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Let $(M,g)$ be a $d$-dimensional Riemannian manifold, $T^*M$ is the cotangent bundle. and the local chart is $$ T^*M|_{U}=U\times R^d=(x^1,...,x^d,p_1,...,p_d) $$ Define $$ H(x,p):=g^{ij}(x)p_ip_j $$ and for some given $\lambda>0$ $$ E_\lambda:=\{(x,p)\in T^*M:H(x,p)=\lambda\} $$

Then, why $M$ is compact ,so is $E_\lambda$ ?

I think it is because that :

Fix $x$, because $H(x,p)$ is continuous , then $E_\lambda$ is compact in $R^d$.

And $M$ is compact ,then $E_\lambda$ is equal to the product of two compact sets. So, it still be a compact set . In fact, the product is not suitable, because it is placed a compact space in every points of $M$.So, how to precisely explain it ?

Enhao Lan
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  • $E_\lambda$ with fixed $x$ is compact not just because $H$ is continuous, but because it is topologically a $d$-dimensional sphere, $S^d$. – lisyarus Oct 17 '16 at 12:01
  • @lisyarus Yes, you are right. Only closed can be got from continuous of $H$, and closed is not compact. Do you know why $E_{\lambda}$ is compact ? See it as $M\times S^d$ ? – Enhao Lan Oct 17 '16 at 12:10
  • $E_\lambda$ definitely should be homeomorphic to $M \times S^d$, but I'm not sure how to make this rigorous. Probably, work in local charts, show local homeomorphism, and then glue the pieces.. – lisyarus Oct 17 '16 at 12:13
  • @lisyarus Seemly, it is troublesome to show $E_\lambda$ homeomorphic to $M\times S^d$. – Enhao Lan Oct 17 '16 at 12:21
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    I'm pretty sure it's troublesome because it's not true. There's a general theorem on fibre bundles that says the total space is compact whenever both the fibre and the base are. Alternatively in this case you could go a more geometric route - choose a metric on $T^* M$ and show $E_\lambda$ is either sequentially compact or totally bounded. – Anthony Carapetis Oct 17 '16 at 12:24
  • @AnthonyCarapetis I agree with your general theorem. But why do you say it's not true ? Do you mean the results of product is not certain ? For example, $S^1\times S^1$, if with the product topology , it is $T^2$. But the Klein bottle can be saw $S^1\times S^1$ with same topology . In fact ,I don't know the difference of structure $T^2$ and Klein bottle as $S^1\times S^1$. – Enhao Lan Oct 17 '16 at 12:41
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    The Klein bottle is certainly not homeomorphic to $S^1 \times S^1$. I see no reason to expect that the unit tangent bundle (your $E_\lambda$) should be trivial. – Anthony Carapetis Oct 17 '16 at 12:44
  • @AnthonyCarapetis Thanks, I understand you. Because we don't know whether the cotangent bundle is trivial bundle, so we don't know whether it is $M\times S^1$ . – Enhao Lan Oct 17 '16 at 12:49

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