Matrices, translations, and lattices

· Rank one Proved here

$$\GLn=\mathrm{GL}_1,\qquad\Wt=\FF^\times/\OO^\times\simeq\ZZ,\qquad\Wa=1,\qquad\Om=\Wt.$$

The valuation map sends \(\unif^j u\) to \(j\), for \(u\in\OO^\times\). The rotation is scalar multiplication.

$$\rot=\unif,\qquad\rot^{-2}\OO=\unif^{-2}\OO,\qquad\ell(\unif^j)=0.$$

The valuation index varies even though there are no root hyperplanes.

· Rank two: a translation and a rotation Proved here

$$s_1=\begin{pmatrix}0&1\\1&0\end{pmatrix},\quad s_0=\begin{pmatrix}0&\unif^{-1}\\\unif&0\end{pmatrix},\quad\rot=\begin{pmatrix}0&1\\\unif&0\end{pmatrix}.$$

Multiply the matrices rather than identify a rotation with a diagonal translation.

$$\rot s_1=\begin{pmatrix}1&0\\0&\unif\end{pmatrix}=\unif^{(0,1)},\qquad\rot^2=\unif\,1_2,$$ $$\rot s_1\rot^{-1}=s_0,\qquad\ell(\rot)=0,\qquad\ell(\unif^{(0,1)})=1.$$

Proof

Apply each product to a vector \((a,b)\): \(\rot s_1(a,b)=(a,\unif b)\); applying \(\rot\) twice gives \((\unif a,\unif b)\). The normalizer calculation in Properties gives length zero; the translation formula gives length one.

· A product down to integers Proved here

Take \(\lam=(1,-1)\), \(\muv=(0,2)\), and \(w=v=s_1\).

$$s_1\muv=(2,0),\qquad\lam+s_1\muv=(3,-1),\qquad s_1^2=1,$$ $$(\unif^{(1,-1)}s_1)(\unif^{(0,2)}s_1)=\operatorname{diag}(\unif^3,\unif^{-1}),\qquad\kap=2,\quad\ell=4.$$

The final length uses \(|3-(-1)|\); it is not obtained by adding the lengths of arbitrary factors.

· Rank three: grow the same chain Proved here

$$\rot(a,b,c)=(b,c,\unif a),\qquad\rot^3=\unif\,1_3,$$ $$L_0=\OO^3,\quad L_1=\OO^2\oplus\unif\OO,\quad L_2=\OO\oplus(\unif\OO)^2,\quad L_3=\unif\OO^3,$$ $$L_4=\unif\OO^2\oplus\unif^2\OO,\qquad L_{-1}=\unif^{-1}\OO\oplus\OO^2.$$

Each rotation advances one step; a scalar factor advances three.