Quote by cesiumfrog
Could you elaborate a little on exactly where those closed timelike curves are in the Kerr solution?

Let me elaborate a bit on what Chris said.
O'Neill, in his book The Geometry of Kerr Black Holes, proves:
there is a closed timelike curve through any event inside the inner (Cauchy) horizon, i.e., through any event for which r < r.
Carroll gives the following simple example. Consider a curve for which [itex]\phi[/itex] varies, and for which [itex]t[/itex], [itex]r[/itex], [itex]\theta[/itex] are held constant. Because of periodicity with respect to [itex]\phi[/itex], any such curve is closed.
Now, the timelike part.
Take [itex]r < 0[/itex] with [itex]r[/itex] small, and [itex]\theta = \pi/2[/itex]. Note [itex]r[/itex] is a coordinate, not a radial distance, and negative [itex]r[/itex] is part of (extended) Kerr. Because [itex]0 = dt = dr = d \theta[/itex], the line element along the curve is
[tex]
ds^2 = \left( r^2 + a^2 + \frac{2Mr a^2}{r^2} \right) d\phi^2
[/tex]
For [itex]r[/itex] negative and small. the last term, whcih is negative, dominates, and thus [itex]ds^2[/itex] is the line element for a timilike curve.