$\newcommand\norm[1]{\lVert#1\rVert}$Given any point $p$ of a smooth Riemannian manifold $M$ there exists $r\in (0,\infty]$ such that the Riemannian exponential is a diffeomorphism in the geodesic ball $B_r(p)$ and for any $\epsilon>0$ the exponential is not a diffeormorphism in $B_{r+\epsilon}(p)$. Such $r$ is called the injectivity radius at $p$ and denoted by $r(p)$. The injectivity radius of the manifold is $i(M):=\inf_{p \in M} r(p)$.

A manifold of bounded geometry is a Riemannian manifold, endowed with the Levi-Civita connection such that the injectivity radius is stricly positive and $\sup_{p \in M}\left(\norm{\nabla^k R}_g(p)\right)$ is bounded for any $k$ (here $R$ is the Riemann curvature tensor and $\norm\cdot_g$ is the pointwise metric of the tensor i.e. $$ \norm T^2_g(p)=g_{i_1,j_1} \dotsm g_{i_r j_r} g^{m^1,n_1} \dotsm g^{m_s,n_s} T^{i_1 \cdots i_r}_{m_1,\dotsc m_s}(p)T^{j_1 \cdots j_r}_{n_1,\dotsc n_s}(p) $$ I'm wondering if it true that any finite dimensional Lie group is of bounded geometry w.r.t. some metric. In particular, I'm interested in the case of linear Lie groups (i.e Lie subgroup of $GL(n,\mathbb{R})$).

I know that the result is true for $\operatorname{Gl}(n,\mathbb{R})$ endowed with the Euclidean metric and that any compact Riemannian manifold is of bounded geometry and any homogeneous space endowed with a $G$-invariant metric also is of bounded geometry.

I think the result can be extended to any Lie group that admits a bi-invariant metric (the curvature and the connection can be expressed in terms of commutators and so are bounded at any point and the injectivity radius is constant because any Lie group acts transitively on itself), but I can't figure out if the result holds for any Lie group endowed with a left invariant metric.

It should be true that the injectivity radius is constant because there is a transitive left action, but the expression of the curvature is more complicated than the bi-invariant case (see e.g Cheeger - Comparison Theorems in Riemannian Geometry, proposition 3.18).

Is it true that in a finite dimensional Lie group endowed with a left invariant metric all the covariant derivatives of the Riemannian curvature tensor are bounded?

  • 1
    $\begingroup$ Your definition of "linear Lie group" is either non-standard, or not clearly defined. Do you mean an arbitrary submanifold with a smooth group law, or a subgroup? $\endgroup$
    – YCor
    Commented May 11, 2023 at 22:04
  • $\begingroup$ As regards the boundedness: for a homogeneous manifold, doesn't follow from compactness? $\endgroup$
    – YCor
    Commented May 11, 2023 at 22:04
  • $\begingroup$ @YCor The linear Lie group is a Lie group that is a submanifold of $M(n,\mathbb{R})$, or equivalently is a Lie subgroup of $GL(n,\mathbb{R})$. I've modified the definition so that is more standard. An homogeneous manifold doesn't need to be compact (the definition of homogeneous space that i know is a space that can be represented as $G/K$ where $G$ is a (connected) Lie group and $K$ a closed subgroup. A basic example of homogeneous space that is not compact is the Euclidean space $E(n)/O(n)$ or the hyperbolic spaces $\endgroup$
    – Marco
    Commented May 11, 2023 at 22:56
  • 3
    $\begingroup$ And I didn't suggest you're assuming compactness of the manifold. But when a Riemannian manifold $M$ is homogeneous, or when, more generally, there's a compact subset $K\subset M$ such thatv$M=\mathrm{Isom}(M)K$, then you should obtain bounded geometry just using compactness of $K$. $\endgroup$
    – YCor
    Commented May 12, 2023 at 4:51
  • 2
    $\begingroup$ Yes, this is my point. $\endgroup$
    – YCor
    Commented May 12, 2023 at 5:50


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