text:
\section{Finite Theta Correspondence}

\subsection{Finite theta correspondence}
Let $(\bfG,\bfG')$ be a \emph{reductive dual pair} consisting of one orthogonal group and
one symplectic group defined over $\bff_q$.
By restricting the \emph{Weil character} with respect to a non-trivial character $\psi$ of $\bff_q$
to $G\times G'$,
we have a decomposition
\[
\omega^\psi_{\bfG,\bfG'}=\sum_{\rho\in\cale(G),\ \rho'\in\cale(G')}m_{\rho,\rho'}\rho\otimes\rho'
\]
where $m_{\rho,\rho'}\in\bbN\cup\{0\}$ and we define
\[
\Theta_{\bfG,\bfG'} =\{\,(\rho,\rho')\in\cale(G)\times\cale(G')\mid m_{\rho,\rho'}\neq 0\,\}.
\]
So now $\Theta_{\bfG,\bfG'}$ is regarded as a relation between $\cale(G)$ and $\cale(G')$.
We say that $\rho\otimes\rho'$ occurs in the \emph{$\Theta$-correspondence} if $(\rho,\rho')\in\Theta_{\bfG,\bfG'}$,
i.e., if $m_{\rho,\rho'}\neq 0$ .
For $\rho\in\cale(G)$,
we define
\[
\Theta_{\bfG'}(\rho) =\{\,\rho'\in\cale(G')\mid (\rho,\rho')\in\Theta_{\bfG,\bfG'}\,\}.
\]
From the definition, it is obvious that $\Theta$ is \emph{symmetric}, i.e.,
$\rho'\in\Theta_{\bfG'}(\rho)$ if and only if $\rho\in\Theta_{\bfG}(\rho')$.

For a classical group $\bfG$,
let $\bfG_n$ denote the group of split rank $n$ in the \emph{Witt series} of $\bfG$, i.e,
\[
\bfG_n=\begin{cases}
\Sp_{2n}, & \text{if $\bfG$ is symplectic};\\
\rmO^+_{2n},\rmO_{2n+1},\rmO^-_{2n+2} & \text{if $\bfG$ is orthogonal}.
\end{cases}
\]
The dual pair $(\bfG,\bfG')=(\bfG_n,\bfG'_{n'})$ is said to be \emph{in stable range} if $n\leq\frac{n'}{2}$.
It is well known that $\Theta$ is \emph{persistent}, i.e.,
for $\rho\in\cale(G_n)$, if $\Theta_{\bfG'_{n'}}(\rho)\neq\emptyset$ for some $n'$,
then $\Theta_{\bfG'_{n''}}(\rho)\neq\emptyset$ for any $n''\geq n'$.
Moreover, it is known that $\Theta_{\bfG'_{n'}}(\rho)\neq\emptyset$ if $n'\geq 2n$.

\subsection{Eta correspondence in stable range}\label{0308}
Now we recall the main result from \cite{gurevich-howe}.
Let $\rho'$ be an irreducible character of $\Sp_{2n'}(q)$.
A notion of the \emph{rank} of $\rho'$ is given in \cite{gurevich-howe}.
Let $\cale(\Sp_{2n'}(q))_{(k)}$ denote the set of irreducible characters of $\Sp_{2n'}(q)$ of rank $k$.

\begin{prop}[Gurevich-Howe]\label{0301}
Consider the dual pair $(\bfG,\bfG')=(\rmO_k^\epsilon,\Sp_{2n'})$ where $k\leq n'$.
Then
\begin{enumerate}
\item[(i)] For $\rho\in\cale(\rmO^\epsilon_k(q))$,
the set $\Theta_{\bfG'}(\rho)$ contains a unique element $\eta(\rho)$ of rank $k$,
and all other elements have ranks less than $k$.

\item[(ii)] The mapping $\rho\mapsto\eta(\rho)$ gives an embedding
$\eta\colon\cale(\rmO^\epsilon_k(q))\rightarrow\cale(\Sp_{2n'}(q))_{(k)}$.
\end{enumerate}
\end{prop}
The mapping $\eta\colon\cale(\rmO^\epsilon_k(q))\rightarrow\cale(\Sp_{2n'}(q))$
is called the \emph{$\eta$-correspondence} for the dual pair $(\rmO_k^\epsilon,\Sp_{2n'})$
in stable range.

\begin{cor}
Consider the dual pair $(\rmO_k^\epsilon,\Sp_{2n'})$ where $k\leq n'$.
Let $\rho'\in\cale(\Sp_{2n'}(q))$.
If $\rho'=\eta(\rho)$ for some $\rho\in\cale(\rmO^\epsilon_{k}(q))$,
then $\rho'$ does not occur in the $\Theta$-correspondence for any dual
pair $(\rmO_{k'}^\epsilon,\Sp_{2n'})$ with $k'<k$.
\end{cor}
\begin{proof}
The assumption $\rho'=\eta(\rho)$ means that $\rho'$ is of rank $k$ by the above proposition.
If $\rho'$ occurs in the $\Theta$-correspondence for a dual pair $(\rmO_{k'}^\epsilon,\Sp_{2n'})$ with $k'<k$,
then the rank of $\rho'$ is less than or equal to $k'$ by the above proposition
and we get a contradiction.
\end{proof}

\subsection{Theta correspondence of unipotent characters}\label{0207}
Let $\lambda=[\lambda_1,\ldots,\lambda_k]$ and $\mu=[\mu_1,\ldots,\mu_l]$ be two partitions.
We may assume that $k=l$ by adding some $0$'s if necessary.
Then we denote
\[
\lambda\preccurlyeq\mu\quad\text{ if \ }\mu_i-1\leq\lambda_i\leq\mu_i\text{ for each }i.
\]
The following lemma is from \cite{pan-finite-unipotent} lemma 2.15:

\begin{lem}\label{0309}
Let $\lambda=[\lambda_i]$ and $\mu=[\mu_i]$ be two partitions.
Then $\lambda^\rmT\preccurlyeq\mu^\rmT$ if and only if $\mu_{i+1}\leq\lambda_i\leq\mu_i$ for each $i$
where $\lambda^\rmT,\mu^\rmT$ denote the dual partitions of $\lambda,\mu$ respectively.
\end{lem}

\begin{lem}\label{0203}
Let $A=\{a_1,\ldots,a_m\}$ and $B=\{b_1,\ldots,b_{m'}\}$ be two $\beta$-sets.
\begin{enumerate}
\item[(i)] Suppose that $m'=m$.
Then $\Upsilon(A)^\rmT\preccurlyeq\Upsilon(B)^\rmT$ if and only if $b_1\geq a_1>b_2\geq a_2>\cdots >b_m\geq a_m$.

\item[(ii)] Suppose that $m'=m+1$.
Then $\Upsilon(A)^\rmT\preccurlyeq\Upsilon(B)^\rmT$ if and only if
$b_1>a_1\geq b_2>a_2\geq \cdots \geq b_m>a_m\geq b_{m+1}$.
\end{enumerate}
\end{lem}
\begin{proof}
The lemma follows from Lemma~\ref{0309} and the definition of $\Upsilon$ in (\ref{0209}) immediately.
\end{proof}

For two symbols $\Lambda$ and $\Lambda'$,
we write $\Upsilon(\Lambda)=\sqbinom{\lambda}{\mu}$ and $\Upsilon(\Lambda')=\sqbinom{\lambda'}{\mu'}$.
Now we define several relations on the set of symbols:
\begin{align}\label{0303}
\begin{split}
\calb^+ &=\{\,(\Lambda,\Lambda')\in\cals\times\cals
\mid\mu^\rmT\preccurlyeq\lambda'^\rmT,\ \mu'^\rmT\preccurlyeq\lambda^\rmT,
\ {\rm def}(\Lambda')=-{\rm def}(\Lambda)+1\,\};\\
\calb^- &=\{\,(\Lambda,\Lambda')\in\cals\times\cals
\mid\lambda^\rmT\preccurlyeq\mu'^\rmT,\ \lambda'^\rmT\preccurlyeq\mu^\rmT,
\ {\rm def}(\Lambda')=-{\rm def}(\Lambda)-1\,\},\\
\calb_{\Sp_{2n},\rmO^\epsilon_{2n'}} &=\calb^\epsilon\cap(\cals_{\Sp_{2n}}\times\cals_{\rmO^\epsilon_{2n'}})
\end{split}
\end{align}
where $\epsilon=\pm$.

For a dual pair $(\bfG,\bfG')$ of a symplectic group and an even orthogonal group,
it is known that the unipotent characters are preserved by the $\Theta$-correspondence,
i.e., if $(\rho,\rho')\in\Theta_{\bfG,\bfG'}$, then $\rho$ is unipotent if and only if $\rho'$ is unipotent.
So let $\omega^\psi_{\bfG,\bfG',1}$ denote the unipotent part of $\omega^\psi_{\bfG,\bfG'}$, i.e.,
$\omega^\psi_{\bfG,\bfG',1}$ is the intersection of $\omega^\psi_{\bfG,\bfG'}$ with
$\calv(G)_1\otimes\calv(G')_1$.
The following proposition is from \cite{pan-finite-unipotent} theorem 3.34:

\begin{prop}\label{0204}
Let $(\bfG,\bfG')$ be a reductive dual pair of a symplectic group and an even orthogonal group.
Then we have
\[
\omega^\psi_{\bfG,\bfG',1}
=\sum_{(\Lambda,\Lambda')\in\calb_{\bfG,\bfG'}}\rho_\Lambda\otimes\rho_{\Lambda'}
\]
\end{prop}

For $\Lambda\in\cals_\bfG$,
we will denote
\[
\Theta_{\bfG'}(\Lambda)=\{\,\Lambda'\in\cals_{\bfG'}\mid(\Lambda,\Lambda')\in\calb_{\bfG,\bfG'}\,\}.
\]
Therefore by the above proposition we have $\rho_{\Lambda'}\in\Theta_{\bfG'}(\rho_\Lambda)$ if and only if
$\Lambda'\in\Theta_{\bfG'}(\Lambda)$.

Suppose that $(\Lambda,\Lambda')\in\calb_{\bfG,\bfG'}$.
From the definition in (\ref{0303}),
we can describe the relation between $\Upsilon(\Lambda)$ and $\Upsilon(\Lambda')$ as follows:
\begin{enumerate}
\item Suppose that $\epsilon=+$.
\begin{enumerate}
\item Suppose that $\bfG=\rmO^+_{2n}$ and $\bfG'=\Sp_{2n'}$.
If $\Lambda$ is of defect $4d$ for some $d\in\bbZ$,
then $\Lambda'$ is of defect $-4d+1$.
Therefore, $\Upsilon(\Lambda)$ is a bi-partition of $n-4d^2$ and $\Upsilon(\Lambda')$ is a bi-partition of
$n'-2d(2d-1)$.

\item Suppose that $\bfG=\Sp_{2n}$ and $\bfG'=\rmO^+_{2n'}$.
If $\Lambda$ is of defect $4d+1$ for some $d\in\bbZ$,
then $\Lambda'$ is of defect $-4d$.
Therefore, $\Upsilon(\Lambda)$ is a bi-partition of $n-2d(2d+1)$ and $\Upsilon(\Lambda')$ is a bi-partition of
$n'-4d^2$.
\end{enumerate}
For both cases,
we have
\[
\|\Upsilon(\Lambda')\|=\|\Upsilon(\Lambda)\|+(n'-n)+2d.
\]
Therefore the Young diagram of $\Upsilon(\Lambda')_*$ can be obtained from the Young diagram of $\Upsilon(\Lambda)^*$
by removing $k$ boxes (for some $k\geq 0$) such that no two boxes are removing from the same column;
and the Young diagram of $\Upsilon(\Lambda')^*$ can be obtained from the Young diagram of $\Upsilon(\Lambda)_*$
by adding $k+(n'-n)+2d$ boxes such that no two boxes are adding to the same column.

\item Suppose that $\epsilon=-$.
\begin{enumerate}
\item Suppose that $\bfG=\rmO^-_{2n}$ and $\bfG'=\Sp_{2n'}$.
If $\Lambda$ is of defect $4d+2$ for some $d\in\bbZ$,
then $\Lambda'$ is of defect $-4d-3$.
Therefore, $\Upsilon(\Lambda)$ is a bi-partition of $n-(2d+1)^2$ and $\Upsilon(\Lambda')$ is a bi-partition of
$n'-(2d+1)(2d+2)$.

\item Suppose that $\bfG=\Sp_{2n}$ and $\bfG'=\rmO^-_{2n'}$.
If $\Lambda$ is of defect $4d+1$ for some $d\in\bbZ$,
then $\Lambda'$ is of defect $-4d-2$.
Therefore, $\Upsilon(\Lambda)$ is a bi-partition of $n-2d(2d+1)$ and $\Upsilon(\Lambda')$ is a bi-partition of
$n'-(2d+1)^2$.
\end{enumerate}
For both cases,
we have
\[
\|\Upsilon(\Lambda')\|=\|\Upsilon(\Lambda)\|+(n'-n)-1-2d.
\]
Therefore the Young diagram of $\Upsilon(\Lambda')^*$ can be obtained from the Young diagram of $\Upsilon(\Lambda)_*$
by removing $k$ boxes (for some $k\geq 0$) such that no two boxes are removing from the same column;
and the Young diagram of $\Upsilon(\Lambda')_*$ can be obtained from the Young diagram of $\Upsilon(\Lambda)^*$
by adding $k+(n'-n)-1-2d$ boxes such that no two boxes are adding to the same column.
\end{enumerate}
For $k\geq 0$, we define
\[
\Theta_{\bfG'}(\Lambda)_k =\begin{cases}
\{\,\Lambda'\in\Theta_{\bfG'}(\Lambda)\mid\|\Upsilon(\Lambda')_*\|=\|\Upsilon(\Lambda)^*\|-k\,\}, & \text{if $\epsilon=+$};\\
\{\,\Lambda'\in\Theta_{\bfG'}(\Lambda)\mid\|\Upsilon(\Lambda')^*\|=\|\Upsilon(\Lambda)_*\|-k\,\}, & \text{if $\epsilon=-$}.
\end{cases}
\]
Clearly, as explained above we have
\begin{equation}\label{0310}
\Theta_{\bfG'}(\Lambda)=\bigsqcup_{k\geq 0}\Theta_{\bfG'}(\Lambda)_k
\end{equation}
and it is known that $\Theta_{\bfG'}(\Lambda)_k=\emptyset$ if either $\epsilon=+$ and $k>\lambda_1$; or
$\epsilon=-$ and $k>\mu_1$.


TITLE: Can evolutionary biology processes be mathematically described?
QUESTION [1 upvotes]: Action movies are full of violent combat, pursuiting and dangerous Scenarios. I think that the Evolution of mankind lead to such a behavior. It may be that selection processes have lead to the fact that under some circumstances human beings can be very violent and do ignore most of risks of the Environment (e.g. large heights). I mention concretely this, because I have seen that a book author has claimed that mathematica Evolution theory would describe the Evolution of violence.
Can evolutionary processes be modelled with mathematics?
I think yes, we can maybe use stochastic process theory. Are there some scientific results about how Evolution can be mathematized? Maybe Fokker-Planck equations?

REPLY [2 votes]: The theory of population genetics summarises microevolution with the Price equation. Game theory can identify the evolutionarily stable strategies for behaviour.


TITLE: Checking whether the image of a smooth map is a manifold
QUESTION [9 upvotes]: I have a specific problem, but would also like to know how to tackle the general case. I will first state the genral question. Let $M$ be an embedded submanifold of $\mathbb{R}^n$ and let $F: \mathbb{R}^n \to \mathbb{R}^n $ be a smooth map. How do I go about checking whether $F(M)$ is a smooth embedded submanifold of $\mathbb{R}^n$ or not? The specific problem I have is the following :- Let $F:\mathbb{C}^2\to \mathbb{C}^2$ the map $(z_1,z_2) \mapsto (z_1+z_2,z_1z_2)$ and let $M$ be the unit sphere in $\mathbb{C}^2$, i.e., $\lbrace (z_1,z_2) : |z_1|^2 + |z|^2 = 1 \rbrace $. Is $F(M)$ an embedded submanifold of $\mathbb{C}^2$ (considered as $\mathbb{R}^4$)?

REPLY [9 votes]: The specific $F(M)$ is not a smooth submanifold. Here is an argument.
To simplify formulas, I renormalize the sphere: let it be the set of $(z_1,z_2)\in\mathbb C^2$ such that $|z_1|^2+|z_2^2|=2$ rather than 1. Then, as Gregory Arone pointed out, $F(M)$ is the set of $(b,c)\in\mathbb C^2$ such that the roots $z_1,z_2$ of the equation $z^2-bz+c$ satisfy $|z_1|^2+|z_2^2|=2$. I claim that it is not a smooth manifold near the point $p:=(2,1)\in F(M)$.
Let us intersect $F(M)$ with two planes: $\{b=2\}$ and $\{c=1\}$. If it is is a smooth submanifold, at least one of the intersections should be locally (near $p$) a 1-dimensional smooth submanifold of the respective plane (otherwise both planes are tangent to $F(M)$ at $p$, but this is impossible since they span the whole space). But this is not the case:
If $b=2$, the equation is $z^2-2z+c=0$, hence $z_1+z_2=2$, then $|z_1|+|z_2|\ge 2$ and therefore $|z_1|^2+|z_2|^2\ge 2$. Equality is attained only for $z_1=z_2=1$, thus the intersection is a single point $c=1$, not a 1-dimensional submanifold.
If $c=1$, the equation is $z^2-bz+1$, hence $z_1z_2=1$, then $|z_1|\cdot|z_2|=1$ and therefore $|z_1|^2+|z_2|^2\ge 2$. The equality is attained if and only if $|z_1|=|z_2|=1$, so $z_1$ and $z_2$ are conjugate to each other. The set of $b$'s for which this happens is the real line segment $[-2,2]$ which is not a submanifold near 2.


TITLE: Fractional Derivatives of Exponential Functions
QUESTION [3 upvotes]: I am studying the generalization/derivation if the $\alpha^{\mbox{th}}$ derivative of $e^{ax}$. I got lost in the third line. Could someone please please fill in the missing lines so that the derivation will be in detail? I really need to figure this thing out.
With $\displaystyle D_x^n f(x)=\lim_{h\to 0}h^{-n}\sum_{m=0}^n(-1)^m{}_nC_m f(x+(n-m)h)$ where $_nC_m=\frac{n!}{m!(n-m)!}$ and with $f(x)=e^{ax}$,
\begin{array}{rcl}
\displaystyle D_x^\alpha e^{ax}&=& \lim_{h\to 0} h^{-\alpha}\sum_{n=0}^\alpha(-1)^n{}
_\alpha C_n e^{a(x+(\alpha-n)h)} \\
&=& e^{ax}\lim_{h\to 0} h^{-\alpha}\sum_{n=0}^\alpha(-1)^n{}_\alpha C_n (e^{ah})^{\alpha-n} \\
&=&  e^{ax}\lim_{h\to 0} h^{-\alpha} (e^{ah}-1)^\alpha \mbox{what happened???} \\
&=& a^\alpha e^{ax}
\end{array}
I didnt get how the third line came up from the second line. Can you plase fill in the missing details for me? Thank you.

REPLY [0 votes]: The formula you used is not correct. The correct formula is the Grunwald-Letnikov derivative.
I cannot write it in this word processor. Send me a mail and I'll send you a paper with the solution.
mdo@fct.unl.pt


TITLE: Converting cos(x-30) to two terms
QUESTION [0 upvotes]: I have a worked example in front of me for a particle kinetics and kinematics question. In the working it has a an equation:
$$0.6=\frac{V_0\sin(60)+V^`cos(x-30)}{10cos(30)}$$
And on the next line they have:
$$5.196=0.866V_0 + 0.866V^`cos(x)+0.5V`sin(x)$$
I can't seem to figure out how they got $$0.866V^`cos(x)+0.5V`sin(x)$$ from $$V^`cos(x-30)$$
Am I missing some information from somewhere else in the question, or is there some sort of logic to this transformation?

REPLY [2 votes]: $\cos(A+B)=\cos A \cos B -\sin A \sin B$
Source: http://en.wikipedia.org/wiki/List_of_trigonometric_identities


TITLE: Newton's method for square roots 'jumps' through the continued fraction convergents
QUESTION [11 upvotes]: I know that Newton's method approximately doubles the number of the correct digits on each step, but I noticed that it also doubles the number of terms in the continued fraction, at least for square roots.
Explanation. If we start Newton's iterations with some partial convergent of the simple continued fraction for the square root, we get another convergent of the same continued fraction on each step, with double amount of CF terms.
Example. Finding the square root of $2$.
1) Start with $\frac{3}{2}$ which is also the first continued fraction convergent. Listing the values on each step and the number of terms in the continued fraction, we get:
$$ \begin{array}( \frac{3}{2} & 1 &  \\ \frac{17}{12} & 3 & +2 \\ \frac{577}{408} & 7 & +4 \\ \frac{665857}{470832} & 15 & +8 \\ \cdots & 31 & +16 \end{array} $$
As you can see, the amount of CF terms (the position of this fraction in the list of all partial convergents) doubles on each step.
2) Start with $\frac{7}{5}$ which is the second continued fraction convergent.
$$ \begin{array}( \frac{7}{5} & 2 &  \\ \frac{99}{70} & 5 & +3 \\ \frac{19601}{13860} & 11 & +6 \\ \cdots & 23 & +12 \end{array} $$
The same happens for other square roots I checked.

How does Newton's method 'jump' through the continued fraction this way, exactly doubling the number of CF terms at each step?
Can we prove this observation using the recurrence relations for the continued fraction?

Just in case, the Newton's method:
$$\frac{p_{n+1}}{q_{n+1}}=\frac{p_n^2+bq_n^2}{2p_nq_n}$$
$$\lim_{n \to \infty} \frac{p_{n}}{q_{n}}=\sqrt{b}$$
And the continued fraction recurrence:
$$\frac{P_{n+1}}{Q_{n+1}}=\frac{a_n P_n+P_{n-1}}{a_n Q_n+Q_{n-1}}$$
$$P_1=a_0,~~~Q_1=1,~~~P_0=1,~~~Q_0=0,~~~P_{-1}=0,~~~Q_{-1}=1$$
$$\lim_{n \to \infty} \frac{P_{n}}{Q_{n}}=a_0+\cfrac{1}{a_1+\cfrac{1}{a_2+\cdots}}$$
How can we even relate these two?

A related question here. But I'm not asking about the digits, I'm asking about continued fraction terms.

REPLY [2 votes]: First some experimentation might be in order. I wrote up a program that could check for chains like this:
program nr
   implicit none
   integer, parameter :: ik16 = selected_int_kind(38)
   integer, parameter :: N = 200
   integer(ik16) p(N), q(N)
   integer D
   integer sqD
   integer r, s, a
   integer i, j, k
   integer(ik16) e, b, c

   write(*,'(a)',advance='no') 'Enter the value of D:> '
   read(*,*) D
   sqD = sqrt(D+0.5d0)
   r = 0
   s = 1
   a = (sqD+r)/s
   p(1) = a
   q(1) = 1
   r = a*s-r
   s = (D-r**2)/s
   a = (sqD+r)/s
   p(2) = a*p(1)+1
   q(2) = a
   do i = 3, N
      r = a*s-r
      s = (D-r**2)/s
      a = (sqD+r)/s
      p(i) = a*p(i-1)+p(i-2)
      q(i) = a*q(i-1)+q(i-2)
      if(p(i) <= p(i-1) .OR. q(i) <= q(i-1)) exit
   end do
   write(*,'(*(g0))') 'There are ',i-1,' convergents computed'
   outer: do j = 1, 5!(i-1)/2
      write(*,'(*(g0))') 'Starting convergent: ', j
      e = p(j)
      b = q(j)
      k = j
      do
         c = e**2+D*b**2
         if(c <= e) exit
         b = 2*e*b
         e = c
         do k = k+1, i-1
            if(b == q(k)) then
               if(e == p(k)) then
                  write(*,'(*(g0))') 'Matching convergent: ', k
                  exit
               else
                  write(*,'(a)') 'Matched denominator but not numerator'
                  stop
               end if
            else if(b < q(k)) then
               write(*,'(*(g0))') 'Missed convergent: ', k-1
               cycle outer
            end if
         end do
      end do
   end do outer
end program nr

So it always seemed to hit chains for $D=2$, $D=5$, $D=10$, $D=26$, $D=37$, and $D=50$ but for other values of $D$ sometimes it hit and sometimes it missed. My program has indices $1$ bigger than in the original question, so I'm seeing $n\rightarrow2n$ rather than $n\rightarrow2n+1$. Even the infamous $D=61$, which has a pretty large nontrivial solution to Pell's equation, has a chain. When another convergent is hit, it always seems to be exactly $2n$, never $2n-2$ or $2n+2$. It can't be $2n-1$ or $2n+1$ because Newton-Raphson approaches roots from above.
To prove a chain once found, it's probably useful to compare to solutoins to Pell's equation. Probably writing out the solutions in exponential form would lead to proof for a given chain.
EDIT: Just after posting it occurred to me that
$$\begin{align}(p^2+Dq^2)^2-D(2pq)^2 & =p^4+2Dp^2q^2+D^2q^4-4Dp^2q^2\\
 & =p^4-2Dp^2q^2+D^2q^4\\
 & =(p^2-Dq^2)^2\end{align}$$
So every solution to Pell's equation $p^2-Dq^2=1$ leads to an infinite chain of solutions via Newton-Raphson and if $|p^2-Dq^2|\ne1$, then this metric grows exponentially so that the roots computed by Newton-Raphson will not be convergents.
EDIT: Relevant link to Newton-Raphson and Pell's equation.


text:
\begin{document}

\title[Multiplicative ergodic theorem on Banach spaces]{A concise proof of the multiplicative ergodic theorem on Banach spaces}
\author{Cecilia Gonz\'alez-Tokman and Anthony Quas}
\address[Gonz\'alez-Tokman]{School of Mathematics and Statistics, University of New South Wales,  Sydney, NSW, 2052, Australia}
\address[Quas]{Department of Mathematics and
  Statistics, University of Victoria, Victoria, BC, CANADA, V8W 3R4}

\begin{abstract}
We give a streamlined proof of the multiplicative ergodic theorem for quasi-compact
operators on Banach spaces with a separable dual.
\end{abstract}

\maketitle

\section{Introduction}

The multiplicative ergodic theorem (MET) is a very powerful result
in ergodic theory establishing the existence of \textsl{generalized eigenspaces}
for stationary compositions of linear operators. It is of great
interest in many areas of mathematics, including
analysis, geometry and applications. The MET was first established by
Oseledets \cite{Oseledets} in the context of matrix cocycles.
The decomposition into generalized eigenspaces is called the \textsl{Oseledets splitting}.

After the original version, the MET was proved by a different
method by Raghunathan \cite{Raghunathan}.
The result was subsequently generalized to compact operators on
Hilbert spaces by Ruelle \cite{Ruelle}.
Ma\~n\'e \cite{Mane} proved a version for compact operators on Banach spaces under some continuity assumptions on the base dynamics and the dependence of the operator on the base
point. Thieullen \cite{Thieullen} extended this to quasi-compact operators.
Recently, Lian and Lu \cite{LianLu} proved a version in the context of
linear operators on separable Banach spaces, in which the continuity assumption was relaxed
to a measurablity condition.

We give a streamlined alternative, hopefully simpler,
proof of Lian and Lu's result, under a mild additional
assumption (the separability of the dual space, rather than the space itself).
At the same time, we remove the assumption of injectivity of the operators.
An important feature of the present approach is its constructive nature.
Indeed, it provides a robust way of approximating the Oseledets splitting,
following what could be considered a \textsl{power method} type strategy.
This makes the work also relevant from an applications perspective.

The approach of this work is similar in spirit to that of Raghunathan,
in that we primarily work with the `slow Oseledets spaces'.
Ma\~n\'e's proof works hard to build the fast space, as
do the subsequent works based on Ma\~n\'e's template.
These proofs rely on injectivity of the operators; some of them make use of natural extensions
to extend the result to non-invertible operators -- this was the strategy in
\cite{Thieullen}, and it was also used by Doan in \cite{Doan} to extend
\cite{LianLu} to the non-invertible context.
In contrast, we establish the non-invertible version first and recover the
(semi-)invertible one, including the `fast spaces', straightforwardly using duality.
Another key simplifying feature of our method
is that we prove measurability at the end of the proof, rather than working to ensure that
all intermediate constructions are measurable.

While Raghunathan's proof uses singular value decomposition and hence relies
on the notion of orthogonality,
we study instead collections of vectors with maximal volume growth.
Another important difference with Raghunathan's approach is that instead of dealing with the
exterior algebra, we work with the Grassmannian. We claim this is
more natural since subspaces correspond  to \emph{rank one} elements of the
exterior algebra (those that can be expressed as $v_1\wedge\ldots\wedge v_k$).
In the Euclidean setting, rank one elements naturally appear as eigenvectors of
$\Lambda^k(A^*A)$, but this does not seem to generalize to the Banach space case.

Section~\ref{sec:volume} contains volume calculations for bounded linear maps
$T$ on a Banach space $X$.
We establish an asymptotic equivalence between $k$-dimensional volume growth under
$T$ and $T^*$, as well as other measures of volume growth.

Section~\ref{sec:rds} deals with random dynamical systems, or linear cocycles.
The non-invertible Oseledets multiplicative ergodic theorem is proved under
the assumptions that $X^*$ is separable and a quasi-compactness condition holds.
Separability allows one to show all the necessary measurability conditions
in an elementary fashion.

The recent semi-invertible version of the MET established in \cite{GTQuas}, building on \cite{FLQ1,FLQ2}, and from which the invertible version of the MET can be recovered, is also presented here.
This result is obtained via duality: complementary spaces to the Oseledets filtration arise as annihilators of the corresponding Oseledets filtration for the dual cocycle.

\section{Volume calculations in Banach spaces}\label{sec:volume}

Let $X$ be a Banach space with norm $\|\cdot\|$. As usual, given a
non-empty subset $A$ of $X$ and a point $x\in X$, we define
$d(x,A)=\inf_{y\in A}d(x,y)$.  We denote by $B_X$ and $S_X$ the unit
ball and unit sphere in $X$, respectively.  The linear span of a
collection $C$ of vectors in $X$ will be denoted by $\lin(C)$ with the
convention that $\lin(\emptyset)=\{0\}$. The dual of $X$ will be
denoted by $X^*$.

We define the $k$-\textsl{dimensional volume} of a collection of
vectors $(v_1,\ldots,v_k)$ by
\begin{equation*}
  \vol_k(v_1,\ldots,v_k)=\prod_{i=1}^k d(v_i,\lin(\{v_j\colon j<i\})).
\end{equation*}

It is easy to see that $\vol_k(\alpha_1v_1,\ldots \alpha_k
v_k)=|\alpha_1|\ldots|\alpha_k|\vol_k(v_1,\ldots,v_k)$. In the case
where the normed space is Euclidean this notion corresponds with the
standard notion of $k$-dimensional volume.  Notice that
$\vol_k(v_1,\ldots,v_k)$ is not generally invariant under permutation
of the vectors.

Given a bounded linear map $T$ from $X$ to $X$, we define
$d_kT(v_1,\ldots,v_k)$ to be $\vol_k(Tv_1,\ldots,Tv_k)$ and
$D_kT=\sup_{\|v_1\|=1,\ldots,\|v_k\|=1}d_kT(v_1,\ldots v_k)$.

\begin{lem}[Submultiplicativity]\label{lem:submult}
  Let $T\colon X\to X$ and $S\colon X\to X$ be linear maps. Then
  $D_k(S\circ T)\le D_k(S)D_k(T)$.
\end{lem}

\begin{proof}
  Let $v_1,\ldots,v_k\in X$. Then one checks from the definition that
  for any collection of coefficients $(\alpha_{ij})_{j<i}$, the
  following holds
  \begin{equation}\label{eq:changeByLinComb}
    \vol_k(v_1,\ldots,v_k)=\vol_k(v_1,v_2-\alpha_{21}v_1,\ldots,v_k-\sum_{j<k}\alpha_{kj}v_j).
  \end{equation}

  Since the linear spans in the definition of volume are
  finite-dimensional spaces, the minima are attained so that $d_k
  T(v_1,\ldots,v_k)=\|T(v_1)\|\|T(v_2)-\alpha_{21}T(v_1)\|\ldots
  \|T(v_k)-\alpha_{k1}T(v_1)-\ldots-\alpha_{k,{k-1}}T(v_{k-1})\|$ for
  appropriate choices of $(\alpha_{ij})_{j<i}$.

  Let $w_j=v_j-\sum_{i<j}\alpha_{ji}v_i$ so that
  $d_kT(v_1,\ldots,v_k)=\|T(w_1)\|\ldots \|T(w_k)\|$ and set
  $u_j=T(w_j)/\|T(w_j)\|$. Using \eqref{eq:changeByLinComb}, we have
  \begin{align*}
    d_k(S\circ T)(v_1,\ldots,v_k)&=\vol_k(ST(v_1),\ldots,ST(v_k))\\
    &=
    \vol_k(ST(w_1),\ldots,ST(w_k))\\
    &=\|T(w_1)\|\ldots\|T(w_k)\|\vol_k(S(u_1),\ldots,S(u_k)) \\
    &\le d_k T(v_1,\ldots,v_k)D_k S.
  \end{align*}

  Taking a supremum over $v_1,\ldots,v_k$ in the unit ball of $X$, one
  obtains the bound $D_k(S\circ T)\le D_k(S)D_k(T)$ as required.
\end{proof}

\begin{lem}\label{lem:unifexpDk}
Let $T\colon X\to X$ be linear. Suppose that $V$ is a $k$-dimensional
subspace and $\|Tx\|\ge M\|x\|$ for all $x\in V$. Then $D_kT\ge M^k$.
\end{lem}

\begin{proof}
Let $v_1,\ldots,v_k$ belong to $V\cap S_X$ and satisfy $d(v_j,\lin(\{v_i\colon i<j\}))=1$.
Then $d_kT(v_1,\ldots,v_k)\ge M^k$.
\end{proof}

We now proceed to compare volume estimates for a linear operator $T$
and its dual.  We introduce a third quantity to which we compare both
$D_k(T)$ and $D_k(T^*)$.  Given linear functionals
$\theta_1,\ldots,\theta_k \in X^*$ and points $x_1,\ldots,x_k \in X$,
we let $U((\theta_i),(x_j))$ be the matrix with entries
$U_{ij}=\theta_i(T(x_j))$ and define
$$
E_k(T)=\sup \left\{\det U\big((\theta_i),(x_j)\big)\colon
  \|\theta_i\|=1\text{ and }\|x_j\|=1\text{ for all }i,j\right\}
$$

\begin{lem}[Relation between volumes for $T$ and $T^*$]\label{lem:primaldual}
  For all $k>0$, there exist positive constants $c_k$ and $C_k$ with
  the following property: For every bounded linear map $T$ from a
  Banach space $X$ to itself,
$$
c_kD_k(T)\le D_k(T^*)\le C_kD_k(T).
$$
\end{lem}

\begin{proof}
  The statement will follow from the following inequalities:
  \begin{align}
    D_k(T)&\le E_k(T)\le k!D_k(T)\label{eq:primal}\\
    D_k(T^*)&\le E_k(T)\le k!D_k(T^*).\label{eq:dual}
  \end{align}

  The second inequality of \eqref{eq:primal} is proved as follows: Let
  $x_1,\ldots,x_k$ and $\theta_1,\ldots,\theta_k$ all be of norm 1 in
  $X$ and $X^*$ respectively. Let
  $\alpha_j=d(Tx_j,\text{lin}(Tx_1,\ldots,Tx_{j-1}))$. Let
  $c^j_1,\ldots ,c^j_{i-1}$ be chosen so that $\|Ty_j\|=\alpha_j$,
  where $y_j$ is defined by
  $y_j=x_j-(c^j_1x_1+\ldots+c^j_{j-1}x_{j-1})$. Note that
  $U'=U((\theta_i),(y_j))$ may be obtained from
  $U=U((\theta_i),(x_j))$ by column operations that leave the
  determinant unchanged. Notice also that $|U'_{ij}|=|\theta_i(Ty_j)|
  \le \alpha_j$. From the definition of a determinant, we see that
  $\det U=\det U'\le k!\alpha_1\ldots \alpha_k$. This inequality holds
  for all choices of $\theta_i$ in the unit sphere of $X^*$. Now,
  maximizing over choices of $x_j$ in the unit sphere of $X$, we
  obtain the desired result.

  The second inequality of \eqref{eq:dual} may be obtained
  analogously.  We let
  $\beta_i=d(T^*\theta_i,\text{lin}(T^*\theta_1,\ldots,T^*\theta_{i-1}))$
  and choose linear combinations $\phi_i$ of the $\theta_i$ for which
  the minimum is obtained. The matrix $U''=U((\phi_i),(x_j))$ is
  obtained by row operations from $U$ and the
  $|U''_{i,j}|=|\phi_i(Tx_j)|=|(T^*\phi_i)(x_j)|\le \beta_i$.

  To show the first inequality of \eqref{eq:primal}, fix
  $x_1,\ldots,x_k$ of norm 1.  As before, let
  $\alpha_j=d(Tx_j,\text{lin}(Tx_1,\ldots,Tx_{j-1}))$. By the
  Hahn-Banach theorem, there exist linear functionals
  $(\theta_i)_{i=1}^k$ of norm 1 such that $\theta_i(x_i)=\alpha_i$
  and $\theta_i(x_k)=0$ for all $k<i$. Now
$$
\det U((\theta_i),(x_j))=\prod \alpha_i.
$$
Maximizing over the choice of $(x_j)$, we obtain $E_k(T)\ge D_k(T)$ as
required.

Finally, for the first inequality of \eqref{eq:dual}, we argue as
follows.  Let $\epsilon>0$ be arbitrary and let
$\theta_1,\ldots,\theta_k$ belong to the unit sphere of $X^*$. We may
assume that $T^*\theta_1,\ldots,T^*\theta_k$ are linearly independent --
otherwise the inequality is trivial. Let $\phi_i=T^*\theta_i-\sum_{k<i}a_{ik}T^*\theta_k$
be such that $\|\phi_i\|=d(T^*\theta_i,\lin(\{T^*\theta_k\colon k<i\})$.
We shall pick $x_1,\ldots,x_k$ inductively in such a way that $\det((\phi_i(x_j))_{i,j\le l})$
is at least $\prod_{i=1}^l(\|\phi_i\|-\epsilon)$ for each $1\le l\le k$.
Suppose $x_1,\ldots,x_{l-1}$ have been chosen.
Then since $\det((\psi_i(x_j))_{i,j<l})$ is non-zero, the rows span $\R^{l-1}$.
Hence there exist $(b_i)_{i<l}$ such that $\psi_l:=\phi_l+\sum_{i<l}b_i\phi_i$
satisfies $\psi_l(x_j)=0$ for all $j<l$. By assumption, $\|\psi_l\|\ge \|\phi_l\|$.
Pick $x_l\in S_X$ such that $\psi_l(x_l)>\|\psi_l\|-\epsilon$. Then the matrix with
a row for $\psi_l$ and a column for $x_l$ adjoined has determinant at
least $\prod_{i=1}^l(\|\phi_i\|-\epsilon)$. The corresponding matrix with $\phi_l$
replacing $\psi_l$ has the same determinant, completing the induction. Maximizing over
the choice of $(\theta_j)_{j\le k}$, letting $\epsilon$ shrink to 0, and observing that
$\det((\phi_i(x_j))_{i,j\le k})=\det((T^*\theta_i(x_j))_{i,j\le k})$ completes the proof.

\end{proof}

A fourth quantity that will play a crucial role in what follows is
$F_k(T)$, defined as
$$
F_k(T)=\sup_{\dim(V)=k}\inf_{v\in V\cap S_X}\|Tv\|.
$$

\begin{lem}[Relation between determinants and $F_k$]\label{lem:Frec}
  Let $T$ be a bounded linear map from a Banach space $X$ to
  itself. Then
$$
E_{k-1}(T)F_k(T)\le E_k(T) \le k2^{k-1}E_{k-1}(T)F_k(T).
$$
\end{lem}

\begin{proof}
  We first show $E_k(T)\le k2^{k-1}E_{k-1}(T)F_k(T)$. We may assume
  $E_k(T)>0$ as otherwise the inequality is trivial.  Let
  $\theta_1,\ldots,\theta_k$ be elements of the unit sphere of $X^*$
  and $x_1,\ldots,x_k$ be elements of the unit sphere of $X$.  Let $U$
  be the matrix with entries $\theta_i(Tx_j)$. Assume that $\det U\ne
  0$.  Since $x_1,\ldots,x_k$ span a $k$-dimensional space, there
  exists a $v=a_1x_1+\ldots+a_kx_k$ of norm 1 such that $\|Tv\|\le
  F_k(T)$. By the triangle inequality, one of the $a$'s, say
  $a_{j_0}$, must be at least $\frac 1k$. Let $\tilde x_j=x_j$ for
  $j\ne j_0$ and $\tilde x_{j_0}=v$ and set $\tilde U$ to be the
  matrix with entries $\theta_i(T\tilde x_j)$. By properties of
  determinants, we see $|\det\tilde U| =a_{j_0}|\det U|\ge \tfrac 1k|\det
  U|$. Next, there exists $i_0$ for which $|\theta_{i_0}(v)|$ is
  maximal, this maximum not being 0 since $|\det\tilde U|$ is
  positive.  Let
  $\bar\theta_i=\theta_i-(\theta_i(v)/\theta_{i_0}(v))\theta_{i_0}$
  for $i\ne i_0$ and $\bar\theta_{i_0}=\theta_{i_0}$, so that
  $\|\bar\theta_i\|\le 2$ and $\bar\theta_i(v)=0$ for $i\ne i_0$.

  Now let $\bar U_{ij}=\bar\theta_i(T\tilde x_j)$, so that $|\det
  U|\le k |\det\tilde U| =k|\det \bar U|$. Finally, the $j_0$th column
  of $\bar U$ has a single non-zero entry that is at most $\|Tv\|\le
  F_k(T)$ in absolute value. The absolute value of the cofactor is
  $\left|\det \big(\bar \theta_i(T(\tilde x_j))\big)_{i\ne i_0,\ j\ne
      j_0}\right| \le 2^{k-1}E_{k-1}(T)$.  Taking a supremum over
  choices of $(\theta_i)$ and $(x_j)$, we have shown $E_k(T)\le
  k2^{k-1}E_{k-1}(T)F_k(T)$.

  For the other inequality, we may suppose that $T$ has kernel of
  codimension at least $k$, otherwise $F_k(T)=0$ and there is nothing
  to prove. Let $\theta_1,\ldots,\theta_{k-1}$ and
  $x_1,\ldots,x_{k-1}$ be arbitrary.  Let $\Delta$ be the determinant
  of the matrix with entries $\theta_i(Tx_j)$.  Let $V$ be a
  $k$-dimensional subspace such that $V\cap \ker T=\{0\}$. Let
  $W=\lin(Tx_1,\ldots,Tx_{k-1})$. Let $z$ be a point in the unit
  sphere of $T(V)$ such that $d(z,W)=1$ (the existence follows from a
  theorem of Gohberg and Krein, see Lemma 211 of \cite{Kato2}).  Let
  $v\in V\cap S_X$ be such that $T(v)$ is a multiple of $z$.  Let
  $\theta_k$ be a linear functional of norm 1 such that
  $\theta_k\vert_W=0$ and $\theta_k(z)=1$ and let $x_k=v$. Now forming
  the $k\times k$ matrix $\big(\theta_i(x_j)\big)_{1\le i,j\le k}$, we
  see the absolute value of the determinant is $\Delta\cdot
  \theta_k(Tv)= \Delta\cdot\|Tv\| \ge\Delta\cdot \inf_{x\in V\cap
    S_X}\|Tx\|$.  Taking suprema over choices of $x$'s, $\theta$'s and
  $k$-dimensional $V$'s, we see that $E_k(T)\ge F_k(T)E_{k-1}(T)$ as
  required.

\end{proof}

\begin{cor}\label{cor:equiv}[of Lemmas \ref{lem:primaldual} and \ref{lem:Frec}]
  For each $k>0$, the quantities $D_k(T)$, $D_k(T^*)$, $E_k(T)$ and
  $\prod_{i\le k}F_i(T)$ agree up to multiplicative factors that are
  independent of the bounded linear map $T$ and the Banach space $X$.
\end{cor}

By definition, for each natural number $k$, one can find sequences
$(\theta_i)_{i\le k}$ and $(x_j)_{j\le k}$ such that
$\det(U((\theta_i),(x_j)))\approx E_k(T)$.  We now show that we can
find infinite sequences $(\theta_i)$ and $(x_j)$ so that, for each
$k$, $\det(U((\theta_i)_{i\le k},(x_j)_{i\le k}))\approx E_k(T)$.

\begin{lem}[Existence of consistent sequences]\label{lem:genseqs}
  For any linear map $T$, there exist $(\theta_i)_{i\ge 1}$
  in $S_{X^*}$ and $(x_j)_{j\ge 1}$ in $S_X$ such that for all $k$,
  \begin{align*}
    &\det\left((\theta_i(Tx_j))_{1\le i,j\le k}\right)\ge
    \tfrac{1}{2^k}\prod_{i\le k}F_i(T)
    \text{; and}\\
    &\|Tx\|\ge 4^{-k}F_k(T)\|x\|\text{ for all $x\in
      \lin(x_1,\ldots,x_k)$.}
  \end{align*}
\end{lem}

\begin{proof}
  The proof is by induction: suppose $(\theta_i)_{i<k}$ and
  $(x_j)_{j<k}$ have been chosen and satisfy the desired inequalities
  at stage $k-1$.  Then pick an arbitrary $k$-dimensional space $V$
  such that $\|Tv\|\ge \frac 12F_k(T)\|v\|$ for all $v\in V$. Using
  the result of Gohberg and Krein, let $x_k\in V\cap S_X$ be such that
  $d(Tx_k,\lin(Tx_1,\ldots,Tx_{k-1}))= \|Tx_k\|$.  Finally choose
  $\theta_k$ of norm 1 such that $\theta_k(Tx_i)=0$ for $i<k$ and
  $\theta_k(Tx_k)=\|Tx_k\|$. The determinant inequality at stage $k$
  follows.

  Let $x=a_1x_1+\ldots+a_kx_k$ be of norm 1. Then
  \begin{equation}\label{eq:rough1}
    \|Tx\|\ge |a_k|d(Tx_k,\lin(Tx_1,\ldots,
    Tx_{k-1}))\geq |a_k|\|Tx_k\|\ge |a_k|F_k(T)/2.
  \end{equation}
  Similarly,
  \begin{equation*}
    \|Tx\|\ge \left\|T(\textstyle\sum_{j<k}a_jx_j)\right\|-|a_k|\|Tx_k\|.
  \end{equation*}
  Averaging the inequalities, we get
  \begin{equation}\label{eq:rough2}
    \|Tx\|\ge \tfrac12 \left\|T(\textstyle\sum_{j<k}a_jx_j)\right\|.
  \end{equation}
  If $|a_k|>\frac 12$, the first inequality yields $\|Tx\|\ge \frac
  14F_k(T)$.  If $|a_k|\le \frac 12$, then $\|\sum_{j<k}a_jx_j\|\ge
  \frac 12$ and the second inequality combined with the inductive
  hypothesis gives $\|Tx\|\ge \frac 14 4^{-(k-1)}F_{k-1}(T)$.

\end{proof}

\begin{lem}[Lower bound on volume growth in a subspace of finite codimension]\label{lem:ExponentsOfReducedCocycle}
  For any natural numbers $k>m$, there exists $C_k$ such that if $X$
  is a Banach space, $T$ is a linear map of $X$ and $V$ is a closed
  subspace of $X$ of codimension $m$, then $D_k(T)\le
  C_kD_m(T)D_{k-m}(T|_V)$.
\end{lem}

\begin{proof}
  Let $\ep>0$.  Let $P$ be a projection from $X$ to $V$ of norm at
  most $\sqrt m+\ep$ (such a projection exists
  by Corollary III.B.11 in the book of Wojtaszczyk \cite{Wojtaszczyk}).
  Then, $\|\mathbf 1 -P\|\leq \sqrt m +\ep+1$.
  Let $x_1,\ldots,x_k$ be a sequence of vectors in $X$ of norm 1.  The
  proof of Lemma \ref{lem:primaldual} shows that there exist
  $\psi_1,\ldots,\psi_k$ in $S_{X^*}$ such that $\det(\psi_i(x_j)) \ge
  d_kT(x_1,\ldots,x_k)$. Write $P_1$ for $P$ and $P_0$ for $\mathbf
  1-P$, which has $m$-dimensional range.  There exists a choice
  $\epsilon_1,\ldots\epsilon_k\in\{0,1\}^k$ such that
  $\det(\psi_i(P_{\ep_j}x_j))>2^{-k}d_kT(x_1,\ldots,x_k)$, by
  multilinearity of the determinant.  At most $m$ of the $\epsilon_j$
  can be 0, as otherwise more than $m$ vectors lie in a common
  $m$-dimensional space, so that at least $k-m$ of them lie in $V$.
  Hence, there exist vectors $y_1,\ldots,y_m$ in $S_X$ and
  $y_{m+1},\ldots,y_k$ in $S_X\cap V$ such that
$$|\det(\psi_i(y_j))|\ge (2(\sqrt m +\ep+1))^{-k}d_kT(x_1,\ldots,x_k).
$$
Using the proof of Lemma \ref{lem:primaldual} again, we deduce that
\begin{align*}
  d_dT(y_1,\ldots,y_m) & d_{k-m}T(y_{m+1},\ldots,y_k)\ge d_kT(y_1,\ldots,y_k)\\
  &\ge (2(\sqrt m +\ep+1))^{-k}/(k!) d_kT(x_1,\ldots,x_k).
\end{align*}
This completes the proof.
\end{proof}

\section{Random dynamical systems}\label{sec:rds}

A closed subspace $Y$ of $X$ is called \textsl{complemented} if there
exists a closed subspace $Z$ such that $X$ is the direct sum of $Y$
and $Z$, written $X=Y\oplus Z$. That is, for every $x\in X$, there
exist $y\in Y$ and $z\in Z$ such that $x=y+z$, and this decomposition
is unique.  The Grassmannian $\mathcal G(X)$ is the set of closed
complemented subspaces of $X$. We denote by $\mathcal G^k(X)$ the
collection of closed $k$-codimensional subspaces of $X$ (these are
automatically complemented).  We equip $\mathcal G(X)$ with the metric
$d(Y,Y')=d_H(Y\cap B_X,Y'\cap B_X)$ where $d_H$ denotes the Hausdorff
distance.  There is a natural map $\perp$ from $\mathcal G(X)$ to
$\mathcal G(X^*)$ given by $Y^\perp=\{\theta\in X^*\colon
\theta(y)=0\text{ for all $y\in Y$}\}$. Recall that if $X^*$ is
separable, then $X$ is separable, as is $\mathcal G^k(X)$.

In this section, we consider \textsl{random dynamical systems}.  These
consist of a tuple $\mathcal{R}=(\Omega,\mathcal F,\mathbb P, \sig, X,
\LL)$, where $(\Omega,\mathcal F,\mathbb P)$ is a probability space;
$\sig$ is a measure
preserving transformation of $\Omega$; $X$ is a separable Banach space; the
\textsl{generator} $\LL\colon \Omega\to B(X,X)$
is strongly measurable (that is for fixed $x\in X$,
$\omega\mapsto \LL_\omega x$ is $(\mathcal F,\mathcal B_X)$-measurable); and
$\log\|\LL_\om\|$ is integrable.

This gives rise to a cocycle of bounded linear
operators $\LL_\om^{(n)}$ on $X$, defined
by $\LL_\om^{(n)}(x)=\LL_{\sigma^{n-1}\om}\circ\dots\circ\LL_\om x$.
We will consider $\mathcal{F}$ and
$\mathbb P$ to be fixed, and thus refer to a random dynamical system
as $\mathcal{R}=(\Omega,\sigma,X,\mathcal L)$.  We say $\mc{R}$ is
ergodic whenever $\sig$ is ergodic.

When the base $\sig$ is invertible, we can also define the dual random
dynamical system $\mc{R}^*=(\Omega,\mathcal F,\mathbb P, \sig^{-1},
X^*, \LL^*)$, where $X^*$ is the dual of $X$ and $\LL^* (\om, \theta)
:= (\LL_{\sig^{-1}\om})^* \theta$.  In this way, $\theta(\LL_\om x)=
\LL^*_{\sig\om}\theta(x)$ and, more generally,
$\LL^{*(n)}_{\sigma^n\omega}\theta(x)=\theta(\LL^{(n)}_\omega x)$ for
every $x\in X, \theta \in X^*$.  Thus, $\LL^{*(n)}_{\sig^n \om}=
(\LL_{\om}^{(n)})^*$.

\begin{lem}[Measurability I]\label{lem:volmeas}
  Given a random dynamical system
  $\mathcal{R}=(\Omega,\sigma,X,\mathcal L)$ acting on a separable
  Banach space, the following functions are measurable:
  \begin{itemize}
  \item $\omega\mapsto D_k(\LL_\omega)$;
  \item $\omega\mapsto \|\LL_\omega\|$;
  \item $\omega\mapsto \ic(\LL_\omega):=
    \inf\Big\{$\parbox{2.5in}{$r>0 : \LL_\om(B_X)$ can be covered by
      finitely many balls of radius $r$} $\Big\}$.
  \end{itemize}
\end{lem}

\begin{proof}
  Fix a dense sequence $x_1,x_2,\ldots$ in the unit ball of $X$. By
  strong measurability, for each fixed $x$, $\omega\mapsto
  \|\LL_\omega x\|$ is measurable.  Then for each $j_1,\ldots,j_i$, we
  have that $f_{j_i|j_1,\ldots,j_{i-1}}(\omega) :=\inf_{q_1,
    \dots,q_{i-1}\in\mathbb Q} \|\LL_\omega x_{j_i}-\sum_{1\le l<i}
  q_l \LL_\omega x_{j_l}\|$ is measurable, so
$$
D_k(\LL_\omega)= \sup_{j_1,\ldots,j_k}\prod_{i\le
  k}f_{j_i|j_1,\ldots,j_{i-1}}(\omega)
$$
is measurable. Similarly,
$\omega\mapsto\|\LL_\omega\|=\sup_{j}\|\LL_\omega x_j\|$ and
$\ic(\LL_\omega)=\lim_n\sup_j\inf_{k\le
  n}\big\|\LL_\omega(x_j)-(2\|\LL_\omega\|) x_k\big\|$ are measurable.

\end{proof}

\begin{rmk}[Random dynamical systems with random domains]
  \label{rmk:diffDomain}
  Let $\mathcal{R}=(\Omega,\sigma,X,\mathcal L)$ be a random dynamical
  system acting on a separable Banach space.  Suppose there exists a
  measurable family of closed, finite codimensional subspaces
  $\mathscr{V}=\{V(\om)\}_{\om \in \Om}$ which is equivariant under
  $\LL$ (that is, $\LL_\om({V(\om}))\subset V({\sig\om})$). Then, one
  can construct a new random dynamical system $\mc{R}|_{\mathscr{V}}$
  by restricting the generator $\LL$ to $\mathscr{V}$. The results of
  this section apply to this more general type of random dynamical
  systems as well.

  Indeed, the fact that the domains of $\LL_\om$ are the same is only
  used in Lemma~\ref{lem:volmeas}, when a countable dense sequence in
  $S_X$ is chosen. In the case of the restriction
  $\mc{R}|_{\mathscr{V}}$, the choice of a sequence of measurable
  functions $\om \mapsto u_j(\om)$ so that $(u_j(\om))_{j>0}$ is dense
  in $V(\om) \cap S_X$ can be made as follows.

  First, for fixed $v\in X, \om \mapsto d(v,V(\om))$ is a measurable
  function, as it is the composition of continuous and measurable
  functions.  Fix a dense sequence $v_1,v_2, \ldots \in S_X$.  Now for
  each $j$, set $u_j^0(\om) = v_j$, and let $u_j^{k+1}(\om) = v_l$,
  where $l=\min \{ m: d(v_m,V(\om)\cap S_X) \leq \tfrac 14
  d(u_j^{k}(\om),V(\om)\cap S_X) \text{ and } d(v_m,u_j^k(\om))\leq 2
  d(u_j^k(\om),V(\om)\cap S_X)\}$.  For each $j$, this is a measurable
  convergent sequence and hence the limit point $u_j^\infty(\om)$ is
  measurable, and belongs to $V(\om)$.  The sequence
  $(u_j^\infty(\om))$ is dense in $V(\om) \cap S_X$ because there are
  $v_j$ arbitrarily close to all points of $V(\om)\cap S_X$.

  One may also consider the quotient spaces $Q(\om):= X / V(\om)$,
  equipped with the quotient norm, $\|\bar x \|_{Q(\omega)}= d(x,
  V(\omega))$, and the induced $\bar{\LL}_\om: Q(\om) \to
  Q({\sig\om})$.  The measurability results of Lemma~\ref{lem:volmeas}
  also apply in this case. Indeed, if $v_1, v_2, \dots$ is a countable
  dense sequence in $S_X$, and $\{u_i(\om)\}_{i\in \N}$ is a sequence
  of measurable functions dense in $V(\om) \cap S_X$ as constructed
  above, then measurability of $\omega\mapsto d(\mathcal
  L_\omega(v_i),V(\omega))$ is straightforward to establish, and hence
  $\omega\mapsto\|\bar{\LL}_\om\|$ is measurable.

  The fact that $\om \mapsto \|D_k(\bar{\LL}_\om)\|_{Q(\om) \to
    Q(\sig\om)}$ is measurable follows similarly from
  Lemma~\ref{lem:volmeas}.
\end{rmk}

When $\mc{R}$ is ergodic, Lemma~\ref{lem:volmeas} (or
Remark~\ref{rmk:diffDomain}) combined with Kingman's sub-additive
ergodic theorem ensures the existence of the \textsl{maximal Lyapunov
  exponent} of $\mc{R}$, defined by
$$
\lam(\mc{R}):= \lim_{n\to\infty}\tfrac1n\log \|\LL^{(n)}_\omega\|,
$$
for $\paeom$.  Similarly, using the fact that the index of compactness
is also sub-multiplicative and bounded above by the norm, we have
existence of the \textsl{index of compactness} of $\mc{R}$, defined by
$$
\ka(\mc{R}):= \lim_{n\to\infty}\tfrac1n\log \ic(\LL^{(n)}_\omega),
$$
with the property that $\ka(\mc{R}) \leq \lam(\mc{R})$.  Furthermore,
we have the following.
\begin{lem}\label{lem:expDefn}
  Given an ergodic random dynamical system $\mc{R}$, there exist
  constants $\Delta_k=\Delta_k(\mc{R})$ such that for almost every
  $\omega\in\Omega$,
$$
\lim_{n\to\infty}\tfrac1n\log D_k(\LL^{(n)}_\omega)=\Delta_k.
$$
Furthermore, $\frac 1n\log E_k(\LL^{(n)}_\omega)\to\Delta_k$.  Define
$\Delta_0=0$ and let $\mu_k=\Delta_k-\Delta_{k-1}$
for each $k\ge 1$.  Then, $\frac 1n\log
F_k(\LL^{(n)}_\omega)\to\mu_k$.
\end{lem}

\begin{proof}
  The first claim follows from Kingman's sub-additive ergodic theorem,
  via Lemma~\ref{lem:volmeas} (or Remark~\ref{rmk:diffDomain}) and
  Lemma~\ref{lem:submult}.  The remaining two claims are consequences
  of Corollary \ref{cor:equiv}.
\end{proof}

The $\mu_k$'s of the previous lemma are called the \textsl{Lyapunov
  exponents} of $\mc{R}$. When $\mu_k>\kappa(\mc{R})$, $\mu_k$ is
called an \textsl{exceptional Lyapunov exponent}.

\begin{thm}[Lyapunov exponents and index of compactness]\label{thm:summ}
  Let $\mc{R}$ be a random dynamical system with ergodic base acting
  on a separable Banach space $X$.  Then
  \begin{itemize}
  \item $\mu_1\ge \mu_2\ge\ldots$;
  \item For any $\rho>\kappa(\mathcal R)$, there are only finitely
    many exponents that exceed $\rho$;
  \item If $\sig$ is invertible, the dual random dynamical system
    $\mc{R}^*$ and $\mc{R}$ have the same Lyapunov exponents.
  \end{itemize}
\end{thm}

\begin{proof}
  That the $\mu_i$ are
  decreasing follows from Lemma \ref{lem:expDefn} and
  the observation that $F_k(T)\le F_{k-1}(T)$.
  That the system and its dual have the same exponents follows from
  Lemma \ref{lem:primaldual} together with the simple result (in
  \cite[Lemma 8.2]{FLQ1}) that if $(f_n)$ is sub-additive and
  satisfies $f_n(\omega)/n\to A$ almost everywhere, then one has
  $f_n(\sigma^{-n}\omega)/n\to A$ also.

  It remains to show that for $\rho>\kappa$, the system has at most
  finitely many exponents that exceed $\rho$.  Let
  $\kappa<\alpha<\beta<\rho$. Since $\log\|\LL_\omega\|$ is
  integrable, there exists a $0<\delta<(\beta-\alpha)/2|\alpha|$ such
  that if $\mathbb P(E)<\delta$, then $\int_E
  \log\|\LL_\omega\|\,d\PP(\omega)<(\beta-\alpha)/2$.  By the
  sub-additive ergodic theorem, there exists $L>0$ such that
  $\PP(\ic(\LL^{(L)}_\omega)\ge e^{\alpha L})<\delta/2$.  If
  $\ic(\LL^{(L)}_\omega)< e^{\alpha L}$, then by definition,
  $\LL^{(L)}_\omega B_X$ may be covered by finitely many balls of size
  $e^{\alpha L}$.  Let $r$ be chosen large enough so that
  $\PP(G)>1-\delta$, where $G$ (the good set) is defined by
$$
G=\{\omega:\LL^{(L)}_\omega B_X\text{ may be covered with $e^{rL}$
  balls of size $e^{\alpha L}$}\}.
$$
We split the orbit of $\omega$ into blocks: if $\sigma^i\omega\in G$,
then the block length is $L$; otherwise, if $\sigma^i\omega$ is bad,
we take a block of length 1. Consider the following iterative process:
start with a ball of radius 1. Then look at the current iterate of
$\omega$, $\sigma^i\omega$.  If it is good, replace the current crop
of balls by $e^{rL}$ times as many balls of size $e^{\alpha L}$ times
their current size.  If the iterate of $\omega$ is bad, replace the
balls by balls inflated by a factor of $\|\LL_{\sigma^i\omega}\|$. In
this way, one obtains a collection of balls that covers the
push-forward of the unit ball.

We claim that for almost all $\omega$, for sufficiently large $N$,
$\LL^{(N)}_\omega(B_X)$ is covered by at most $e^{rN}$ balls of size
$e^{\beta N}$. For large $N$, the combined inflation through the bad
steps is less than $e^{(\beta-\alpha)N/2}$.
If $\alpha\ge 0$, then for large $N$, through
the good steps, the balls are inflated by a factor at most $e^{\alpha
  N}$. If $\alpha<0$, then combining the good
blocks, the balls are scaled by a factor of $e^{\alpha(1-\delta)N}
<e^{(\alpha+\beta)N/2}$ or
smaller. In both cases, we see that overall, balls are
scaled by at most $e^{\beta N}$. The splitting only takes place
in the good blocks, and yields at most $e^{rN}$ balls.

Now suppose that $\mu_k> \rho$. For almost all $\omega$, we have that
for all large $N$, we have $D_k(\LL_\omega^{(N)})>e^{kN\rho}$.  Fix
such an $N$, and suppose that $x_1,\ldots,x_k$ have the property
$\prod_{i\le k}D_i>e^{kN\rho}$ where
$$
D_i=d(\LL^{(N)}_\omega x_i,\lin(\{\LL^{(N)}_\omega x_j\colon j<i\})).
$$

Let $T_i=\{0,1,\ldots,\lfloor D_i/(2ke^{\beta N})\rfloor\}$ and notice
that $|T_1\times \cdots\times T_k|\ge e^{kN(\rho-\beta)}/(2k)^k$.  For
$(j_1,\ldots,j_k)\in T_1\times\cdots\times T_k$, define
$$
y_{j_1,\ldots,j_k}=\sum_{i=1}^k \frac {2j_ie^{\beta
    N}}{D_i}\LL^{(N)}_\omega x_i
$$
It is not hard to see that all of these points belong to the image of
the unit ball of $X$ under $\LL^{(N)}_\omega$. Further, from the
definition of $D_i$, one can check that these points are mutually
separated by at least $2e^{\beta N}$, so that one requires at least
$e^{kN(\rho-\beta)}/(2k)^k$ balls to cover $\LL^{(N)}_\omega(B_X)$.

Hence we obtain
$$
\frac {e^{kN(\rho-\beta)}} {(2k)^k}\le e^{rN}.
$$
Since this holds for all large $N$, we deduce $k\le r/(\rho-\beta)$ as
required.
\end{proof}

\begin{lem}[Measurability II]\label{lem:grassmeas}
  Suppose that $X$ is a Banach space with separable dual.
  Suppose further that $\mathcal R$ is an ergodic random dynamical
  system acting on $X$.

  Assume there exist $\lambda'>\lambda\in\R$ and $d\in\mathbb N$ such that
  for $\PP$-almost every $\omega$, there is a $d$-codimensional
  subspace $V(\omega)$ of $X$ such that for all $v\in V(\omega)$,
  $\limsup_{n\to\infty} \frac 1n\log\|\mathcal L^{(n)}_\omega
  v\|\le\lambda$; and for $v\not\in V(\omega)$ and arbitrary $\ep>0$,
    $\|\mathcal L^{(n)}_\omega v\|\ge d(v,V(\om))e^{n(\lambda'-\ep)}$
    for all large $n$.
   Then $\omega\mapsto V(\omega)$ is measurable.
\end{lem}

\begin{proof}
  Given $V\in \mathcal G^d(X)$ and vectors $w_1,\ldots,w_d$ such that
  $V\oplus\lin(w_1,\ldots,w_d)=X$, define a neighbourhood of $V$ by
$$
N_{V;w_1,\ldots,w_d;\gamma}= \{U\in \mathcal G^d\colon
\|\proj_{\lin(w_i)\parallel U\oplus \lin(\{w_j\colon j\ne
  i\})}|_V\|\le \gamma\text{ for $1\le i\le d$}\}.
$$
Since $\mathcal G^d(X)$ is separable, one can fix a countable
collection of $V$'s dense in $\mathcal G^d(X)$ and fix for each such
$V$ a complementary sequence of $w$'s.  By restricting to rational
$\gamma$'s, we obtain a countable collection of neighbourhoods which
generates the Borel $\sigma$-algebra on $\mathcal G^d(X)$.  (To see
this, notice that each open set is the union of the neighbourhoods in
the collection that it contains).
Hence to show the desired measurability, it suffices to show that for
each $N =N_{V;w_1,\ldots,w_d;\gamma}$, $\{\omega\colon V(\omega)\in
N\}$ is measurable.

Fix a dense set $v_1,v_2,\ldots$ in the unit sphere of
$V$.  We claim that (provided $\omega$ lies in the set of full measure
on which the growth rates are $\le \lambda$ and $\ge\lambda'$ and
where the dimension is correct), $V(\omega)$ lies in $N$ if and only
if the following condition holds:

\begin{quote}
For each rational $\epsilon>0$, there is $n_0>0$ such that for
each $n\ge n_0$, for each $j\in\N$, there exist rational
$a^j_1,\ldots,a^j_d$ in $[-\gamma,\gamma]$ such that $
\|\mathcal L_\omega^{(n)}(v_j-\sum_{i=1}^da^j_iw_i)\| \le
e^{(\lambda+\epsilon)n}$.
\end{quote}

Indeed, the `only if' direction is straightforward.
The `if' direction follows from the fact that if
$V(\om) \not\in N$, then either (i)
$V(\om) + \lin(\{w_j\colon 1\leq j \leq d\})=:Y \neq X$ and therefore
there exists $j\in\N$ such that $d(v_j,Y)>0$;
or (ii) there exists $j\in\N$ such that
$v_j=v+\sum_{i=1}^d b_iw_i$, with $v\in V(\om)$ and $|b_k|>\gamma$
for some $1\leq k \leq d$.
In the first case, for every
$a_1,\dots, a_d \in [-\ga,\ga]$, $d(v_j-\sum_{i=1}^d a_i w_i,V(\om))\geq d(v_j,Y)$.
In the second case, there exists $\eta>0$ such that for every
$a_1,\dots, a_d \in [-\ga,\ga]$, $d(v_j-\sum_{i=1}^d a_i w_i,V(\om))>\eta$.
Hence, in both cases the condition above fails.

Since this condition is obtained by taking countable unions and
intersections of measurable sets, the measurability of $\{\omega\colon
V(\omega)\in N\}$ is demonstrated.
\end{proof}

\begin{lem}\label{lem:tempered}
  Let $\sig$ be an ergodic measure-preserving transformation of a
  probability space $(\Om,\PP)$. Let $g$ be a non-negative measurable
  function and let $h\geq0$ be integrable.  Suppose further that
  $g(\om)\le h(\om)+g(\sig(\om))$, $\PP$-a.e.  Then $g$ is tempered;
  that is $\lim_{n\to\infty}g(\sig^n\om)/n=0$, $\PP$-a.e.
\end{lem}

\begin{proof}
  Let $\epsilon>0$ and let $K>\int h$. By the maximal ergodic
  theorem, $B_1=\{\om\colon h(\om)+\ldots+h(\sigma^{n-1}\om)<nK,
  \text{ for all $n$}\}$ has positive measure.
  Let $M$ be such that $B_2:=\{\om\colon g(\om)<M\}$
  has positive measure,
  As a consequence of the
  Birkhoff ergodic theorem, for any measurable set $B$ with
  $\PP(B)>0$, for $\PP$-a.e. $\om$, for all sufficiently large $k$,
  there exists $j\in [(1+\ep)^k,(1+\ep)^{k+1})$ such that
  $\sig^j(\om)\in B$. Now for $\om\in\Om$, let $k_0$ be such that for
  all $k\ge k_0$, there exist $j\in [(1+\ep)^k,(1+\ep)^{k+1})$ such
  that $\sig^j\om\in B_1$ and $j'\in((1+\ep)^{k+2},(1+\ep)^{k+3})$
  such that $\sig^{j'}\om\in B_2$.  If $n>(1+\ep)^{k_0+1}$, then $n\in
  [(1+\ep)^{k+1},(1+\ep)^{k+2})$ for some $k\ge k_0$.  Let $j\in
  [(1+\ep)^k,(1+\ep)^{k+1})$ and $j'\in [(1+\ep)^{k+2},(1+\ep)^{k+3})$
  be as above. Then $g(\sig^n\om)\le \sum_{k=n}^{j'-1}h(\sigma^k\om)
   + g(\sig^{j'}\om) \le \sum_{k=j}^{j'-1}h(\sig^k\om)
  +g(\sig^{j'}\om)\le K(j'-j)+M$, so that $\limsup
  g(\sig^n\om)/n\le 4\ep$. Since $\ep$ is arbitrary, the conclusion
  follows.
\end{proof}

\begin{thm}[The Oseledets filtration]\label{thm:Osel}
  Let $\mathcal R$ be an ergodic random dynamical system acting on a
  Banach space $X$ with separable dual. Suppose that
  $\kappa(\mathcal R)<\lambda(\mathcal R)$. Then there exist $1\le
  r\le\infty$\footnote{If $r=\infty$, the conclusions must be replaced
    by: $\lambda(\mathcal
    R)=\lambda_1>\lambda_2>\ldots>\kappa(\mathcal R)$,
    $\mult_1,\mult_2,\ldots\in \N$ and $X=V_1(\om)\supset V_2(\om)
    \supset \dots \supset V_\infty(\om)$.}  and:
  \begin{itemize}
  \item a sequence of exceptional Lyapunov exponents $\lambda(\mathcal
    R)=\lambda_1>\lambda_2>\ldots>\lambda_r>\kappa(\mathcal R)$;
  \item a sequence $\mult_1,\mult_2,\ldots,\mult_r$ of positive
    integers; and
  \item an equivariant measurable filtration $X=V_1(\om)\supset
    V_2(\om) \supset \dots \supset V_r(\omega)\supset V_\infty(\om)$
\end{itemize}
such that for $\PP$-a.e.\ $\omega$,  $\codim V_\ell(\omega)=\mult_1+\dots+\mult_{\ell-1}$;
for all $v\in V_\ell(\omega)\setminus V_{\ell+1}(\omega)$, one has
$\lim\frac 1n\log\|\LL^{(n)}_\omega v\|= \lambda_\ell$; and for
$v\in V_\infty(\omega)$, $\limsup\frac 1n\log\|\LL^{(n)}_\omega v\|\le \kappa(\mathcal R)$.
\end{thm}

\begin{proof}
  Let $\mu_1\ge \mu_2\ge\ldots$
  be as in Theorem \ref{thm:summ}. Let
  $\lambda_1>\lambda_2>\ldots$ be the decreasing enumeration of the
  distinct $\mu$-values that exceed $\kappa(\mathcal R)$ (if this an
  infinite sequence, then Theorem \ref{thm:summ} establishes that
  $\lambda_i\to\kappa(\mathcal R)$).  The fact that $\lambda(\mathcal
  R)=\lambda_1$ is straightforward from the definitions.  Let
  $\mult_\ell$ be the number of times that $\lambda_\ell$ occurs in
  the sequence $(\mu_i)$ and let $\D_\ell=\mult_1+\ldots +\mult_\ell$,
  so that $\mu_{\D_{\ell-1}}=\lambda_{\ell-1}$ and
  $\mu_{1+\D_{\ell-1}}=\lambda_{\ell}$.

  We now turn to the construction of $V_\ell(\omega)$.  For a fixed
  $\omega$, let the sequences $(\theta^{(n)}_i)_{i\ge 1}$ and
  $(x^{(n)}_j)_{j\ge 1}$ be as guaranteed by Lemma \ref{lem:genseqs}
  for the operator $\mathcal L^{(n)}_\omega$.  We let
  $V_\ell^{(n)}(\omega)$ be
  $\lin(\theta^{(n)}_1,\ldots,\theta^{(n)}_{\D_{\ell-1}})^\perp$, let
  $Y_\ell^{(n)}(\omega)$ be $\lin(\{x^{(n)}_j\colon j\le
  \D_\ell\})$. Thus, $X=V_\ell^{(n)}(\omega)\oplus
  Y_{\ell-1}^{(n)}(\omega)$.
  All of these depend on the choice of $\theta$'s and $x$'s.  No claim
  of uniqueness or measurability is made. The space $V_\ell^{(n)}(\om)$ is
  an approximate slow space.

  The proof will go by the following steps:
  \begin{enumerate}[(a)]
  \item \label{it:nrate} For almost all $\omega$, for arbitrary $\epsilon>0$
    and for sufficiently large $n$, $\|\mathcal
    L^{(n)}_\omega x\|\le e^{(\lambda_\ell+\epsilon)n}\|x\|$ for all
    $x\in V_\ell^{(n)}(\omega)$;
  \item \label{it:Cauchy} $V_\ell^{(n)}(\omega)$ is a Cauchy sequence
    for almost all $\omega$ -- we define the limit to be
    $V_\ell(\omega)$;
  \item \label{it:equivar} The $V_\ell(\omega)$ are equivariant: $\mathcal L_\omega(V_\ell(\omega))\subseteq
    V_\ell(\sigma(\omega))$;
  \item \label{it:liminfrate} If $x\not\in V_{\ell+1}(\omega)$, then
  $\| \mathcal L^{(n)}_{\om} v \| > e^{(\lambda_\ell-\ep)n} d(v,V_{l+1}(\om))$
  for large $n$.
\item \label{it:unifliminf} For all $a>0$ and $\epsilon>0$, there exists $n_0$ so that
for all $n\ge n_0$ and all $x\in S_X$ such that $d(x,V_{\ell+1}(\om))\ge a$, one has
$\|\LL_\om^{(n)}x\|\ge e^{(\lambda_\ell-\ep)n}$.
    \setcounter{saveenum}{\value{enumi}}
  \end{enumerate}
  The remaining steps are proved by induction on $\ell$.
  \begin{enumerate}[(a)]
    \setcounter{enumi}{\value{saveenum}}
  \item \label{it:limsuprate} If $x\in V_\ell(\omega)$, then
    $\limsup\frac 1n\log\|\mathcal L^{(n)}_\omega x\|\le
    \lambda_\ell$;
  \item \label{it:mble}$\omega\mapsto V_\ell(\omega)$ is measurable;
  \item \label{it:inducexpts} The restriction, $\mathcal R_\ell$, of
    $\mathcal R$ to $V_\ell(\omega)$ has the same exponents as
    $\mathcal R$ with the initial $\D_{\ell-1}$ exponents removed.
  \end{enumerate}

\paragraph{\emph{Proof of} \eqref{it:nrate}}
Note that by construction
$$
\det\left( (\theta^{(n)}_i(\mathcal L^{(n)}_\omega x^{(n)}_j))_{1\le
    i,j\le \D_{\ell-1}} \right) \ge
2^{-\D_{\ell-1}}E_{\D_{\ell-1}}(\mathcal L_\omega^{(n)}).
$$
For an arbitrary $x\in V_\ell^{(n)}\cap S_X$, let $\phi \in S_{X^*}$
be such that $\phi(\mathcal L^{(n)}_\omega x)=\|\mathcal
L^{(n)}_\omega x\|$.  Then, adding a column for $x$ and a row for
$\phi$ to the matrix $U((\theta_i^{(n)}),(x_j^{(n)}))_{1\le i,j\le
  \D_{\ell-1}}$, we see that the $x$ column has all 0 entries except
for the $1+\D_{\ell-1}$-st (by definition of $V^{(n)}_\ell(\omega)$),
and so we arrive at the bound (uniform over $x\in S_X \cap
V_\ell^{(n)})$,
\begin{equation}\label{eq:unifexprate}
  {2^{-\D_{\ell-1}}}E_{\D_{\ell-1}}(\mathcal L^{(n)}_\omega)
  \|\mathcal L^{(n)}_\omega x\|
  \le E_{1+\D_{\ell-1}}(\mathcal L^{(n)}_\omega).
\end{equation}
The conclusion follows from Lemma~\ref{lem:expDefn}.

\paragraph{\emph{Proof of} \eqref{it:Cauchy}}
Let us assume that $n_0$ is chosen large enough that for all $n\ge
n_0$, the following conditions are satisfied: $\|\mathcal L_\omega\|$,
$\|\mathcal L_{\sigma^n\omega}\|$ are less than $e^{\epsilon n}$;
$\|\mathcal L^{(n)}_\omega x\|\leq e^{(\lambda_\ell+\epsilon)n}\|x\|$
for all $x\in V_\ell^{(n)}$; and $\|\mathcal L_\om^{(n)}x\|\geq
e^{(\lambda_{\ell-1}-\epsilon)n}\|x\|$ for all $x\in Y_{\ell-1}^{(n)}(\om)$
(using integrability of $\log\|\LL_\om\|$; \eqref{it:nrate}; and Lemma \ref{lem:genseqs}).
Let $n\ge n_0$.  Let $x\in
V_\ell^{(n)}(\om)\cap S_X$ and write $x=u+w$ where $u\in
V_\ell^{(n+1)}(\om)$ and $w\in Y^{(n+1)}_{\ell-1}(\om)$.  Now we have
$$
\|\mathcal L_\om^{(n+1)}x\|\le e^{(\lambda_\ell+\epsilon)n} \|\mathcal
L_{\sigma^n\omega}\| \le e^{(\lambda_\ell+2\epsilon)n}.
$$
We also have $\|u\|\le 1+\|w\|$, $\|\mathcal L^{(n+1)}_\omega w\|\ge
e^{(\lambda_{\ell-1}-\epsilon)(n+1)}\|w\|$ and $\|\mathcal
L^{(n+1)}_\omega u\|\le e^{(\lambda_\ell+\epsilon)(n+1)}(1+\|w\|)$.
Manipulation with the triangle inequality yields
\begin{equation}\label{eq:wbound}
  \|w\|\le e^{-n(\lambda_{\ell-1}-\lambda_\ell-4\epsilon)}.
\end{equation}
Hence, each point in the unit sphere of $V_\ell^{(n)}(\omega)$ is
exponentially close to $V_\ell^{(n+1)}(\omega)$. Since the two spaces
have the same codimension, one obtains a similar inequality in the
opposite direction (possibly with the constant increased by a factor
of at most 2) - see Lemma 233 of \cite{Kato2}.  This establishes that
$V_\ell^{(n)}(\omega)$ is a Cauchy sequence.

\paragraph{\emph{Proof of} \eqref{it:equivar}}
We argue essentially as in \eqref{it:Cauchy}. For
large $n$, we take $v\in V^{(n+1)}_\ell(\omega)\cap S_X$. We write
$\mathcal L_\omega(v)$ as $u+w$ with $u\in
V^{(n)}_\ell(\sigma(\omega))$ and $w\in
Y^{(n)}_{\ell-1}(\sigma(\omega))$.  We have
bounds of the form $\|\mathcal L^{(n+1)}_\omega v\|\lesssim
e^{\lambda_\ell n}$; $\|u\|\lesssim 1+\|w\|$, $\|\mathcal
L^{(n)}_{\sigma(\omega)}u\|\lesssim e^{\lambda_\ell n}(1+\|w\|)$ and
$\|\mathcal L^{(n)}_{\sigma(\omega)}w\|\gtrsim
e^{\lambda_{\ell-1}n}\|w\|$ (here $\lesssim$ means `is smaller up to
sub-exponential factors').  Combining the inequalities as before, one
obtains a bound $\|w\|\lesssim e^{-(\lambda_{\ell-1}-\lambda_\ell)n}$.
Taking a limit, we obtain $\mathcal L_\omega V_\ell(\omega)\subset
V_\ell(\sigma(\omega))$ as required.

\paragraph{\emph{Proof of} \eqref{it:liminfrate}}
Let $x\not\in V_{\ell+1}(\omega)$, with $\|x\|=1$.
For large $n$, if $x$ is written as
$u_n+v_n$ with $u_n\in V^{(n)}_{\ell+1}(\omega)$ and
$v_n\in Y^{(n)}_\ell(\om)$, then $\|v_n\|\ge
\frac 12d(x,V_{\ell+1}(\om))$ and $\|u_n\|\le 1+\|v_n\|$. By \eqref{eq:unifexprate},
$\|\mathcal L^{(n)}_\omega u_n\|\le
e^{(\lambda_{\ell+1}+\epsilon)n}(1+\|v_n\|)$ for large $n$, while
Lemma \ref{lem:genseqs}
gives $\|\mathcal L^{(n)}_\omega v_n\|\ge
e^{(\lambda_{\ell}-\epsilon)n}\|v_n\|$ for large $n$. The
conclusion follows. The proof of \eqref{it:unifliminf} is the same,
using the uniformity in Lemma \ref{lem:genseqs}.

For the inductive part, notice that the case $\ell=1$ is trivial.
Suppose that the claims have been established for all $k<\ell$.

\paragraph{\emph{Proof of} \eqref{it:limsuprate}}
Let $V_{2,\ell-1}^{(n)}(\omega)$ be the `$V_2$ space' for the random
dynamical system $\mathcal R_{\ell-1}$, that is the original system
restricted to $V_{\ell-1}(\omega)$.  Call the limit
$V'_{\ell}(\omega)$.  From \eqref{eq:wbound} and the inductive step
\eqref{it:inducexpts}, we see that the distance between
$V'_\ell(\omega)$ and $V_{2,\ell-1}^{(n)}(\omega)$ is at most
$e^{-(\lambda_{\ell-1}-\lambda_\ell-4\epsilon)n}$ for sufficiently
large $n$. Given $v\in V'_\ell(\omega)\cap S_X$, it can be
written as $u+w$ with $u\in V_{2,\ell-1}^{(n)}(\omega)$ and $w\in V_{\ell-1}(\om)$
with
$\|w\|\le e^{-(\lambda_{\ell-1} -\lambda_\ell-4\epsilon)n}$.  Then,
$\|\mathcal
L^{(n)}_\omega u\|\lesssim e^{n\lambda_\ell}$ by \eqref{it:nrate}
and \eqref{it:inducexpts} for $\mathcal R_{\ell-1}$;
and $\|\mathcal L^{(n)}_\omega w\|\lesssim e^{-(\lambda_{\ell-1}-\lambda_\ell)n}
\cdot e^{\lambda_{\ell-1}n}$. So $\|\mathcal L^{(n)}_\om v\|\lesssim e^{\lambda_\ell n}$
by the triangle inequality.  From \eqref{it:liminfrate},
we deduce $V'_\ell(\omega) \subseteq
V_\ell(\omega)$.  Since, by \eqref{it:inducexpts} (applied to
 $\mathcal R_{\ell-1}$), $V_\ell(\omega)$
and $V'_\ell(\omega)$ have the same finite co-dimension as subspaces of
$V_{\ell-1}(\omega)$, $V_\ell(\omega)= V'_\ell(\omega)$ and
\eqref{it:limsuprate} follows.

\paragraph{\emph{Proof of} \eqref{it:mble}}
From \eqref{it:limsuprate} and \eqref{it:liminfrate}, we see that
$V_\ell(\omega)=\{v\colon \limsup \frac 1n\log \|\mathcal
L^{(n)}_\omega v\|\le \lambda_\ell\}$ and the assumptions of Lemma \ref{lem:grassmeas} hold. Measurability of $V_\ell(\omega)$ follows.

\paragraph{\emph{Proof of} \eqref{it:inducexpts}}
Let $W(\omega)=\{w\in V_{\ell-1}(\omega)\colon \|w\|=1, \
d(w,V_\ell(\omega))\geq \frac 12\}$.  Let $\epsilon>0$. We claim that
for sufficiently large $n$,
\begin{equation}\label{eq:minang}
  d(w',v')>e^{-\epsilon n}\text{ for all
    $w'\in S_X\cap \mathcal L^{(n)}_\omega W(\omega)$
    and $v'\in V_\ell(\sigma^n\omega)$.}
\end{equation}

Let $\delta < \frac{\ep (\lambda_{\ell-1}-\lambda_\ell)}
{4(\lambda_{\ell-1}-\lambda_\ell+\ep)}$.  Let $g_\ell(\om)=\sup_{p\in
  \N}e^{-p(\lambda_\ell+\delta)} \left\|\mathcal
  L^{(p)}_{\omega}|_{V_{\ell}(\om)}\right\|$.  Notice that $\log^+
g_\ell(\om) \leq \log^+ \|\LL_\om \| +\max(-\lam_\ell -\del,0) +
\log^+ g_\ell(\sig\om)$.  Lemma \ref{lem:tempered} ensures that
$\lim_{n\to\infty}\frac 1n \log^+ g_\ell(\sig^{n}\om)=0$.

Then, there exists $n_0(\omega)$ such that for $p,n\ge n_0$, one has
\begin{equation}\label{eq:growthbounds}
  \begin{split}
    \|\mathcal L^{(p)}_{\sigma^n\omega}z\|& \le \tfrac13 \exp(n\tfrac
    \epsilon 2 + p(\lambda_{\ell-1}+\delta))\|z\|\text{ for all $z\in
      V_{\ell-1}(\sig^n\om)$;}
    \\
    \|\mathcal L^{(p)}_{\sigma^n\omega}v\|& \le \tfrac13 \exp(n\tfrac
    \epsilon 2 + p(\lambda_{\ell}+\delta))\|v\| \text{ for all $v\in
      V_\ell(\sigma^n\omega)$}.
  \end{split}
\end{equation}
Additionally by \eqref{it:unifliminf}, $n_0$ may be
chosen so that
\begin{equation}\label{eq:growthbounds2}
  e^{n(\lambda_{\ell-1}-\delta)}<\|\mathcal L^{(n)}_\omega w\|
  <e^{n(\lambda_{\ell-1}+\delta)}\text{ for all $w\in W(\omega)$ and $n\ge n_0$.}
\end{equation}

We will show \eqref{eq:minang} by contradiction.  Suppose $d(\mathcal
L^{(n)}_\omega w,V_\ell(\sigma^n\omega)) <e^{-\epsilon n}\|\mathcal
L^{(n)}_\omega w\|$ for some $n>n_0$, and such that $\epsilon
n/(\lambda_{\ell-1}-\lambda_\ell)>n_0$.  Write $\mathcal
L^{(n)}_\omega w=v+z$, with $v\in V_\ell(\sig^n\omega)$ and
$\|z\|<e^{-\epsilon n}\|v\|$. Now $\mathcal L^{(p+n)}_\omega w
=\mathcal L^{(p)}_{\sigma^n\omega}v + \mathcal
L^{(p)}_{\sigma^n\omega}z$.  Taking $p=\epsilon
n/(\lambda_{\ell-1}-\lambda_\ell)$, the bounds on the two terms coming
from \eqref{eq:growthbounds} agree, giving
$$
\|\mathcal L^{(n+p)}_\omega w\|\le e^{-\epsilon
  n/2}e^{(p+n)(\lambda_{\ell-1}+\delta)}.
$$
One checks, however, that by the choice of $\delta$, this is smaller
than $e^{(p+n)(\lambda_{\ell-1}-\delta)}$, contradicting
\eqref{eq:growthbounds2}. This establishes claim \eqref{eq:minang}.
Notice that combining \eqref{eq:minang} and \eqref{eq:growthbounds2},
we see that the restriction of the random dynamical system to the
equivariant family
$Q_{\ell-1}(\omega)=V_{\ell-1}(\omega)/V_\ell(\omega)$ satisfies for
all sufficiently large $n$,
\begin{equation}\label{eq:quotexp}
  \|\bar{\mathcal L}^{(n)}_\omega \bar w\|_{Q_{\ell-1}(\sig^n\omega)}
  \ge e^{(\lambda_{\ell-1}-\epsilon)n}\|\bar w\|_{Q_{\ell-1}(\omega)}
  \text{ for all $\bar w\in Q_{\ell-1}(\omega)$},
\end{equation}
where $\bar{\mathcal L}_\omega$ denotes the induced action of
${\mathcal L}_\omega$ on $Q_{\ell-1}(\omega)$.

Set $\mult=\mult_{\ell-1}$. To complete the proof of
\eqref{it:inducexpts}, let $n>n_0$ be arbitrary; let
$v_1,\ldots,v_{k-\mult}$ be unit vectors in $V_\ell(\omega)$ and
$w_1,\ldots,w_{\mult}$ be unit vectors in $V_{\ell-1}(\omega)$. Then
\begin{align*}
  &d_k\mathcal L^{(n)}_\omega(v_1,\ldots,v_{k-\mult},w_1,\ldots,w_{\mult})\\
  &\ge d_{k-\mult}\mathcal L^{(n)}_\omega(v_1,\ldots,v_{k-\mult})
  d_{\mult}\bar{\mathcal L}^{(n)}_\omega(\bar w_1,\ldots,\bar
  w_{k-\mult}),
\end{align*}
where $\bar w_i$ is $w_i+V_\ell(\omega)$.  We therefore see
$$
D_k\mathcal L_\omega^{(n)}|_{V_{\ell-1}(\omega)} \ge
D_{k-\mult}\mathcal L_\omega^{(n)}|_{V_\ell(\omega)} \cdot D_\mult
\bar{\mathcal L}_\omega^{(n)}.
$$
By \eqref{eq:quotexp} and Lemma \ref{lem:unifexpDk}, $D_\mult \bar{\mathcal L}_\omega^{(n)}\gtrsim
e^{\lambda_{\ell-1}mn}$.

This gives a matching upper bound for $D_{k-\mult}\mathcal
L_\omega^{(n)}|_{V_\ell(\omega)}$ to the lower bound that we obtained
in Lemma \ref{lem:ExponentsOfReducedCocycle}.  Hence we deduce the
first $k$ exponents of $\mathcal R_{\ell-1}$ are
${\mult}={\mult}_{\ell-1}$ repetitions of $\lambda_{\ell-1}$ followed
by the first $k-{\mult}$ exponents of $\mathcal R_\ell$, establishing
\eqref{it:inducexpts}.
\end{proof}

The next corollary provides a \textsl{splitting} when the base $\sig$
is invertible.

\begin{cor}[The Oseledets splitting]\label{cor:splitting}
  Let $\mathcal R$ be a random dynamical system as in
  Theorem~\ref{thm:Osel}.  Suppose that the base $\sig$ is
  invertible. Then there exist $1\le r\le\infty$ and exceptional Lyapunov exponents
  and multiplicities as in Theorem \ref{thm:Osel}. Furthermore, there is
  an equivariant measurable direct sum decomposition
  \begin{equation*}
  	X=Z_1(\omega)\oplus \dots \oplus Z_r(\omega)\oplus V_\infty(\omega),
  \end{equation*}
  such that for $\PP$-a.e.\ $\omega$, $\dim Z_i(\omega)=m_i$ and
  $v\in Z_i(\omega)\setminus\{0\}$ implies $\lim_{n\to\infty}
  \frac1n\log\|\LL^{(n)}_\omega v\|=\lambda_i$; $v\in V_\infty(\omega)$
  implies $\limsup_{n\to\infty} \frac1n\log\|\LL^{(n)}_\omega v\|\le \kappa(\mathcal R)$.
\end{cor}

The deduction of this in general starting from a non-invertible filtration
theorem, including uniqueness of the complement,
appears in \cite{GTQuas}. We give a simple proof here, based on
duality, in the special case where $X$ is reflexive.  We make use of
the following facts valid for reflexive Banach spaces.  If $X$ is
reflexive and $\Theta$ is a closed subspace of $X^*$ of codimension
$k$, then its annihilator, $\Theta^\circ$ is $k$-dimensional. Further
if $\theta$ is a bounded functional such that
$\theta|_{\Theta^\circ}=0$, then $\theta\in\Theta$.

\begin{proof}
  Let $\mc{R}^*$ be the dual random dynamical system to $\mc{R}$ as
  defined above.  Applying Theorem~\ref{thm:Osel} to $\mc{R}^*$, and
  recalling from Theorem~\ref{thm:summ} that the Lyapunov exponents
  and multiplicities of $\mc{R}$ and $\mc{R}^*$ coincide, yields a
  $\sig^{-1}$ equivariant measurable filtration $X^*=V^*_1(\om)\supset
  \dots \supset V^*_r(\omega)\supset V^*_\infty(\om)$, with the same
  codimensions as those of $\mc{R}$.

  Let $Y_\ell(\omega)={V_\ell^*(\omega)}^\circ$. Notice that $\dim
  Y_\ell(\om)=\D_{\ell-1}$.  Since $V^*_{\ell}(\om)$ is measurable and
  $(\cdot) ^ \circ : \mc{G}(X^*) \to \mc{G}(X)$ is continuous \cite[IV
  \S2]{Kato}, then $Y_\ell(\om)$ is measurable.  Also, for every $\psi
  \in V^*_{\ell}(\sig\om)$, we have $\mathcal L_{\sig\om}^*\psi \in
  V^*_{\ell}(\om)$ by equivariance of $V^*_{\ell}(\cdot)$.  Hence, for
  every $y \in Y_\ell(\om)$, $0=\mathcal L_{\sig\om}^* \psi(y)=\psi
  (\mathcal L_{\om} y)$. Thus, $\mathcal L_{\om} Y_\ell(\om) \subset
  Y_\ell(\sig\om)$, yielding equivariance.

  \newcommand{\bigspace}{V_{\ell-1}(\om)} We define $Z_\ell(\omega)=
  Y_{\ell+1}(\omega)\cap V_\ell(\omega)$. It remains
  to show that $V_{\ell-1}(\om)=V_{\ell}(\omega)\oplus
  Z_{\ell-1}(\om)$, which would follow from showing
  $V_{\ell}(\omega)\oplus Y_{\ell}(\omega)=X$.
  Suppose this is not the
  case. Then, there exists $w\in V_\ell(\omega)\cap Y_\ell(\om)\cap
  S_X$.  Let $\theta \in S_{X^*}$ be such that $\theta(w)=1$.  Let
  $\bar{\theta}$ be the equivalence class of $\theta$ in
  $Q^*_{\ell-1}(\om)=V^*_{\ell-1}(\om)/V^*_\ell(\om)$.

  Since $Q^*_{\ell-1}(\omega)$ is a finite-dimensional space, and
  $\bar{\mathcal L}^*_\omega$ acts injectively on it by \eqref{eq:quotexp},
  we see $\bar{\mathcal L}^*_\omega$ is bijective.
  Also the quantity $C(\omega)=\inf_{n\in\mathbb
    N;\,\bar w\in S_X\cap Q^*_{\ell-1}(\omega)}
  e^{-(\lambda_{\ell-1}-\epsilon)n} \|\bar{\mathcal L}^{*(n)}_\omega
  \bar w\|_{Q^*_{\ell-1}(\omega)}$ is positive. Arguing as in
  Remark~\ref{rmk:diffDomain}, we see that $C(\omega)$ is measurable
  and hence exceeds some quantity $c$ on a set of positive measure.

  Let $\bar{\phi}_n\in Q^*_{\ell-1}(\sig^n\om)$ be such that
  $\bar{\LL}^{*(n)}_{\sig^n\om} \bar \phi_n = \bar \theta$.  Then,
  $\|\bar{\LL}^{*(n)}_{\sig^n\om} \bar \phi_n\|\geq
  C(\sig^n\om)e^{(\lam_{\ell-1}-\ep)n}\|\bar \phi_n\|$.  By
  ergodicity, there exist arbitrarily large values of $n$ for which
  $\|\bar \phi_n\|\leq c^{-1}e^{-(\lam_{\ell-1}-\ep)n}\|\bar
  \theta\|$.  On the other hand, if $\phi_n\in X^*$ is a
  representative of $\bar\phi_n$, then for every $\psi \in
  V_\ell^*(\sig^n\omega)$, one has $\mathcal L^{(n)}_\omega w\in
  Y_\ell(\sigma^n\omega)$, so that $\psi({\LL}^{(n)}_{\om}w)=0$. Hence
  $1=(\phi_n+\psi) ({\LL}^{(n)}_{\om}w)\leq \|\phi_n + \psi\|
  e^{(\lam_\ell +\ep)n}$.  Thus, $\|\bar \phi_n\| \geq e^{-(\lam_\ell
    +\ep)n}$, giving a contradiction.  Hence,
  $V_{\ell-1}(\om)=V_\ell(\omega)\oplus Z_{\ell-1}(\om)$ as required.
\end{proof}

\subsubsection*{Acknowledgments}
CGT acknowledges support from Australian Research Council Discovery
Project DP110100068. AQ acknowledges support from the Canadian NSERC,
and thanks the Universidade de S\~ao Paulo for the invitation to
deliver a mini-course from which this work originated.

config: arxiv


TITLE: Where does one more '$\rm m$' come from in the units?
QUESTION [4 upvotes]: $$\nabla \times A = B$$
$A$ is vector magnetic potential, $\mathrm{Wb/m}$
$B$ is magnetic field intensity, $\mathrm{Wb/m^2}$
Where does one more m come from for $B$? Is that from the gradient operator so it is in meter or something?

REPLY [3 votes]: The del operator is
$$\nabla = [\frac{d}{dx},\frac{d}{dy},\frac{d}{dz}] $$
and ends up taking derivatives of A with respect to x,y,z axis, which are  displacements measured in meters.
Since a derivative is a rate of change, you are finding the change in A(Wb/m) per meter, hence $$\frac{Wb/m}{m}=Wb/m^2$$.

REPLY [3 votes]: As you suggested the del operator provides the additional “per/meter”. The del is the rate of change with respect to position. So just like taking a derivative with respect to time divides the units by time so also taking a derivative with respect to position divides the units by distance.


TITLE: finding csc given sin
QUESTION [1 upvotes]: $\sin \theta= -\frac{3}{7}$.
Find $\csc\theta$.

I don't get it, is $x$ or $r$ the negative?
There just doesn't seem to be enough information given.

REPLY [2 votes]: $\sin$ is a function which takes any real argument, so in general you don't really need to think of it in terms of triangles.
But let's say that you want to think of it exclusively in terms of triangles, so that $\sin(\theta)$ is defined as follows: draw a right triangle with angle $\theta$; then $\sin(\theta)$ equals the length of the opposite side divided by the length of the hypothenuse; if we place the adjacent side on the positive $x$-axis, then length is taken to be "positive" if it goes to the right or up, and negative if it goes left or down. Then, it doesn't matter if you are considering the opposite side or the radius to be "negative", what matters is:

The value of $\sin\theta$; and
The relationship between $\sin\theta$ and $\csc\theta$.

Remember that, as long as $\sin\theta\neq 0$, then you have that
$$\csc\theta = \frac{\text{hypothenuse}}{\text{opposite}} = \frac{1}{\quad\frac{\text{opposite}}{\text{hypothenuse}}\quad} = \frac{1}{\sin\theta}.$$
So, since you know what $\sin\theta$ is, how much is $\csc\theta$?