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Suppose $M$ is a parallelizible Riemannian manifold with metric tensor $g_x(\cdot,\cdot)$ Let $F_x(\cdot,\cdot)$ denote the flat metric on $M$ that we get from parallelization. Is it true that there exist c, C such that for any $x$ and $v$ in $T_xM$, $cg_x(v,v)\leq F_x(v,v)\leq Cg_x(v,v)$?

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If $M$ is not compact then NO, take $M=R^1$, $g=e^x dx^2$. If $M$ is compact then yes. – Dmitri Sep 2 '10 at 7:51
Thank you, Dmitri. Could you please give idea of the proof, or point me in the right direction? I'd appreciate it! Thanks! – William Sep 2 '10 at 8:50
Parallelizable doesn't imply existence of a flat metric : think of $S^3$ for instance (or any other compact non commutative Lie group, or $S^7$). I suppose you mean a constant metric in a trivialization of the tangent bundle. Anyway, as remarked by Dmitri, any two (continuous) riemann metrics on a compact manifold are Lipschitz equivalent, and this is false on any non compact manifold. – BS. Sep 2 '10 at 8:50
PS. I suppose we can drop the restriction that the second metric is the flat metric, and simply speak of two Riemannian metrics on a compact manifold, and ask whether they are equivalent (in the above sense). Again, I'd appreciate an idea for the proof. – William Sep 2 '10 at 8:52
@BS: Thank you for your comments. Well, suppose $X_1,..., X_n$ are vector fields on $M$ that parallelize $M$. Can't we define $F_p(v,u)$ = \left\langle (a_1,...,a_n), (b_1,...,b_n)\right\rangle$ where $v = a_1X_1(p) + \cdots + a_nX_n(p)$ and $u = b_1X_1(p) + \cdots + b_nX_n(p)$? – William Sep 2 '10 at 8:56
up vote 7 down vote accepted

As Dmitri says any two Riemannian metrics on a compact manifold are Lipschitz equivalent. The proof is quite simple.

Consider $g$ and $h$ two metrics on $M$, Let $UM$ be the unit tangent bundle, since $M$ is compact, $UM$ is compact. Then you see that $f:UM\to \mathbb{R}$ defined by $f(x)=\frac{g(x,x)}{h(x,x)}$ is continuous and strictly positive. By compactness, it is bounded above and below by positive constants.

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