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I am trying to motivate spinors by making sure the Dirac equation is relativistically invariant (and it suffices to discuss just the Dirac operator).

Let $\{ e_i \}$ be an orthonormal frame and $x^i$ its associated global inertial coordinates. The Dirac operator is then $$D = \sum_{i = 0}^3 \gamma(e_i) \frac{\partial}{\partial x^i}. $$ To satisfy $D^2 = \Delta$, we need that $$\gamma(e_i) \gamma(e_j) + \gamma(e_j) \gamma(e_i) = 2 \eta(e_i,e_j) \mathbb{I}$$ i.e. a matrix representation of $Cl(1,3)$. Since a complex representation of this algebra is the same as a complex representation of the complex Clifford algebra $\mathbb{C}l(4)$, which itself is isomorphic to $\mathrm{End}(\mathbb{C}^4)$, we are led to the gamma matrices and in fact told by the structure theorem of complex clifford algebras how to get them from the Pauli matrices (The isomorphism singles out a basis so somehow this will need to be reflecting in going to another reference frame). Simultaneously, we establish a Dirac spinor field $\Psi(x^i)$ as a map from Minkowski space to $\mathbb{C}^4$.

Now suppose another inertial frame wants to write down the Dirac operator. They would write $$D = \sum_{j = 0}^3 \gamma(\tilde{e_j}) \frac{\partial}{\partial \tilde{x}^j}.$$ Now by the linearity of the Clifford map $\gamma: T_p M \to Cl(T_p M)$, we have $$\gamma(\tilde{e_{j}}) = \Lambda_{j}^i \gamma(e_i).$$ Since the spinor representation is just the identity, going to the level of representation (i.e. actual matrices), it must be that $$\tilde{\gamma^j} = \Lambda_{j}^i \gamma^i.$$ Now what I don't understand in the notes I'm following is that there must exist some spacetime dependent matrix $S(x^i)$ such that $$S \gamma^i S^{-1} = \tilde{\gamma^i}$$ and thus for relativistic invariance we must have $$\tilde{\Psi}(\tilde{x^i}) = S(x^i) \Psi(x^i)$$ and that we already know that the $S$ matrix must pointwise belong to $\mathrm{Spin}(1,3)$. Can someone help me finish my mathematical physics motivation for the Dirac spinor representation?

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  • $\begingroup$ Are you talking about Global or Local Lorentz transformation? By $\tilde{\Psi}(\tilde{x^i}) = S(x^i) \Psi(x^i)$ you seem to talk about Local ($x^i$ dependent $S(x^i)$) Lorentz transformation. But Local Lorentz transformation does NOT change the coordinates, as opposed to your $\tilde{x^i}$ which changes coordinates. Only the Global ($x^i$ independent)) Lorentz transformation changes the coordinates. So there is a contradiction. $\endgroup$
    – MadMax
    Commented May 6 at 17:17
  • $\begingroup$ For the differences between Global and Local Lorentz transformations, see the answer here: physics.stackexchange.com/questions/502982/… $\endgroup$
    – MadMax
    Commented May 6 at 17:28
  • $\begingroup$ The correct Global Lorentz transformation should be written as $\tilde{\Psi}(\tilde{x^i}) = S \Psi(\tilde{x^i})$. $\endgroup$
    – MadMax
    Commented May 6 at 17:36
  • $\begingroup$ If you insist on spacetime dependent $S(x^i)$, then your version of Dirac operator $D = \sum_{j = 0}^3 \gamma(\tilde{e_j}) \frac{\partial}{\partial \tilde{x}^j}$, is not covariant at all. You have to include the spin connection as part of the Dirac operator. $\endgroup$
    – MadMax
    Commented May 6 at 17:44
  • $\begingroup$ These are some helpful comments. To your last point, the spin connection on Minkowski space is flat, so just the partial derivative is indeed the spin connection: I think a part of the problem is notation. What I'm trying to do is motivate why the spinor field pointwise transforms in the way it does. To that end, my Lorentz transformation is the following: a change of frame at each point in spacetime (change of trivialization of orthonormal frame bundle). This is close to your concept of a local Lorentz transformation. $\endgroup$ Commented May 6 at 23:04

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