Given the following 2+1 dimensional metric:
$$ds^{2}=2k\left(dr^{2}+\left(1-\frac{2\sin\left(\chi\right)\sin\left(\chi-\psi\right)}{\Delta}\right)d\theta^{2}\right)-\frac{2\cos\left(\chi\right)\cos\left(\chi-\psi\right)\sinh^{2}\left(r\right)}{\Delta}dt^{2}+\frac{4\sqrt{2k\cos\left(\chi\right)\cos\left(\chi-\psi\right)}\sinh^{2}\left(r\right)\sin\left(\frac{\psi}{2}\right)}{\Delta}drdt$$
Where: $$\Delta=1+\cosh^{2}\left(r\right)\cos\left(\psi\right)+\sinh^{2}\left(r\right)\cos\left(2\chi-\psi\right)$$
$$\theta\sim\theta + 2\pi$$
where $k,\psi,\chi$ are constants.
The induced metric on some sphere (circle) at radius $r=\text{const}$ in a time slice of $t=\text{const}$ should be: $$ d\sigma^{2}=2k\left(1-\frac{2\sin\left(\chi\right)\sin\left(\chi-\psi\right)}{\Delta}\right)d\theta^{2} $$
I want to calculate the extrinsic curvature over this sphere, which is given by the following formula:
$$K_{ij} = \frac{1}{2}\sigma_{ca}n^c\sigma^{ab}\partial_{b}\sigma_{ij}$$ Where $n^{c}$ is the normal vector to the surface of interest, in our case this is the sphere at some constant radius in a given constant time slice: $$n^{c}=n^{r}\equiv (n^t,n^r,n^{\theta})=(0,1,0)$$ plugging this inside the formula for $K_{ij}$ will yield to zero. because: $$\sigma_{ca}n^c=\sigma_{ra}n^r=\sigma_{ra}$$ and: $$\sigma_{ra}=0,\quad \forall a$$
This answer seems to be false, what am I interpretating wrong about the formula for the extrinsic curvature?
