MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Cheeger's splitting theorem says "Let $\left( {M_i^n,{p_i}} \right)\mathop \to \limits^{G - H} \left( {X,p} \right)$ with $Ric\left( {{M_i}} \right) \ge - \left( {n - 1} \right){\varepsilon _i}$ $\left( {{\varepsilon _i} \to 0} \right)$.Suppose X has a line.Then X splits isometrically as $X \cong Y \times R$.Then proof is to construct harmonic functions ${b_i}$ on ${{M_i}}$.And prove that $$\frac{1}{{Vol\left( {{B_R}\left( p \right)} \right)}}{\int_{{B_R}\left( p \right)} {\left| {\nabla {b_i} - 1} \right|} ^2} + {\left| {Hess{b_i}} \right|^2}d{\mu _i} \to 0$$ for any R>0.Then how does he prove the theorem from this integral?And for manifolds $\left( {M_i^n,{p_i}} \right)\mathop \to \limits^{G - H} \left( {X,p} \right)$,if there are harmonic functions $b_i^1,b_i^2$on ${{M_i}}$.And the following holds:$$\frac{1}{{Vol\left( {{B_R}\left( p \right)} \right)}}{\int_{{B_R}\left( p \right)} {\left| {\left\langle {\nabla b_i^j,\nabla b_i^l} \right\rangle - {\delta _{j,l}}} \right|} ^2} + {\left| {Hess{b_i}} \right|^2}d{\mu _i} \to 0$$ for j,l=1,2.Then X splits isometrically as $X \cong Y \times {R^2}$?

share|cite|improve this question
You should read their paper directly to see why it is true. btw this is proved by Cheeger-Colding you should not call it "Cheeger's splitting" theorem. Even for manifold case, it is Cheeger-Gromoll's splitting theorem. – J. GE Jun 27 '13 at 20:44
Thank you!I will not make the mistake again. – jiangsaiyin Jun 28 '13 at 3:29

Your Answer


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

Browse other questions tagged or ask your own question.