Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $T_xM$, $cg_x(v,v)\leq F_x(v,v)\leq Cg_x(v,v)$?

share|improve this question
3  
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
add comment

1 Answer 1

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.