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I'm reading Hamilton's Mathematical Gauge Theory. I'm currently on section 6.10, about the spin covariant derivative. Letting $S$ be the (Dirac) spinor bundle, a section $\Psi \in \Gamma(S)$ can be written locally as $\Psi = [\epsilon,\psi]$, where $\epsilon$ is a local trivialization of the spin frame bundle $\mathrm{Spin}^+(M)$ in the contractible set $U$ and $\psi \colon U \to \Delta$, with $\Delta$ being the vector space of the Dirac spinor representation. Hamilton says on page 386 that the spin covariant derivative can then be written locally as $$\nabla_X \Psi = [\epsilon, \nabla_X \psi]$$ with $$\nabla_X \psi = \mathrm{d}\psi(X) + A_{\mathrm{Spin}}^{\epsilon}(X) \cdot \psi.$$ $A_{\mathrm{Spin}}^{\epsilon}$ is the local connection one-form on $U$. However, I do not know what $\mathrm{d}\psi(X)$ means. The notation seems to suggest this is some sort of an exterior derivative of a spinor, but the exterior derivative is only defined for forms, so I'm confused about how this object is defined.

What is the meaning of $\mathrm{d}\psi(X)$?

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$\psi$ is the coordinate of the spinor $\Psi$ in the local basis $\epsilon$. As such, I would say that locally $\psi$ behaves a function $\psi: U \to \mathbb{C}^4$ i.e. a zero form, from which it is possible to take the exterior derivative $d\psi$. Then, as usual ${\rm d} \psi$ can be seen as a 1-form acting on vector fields $X$ to give back the function ${\rm d}\psi(X)=X(\psi)$.

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  • $\begingroup$ I know there are complications with defining the wedge product of forms with values on a vector space. Does your answer also means that the exterior derivative of a form valued on a vector space is well-defined? $\endgroup$ Commented Apr 23 at 18:20
  • $\begingroup$ That's a very good question. I always assumed so, but perhaps I am wrong. It is quite common to take the exterior derivative of vector space valued forms right? A similar case happen when you compute the curvature of a connection e.g. $F\propto{\rm d}A$ (where $A$ is a 1-form valued in a Lie Algebra). Or am I missing something? $\endgroup$ Commented Apr 23 at 18:26
  • $\begingroup$ That is a good point. Maybe I'll ask it over at Math SE to be sure $\endgroup$ Commented Apr 23 at 18:58
  • $\begingroup$ Amazing thanks! Could you link here your post to math SE? I am very curious to see their reply. $\endgroup$ Commented Apr 23 at 19:22
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    $\begingroup$ Here is the question: math.stackexchange.com/q/4904425/479421 $\endgroup$ Commented Apr 23 at 20:46
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Spinor $\psi(X)$ is a 0-form. Therefore, the exterior derivative $\mathrm{d}\psi(X)$ can be defined as such.

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    $\begingroup$ No, $\psi$ is a zero form, not $\psi(X)$. $X$ is a vector field. $d\psi$ is a 1-form which can act on the vector such that $d\psi(X)$ is a 0-form/function. $\endgroup$ Commented Apr 23 at 18:08
  • $\begingroup$ $\psi(X)$ is 0-form, and $d\psi(X)$ is 1-form. Nothing complicated at all. $\endgroup$
    – MadMax
    Commented Apr 23 at 18:10
  • $\begingroup$ Then what kind of object would be $\psi$ in your interpretation of the covariant derivative? $\endgroup$ Commented Apr 23 at 18:13
  • $\begingroup$ I am curious. $A_{\mathrm{Spin}}^{\epsilon}$ is the local connection one-form. What kind of object would $A_{\mathrm{Spin}}^{\epsilon}(X)$ be in your interpretation? $\endgroup$
    – MadMax
    Commented Apr 23 at 18:38
  • $\begingroup$ Well, as $A_{\mathrm{Spin}}^{\epsilon}$ is the local (Lie-Algebra valued) connection one-form, after acting on a vector field $X$, $A_{\mathrm{Spin}}^{\epsilon}(X)$ gives a Lie-Algebra element which can act on $\psi$. $\endgroup$ Commented Apr 23 at 18:48

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