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

Question

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.

Answer 1

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).

Answer 2

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 $B=I$.

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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 $(2,\sqrt 3,1),(-2,\sqrt 3,1),(0,-\frac{1}{\sqrt{3}},1)$).

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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?