Cosmological constant in GR can induce vector field tachyons?

A recent question on Light dispersion in gravitational theories, led to an implication that confuses me. It appears that the field equation for a massless vector field travelling in a background with non-zero cosmological constant $\Lambda$ becomes: $$(\nabla^b \nabla_b - \Lambda) A^{a} = 0 \tag{1}$$

This looks a lot like the Proca equation. Meaning the cosmological constant can induce an effective mass on the field? And depending on the sign of $\Lambda$, this can look like a tachyon?

Can someone please explain what the error is here, or if not, then help explain what is going on here?

Here are the details in case I'm just making a stupid computational error:

Starting with Einstein's field equations with a cosmological constant $$R_{\mu \nu} - \frac{1}{2} R \, g_{\mu \nu} + \Lambda g_{\mu \nu} = \frac{8 \pi G}{c^4} T_{\mu \nu} \tag{2}$$ we can solve for vacuum $(T_{\mu \nu}=0)$ and find that $$R_{\mu \nu} = \Lambda g_{\mu \nu} \tag{3}$$

The field equations for a massless vector field evolving in a curved space-time are $$\nabla^b \nabla_b A^{a} - \nabla^a \nabla_b A^b = {R^a}_{b} A^b \tag{4}$$ which if we consider it moving in the cosmological constant background as described by GR, the right hand side becomes $${R^a}_{b} A^b = {g^{a}}_b \Lambda A^b = \Lambda A^a \tag{5}$$ In the Lorenz gauge, $\nabla_b A^b = 0$, the field equation can then be simplified to $$(\nabla^b \nabla_b - \Lambda) A^{a} = 0 \tag{6}$$

Actually, the vacuum implies $R_{\mu\nu}=(\frac{R}{2}-\Lambda)g_{\mu\nu}$, so all your uses of $\Lambda$ should have read $\overline{\Lambda}:=\frac{R}{2}-\Lambda$ or some equivalent result. In $n$-dimensional spacetime $g_{\mu\nu}g^{\mu\nu}=n$ so $R=n(\frac{R}{2}-\Lambda)$ and $\overline{\Lambda}=\frac{R}{n}=\frac{2\Lambda}{n-2}$ provided $n\neq 2$.
Your Eq. (6) therefore becomes $(\square -\frac{R}{n})A^a = (\square -\frac{2\Lambda}{n-2})A^a = 0$. You can think of the tachyon question in terms of the sign of $R$ or the sign of $\Lambda$, depending on your preference.
• In 4d space-time with a cosmological constant, GR predicts $R_{\mu\nu} = \Lambda g_{\mu\nu}$ in vacuum. Your math looks find up to $R = n(\frac{R}{2} - \Lambda)$, but that gives $R = \frac{2n}{n - 2}\Lambda$, or simply $R=4\Lambda$ in our usual 4d space-time. Plugging this into the Einstein field equations in vacuum, $R_{ab} -\frac{1}{2}R g_{ab} + \Lambda g_{ab} = 0$, results in $R_{ab}=\Lambda g_{ab}$. – PPenguin Sep 17 '17 at 23:24
• To make it clear, I am only interested in 4d space-time here. So if we do agree, then all you have actually done is point out that $R_{ab} = \Lambda g_{ab}$ means I could replace $\Lambda$ with $R/4$ in the equations. That's true, but I'm not sure how that observation is useful and is more of a comment than an answer. – PPenguin Sep 18 '17 at 8:57