= Solution
The gravitational variable and the fields held fixed in a variation must be specified. For bosonic matter a second-order <metric tensor> formulation is sufficient. For spinorial matter a <vierbein> and a <spin connection> are needed to define the local Lorentz representation and its <covariant derivative>. “Second order” refers to a connection already expressed through derivatives of the <vierbein>; “first order” treats that connection as an independent variable.
For a <metric tensor>, a <scalar field> and an Abelian <gauge field>, consider the <Einstein-Hilbert action> with matter,
$$
S=\int d^4x\,e\left[\frac{R-2\Lambda}{2\kappa^2}-\frac12g^{\mu\nu}\partial_\mu\phi\partial_\nu\phi-V(\phi)-\frac14F_{\mu\nu}F^{\mu\nu}\right],\qquad F=dA.
$$
Here $e=\sqrt{-g}$ and the connection in $R$ is Levi-Civita. The <metric tensor> variation uses $\delta e=-eg_{\mu\nu}\delta g^{\mu\nu}/2$ and
$$
\delta R=R_{\mu\nu}\delta g^{\mu\nu}+g^{\mu\nu}\delta R_{\mu\nu},\qquad
\delta R_{\mu\nu}=\nabla_\rho\delta\Gamma^\rho{}_{\mu\nu}-\nabla_\nu\delta\Gamma^\rho{}_{\mu\rho}.
$$
<Metric compatibility> makes the last two terms a divergence. Thus
$$
\delta S=\frac12\int e\left[\kappa^{-2}(G_{\mu\nu}+\Lambda g_{\mu\nu})-T_{\mu\nu}\right]\delta g^{\mu\nu}+\text{matter variations}+\text{boundary},\qquad
T_{\mu\nu}=-\frac2e\frac{\delta S_{\rm matter}}{\delta g^{\mu\nu}}.
$$
Stationarity for arbitrary compactly supported variations gives
$$
\boxed{G_{\mu\nu}+\Lambda g_{\mu\nu}=\kappa^2T_{\mu\nu}.}
$$
Explicitly,
$$
T^{\phi}_{\mu\nu}=\partial_\mu\phi\partial_\nu\phi-g_{\mu\nu}\left[\frac12(\partial\phi)^2+V\right],\qquad
T^{A}_{\mu\nu}=F_{\mu\rho}F_\nu{}^\rho-\frac14g_{\mu\nu}F^2.
$$
<Scalar field> and gauge-potential variations give $\nabla^2\phi-V'=0$ and $\nabla_\mu F^{\mu\nu}=0$. Their equations imply <stress-energy conservation>, consistently with the contracted <Bianchi identity>. If the boundary <metric tensor> is fixed instead of using compactly supported variations, add the <Gibbons–Hawking–York boundary term>, $\kappa^{-2}\int_{\partial M}\varepsilon_{\partial M}\sqrt{|h|}K$, with the outward-normal orientation. Fixing the <metric tensor> alone does not otherwise remove the normal derivatives of its variation.
For spinorial matter, write $g_{\mu\nu}=e_\mu{}^ae_\nu{}^b\eta_{ab}$ and $\nabla_\mu\chi=\partial_\mu\chi+\omega_{\mu ab}\gamma^{ab}\chi/4$. In the second-order theory set $\omega=\omega(e)$ before varying. A symmetrized Dirac <action>, in the same no-$i$, mostly-plus convention as the earlier solutions, is
$$
S_\chi=\int e\left[-\frac12\left(\bar\chi\gamma^\mu\nabla_\mu\chi-(\nabla_\mu\bar\chi)\gamma^\mu\chi\right)-m\bar\chi\chi\right].
$$
Varying $\bar\chi$ and integrating by parts gives $(\not\nabla+m)\chi=0$; varying $\chi$ gives the adjoint equation. The spinor components held fixed in a gravitational variation are components in the <Lorentz frame>. Both $\gamma^\mu=e_a{}^\mu\gamma^a$ and $\omega(e)$ vary with the <vierbein>. The former gives the canonical stress, and integrating the spin-connection variation supplies its spin-current improvement. On the <Dirac equations> the resulting symmetric stress <tensor> is
$$
T^\chi_{\mu\nu}=\frac14\left[\bar\chi\gamma_\mu\overleftrightarrow\nabla_\nu\chi+\bar\chi\gamma_\nu\overleftrightarrow\nabla_\mu\chi\right],\qquad
\bar\chi\gamma_\mu\overleftrightarrow\nabla_\nu\chi=\bar\chi\gamma_\mu\nabla_\nu\chi-(\nabla_\nu\bar\chi)\gamma_\mu\chi.
$$
The off-shell expression also contains $g_{\mu\nu}\mathcal L_\chi/e$; this vanishes on the matter equations for the displayed symmetrized <action>. Treating the coordinate gamma matrices as fixed while varying the <metric tensor> would miss the displayed stress. Local Lorentz invariance removes the independent antisymmetric <vierbein> equation, leaving the symmetric <Einstein field equations> with this matter source.
In a first-order <vierbein> formulation, $e^a$ and an antisymmetric Lorentz connection $\omega^{ab}$ are independent. Define
$$
T^a=de^a+\omega^a{}_b\wedge e^b,\qquad R^{ab}=d\omega^{ab}+\omega^a{}_c\wedge\omega^{cb},
$$
and use $R(e,\omega)=e_a{}^\mu e_b{}^\nu R_{\mu\nu}{}^{ab}$. The <curvature form of a connection> variation is $\delta R^{ab}=D_\omega\delta\omega^{ab}$. For the Einstein-Hilbert term, <integration by parts> in form notation gives a connection equation proportional to
$$
\varepsilon_{abcd}\,T^c\wedge e^d=0
$$
when matter has no connection dependence. For an invertible <vierbein> this forces $T^a=0$, so $\omega=\omega(e)$. The <vierbein> equation then becomes the second-order <Einstein field equations>. This applies to the scalar and Maxwell examples: use $F=dA$, which is independent of $\omega$ and remains gauge invariant even if torsion is present.
A <Dirac field> changes the connection equation. Directly varying its two <covariant derivatives> gives
$$
\delta_\omega\mathcal L_\chi=-\frac e8\bar\chi\{\gamma^\mu,\gamma^{ab}\}\chi\,\delta\omega_{\mu ab}
=-\frac e4\bar\chi\gamma^{\mu ab}\chi\,\delta\omega_{\mu ab}.
$$
Thus its <spin> density sources totally antisymmetric torsion. This is <Einstein-Cartan theory>, and the connection equation is algebraic rather than a propagation equation for an extra field. To see both its solution and its effect, write
$$
\omega_{\mu ab}=\omega(e)_{\mu ab}+K_{\mu ab},\qquad K_{abc}=e_a{}^\mu K_{\mu bc},\qquad B^{abc}=\bar\chi\gamma^{abc}\chi.
$$
For the totally antisymmetric component sourced by this field, the <Ricci scalar> is $R(e,\omega)=R(e)-K_{abc}K^{abc}$ up to a divergence. The other irreducible contorsion components have no spinor source and their connection equations set them to zero. The auxiliary part of the <Lagrangian> is therefore
$$
\mathcal L_K=-\frac e{2\kappa^2}K_{abc}K^{abc}-\frac e4K_{abc}B^{abc}.
$$
Its algebraic variation gives the <algebraic torsion of a Dirac field>,
$$
\boxed{K_{abc}=-\frac{\kappa^2}{4}B_{abc},\qquad T_{abc}=-2K_{abc}=\frac{\kappa^2}{2}B_{abc}.}
$$
Here the first index of $K_{abc}$ labels the derivative, and the torsion convention is $T_{abc}=K_{bac}-K_{cab}$; specifying this convention fixes the sign. Eliminating $K$ leaves the second-order effective <action> with contact interaction
$$
\boxed{\mathcal L_{\rm contact}=\frac{e\kappa^2}{32}(\bar\chi\gamma^{abc}\chi)(\bar\chi\gamma_{abc}\chi).}
$$
It is a local axial-current four-fermion interaction. Hence first-order gravity with Dirac <spin> density is equivalent to torsion-free second-order gravity with this additional interaction. It is not equivalent to simply declaring the <spin connection> torsion-free and dropping the contact term. In the effective formulation the <Einstein field equations> have the stress <tensor> of the full effective matter <action>. Before eliminating the connection, the antisymmetric stress and its <spin> current are related by the local Lorentz Noether identity.
Finally, the <1.5-order formalism> is a convenient way to vary this effective <action>. Let $\widehat\omega(e,\chi)$ solve the algebraic connection equation and let $S_2(e,\chi)=S_1(e,\widehat\omega(e,\chi),\chi)$. The chain rule gives
$$
\delta S_2=\left.\frac{\delta S_1}{\delta e}\right|_{\widehat\omega}\delta e+\left.\frac{\delta S_1}{\delta\chi}\right|_{\widehat\omega}\delta\chi+\left.\frac{\delta S_1}{\delta\bar\chi}\right|_{\widehat\omega}\delta\bar\chi+\underbrace{\left.\frac{\delta S_1}{\delta\omega}\right|_{\widehat\omega}}_{0}\delta\widehat\omega.
$$
Thus \b[vary as in first order, then insert the solved connection]. One does not need to calculate the complicated induced connection variation. This procedure is justified by its connection equation, not by discarding that variation arbitrarily. In <supergravity> the <gravitino> sources a contorsion quadratic in <fermions>; substituting it generates the four-fermion terms, while the 1.5-order method simplifies the local <supersymmetry> cancellation through the lower orders considered above.
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