Riemannian metrics with discrete isometry group

Question

Does anybody knows an example of a Riemannian metric on a manifold (other than the compact nonpositive curvature case) whose isometry group is discrete? How about such a Riemannian metric with postive curvature?

Answer 1

A kind of dumb example (i.e. it is easily broken by requiring your manifold to be complete): You can take $X =S^2$ with the usual round metric, and then remove 3 linearly independent vectors (in the ambient space) from it $\{x,y,z \}= A$. $Y = X \setminus A$.

Note that each isometry of $Y$ extends to an isometry of $S^2$. (Proof: If you have an isometry $f : Y \to Y$, you get $g = i \circ f : Y \to S^2$, by inclusion $i : Y \to S^2$. Next, since $S^2$ is complete and $i \circ f$ is an isometry, this extends continuously to a map $h : S^2 \to S^2$ (universal property of the completion of a metric space). Now, if $h$ was not an isometry, we could find $x,y \in S^2$ so that $d( h(x) , h(y) ) \not = d(x,y)$. But we can take $x_n, y_n \in Y$ that converge to $x,y$ repectively, and we know that $d(h(x_n),h(y_n)) = d(x_n,y_n)$, letting $n \to \infty$ gives a contradiction. The fact that $h$ is surjective now follows because $Y$ is dense in $S^2$ and the image of $S^2$ is compact.)

In particular, since all isometries of $S^2$ are linear, it follows that all isometries of $Y$ are restrictions of linear maps.

Moreover, an isometry of $Y$ (thought of as an extension to $S^2$) must preserve $A$, and since all the isometries are linear and $A$ contains a set of linearly independent vectors, each isometry is determined by the restriction to $A$.

This gives an injection $isom(Y) \to S_3 = Aut(A)$, which implies that the former is finite, and in particular discrete.

(So, you may want to require that the manifold be complete, which breaks this example.)