general relativity - Why is the cosmological constant a scalar?

1. 
   

   Maybe my understanding is just off, but the cosmological constant is just a scalar, right?

   
   


2. 
   

   What are it's units?

   
   


3. 
   

   Why a scalar? - was a tensor 'cosmological constant' ever considered or is it just not conformable to the math or physics of the field equations?

   
   

Answer

As mentioned in another answer, it was first added by Einstein to make the universe static. As to why it is a scalar, I paraphrase from Straumann (2013).

It is natural to postulate that the ten potentials $g_{\mu\nu}$ satisfy equations of the form

$$\mathcal{D}_{\mu\nu}[g]=T_{\mu\nu}$$ where $\mathcal{D}$ is a tensor field constructed pointwise from $g$ and its first and second partial derivatives. Furthermore, this tensor field should satisfy $$\nabla^\mu\mathcal{D}_{\mu\nu}=0$$ in order to guarantee $\nabla^\mu T_{\mu\nu}=0$. [Some remarks follow.]

Theorem (Lovelock) A tensor $\mathcal{D}_{\mu\nu}$ with the required properties is in four dimensions

$$\mathcal{D}_{\mu\nu}=aG_{\mu\nu}+bg_{\mu\nu}$$ where $a,b\in \mathbb{R}$.

In four dimensions, the field equations are quite unique. $a=(8\pi G)^{-1}$ is determined by taking the Newtonian limit. $b=\Lambda/8\pi G$ is chosen such that the field equations have the simple form you are used to. In four dimensions, it is not possible to have anything else other than a cosmological constant scalar.

A more physical interpretation is this: a constant tensor in spacetime would violate cosmological isotropy. It would pick out a preferred direction in a given coordinate system.

I should note that there exists a generalization of the Lovelock theorem to higher dimensions, but I am unaware of the proof. In any case, the associated length scales of something like Kaluza-Klein extra dimensions must be very small such that we needn't worry about more than 4 dimensions in standard GR. (See any text on string theory for extra-dimensional stuff.)