# Finding explicit elements in $\text{SL}(3,\mathbb{R})/\text{SO}(3)$ whose simultaneous stabilizer in $\text{SL}(3,\mathbb{R})$ is the identity

Consider the action of $\text{SL}(3,\mathbb{R})$ on $M=\text{SL}(3,\mathbb{R})/\text{SO}(3)$ (via left multiplication).

I want to find explicitly a minimal number of elements $s_1,\ldots,s_k \in M$ with the property that the simultaneous stabilizer of all the $s_i$ in $\text{SL}(3,\mathbb{R})$ is the identity matrix.

Robert Bryant commented here that probably $k=3$ is the minimal number, but I am not sure of this.

(I am interested in the representatives $A_i$ s.t $s_i=A_i\text{SO}(3)$. In fact, I am only interested in $A_iA_i^T$ since this is the information needed for constructing a left invariant metric on $\text{SL}(3,\mathbb{R})$- context is provided below if you are interested).

Motivation:

I am trying to realize concretely Robert's idea to construct a left invariant metric on $\text{SL}(n,\mathbb{R})$.

The smallest non-trivial dimension is $n=2$. However, as mentioned by Robert, we must embed $\text{SL}(2,\mathbb{R})$ in $\text{SL}(3,\mathbb{R})$ for his approach to work (since $-I$ acts trivially on $M$).

The reason I am interested in left invariant metrics on $\text{SL}(n,\mathbb{R})$, is that such metrics can be used to construct left invariant metrics on $\text{GL}(n,\mathbb{R})$, as explained in this unanswered question.

You can (equivariantly) identify the quotient $SL(3,R)/SO(3)$ with the space of positive-definite quadratic forms in 3 variables with determinant (of the associated symmetric matrix) equal to $1$. The identification comes from the fact that $SL(3,R)$ acts on quadratic forms via the standard change of variables, so that the stabilizer of the Euclidean quadratic form is $SO(3)$. More explicitly, a matrix $A\in SL(3, R)$ corresponds to the quadratic form $q_M$ associated with the symmetric matrix $M=A^TA$. (The same works in all dimensions.)

With this identification, you can take $q_1=x^2+y^2+z^2$, $q_2=2x^2 + y^2 + \frac{1}{2} z^2$ and for $q_3$ take the quadratic form $$q_3=x^2 + 2xy + 2y^2 + 2yz + 2z^2,$$ corresponding to the matrix $$A= \left[\begin{array}{ccc} 1&1&0\\ 0&1&1\\ 0&0&1\end{array}\right].$$
The reason why it works is that the stabilizer of $q_1$ is $SO(3)$; the common stabilizer of $q_1, q_2$ is the subgroup of diagonal matrices with $\pm 1$ on the diagonal and determinant $1$. None of the latter can stabilize $q_3$ except for the identity matrix.

From this you can figure out how things work of $SL(n,R)$ (again, 3 quadratic forms will suffice, one of which is Euclidean, one is diagonal with distinct diagonal entries and the last one corresponds to the upper-triangular band matrix with $1$'s on the diagonal and just above the diagonal).

• Thanks! This is very nice. I was actually writing "my" answer while you edited yours... I still have some questions and comments: (1) Minor typo: I think $q_3=x^2 + 2xy + 2y^2 + 2yz + 2z^2$(since $A^TA=\begin{pmatrix} 1 & 1 & 0 \\\ 1 & 2 & 1 \\ 0 & 1 & 2 \end{pmatrix}$. (2) I wonder how did you come up with your matrix $A$ (or equivalently with $q_3$). I tried to follow your idea to find a matrix with distinct eigenvalues and eigenvectors different from the standard basis - so the natural thing to me was to use rotations... (You can see my answer below)... – Asaf Shachar Mar 22 '17 at 16:50
• (3) Do you still think that your idea (to take a matrix with these properties) should work in general? (4) Do you see why two quadratic forms won't ever be enough? Thanks. – Asaf Shachar Mar 22 '17 at 16:51
• Yes, two forms are not enough (but "almost" enough: The common stabilizer will be finite). The reason is the simultaneous diagonalization theorem: You can always assume after some change of variables that one form is standard and the second is diagonal. After that, the common automorphisms are sign changes of the coordinates. Then you need to add one more "generic" form to eliminate these. That's what I did. – Moishe Kohan Mar 22 '17 at 18:18

This is an attempt to write a detailed answer based on the suggestion of Moishe Cohen:

We want to find $3$ matrices $A_i \in \text{SL}(3,\mathbb{R})$, such that for any $B \in \text{SL}(3,\mathbb{R})$,

if $BA_i\text{SO}(3)=A_i\text{SO}(3)$ for $i=1,2,3$ then $B=I$.

Note that $BA_i\text{SO}(3)=A_i\text{SO}(3) \iff BA_iA_i^TB^T=A_iA_i^T$.

Denote $\tilde A_i=A_iA_i^T$, Moishe's suggestion was to take:

$\tilde A_1=I$,

$\tilde A_2 = \begin{pmatrix} 2 & 0 & 0 \\\ 0 & 1 & 0 \\ 0 & 0 & \frac{1}{2} \end{pmatrix},$

and to choose $\tilde A_3$ to be a symmetric positive-definite matrix, with distinct eigenvalues and unit determinant, whose three eigenvectors are all different from the standard basis.

I started from $\tilde A_2$, than rotated it in the $[xy]$ plane in $45^0$, and then in the $[yz]$ plane in $30^0$.

I got

$\tilde A_3 = \frac{1}{4} \begin{pmatrix} 6 & \sqrt{3} & 1 \\\ \sqrt{3} & 5 & \sqrt{3} \\ 1 & \sqrt{3} & 3 \end{pmatrix}$.

(A calculation shows that the eigenvalues of $\tilde A_3$ are $2,1,\frac{1}{2}$, with corresponding eigenvalues $(2,\sqrt 3,1),(-2,\sqrt 3,1),(0,-\frac{1}{\sqrt{3}},1)$).

The question is how to see that $\, B\tilde A_iB^T=\tilde A_i \Rightarrow B=I$:

Plugging $\tilde A_1=I$, we get that $B \in \text{O}(3)$, and since $B \in \text{SL}(3,\mathbb{R})$, it follows that $B \in \text{SO}(3)$.

Is there an elegant way to see why $B \in \text{SO}(3)$ plus the other two conditions (for $i=2,3$) imply that $B$ is the identity matrix?

Can we do it without knowing the explicit form of $\tilde A_3$, but only the properties mentioned by Moishe, i.e that it has distinct eigenvalues and unit determinant, and all its eigenvectors are different from the standard basis?