!newcal U
!newfrak g
!newfrak m
!newop st
!newop mod
!newop Ad
!newop SO

\renewcommand{\O}{\mathcal{O}}
\renewcommand{\v}{\operatorname{v}}

\renewcommand{\i}{\underline{i}}
\renewcommand{\j}{\underline{j}}
\renewcommand{\k}{\underline{k}}

\newcommand{\Goo}{G^{00}}
\newcommand{\gm}{\g_\m}

\newcommand{\inner}[1]{\left<#1\right>}

\newcommand{\CHECKME}{\textbf{CHECKME}}
%\newcommand{\FIXME}[1]{\textbf{FIXME: #1}}
%\newcommand{\TODO}[1]{\textbf{TODO: #1}}

# Some non-standard analysis on compact Lie groups
Let $R := \R^{\U}$ be an ultrapower of the reals over a non-principal 
ultrafilter $\U$ on $\N$.
Let $\st : \O_R --> \R$ be the standard part map,
the domain $\O_R (= R$ of which is a valuation ring in $R$.
Let $\m = \st^{-1}(0)$ be the maximal ideal of $\O_R$,
and let $\v : R --> \Gamma\cup\{\infty\}$ be the corresponding valuation.

Let $G$ be a compact connected algebraic group over $\R$.
Then $G(\R)$ is naturally a Lie group,
and in fact any compact real Lie group is of this form.
% follows from Peter-Weyl - see
% https://www.math.ubc.ca/~cass/research/pdf/Compact-algebraic.pdf
Let $\g_\R$ be the Lie algebra of $G(\R)$.

We have the group $G(R) ~= G(\R)^{\U}$.
Since $G$ is compact,
$\st : G(R) --> G(\R)$ is a totally defined homomorphism.
% Note: in terms of \U, \st is the ultralimit map,
% so totality follows from the general topological Bourbakian fact that 
% ultrafilters converge on compact spaces.
Let $\Goo = \st^{-1}(e)$ be the kernel.
Note that $\Goo$ is the smallest subgroup of bounded index which is 
type-definable in the field structure.
\TODO{Cite something on this?}.

Let $\g_R$ be the non-standard Lie algebra
$\g_R = (\g_\R)^{\U}$.
Then $\g_R$ has the structure of a ``Lie algebra over $R$'',
i.e. a finite-dimensional $R$-vector space equipped with a Lie bracket,
with $\g_\R$ embedded as an $\R$-subalgebra.
Define the valuation $\v : \g_R --> \Gamma$ in the natural way,
namely such that $v(\st^{-1}(\g_\R \\ \{0\})) = \{0\}$ and
$\v(\lambda x) = \v(\lambda)+v(x)$.
Note that $\v([X,Y]) >= v(X) + v(Y)$.
Define $\gm := \{ g (- \g_R : \v(g) > 0 \} = \st^{-1}(0)$.
So picking an $\R$-module isomorphism $g ~= \R^n$ yields a valued $R$-module 
isomorphism $g_R ~= R^n$ (where $\v : R^n --> \Gamma$ is defined by 
$v((a_1,...,a_n)) = \min_i(\v(a_i))$), restricting to a valued $\O_R$-module 
isomorphism $\gm ~= \m^n$.

Let $\exp_R : \g_R -> G(R)$ be the non-standard exponential map $\exp_R := 
\exp^\U$.

By Lie theory, $\exp$ defines a homeomorphism between neighbourhoods of $0 (- 
\g_\R$ and $e (- G(\R)$.
It follows that $\exp_R$ restricts to a bijection $\exp_R : \gm -> \Goo$.

Write $(*)$ for the group operation pulled back to $\gm$ from $\Goo$,
i.e. that defined by
\[ \exp_R(X*Y) = \exp_R(X)\exp_R(Y) .\]
So the group $(\gm;*)$ is isomorphic to $\Goo$,
and we will work exclusively with $\gm$ from here on.

Note that the group identity is $0$,
and the group inverse precisely agrees with negation,
\[ (-X)*X = 0 .\]

Note also that group conjugation agrees with the adjoint action of $G(R)$ on 
$\g_R$,
\[ X^Y := Y*X*(-Y) = Ad_{\exp_R(Y)}(X) .\]
Indeed, this follows by transfer from the following formula for standard $X$ 
and $Y$,
which holds by definition of $\Ad_g$ as the derivative of conjugation by $g$:
\[ \exp(Y)\exp(X)\exp(-Y) = \exp(Ad_{\exp(Y)}(X)) .\]

Write $[X,Y]^*$ for the group commutator
\[ [X,Y]^* := (-X)*(-Y)*X*Y .\]

Let $\Theta(X,Y) := [X,Y]^* * [X,-Y]^*$.

By a _Lie word_ in an alphabet $\Sigma$ of order $n$ we mean a formal 
expression
    \[ [\sigma_1,[\sigma_2,[...,[\sigma_{n-1},\sigma_n]...]]] \]
where $\sigma_i (- \Sigma$. The \defnstyle{degree in $\sigma$} is the number 
of occurences of $\sigma$.

Lemma{thetaApprox}:
    Let $X,Y (- \gm$. Then
    \[ \Theta(X,Y) ~~ [Y,[Y,X]] \mod 2v(Y) + v(X) .\]
    .
Proof:
    Fix an $\R$-vector space norm $||\cdot||$ on $\g_\R$,
    and the corresponding norm $||(X,Y)|| := \max(||X||,||Y||)$ on $\g_\R^2$.

    Write $(*)$ for the locally defined operation on $\g_\R$ pulled back via 
    $\exp$ from the group operation on $G(\R)$, and again define
    $\Theta(X,Y) := [X,Y]^* * [X,-Y]^*$.

    Claim:
        There exists a series representation
        \[ \Theta(X,Y) = [Y,[Y,X]] + \Sigma_{i=0}^\infty a_iw_i(X,Y) \]
        where $a_i (- \R$ and each $w_i(X,Y)$ is a Lie word in $\{X,Y\}$ of 
        order at least 4 with degree in $Y$ at least 2,
        such that for $(X,Y)$ in some $||\cdot||$-ball $B$ around $(0,0)$,
        the series is $||\cdot||$-absolutely convergent
        and the stated equality holds.
        .
    Proof:
        Recall the Baker-Campbell-Hausdorff power series
        \[ X + Y + \frac12 [X,Y] + \frac1{12} ([X,[X,Y]] + [Y,[Y,X]]) + ... 
        ,\]
        which converges absolutely to $X*Y$ on a ball around $(0,0)$ 
        \cite[II.7.2~Proposition~1]{Bourbaki-Lie_1-3}.

        Now we express $\Theta(X,Y)$ as the composition of such series up to 
        order 3:
        *\[a* [X,Y]^*
          &= &&(-X-Y+\frac12[X,Y] - \frac1{12}([X,[X,Y]]+[Y,[Y,X]] + ...) \\\
          &   &&*(X+Y+\frac12[X,Y] + \frac1{12}([X,[X,Y]]+[Y,[Y,X]])) + ... \\\
          &= &&[X,Y] + \frac12 ([Y,[Y,X]] - [X,[X,Y]]) + ...
          ,*a\]*
        and hence
        \[ [X,-Y]^*
          = -[X,Y] + \frac12 ([Y,[Y,X]] + [X,[X,Y]]) + ... ;\]
        so
        \[ \Theta(X,Y) = [Y,[Y,X]] + ... .\]

        Moreover, one observes from this calculation that no term in this 
        series obtained for $\Theta(X,Y)$ has degree in $Y$ less than two, 
        since every term in $[X,Y]^*$ has degree at least one in $Y$,
        and those terms with degree $1$ in $Y$ occur with the opposite sign in 
        $[X,-Y]^*$, and so cancel in the series for the product.

        Now note that if $F(X,Y)$, $G(X,Y)$, and $H(X,Y)$ are series with zero 
        constant term which converge absolutely on balls around $(0,0)$,
        then also the composition $H(F(X,Y),G(X,Y))$ converges absolutely on 
        some such ball.
        Hence the series for $\Theta(X,Y)$ converges absolutely on some such 
        ball.
        .
    Now let $X,Y (- \gm$,
    and say $a,b (- R$ are such that $X' := \st(aX)$ and $Y' := \st(bY)$ are 
    non-zero elements of $\g_\R$,
    and $(X',Y') (- B$.
    So $v(a)=-v(X)$ and $v(b)=-v(Y)$.

    Let $e := \Sigma_{i=0}^\infty ||{ a_iw_i(X',Y') }|| (- \g_\R$.
    By the stated properties of the $w_i$ and bilinearity of the Lie bracket,
    for $a',b' (- [0,1] (= \R$ we have
    *\[a* ||{ \Theta(a'X',b'Y') - [b'Y',[b'Y',a'X']] }||
      &= ||{ \Sigma_{i=0}^\infty a_iw_i(a'X',b'Y') }|| \\\
      &<= \Sigma_{i=0}^\infty ||{ a_iw_i(a'X',b'Y') }|| \\\
      &<= \max(a'^2b'^2,a'b'^3)e .*a\]*

    In terms of valuations, then, we have
    *\[a* v(\Theta(X,Y) - [Y,[Y,X]])
    &= v(\Theta(a^{-1}X',b^{-1}Y') - [b^{-1}Y',[b^{-1}Y',a^{-1}X']] \\\
    &>= \min(v(a^{-2}b^{-2}),v(a^{-1}b^{-3})) \\\
    &> 2v(Y) + v(X) .*a\]*
    .

For $X (- \gm$,
let $C_X$ be the centraliser in the sense of the group,
or equivalently in the sense of the Lie algebra:
\[ C_X := \{ Y (- \gm : Y*X = X*Y \} = \{ Y (- \gm : [X,Y]=0 \}. \]


# Finding a group interval in $\SO_3^{00}$

Now specialise to the case $G=\SO_3$.

\TODO{Generalise the arguments of this section to arbitrary semisimple compact 
$G$ (or maybe just to simple?); use of $|\cdot|$ should be replaced with use 
of the (negative definite) Killing form, but we may have difficulties with 
seeing exactly what conjugation yields, and with centralisers being of higher 
dimension.}

In this case, $\g_\R$ has a natural Euclidean inner product $\inner{X,Y}$;
intrinsically, this is the negation of the Killing form.
So $\g_R$ and $\gm$ inherit positive definite inner products
and norms $||X|| = \inner{X,X}^{\frac12}$,
taking values in $R$ and $\m$ respectively

Explicitly, taking the standard basis $(\i,\j,\k)$ for $\g_R$ where 
$[\i,\j]=\k$ and $[\j,\k]=\i$ and $[\k,\i]=\j$, the inner product corresponds to the 
standard Euclidean inner product on $R^3$,
\[ \inner{a\i+b\j+c\k,a'\i+b'\j+c'\k} = aa'+bb'+cc' .\]

Note that for $X (- \gm \\ \{0\}$, the centraliser of $X$ is just its linear 
span, $C_X = RX \cap \gm$.

Lemma{conjClass}:
    For $X (- \gm$,
    \[ X^{\gm} := \{ X^Y : Y (- \gm \} =
        \{ Z : ||Z|| = ||X||\ \text{and}\ Z ~~ X \mod v(X) \} .\]
    .
Proof:
    Rather obvious if you think about matrices.
    \TODO{Actual proof}.
    .

Lemma{halfInterval}:
    Let $a (- \m$ with $a>0$.
    Then
    \[ \{ \Theta(a\k,b\i)^Z : b (- \m, Z (- \gm \} \cap C_{a\k}
      = \{ c\k : c (- \m \;\text{and}\; c <= 0 \;\text{and}\; v(c) > v(a) \}
    .\]
    .
Proof:
    Let $b (- \m$, and let $W := \Theta(a\k,b\i)$.
    Then $[b\i,[b\i,a\k]] = -ab^2\k$,
    so by Lemma~\ref{thetaApprox},
    \[ W ~~ -ab^2\k \mod v(ab^2) .\]

    In particular $v(W) = v(ab^2)$.

    Now by Lemma~\ref{conjClass},
    if $Z (- \gm$ and $W^Z = c\k$,
    then $c ~~ -ab^2 \mod v(ab^2)$,
    and so $c <= 0$ and $v(c) = v(ab^2) > v(a)$.

    Conversely, we have
    \[ ||W|| ~~ ab^2 \mod v(ab^2) ,\]
    and so
    \[ W ~~ ||W||\k \mod v(W) ,\]
    and so by Lemma~\ref{conjClass} there exists $Z (- \gm$ such that
    $W^Z = ||W||\k$.
    We conclude on noting that $v(||W||) = v(ab^2)$ takes all values greater 
    than $v(a)$ as $b$ varies in $\m$.

    \TODO{More abstract proof which is more likely to generalise?
    Do we always have $K([Y,[Y,X]],X) <= 0$?}
    .


# Acknowledgements
Thanks to Mohammed Bardestani for helpful conversations.
