to the above consideration, we now give a comment on another type of exceptional mode. As is well known, the Kerr solution can be extended to a larger regular family by introducing the NUT parameter $\nu$. This family reduces to the Schwarzschild solution in the simultaneous limit $J\tend 0$ and $\nu\tend 0$. In particular, the Taub-NUT solution with $J=0$ expressed as with can be treated as a perturbed form of the Schwarzschild solution when $|\nu|$ is small. This perturbation is of the vector type and given by $h_{ti}=-2\nu f \VHB_i$, with It is easy to see that this $\VHB_i$ satisfies the divergence-free condition and that $f_t=f/r$ is a solution to the Einstein equations satisfying the regularity condition at the horizon. (Of course, the NUT solution is not asymptotically flat and does not satisfy the above asymptotic condition at infinity.) Further, $\VHB_i$ is also harmonic. However, the corresponding eigenvalue is given by $k_V^2=-1$, i.e., $l=0$. This apparently contradicts our previous claim that $k_V^2>0$ for $K=1$. This contradiction arises simply because $\VHB^i$ is singular and not square integrable on $S^2$.

This peculiarity of the solution arises from two pathological features of the Taub-NUT solution. Firstly, each $r=\const$ submanifold is diffeomorphic to $S^3$ 