I would suggest thinking in terms of the $\gamma_0$ stuff, if you've seen the Dirac equation where they derive scalar, pseudo-scalar vector, etc invariants after showing the Dirac equation is Lorentz invariant that's probably the first time most people see this stuff
Okay, so let start from the basics. Our starting point is that $\psi$ is an $SU(2)$ spinor with an inner product $\epsilon_{\alpha \beta}\psi^\alpha \psi^\beta$. We want to define invariants of SU(2) so that out interacting Lagrangian is real and invariant under $SU(2)$
One invariant that satisfies this is $\epsilon_{\alpha \beta}\psi^*{}^\alpha \psi^\beta$
From the beginning, you can start from spinors of $\mathrm{SU}(2)$. You then note that a determinant whose columns are two different spinors is invariant under conjugation by $\mathrm{SU}(2)$ which is just an $\mathrm{SU}(2)$ transform of the spinors. This gives you a way to form scalars from two spinors. Of course this should agree with standard Clifford algebra terminology, so you can realize it can be related to the usual $\gamma_0$ thing familiar from the Dirac equation
If I have $F_{\alpha \beta}$ and transpose it, is it $-F^T{}_{ \beta \alpha}$ because the objects are Grassmann in their indices I think
I'm not sure, I keep getting $\epsilon _{\alpha \beta} \psi^*{}^\alpha \psi^\beta \rightarrow - \epsilon_{\alpha \beta}\psi^*{}^\alpha \psi^\beta$ as the transformation but I'm not sure if Ive made a mistake
@Slereah I'm not sure what you mean by "dual" there, but if you pass from a foliation to the associated distribution (its set of tangent vectors, essentially), you can ask for an orthogonal distribution. If this orthogonal distribution is integrable, then there's also a foliation associated with it that will intersect the original foliation "orthogonally". Is that what you're looking for?
If you have some timelike vector field on a Cauchy surface and then do the Hamiltonian flow it gives you an appropriate foliation of the manifold into timelike curve
not currently pingable, but I see him drop in from time to time
@Slereah the problem is likely that most of these names were made up by people that couldn't google whether someone else already named their stuff the same :P
@Slereah probably it determines at least the curvature, since curvature "is infinitesimal holonomy" and the angles/lengths tell you how parallel transport along that triangle works
If this equation is true $D = \Epsilon_0 E + P$. Ofc in order for polarisation to exist in an object you need to expose it in an external electric field, lets call it E_0. Now if the electric field
Inside the material is the one in the initial equation
then if we go outside the material the P vanishes
leaving D=epislon_0 E
And E is the electric field inside
but outside the material E=D
Then D=E but the field outside shouldn't be the same with E_0 ?
In these handwritten notes I've found, first he defines $\overline{\psi}{}_{\dot{\alpha}} = (\psi_{\alpha})^*$ and then he says now that we have Grassmann fields we define $\overline{\psi}{}_{\dot{\alpha}} = (\psi_\alpha)^\dagger$
For complex numbers you just take the complex conjugate, when you quantize you promote them to operators so you now use the Hermitian conjugate rather than just complex conjugation
@Slereah I said "at least the curvature", but I now realize (see e.g. mathoverflow.net/a/243517, if I'm reading it right) that "LC" also fixes the torsion and so there probably generically isn't more than one LC connection with the same curvature
so your triangles have a decent chance at fixing the LC connection in the generic case
But we have $(\sigma_m)_{\alpha \dot{\alpha}}$ and for $m=0$, this is essentially $\mathbb{I}_{\alpha \dot{\alpha}}$ so why cant we have $\delta_{\alpha \dot{\alpha}}$?
