Spin derivative on a submanifold I am having some trouble proving the following expression. For a 3-manifold $Y$, spinor bundle $S$, codimension 1 submanifold $M\subset Y$, subbundle $S_M$, and normal vector $N$, the covariant derivative is
$$\nabla_X ^S \phi=\nabla_X^{S_M} \phi|_M -\frac{1}{2}\nabla_X^{M}N\cdot N\cdot \phi,$$
where the $\cdot$ are Clifford products, $X\in TM$ and $\phi \in \Gamma(S)$. I first saw this here, which references here, which references something in German.
So what I've tried so far: That book (Baum et al) gives the following formula for the spin derivative:
$$\nabla_X^S \phi=X(\phi)+\frac{1}{2}\sum_{1\leq k < l\leq n} \omega_{kl} s_k\cdot s_l\cdot \phi$$
that spin connection is defined as $\omega_{kl}=g(\nabla ^M s_k,s_l)$ with orthogonal coordinates ("tetrads", I think) $(s_1,s_2,...)$ I think I understand those like
$$\omega_{jk}=<\nabla_X e_j,e_k>=<\omega^i_j e_i,e_k>,$$
where the $\omega^i_j$ are what I am more used to seeing, i.e. from tetrads, and the inner product is over the manifold coordinates. If I use this expression to try and directly derive what I can't get from above, I get
$$\nabla_X ^S \phi-\nabla_X^{S_M} \phi|_M =\frac{1}{2}\sum_{1\leq k < l\leq n} \omega_{kl}s_k\cdot s_l\cdot \phi-\frac{1}{2}\sum_{1\leq k < l\leq n-1} \omega_{kl}s_k\cdot s_l\cdot \phi|_M$$
So a bunch of terms on the right cancel, and I think you are left with
$$\frac{1}{2}(\omega_{1,3}s_1+\omega_{2,3}s_2)\cdot s_3\cdot \phi$$
just two terms, and the spinor is the one from $S$. So I want to relate this to the covariant derivative of the normal vector, which I guess I want to express in the tetrad basis (say $N=N_3 s_3$ and $X=X_aS_a$ picking the normal and tangent directions so that a,b,c=1,2),
$$ \nabla_X N=X_a\nabla_a (N_3 s_3)=X_a N_3 (\partial_a s_3+\omega_{a3b} s_b)=X_a N_3 \omega_{a3b}\cdot s_b=\omega_{3b} N_3\cdot s_b$$
(repeated indices summed, and that partial vanishes since $a,b\neq 3$). So this is kinda close, since the sum indeed gives me two terms, but I don't actually have the right number of spinors - $N_3$ is just the component here.
So there are several steps here I am a bit uncomfortable with, but if I had gotten down to something like $\omega_{23}$ I would think I was basically doing it correctly. Can anyone help?
 A: I guess I will answer this, since I've been working on it and no one else seems too interested!
The thing I want to start with comes from the definition of a spin derivative in the original question:
$$\nabla_X^S \phi - \nabla_X^F \phi |_F = X(\phi) - X(\phi |_F) +\frac{1}{2}\sum_{1\leq i<j\leq n}\omega_{ij}s_i\cdot s_j\cdot \phi -\frac{1}{2}\sum_{1\leq i<j\leq n-1}\omega_{ij}s_i\cdot s_j\cdot \phi|_F$$
Here $F$ is a codimensional 1 subbundle of spin bundle $S$, $N$ is the normal to the submanifold $\Sigma$ and $X\in T\Sigma$. The first trick is to identify that there is an isomorphism between a $n$-dimensional spin representation and a $n-1$-dimensional representation. This comes from that link above (Baum et al 1991), but it's essential an inclusion, with the extra dimension either
$$e_{2m+1}(u^+ \oplus u^-)=(-1)^m i (u^+ - u^i)$$
or
$$e_{2m+2}(u\oplus \hat{v})=(-1)^m i (v\oplus \hat{u})$$
depending on the dimension. That isomorphism lets us set $X(\phi)=X(\phi |_F)$, so the first terms in the expression vanishes.
The next two terms mostly cancel each other - only terms in the first with $j=n$ survive. So we can write that as
$$\frac{1}{2}\sum_{1\leq i < n}\omega_{in}s_i\cdot s_n \cdot \phi$$
We need to make that look like a derivative of the normal vector times (Clifford Product) the normal vector times the spinor. Presumably we'll have to assume the normal vector is unit...so then $s_n=N$. The covariant derivative of the normal vector will be
$$ \nabla_X^M N = X(N) + N^a\nabla_X (s_a)$$
First, $X(N)=0$ since they exist in orthogonal spaces. Now (I think) the traditional definition of that covariant derivative is
$$\nabla_X (s_a)=\sum_b \omega(X)_{ab}s_b$$
Which corresponds to Baum et al when they write
$$\omega_{kl}=\langle \nabla s_k,s_l\rangle=\nabla s_{k\mu} s^{\mu}_l$$
But really this should be written $\nabla_{\nu}s_{k\mu} s^{\mu}_l$ and contracts with the components of the vector $X^{\nu}$. So the derivative of the normal vector is
$$\nabla_X^M N = N^a\sum_b \omega(X)_{ab}s_b=N^n\sum_b \omega(X)_{nb}s_b =-N^n\sum_b \omega(X)_{bn}s_b$$
using the antisymmetry of $\omega_{ab}$. Now with $N^n=1$ we have
$$\frac{1}{2}\sum_{1\leq i < n}\omega_{in}s_i\cdot s_n \cdot \phi =-\frac{1}{2} \sum_b \omega(X)_{bn}s_b \cdot N \cdot \phi = -\frac{1}{2}\nabla^M_X N \cdot N\cdot \phi$$
and we're done!
$$\nabla_X^S \phi=\nabla_X^F \phi|_F-\frac{1}{2}\nabla^M_X N \cdot N\cdot \phi$$
I will certainly take comments - it seems a bit of an awkward proof to be left to "can be verified", but maybe the authors also didn't like what it looked like.
