TITLE: Is my proof that nonnegative polynomials on $[0,1]$ form a convex set correct?
QUESTION [2 upvotes]: I want to prove that the set $K = \{c \in R^n\mid c_{1} + c_{1}t +\dotsb+ c_{n}t^{n-1} ≥ 0 \forall t \in [0,1]\}$ is a cone i.e that for $x \in K$ and $\theta \ge 0$, $\theta x\in K$.
Is the follow attempt a correct proof?

Let's consider $x \in K$.
We have $\sum\limits_{i=1}^n{x_{i} t^{i-1}} \ge 0, \space t ∈ [0,1]$
For $\theta \ge 0$, we then have
$\sum\limits_{i=1}^n{\theta x_{i} t^{i-1}} = \theta \sum\limits_{i=1}^n{x_{i} t^{i-1}}$
since both $\theta$ and the polynomial $\sum\limits_{i=1}^n{x_{i} t^{i-1}}$ are greater than or equal to $0$, we must have
$\sum\limits_{i=1}^n{\theta x_{i} t^{i-1}} = \theta \sum\limits_{i=1}^n{x_{i} t^{i-1}} \ge 0$
Thus $x \in K$, $\space$ $\theta \ge 0$ $\rightarrow$ $\theta x \in K$.
Hence $K$ is a cone. $\square$

REPLY [4 votes]: In general, if a question on Math SE may be answered with a simple yes-or-no, then the question is almost certainly off-topic.  There seems to be a consensus in the community that these kinds of proof verification problems are not off-topic, hence a correct answer to this question cannot be a simple yes-or-no.  As such, I must assume that the question is really more about the style of the presentation, not the actual technical details.
I'll note that much of my answer is a matter of opinion; other author's may have differing thoughts on specific matters of style.  Thus if there is some pedantic point below with which you disagree, please disregard it.

Starting from the top, I find the statement of the result a little hard to follow, and would rewrite it as

Exercise: Show that the set
  $$ K = \{ c \in \mathbb{R}^n \mid c_1 + c_2t + \dotsb + c_n t^{n-1} \ge 0 \forall t \in [0,1] \} $$
  is a cone.  That is, for all $x\in K$ and all $\theta \ge 0$, show that  $\theta x \in K$.

The very minor edits here a mostly for readability:  the displayed equation is a little easier to follow, and restructuring the final sentence to avoid notation immediately following a comma is (in my opinion) an improvement.

As to your proof (which is correctly argued), here is your presentation, with some notes:

Let's consider $x \in K$.

Personally, I don't like the phrase "Let's consider..." I find it overly casual, and also kind of meaningless.  I prefer to write in a more imperative mode and to use more precise verbs.  I think it is better to say "Let $x \in K$," or "Fix an arbitrary $x\in K$," or something similar.  You could also say "Consider $x\in K$," if you are really attached to that verb.
You might also want to fix a value of $\theta$ here, too.  Remember you want to show that the desired result holds for any $x\in K$ and any $\theta \ge 0$.  Why not fix both of those values right at the start?

We have $\sum\limits_{i=1}^n{x_{i} t^{i-1}} \ge 0, \space t ∈ [0,1]$

Again criticizing style, I don't really like the use of the "mathematical 'we'".  I know that lots of people use it---I even use it quite a lot when I am not terribly worried about the quality of the presentation, or when I am writing quickly (by way of example, I spent a couple of hours last week removing every instance of "we" from my thesis).  I think that it is better to be more direct.  Maybe an "if-then" statement, such as "If $t \in [0,1]$, then...".  It might even be better to explain where the inequality comes from, i.e. "Since $x \in K$..."

For $\theta \ge 0$, we then have
$\sum\limits_{i=1}^n{\theta x_{i} t^{i-1}} = \theta \sum\limits_{i=1}^n{x_{i} t^{i-1}}$

If you follow my previous suggestion and fix $\theta$ earlier in the argument, then this entire line is unnecessary.  Also, it should be "displayed" using $$; this avoids the use of \limits, which is poor style for inline (rather than displayed) mathematics.
Also, what is $t$?  (see below)

since both $\theta$ and the polynomial[1] $\sum\limits_{i=1}^n{x_{i} t^{i-1}}$[2] are greater than or equal to $0$, we[3] must have
$\sum\limits_{i=1}^n{\theta x_{i} t^{i-1}} = \theta \sum\limits_{i=1}^n{x_{i} t^{i-1}} \ge 0$[4]

A couple of comments here:

I am a little uncomfortable calling the sum a polynomial, then declaring that the polynomial is nonnegative.  Generally, a polynomial is an abstract object, and $t$ is a formal variable.  Alternatively, you can see the sum as a polynomial function, which is being evaluated at a particular value of $t$.  I think that it would be better to just call it a "sum".
What is $t$?  Where have you declared that $t \in [0,1]$?  You say at the beginning that if $t\in [0,1]$, then other nice things happen.  This doesn't clearly describe every subsequent use of $t$ in the rest of the argument.
Again, drop the mathematical "we". :\
Again, display the math.  Also, sentences end with periods.

Thus $x \in K$, $\theta \ge 0$ $\rightarrow$ $\theta x \in K$.
Hence $K$ is a cone. $\square$

A lot of this is redundant.  Both $x$ and $\theta$ are defined above, and you have shown that $\theta x \in K$.  Done.  Everything else is superfluous.  I also rather object to the overuse of notation in an inline environment.  If you are going to write out all of these details, please write them out:  "Thus if $x \in K$ and $\theta \ge 0$, then $\theta x \in K$.  Therefore $K$ is a cone.

My presentation of your proof is as follows:

Proof:  Let $x = (x_1, \dotsc, x_n) \in K$ and fix $\theta > 0$.  By definition of $K$,
  $$ \sum_{j=1}^{n} x_j t^{j-1} \ge 0 $$
  for all $t \in [0,1]$.  Then
  $$ \sum_{j=1}^{n} (\theta x_j) t^{j-1}
= \theta \sum_{j=1}^{n} x_j t^{j-1}
\ge 0 $$
  for any $t\in [0,1]$, as both $\theta$ and the sum $\sum x_jt^{j-1}$ are nonnegative.  Therefore $\theta x = (\theta x_1, \dotsc, \theta x_n) \in K$.

I will note that in the second-to-last sentence, I have written $\sum x_jt^{j-1}$ (and have omitted the limits of summation).  This is an abuse of notation, but should be clear from context.  I have also replaced the indices of summation with $j$s throughout, as I like to reserve $i$ for the imaginary unit.  This is a completely pedantic and unnecessary change, but I think it makes my life better.


TITLE: Something wrong with this definition of absolute value function?
QUESTION [1 upvotes]: In Spivak's book on calculus, the absolute value function has been defined as:
$$|x| =
\left\{
 \begin{array}{ll}
  x  & \mbox{if } x \geq 0 \\
  -x & \mbox{if } x \leq 0
 \end{array}
\right.
$$
The use of equality in both the parts makes me uncomfortable. Up to now in my life, I've always seen this function defined as $-x$ if $x < 0$. As I see it, this only introduces ambiguity when $x=0$ (yes, one can argue that $0$ and $-0$ are the same thing, but why complicate matters? I'm wondering if this is a typo or something rigorous that is escaping me.

REPLY [0 votes]: The point is that $|x|$ is not differentiable at $x=0$. This is not a typo.


text:
\begin{document}

\maketitle

\begin{abstract}
We study $C^1$-generic diffeomorphisms with a homoclinic class with non
empty interior and in particular those admitting a codimension one
dominated splitting. We prove that if in the finest dominated
splitting the extreme subbundles are one dimensional then the
diffeomorphism is partially hyperbolic and from this we deduce that
the diffeomorphism is transitive.
\end{abstract}

\section{Introduction}

It is a main problem in generic dynamics to understand the structure of homoclinic classes, this has became most important after some results in \cite{BC} which raised the interest in the study of chain recurrence classes and in particular homoclinic classes which are, generically, the chain recurrence classes containing periodic points. This results, as most of the results (including the ones here presented) are in the $C^1$ category, very little is known about $C^r$ generic diffeomorphisms with $r>1$ (see \cite{Pujals}).

A lot is known in the case of a isolated homoclinic class $H(p,f)$
of a hyperbolic periodic point $p$ (i.e., it is maximal invariant in
a neighborhood) of a \textit{generic diffeomorphism $f$}, see for
instance  \cite{beyondhip}, chapter 10 (we remark here  that the
genericity of $f$ implies that the continuation $H(p_g,g)$ is also
isolated for $g$ in a neighborhood of $f$). The key point of knowing
that the class is isolated is that, after perturbation, orbits that
remains in a neighborhood must belong to the continuation of the
homoclinic class $H(p_g,g).$ However, if one does not know
\textit{apriori} that the homoclinic class is isolated (in other
words, the class might be wild, i.e., non isolated) only very sparse
results have been obtained (see for example \cite{ABCDW}
 where they prove that generic homoclinic classes are index complete). The main difficulty is to
 overcome the fact that after perturbations one cannot ensure that the perturbed points remain in the class.

For example, it is not known whether the non-wandering set of a $C^1$-generic diffeomorphism may have nonempty interior and not coincide with the whole manifold. We treat this problem in this paper solving it in some particular cases and also giving some results which may help to obtain a general solution. We remark that it is not difficult to construct examples of diffeomorphisms such that its non-wandering set has non empty interior and doesn't coincide with the whole manifold.

\subsection{Definitions and statement of results}

\smallskip

Let $M$ be a compact connected boundaryless manifold of dimension
$d$ and let $\diff 1$ be the set of diffeomorphisms of $M$ endowed
with the $C^1$ topology. We shall say that a property (or a diffeomorphism) is \emph{generic} if
and only if there exists a residual ($G_{\delta}$-dense) set $\Res$ of $\diff 1$ for
which for every $f \in \Res$ satisfies that property.

The main result of this paper concerns the following conjecture of
\cite{ABD}, we shall denote $\Omega(f)$ to the nonwandering set of $f$ :

\begin{conj}
 There is a residual set $\Res \en \diff 1$ such that if $f\in \Res$ and $int(\Omega(f))\neq \emptyset$ then $f$ is transitive.
\end{conj}

For a hyperbolic periodic point $p \in M$ of some diffeomorphism $f$
we denote its \emph{homoclinic class} by $H(p,f)$, defined as the
closure of the transversal intersections between the stable and
unstable manifolds of the orbit of $p$. For generic diffeomorphisms, if the nonwandering set has nonempty interior, then there is a homoclinic class with nonempty interior (see \cite{ABD} or \cite{BC}). So, the conjecture is reduced to the study of homoclinic classes with nonempty interior.

Some progress has been made towards the proof of this conjecture
(see \cite{ABD} and \cite{ABCD}), in particular, it has been proved
in \cite{ABD} that isolated homoclinic classes as well as homoclinic
classes admitting a strong partially hyperbolic splitting (we shall define this concept later) verify the
conjecture. Also, they proved that a homoclinic class with non empty
interior must admit a dominated splitting (see Theorem 8 in
\cite{ABD}). In \cite{ABCD} the conjecture was proved for surface
diffeomorphisms.

In \cite{ABD} the question about whether within the finest dominated
splitting the extremes subbundles should be volume hyperbolic was
posed. We give a positive answer when the class admits codimension
one dominated splitting. This gives also new situations where the
above conjecture holds and weren't known.

Let us recall the definition of dominated splitting: a compact set
$H$ invariant under a diffeomorphism $f$ admits dominated splitting
if the tangent bundle over $H$ splits into two $Df$ invariant
subbundles $T_HM=E\oplus F$ such that there exist $C>0$ and
$0<\lambda<1$ such that for all $x\in H:$
$$\|Df^n_{/E(x)}\|\|Df^{-n}_{/F(f^n(x))}\|\le C\lambda^n$$

\noindent we say in this case that $F$ dominates $E$.

Let us remark that Gourmelon (\cite{GourmelonAdaptada}) proved that
there  always exists an adapted metric for which $C=1.$

We shall say that a bundle $F$ is uniformly expanding (contracting) if there exists $n_0<0$ ($n_0>0$) such that $\|Df^{n_0}_{/F(x)}\|< 1/2$ $\forall x \in H$.

The main theorem of this paper is the following

\begin{teo}\label{maintheorem}
 Let $f$ be a generic diffeomorphism with a homoclinic class $H$ with non empty interior
 and
 admitting a codimension one dominated splitting $T_H M=E \oplus F$
 where $dim(F)=1$. Then, the bundle $F$ is uniformly expanding for $f$.
\end{teo}

As a consequence of our main theorem we get the following easy corollaries.

Recall that a compact invariant set $H$ is strongly partially hyperbolic if it admits a three ways dominated splitting $T_HM= E^s \oplus E^c \oplus E^u$ (that is, $E^s\oplus E^c$ is dominated by $E^u$ and $E^s$ is dominated by $E^c\oplus E^u$), where $E^s$ is non trivial and uniformly contracting and $E^u$ is non trivial and uniformly expanding.

\begin{cor}\label{Cor1} Let $H$ be a homoclinic class with non empty interior
for a generic diffeomorphism $f$ such that $T_HM= E^1 \oplus E^2
\oplus E^3$ is a dominated splitting for $f$ and
$dim(E^1)=dim(E^3)=1$. Then, $H$ is strongly partially hyperbolic and $H=M$.
\end{cor}

\dem{} The class should be strongly partially hyperbolic because of
the previous theorem (applied to $f$ and to $f^{-1}$).  Corollary 1
of \cite{ABD} (page 185)  implies that $H=M$. \lqqd

We say that a homoclinic class $H$ is far from tangencies if there
is a neighborhood of $f$ such that there are no homoclinic
tangencies associated to periodic points in the continuation of $H$. The tangencies are of index $i$ if they are associated to a periodic point of index $i$, that is, its stable manifold has dimension $i$.
We  get the following result following the generalization of the results of \cite{W} in \cite{Gou2} (see also
\cite{ABCDW}):

\begin{cor} Let $H$ be a homoclinic class with non empty interior
for a generic diffeomorphism $f$ such that $H$ is far from tangencies of index $1$ and $n-1$ and has index $1$ and
$n-1$ periodic points. Then, $H=M$.
\end{cor}

\dem{} Since the class is far from tangencies, and the classes for
generic diffeomorphisms either  coincide or are disjoint (see
\cite{BC}) we have that using \cite{Gou2} the class must admit a
dominated splitting with one dimensional extremal subbundles (see
also \cite{ABCDW} Corollary 3), thus, by using Corollary \ref{Cor1}
we get the result.

\lqqd

In fact, the previous corollary can be compared to a corollary of a new result of Yang (\cite{Y}) on Lyapunov stable homoclinic classes far from tangencies. Yang's result (Theorem 3 in \cite{Y}) implies that a generic homoclinic class with nonempty interior and far away from (any) tangencies must be strongly partially hyperbolic or contained in the closure of the set of sinks and sources, and thus (using the results of \cite{ABD}) the whole manifold.

Incidentally, we also give a new proof in the two dimensional case:

\begin{cor}
Let $f$ be a generic surface diffeomorphism having a homoclinic
class with nonempty interior. Then $f$ is conjugated to a linear Anosov diffeomorphism in $\T^2$.
\end{cor}

\dem{} Since the class must admit dominated splitting (Theorem 8 of \cite{ABD}), this should be into 2 one dimensional subbundles. So, the class must be hyperbolic and thus, since the conjecture holds for hyperbolic homoclinic classes $f$ is Anosov (the rest follows from classical theory of Anosov diffeomorphisms).
\lqqd

\subsection{Idea of the proof}

\smallskip

The idea of the proof is the following.

First we prove that if the homoclinic class has interior, the
periodic points in the class (which are all saddles) should have
eigenvalues (in the $F$ direction) exponentially (with the period)
far from $1$. Otherwise we manage to obtain a sink or a source
inside the interior of the class and thus contradicting the fact
that the interior of the homoclinic class for generic
diffeomorphisms is, roughly speaking, robust (Theorem 4 of
\cite{ABD}).

Then, using the previous fact and some results of \cite{pablito} and
\cite{integrability} we manage to prove that  the center manifolds
integrating  a one dimensional extreme subbundle  should have nice
dynamical properties. For this we also use the connecting lemma for
pseudo orbits of \cite{BC}.

Finally, in the event that the extreme subbundle is not hyperbolic,
we manage to obtain (using dynamical properties and a Lemma of Liao)
periodic points near the class with bad contraction or expansion in
those extreme subbundles. Using Lyapunov stability of the homoclinic
class (which is generic, see \cite{ABD} and \cite{CMP}) we  ensure
that the periodic points we found belong to the class and thus reach
a contradiction.

\medskip

\textit{Acknowledgements:} We would like to thank Andr\'es Sambarino for
motivating us to study this problem. Also, we thank Sylvain Crovisier for reading a draft, explaining us Yang's results and \cite{C}.
Finally, we would like thank the referee for many valuable comments that helped improve the presentation of this paper.

\section{Preliminary results}

In this section we shall state some results we are going to use in
the proof of the main theorem. It can be skipped and used as
reference when the results are used.

Some generic properties of diffeomorphisms are contained in the
following Theorem (see \cite{ABD} and references therein):

\begin{teo}\label{propiedadesgenericas}
 There exists a residual subset $\Res$ of $\diff 1$ such that if $f \in \Res$
\begin{itemize}
 \item[a1)] $f$ is Kupka Smale (that is, all its periodic
  points are hyperbolic and their invariant manifolds intersect transversally).
 \item[a2)] The periodic points of $f$ are dense in the chain
 recurrent set of $f$(\footnote{The chain recurrent set is
 the set of points $x$ satisfying that for every $\eps>0$
 there exist an $\eps$-pseudo orbit form $x$ to $x$, that is,
 there exist points $x=x_0, x_1, \ldots x_k=x$, $k>0$ such that
 $d(f(x_i),x_{i+1})< \eps$.}). Moreover, homoclinic classes
 coincide with those chain recurrent classes which contain periodic points.
 \item[a3)] Every homoclinic class with non empty interior of
 $f$ is Lyapunov stable for $f$ and $f^{-1}$ (\footnote{Lyapunov
 stability of $\Lambda$ means that $\forall U$ neighborhood of
 $\Lambda$ there is $V\en U$ neighborhood of $\Lambda$ such that
 $f^n(V) \en U$ $\forall n\geq 0$.}). This implies that the stable and unstable set of any point in the class is contained in the class.
 \item[a4)] For every periodic point $p$ of $f$,
 $H(p,f) =\overline{W^s(p)} \cap \overline{W^u(p)}$.
  \item[a5)] Given a homoclinic class $H$ of a periodic point $p$, if $U$ is an
 open set such that $\overline{U} \en int(H)$
then there exists $\U$ neighborhood of $f$ such that for every $g
\in \U \cap \Res$, $U \en H(p_g,g)$ is satisfied (where $p_g$ is the
continuation of $p$ for $g$).
  \item[a6)] Homoclinic classes vary continuously with the Hausdorff distance with respect to $f$. This means, that given $p\in H$ a periodic point and $\eps>0$, there exists $\U$ a neighborhood of $f$ such that for every $g\in \U$, the homoclinic class of the continuation of $p$ lies within less than $\eps$ from $H$ in the Hausdorff distance.
\end{itemize}
\end{teo}

To obtain dynamical properties of the center manifolds we shall use
the following results from \cite{pablito} and \cite{integrability}.
First recall that if $T_H M= E \oplus F$ is a dominated
splitting then, Theorem 5.5 of \cite{invariant} gives us a local
$f-$invariant manifolds $W^F_{\eps}$ tangent to $F$.

Local $f-$invariance means that $\forall \eps>0$ there exists
$\delta>0$ such that $f^{-1}(W^F_\delta(x)) \en W^F_\eps(f(x))$. Taking
$f^{-1}$ we have an analog for $E$.

\begin{teo}[Main Theorem of \cite{pablito}]\label{pablito}
Let $\Lambda$ a compact invariant set of a generic diffeomorphism
$f$ admitting a codimension one dominated splitting $T_\Lambda M=
E \oplus F$ with $dim(F)=1$. Assume that
$\overline{Per(f_{/\Lambda})}=\Lambda.$ Then, $\forall x \in \Lambda$ and
$\forall \eps>0$ there exists $\delta>0$ such that

\[ f^{-n}(W^F_{\delta}(x)) \en W^F_{\eps}(f^{-n}(x))  \ \ \forall n \geq
0 \]

In particular, $W^F_{\delta}(x) \en \{ y \in M \ : \ d(f^{n}(x),
f^n(y))\leq \eps \}$.
\end{teo}

Remark 2.8 of \cite{C} gives a similar result that is enough in our context.

If there is a dominated splitting for $H$ of the form $T_H M=
E\oplus F$, then, there exists $V$, a neighborhood of $H$ such that
if a point $z$ satisfies that $f^n(z) \in V$ $\forall n \in \Z$ then
we can define the splitting for $z$ and it will be dominated (see
\cite{beyondhip}). Such a neighborhood will be called adapted.

If $I$ is an interval, we denote by $\omega(I) = \bigcup_{x\in I}
\omega(x)$, and by $W^{ss}_\eps(I)= \bigcup_{x\in I} W^{ss}_\eps
(x)$ its strong stable manifold. Also $\ell(I)$ denote its length.
We shall state the following result which is an immediate Corollary
of Theorem 3.1 of \cite{integrability} for generic dynamics.

\begin{teo}[\cite{integrability}]\label{pelado} Let $f\in \diff 1$ a
generic diffeomorphism and $\Lambda$ compact invariant set admitting
a codimension one dominated splitting  $T_\Lambda M= E  \oplus
F$ (where $dim(F)=1$). Then, there exists $\delta_0$ such that
if $I$ is an interval integrating the subbundle $F$ satisfying
$\ell(f^n(I))< \delta <\delta_0$ $\forall n\geq 0$ and that its
orbit remains in an adapted neighborhood $V$ of $\Lambda$, then, only
one of the following holds:

\begin{enumerate}
\item $\omega(I)$ is contained in the set of periodic points
of $f$ restricted to $V$ of $\Lambda$ one of which is an attractor.
\item $I$ is wandering (that is, $W^{ss}_\eps(f^n(I)) \cap W^{ss}_\eps(f^m(I))=
\emptyset$ for all $n\neq m$). This implies that $\ell(f^n(I)) \to
0$ as $|n|\to \infty$.
\end{enumerate}
\end{teo}

Other result we shall use is the following well known Lemma of
Franks:

\begin{teo}[Frank's Lemma \cite{frankslema}]\label{frankslema}
Let $f \in Diff^1(M)$. Given $\mathcal U (f)$, a $C^1$ neighborhood
of $f$, then $\exists \;\mathcal U_0 (f)$ and $\eps>0$ with the
following property: if $g \in \mathcal U_0 (f)$, $\theta= \{
x_1,\ldots, x_m\}$ and

\[ L:\bigoplus_{x_i \in \theta} T_{x_i}M \to \bigoplus_{x_i \in
\theta}T_{g(x_i)} M  \ \ \text{such that} \ \ \left\|L-
Dg|_{\bigoplus T_{x_i}M} \right\|< \eps \]

\noindent are given, then there exists $\tilde{g} \in \mathcal U(f)$
such that $D\tilde{g}_{x_i} = L|_{T_{x_i}M}$. Moreover if $R$ is a
compact set disjoint from $\theta$ we can consider $\tilde{g}=g$ in
$R \cup \theta$.
\end{teo}

Finally we state the following Lemma of Liao. A proof can be found
(with the same notation) in \cite{wen}. We shall state the result in
the particular case of index one dominated splitting with an adapted
metric (which always exist because of \cite{GourmelonAdaptada}), but
it holds in a wider context. Recall also that for linear maps $A_i$
in one dimensional spaces it holds that $\prod_i \|A_i\|= \|\prod_i
A_i \|$.

\begin{lema}[Liao \cite{liao}]\label{lemaliao} Let $\Lambda$ be a compact
invariant set of $f$ with dominated splitting $T_H M= E \oplus
F$ such that $\|Df_{/E(x)}\|\|Df^{-1}_{/F(x)}\|< \gamma$
$\forall x  \in \Lambda$ and $dim (F)=1$. Assume that
\begin{enumerate}
 \item There is a point $b\in \Lambda$ such that $\|Df^{-n}|_{F(b)}\| \geq 1$
 $\forall n\geq 0$.
 \item There exists $\gamma<\gamma_1<\gamma_2<1$ such that given $x\in \Lambda$
  satisfying $\|Df^{-n}_{/F(x)}\|\geq \gamma_2^n$ $\forall n\geq 0$ we have that
   there is $y \in \omega(x)$ satisfying $\|Df^{-n}_{/F(y)}\|\leq \gamma_1^n$
    $\forall n\geq 0$.
\end{enumerate}

Then, for any $\gamma_2<\gamma_3<\gamma_4<1$ and any neighborhood
$U$ of $\Lambda$ there exists a periodic point $p$ of $f$ whose
orbit lies in $U$, is of the same index as the dominated splitting
and satisfies $\|Df^{-n}_{/F(p)}\| < \gamma_4^n$ $\forall n\geq 0$
and $\|Df^{-n}_{/F(p)}\| \geq \gamma_3^n$ $\forall n \geq 0$.

\end{lema}

\section{Proof of the main theorem}

For $p\in Per(f)$, $\pi(p)$ denotes the period of $p$.

\begin{lema}\label{normadiferencial} Let $H$ be a homoclinic class
with interior of a generic diffeomorphism $f$ such that $T_HM=E\oplus F$ is a dominated splitting with $dim F=1$ . Then, there exists $\lambda<1$ such that for all $p \in Per(f|_H)$ the following holds:

$$\|Df_{/F(p)}^{-\pi(p)}\|\le \lambda^{\pi(p)}$$

\end{lema}

\dem{}
Arguing by contradiction assume that the conclusion does not hold, that is,
 for every $\lambda<1$ there exists $p \in
Per(f_{/H})$  such that $\|Df_{/F(p)}^{-\pi(p)}\|\ge
\lambda^{\pi(p)}$ which is equivalent to
$\|Df_{/F(p)}^{\pi(p)}\|\le \la^{-\pi(p)}$ since $F$ is one
dimensional. When the class is isolated this is enough since one can perturb the orbit in order to create a sink contradicting the isolation. Here, since the class may be wild, the creation of the sink represents no contradiction, so we must use the persistence of the interior given by generic property $a5)$ of Theorem \ref{propiedadesgenericas} and create a sink there.

Let $U$ be an open set such that $\overline{U}\en int(H)$. Since $f$
is generic, property $a5)$ of Theorem \ref{propiedadesgenericas}
ensure us the existence of a neighborhood $\U$ of $f$ such that for
every $g$ in a residual subset of $\U$ we have $U \en H_g$ ($H_g$ is
the continuation of $H$ for $g$, from $a5)$ of Theorem \ref{propiedadesgenericas} this continuation makes sense since it will be the only class containing $U$).

Frank's Lemma implies the existence of $\eps>0$ such that if we fix
an arbitrary finite set of points, we can perturb the diffeomorphism
as near as we want of those points obtaining a new diffeomorphism
with arbitrary derivatives ($\eps-$close to the originals) in those points and such that the diffeomorphism lies inside
$\U$.

Let us fix $1>\lambda> 1-\eps/2$  and let $p \in Per(f_{/H})$ as
before.
 Since $f$ is generic, the periodic points of
the same index as $p$ are dense in $H$ so, we can choose $q\in U
\cap Per(f)$ homoclinically related to $p$.

Let $x \in W^s(p)\cap W^u(q)$ and $y \in W^s(q) \cap W^u(p)$, we get
that the set $\Lambda= \mathcal O (p) \cup  \mathcal O (q) \cup
\mathcal O (x) \cup   \mathcal O (y)$ hyperbolic.

Consider  the following periodic pseudo orbit contained in
$\Lambda$, $$\{...,p, f(p),...,f^{N\pi(p)-1}(p), f^{-n_0}(y), \ldots
,f^{n_0}(y), f^{-n_0}(x), \ldots, f^{n_0}(x),p,...\}$$ which we
shall denote as $\wp^N$. Clearly, given $\beta>0$ there exists $n_0$
such that  $\wp^N$ is a $\beta$-pseudo orbit. At the same time, if
we choose $N$ large enough we  obtain a pseudo orbit which stays
near $p$ much longer than of $q$ and then inherit the behavior of
the derivative of $p$ rather than that of $q$.

The shadowing lemma for hyperbolic sets (see \cite{shub}) implies
that for every $\alpha>0$ there exists $\beta$ such that every
closed $\beta$-pseudo orbit is $\alpha$-shadowed by a periodic
point. So, let us choose $\alpha$ in such a way that the following
conditions are satisfied:

\begin{itemize}
\item[(a)] $B_{2\alpha}(q) \en U$.
\item[(b)] If $d(z,w)<\alpha$ and $x,y$ are
in an adapted neighborhood of $H$ then,
$$\frac{\|Df_{/F(z)}\|}{\|Df_{/F(w)}\|}< 1+c$$ ($c$ verifies
$(1+c)(1-\frac{\eps}{2})^{-1}< 1+\eps$).
\end{itemize}

Let $\beta<\alpha$ be given from  the Shadowing Lemma for that
$\alpha$ and let $n_0$ be such that $\wp^N$ is a $\beta$-pseudo
orbit. Therefore there exists a periodic orbit $r$ of period
$\pi(r)=N\pi(p)+4n_0$ which $\alpha$- shadows $\wp^N.$
Therefore, setting $k=\sup_{x\in M} \|Df_x\|$, we have
\begin{eqnarray*}
\|Df_{/F(r)}^{N\pi(p)+4n_0}\|&\le
&k^{4n_0}(1+c)^{N\pi(p)}\|Df_{/F(p)}^{\pi(p)}\|^N\\
&\le &
k^{4n_0}\left((1+c)(1-\frac{\eps}{2})^{-1}\right)^{N\pi(p)}<(1+\eps)^{\pi(r)}
\end{eqnarray*}
where the last inequality holds provided  $N$ is large enough.
Notice that the orbit of $r$ passes through $U.$ On the other hand,
by domination, we have that
$\|Df_{/E(r)}^{\pi(r)}\|<\|Df_{/F(r)}^{\pi(r)}\|.$ Since $E$
and $F$ are invariant we conclude that any eigenvalue of
$Df_r^{\pi(r)}$ is less than $(1+\eps)^{\pi(r)}.$

Now, if we compose in the orbit of $r$ its derivatives with
homoteties of value $(1+\eps)^{-1}$  we obtain, by using Frank's
Lemma, a diffeomorphism $g$ so that all the eigenvalues associated
to the periodic orbit $r$ are less than $1$, that is, $r$ is a
periodic attractor (sink). This contradicts the generic assumption,
since the sink is persistent, so every residual $\Res \in \U$ will
have diffeomorphisms with a sink near $r$, thus contained in $U$,
and thus contradicting that the interior is persistent.

\lqqd

\begin{lema}\label{variedadesdinamicamentedef}
Let $H$ be a homoclinic class with non empty interior for a generic
diffeomorphism $f$ such that $T_HM= E \oplus F$ is a dominated
splitting with $dim(F)=1$. Then, there exists $\eps_0$ such
that for all $\eps\le \eps_0$ there exists $\delta$ such that
$\forall x \in H$, $$W^F_{\delta}(x) \en W^{uu}_{\eps}(x):= \{ y \in M
: \ d(f^{-n}(x), f^{-n}(y))\leq \eps \ ; \ d(f^{-n}(x), f^{-n}(y))\to 0\}$$.
\end{lema}

\dem{} First we shall prove the Lemma for periodic points and then,
using this fact prove the general statement. Let $\eps_0>0$ such
that $B_{\eps_0} (H)$ is contained in the adapted neighborhood of
$H$ and such that if $d(x,y)<\eps_0$ then

\[\frac{\|Df^{-1}_{/F(x)}\|}{\|Df^{-1}_{/F(y)}\|} < \lambda^{-1} \]

\noindent where $\lambda$ is given by Lemma \ref{normadiferencial}.
Let $\eps\le \eps_0$ and let $\delta>0$ from Theorem \ref{pablito}
corresponding to this $\eps.$

Let $p \in Per(f_{/H})$ for which there is $y \in W^F_\delta(p)$ such
that $d(f^{-n}(y), f^{-n}(p)) \nrightarrow 0$. Since $W^F_\delta(p)$
is one dimensional, $W^F_\delta(p) \backslash \{p\}$ is a disjoint
union of two intervals. Denote $I_\delta$ the connected component of
 $W^F_\delta(p) \backslash \{p\}$ that contains $y.$ By Theorem
 \ref{pablito} we have either $f^{2\pi(p)}(I_\delta)\subset
 I_\delta$ or $f^{2\pi(p)}(I_\delta)\supset
 I_\delta.$ In any event, since $y\in I_\delta$ we conclude that
 there exits a point $z_0\in W_\eps^F(p)$ fixed under $f^{2\pi(p)}$
 and such that $\|Df_{/F(z_0)}^{2\pi(p)}\|\le 1.$

This contradicts the previous Lemma, since by the way  $\eps$ was
chosen we get (since  we know that $d(f^i(p), f^i(z_0))<\eps$ for
all $i$) that

\[ \|Df^{2\pi(p)}_{/F(p)}\| = \prod_{i=0}^{2\pi{p}-1} \|Df_{/F(f^i(p))}\|
< \lambda^{-2\pi(p)} \prod_{i=0}^{2\pi{p}-1} \|Df_{/F(f^i(z_0))}\|
=
\]\[= \lambda^{-2\pi(p)} \| Df^{2\pi(p)}|_{F(z_0)}\| < \lambda^{-2\pi(p)} \]

Now, lets prove the general statement. Let us suppose that for every
$\eps>0$ there exist $x\in H$ and a small arc $I\en W^F_\delta(x)$
containing $x$ such that $\ell(f^{-n}(I))\nrightarrow 0$. We know,
because of Theorem \ref{pablito} that $\ell(f^{-n}(I))\leq \eps$,
then, taking $n_j\to +\infty$ such that $\gamma\leq
\ell(f^{-n_j}(I))\leq \eps$ and taking limits, we obtain an arc $J$
integrating $F$ such that $\ell(f^{n}(J))\leq \eps$ $\forall n
\in \Z$ and containing a point $z \in J\cap H$ (a limit point of
$f^{-n_j}(x)$).

Now, we shall use Theorem \ref{pelado} to reach a contradiction. It
is not difficult to discard the first possibility in the Theorem
because it will contradict what we have proved for periodic points.

On the other hand, if $J$ is wandering, we know that it can not be
accumulated by periodic points. Since $f$ is generic, we reach a
contradiction if we prove that the points in $J$ are chain recurrent
(see property $a2)$ of Theorem \ref{propiedadesgenericas}). Theorem
\ref{pelado}, implies that, $\ell(f^n(J)) \to 0$ ($|n| \to
+\infty$), then, since $z \in H\cap J$, if we fix $\epsilon$, and $y
\in J$, then, for some future iterate  $k_1$ and a past one $-k_2$,
we know that $f^{k_1}(y)$ is $\epsilon$-near of $f^{k_1}(z)$ and
$f^{-k_2}(y)$ is $\epsilon$-near  $f^{-k_2}(z)$. Since homoclinic
classes are chain recurrent classes, there is an $\epsilon$ pseudo
orbit from $f^{k_1}(z)$ to $f^{-k_2}(z)$ and then, $y$ is chain
recurrent, a contradiction.

\lqqd

\begin{cor}  Let $H$ be a homoclinic class with non empty interior
for a generic diffeomorphism $f$ such that $T_HM= E \oplus F$ is
a dominated splitting with $dim(F)=1$. Then, $F$ is
uniquely integrable. \end{cor}

\dem{} It follows from the fact that the center stable manifold is
dynamically defined (see  \cite{invariant}). \lqqd

Uniqueness of the center manifolds imply that one can know that if you have a point $y\in W^F_\delta (x) \cap H$ then there exists $\gamma<\delta$ such that $W^F_\gamma (y)\en W^F_\delta (x)$.

\begin{cor}\label{vaacero} Let $H=H(p,f)$ be a homoclinic class with non empty interior
for a generic diffeomorphism $f$ such that $T_HM=E \oplus F$ is
a dominated splitting with $dim(F)=1$. Then, for all $L>0$
and $l>0$ there exists $n_0$ such that if $I$ is a compact arc
integrating $F$ whose length is smaller than $L$, then
$\ell(f^{-n}(I))< l $ $\forall n>n_0$.
\end{cor}

\dem{} It is easy to see that every compact arc integrating $F$
should have its iterates of length going to zero in the past because
of Theorem \ref{variedadesdinamicamentedef} (it is enough to
consider a finite covering of $I$ where the Theorem applies).

Lets suppose then that there exists $L$ and $l$ such that for every
$j>0$ there is an arc $I_j$ integrating $F$ of length smaller than
$L$ and $n_j>j$ such that $\ell(f^{-n_j}(I_j) \geq l$. We can
suppose without loss of generality that $\ell(I_j) \in (L/2,L)$.

Also, we can assume (maybe considering subsequences) that $I_j$
converges uniformly to an arc $J$ integrating $F$ and verifying
$L/2\leq \ell(J) \leq L$.

Since the length of $J$ is finite and it integrates $F$ we know
that $\ell(f^{-n}(J)) \to 0$ with $n\to +\infty$.

Let $\eps= l/2$ and $\delta$ given by Theorem \cite{pablito} which
ensures that $W^F_\delta(x) \en W^u_\eps(x)$ $\forall x$.

Let $n_0$ such that $\forall n \geq n_0$ we have $\ell(f^{-n}(J)) <
\delta/4$. Let also be $\gamma$ small enough such that if $x\in
B_{\gamma}(J)$ then $d(f^{-k}(x),f^{-k}(J))< \delta/4$ $\forall
0\leq k \leq n_0$.

Now, if we consider $j$ large enough (in particular $j>n_0$) such
that $I_j \en B_\gamma(J)$ we obtain $\ell(f^{-n_0}(I_j)) < \delta$
and so $\ell(f^{-n}(I_j)) < \eps < l$ $\forall n \geq n_0$, so,
$n_j< n_0$ which is a contradiction.

\lqqd

We are ready to give the proof of our main theorem:

\begin{teo} Let $H$ be a homoclinic class with non empty interior
for a generic diffeomorphism $f$ such that $T_HM= E \oplus F$ is
a dominated splitting with $dim(F)=1$. Then, $F$ is
uniformly expanding

\end{teo}

\dem{} Because of the existence of an adapted norm for the
dominated splitting (see \cite{GourmelonAdaptada}) we can assume
that $\|Df|_{E(x)}\|\|Df^{-1}|_{F(f(x))}\| < \gamma$ (for the
sake of simplicity).

Suppose the theorem is not true. Thus,  for every $0<\nu<1$ there
exists some $x\in H$ such that $\|Df^{-n}_{/F(x)}\| \geq \nu$,
$\forall n\geq 0$ (otherwise for every $x$ there would be some
$n_0(x)$ which would be the first one for which $\|Df^{-n}_{/F(x)}\|<
\nu$ and by compactness, the $n_0(x)$ are uniformly bounded, then $F$
would be hyperbolic). If we choose points $x_m$ satisfying
$\|Df^{-n}_{/F(x)}\|\geq 1 - 1/m$ $\forall n \geq 0$, a limit
point $x$ will satisfy $\|Df^{-n}_{/F(x)}\|\geq 1$ $\forall n \geq 0$.

First of all, we consider the case where we cannot use the Shifting
Lemma of Liao (Lemma \ref{lemaliao}). It is not difficult to see
that this implies (using Pliss' Lemma, see also \cite{wen}) that
$\forall \gamma <\gamma_1< \gamma_2 < 1$, there exists $x\in H$ such that

\[ \|Df^{-n}_{/F(x)}\| \geq \gamma_2^n \qquad \forall n \geq 0 \]

\noindent but, $\forall y \in \omega(x)$ we have that

\[ \|Df^{-n}_{/F(y)}\| \geq \gamma_1^n \qquad \forall n \geq 0 \]

So, if we work in $\omega(x)$ which is a closed invariant set, we
have that the subbundle $E$ will be hyperbolic since the dominated
splitting implies that $\forall z \in \omega(x)$

\[ \|Df_{/E(z)}\| < \frac{\gamma}{\|Df^{-1}_{/F(f(z))}\|}<
\frac{\gamma}{\gamma_1} < 1 \]

This implies that, since we have dynamical properties for the
manifolds integrating the subbundle $F$, that we can  shadow
recurrent orbits. Indeed, if we have a recurrent point $y \in
\omega(x)$, for every small $\eps$ (in particular, such that the
stable and unstable manifolds of $y$ are well defined) we can
consider $n$ large enough so that $d(f^n(y),y) \leq \eps/3$,
$f^n(W^E_\eps (y)) \en W^E_{\eps/3}(f^n(y))$ and
$f^{-n}(W^F_{\eps}(f^n(y))) \en W^F_{\eps/3}(y)$ which gives us
(using classical arguments) a periodic point $p$ of $f$ which
verifies that has period $n$ and remains $\eps$-close to the first
$n$ iterates of $y$. It is not difficult to see that we can consider
this periodic point to be of index $1$ and such that its stable
manifold intersects the unstable manifold of $y$. So, $p \in \overline{ W^u(H)}$
and using Lyapunov stability of $H$  we know
$\overline{W^u(H)} \en H$ (see \cite{CMP} Lemma 2.1), so $p \in H$.

Since $\gamma_1$ was arbitrary, we can choose it to satisfy
$\gamma_1>\lambda$ where $\lambda$ is as in Lemma
\ref{normadiferencial}. Also, we can choose $\eps$ small so that
$\|Df^{-n}_{/F(p)}\|> \lambda^n$ contradicting Lemma
\ref{normadiferencial}.

Now, we shall study what happens if Liao's shifting Lemma can be
applied. That is, there exists $\gamma<\gamma_1<\gamma_2<1$ such
that for all $x \in H$ satisfying

\[ \|Df^{-n}_{/F(x)}\| \geq \gamma_2^n \qquad \forall n \geq 0 \]

\noindent there exists $ y \in \omega(x)$ such that

\[ \|Df^{-n}_{/F(x)}\| \leq \gamma_1^n \qquad \forall n \geq 0 \]

So, using the Shifting Lemma we have that for every
$\gamma_2<\gamma_3< \gamma_4<1$ we have a periodic orbit $p_U$ of
$f$ contained in any neighborhood $U$ of $\Lambda$ and satisfying
that

\[ \|Df^{-n}_{/F(p)}\|\leq \gamma_4^n \]
\[ \|Df^{-n}_{/F(f^i(p))}\| \geq \gamma_3^n \]

\noindent for some $i \in 0, \ldots, \pi(p)$ (remember that $F$ is
one dimensional, so the product of norms is the norm of the
product). But since this periodic points are not very contracting in
the direction $F$, if we choose $\gamma_3 > \lambda$ (as before)
and $U$ sufficiently small to ensure that the stable manifold of
some periodic point will intersect the unstable one of a point in $H$
we reach the same contradiction as before.

\lqqd

config: arxiv


text:
\begin{document}

\maketitle

\begin{abstract}
The goal of this paper is to demonstrate the use of techniques from
hyperbolic geometry to compute generating sets of certain subgroups
of $SL^+(2,\mathbb{C})$; specifically, $SO^+(Q,\mathbb{Z})$ for $Q$
some integral quadratic form of signature $(3,1)$ that does not
represent $0$.
\end{abstract}

\section*{Introduction}

Hyperbolic space $\mathbb{H}^3$ can be thought of as the subset of
$\mathbb{R}^{4}$ that satisfies the equation $Q_1(\mathbf{x}) =
x_1^2+x_2^2+x_3^2-x_4^2=-1$ and the condition $x_4>0$, together with
the Riemannian metric induced by $Q_1$.  The group of isometries of
$\mathbb{H}^3$ is the space of $4$ by $4$ real matrices $A$ such
that $A^tQ_1A=Q_1$ and $e_4\cdot Ae_4>0$, where $e_4$ is the unit
vector in the $x_4$ direction.\footnote{The expression $A^tQ_1A=Q_1$
is using the fact that any quadratic form $Q$ can be written as a
symmetric matrix $\tilde{Q}$ such that
$Q(\mathbf{x})=\mathbf{x}^t\tilde{Q}\mathbf{x}$.  Throughout this
paper, the same notation will be used for a quadratic form and its
associated matrix.} It is denoted $SO^+(Q_1,\mathbb{R})$. This group
contains the discrete subgroup $SO^+(Q_1,\mathbb{Z})$ of integer
valued matrices.

Now consider a different quadratic form on $\mathbb{R}^4$; we will
be particularly interested in
\[Q_7(x_1,x_2,x_3,x_4)=x_1^2+x_2^2+x_3^2-7x_4^2.\]

Over $\mathbb{R}$, $Q_7$ and $Q_1$ are isomorphic; that is, there
exists an isometry $M:\mathbb{R}^4\rightarrow\mathbb{R}^4$ such that
$M^tQ_7M=Q_1$.  Therefore, $SO^+(Q_1,\mathbb{R})$ is isomorphic to
$SO^+(Q_7,\mathbb{R})$. However, there is no reason for there to be
an isomorphism between $SO^+(Q_7,\mathbb{Z})$ and
$SO^+(Q_1,\mathbb{Z})$.  What is true is that $SO^+(Q_7,\mathbb{Z})$
does act on $\mathbb{H}$ as a subgroup of
$SO^+(Q_7,\mathbb{R})=SO^+(Q_1,\mathbb{R})$.

All of the above is true for any integral quadratic form of
signature $(3,1)$.  Why, then, focus on the form $Q_7$?  The
following theorem sheds some light on this.

\begin{thm}\cite{Borel}.
$Q(\mathbf{x})=0$ has a non-zero integral solution if and only if
$\mathbb{H}^3/SO^+(Q,\mathbb{Z})$ is not compact.
\end{thm}

$Q_7$ does not have such a solution.  To see this, assume it does,
and furthermore, assume that the coordinates of $\mathbf{x}$ have no
common factor (no loss of generality because a common factor can
just be pulled out). Then consider the above equation mod 8.
\[0\equiv x_1^2+x_2^2+x_3^2-7x_4^2\equiv x_1^2+x_2^2+x_3^2+x_4^2
\mod 8\] However, mod 8, there are only three possible squares: 0, 1
and 4. In the above equation, none of the summands can be 1, because
then the rest would have to also be 1 to satisfy the equation mod 4,
which is not a solution mod 8.  Therefore, all the summands are
either 0 or 4, meaning $x_1$, $x_2$, $x_3$ and $x_4$ are all even,
contradicting the assumption of having no common factor.

Thus, the quotient orbifold $\mathbb{H}^3/SO^+(Q_7,\mathbb{Z})$ is
compact, and the goal now is to use this to understand the algebraic
structure of $SO^+(Q_7,\mathbb{Z})$.

It is important to note that the techniques used here to compute a
generating set for this group can be used to find similar generating
sets for $SO^+(Q_n,\mathbb{Z})$, with $Q_n=x_1^2+x_2^2+x_3^2-nx_4^2$
and $Q_n(\hat{x})=0$ having no non-zero solutions\footnote{The proof
that $Q_7$ was such a form also proves that $Q_n$ will also work,
for any $n=8m-1$, for $m>0$.}.  The basic algorithm will, in fact,
work for any integral quadratic form of signature $(3,1)$ which
yields a compact quotient orbifold, although the details will
require some work.

\section*{Dirichlet Fundamental Domains}

To further understand $\mathbb{H}^3/SO^+(Q_7,\mathbb{Z})$, the next
step will be to create a fundamental domain for it.  The fundamental
domain of choice will be a Dirichlet fundamental domain.

Constructing a Dirichlet fundamental domain $D$ for the action of a
group $G$ begins with choosing a point $x$ in hyperbolic space.
Then, determine two open sets, $U$ and $V$.

Let $U$ be a fundamental domain for the stabilizer of $x$.  In the
disc model, if the point $x$ is the central point, then the
stabilizer consists entirely of rotations, and so the fundamental
domain $U$ can be taken to be a cone.

Let $V$ be the set of points in hyperbolic space that are closer to
$x$ than any other point in $Gx$, the orbit of $x$.  This is often
called the 'Voronoi cell' of $x$. In practice, Voronoi cells can be
constructed in the following way. Let $A\in G$, and let $H_A$ denote
the half-space consisting of points that are closer to $x$ than to
$Ax$.  Then, the Voronoi cell is the intersection of $H_A$ over all
$A\in G$.

Then, the Dirichlet fundamental domain $D$ is $U\cap V$.

When the quotient orbifold is compact, a Dirichlet fundamental
domain can be computed algorithmically.

\textbf{Step 1)} Choose a fundamental domain $U$ for the stabilizer
of $x$.

\textbf{Step 2)} Produce $G_L$, the set of elements of $G$ that move
the point $x$ less than some distance $L$.

\textbf{Step 3)} Let $V_L$ be the intersection of $H_A$ over all
$A\in G_L$. If $V_L$ is compact, then there is some distance $M$
such that no point in $V_L$ is more than $M$ away from $x$.

\textbf{Step 4)} If $M<L/2$, then $V_L$ is the same as $V$, the
Voronoi cell of all of $G$, because $V_L\subset H_A$ for all $A$
that move $x$ more that $2M$. If $M\geq L/2$ or $V_L$ is not
compact, increase $L$ and do it again.

This will work eventually, because it must work when $L$ is greater
than the diameter of the orbifold.

How does this look for $\mathbb{H}^3/SO^+(Q_7,\mathbb{Z})$?  The
point $x$ will be $(0,0,0,7^{-1/2})$, and its stabilizer consists of
$SO^+(3,\mathbb{Z})$ acting on the first 3 coordinates.  It is a
group of order 24, and consists of compositions of permutations and
reflections along axes which are orientation preserving.
Fortunately, their action on the disc model with $x$ in the center
is exactly the same as their action on the first 3 coordinates of
the hyperboloid model, and so creating a fundamental domain is easy.
We will take the fundamental domain $U$ to be the set of points in
the disc which have positive Euclidean inner-product with the
following three vectors:
\[v_1=(1,-1,0,0);\;\;\;v_1=(0,1,-1,0);\;\;\;v_2=(0,0,1,0)\]

\subsection*{Generating Elements of $SO^+(Q_7,\mathbb{Z})$.}

The hard part about producing the rest of the fundamental domain is
producing all elements of $SO^+(Q_7,\mathbb{Z})$ which move $x$ less
than some fixed distance. To do this, several facts about elements
of $SO^+(Q_7,\mathbb{Z})$ will be helpful. Let $A$ be an element of
$SO^+(Q_7,\mathbb{Z})$, and let $a_{ij}$ denote the $ij$th
coefficient of the matrix $A$.

\begin{fact} The first three columns of the matrix $A$ have pseudolength
\footnote{The term 'pseudolength' of a vector $\mathbf{x}$ in
$\mathbb{R}^4$ just means $Q_7(\mathbf{x})$.  The term emphasizes
the idea of $Q_7$ giving $\mathbb{R}^4$ a pseudometric.} 1, the last
column has pseudolength -7, and they are all pairwise orthogonal.
Furthermore, any integral matrix that satisfies these properties is
automatically in $SO(Q_7,\mathbb{Z})$.
\end{fact}

\begin{fact} Let $x$ be an element of $\mathbb{H}^3$.
The distance in $\mathbb{H}^3$ from $x$ to $Ax$ is $\cosh(a_{44})$,
where $a_{44}$ is the bottom right entry in the matrix.
\end{fact}

\begin{fact} The entries $a_{14}$, $a_{24}$ and $a_{34}$, the right
column except for the bottom right entry, are all divisible by $7$.
\end{fact}

\begin{fact} The entry $a_{44}$ is either 1 more or 1 less than a
multiple of 7.
\end{fact}

\begin{fact} The absolute values of $a_{41}$, $a_{42}$ and $a_{43}$, the
bottom entries save the bottom right entry, are all bounded above by
$a_{44}/\sqrt{7}$.
\end{fact}

\begin{proof}
1) This is just a different way of saying $A^tQ_7A=Q_7$.

3) $A$ satisfies the equation $A^tQ_7A=Q_7$. $(Q_7)^{-1}A^tQ_7A=1$,
and so $A^{-1}=(Q_7)^{-1}A^tQ_7$.  Given the simple nature of $Q_7$
written as a symmetric matrix, this has a simple form:
\[A^{-1}=\left(
           \begin{array}{cccc}
             a_{11} & a_{21} & a_{31} & -7a_{41} \\
             a_{12} & a_{22} & a_{32} & -7a_{42} \\
             a_{13} & a_{23} & a_{33} & -7a_{43} \\
             -a_{14}/7 & -a_{24}/7 & -a_{34}/7 & a_{44} \\
           \end{array}
\right)\] Since this matrix also consists of integers, the specified
entries are divisible by 7.

4) The right column has pseudolength $-7$, which means
$a_{14}^2+a_{24}^2+a_{34}^2-7a_{44}^2=-7$. Using their divisiblity
by 7, let $a_{14}=7\alpha$, $a_{24}=7\beta$ and $a_{34}=7\gamma$.
Then $49(\alpha^2+\beta^2+\gamma^2)-7a_{44}^2=-7$, and so
$7(\alpha^2+\beta^2+\gamma^2)=a_{44}^2-1=(a_{44}-1)(a_{44}+1)$.  $7$
must divide $a_{44}-1$ or $a_{44}+1$, proving the statement.

5) Since $A$ is the inverse of some other matrix in
$SO^+(q,\mathbb{Z})$, its entries satisfy the equation
$7(a_{41}^2+a_{42}^2+a_{43}^2)=a_{44}^2-1$.  Therefore,
$\max(|a_{41}|,|a_{42}|,|a_{43}|)\leq a_{44}/\sqrt{7}$.
\end{proof}

\subsection*{The Algorithm for Generating Such Elements.}

With these facts established, elements of $SO^+(Q_7,\mathbb{Z})$
which move $x$ less than some fixed distance $\cosh(L)$ can be
generated.

\textbf{Step 1)} Generate a list of pseudolength 1 vectors whose
fourth component is bounded above by $L/\sqrt{7}$.  This list is
finite since the sum of the squares of the first three coordinates
is bounded above by $L^2+1$.

\textbf{Step 2)} For every $0\leq n:=(7i\pm1)\leq L$ (for $i$ an
integer) find all the solutions of $(7a)^2+(7b)^2+(7c)^2-7n^2=-7$,
and correspondingly, all the possible right columns of elements of
$SO^+(Q_7,\mathbb{Z})$.  Again, there are only a finite number of
possibilities.

\textbf{Step 3)} For each possible right column found in Step 2,
compare it to every vector on the list from Step 1, to find the
vectors which are orthogonal to the chosen right column.  Of the
vectors which are orthogonal, compare them to each other and find
which pairs are orthogonal to each other.  Then, any triple of
vectors which are orthogonal to the chosen right column and each
other constitutes an element of $SO^+(Q_7,\mathbb{Z})$.

Once the elements have been generated, the half planes they generate
must be drawn, and compactness and maximum distance must be checked.

\section*{The Generators}

So what does this algorithm yield?  First, in the interest of coming
up with a generating set for $SO^+(Q_7,\mathbb{Z})$, we need a
generating set of the stabilizer of the distinguished point.  As was
mentioned before, this group is $SO^+(3,\mathbb{Z})$, which is known
to be isomorphic to $S_4$.

To see this, notice that this group is the symmetry group of the
cube, and that a symmetry of the cube is determined by its
permutation of the diagonals of the cube.  There are four diagonals
in a cube, and every permutation of the diagonals is possible.
Generators for this group as a subgroup of $SO^+(Q_7,\mathbb{Z})$
are:

\[(12)=\left(
  \begin{array}{cccc}
    0 & 1 & 0 & 0 \\
    1 & 0 & 0 & 0 \\
    0 & 0 & -1 & 0 \\
    0 & 0 & 0 & 1 \\
  \end{array}
\right),\;\;\;(1234)=\left(
  \begin{array}{cccc}
    0 & 0 & 1 & 0 \\
    0 & 1 & 0 & 0 \\
    -1 & 0 & 0 & 0 \\
    0 & 0 & 0 & 1 \\
  \end{array}
\right)\]

The above algorithm yields the following as a sufficient set of
generators.  There is some ambiguity in which matrices to choose, up
to multiplication by an element of the stabilizer.  These
particularly matrices have been chosen so that they move the
Dirichlet fundamental domain to an adjacent region.

\[A=\left(
  \begin{array}{cccc}
    -2 & -2 & 0 & 7 \\
    -5 & -2 & 0 & 14 \\
    0 & 0 & -1 & 0 \\
    -2 & -1 & 0 & 6 \\
  \end{array}
\right) A^{-1}=\left(
  \begin{array}{cccc}
    -2 & -5 & 0 & 14 \\
    -2 & -2 & 0 & 7 \\
    0 & 0 & -1 & 0 \\
    -1 & -2 & 0 & 6 \\
  \end{array}
\right)\]

\[B=\left(
  \begin{array}{cccc}
    0 & -1 & 0 & 0 \\
    -8 & 0 & 0 & 21 \\
    0 & 0 & -1 & 0 \\
    -3 & 0 & 0 & 8 \\
  \end{array}
\right) C=\left(
  \begin{array}{cccc}
    -4 & -3 & -2 & 14 \\
    -3 & -4 & -2 & 14 \\
    -2 & -2 & 0 & 7 \\
    -2 & -2 & -1 & 8 \\
  \end{array}
\right) B^{-1}=\left(
  \begin{array}{cccc}
    0 & -8 & 0 & 21 \\
    -1 & 0 & 0 & 0 \\
    0 & 0 & -1 & 0 \\
    0 & -3 & 0 & 8 \\
  \end{array}
\right)\]

\[D=\left(
  \begin{array}{cccc}
    -4 & -3 & -2 & 14 \\
    -9 & -4 & -4 & 28 \\
    -4 & -2 & -3 & 14 \\
    -4 & -2 & -2 & 13 \\
  \end{array}
\right) D^{-1}=\left(
  \begin{array}{cccc}
    -4 & -9 & -4 & 28 \\
    -3 & -4 & -2 & 14 \\
    -2 & -4 & -3 & 14 \\
    -2 & -4 & -2 & 13 \\
  \end{array}
\right)\]

\[E=\left(
  \begin{array}{cccc}
    -7 & -8 & 0 & 28 \\
    -8 & -7 & 0 & 28 \\
    0 & 0 & -1 & 0 \\
    -4 & -4 & 0 & 15 \\
  \end{array}
\right)\]

\[F=\left(
  \begin{array}{cccc}
    -10 & -12 & -10 & 49 \\
    -2 & -4 & -3 & 14 \\
    -3 & -4 & -2 & 14 \\
    -4 & -5 & -4 & 20 \\
  \end{array}
\right) F^{-1}=\left(
  \begin{array}{cccc}
    -10 & -2 & -3 & 28 \\
    -12 & -4 & -4 & 35 \\
    -10 & -3 & -2 & 28 \\
    -7 & -2 & -2 & 20 \\
  \end{array}
\right)\]

Notice that a left-right reflection of this list sends matrices to
their inverses.

Which of these matrices are necessary to generate
$SO^+(Q_7,\mathbb{Z})$?  The following relations demonstrate that
only $A$, $C$ and the stabilizer group are necessary.
\[B=(1234)^2(12)A^{-2}(12)\]
\[D=(1234)^3(12)C(1234)^3\]
\[E=A^{-1}(1234)^2(12)A^{-1}(12)\]
\[F=C(12)(1234)^2A^{-1}(12)\]

Therefore, the four matrices $(12)$, $(1234)$, $A$ and $C$ are
enough to generate the entire group $SO^+(Q_7,\mathbb{Z})$.


text:
\section{A criterion for the failure of super-rigidity}

The orbifoldization process from \autoref{Def_OrbifoldRiemannSurface} does not affect the kernel and cokernel of real Cauchy--Riemann operators.

\begin{definition}
  \label{Def_AssociatedOrbifoldCROperator}
  Let $(\Sigma,j)$ be a Riemann surface with a multiplicity function $\nu$.
  Given a complex vector bundle $E$ over $\Sigma$ and a real Cauchy--Riemann operator $\fd \co \Gamma(E) \to \Omega^{0,1}(\Sigma,E)$,
  set
  \begin{equation*}
    E_\nu \coloneq \beta_\nu^*E
    \qandq
    \fd_\nu \coloneq \beta_\nu^\star\fd
  \end{equation*}
  with $\beta_\nu^\star\fd \co \Gamma(E_\nu) \to \Omega^{0,1}(\Sigma_\nu,E_\nu)$ as in \autoref{Def_PullbackCauchyRiemannOperator}.
\end{definition}

\begin{prop}
  \label{Prop_KernelOfOrbifoldCROperator}
  If $(\Sigma,j)$ is a closed Riemann surface with a multiplicity function $\nu$,
  $E$ is a complex vector bundle over $\Sigma$, and
  $\fd \co \Gamma(E) \to \Omega^{0,1}(\Sigma,E)$ is a real Cauchy--Riemann operator,
  then
  \begin{equation*}
    \ker \fd_\nu \iso \ker \fd \qandq
    \coker \fd_\nu \iso \coker \fd.
  \end{equation*}
\end{prop}

\begin{proof}
  The pullback map $\beta_\nu^* \co \Gamma(E) \to \Gamma(E_\nu)$ induces an injection $\ker \fd \incl \ker \fd_\nu$.
  In fact, this map is an isomorphism.
  To see this, the following local consideration suffices.
  Let $x \in \Sigma$ and set $k \coloneq \nu(x)$.
  Define $\beta\co \bD \to \bD$ by $\beta(z) \coloneq z^k$.
  Choose a holomorphic chart $\phi\co \bD \to \Sigma$ with $\phi(0) = x$ and a trivialization of $E$ over $\phi(\bD)$.
  With respect to these $\fd$ and $\fd_\nu$ can be written as
  \begin{equation*}
    \fd = \delbar + \fn \qandq \fd_\nu = \delbar + \beta^*\fn
  \end{equation*}
  for some $\fn \in \Omega^{0,1}(\bD,\End_\R(\C^r))$.
  If $\tilde s \in C^\infty(\bD,\C^r)$ is $\mu_k$--invariant,
  then there is a bounded map $s \in C^\infty(\bD\setminus\set{0},\C^r)$ such that $\tilde s = s\circ \beta$.
  If $(\delbar + \beta^*\fn)\tilde s = 0$,
  then $(\delbar + \fn)s = 0$;
  hence, $s$ extends to $\bD$ by elliptic regularity.

  Since $\coker \fd \iso (\ker \fd^\dagger)^*$ and similarly for $\fd_\nu$,
  it suffices to produce an isomorphism $\ker \fd^\dagger \iso \ker \fd_\nu^\dagger$.
  The formal adjoints $\fd^\dagger \co \Omega^{1,0}(\Sigma,E^*) \to \Omega^{1,1}(\Sigma,E^*)$ and $\fd_\nu^\dagger \co \Omega^{1,0}(\Sigma_\nu,E_\nu^*) \to \Omega^{1,1}(\Sigma_\nu,E_\nu^*)$ are real Cauchy--Riemann operator acting on $(1,0)$--forms and locally of the form $\fd^\dagger = \delbar + \fn$ and $\fd_\nu^\dagger = \delbar + \beta^*\fn$.
  The pullback map $\beta_\nu^* \co \Omega^{1,0}(\Sigma,E) \to \Omega^{1,0}(\Sigma_\nu,E_\nu)$ induces an injection $\ker \fd^\dagger \incl \ker \fd_\nu^\dagger$.
  This map is an isomorphism by the following local consideration.
  If $\tilde s \in C^\infty(\bD,\C^r)$ is such that $\tilde s \,\rd z$ is $\mu_k$--invariant,
  then there is a map $s \in C^\infty(\bD\setminus\set{0},\C^r)$ such that $\tilde s = kz^{k-1}s\circ \beta$.
  If $(\delbar + \beta^*\fn)\tilde s = 0$,
  then $(\delbar + \fn)s = 0$ and a consideration of the Taylor expansion of $\tilde s$ shows that $s$ is bounded.
  Therefore, $s$ extends to $\bD$ and $\tilde s \ \rd z = \beta^*(s \,\rd z)$.
\end{proof}

This together with the discussion in \autoref{Sec_BranchedCoveringMapsAsOrbifoldCoveringMaps} leads to the following criterion for the failure of super-rigidity.

\begin{definition}
  \label{Def_MonodromyAroundX}
  Let $(\Sigma,j)$ be a Riemann surface with a multiplicity function $\nu$ and
  let $x_0 \in \Sigma\setminus Z_\nu$.
  For every $x \in Z_\nu$ there is a conjugacy class of a subgroup $\mu_{\nu(x)} \leq \pi_1(\Sigma_\nu,x_0)$,
  generated by the homotopy class of  a loop in $\Sigma\setminus Z_\nu$ based at $x_0$ which is contractible in $(\Sigma\setminus Z_\nu) \cup \set{x}$ and goes around $x$ once.
  If $\uV$ is a Euclidean local system on $\Sigma_\nu$,
  then its \defined{monodromy around $x$} is the representation $\mu_{\nu(x)} \to \O(V)$ induced by the monodromy representation.
\end{definition}

\begin{prop}
  \label{Prop_FailureOfSuperRigidity}
  Let $u \co (\Sigma,j) \to (M,J)$ be a simple $J$--holomorphic map.
  If $u$ is not super-rigid,
  then there are a multiplicity function $\nu \co \Sigma \to \N$ and
  an irreducible Euclidean local system $\uV$ on $\Sigma_\nu$ such that:
  \begin{enumerate}
  \item
    the monodromy representation of $\uV$ factors through a finite quotient of $\pi_1(\Sigma_\nu,x_0)$,
  \item
    $\uV$ has non-trivial monodromy around every point of $Z_\nu$, and
  \item
    the twist
    \begin{equation*}
      \fd_\nu^\uV \co \Gamma((Nu)_\nu\otimes\uV) \to \Omega^{0,1}(\Sigma_\nu, (Nu)_\nu \otimes\uV)
      \qwithq
      \fd \coloneq \fd_{u,J}^N
    \end{equation*}
  has non-trivial kernel.
  \end{enumerate}

\end{prop}

\begin{proof}
  Let $(\tilde\Sigma,\tilde j)$ be a closed Riemann surface and $\pi \co (\tilde\Sigma,\tilde j) \to (\Sigma,j)$ a non-constant holomorphic map such that $\tilde u \coloneq u\circ\pi$ is not rigid, that is: $\ker \fd_{\tilde u,J}^N$ is non-trivial.

  Let $\hat\pi \co (\tilde \Sigma_{\tilde \nu},\tilde j_{\tilde \nu}) \to (\Sigma_\nu,j_\nu)$ be the corresponding holomorphic covering map between orbifold Riemann surfaces constructed in \autoref{Prop_BranchedCoversOfRiemannSurfacesAsOrbifoldCovers}.
  Set $\fd \coloneq \fd_{u,J}^N$ and
  $\tilde\fd \coloneq \fd_{u\circ\pi,J}^N$.
  By
  \autoref{Prop_KernelOfOrbifoldCROperator},
  \autoref{Prop_PullbackGeneralizedNormalBundle_Pre}, and
  \autoref{Prop_Pullback=Twist},
  \begin{equation*}
    \ker \tilde\fd
    \iso
    \ker \tilde\fd_{\tilde \nu}
    \iso
    \ker \hat\pi^*\fd_\nu
    \iso
    \ker \fd_\nu^{\hat\pi_*\ubR}.
  \end{equation*}
  Therefore, $\ker \fd_\nu^{\hat\pi_*\ubR}$ is non-trivial.

  Since $\hat\pi_*\ubR$ decomposes into irreducible local systems,
  there is an irreducible local system $\uV$ such that $\ker \fd_\nu^\uV$ is non-trivial.
  Define the multiplicity function $\nu'\co \Sigma \to \N$ by
  \begin{equation*}
    \nu'(x) \coloneq
    \begin{cases}
      \nu(x) & \text{if $\uV$ has non-trivial monodromy around $x$} \\
      1 & \text{otherwise}.
    \end{cases}
  \end{equation*}
  $\uV$ descends to an irreducible local system $\uV'$ on $\Sigma_{\nu'}$ with non-trivial monodromy around every $x \in Z_{\nu'}$.
  By \autoref{Prop_KernelOfOrbifoldCROperator},
  $\ker\fd_{\nu'}^{\uV'} \iso \ker\fd_\nu^\uV$.
\end{proof}

The following index formula is the final preparation required for the proof of \autoref{Thm_SuperRigidity}.
Its proof is presented in \autoref{Sec_OrbifoldRiemannRoch}.

\begin{prop}
  \label{Prop_IndexOfNormalCROperatorOfBranchedCover}
  Let $(\Sigma,j)$ be a closed Riemann surface with a multiplicity function $\nu$,
  let $E$ be a complex vector bundle over $\Sigma$, and
  let $\fd\co \Gamma(E) \to \Omega^{0,1}(\Sigma,E)$ a real Cauchy--Riemann operator on $E$.
  If $\uV$ on $\Sigma_\nu$ is a Euclidean local system,
  then
  \begin{equation*}
    \ind \fd_\nu^\uV = \dim V \ind \fd - \rk_\C E \sum_{x \in Z_\nu}\dim(V / V^{\rho_x}).
  \end{equation*}
  Here $\rho_x$ denotes the monodromy of $\uV$ around $x$.
\end{prop}


TITLE: Is it correct that if (1) then (2)?
QUESTION [1 upvotes]: Is it correct that if
$$3 < a < b$$
then
$$a^b > b^a?$$

REPLY [1 votes]: Because $\ln$ is strictly increasing function, if we prove that $\ln a^b > \ln b^a$ we prove that $a^b > b^a$.  $\ln a^b > \ln b^a \rightarrow \frac{b}{\ln b} > \frac{a}{\ln a}$ which is true because $(\frac{x}{\ln x})'=\frac{\ln x -1}{\ln^2 x}>0$ if $x > e$


TITLE: Decide whether there exists an inversion that transforms one onto the other.
QUESTION [2 upvotes]: Question: For the following pairs of curves, Decide whether there exists an inversion that transforms one onto the other. Identify the inversion    if it exists.

The circles $x^2+y^2=16$ and $(x-34)^2+y^2=900$

The first circle with the equation $x^2+y^2=16$ shall be call $C$.
 $C$ is centered at the origin $(0,0)$ and has a radius of $4$.
The second circle with the equation $(x-34)^2+y^2=900$ shall be call $D$. $D$ is centered at $(34,0)$ and has a radius of $30$.
That is the amount of information I was able to obtain by my general knowledge of circles. Now, if anyone can help me determine how to know if there exist an inversion?

REPLY [2 votes]: Your setup is symmetric about the $x$ axis. So your circle of inversion has to have its center on that axis as well. Furthermore it has to map the pair of points where $C$ intersects the $x$ axis onto the pair where $D$ intersects the axis. Which means

either $(-4,0)\mapsto(4,0)$ and $(4,0)\mapsto(64,0)$
or $(-4,0)\mapsto(64,0)$ and $(4,0)\mapsto(4,0)$

An inversion is an involution, so applying the inversion twice will give the identity transformation. This means the first alternative is not an option, and the second one is the one you need. So you need a circle of inversion with center $(x,0)$ and radius $r$ which satisfies
$$(x+4)(x-64)=(x-4)^2=r^2$$
I believe you can take it from here.


TITLE: When did people start thinking of elliptic curves as groups?
QUESTION [28 upvotes]: I have been reading some old papers of Cassels and Selmer from around 1950, and they talk about generators of rational solutions to elliptic curves, in the sense of Mordell–Weil, but do not appear to use the word group.  (Edit: Taking another look, at least some of Cassels' papers from this period do use the word group.)
Weil - L'arithmétique sur les courbes algébriques (footnote 1, p. 281) says:

Afin de réserver le mot de groupe au sens qu'il a pris depuis Galois, je parlerai toujours de systèmes de points, bien qu'on air l'habitude en géométrie algebrique de parler de groupes de points sur une courbe.

Question: When did it become common parlance to call elliptic curves groups?

REPLY [18 votes]: The first mathematician who talked about groups of points on elliptic curves (in the sense of Galois, i.e., in the modern sense of the word group) was
Juel [Ueber die Parameterbestimmung von Punkten auf Curven zweiter
und dritter Ordnung. Eine geometrische Einleitung in die Theorie der
logarithmischen und elliptischen Funktionen, Math. Ann. 47 (1896), 72-104].
Poincare, in his important article referred to by ThiKu, does not use the concept of groups (this is the point of Schappacher's article). Even Mordell proved his theorem, namely that the group of rational points on an elliptic curve is finitely generated, without using the notion of a group: this was only done by Weil.
For a long time, most people interested in elliptic curves regarded them as
a variety of diophantine equations, and preferred thinking about secant and
tangent methods instead of group operations. Only when it became clear in
the 1960s that the rank of an elliptic curve could be computed by applying
all kinds of homomorphisms and determining the orders of kernels and images of such homomorphisms, the group theoretic point of view became indispensible.


TITLE: Show this assertion
QUESTION [0 upvotes]: I am stuck on the following task:
Fix a vector $y \in C^n$ and let $f(x,y)$ be a function defined on $C^n \times C^n$ such that $f(0,y)=1$ and its $n$th partial derivative on $x$ satisfies $\frac{\partial^n{f}(0,y)}{\partial{x}^n}=1.$  Show that $f$ has an expension  entiere series on $x$.
In addition,   have we the following $$\parallel f(x,y)\parallel \leq \exp(\parallel x \parallel \times \parallel y \parallel)$$?
Any help please and thanks in advance?

REPLY [0 votes]: Hint: Using Taylor's theorem aroung $0$, in fact, I assume that the function $h(x)=f(x,y)$ is differentiable at every point $x$ because $n$th derivative on $x$ is equal to $y^nf(x,y)$ and $f(0,y)=1$, then $$h(x)=\sum_{n=0} ^{\infty} \frac{h^{(n)}(0)}{n!}x^n =\sum_{n=0} ^{\infty} \frac{y^{n}f(0,y)}{n!}x^n .$$
For other hand, assuming $a_n=\frac{f^{(n)}(0,y)}{n!}$, is not easy to prove  $||\sum a_n x^n || < \sum ||a_nx^n||$ because $f^{(n)}(0,y)$ is not a number, it is in a Dual space, really, we have that $$\begin{array}{ccc}
f^{(n)}(0,y)\in & \underbrace{\mathcal{L}(\mathbb{R}^{2},\mathcal{L}(\mathbb{R}^{2},\cdots,\mathcal{L}(\mathbb{R}^{2}} & ,\mathbb{R})\cdots)).\\
 & n-\mbox{times}
\end{array}$$
Note that $\mathcal{L}\left(\mathbb{R}^{2},\mathcal{L}\left(\mathbb{R}^{2},\cdots\mathcal{L}\left(\mathbb{R}^{2},\mathbb{R}\right)\cdots\right)\right)$ and $\mathcal{L}\left(\mathbb{R}\times\cdots\times\mathbb{R},\mathbb{R}\right)$ are homeomorphic, this homeomorphism is linear and isometry.
Remark: I assume that the function $f$ is of real value.


TITLE: Getting stuck on difficult problems.
QUESTION [13 upvotes]: First, a little background: I hope to go to graduate school in mathematics, but for financial reasons I will be unable to go back to school any sooner than the fall of 2016. However, since I feel quite sure I would like to undertake further studies in math, and because I really enjoy the subject, I have been self-studying on my own in my free time.
The issue I am having is this: When I tackle a problem, I am very determined, and I don't mind spending a long time working on any given problem. But, sometimes I get stuck on a difficult problem, and I work on it for days, only to end up frustrated by not finding a solution. Then, I find it difficult to skip that problem and move on to other problems, because the skipped problem keeps nagging at me. I worked nearly all the problems in Blue Rudin, chapters 1-8, but there are a few problems I never got that are still bothering me. Sometimes, I think there must be some fact that I didn't absorb well enough for that particular problem, and it seems like a problem I "should" be able to solve.
Does anyone else feel frustrated or dismayed when there is a single problem in a chapter's exercises that you just can't seem to get no matter how long you work on it? During my undergrad math studies, I did very well, so I don't think it is that I have a lack of ability. Does this sort of thing even happen to great mathematicians? I feel that I should somehow come to terms with this experience, as I am sure it will keep coming up as I continue to study math. I would very much like to hear others' thoughts and experiences with this sort of thing.

REPLY [2 votes]: Look at the video here. If you do not speak French, use the subtitles. The state of being stuck is essential to getting your mind moving. Remember, rewiring your brain hurts!!
On a related note (again pointing to Connes), he says:
"I like to work on this problem because it is a test of myself that I cannot escape from. It’s not like you build a new theory, and then you can think you are the greatest. In mathematics, there is no better way to progress than to be confronted with a problem that you cannot solve. If you work on a problem that you can solve, it means that it’s not the right problem. Fighting with a very hard problem is a much better way to build a mental picture than when you are working on an easy problem. When the mind is blocked, it has much more power to build and to conceive. I see a problem like that as a gift."
This is in reference to working on the Riemann Hypothesis. Of course, Connes is more ready for that problem than the rest of us, but the idea holds weight.


TITLE: Subdivision of nonplanar graph is nonplanar?
QUESTION [0 upvotes]: If $G$ is a nonplanar graph, does it follow that every subdivision of $G$ is nonplanar?

REPLY [1 votes]: Yes, because Kuratowski's Theorem states that a graph $G$ is nonplanar if and only if it contains a subgraph $H$ that is a subdivision of $K_{3,3}$ or $K_5$.
If $G$ is nonplanar, let $H$ be such a subgraph. A subdivision $G'$ of $G$ induces a subdivision $H'$ of $H$; since $H$ is a subdivision of either $K_{3,3}$ or $K_5$, so is $H'$, and so $G'$ is nonplanar.
There is an alternative (basically identical) proof using Wagner's theorem.
A third, more direct proof: If $G' = (V',E')$ is a subdivision of $G = (V,E)$, and $G'$ is planar, then draw $G'$ without intersecting edges. Now remove the vertices in $V' \setminus V$ but keep the edges. Now you have a drawing of $G$ without intersecting edges, so $G$ is planar.


text:
\section{A Short Introduction to Asymptotic Freeness for Machine Learning}
\newcommand{\C}{{\mathbb{C}}}

\subsection{Comparison with Classical Probability Theory}
Asymptotic freeness is the vital notion in free probability theory.
In order to introduce asymptotic freeness to readers in the machine learning community, we explain the freeness by comparing free probability theory to classical probability theory.
We refer readers to \citep{Mingo2017free} or \citep{voiculescu1992free} for the detail.

\begin{table}[h]
\centering
\begin{tabular}{|l|l|l|l|l|l}
\hline
            & r.v.        &  moments  & for multiple r.v.s  & independence  \\ \hline
classical   &  $X$             &  $\E[X^k]$ &  joint distribution    & decomposition of joint distribution     \\
free        &  $A$             &  $\tr[A^k]$ &  joint moments &   decomposition of joint moments \\
\hline
\end{tabular}
\caption{Comparison of free and classical probability theory.}
\label{table:free-classical}
\end{table}

Firstly, consider a matrix $A \in M_M(\C)$. We denote by $A^*$ the adjoint matrix, that is, complex-conjugate transpose matrix, of $A$.  Assume that $A$ is self-adjoint, that is, $A^*=A$.
Then the spectral distribution, denoted by $\mu_A$,  is given by
\begin{align}
    \mu_A = \frac{1}{N}\sum_{\lambda \in \sigma(A) } \delta_{\lambda},
\end{align}
where $\sigma(A) = \{ \lambda \in \C \mid  A - \lambda I \text{ \ is invertible }  \} \subset \R$, and $I$ is the identity operator.
We emphasize that the spectral distribution is determined by its moments. That is,  a distribution $\mu$ is equal to $\mu_A$ if and only if
\begin{align}\label{align:moments-mat}
     \tr(A^k)  =   \int x^k \mu(dx)  \ (k \in \N),
\end{align}
where $\tr$ is the normalized trace so that $\tr(I) = 1$.
In other words, the family of moments  $\tr(A^k)$  $(k \in \N)$ has the same information as the spectral distribution $\mu_A$.
Now consider  a counterpart of the spectral distribution at classical probability theory.
Let $X$ be a real random variable and $\mu_X$ be the distribution of $X$.
If $\mu_X$ is compactly supported, the distribution $\mu_X$ is determined by the moments:
\begin{align}\label{align:moments-rv}
    \E[X^k] = \int x^k \mu_X(dx) \ (k \in \N).
\end{align}
By comparing \eqref{align:moments-mat} and \eqref{align:moments-rv}, we see that the self-adjoint operator $A$ corresponds to a real random variable and the spectral distribution $\mu_A$ corresponds to the distribution of the random variable.

Secondly, consider multiple matrices.
Let $A_1, A_2 \in M_M(\C)$ be non-commuting self-adjoint matrices.  As an example, consider the distribution of the sum of them. The spectral distribution of the sum $A_1 + A_2$ is determined by its moments
\begin{align}\label{align:moment-sum-mat}
\tr[ (A_1+A_2)^k ] =  \sum_{i_1, i_2 ,\dots, i_k \in \{1,2\} } \tr[ A_{i_1} A_{i_2} \cdots A_{i_k} ]   \ (k \in \N).
\end{align}
Hence the trace of all words on $A_1$ and $A_2$ determines the spectral distribution $\mu_{A_1 + A_2}$.
Now consider its counterpart at classical probability theory. Let $X_1$ and $X_2$ be real random variables.  Then
\begin{align}\label{align:moment-sum-rv}
\E[ (X_1+X_2)^k ] =  \sum_{m=1}^k \begin{pmatrix} k \\ m\end{pmatrix} \tr[ X_1^m X_2^{k-m} ]   \ (k \in \N).
\end{align}
By comparing \eqref{align:moment-sum-mat} and \eqref{align:moment-sum-rv}, we need much more information to determine the distribution for non-commuting matrices  than for commuting real random variables.
Therefore, we extend the definition of the joint distribution in a different way from that in the classical probability theory.
For a family of matrices $(A_j)_{j \in J}$, its \emph{joint moments} are trace of all words in the family, which are given by
\begin{align}
    \tr[ A_{j_1} A_{j_2} \cdots A_{j_k}]    \  (j_1, j_2, \dots, j_k \in J, \ k \in \N).
\end{align}
Note that we use the joint moments as a counterpart of the joint distribution, which is a probability distribution in classical probability theory and does not exist for non-commuting matrices.

Lastly, consider the notion of independence.
The independence in the classical probability theory means that the joint distribution of multiple random variables decomposed to the product of marginal distributions of each random variable.
Since we consider the joint moments for multiple operators,  we extend the concept of independence as a decomposition law of joint moments to each operator's moments.
The freeness is one of the decomposition laws of joint moments (see \cref{table:free-classical}).

\subsection{Freeness}
Summarizing above, we formulate the free algebra, the tracial state,  and introduce the freeness.

\begin{defn}
A \emph{unital $*$-algebra over $\C$} is a nonempty set  $\mc{A}$ equipped with a triplet $( +, \cdot, *)$ satisfying  the following conditions.
\begin{enumerate}
    \item  $(\mc{A}, +, *)$ is an associative but possibly noncommutative algebra over $\C$, where $+$ is the addition, $\cdot$ is the multiplication, $0 \in \C$ is equal to the additive identity, and $1 \in \C$ is equal to the multiplicative identity.
    We omit the symbol $\cdot$ and simply write $x \cdot y$ as $xy$ for  $x,y \in \mc{A}$.
    \item  The map $* \colon \mc{A} \to \mc{A}$ satisfies the following:
    \begin{align}
        (xy)^* &= y^*x^*, \\
        (\alpha x + \beta y)^* &= \overline{\alpha} x^* + \overline{\beta}y^*,
    \end{align}
    for any $x,y \in \mc{A}$, $\alpha, \beta \in \C$, where $\overline{\alpha}$ is the complex conjugate of $\alpha$.
\end{enumerate}
\end{defn}

\begin{example}
The matrix algebra $M_n(\C)$ of $n \times n$ matrices over $\C$ is a unital $*$-algebra for any $n \in \N$, where $A*$ is the adjoint matrix of  $A  \in M_n(\C)$, which is the $n \times n$ matrix obtained from $A$ by taking the transpose and the complex conjugate of the entries.
The matrix algebra $M_n(\R)$ of $n \times n$ matrices over $\C$ is a subalgebra of $M_n(\C)$ as an algebra over $\R$.  Now for $A \in M_n(\R)$,  $A^* = A^\top$.
\end{example}

\begin{example}
Let us denote by  $\C \langle Z_j \mid j \in J \rangle$   the free $\C$-algebra of indeterminates $(Z_j)_{j \in J}$, which is the algebra of polynomials of noncommutative variables given by the weighted sum of words as follows:
\begin{align}
    \C \langle Z_j \mid j \in J \rangle  = \{ \alpha_\emptyset 1 +  &\sum_{k =1}^\infty \sum_{j_1, \dots, j_k \in J } \alpha_{j_1, j_2, \dots, j_k} Z_{j_1} Z_{j_2}\dots Z_{j_k},\\  &\text{\  where $\alpha_* \in \C$ is zero except for finite number of indices } \} .
\end{align}
We introduce an adjoint operation $*$ on $\C \langle Z_j \mid j \in J \rangle$ using universality of the free product (see \citep{voiculescu1992free}) by
\begin{align}
    (\alpha_{j_1, j_2, \dots, j_k} Z_{j_1} Z_{j_2}\cdots Z_{j_k} )^* = \overline{\alpha_{j_1, j_2, \dots, j_k}}  Z_{j_k} \cdots Z_{j_2}Z_{j_1}.
\end{align}
\end{example}

Next, we introduce the tracial state, which is an abstracted notion of the normalized trace of matrices and the expectation operator to random variables.
\begin{defn}
A  \emph{tracial state} $\tau$ on a unital $*$-algebra $\mc{A}$ is a $\C$-valued map  satisfying the following conditions.
\begin{enumerate}
    \item  $\tau(1) = 1$.
    \item $\tau(\alpha a + \beta b) = \alpha \tau(a) + \beta \tau(b)$  $(a,b \in \mc{A}, \alpha,\beta \in \C)$.
    \item $\tau(a^*) = \overline{\tau(a)}$   $(a \in \mc{A})$.
    \item $\tau(a^*a) \geq 0$  $(a \in \mc{A})$
    \item $\tau(ab) = \tau(ba)$  $(a,b \in \mc{A})$.
\end{enumerate}
\end{defn}
The first condition is the normalization so that the total volume becomes $1$.
The last one is called the tracial condition.

\begin{defn}
A \emph{noncommutative probability space} (NCPS, for short) is a pair of a unital $*$-algebra $\mc{A}$ and a tracial state $\tau$ on $\mc{A}$.
\end{defn}

Here we have prepared to introduce the freeness. For readers' convenience, we introduce a simpler version of the freeness that is the minimum necessary to read the primary material.
\begin{defn}\label{defn:freeness} (Freeness)
Given two families $a = (a_j)_{j \in I}$  and $b = (b_j)_{j \in J}$ of elements in $\mc{A}$ are said to be $\emph{free}$ (or \emph{free independent} ) with respect to $\tau$ if the following decomposition of joint moments  follows:
For any $k \in \N$, any $p_1, p_2, \dots, p_k \in \C \langle X_i \mid i \in I \rangle $, and any $q_1, q_2, \dots, q_k \in \C \langle Y_j \mid j \in J \rangle$, it holds that
\begin{align}
    \tau[ p_1(a)q_1(b)p_2(a) q_2(b)  \cdots p_k(a) q_k(b)] = 0
\end{align}
if $\tau[p_m(a)]=\tau[q_m(b)] = 0$ ($m=1, 2, \dots, k$).
\end{defn}
See \citep{voiculescu1992free} for the full definition of the freeness for more general cases.

\begin{example}
Assume that  $a$ and $b$ are free elements in a NCPS ($\mc{A}, \tau$), that is, we consider the case $|I|=|J|=1$ in \cref{defn:freeness}. Write $x^\circ = x - \tau(x)$ for $x \in \mc{A}$.
Then
\begin{align}
\mathrm{Cov}(a,b):= \tau(ab) - \tau(a)\tau(b) = \tau(a^\circ b^\circ)  + \tau(a^\circ)\tau(b) + \tau(a) \tau(b^\circ) = 0 .
\end{align}
Here we use the freeness to eliminate the term $\tau(a^\circ b^\circ)$.
From this equation, we see that free variables are uncorrelated.
The difference between freeness and classical independence appears in the decomposition of higher moments such as $\tau(abab)$:
\begin{align}
    \tau(abab) = \V[a]\E[b]^2 + \E[a]^2\V[b] + \E[a]^2\E[b]^2,
\end{align}
where $\E[a]=\tau(a)$ and $\V[a]=\tau(a^2) - \tau(a)^2$. The decomposition rule is different from that in the classical probability given by  $\E[XYXY] = \E[X^2Y^2] = \E[X^2]\E[Y^2]$ for independent random variables $X$ and $Y$.
\end{example}

\subsection{Infinite Dimensional Approximation of Random Matrices}

Here we introduce the relation between freeness and random matrices.

\begin{defn}
Let $ A_i(M), B_j(M) \in M_M(\C)$ ($M \in \N, i \in I, j \in J$).
Then the families $(A_i)_{i \in I}$ and $(B_j)_{j \in J}$ are said to be \emph{asymptotically free} as $M \to \infty$ if there exists $\mc{A} ,\tau$, and $a_i \in \mc{A} \ (i \in I)$ and $b_j \in \mc{A} \ (j \in J)$  so that
\begin{align}
    &\lim_{M \to \infty}\tr(A_i(M)^k) = \tau(a_i^k)  \ (k \in \N, i \in I),\\
    &\lim_{M \to \infty}\tr(B_j(M)^k) = \tau(b_j^k)  \ (k \in \N, j \in J),\\
    &\text{and, $(a_i)_{ i \in I}$ and $(b_j)_{j \in J}$ are free.}
\end{align}
\end{defn}

Here we introduce a known result in free probability theory.
\begin{prop}[{\cite[Prop.\,3.5]{Hiai2000asymptotic}} ]\label{prop:asymptotic-free}
For each $M \in \N$, consider the following matrices.
Let $U(M)$ be random matrix uniformly distributed on  $M \times M$ unitary matrices (resp.\,orthogonal matrices).
Let $A(M)$ and $B(M)$ be complex (resp.\,real) self-adjoint random matrices independent of $U(M)$.
Assume that there exist two compactly supported distributions $\mu$ and $\nu$ such that the following limits hold almost surely.
\begin{align}
    \lim_{M\to\infty}\tr[A(M)^k]&=  \int x^k \mu(dx) \  (k \in \N), \\
    \lim_{M\to\infty}\tr[B(M)^k]&=  \int x^k \nu(dx) \  ( k \in \N).
\end{align}
Under the above conditions, it holds that $B(M)$ and $U(M)^* A(M) U(M)$ are asymptotically free as $M \to \infty$ almost surely.
Furthermore, when $A(M)$ and $B(M)$ are positive definite, then the limit distribution of $B(M)^{1/2}U(M)^*A(M)U(M)B(M)^{1/2}$ is the multiplicative free convolution $\mu \boxtimes \nu$.
\end{prop}

Note that random matrices $A(M)$ and $B(M)$ do not have to be independent in \cref{prop:asymptotic-free}.

\subsection{Application to the FIM}
Recall that the propagation of the conditional FIM is given by the following equation:
\begin{align}
\Hmltn_{\ell +1 } =   \hat{q}_{\ell}I + W_{\ell+1} D_{\ell}\Hmltn_{\ell}D_{\ell}W_{\ell+1}^\top.
\end{align}
Let $A_\ell := W_\ell^* H_\ell W_\ell  = \hat{q}_{\ell-1}I + D_{\ell-1}\Hmltn_{\ell-1}D_{\ell-1}$. Note that $W_\ell^*=W_\ell^\top$. Then, firstly, $W_{\ell}$ and $A_\ell$ are independent.
Secondly, recall that $D_\ell$ and $(W_\ell, W_\ell^*)$ are assumed to be asymptotic free (see Assumption~3.1). Thirdly, each $W_\ell$ is uniformly distributed on orthogonal matrices.
By the above conditions,  it holds that $W_\ell A_\ell W_\ell^*$ and $D_\ell^2$ are asymptotic free as the limit $M \to \infty$ by \cref{prop:asymptotic-free}.
Then the limit spectral distribution of $D_\ell H_\ell D_\ell = D_\ell W_\ell A_\ell W_\ell^* D_\ell$  is equal to $\mu_\ell \boxtimes  \nu_\ell$, where $\mu_\ell$ (resp\,$\nu_\ell$) is the limit spectral distribution of $H_\ell$ (resp.\,$D_\ell^2$).
Then we get the following desired recursive equation:
\begin{align}
    \mu_{\ell+1} = \affine{\sigma_{\ell+1}^2}{q_\ell}(\nu_\ell \boxtimes \mu_\ell).
\end{align}


TITLE: Group completion of $\mathbb{E}_{\infty}$-monoids via tensor products
QUESTION [9 upvotes]: $\newcommand{\K}{\mathrm{K}}$The abelian group completion functor $\K_0\colon\mathsf{CMon}\to\mathsf{Ab}$ satisfies
$$
\K_0(A)
\cong
\mathbb{Z}\otimes_{\mathbb{N}}A,
$$
naturally in $A\in\mathrm{Obj}(\mathsf{CMon})$, where

$\mathbb{Z}$ is the additive monoid of integers (i.e. $\K_0(\mathbb{N})$, the group completion of $\mathbb{N}$);
$\otimes_\mathbb{N}$ is the tensor product of commutative monoids.

Question. Does the $\mathbb{E}_{\infty}$-group completion functor $\K_0\colon\mathsf{Mon}_{\mathbb{E}_\infty}(\mathcal{S})\to\mathsf{Grp}_{\mathbb{E}_\infty}(\mathcal{S})$ similarly satisfies
$$\K_0(X)\cong QS^{0}\otimes_\mathbb{F}X,$$
where now

$QS^0$, the stabilization of $S^0$, is the $\mathbb{E}_{\infty}$-group completion of $\mathbb{F}=\coprod_{n\in\mathbb{N}}\mathbf{B}\Sigma_{n}$, the groupoid of finite sets and permutations;
$\otimes_{\mathbb{F}}$ is the tensor product of $\mathbb{E}_{\infty}$-spaces?

REPLY [9 votes]: Yes, for the same reason. Let me sketch a proof.
1- $QS^0\otimes X$ is group-complete. Indeed, its $\pi_0$ is $\mathbb Z\otimes \pi_0(X)$, and that's a group for the usual reasons. Another way to prove it is to prove that the shear map for $X\otimes Y$ is (the shear map of $X)\otimes Y$, which can be seen by noting that $\otimes$ commutes with coproducts and hence finite products in each variable.
2- There is a natural transformation $X\to QS^0\otimes X$ given by tensoring $\mathbb F\to QS^0$ by $X$, and this induces a natural transformation $X^{gp}\to QS^0\otimes X$.
3- Both sides commute with colimits (a colimit of grouplike $E_\infty$-spaces is grouplike so I don't have to worry about whether I'm talking about colimits in monoids or grouplike monoids), therefore to check that this map is an equivalence, it suffices to do so for $X= \mathbb F$, and for that one it is a tautology.
Another way to phrase this is to use the following sequence of natural equivalences (and using point 1- for the last one):
$X^{gp} = QS^0\otimes_{QS^0} X^{gp} = (QS^0\otimes_\mathbb F X)^{gp}= QS^0\otimes X$
The second natural equivalence comes from the fact that group completion is symmetric monoidal, and $(QS^0)^{gp}\simeq QS^0$.