The goal here is to find an exact solution to Einstein’s general relativity equations for empty space, save one infinite cosmic string. The string is massless but under tension. (It turns out I was wrong about massless.) Tension amounts to stress as in “stress energy tensor”. I thought that there would be an angular excess about the string. Here is what I learned.

Search for “fluid in equilibrium” here. It claims that T11 is positive for ordinary positive pressure.

First we describe a metric for a 2D space with a uniform positive curvature in a disk of radius tr, and flat outside the disk. We use polar coördinates, (r, θ) = (x1, x2).
gij =
 1 0 0 f(r)
.
0≤r; 0≤θ≤2π; identify (r, 0) with (r, 2π). Identify all points of form (0, θ) with each other. We choose f so that we have curvature 1 out to radius tr and 0 beyond.
f(r) is roughly proportional to r2. Now we choose f to meet our goal. I think that for 0≤r≤tr, f(r) = (sin r)2. (This is the embedding map for a sphere.)

For r ≥ tr, f(r) =(((sin tr)−(cos tr)tr)+(cos tr)r)2
Perhaps f(r) = (sinh r)2 gives us uniform negative curvature.

Γk ij= ½(−∂gij/∂xk +∂gik/∂xj +∂gkj/∂xi)

∂gij/∂xk is 0 except for ∂g22/∂x1 which is f'(r).
Γk ij is 0 except for Γ1 22 = −½f'(r) and Γ2 12 = Γ2 21 = ½f'(r).
Note that the two formulae for f agree at the splice point. For r < tr the space is isometric to a slice of a 3D of sphere radius 1. For r > tr the space is isometric to a truncated cone where R𝜌σμν = 0.

Γkij = gknΓn ij.
gij =
 1 0 0 1/f(r)
.

Γkij is 0 except for Γ122 = −½f'(r) and Γ212 = Γ221 = ½f'(r)/f(r).

R𝜌σμν = ∂μΓ𝜌νσ − ∂νΓ𝜌μσ + Γ𝜌μλΓλνσ − Γ𝜌νλΓλμσ

See this code with this map:
 here tr r gij gij Γkij ∂μΓ𝜌νσ R𝜌σμν Rijkl Rij R Gij there tr r gl gu gam delgam rie rr rt bigr bigg
Here is the output. (This helped.)

μΓ𝜌νσ is zero except for ∂1Γ122 = −½f''(r)
and ∂1Γ212 = ∂1Γ221 = ½(f''(r)/f(r) − (f'(r)/f(r))2).
(In each case the factor with λ=1 is 0.)

Rijkl = ginRnjkl.
Rij = Rkikj.
R = gijRij.
Gjk = Rjk − ½gjkR.