability measure $p_{X\mid X\in A}(x)$ will be a probability density function. $$p_{\small X\mid X\in A}(x)=\dfrac{p_{\small X}(x)\mathbf 1_{x\in A}}{\int_A p_{\small X}(s)\mathsf d s}=\dfrac {\mathbf 1_{x\in A}}{\int_A \mathsf d s}$$
If $X$ has a probability density of $3$ in an interval of finite many points, say $B$, then the conditional probability measure $p_{X\mid X\in B}(x)$ will be a probability mass function.
$$p_{\small X\mid X\in A}(x)=\dfrac{p_{\small X}(x)\mathbf 1_{x\in B}}{\sum_{x\in B} p_X(s)}=\dfrac {\mathbf 1_{x\in B}}{\lvert B\rvert}$$

For instance, if $X$ has density of $3$ at only one point, say at $x_3$, then the conditional probability that $X=x_3$ when $X=x_3$ will most surely be $1$.

In other cases it may not even be sensible.


TITLE: Rigidity and total swappability
QUESTION [2 upvotes]: Let $(X,\tau)$ be a topological space. Let us call $x,y\in X$ swappable if there is $f:X\to X$ continuous such that $f(x)=y$ and $f(y)=x$. This relation is obviously reflexive and symmetric, but not necessarily transitive.
Moreover, we call $(X,\tau)$ rigid if the identity is the only homeomorphism from $X$ to itself.
Is there a rigid space $(X,\tau)$ with $|X| > \aleph_0$ such that for all $x,y\in X$ we have that $x,y $ are swappable?

REPLY [4 votes]: The answer is yes by a 1951 result of Miroslav Katětov, who proved that there is an (uncountable) rigid totally disconnected compact space. Any such space $X$ is rigid and has the property that any two points are swappable by continuous functions, since if $x\neq y$, we can find an open separation $X=U\sqcup V$ with $x\in U$ and $y\in V$, and then let $f$ be constant on each piece with value $y$ or $x$, respectively. This is a continuous function swapping $x$ and $y$.


TITLE: Does a cooler, submerged in water, keep beer colder, longer?
QUESTION [2 upvotes]: Looking around the web, I see some submerged bucket-like things, without insulation, and some floating coolers, where the actual cooler is not submerged.
Given that your water temperature is lower than the air temperature, would a mostly submerged, insulated cooler keep cold, inside, longer?  Would the heat exchange be hastened by the conductivity of the water or slowed by the improved temperature differential between inside/outside the cooler?
Summertime sea temp near Seattle is 52$^{\circ}$F.  Lakes are warmer, maybe 60-65$^{\circ}$F.  On a summer day, the air temp might be 65-75$^{\circ}$F.
To clarify, I have a cooler with beer and ice inside, hopefully around 40$^{\circ}$F.  I want to keep that beer as cold as possible, for as long as possible.

REPLY [1 votes]: A cooler that has ice and water in it will be held at 32 degrees Fahrenheit until all the ice is melted. The rate at which the ice melts depends on the rate at which heat can enter the container.
The rate at which heat crosses any thermal boundary can be modeled as:
$$\dot Q=\frac{\Delta T}{\sum 1/h_i}$$
Where $h_i$ represent the thermal conductivity of each thermal resistance. In the case of the cooler there will be a thermal resistance due to the cooler. A thermal resistance due to the transition from liquid inside the cooler to the cooler wall, and finally an transition between the cooler outside and the outside medium.
This outside thermal resistance changes depending of whether the medium is air or water. A typical value for still air is around $5 \frac{W}{m^2K}$, while a typical value for water is $100\frac{W}{m^2 K}$.
Now if we assume the cooler can be approximated as 4cm thick Styrofoam then it would have a h of $0.8 \frac{W}{m^2 K}$. So the thermal resistances of the cooler in the two scenarios are:
$$\frac1{100}+\frac1{0.8}+\frac1{5}=0.01+1.25+0.2=1.45$$
$$\frac1{100}+\frac1{0.8}+\frac1{100}=0.01+1.25+0.01=1.26$$
So the air version would insulate 1.15 times better.
The temperature differences however would drive the heat with $\frac{75-32}{52-32}=2.15$ times the temperature difference. So on a warm summer day the sea would keep your cooler contents cool 1.9 times longer.
Note that a 4cm thick Styrofoam cooler is actually quite a nice cooler. If you don't have such a nice cooler, it might be better to keep the cooler out of the water, but if you don't have a nice cooler, the ice is gonna melt pretty fast no matter where you put it.


TITLE: Prove that $\angle{BED} = \angle{CFD} = 30^{\circ}$
QUESTION [4 upvotes]: Let $D, E$ and $F$ be on $\overline{BC}, \overline{AC},$ and $\overline{AB},$ respectively, such that $\overline{AD}, \overline{BE},$ and $\overline{CF}$ are the angle bisectors of $\triangle {ABC}$. If $\angle{EDF}=90^{\circ}$, prove that $\angle{BED} = \angle{CFD} = 30^{\circ}$.

I have a few conjectures on how prove this. Firstly, it seems that $DE$ is an angle bisector of $\angle{CDA}$ and similarly $DF$ is an angle bisector of $\angle{BDA}$. Also it seems that $\overline{EF} \parallel \overline{BC}$, so that might help, but I don't see how to show $\angle{BED} = \angle{CFD} = 30^{\circ}$.

REPLY [3 votes]: Note that the quadruple $C,K,G,F$ is harmonic. This can be easily seen via Menelaos and Ceva theorem.
Since $\angle GDF = \frac \pi 2$ and $C,K,G,F$ is harmonic, it follows that $DG, DF$ are bisectors of the angle between lines $CB, AD$. To see that, let $C'$ be a point on $CK$ such that $DK, DC'$ are symmetric with respect to $DG$. Then $DF, DG$ are angle bisectors of the angle between $DK, DC'$. Using angle bisector theorem we find that $\frac{KG}{GC'} = \frac{KD}{DC'} = \frac{KF}{FC'}$ and so $C',K,G,F$ is harmonic. Therefore $C'=C$.
Since $F$ lies on the internal bisector of angle $ECD$ and on the external bisector of angle $CDE$, it follows that $F$ is $C$-excenter of triangle $CAD$. This means that $\angle FAD = \frac{\pi - \angle DAC}2 = \frac{\pi - \angle FAD}2$. Therefore $\angle FAD = \frac \pi 3= \angle DAC$. Now, angle chasing gives
$$\angle DFC = \pi - \angle FDC - \angle ADF = \pi - \frac{\angle ACD}2 - \angle CDA - \frac{\pi - \angle CDA}2 = \frac \pi 2 - \frac{\angle ACD + \angle CDA}2 = \frac \pi 2 - \frac{\pi - \angle DAC}2 = \frac \pi 6.$$


TITLE: What is the intersection of real numbers' set and span $\{\sin^2x,\cos^2x\}$?
QUESTION [2 upvotes]: I know intersection of real numbers' set and span $\{\sin x,\cos x\}$ is $\{0\}$.
Here in this problem $\sin^2x,\cos^2x$ are linearly independent like as $\sin x,\cos x$ .
If $a\sin^2x+b\cos^2x=k$, which is not equal to $0$.
Then this is absurd.
So $a\sin^2x+b\cos^2x=0$, since $\sin^2x,\cos^2x$ are linearly independent also.
Hence the answer is $\{0\}$.
But my book says it is real numbers' set.
Why? I don't understand.
Please help me!

REPLY [4 votes]: I am interpreting this as taking place within $\scr{C}^0(\mathbb{R},\mathbb{R})$, i.e.  the space of continuous functions from $\mathbb{R}$ to $\mathbb{R}$. If we do  that, then we view $\mathbb{R}$ as being identified with the subspace of constant functions. Now, $\operatorname{span}(\cos^2 x,\sin^2 x)\cap \mathbb{R}=\mathbb{R}$ because as functions, $\cos^2x+\sin^2x=1$. So, taking any $\lambda \in \mathbb{R}$, we get $\lambda\cos^2x+\lambda \sin^2x=\lambda$. So, $\mathbb{R}\subseteq \operatorname{span}(\cos^2x ,\sin^2x)\cap \mathbb{R}\subseteq \mathbb{R}$. Hence,
$$ \boxed{\operatorname{span}(\cos^2x,\sin^2x)\cap \mathbb{R}=\mathbb{R}.}$$