Qfyilyi

What is the correct metric to use for two dimensional de Sitter? If one starts with the following metric, which looks similar to de Sitter in 4 dimensions:

$$ds^2 = -dt^2 + e^{2H t} dx^2,$$

one can calculate $R = 2H^2$, and $R_{00} = -H^2$, which gives the $Lambda = 0$, which is not the solution one is looking for. What should be the correct metric to use for the same?

In two-dimensional spacetime, the Einstein tensor $R_{ab}-frac{1}{2}g_{ab}R$ is identically zero (https://en.wikipedia.org/wiki/Einstein_tensor), which explains why you get $Lambda=0$.

In any number $D$ of spacetime dimensions, including $D=2$, de Sitter spacetime can be constructed like this. Start with the $D+1$ dimensional Minkowski metric
$$
-(dX^0)^2+sum_{k=1}^D(dX^k)^2.
tag{1}
$$
The submanifold defined by the condition
$$
sum_{k=1}^D(X^k)^2=L^2+(X^0)^2
tag{2}
$$
is $D$-dimensional de Sitter spacetime. The length parameter $L$ is related to the cosmological constant $Lambda$ by
$$
Lambda = frac{(D-2)(D-1)}{2L^2}.
tag{3}
$$
This is equation (4) in "Les Houches Lectures on de Sitter Space" (https://arxiv.org/abs/hep-th/0110007). Setting $D=2$ recovers your result $Lambda=0$.

By the way, equations (13)-(14) in the same paper show how to derive the de Sitter metric in the form
$$
-dt^2+e^{2t}sum_{k=1}^{D-1}(dx^k)^2
$$
starting from equations (1)-(2). For $D=2$, this reduces to the form shown in the question.