Help following step in spin geometry paper.

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I was reading a paper regarding the Dirac operator on embedded hypersurfaces and can't follow a step taken on page 5. They're examining the Schroedinger-Lichnerowicz equation on a manifold $\Omega$ with boundary $\Sigma$.

$$D^{*}D\psi=\bar{\nabla}^{*}\bar{\nabla}\psi+\frac{1}{4}\bar{R}\psi$$

Consider the 1-forms:

$$\alpha(X)=\langle\gamma(X)\bar{D}\psi,\psi\rangle,\beta(X)=\langle\bar{\nabla}_{X}\psi,\psi\rangle$$

for any $X\in\Gamma\left(T\Omega\right)$. It is clear that:

$$\delta\alpha=\langle\bar{D}^{2}\psi,\psi\rangle-|\bar{D}\psi|^{2}$$

because $\gamma$ acts by skew-symmetric endomorphisms for $\langle,\rangle$. Also

$$\delta\beta=-\langle\bar{\nabla}^{*}\nabla\psi,\psi\rangle+|\bar{\nabla}\psi|^{2}$$

hence, we obtain

$$\delta\alpha+\delta\beta=|\bar{\nabla}\psi|^{2}-|\bar{D}\psi|^{2}+\frac{1}{4}\bar{R}|\psi|^{2}$$Integrating on $\Omega$ and applying the divergence theorem

$$-\intop_{\Sigma}\langle\gamma(N)\bar{D}\psi+\bar{\nabla}_{N}\psi,\psi\rangle d\Sigma=\intop_{\Omega}\left(|\bar{\nabla}\psi|^{2}-|\bar{D}\psi|^{2}+\frac{1}{4}\bar{R}|\psi|^{2}\right)d\Omega$$

Now initially I had thought that $\delta$ applied to the 1-forms $\alpha$ and $\beta$ meant taking the variance as that seems to fit. If that's the case however I don't see how:

$$\intop_{\partial\Sigma}\left(\hat{n}\cdot\alpha\right)d\Sigma=\intop_{\Omega}\left(\delta\alpha\right)d\Omega$$

I then considered that maybe $\delta$ is meant to the the codifferential, which fits the last equation but doesn't seem to fit the definitions of $\delta\alpha$ and $\delta\beta$ above. Can someone please enlighten me as to what exactly they're doing here?