As per I can visualize the eigen value $\lambda$ of a linear map $T:V \rightarrow V$, defined by $Tv=\lambda v$, is actually the scaling factor of the vector in the same direction as of $v$.My question is how to visualize the eigenvalues of a Ricci tensor on a Riemannian manifold? Then what kind of estimate one can obtain from the lowest eigenvalue?

  • $\begingroup$ There's a theorem that self-adjoint linear maps can be diagonalized via an orthogonal matrix. You can re-interpret this as a theorem about bilinear functions, that they can be diagonalized orthogonally. That gives you your interpretation. $\endgroup$ Commented Aug 5, 2015 at 17:59
  • $\begingroup$ But what geometric estimate does those eigenvalues gives? $\endgroup$ Commented Aug 5, 2015 at 18:22
  • $\begingroup$ If you fix a vector $v$, then, up to a scalar factor, $Rc(v,v)$ is the average sectional curvature of tangent 2-planes containing $v$ $\endgroup$
    – Deane Yang
    Commented Aug 5, 2015 at 19:26
  • 1
    $\begingroup$ The lowest eigenvalue gives you a lower bound of the Ricci tensor's values (with input on the unit sphere). Is that what you are interested in? $\endgroup$ Commented Aug 5, 2015 at 20:00
  • $\begingroup$ yes I got it.Can you give me some reference where this is clarified vastly. $\endgroup$ Commented Aug 6, 2015 at 4:08