The Killing vector at the bifurcation surface of a stationary black hole

In the paper Black Hole Entropy is Noether Charge, Wald related the black hole entropy to the Neother charge using the covariant phase space formalism. In proving this relation, Wald noticed that on the bifurcation surface, the Killing vector field $$\xi^a$$ for a stationary black hole vanishes, and at the same time, $$\nabla_a\xi_b=\kappa\epsilon_{ab}$$.

Here, $$\kappa$$ is the surface gravity, and satisfies $$\xi^b\nabla_b\xi^a=\kappa\xi^a$$. $$\epsilon_{ab}$$ is the binormal to the bifurcation surface, whose explicit expression was not given in the paper. My understanding is that I want to introduce another null vector field $$n^a$$ on the horizon such that $$n^a\xi_a=-1$$. $$n^a$$ is also normal to the cut of the horizon. Since $$\xi^a$$ is a normal vector on the horizon, $$\xi_{[a}\nabla_b\xi_{b]}=0$$, which implies that

$$\nabla_a\xi_b=\xi_bv_a-\xi_av_b$$

where $$v_a=n^b\nabla_b\xi_a$$. Furthermore, one can decompose $$v_a=-\kappa n_a+\hat v_a$$ such that $$\xi^a\hat v_a=n^a\hat v_a=0$$. Therefore,

$$\nabla_a\xi_b=2\kappa\xi_{[a}n_{b]}+2\xi_{[b}\hat v_{a]}$$

Now, this equation is valid everywhere on the horizon. Apply it to the bifurcation surface where $$\xi^a=0$$, then

$$\nabla_a\xi_b=0!$$

Since $$\xi^a=0$$, $$\xi_a=g_{ab}\xi^b=0$$ should be correct, as on the horizon, the metric behaves well, although it might have components blowing up in some coordinates.

Please help me out with this! There must be something wrong with my calculation.

I'll be using the notation followed by Poisson in his book, "A Relativist's Toolkit". Let $$\Sigma$$ denote the bifurcation 2-surface, which is null. As you have already mentioned, it will have two normal null vectors, $$\xi^a$$ (a Killing vector) and $$n^a$$, along with two tangential spacelike vectors, $$e^a_A$$, ($$A={1,2}$$). The vectors $$\xi^a$$ and $$n^a$$ are normalized such that, $$\xi^a\xi_a=n^an_a=0$$ and, $$n^a\xi_a=-1$$. This must be true everywhere on $$\Sigma$$. Hence, even though $$\xi$$ goes to zero, because of the normalization, $$n$$ must go to infinity. Therefore their product remains constant. Here's where I think you made an error, you considered $$n$$ to be finite in your second last equation and therefore took $$\nabla_a\xi_b$$ to zero as $$\xi$$ went to zero.
To prove the relation, first, an appropriate definition of the binormal tensor is required. I looked all over the internet but could only find the definition for binormal vectors, which are defined for smooth curves in 3 dimensions. The only information about the binormal tensor available was that it's antisymmetric and normalized, such that, $$\epsilon_{ab}\epsilon^{ab}=2$$. Thus, I took the definition of the binormal tensor for any 2-surface to be, $$\epsilon_{ab}=n_{[a}k_{b]}$$, where $$n$$ and $$k$$ are null vectors that are normal to the surface. This choice for $$\epsilon$$ makes it normalized and antisymmetric, as required. It is non-zero on $$\Sigma$$ because although $$\xi$$ goes to zero on $$\Sigma$$, $$n$$ goes to infinity in such a way that their product remains finite. This definition can be generalized further perhaps, but for this calculation, it serves the purpose. I'll also invoke the geodesic equation for a non-affinely parametrized geodesic, $$\xi^a\nabla_a\xi_b=-\xi^a\nabla_b\xi_a=\kappa\xi_b$$. The second relation holds because $$\xi$$ is a Killing vector. Now $$\nabla_a\xi_b$$ can be expanded as follows,$$\nabla_a\xi_b=a\xi_a\xi_b+bn_an_b+mn_a\xi_b+n\xi_an_b+c_A\xi_ae_b^A+d_A\xi_be_a^A+f_An_ae_b^A+g_An_be_a^A+j_{AB}e_a^Ae_b^B$$. This can be done because $$(n,\xi,e_1,e_2)$$ form a complete basis. The first two terms, being symmetric, cancel out. Since the L.H.S. is proportional to $$\xi$$, the coefficients of the last five terms in the R.H.S also turn out to be zero because they contain components in the perpendicular directions. Hence, after antisymmetrizing the two remaining terms, $$\nabla_a\xi_b=\alpha n_{[a}\xi_{b]}=\alpha\epsilon_{ab}$$. To determine $$\alpha$$, just take an inner product of the L.H.S. with $$\xi^a$$, equate it to $$\kappa\xi_b$$ and since only one term in the R.H.S. contributes ($$\xi$$ is a null vector), you get $$\kappa=\alpha$$ (or $$-\alpha$$, in that case, just flip one of the vectors in the commutator).