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1

It is nearly correct. First, $F_x$ is the space of oriented and orthonormal frames of $T_x M$. Furthermore, you gave the correct definition of the $Spin(1,1)$ bundle but confused a little bit its meaning. The $Spin(1,1,)$ bundle is defined as the lift of the $SO(1,1)$ bundle to $Spin(1,1)$. This means we have a principal $Spin(1,1)$-bundle $P$ and a ...


0

The answer to this question is no assuming dimensions higher than 1+1. This can be seen observing that the equation of motion for the fermionic field is just the limit of mass going to infinity of a scalar field coupled to a fermionic field. This can be seen in the following way. Consider the Lagrangian $$ L = ...


1

The Clifford algebra $\mathrm{Cl}(\mathbb{R}^{1,3})$ is the algebra generated by endowing the vectors in $\mathbb{R}^{1,3}$ with a free algebra multiplication and then imposing the constraint given by $v \circ v = \eta^{\mu\nu}v_\mu v_\nu$ with $\eta$ as the Minkowski metric on $\mathbb{R}^{1,3}$. Any Clifford algebra has a natural connection to the ...


2

Just to clarify to Robin Ekman's answer, superpositions of the Pauli matrices exponentiate to $SU(2)$, not $SO(3)$, but both these Lie groups have $\mathfrak{so}(3)$ as their Lie algebra - but I am sure you already know this. Also, there is another way to look at the problem that you might find helpful, even though it is a mathematical insight rather than a ...


4

Yes. This commutation relation is that of the Lie algebra $\mathfrak{so}(3)$ corresponding to the rotation group in three dimensions. Thus the commutation relation states that the Pauli matrices generate rotations. To understand why this is the commutation relation of $\mathfrak{so}(3)$, one can draw a diagram showing that the commutator of two ...


2

We are not entirely sure what OP's question (v4) is asking, but here are some comments: I) The Dirac belt trick demonstrates that the Lie group $SO(3)$ of 3D rotations is doubly connected, $$ \pi_1(SO(3))~=~\mathbb{Z}_2. $$ II) As for the title question Are spinors somehow connected to spacetime? one answer could be: Yes, in the sense that the mere ...


0

You might equally well ask, "How does the physical belt in the Dirac trick sense the topology?" This question is, when you think about it, no less mysterious than yours. The answer, by experiment, is that it simply does. And ultimately, if something transforms "compatibly" with the Lorentz group, or with $SO(3)$, then there is really only a one-bit question ...


1

Take a self-adjoint matrix $A$. There exists a unitary matrix $U$ such that $UAU^*$ is diagonal. Take the square root of every diagonal element in order to define $\sqrt{UAU^*}$ (now you are allowing for roots of negative numbers, so imaginary numbers as well). Then rotate the matrix back with $U^*$ and set $$\sqrt A := U^*\sqrt{UAU^*}U.$$ Both of your ...



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