A step in the proof of Proposition 3.3 (p.6) in [this paper](http://arxiv.org/abs/1107.0606) argues the following:


Let $M^n$ be a topological sphere with a Riemannian metric $g$. Now assume further $(M^n,g)$ is locally symmetric, then it must be the standard round sphere by some results from [a paper by Borel](http://www.ams.org/journals/bull/1949-55-06/S0002-9904-1949-09251-0/home.html). I just can't follow Borel's paper, it might be something like 'For the sphere in the form $G/H$ with $G$ the isometric group, even-dim sphere has Euler number 2, which after some arguments can determine G and H'. But I am not sure how it really works or what happens for odd-dim sphere. Maybe the whole question follow from the classification theory for symmetric space of compact type, which I know almost zero. 


Now I was wondering if a geometric way possible, or if anyone knows how exactly Borel's aurgment works. 



How to show 'symmetric topological sphere must be round'?
Perhaps the following is not the quickest way, but it works. A locally symmetric simply-connected manifold is symmetric, and hence homogeneous. Homogeneous simply-connected rational cohomology spheres are all classified. References are listed in Section 2.4 in the thesis (<http://opus.bibliothek.uni-wuerzburg.de/volltexte/2002/399/pdf/bletzdiss.pdf>‎) of Oliver Bletz-Siebert. These results give a list of pairs $(G,H)$ such that $G/H$ is homeomorphic to a sphere. One can go down the list, compare it with the classification of symmetric spaces, and see that the only possibility is $SO(n+1)/SO(n)$, the round sphere.


