# Size of the universe

This is really a follow up to the Shape of the universe? question.

In the first answer to the question, Ted Bunn says:

However, the best available data seem to indicate that the Universe is very nearly spatially flat. This means that, if we do live in a 3-sphere Universe, the radius of the sphere is very large, and the distance you'd have to travel is much larger than the size of the observable Universe.

My question is, can the available data be used to set a lower bound to the size of the Universe?

In other words, suppose the Universe was actually shaped like a 3 sphere. Then based on our best data, what would the "radius" of that sphere be? How much larger is that than the observable universe?

However, cosmological measurements such as WMAP have measured the density to be very close to the critical density so the curvature to be something like $1.02\pm 0.02$. So if we assume that the topology is a sphere, its own curvature radius has to be about 50 times larger than the radius of the visible universe - or, approximately, the characteristic curvature radius of spacetime as measured now.
I don't know the exact figure - which sensitively depends on the error margin you assign to the measured total $\Omega$ - but it's of order 1 trillion light years.
Also, note that the lower bound on the 3-sphere radius doesn't mean that the minimum volume has to be the volume of the 3-sphere even if the 3-sphere the correct local shape. There are other topologies that are locally 3-sphere, such as the "lens spaces" - the quotient of the 3-sphere with respect to the discrete $ADE$ groups such as the symmetry group of an icosahedron (translated from a subgroup of $SO(3)$ to a subgroup of $SU(2)$ which is embedded to $SO(4)$ so that it can naturally act on a 3-sphere). In those ADE cases, the total volume would be divided by the order of the group.