In an article from the University of Chicago, July 17, 2020, it is stated that

"Judging cosmic distances from Earth is hard. So instead, scientists measure the angle in the sky between two distant objects, with Earth and the two objects forming a cosmic triangle. If scientists also know the physical separation between those objects, they can use high school geometry to estimate the distance of the objects from Earth."

That seems straightforward, except for the fact that high school geometry only works in flat space where the angles enclosed by a triangle add up to precisely 180 degrees. In a curved universe, a triangle can enclose either more or less than 180 degrees. Unless the curvature is known, triangulation shouldn't work reliably in a curved space.

So my question is: in measurements of the Hubble Constant by the triangulation method, what assumptions are made about curvature of the universe? And, how well-founded are those assumptions?

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    $\begingroup$ en.wikipedia.org/wiki/… and en.wikipedia.org/wiki/Lambda-CDM_model $\endgroup$ – Umaxo Jul 18 at 5:01
  • $\begingroup$ That link contains a lot of interesting information, but does not seem to contain an answer to my specific question. $\endgroup$ – S. McGrew Jul 18 at 5:45
  • $\begingroup$ How so? You asked about assumptions about curvature of the universe and especially the first link contains them (homogenity and isotropy) together with the general form of the metric possessing these symmetries and resulting field equations. The second link then elaborates on field equations themselves. $\endgroup$ – Umaxo Jul 18 at 8:17
  • $\begingroup$ Perhaps I've missed something, but it seems that assuming homogeneity and isotropy does not restrict curvature other than to say that the curvature is the same everywhere. My question is about how curvature affects interpretation of results when measuring the Hubble constant. I get the impression that, usually, space is assumed to be flat (have zero curvature). $\endgroup$ – S. McGrew Jul 18 at 14:52
  • $\begingroup$ @S.McCrew "homogeneity and isotropy does not restrict curvature other than ..." It is pretty strong restriction though. The spatial geometry is given up to one constant (and scale/expansion function).. "usually, space is assumed to be flat" because it is. en.wikipedia.org/wiki/Shape_of_the_universe. $\endgroup$ – Umaxo Jul 18 at 16:16

I guess you are looking for angular diamater distance. For different curvature the equation takes different forms. See here https://en.wikipedia.org/wiki/Angular_diameter_distance

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What appears to be a sufficient answer to the question can be found in this SE answer by @JohnRennie, combined with a few other articles. I had confused "flat space" with "flat spacetime". As John Rennie said in that answer, spacetime is not flat in an expanding universe, but space can be flat. So, indeed, it is necessary to account for the expansion of space when measuring the Hubble constant via triangulation.

The link given by @Layla provides the formula used to relate distance, physical separation, and angular separation, and space curvature. The formula is based on the FLRW model, described in this link provided by @Umaxo This NASA article outlines the discrepancies between results obtained by different measurement methods.

Various approaches are taken to measure the curvature of space (space, not spacetime), including this article which describes a method using gravitational lensing.

So, the answer is this: generally, when calculating the Hubble constant it is assumed that the universe is spatially flat but expanding at a rate that may change over time. The assumption of spatial flatness is well-founded, at least to a close approximation, on several types of astronomical observations.

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