# Quotient of $GL(\mathbb{R}^n)$ by $O(\mathbb{R}^n)$

I suppose the quotient $$GL(\mathbb{R}^n)/O(\mathbb{R}^n)$$ has manifold structure. Is there a name for this manifold? Google isn't helping find it.

• How about the case $n=1$ ? May 18 at 19:48
• Polar decomposition does it.
– Ruy
May 18 at 20:11
• By the QR decomposition (or RQ decomposition if you want to consider the right action of $O(n)$ on $GL(\mathbb{R}^n)$) the quotient space is in bijection with the set of upper-triangular matrices with positive terms on diagonal. Set of such matrices is just $(\mathbb{R}_{> 0})^n \times (\mathbb{R})^{\frac{n(n-1)}{2}} \cong \mathbb{R}^{\frac{n(n+1)}{2}}$.
– pmp
May 18 at 20:30

You are almost surely familiar with the case $$n=2$$, where the quotient space $$GL(\mathbb R^2) / O(\mathbb R^2)$$ is the hyperbolic plane.
First, for any Lie group $$G$$ and any closed subgroup $$H < G$$, the quotient $$G/H$$ is a manifold, moreover the left action of $$G$$ on itself descends to a left action on $$G/H$$, and with respect to this action there is an invariant Riemannian metric (unique up to rescaling). This Riemannian manifold is called a homogeneous space for $$G$$. In the special case $$GL(\mathbb R^2) / O(\mathbb R^2)$$, this Riemannian manifold is isometric to the hyperbolic plane $$\mathbb H^2$$ (up to rescaling).
Second, $$GL(\mathbb R^n)$$ is a semisimple Lie group. For any semisimple Lie group $$G$$, up to conjugacy there is a unique maximal compact subgroup $$H < G$$. The quotient space $$G/H$$ is a special case of a homogeneous space, called the symmetric space of $$G$$. For your example, $$O(\mathbb R^n)$$ is indeed the maximal compact subgroup of $$GL(\mathbb R^n)$$, and so $$GL(\mathbb R^n) / O(\mathbb R^n)$$ is the symmetric space of $$GL(\mathbb R^n)$$.
As it turns out one can also kinda/sorta give a name to the particular symmetric space $$GL(n,\mathbb R) / O(n,\mathbb R)$$: it is the space of normalized positive definite quadratic forms on $$\mathbb R^n$$, or if you like the space of volume 1 ellipsoids in $$\mathbb R^n$$, centered on the origin. One way to see this is that $$GL(n,\mathbb R)$$ acts transitively on this set of ellipsoids, and the stabilizer of the roundest ellipsoid, that is the unit volume round ball centered on the origin, is $$O(n,\mathbb R)$$.
• Quibbles: of course $GL_2(\mathbb R)$ is officially reductive, not semi-simple. Also, it's $SL_2(\mathbb R)/SO(2,\mathbb R)$ that is isomorphic to the upper half-plane. Yes, $GL_w$ contains the slightly larger $O(2,\mathbb R)$, but/and the quotient is the Cartesian product of the upper half-plane with a ray (isomorphic to the positive diagonal matrices...) May 18 at 21:16