Warm-up: \(\SL_2/\SO_2\)
Two-by-two symmetric matrices of determinant one form a surface you can draw. It has one of everything that \(\Esix/\Ffour\) has several of: one direction of size, one direction of turning, one root.
The groups and the spaceWhich matrices, and what does the group do to them?
(the groups) Definition
The special linear group and the special orthogonal group are
\[\SL_2(\R)=\{g\in M_2(\R):\ \det g=1\},\] \[\SO_2=\{k\in\SL_2(\R):\ kk^T=I\}.\]“Special” means \(\det=1\); “orthogonal” means \(kk^T=I\), i.e. \(k\) preserves lengths, \(|kv|=|v|\).
(\(\SO_2\) is the rotations) Proved here
\[\SO_2=\{k_\theta:\ \theta\in\R\},\qquad k_\theta=\begin{pmatrix}\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\end{pmatrix}.\]Proof
- \(k^Tk=I\) (equivalent to \(kk^T=I\) for square matrices) says that the columns of \(k\) are orthonormal.
- A unit first column is \((\cos\theta,\sin\theta)^T\) for some \(\theta\); a unit vector orthogonal to it is \(\pm(-\sin\theta,\cos\theta)^T\).
- \(\det k=1\) picks the sign \(+\), which gives \(k=k_\theta\). Conversely each \(k_\theta\) satisfies \(k_\theta k_\theta^T=I\) and \(\det k_\theta=\cos^2\theta+\sin^2\theta=1\).
Geometrically, \(\SL_2(\R)\) is the linear maps of the plane that preserve area and orientation, and \(\SO_2\) is those that also preserve length: the rotations.
(the space) Definition
Let \(\Pd{2}\) be the set of real symmetric \(2\times2\) matrices of determinant one with positive upper-left entry,1
\[\Pd{2}=\left\{S=\begin{pmatrix}a&b\\ b&c\end{pmatrix}:\ a,b,c\in\R,\ \ ac-b^2=1,\ \ a>0\right\}.\]The group \(\SL_2(\R)\) acts on it by
\[\SL_2(\R)\times\Pd{2}\to\Pd{2},\qquad (g,S)\mapsto gSg^T .\]The action is well defined:
\[(gSg^T)^T=gS^Tg^T=gSg^T,\] \[\det(gSg^T)=\det(g)^2\det S=1,\] \[v^T(gSg^T)v=(g^Tv)^TS(g^Tv)>0\quad(v\ne0).\]Three entries \(a,b,c\) and one equation leave a surface, of dimension \(3-1=2\).
- Why \(a>0\). From \(ac-b^2=1\) we get \(ac=1+b^2>0\), so \(a\) and \(c\) have the same sign. The eigenvalues \(\lambda_1,\lambda_2\) of \(S\) satisfy \(\lambda_1\lambda_2=\det S=1\) and \(\lambda_1+\lambda_2=a+c\), so \(a>0\) says exactly that both eigenvalues are positive: \(S\) is positive definite, \(v^TSv>0\) for \(v\ne0\). Without the condition we would also get \(-S\) for every \(S\) (for \(2\times2\) matrices \(\det(-S)=\det S\)). We keep \(a>0\) because the matrices \(gg^T\) are always positive definite, \(v^Tgg^Tv=|g^Tv|^2>0\), and these are the ones that the quotient \(\SL_2(\R)/\SO_2\) produces () and the ones that are ellipses. The negative ones return over \(\C\) in the last section. ↩
Coordinates: from \(a,b,c\) to \(t,x,y\)How do three matrix entries become a point on a surface?
Replace the entries \(a,b,c\) by the average of the diagonal, half its difference, and the off-diagonal entry:
\[t=\frac{a+c}{2},\qquad x=\frac{a-c}{2},\qquad y=b,\] \[a=t+x,\qquad c=t-x,\qquad b=y .\]So every symmetric matrix is
\[S=\begin{pmatrix}t+x&y\\ y&t-x\end{pmatrix}=t\begin{pmatrix}1&0\\0&1\end{pmatrix}+x\begin{pmatrix}1&0\\0&-1\end{pmatrix}+y\begin{pmatrix}0&1\\1&0\end{pmatrix},\]and the determinant becomes a difference of squares:
Hence \(\det S=1\) is the hyperboloid \(t^2=1+x^2+y^2\), and \(a>0\) is \(t>0\), its upper sheet. Over each point \((x,y)\) of the plane there is exactly one \(t=\sqrt{1+x^2+y^2}\): the surface is a bowl lying over the \((x,y)\)-plane, with \(t\) its height.
| \(S\) | \((t,x,y)\) | where on the bowl |
|---|---|---|
| \(I\) | \((1,0,0)\) | the bottom |
| \(\begin{pmatrix}e^{2h}&0\\0&e^{-2h}\end{pmatrix}\) | \((\cosh2h,\ \sinh2h,\ 0)\) | up the \(x\)-direction |
| \(\begin{pmatrix}\cosh2h&\sinh2h\\ \sinh2h&\cosh2h\end{pmatrix}\) | \((\cosh2h,\ 0,\ \sinh2h)\) | up the \(y\)-direction |
Three pictures of one pointWhat does a point of \(\Pd{2}\) look like?
- (a) The hyperboloid is the surface itself, in the coordinates \((t,x,y)\) above.
- (b) The disk flattens the bowl, as the literature usually draws it (the Poincaré disk): \[w=\frac{x+\mathrm iy}{1+t}\in\Disk=\{w\in\C:\ |w|<1\}.\]
- (c) The ellipse of \(S\): \[\Ell{S}=\{v\in\R^2:\ v^TS^{-1}v\le1\},\qquad \text{area}(\Ell{S})=\pi\sqrt{\det S}=\pi .\]
Why picture (c). Pictures (a) and (b) show where \(S\) is; picture (c) shows what \(S\) does. Two reasons to keep it:
- The action becomes visible. By (d), \(\Ell{gSg^T}=g(\Ell{S})\): moving \(S\) is applying the linear map \(g\) to its ellipse, and \(\SO_2\) is exactly the maps that leave the round disk \(\Ell{I}\) in place. On the hyperboloid the same \(g\) is a Lorentz transformation, which is hard to see.
- It is the only one of the three that survives one size up. The \(3\times3\) analogue \(\Pd{3}\) is 5-dimensional, with no picture like (a) or (b), but each of its points is still an ellipsoid in \(\R^3\) (next page).
(the three pictures agree) Proved here
(a) \(S\mapsto(t,x,y)\) is a bijection from \(\Pd{2}\) onto the upper sheet \(\{t^2-x^2-y^2=1,\ t>0\}\); each \(g\in\SL_2(\R)\) acts on it by a linear map preserving \(t^2-x^2-y^2\).
(b) \(S\mapsto w\) is a bijection onto the open unit disk \(\Disk\).
(c) \(S\mapsto\Ell{S}\) is a bijection onto the centred ellipses of area \(\pi\).
(d) The action is the action on ellipses:
\[\Ell{gSg^T}=g(\Ell{S}).\]Proof
- (a) \(a>0\) and \(ac-b^2=1\) give \(c>0\), so \(t=\tfrac{a+c}{2}>0\); conversely \(t>0\) and \(t^2=1+x^2+y^2\) give \(a=t+x>0\) since \(t>|x|\). \(S\mapsto gSg^T\) is linear in \((a,b,c)\), hence in \((t,x,y)\), and it preserves \(\det S=t^2-x^2-y^2\).
- (b) On the sheet, \(t=\cosh\rho\) and \(x+\mathrm iy=\sinh\rho\,e^{\mathrm i\psi}\) for some \(\rho\ge0\), so \[w=\frac{\sinh\rho}{1+\cosh\rho}\,e^{\mathrm i\psi}=\tanh\tfrac\rho2\,e^{\mathrm i\psi},\] and \(\rho\mapsto\tanh\frac\rho2\) maps \([0,\infty)\) onto \([0,1)\).
- (c) A positive definite \(S\) is determined by \(\Ell{S}\), whose boundary is the conic \(v^TS^{-1}v=1\); and \(\Ell{S}=S^{1/2}(\Disk)\) has area \(\pi\det S^{1/2}=\pi\).
- (d) \(v\in g(\Ell{S})\iff (g^{-1}v)^TS^{-1}(g^{-1}v)\le1\iff v^T(gSg^T)^{-1}v\le1\).
\(\SL_2\), \(\SO_2\) and the quotientHow is a quotient of groups the same thing as a set of ellipses?
Read each object as what it does to the unit disk \(\Disk\subset\R^2\):
| object | what it is | dimension |
|---|---|---|
| \(\SL_2(\R)\) | linear maps of the plane preserving area; \(g(\Disk)\) is an ellipse of area \(\pi\) | 3 |
| \(\SO_2\) | the rotations: the \(g\) with \(g(\Disk)=\Disk\) | 1 |
| \(\SL_2(\R)/\SO_2\) | \(g\) up to a rotation first, \(g\sim gk\): only the ellipse \(g(\Disk)\) is remembered | \(3-1=2\) |
(the quotient) Proved here
The map \(g\mapsto gg^T\) induces a bijection
Proof
- \(v\in g(\Disk)\iff |g^{-1}v|\le1\iff v^T(gg^T)^{-1}v\le1\), so \(\Ell{gg^T}=g(\Disk)\).
- \(gg^T=g'g'^T\iff (g^{-1}g')(g^{-1}g')^T=I\iff g^{-1}g'\in\SO_2\) (determinant one and orthogonal).
- Onto: \(S\in\Pd{2}\) has the square root \(g=S^{1/2}\in\SL_2(\R)\), with \(gg^T=S\).
So the stabilizer of the point \(I\) (the round disk) is \(\SO_2\), and \(\SL_2(\R)\) moves \(I\) to every point: this is the pattern “space \(=G/K\)” of every page that follows.
Size and angleWhat are the two sliders?
(polar coordinates) Proved here
Every \(S\in\Pd{2}\) is
\[S=k_\varphi\,\diag(e^{2h},e^{-2h})\,k_\varphi^T,\qquad h\ge0,\]with \(k_\varphi\) the rotation of , \(h\) unique, and \(\varphi\) unique modulo \(\pi\) when \(h>0\).
Proof
The spectral theorem gives \(S=kDk^T\) with \(k\in\SO_2\) and \(D=\diag(\lambda_1,\lambda_2)\), \(\lambda_1\ge\lambda_2>0\); \(\lambda_1\lambda_2=1\) gives \(\lambda_1=e^{2h}\), \(h\ge0\). The eigenvalues fix \(h\). If \(h>0\) the eigenline of \(e^{2h}\) is the line at angle \(\varphi\), which fixes \(\varphi\) modulo \(\pi\).
In picture (c), \(h\) is the size of the deformation (semi-axes \(e^{h}\) and \(e^{-h}\)) and \(\varphi\) is the direction of the long axis. In picture (a), \(t=\cosh 2h\), so \(h\) is the height on the sheet and \(\varphi\) is half the angle around the axis.
- The green line \(\{\diag(e^{2h},e^{-2h})\}\) meets every orbit of \(\SO_2\): it plays the role of the Cartan subspace.
- Each orange circle \(\{kSk^T\}\) is an orbit of \(\SO_2\), of dimension 1, except the point \(I\), where it shrinks to dimension 0.
Size and angle are exactly the two kinds of directions that return on every later page: size becomes the Cartan subspace \(\af\), angle becomes the orbit of \(K\).
One root of multiplicity oneHow do size and angle look in the Lie algebra?
Differentiate at \(I\). The involution \(\theta(g)=(g^T)^{-1}\) has derivative \(Y\mapsto-Y^T\), and
\[\slf_2=\kf\oplus\pf,\qquad \kf=\so_2=\C\,Z_{12},\quad Z_{12}=\begin{pmatrix}0&1\\-1&0\end{pmatrix},\qquad \pf=\{Y=Y^T,\ \tr Y=0\}=\af\oplus\C\,Y_{12},\] \[\af=\{H=\diag(h,-h)\},\qquad Y_{12}=\begin{pmatrix}0&1\\1&0\end{pmatrix}.\]For \(H\in\af\),
\[[H,Y_{12}]=2h\,Z_{12},\qquad [H,Z_{12}]=2h\,Y_{12} .\]So \(\operatorname{ad}H\) swaps the angle directions \(Y_{12}\) and \(Z_{12}\) with factor \(\alpha(H)=2h=h_1-h_2\) (where \(H=\diag(h_1,h_2)\)). This \(\alpha\) is the one restricted root, up to sign, and its multiplicity is \(\mult=\dim\C Y_{12}=1\).
| directions | count |
|---|---|
| size: \(\af\) | rank \(=1\) |
| angle: one root, multiplicity \(\mult\) | \(\mult=1\) |
| \(\dim\Pd{2}\) | \(1+\mult=2\) |
The pattern for later: \(\dim=\text{rank}+\mult\cdot(\text{number of positive roots})\). For \(\SL_3/\SO_3\) it reads \(2+1\cdot3=5\), for \(\Esix/\Ffour\) \(2+8\cdot3=26\).
Where this surface sits nextHow does \(\Pd{2}\) appear inside the \(3\times3\) story?
Put a \(2\times2\) block in rows and columns \(i,j\) and a \(1\) elsewhere:
\[\Pd{2}\hookrightarrow\Pd{3},\qquad \begin{pmatrix}a&b\\b&c\end{pmatrix}\mapsto\begin{pmatrix}a&b&0\\b&c&0\\0&0&1\end{pmatrix}\quad(i,j)=(1,2),\]and the same for \((1,3)\) and \((2,3)\). The three copies of this surface through \(I\) correspond to the three roots \(\alpha_{12},\alpha_{13},\alpha_{23}\) of the roots page. For \(\Esix/\Ffour\) each copy becomes 9-dimensional: one size direction and \(\mult=8\) angle directions.
Over \(\C\). The pages work with the complex variety \(\{S\in\mathrm{Sym}_2(\C):\det S=1\}\), on which \(\SL_2(\C)\) acts with stabilizer \(\SO_2(\C)\). Its real points with \(S\) positive definite are \(\Pd{2}\); the lower sheet \(t<0\) (negative definite \(S\)) is a second real piece, and over \(\C\) the two are joined: \(g=\diag(\mathrm i,-\mathrm i)\) has \(\det g=1\) and \(g(-I)g^T=I\). The real pictures show the shape; the complex variety is what the algebraic statements (roots, Borel subgroups) are about.