# The physical meaning of the cross-term of Kerr metric

I have been experimenting with different types of metric tensors in General Relativity. I decided to try my hand with the Kerr Metric. When I did, I found an odd term in it: namely, a cross product of $dt$ and $d\phi$. My question is, why sort of physical meaning does this cross product have? What does it mean?

Here is the full Kerr Metric for reference:\begin{aligned}c^{2}d\tau ^{2}=&\left(1-{\frac {r_{s}r}{\rho ^{2}}}\right)c^{2}dt^{2}-{\frac {\rho ^{2}}{\Delta }}dr^{2}-\rho ^{2}d\theta ^{2}\\&-\left(r^{2}+\alpha ^{2}+{\frac {r_{s}r\alpha ^{2}}{\rho ^{2}}}\sin ^{2}\theta \right)\sin ^{2}\theta \ d\phi ^{2}+{\frac {2r_{s}r\alpha \sin ^{2}\theta }{\rho ^{2}}}\,c\,dt\,d\phi \end{aligned} $r_{s}$ is the Schwarzschild Radius, $r,\theta,\phi$ are the standard spherical coordinate system, and where $\alpha,\rho,\Delta$ have been introduced for brevity. $${\displaystyle \rho ^{2}=r^{2}+\alpha ^{2}\cos ^{2}\theta }$$ $${\displaystyle \alpha ={\frac {J}{Mc}}}$$ $${\displaystyle \Delta =r^{2}-r_{s}r+\alpha ^{2}}$$

The Kerr metric is asymptotically flat. An observer stationary with respect to the asymptotically flat boundary will have 4-vector $\partial_{t}$. If you define a frame that is stationary with respect to this observer, and extend it close to the horizon, it will be clear that the time translation of this observer will have nonzero inner product with $\partial_{\phi}$. This means that the timelike direction at infinity is dragged a bit in the angular direction by the motion of the black hole horizon. In fact, there is a region called the ergosphere of the black hole at whose surface $\partial_{t}$ becomes null, and a true unit timelike observer necessarily must rotate around the hole relative to this observer at infinity.
A quick answer is that the cross term means the angular momentum vector is a constant of the motion with the Killing vector $K_t$. This means the angular momentum is a form of "hair" on the black hole that is conserved.