Undergrad Density parameter and curvature index

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In cosmology, when the density parameter Ω equals 1, the curvature index κ is defined as 0, indicating a flat universe. If Ω exceeds 1, κ must be positive, reflecting a closed universe, while if Ω is less than 1, κ is negative, indicating an open universe. The value of κ is intrinsically linked to Ω, meaning it cannot exist independently. The curvature index can be defined using different conventions, such as the (-1,0,1) system, which simplifies understanding its relationship with Ω. Thus, κ's value is directly influenced by the density parameter Ω.
Ranku
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For Ω=1, κ=0. Does the value of κ simply follow from the value of Ω, or can its value have an independent existence? So if Ω>1, does κ have to be 1?
 
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Context would be useful. I don't want to guess what you mean.
 
Ranku said:
So if Ω>1, does κ have to be 1?
Yes, by definition. Since Ω=1 is the critical density - the density at which the universe is flat - any other value of Ω necessitates that the k parameter is not 0 and has the same sign as Ω-1.
 
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Likes Jaime Rudas and Ranku
Define k for me.
 
The definition of k seems to have more than one option. For example the (-1,0,1) option simplifies your question.
 

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