= Solution
Write $\beta=1/(k_BT)$, with $k_B$ the <Boltzmann constant>. For independent <Ising spins>, the one-site <partition functions> on the two sublattices are $z_B=2\cosh\beta(h+g)$ and $z_W=2\cosh\beta(h-g)$. Differentiating each logarithm with respect to its local field gives
$$
\boxed{M_B=\tanh\beta(h+g),\qquad M_W=\tanh\beta(h-g).}
$$
At $h=0$, the two sublattices experience opposite fields, so spin reversal combined with sublattice exchange leaves the <Hamiltonian> invariant and gives $M_W=-M_B$.
For the <antiferromagnetic Ising model>, take $N$ sites, equal sublattice sizes and $q=2D$ neighbors per site. There are $Nq/2$ bonds counted once. The <two-sublattice Ising mean-field approximation> expands each bond as
$$
\sigma_B\sigma_W=M_BM_W+M_W(\sigma_B-M_B)+M_B(\sigma_W-M_W)+(\sigma_B-M_B)(\sigma_W-M_W),
$$
and drops only the product of fluctuations. Summing the remaining terms gives
$$
\mathcal H_{\rm MF}=-\frac{NqJ}{2}M_BM_W-f_B\sum_B\sigma_B-f_W\sum_W\sigma_W,\qquad f_B=h+g-qJM_W,\quad f_W=h-g-qJM_B.
$$
The printed constant omits $N$; for a total <Hamiltonian> with unnormalized spin sums, this factor is necessary. Applying the independent-spin calculation now gives
$$
\boxed{M_B=\tanh\beta f_B,\qquad M_W=\tanh\beta f_W.}
$$
The <mean-field approximation> is favored by large coordination number: averaging many neighbors reduces local relative fluctuations. The size of critical fluctuations still has to satisfy the <Ginzburg criterion>; high dimension is a reason for validity, not permission to discard every correlation in every regime.
\b[Antisymmetric solutions and the transition.] At $h=0$, putting $M_B=-M_W=M$ makes both equations reduce to $M=\tanh\beta(g+qJM)$. A zero-field continuous transition concerns $g=0$ as well: nonzero $g$ explicitly selects the staggered sign and rounds this critical bifurcation. Put $K=\beta qJ$. The graphical intersections of $M$ and $\tanh(KM)$ give only $M=0$ for $K\leq1$. For $K>1$ there are additionally two nonzero intersections $\pm M_*$: on $M>0$, $\tanh(KM)-M$ is strictly concave, initially increases, and is negative at $M=1$, so there is exactly one positive root. Spin reversal supplies the negative root. The origin becomes unstable in the staggered direction, whereas the nonzero roots have $K(1-M_*^2)<1$ and are stable. Expanding about the bifurcation gives
$$
0=(K-1)M-\frac{K^3M^3}{3}+O(M^5),\qquad M_*^2=\frac{3(K-1)}{K^3}+O((K-1)^2)\sim3\frac{T_c-T}{T_c}.
$$
Thus $M_-=(M_B-M_W)/2$ turns on continuously, with \b[$k_BT_c=qJ$] and <order-parameter critical exponent> $1/2$.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2005/iii/paper-51-antiferromagnetic-mean-field.png]
{title=Antiferromagnetic mean-field graphical solutions and continuous onset of staggered magnetization}
\b[Uniform solutions.] At $g=0$, imposing $M_B=M_W=M$ instead gives $M=\tanh\beta(h-qJM)$. The function $M-\tanh\beta(h-qJM)$ has derivative $1+\beta qJ\operatorname{sech}^2\beta(h-qJM)>0$, is negative at $M=-1$ and positive at $M=1$, and consequently has exactly one zero for every positive temperature. At $h=0$ this zero is $M=0$. Antiferromagnetic exchange opposes uniform polarization, so the uniform branch has no ferromagnetic ordering bifurcation. This does not remove the antisymmetric solutions below $T_c$, or exclude transitions of staggered order in the full problem at nonzero uniform field. In particular, uniqueness on the diagonal $M_B=M_W$ is not uniqueness of all solutions.
\b[Free energy and stability.] The appropriate off-shell <two-sublattice Ising variational free energy> is the energy minus temperature times the independent-spin <entropy>. Define
$$
s(m)=\frac{1+m}{2}\ln\frac{1+m}{2}+\frac{1-m}{2}\ln\frac{1-m}{2}.
$$
There are $Nq/2$ bonds and $N/2$ spins of each kind, so the <free-energy density> per site is
$$
A(M_B,M_W)=\frac{qJ}{2}M_BM_W-\frac h2(M_B+M_W)-\frac g2(M_B-M_W)+\frac{k_BT}{2}[s(M_B)+s(M_W)].
$$
Since $s'(m)=\operatorname{atanh}m$, stationarity gives
$$
2A_{M_B}=qJM_W-h-g+k_BT\operatorname{atanh}M_B=0,
$$
and the analogous equation for $M_W$; these are exactly the two <mean-field self-consistency equations>. On a stationary solution, the identity $s(m)=m\operatorname{atanh}m-\ln(2\cosh(\operatorname{atanh}m))$ gives the equivalent on-shell expression
$$
A=-\frac{qJ}{2}M_BM_W-\frac{k_BT}{2}[\ln(2\cosh\beta f_B)+\ln(2\cosh\beta f_W)].
$$
The entropy form must be used to compare off-shell variations; treating this latter auxiliary expression as the variational potential would give spurious stability conclusions.
Using $s(m)=-\ln2+m^2/2+m^4/12+O(m^6)$ and $M_B=M_++M_-$, $M_W=M_+-M_-$ gives
$$
A=-k_BT\ln2+\frac{k_BT+qJ}{2}M_+^2+\frac{k_BT-qJ}{2}M_-^2-hM_+-gM_-+\frac{k_BT}{12}(M_+^4+6M_+^2M_-^2+M_-^4)+O(M_\pm^6).
$$
This includes the requested quadratic expansion and first-order field terms. The positive quartic coefficient verifies a <continuous phase transition> in $M_-$ when the staggered quadratic coefficient changes sign at $k_BT=qJ$. The uniform quadratic coefficient remains positive.
\b[Susceptibilities and a printed error.] On a stable zero-field ordered branch let $M_B=-M_W=M$ and $a=\beta(1-M^2)$. Since the derivative of $\tanh x$ is $1-\tanh^2x$, differentiating the two equations with respect to the uniform field gives
$$
\chi_B=a(1-qJ\chi_W),\qquad\chi_W=a(1-qJ\chi_B).
$$
Their sum and difference yield
$$
\boxed{\chi_+=\frac{\beta(1-M^2)}{1+\beta qJ(1-M^2)},\qquad\chi_-=0\quad(T<T_c).}
$$
The difference equation is $(1-qJa)\chi_-=0$; stability implies $qJa<1$ for $T<T_c$. Symmetry also explains the zero result: a uniform perturbation changes the two sublattice <magnetizations> equally to first order, and hence does not linearly change the <staggered magnetization>. This $\chi_-$ is a cross response to $h$, not the <magnetic susceptibility> to the staggered field $g$.
The displayed $\chi_+$ in the source instead contains $1+M^2$. It cannot follow from differentiating the stated equations. A decisive limiting check is $T\to0$: $1-M^2$ vanishes exponentially and the actual uniform response tends to zero, whereas the printed expression tends to $1/(qJ)$. Above $T_c$, $M=0$ and the same linear system gives
$$
\boxed{\chi_+=\frac{\beta}{1+\beta qJ}=\frac1{k_BT+qJ},\qquad\chi_-=0\quad(T>T_c).}
$$
For comparison, the actual staggered susceptibility is $\partial M_-/\partial g=a/(1-qJa)$, becoming $1/(k_BT-qJ)$ above $T_c$. Its divergence, rather than a divergence of the uniform response, detects antiferromagnetic criticality.
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