$\DeclareMathOperator\End{End}$Following the deduction by John W. Morgan in his book [The Seiberg–Witten equations and applications to the topology of smooth four manifolds](https://press.princeton.edu/books/paperback/9780691025971/the-seiberg-witten-equations-and-applications-to-the-topology-of), an almost complex manifold $X$ has a natural $\operatorname{spin}^\mathbb{C}$ structure: $S\_{\mathbb{C}}(\tilde{P}\_X)\mathrel{:=}\bigoplus\_q\bigwedge^{0,q}(X;\mathbb{C})$, $S\_{\mathbb{C}}^+(\tilde{P}\_X)=\bigoplus\_q\bigwedge^{0,2q}(X;\mathbb{C})$, and $S\_{\mathbb{C}}^-(\tilde{P}\_X)=\bigoplus\_q\bigwedge^{0,2q+1}(X;\mathbb{C})$. And we can view $1$-forms acting on the spinors by Clifford multiplication using the following formula: For a one form $\alpha$ and $\nu$ a spinor we have
\begin{equation\*}
\alpha\cdot\nu=\sqrt{2}\big(\pi^{0,1}(\alpha)\wedge\nu-\pi^{0,1}(\alpha)\angle\nu\big)
\end{equation\*}
where the contraction is defined as follows:
\begin{equation\*}
v\angle(a^1\wedge\cdots\wedge a^t)=\sum\limits\_{i=1}^t(-1)^{i-1}\langle a^i,v\rangle a^1\wedge\cdots\wedge\widehat{a^i}\wedge\cdots \wedge a^t
\end{equation\*}
These discussions are in pages 51, 52 and 109.


Now, if I am not wrong $T^\*X$ injects inside $\End(S^+,S^-)$ by clifford multiplication and hence I expect the complex linear extension to give an injection $T^\*X\otimes\mathbb{C}$ into $\End(S^+,S^-)$. But the formula says that the $(1,0)$ part of a covector acts trivially on the spinors. So, the formula seems a bit odd, can someone please clear this confusion?



Clifford multiplication formula on an almost complex manifold
I think I figured this out and the calculations in the Kähler case in the book are misleading in a sense and the multiplication formula mentioned in page 109 is wrong.


As calculated in page 52, the action of a REAL one form $\alpha\in\Omega^1(X;\mathbb{R})$ on a spinor $\nu$ is indeed given by the formula:
\begin{equation\*}
\alpha\cdot\nu=\sqrt{2}\big(\pi^{0,1}(\alpha)\wedge\nu-\pi^{0,1}(\alpha)\angle\nu\big)
\end{equation\*}
Now we need to calculate its complex-linear extension to the complexified forms very carefully.
So, for $a,b\in\Omega^1(X;\mathbb{R}),$
\begin{align\*}(a+ib)\cdot\nu&=a\cdot\nu+ib\cdot\nu\\
&=\sqrt{2}\big(\pi^{0,1}(a)\wedge\nu-\pi^{0,1}(a)\angle\nu\big)+i\sqrt{2}\big(\pi^{0,1}(b)\wedge\nu-\pi^{0,1}(b)\angle\nu\big)
\end{align\*}
The important point is contraction is complex anti-linear in the first variable.


So, $i(\pi^{0,1}(b)\angle\nu)=(-i\pi^{0,1}(b))\angle\nu.$ Hence,
\begin{align\*}\sqrt{2}\big(\pi^{0,1}(a)\wedge\nu-\pi^{0,1}(a)\angle\nu\big)+i\sqrt{2}\big(\pi^{0,1}(b)\wedge\nu-\pi^{0,1}(b)\angle\nu\big)
&=\sqrt{2}\big((\pi^{0,1}(a)+i\pi^{0,1}(b))\wedge\nu-(\pi^{0,1}(a)-i\pi^{0,1}(b))\angle\nu\big)\\
&=\sqrt{2}\big(\pi^{0,1}(a+ib)\wedge\nu-\overline{\pi^{1,0}(a+ib)}\angle\nu\big)
\end{align\*}
I believe this is the right formula for a complexified form, let me know if I have made any mistake.


