It was famously shown by Carter that the Kerr metric possesses a 4th non-obvious constant of the motion, derived from the separability of the Hamiltonian. This constant is related to a Killing tensor.
My question is, is it possible to show apriori the existence and value of this Killing tensor from the Kerr metric, before considering the separability of the Hamiltonian?
The Killing tensor is defined as a symmetric tensor $K_{\alpha \beta}$ whose total symmetrization of the covariant gradient vanishes $$K_{(\alpha \beta;\gamma)} = K_{\alpha \beta;\gamma} + K_{ \beta\gamma;\alpha}+ K_{\gamma \alpha;\beta}= 0$$ This can be seen as a higher-order generalization of the Killing vector $\xi_\mu$ and the Killing equation which says that the symmetrization of its covariant gradient vanishes $$\xi_{(\mu;\nu)} = \frac{1}{2}(\xi_{\mu;\nu} + \xi_{\nu;\mu})=0$$ In special coordinates where $\xi_\mu$ is tangent to the lines of some coordinate $X$, we can show that the Killing equation is equivalent to the requirement that the metric is independent of $X$.
The equations fulfilled by a Killing vector or a Killing tensor are "just" homogeneous linear partial differential equations with non-constant coefficients. I.e., we can solve them directly by brute force. Consider Minkowski space-time in Cartesian coordinates. There we have the covariant gradient equal simply to the coordinate gradient and a trivial example fulfilling the Killing equation is a constant vector $\xi_\mu$. Similarly, a trivial example of a Killing tensor in Minkowski is any constant symmetric tensor $K_{\mu \nu}$. I.e., we do not need to generally consider the separability of the Hamilton-Jacobi equation to obtain the Killing tensor.
On the other hand, in the case of a generally curved space-time (including Kerr), it is virtually impossible to find analytical solutions to the Killing-tensor equation by brute force. You have to use special algebraic techniques such as the spinor techniques used by Walker and Penrose in 1970 to derive (or guess) the Killing tensor without reference to the separability of equations of motion. In the light of these procedures, it does not seem as such a stretch to use the Hamilton-Jacobi equation to indirectly look for a Killing tensor.
Nevertheless, looking just at the separability of the Hamilton-Jacobi equation of a geodesic is not a surefire way to see whether the space-time has a Killing tensor or not. It has been shown that the Hamilton-Jacobi equation separates if and only if you are in a special set of adapted coordinates (Boyer-Lindquist and similar in Kerr, I like the way this is reviewed by Chervonyii and Lunin). If you are in a generic set of coordinates, your Hamilton-Jacobi equation does not separate even though you have a Killing tensor.
Furthermore, another important object often called the ``square root" of the Killing tensor, the Killing-Yano tensor, cannot be found by the Hamilton-Jacobi method. The Killing-Yano tensor of rank two is an antisymmetric tensor $Y_{\alpha \beta}$ whose covariant gradient is also antisymmetric $$Y_{\alpha (\beta;\gamma)} = 0$$ You can then show that $K_{\alpha \beta} = Y_{\alpha \gamma} Y^\gamma_{\;\; \beta}$ is a Killing tensor. The existence of the Killing-Yano tensor implies separability of the Dirac equation on the curved background. This means that in principle you can circumvent the Hamilton-Jacobi equation by instead looking for the Killing-Yano tensor from the separability of the Dirac equation, and squaring it to obtain the Killing tensor. However, the construction of the Dirac equation on curved background is not trivial.
Thus, it still seems to me that looking for the separation constant of the Hamilton-Jacobi equation for a geodesic is the easiest trick in trying to find the Killing tensor.
If you are looking more for a pragmatic way to simply read of the Killing tensor from the Kerr metric and any metric with a Killing tensor, the recipe goes as follows.
You have to be in coordinates $\xi, \eta$ (plus the ones corresponding to Killing-vector directions) where the inverse metric components acquire the form $$g^{\mu \nu} = \frac{1}{f_\xi(\xi) - f_\eta(\eta)} (X^{\mu \nu}(\xi) + Y^{\mu\nu}(\eta))\,,$$ where $X^{\eta \nu} = Y^{\xi \nu}=0$. You can see that the Kerr metric in Boyer-Lindquist coordinates $r,\vartheta$ has exactly this property.
Your Killing tensor then is $$K^{\mu\nu} = -\frac{f_\xi Y^{\mu \nu} + f_\eta X^{\mu \nu}}{f_\xi - f_\eta}$$ This can be verified by brute-force calculation.
However, the derivation of this formula would be once again easiest by considering the Hamilton-Jacobi equation because these $\xi$ and $\eta$ are once again coordinates in which the HJ equation separates.
The Killing equation is a overdetermined system of PDEs of finite type, as a result, there is a algorithm to compute all Killing tensors for a given metric. See for example arXiv 1704.02074.
