# What is topology of all Square matrices such that matrix times it transpose is diagonal

What is the topology of the subspace $$X_n\subset M_n(\mathbb R)$$ consisting of all non-zero $$n\times n$$ matrices $$A$$ such that $$A^t A$$ is diagonal ? For example $$X_2$$ is the product of a torus and an interval. Is $$X_n$$ a manifold ? Reference sought.

• Isn't it just some sort of product of the space of orthogonal matrices with $\mathbb{R}^{n} \backslash \{0\}$ ? – Geoff Robinson Jun 19 '19 at 22:17
• That is true for $X_n\cap GL(n,\mathbb R)$ – Daryl Cooper Jun 19 '19 at 22:45
• In that case you want diagonal matrices with all non-zero entries. I allow some ( but not all) zero entries in the diagonal matrix, but uniqueness of the product is admittedly lost when $A$ is singular. – Geoff Robinson Jun 20 '19 at 6:21
• I should have said above that $A$ has the form $CD$ where $C$ is orthogonal and $D$ is a non-zero diagonal matrix (this product is not unique, though, when $A$ is singular. Actually, I suppose that even when $A$ is invertible, we are allowed to change the signs of the diagonal entries of $D$, but I think we obtain some sort of uniqueness in the invertible case if we insist that the non-zero entries of $D$ are positive). – Geoff Robinson Jun 20 '19 at 8:30
• Yes, so $X_n$ is the orbit in $M_n(\mathbb R)$ under the action of the orthogonal group of the set of non-zero diagonal matrices. Thus $X_n$ is stratified by rank. I am particularly interested in how these strata fit together. It is closely related to a moduli space problem for certain projective manifolds. – Daryl Cooper Jun 20 '19 at 15:01