Isotropic left invariant Riemannian metric on $GL_n^+$?

I am trying to see if it's possible to construct a left invariant isotropic Riemannian metric on $GL_n^+$. (the group of $n \times n$ invertible real matrices with positive determinant)

(When by isotropic I mean that the sectional curvature $k(p,\sigma)$ for $\sigma \subseteq T_pM$ does not deped on $\sigma$. By left invariance it does not depend on $p$ as well).

Question:

Is there a simple proof such a metric cannot exist?

Here is a possible explanation, but I suspect there is a much simpler argument:

Suppose such a metric $g$ exists. Then the left invariance implies $(GL_n^+,g)$ is complete. Hence it's universal covering space is complete*, simply connected, and with constant curvature , so it must be diffeomorphic to $\mathbb{R}^{n^2}$ or $\, \mathbb{S}^{n^2}$.

Since compact spaces can cover only compact spaces, this rules out $\mathbb{S}^{n^2}$.

So the universal cover of $GL_n^+$ must be $\mathbb{R}^{n^2}$.

However, Milnor states here that the universal cover of a Lie group $G$ is not homeomorphic to Euclidean space iff $G$ contains a compact non-abelian subgroup**. Since $GL_n^+$ contains $SO(n)$ we have a contradiction.

*I am not entirely sure why completeness of a Riemannian manifold implies completeness of the universal cover. Do we lift the geodesics?

**I actually don't know how to prove this. Any hints or references would be welcome. Milnor states this in theorem 3.4.

• Well, The original argument that required using the result on non-parallelizability of spheres was eliminated quite trivially (by me). However, I would still like to see if there is any simpler way of showing this. (without using the classification of simply connected, complete, constant curvature manifolds) – Asaf Shachar Nov 8 '15 at 21:57

$GL_n^{+}$ deformation retracts onto $SO(n)$. The universal cover of this (for $n \ge 3$) is a double cover, since $\pi_1(SO(n)) = \mathbb{Z}_2$. It's known as the Spin group $Spin(n)$, and in particular, because it's a double cover it's compact. Hence it's certainly not homotopy equivalent to a Euclidean space (e.g. because its top $\mathbb{Z}_2$ homology is nontrivial), which means the universal cover of $GL_n^{+}$ can't be either.

Various other arguments are possible. For example, we also know that (again, for $n \ge 3$) $\pi_3(Spin(n)) \cong H_3(Spin(n)) \cong \mathbb{Z}$, so the same is true of the universal cover of $GL_n^{+}$.

• 1) I guess you are using the fact that homotopy equivalence of the spaces implies homotopy equivalence of the covers? 2) Why the top $Z_2$ homology detects compactness? ( I only know about cohomology in this sense mathoverflow.net/questions/101554/…) – Asaf Shachar Nov 11 '15 at 23:57
• 1) Yes. 2) Every closed manifold is orientable over $\mathbb{Z}_2$. – Qiaochu Yuan Nov 12 '15 at 0:33

Here is a topological argument ruling out covering by $\mathbb R^{n^2}$. However, it might be an overkill for this statement.

Assume $n>2$. If $\mathbb R^{n^2}$ were the universal cover of $GL^+_n$, we have $GL^+_n=K(\pi, 1)$, where $\pi=\pi_1(GL_n^+)$. At the same time, $\pi_1(GL_n^+)=\pi_1(SO(n))=\mathbb Z_2$, so $K(\pi, 1)=\mathbb R P^{\infty}$, which is clearly not homotopy equivalent to a finite dimensional space.

• What's $K(\pi, 1)$ ? – Bey Alexander Jul 23 '18 at 3:49
• $K(\pi,1)$ is a topological space with $\pi_1=\pi$ and $\pi_k=0$ for $k\ge 2$. This space is unique up to homotopic equivalence in the category of CW complexes. – Yury Ustinovskiy Jul 25 '18 at 11:13