# On the local form of the spin covariant derivative: is this an exterior derivative of a spinor?

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 $$\mathrm{d}\psi(X)$$?

$$\psi$$ 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 $${\rm d}\psi(X)=X(\psi)$$.

• 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? Commented Apr 23 at 18:20
• 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? Commented Apr 23 at 18:26
• That is a good point. Maybe I'll ask it over at Math SE to be sure Commented Apr 23 at 18:58
• Amazing thanks! Could you link here your post to math SE? I am very curious to see their reply. Commented Apr 23 at 19:22
• Here is the question: math.stackexchange.com/q/4904425/479421 Commented Apr 23 at 20:46

Spinor $$\psi(X)$$ is a 0-form. Therefore, the exterior derivative $$\mathrm{d}\psi(X)$$ can be defined as such.

• 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. Commented Apr 23 at 18:08
• $\psi(X)$ is 0-form, and $d\psi(X)$ is 1-form. Nothing complicated at all. Commented Apr 23 at 18:10
• Then what kind of object would be $\psi$ in your interpretation of the covariant derivative? Commented Apr 23 at 18:13
• 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? Commented Apr 23 at 18:38
• 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 $\psi$. Commented Apr 23 at 18:48