Classical model · page 2 of 3

\(\SL_3/\SO_3\): ellipsoids and the matrix lab

The same construction as the warm-up \(\SL_2/\SO_2\), one size up: symmetric \(3\times3\) matrices of determinant one, which are the ellipsoids of volume \(\tfrac43\pi\). Every step can still be computed by hand.

Roadmap of this page

The space and the actionWhat does the group do to a symmetric matrix?

(the classical model) Definition

As in the warm-up, one size up:

\[\SL_3(\R)=\{g\in M_3(\R):\ \det g=1\},\qquad \SO_3=\{k\in\SL_3(\R):\ kk^T=I\},\] \[\Pd{3}=\{S\in M_3(\R):\ S^T=S,\ \det S=1,\ v^TSv>0\ \text{for}\ v\ne0\},\]

with the action \((g,S)\mapsto gSg^T\). Letting the entries be complex gives the variety

\[\Xs{\R}=\{S\in\mathrm{Sym}_3(\C):\ \det S=1\},\qquad \SL_3(\C)\times\Xs{\R}\to\Xs{\R},\quad (g,S)\mapsto g\,S\,g^T,\]

where \(\mathrm{Sym}_3(\C)=\{S\in M_3(\C):S^T=S\}\), a vector space of dimension 6. \(\Pd{3}\subset\Xs{\R}\) is its real positive part.

\(\Pd{3}\) is what the pictures show; \(\Xs{\R}\) is what the algebraic statements (transitivity, roots, Borel subgroups) are about, since those need square roots of arbitrary numbers. The action is well defined because

\[(gSg^T)^T=gS^Tg^T=gSg^T,\] \[\det(gSg^T)=\det(g)^2\det S=\det S .\]

Both equalities use that matrix multiplication is associative and that \(\det\) is multiplicative. Keep an eye on this: it is exactly what needs repair on the Albert algebra page.

Why \(\Xs{\R}\) and not \(\Xs{\C}\)

The coordinates of \(\Xs{\R}\) are complex, yet the subscript is \(\R\). The subscript names the algebra \(\K\) whose Hermitian matrices (\(\bar X^T=X\)) give the shape of the matrix; only afterwards are the coordinates allowed to be complex. For \(\K=\R\), Hermitian means symmetric, so the shape is \(S^T=S\):

\[\Xs{\K}=\{X\in\JK{\K}:\ \det X=1\},\qquad \JK{\R}=\mathrm{Sym}_3(\C).\]
Table 1. The subscript is the entry algebra, before the coordinates are complexified.
\(\K\)Hermitian over \(\K\)with complex coordinates, \(\JK{\K}\)\(\dim\Xs{\K}\)
\(\R\)real symmetric\(\mathrm{Sym}_3(\C)\)5
\(\C\)complex Hermitian, \(A+\mathrm iB\) with \(A\) symmetric, \(B\) antisymmetricall of \(M_3(\C)\)8
\(\Ha\)quaternionic Hermitian\(\bigwedge^2\C^6\)14
\(\Oc\)octonionic Hermitianthe Albert algebra \(\Jalg\)26

So \(\Xs{\C}\) is a different space: \(\{X\in M_3(\C):\det X=1\}=\SL_3(\C)\) itself, of dimension 8. The whole row \(\K=\R,\C,\Ha,\Oc\) is on the \(A_2\) row page, and why complex coordinates are used at all is on octonions over \(\C\).

What \(\SL_3\), \(\SO_3\) and the quotient areWhat does a point of \(\SL_3/\SO_3\) look like?

As in the warm-up, read each object as what it does to the unit ball \(\Ball\subset\R^3\).

Table 2. The three objects, as maps of space.
objectwhat it isdimension
\(\SL_3(\R)\)linear maps preserving volume; \(g(\Ball)\) is an ellipsoid of volume \(\tfrac43\pi\)8
\(\SO_3\)the rotations: the \(g\) with \(g(\Ball)=\Ball\)3
\(\SL_3(\R)/\SO_3\)\(g\) up to a rotation first, \(g\sim gk\): only the ellipsoid \(g(\Ball)\) is remembered\(8-3=5\)

(ellipsoids) Proved here

The map \(g\mapsto gg^T\) induces a bijection

\[\SL_3(\R)/\SO_3\ \xrightarrow{\ \sim\ }\ \Pd{3},\qquad g\,\SO_3\mapsto gg^T,\qquad g(\Ball)=\Ell{gg^T}=\{v:\ v^T(gg^T)^{-1}v\le1\}.\]

Every \(S\in\Pd{3}\) is \(S=k\,\diag(e^{2h_1},e^{2h_2},e^{2h_3})\,k^T\) with \(k\in\SO_3\) and \(h_1+h_2+h_3=0\); the ellipsoid \(\Ell{S}\) has semi-axes \(e^{h_1},e^{h_2},e^{h_3}\) along the columns of \(k\).

Proof
  1. \(v\in g(\Ball)\iff|g^{-1}v|\le1\iff v^T(gg^T)^{-1}v\le1\).
  2. \(gg^T=g'g'^T\iff(g^{-1}g')(g^{-1}g')^T=I\iff g^{-1}g'\in\SO_3\).
  3. Onto: \(g=S^{1/2}\). The second sentence is the spectral theorem, with \(e^{2(h_1+h_2+h_3)}=\det S=1\).
The point \(g\,\SO_3\) as the ellipsoid \(g(\Ball)\) · drag the picture to turn the view

\(g(\Ball)\) for \(g=k_{\text{left}}\,\diag(e^{h_1},e^{h_2},e^{h_3})\,k_{\text{right}}\). Dashed: the unit sphere. Orange: curves painted on \(\Ball\), carried along by \(g\).

Change

Shape: semi-axes \(e^{h_i}\), with \(h_3=-h_1-h_2\)

\(k_{\text{left}}\): turns the ellipsoid, a new point of \(\Pd{3}\)

\(k_{\text{right}}\): \(g\mapsto gk\), the same point; only the paint moves

The point, live

The two rotation sliders are the quotient in action: \(k_{\text{right}}\) changes \(g\) but not the point \(g\,\SO_3\), while \(k_{\text{left}}\) moves the point along its orbit under \(\SO_3\). The orbit has dimension 3 for three different axes, 2 for a spheroid (turning about its own axis does nothing), and 0 for the ball. These are the numbers \(\mult\cdot(3-\#\text{walls})\) with \(\mult=1\) of the roots page, whose plane is the plane of the \((h_1,h_2,h_3)\).

The surface \(\Pd{2}\) sits inside three times: the ellipsoids with one semi-axis equal to 1 and the other two in a coordinate plane, one copy per pair of axes \(\{i,j\}\).

Microworld: make the ellipsoid roundCan you move \(S_0\) to \(I\) using only moves with \(\det g=1\)?
Matrix lab · state \(S\), moves \(S\mapsto gSg^T\) with \(\det g=1\)
Predict first: the volume of \(\Ell{S}\) after a move
  1. clear column 1
  2. clear entry \((2,3)\)
  3. make the diagonal 1

Change: click an entry

What the move did

Free moves: choose \(i\), \(j\) and a number yourself
Bookkeeping: the product \(g\) and the list of moves

Challenge A. Reach \(I\) from \(S_0\), then from a random start. Clicking an off-diagonal entry clears it with a shear; once \(S\) is diagonal, clicking a diagonal entry rescales it to 1.

A real \(g\) preserves the signature of \(S\): the numbers of positive and negative diagonal entries after diagonalization. \(\diag(-1,-1,1)\) has signature \((1,2)\), \(I\) has \((3,0)\). Over \(\C\), \(g=\diag(i,-i,1)\) has \(\det g=1\) and \(g\,\diag(-1,-1,1)\,g^T=I\).

Every point is reachedDid Challenge A always succeed, and why?

(transitivity) Proved here

\(\SL_3(\C)\) acts transitively on \(\Xs{\R}\): for every \(S\in\Xs{\R}\) there is \(g\in\SL_3(\C)\) with \(gSg^T=I\).

Proof

Proof (the lab's algorithm)

  1. If some diagonal entry is nonzero, a signed swap puts it at position \((1,1)\). If all diagonal entries vanish but \(S_{12}\ne0\), the shear \(I+E_{12}\) makes \(S_{11}=2S_{12}\ne0\).
  2. With \(S_{11}\ne0\), the shears \(I-\tfrac{S_{i1}}{S_{11}}E_{i1}\) (\(i=2,3\)) clear the first column and, by symmetry, the first row. Repeat on the lower \(2\times2\) block. Now \(S=\diag(d_1,d_2,d_3)\) with \(d_1d_2d_3=\det S=1\).
  3. Choose square roots \(\sqrt{d_k}\in\C\) and set \(g=\diag(d_1^{-1/2},d_2^{-1/2},d_3^{-1/2})\). Then \(gSg^T=I\) and \((\det g)^2=1/(d_1d_2d_3)=1\).
  4. If \(\det g=-1\), replace \(g\) by \(-g\): \(\det(-g)=(-1)^3\det g=1\) and \((-g)S(-g)^T=gSg^T\).

Step 3 is where \(\C\) is needed; it is the “finish over \(\C\)” button.

What fixes the identityWhich \(g\) satisfy \(gIg^T=I\)?

Challenge B. Predict, then test.

(stabilizer) Proved here

The stabilizer of \(I\) is \(\SO_3(\C)=\{g\in\SL_3(\C):gg^T=I\}\), so the orbit map gives an isomorphism of varieties

\[\SL_3(\C)/\SO_3(\C)\ \xrightarrow{\ \sim\ }\ \Xs{\R},\qquad g\,\SO_3(\C)\mapsto gg^T .\]
Proof

\(gIg^T=gg^T\), so \(g\) fixes \(I\) iff \(gg^T=I\). The orbit map is surjective by ; its fibres are the cosets of the stabilizer. Over \(\C\) a bijective orbit map from a homogeneous space onto a smooth variety is an isomorphism (characteristic 0).

The last candidate shows that \(\SO_3(\C)\) is not compact: \(\cosh t\) is unbounded. Dimension check:

\[\dim\Xs{\R}=\dim\SL_3-\dim\SO_3=8-3=5=\dim\mathrm{Sym}_3(\C)-1 .\]

Why “symmetric”Which structure makes \(\SO_3\) special inside \(\SL_3\)?

The stabilizer is the fixed group of an involution of \(\SL_3(\C)\):

\[\theta:\SL_3(\C)\to\SL_3(\C),\qquad \theta(g)=(g^T)^{-1},\qquad \theta^2=\Id,\qquad \SL_3(\C)^\theta=\SO_3(\C).\]

(symmetric variety) Definition

Let \(G\) be a connected reductive group over \(\C\) and \(\theta:G\to G\) an involution (an automorphism with \(\theta^2=\Id\)). If \(K\subset G^\theta\) is a subgroup containing the identity component of \(G^\theta=\{g:\theta(g)=g\}\), then \(G/K\) is a symmetric variety.

Differentiating \(\theta\) at the identity gives \(d\theta:\slf_3\to\slf_3\), \(Y\mapsto -Y^T\), with eigenvalues \(\pm1\):

\[\slf_3=\kf\oplus\pf,\qquad \kf=\{Y:Y^T=-Y\}=\so_3\ (\dim 3),\qquad \pf=\{Y:Y^T=Y,\ \tr Y=0\}\ (\dim 5).\]

\(\pf\) is the tangent space of \(\Xs{\R}\) at \(I\): the curve \(e^{tY}e^{tY^T}=I+t(Y+Y^T)+O(t^2)\) moves \(I\) in the symmetric direction \(Y+Y^T\). The word “symmetric” refers to \(\theta\), not to symmetric matrices.

No. With \(G=\SL_3\times\SL_3\) and \(\theta(g,h)=(h,g)\), the fixed group is the diagonal \(\{(g,g)\}\) and \(G/G^\theta\cong\SL_3\) via \((g,h)\mapsto gh^{-1}\): a group, not a space of symmetric matrices. It reappears on the A₂ row as the case \(\K=\C\).

What the lift must changeWhich steps above would fail with octonion entries?
Table 3. Each step of this page and what it used.
stepuseswith entries in \(\Oc\)
\(gSg^T\) is defined\((gS)g^T=g(Sg^T)\)fails for octonionic \(g\); scalar \(g\) still fine
\(\det(gSg^T)=\det(g)^2\det S\)\(\det\) multiplicativeno multiplicative \(\det\) on octonion matrices; a cubic \(\det\) on Hermitian ones survives
stabilizer \(=\{gg^T=I\}\)the group is made of matricesthe group \(\Esix\) is not a group of \(3\times3\) matrices

The repair keeps the space (Hermitian matrices with a determinant) and changes the group: instead of \(\SL_3\) acting by matrices, take all linear maps of the space that preserve the determinant. For real entries the identity component of that group is the image of \(\SL_3(\C)\) acting by \(gSg^T\), so nothing above changes; for octonion entries it is \(\Esix\) (the Albert algebra page). Next, the tangent space \(\pf\) is organized by roots.