Parallel Transport on Hypersurface Spinor Bundle

So this has been driving me up a wall. I'm trying to digest parts of the Parker & Taubes paper, "On Witten's Proof of the Positive Energy Theorem." Here's a link:

My current confusion is centered on page 227 (page 5 of the document), the paragraph beginning, "The metric connection on..."

Here's the situation (explained in the last paragraph of page 226):

We have a spacetime manifold $N$ and a spacelike hypersurface $i:M\hookrightarrow N$. One can pull the frame bundle $F(N)$ back to an $SO(3,1)$ bundle over $M$: $i^*F(N)=F(M)\times_{\alpha}SO(3,1)$ ($\alpha$ is defined in the paper; it's the obvious inclusion of $SO(3)$ into $SO(3,1)$). There are obvious spin structures on this bundle over $M$ by choosing a lift $\widetilde{F(M)}\rightarrow F(M)$ and using the double cover $SL(2,\mathbb{C})\rightarrow SO(3,1)$, i.e. $\widetilde{i^*F(N)}=\widetilde{F(M)}\times_{\hat\alpha}SL(2,\mathbb{C})$. (The map $\hat\alpha$ covers the map $\alpha$.)

Now we can define a bundle $S$ of spinors on $M$ using the representation space $(S,\rho)$ (the Dirac spinors) of $SL(2,\mathbb{C})$ (note the abuse of notation with the letter $S$). That is, $S=\widetilde{i^*F(N)}\times_{\rho}S$. So to observe a spinor on $M$, we choose a (lift of a) Lorentz frame and an element of the vector space $S$. But we have the freedom to choose our Lorentz frame so that we have 3 orthonormal spacelike vectors tangent of $M$ and the fourth is a timelike vector orthogonal to the hypersurface $M$. That is, $S=\widetilde{F(M)}\times_{\bar\rho}S$. ($\bar\rho=\rho\circ\hat\alpha$) We can explicitly write a map $$\widetilde{F(M)}\times_{\bar\rho}S\rightarrow\widetilde{i^*F(N)}\times_{\rho}S$$ $$[(p,\Psi)]\mapsto [([(p,I)],\Psi)]$$ formalizing this.

These two ways of looking at the spinor bundle over $M$ gives two inner products on $S$. If we glue in the representation space $S$ with the $SL(2,\mathbb{C})$-action we can transfer the Hermitian inner product $(\cdot,\cdot)$, and if we glue $S$ in with the $Spin(3)$-action we can transfer the positive-definite product $<\cdot,\cdot>$ (see the paper for definitions).

Now, if we do parallel transport along a curve in $M$ in the spinor bundle $S=\widetilde{i^*F(N)}\times_{\rho}S$ of a vector $\xi=[([(p,I)],\Psi)]$ using the Riemannian structure of $N$, we first parallel transport $[(p,I)]$, which will be $[(p(t), A(t))]$ along the curve, where $A(t)\in SL(2,\mathbb{C})$ indicates that the initially spacelike vectors tangent to $M$ don't necessarily remain so. The parallel translation of $\xi$ is then $\xi(t)=[([(p(t), A(t))],\Psi)]$ along the curve, the key point being that $\Psi$ remains constant. The inner product $(\xi(t),\eta(t))$ remains constant along the curve since we can compute the inner product in any frame and the vector space component remains constant. Thus this connection is compatible with $(\cdot,\cdot)$. It is not compatible with $<\cdot,\cdot>$, since we have to recalibrate the frame first and write $\xi(t)=[([(p(t), I)],\rho(A(t))\Psi)]$, and $<\rho(A(t))\Psi,\rho(A(t))\Phi>\not=<\Psi,\Phi>.$

Now do parallel transport along a curve in $M$ with respect to the Riemannian structure of $M$. Then parallel transport of $\xi$ is $\xi(t)=[([(p(t), I)],\Psi)]$ since only the spacelike vectors tangent to $M$ are affected by the parallel transport. But then it seems like $\textit{both}$ inner products are compatible because the vector component remains constant and we don't have to recalibrate like we did above. Parker & Taubes state, however, that $(\cdot,\cdot)$ is in fact not compatible with the connection coming from $M$. I'm inclined to believe them, and thinking about the spinorial Gauss formula (see page 7 here: http://arxiv.org/pdf/math/0101111v1.pdf) seems to indicate that they are correct.

But then where does my thinking above go wrong? I'm really hoping to see where my thinking is wrong at the level of parallel transport as I detailed here, rather than at the infinitesimal level of covariant derivatives (for which the spinorial Gauss formula would be sufficient). I really hope this post isn't too overwhelming or confusing for people to make sense of; please let me know if I can improve it.

• The two products are related by $\langle\cdot,\cdot\rangle=(\cdot,e_0\cdot)$, where $e_0$ is a timelike vector. I understand that it is normal to $M\subset N$. Unless $M$ is totally geodesic in $N$, this vector field $e_0$ will not be parallel with respect to the Levi-Civita connection on $N$. However, it is always parallel with respect to the Riemannian structure on $M$. So I think you are right. Your view would be compatible with the spinorial Gauss formula if the difference of the two spin connection is skew w.r.t $Spin(3,1)$. Have you checked this? – Sebastian Goette Jan 30 '16 at 20:28
• ... last line of my comment above doesn't make sense: I mean "is skew w.r.t the $Spin(3,1)$-invariant scalar product $(\cdot,\cdot)$." I am not familiar with Lorentz-Clifford algebras. In the Riemannian setting, this would be true (but then $(\cdot,\cdot)$ would not make much sense). – Sebastian Goette Jan 30 '16 at 20:44
• @SebastianGoette Thank you for the comment. It got me thinking about the spinorial Gauss formula, and I realized I made a trivial mistake when checking compatibility (which caused a lot of undue anguish for me). I posted an answer with details below. – Brian Klatt Feb 1 '16 at 21:27

I now believe that the statement in question (from the paper), "... the Riemannian connection $\bar\nabla$ of $M$... is compatible with $<\,,\,>$ but not with $(\,,\,)$" is false. It seems that both products are compatible with the connection $\bar\nabla$. As far as I can tell, the argument about parallel transport that I outlined above is correct, and the source of my confusion was a relatively trivial one with the spinorial Gauss equation.
I'd like to outline how the spinorial Gauss equation is consistent with the parallel transport picture. I'll reference the arxiv document, page 7, that was linked at the end of the original post. The spinorial Gauss equation as stated in this link (and using the notation from the Parker/Taubes paper; sorry for the possible annoyance) is $$\nabla_X\phi=\bar\nabla_X\phi-\displaystyle\frac{1}{2}h(X)\cdot e_0\cdot\phi$$ Here $h$ is the second fundamental form, thought of as an endomorphism of the tangent bundle, i.e. the shape operator. The minus sign (as opposed to the $+$ in the link; hope I have this right, though it shouldn't affect the remainder of the argument) is due to the fact that $e_0$ is lightlike, which doesn't factor in to the Riemannian situation.
Now, we want to see if $\bar\nabla$ is compatible with $(\,,\,)$. First note that $\nabla$ is compatible with $(\,,\,)$, so $$X(\phi,\psi)=(\nabla_X\phi,\psi)+(\phi,\nabla_X\psi)=(\bar\nabla_X\phi,\psi)+(\phi,\bar\nabla_X\psi)-\displaystyle\frac{1}{2}[(h(X)\cdot e_0\cdot\phi,\psi)+(\phi,h(X)\cdot e_0\cdot\psi)]$$ which shows compatibility if $(h(X)\cdot e_0\cdot\phi,\psi)+(\phi,h(X)\cdot e_0\cdot\psi)=0$. To this end, we recall (see the Parker/Taubes paper) that Clifford multiplication is Hermitian with respect to $(\,,\,)$. Thus $$(h(X)\cdot e_0\cdot\phi,\psi)=(e_0\cdot\phi,h(X)\cdot\psi)=(\phi,e_0\cdot h(X)\cdot\psi)=-(\phi,h(X)\cdot e_0\cdot\psi)$$ which cancels out the other term. This establishes that $\bar\nabla$ on $M$ is compatible with $(\,,\,)$.