How to show 'symmetric topological sphere must be round'?

A step in the proof of Proposition 3.3 (p.6) in this paper 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. 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.

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Being simply connected, it must be globally symmetric, a first step. –  Ben McKay Sep 21 '13 at 7:06

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.