By this useful discussion I know that in finite time Ricci Flow does not change the symmetry of manifold; i.e. $$\def\isom{\operatorname{Isom}}\isom(g_0)=\isom(g_t),\qquad \forall t\in[0,T].$$
I am arguing something like (not exactly) Achilles paradox: (Suppose $M$ is close (compact and without boundary) and has no singularity during RF.)
RF preserve symmetry $\forall t\in[0,T]$. Now lets to go further, RF preserve symmetry $\forall t\in[T,2T]$ So one can argue that: RF preserve symmetry $\forall t\in[kT,(k+1)T]$, consequently $$\isom(g_0)=\isom(g_t),\qquad \forall t\in[0,(k+1)T].$$ and for arbitrary $k$. Therefore
"An ugly sphere cannot be rounded using RF at all." by taking $k\to \infty$.
Question: What is the wrong with this paradox? I think the paradox point is $k\to \infty$.
Best Answer
While it's true that $g_t$ has the same isometry group for any $t\in[0,\infty)$, it may not be the case that the normalized limit $\lim_{t\to\infty}g_t$ has the same isometry group. As a simpler analogy, consider the family of functions $f_t(x)=e^{-t}f_0(x)$. $f_t$ may not be translation invariant for any $t$, but $\lim_{t\to\infty}f_t$ is. The previous statements doesn't allow anything to be said about the isometries of this limit since isometry groups do not depend "continuously" on the metric.
What the discussion of isometry groups essentially says is that the "features" of an "ugly" space do not disappear in finite time, but the convergence results say that these features do get "arbitrarily small" in some suitable sense. In order to get rid of the features entirely, thereby introducing new symmetries, a limit is required.