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I am currently trying to teach myself a little about spinors, and I am having difficulty understanding the motivation behind the definition of the spinor bundle. I am approaching the topic from differential geometry; my background in physics is fairly limited.

My main difficulty is understanding why the spin group appears at all.

A common motivation I have seen is that one wants to define a first-order differential operator $D$ whose square is the Laplacian. I tried to look at a simple example and see how it might generalize. Since (as far as I understand) physicists often work with spinors on flat spaces, I would expect the spin group to already show up here.

Consider the linear map $$ c : T_{(x,y)}\mathbb{R}^2 \cong \mathbb{R}^2 \to \mathrm{End}(\mathbb{C}^2) $$ defined by $$ c(e_1)=\begin{pmatrix} 0 & i\\ i & 0 \end{pmatrix}, \quad c(e_2)=\begin{pmatrix} 0 & 1\\ -1 & 0 \end{pmatrix}. $$ Then $c$ satisfies the Clifford relations.

We can then define an operator $$ D: C^\infty(\mathbb{R}^2,\mathbb{C}^2)\to C^\infty(\mathbb{R}^2,\mathbb{C}^2) $$ by $$ D\psi = c(e_1)\nabla_{e_1}\psi + c(e_2)\nabla_{e_2}\psi. $$ (One can check that this is essentially the Cauchy--Riemann operator.) Moreover, this operator does not depend on the choice of orthonormal basis $\{e_1,e_2\}$ of $T_x\mathbb{R}^2$: if $A \in SO(2)$, then one checks that $$ \sum_i c(Ae_i)\nabla_{Ae_i}\psi = \sum_i c(e_i)\nabla_{e_i}\psi. $$ Thus this seems to give a perfectly good ``square root'' of the Laplacian without introducing any additional structure. In particular, I do not see where a double cover of $SO(2)$ is required in this picture.

One way to motivate the double cover is to require that $c$ be equivariant under the action of $SO(2)$, i.e.\ to ask for a representation $$ \rho : SO(2) \to GL(\mathbb{C}^2) $$ such that $$ \rho(A)\,c(v)\,\rho(A)^{-1} = c(Av). $$ I believe one can show that such a $\rho$ does not exist (though I have not checked myself yet).

My main question is: why should one impose such an equivariance condition in the first place? In other words, what role does the action $\rho$ on $\mathbb{C}^2$ play in the construction? Do I run into any problems without it in my example? Or is this something my example does not capture (and only shows up in change-of-charts, say)?

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  • $\begingroup$ In differential geometry, you either have to define an object globally on a smooth manifold, or locally together with its transformation rules. Otherwise there's no globally defined object. You have defined the local object, but you haven't specified its transformation rules. $\endgroup$ Commented 19 hours ago
  • $\begingroup$ @QuaereVerum ah so you need the equivariance to make the chart-changes consistent? So since I have a global chart on $\mathbb{R}^2$, I am not "forced" to consider this representation? Then my example would indeed have been too simple $\endgroup$ Commented 19 hours ago

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The complex algebra of the Gaussian plane is the simplest Clifford algebra but lacks the full mirror group, so you need to take the complex conjugate * and two spaces $\mathbb C^2$ in order to implement the full euclidean group.

The same is true for the Pauli algebra with the invariant Pauli operator $$H = \vec \sigma \nabla $$

The fact that the Pauli algebra has no mirror operator countering $$ S_x :(x\mapsto -x, \partial_x \mapsto -\partial_x )$$ is obscured by the fact, that as a tensor product representation of many electron states with even and odd spins, it is seen always in the context of complex Hilbert spaces and repesentations of the symmetric group with denumerable infinitely many variables in the infinite sum of the free tensor product $$\mathcal H = e^{\otimes\mathbb C^2* \mathcal L^2}$$ of which only the complete symmetric and antisymmetric subspaces are used in physics to represent ensembles of identical elementary particles.

This is the famous spin and statistics theorem, the leading principle in the evolution from Boltzmanns probability theory of ideal gases into quantum theory.

It was a real surprise to mathematicians, that the two distinct mirrorless representation of $O(3,\frac{1}{2})$ is realized in physics: the neutrino and antineutrino states, electron and positron states stripped of mass and electric charge.

So its not the Laplacian, that has to be the square, but the square of the full 4d relativistc Dirac operator with mass yields the massive d'Alembert operator. It contains a pair of mirrorless 3d representation as a representationof the full euclidean $O(3)$ group. As a side effect it includes even time reversal.

$$SO(3,1,\frac{1}{2}): \quad D \ \psi = e_i \nabla_i \ \psi = \frac{mc}{\hbar} \psi \quad D^2 = \square = -\partial_{ct}^2 + \Delta_x = \left(\frac{mc}{\hbar}\right)^2 $$

The mass is a Casimir operator tied to the repressentation of the translation group. The subtlety of the Dirac operator is the electric charge, induced as a nontrivial gauge transformation of the generator of translations

$$\partial_k \mapsto \partial_k + i \frac{e}{\hbar c} A_k(x)$$ by

$$\psi \mapsto e^{i \Phi(x)} \ \psi(x) \quad A_i = \partial_i \Phi$$ if $A$ is not a gradient and redefines the local zero of the momentum space in an observable way: charged massive paricles reduce their current velocity in regions of high potentials: they avoid them.

The gauge group of the spinor bundle occurs naturally by the fact, tht only sesquilinear Hilbert space products yield observable quantities, where complex unitary factors on both factors drop out.

Complex unitary basis transformations in vector bundles are a natural ingredient, they make observable effects by their commutator with derivative operators, i.e. by their inexactness as differential forms with values in the representation vector space.

The knot of algebraic equations tieing together spin and coordinate rotations, Lorentz transforms and the full mirror group is difficult to understand: Consult Perti Lounesto's book about Clifford algebra. Lounesto was one of thr very active members in the early times of the sci.physics.reseach newsgroup, because, as he wrote, a professor in Finland has much free time.

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