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The answer is yes: they are related. The quantum field theory estimate is wrong by some huge factor (e.g. 120 orders of magnitude) but this could be avoided if it were possible to simply assert that the cosmological constant $\Lambda$ is zero. But the astrophysical observations suggest $\Lambda$ is not zero. In a bit more detail, the situation is as follows....


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One thing to note first is that pure gravity (i.e. $\Lambda = 0$) is a topological theory. What this means is that on any manifold, the Einstein action $S_E = \int d^2 x \sqrt{|g|} R$ only depends on quantities on the boundary. In fact, due to the Gauss-Bonnet theorem, if your manifold is compact and has no boundary, then \begin{equation} \tag{1} \int d^2 x ...


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Minkowski spacetime is flat; it has zero curvature. De Sitter spacetime is curved; specifically, it has the same positive scalar curvature at every point. There is also anti-de Sitter spacetime, which has the same negative scalar curvature at each point. These three spacetimes thus exhaust the possibilities for spacetimes of uniform scalar curvature. ...


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There is a potential tension between low redshift probes of mass clustering and Planck data (CMB measurements). This ongoing speculation might be evidence of new physics or even modifications of general relativity. However, the author of the article you cited seems to have a confusion between the cosmological constant (no tension discussed in the literature) ...


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No. Evidence for the accelerating expansion of the universe comes from multiple angles: supernovae data, Baryon acoustic oscillations, the mass functions of galaxy clusters, etc. That the universe's expansion is accelerating is not in doubt; the question is by how much.


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Cosmologists believe the universe is expanding at an accelerating rate because the measured value of the cosmological is positive. The "tension" is that the two different methods give different values of the Hubble parameter (from which the cosmological constant can be calculated). There is a low probability ($4.4 \sigma$ according to Wikipedia) that these ...


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