When i try to read Hamilton’s famous thesis “Three-manifolds with positive Ricci curvature” on section 10 page 283, it says this expression will vanish for any symmetric metric: $$(\lambda^3+\mu^3+v^3)-(\lambda^2\mu+\lambda\mu^2+\lambda^2v+\lambda v^2+\mu^2v+\mu v^2)+3\lambda\mu v.$$
In addition, We first give a Riemannina manifold $(M,g)$ and if we regard the Ricci tensor as a bilinear function on the tangent space of $M$. Then $Ric=R_{ij}dx^idx^j$, the matrix $(R_{ij})$ has eigenvalues $\lambda, \mu, v$.
The symmetric metric means the symmetric space, that is for any $x\in M$, there exists an isometry $f$, such that $f(x)=x,~ f\circ f=Id$. For example $R^n,S^n$ are examples of symmetric spaces.
However, I can connect this with the eigenvalues of Ricci. Hamilton says in 3-dimension symmetric space either has all it’s eigenvalues equal or has two equal eigenvalues and the third is zero. If so, the above expression is zero. But I fail to understand this. How can I get the eigenvalue if it is a symmetric space in three-dimension?