# Schwarzschild Horizon being Killing Horizon

The Schwarzschild horizon $$r=2GM$$ should have normal vectors proportional to $$\nabla^\mu r=g^{\mu \nu}\delta^r_\nu$$ Isn't it? Then I fail to understand how could one have Killing vector field $$\partial_t$$ being normal to the surface.
More generally, given a surface $$\Sigma:f(x)=0,$$ how could we tell if there is a Killing vector field which is normal to the surface at every spacetime point and find out that Killing vector field?

• The normal to a $f=$constant surface is $n_\mu = \partial_\mu f$ (lowered index, not raised). $\partial_t$ is not supposed to normal to the surface. It generates the null geodesics (so it is null) and also happens to be a Killing vector field. Commented Apr 24, 2023 at 7:12

Now we are given the Killing vector $$\partial_t$$ so we compute its norm which is simply $$g_{tt}$$ of our usual Schwarzschild metric $$g_{tt}=-\left(1-\frac{2GM}{r}\right)$$
The requirement of this Killing vector field being null corresponds to $$r=2GM$$ which is our event horizon.
• I think you have answered my recent question "Does condition $g_{00}(r_{0})=0$ define the event horizon on $r_{0}$?" (physics.stackexchange.com/q/759098/281096), too. Would you mind to write your answer there?