Suppose g_{ab} is a metric on a manifold M, ∇ is the derivative operator on M compatible with g_{ab}, and R^{a}_{bcd} is associated with ∇. Then R^{a}_{bcd} (= g_{am} R^{m}_{bcd}) satisfies the following conditions.

(1) R_{ab(cd)} = 0.

(2) R_{a[bcd]} = 0.

(3) R_{(ab)cd} = 0.

(4) R_{abcd} = R_{cdab}.

Conditions (1) and (2) follow directly from clauses (2) and (3) of proposition, which goes like

Suppose ∇ is a derivative operator on the manifold M. Then the curvature tensor field R^{a}_{bcd} associated with ∇ satisfies the following conditions:

(1) For all smooth tensor fields α^{a1…ar}_{b1 …bs} on M,

2∇[_{c}∇_{d}] α^{a1…ar}_{b1 …bs} = α^{a1…ar}_{nb2…bs} R^{n}_{b1cd} +…+ α^{a1…ar}_{b1…bs-1n} R^{n}_{bscd} – α^{na2…ar}_{b1…bs} R^{a1}_{ncd} -…- α^{a1…ar-1n}_{b}1…b_{s} R^{ar}_{ncd}.

(2) R^{a}_{b(cd)} = 0.

(3) R^{a}_{[bcd]} = 0.

(4) ∇_{[m}R^{a}_{|b|cd }(* Bianchi’s identity*).

And by clause (1) of that proposition, we have, since ∇_{a}g_{bc} = 0,

0 = 2∇[_{c}∇_{d}]g_{ab} = g_{nb}R^{n}_{acd} + g_{an}R^{n}_{bcd} = R_{bacd} + R_{abcd}.

That gives us (3). So it will suffice for us to show that clauses (1) – (3) jointly imply (4). Note first that

0 = R_{abcd} + R_{adbc} + R_{acdb}

= R_{abcd} − R_{dabc} − R_{acbd}.

(The first equality follows from (2), and the second from (1) and (3).) So anti-symmetrization over (a, b, c) yields

0 = R_{[abc]d} −R_{d[abc]} −R_{[acb]d}.

The second term is 0 by clause (2) again, and R_{[abc]d} = −R_{[acb]d}. So we have an intermediate result:

R_{[abc]d} = 0 —– (1)

Now consider the octahedron in the figure below.

Using conditions (1) – (3) and equation (1), one can see that the sum of the terms corresponding to each triangular face vanishes. For example, the shaded face determines the sum

R_{abcd} + R_{bdca} + R_{adbc} = −R_{abdc} − R_{bdac} − R_{dabc} = −3R_{[abd]c} = 0

So if we add the sums corresponding to the four upper faces, and subtract the sums corresponding to the four lower faces, we get (since “equatorial” terms cancel),

4R_{abcd} −4R_{cdab} = 0

This gives us (4).

We say that two metrics g_{ab} and g′_{ab} on a manifold M are projectively equivalent if their respective associated derivative operators are projectively equivalent – i.e., if their associated derivative operators admit the same geodesics up to reparametrization. We say that they are conformally equivalent if there is a map : M → R such that

g′_{ab} = Ω^{2}g_{ab}

is called a conformal factor. (If such a map exists, it must be smooth and non-vanishing since both g_{ab} and g′_{ab} are.) Notice that if g_{ab} and g′_{ab} are conformally equivalent, then, given any point p and any vectors ξ^{a} and η^{a} at p, they agree on the ratio of their assignments to the two; i.e.,

(g′_{ab} ξ^{a} ξ^{a})/(g_{ab} η^{a}η^{b}) = (g_{ab} ξ^{a} ξ^{b})/(g′_{ab} η^{a}η^{b})

(if the denominators are non-zero).

If two metrics are conformally equivalent with conformal factor, then the connecting tensor field C^{a}_{bc} that links their associated derivative operators can be expressed as a function of Ω.

I don’t mean to be dense, naively reductive. Or simple minded demanding. Lol. But…

This pure math stuff like this post I really have no ability to penetrate.

Just as an exercisemaybe and just really out of curiosity, can you briefly and in layman terms tell me not just what you were talking about, but what was the point behind the post or what are you trying to say or what are you saying mathematically or about space or about… I mean I really have no way of penetrating what that is. Is it possible to tell me or is it something just strictly for mathematicians?

Conformal factors are transformations, which are critically important in any analysis of the global structure of a spacetime – the strategy is usually to use a conformal structure to make a topologically open spacetime conformally equivalent to one with a finite area in coordinate space, and then to look at the boundary of this conformally equivalent spacetime. One can map a formally infinite spacetime to a compact interval, and study its properties there.

I am doing a didactic series of metrics, and I guess I’d do two more.

Wow. 😂. Are you a quantum physicist ?

No sir. I studied physics and pure mathematics more than a couple of decades ago in a\the formal sense of the term. All I do now is analyze finance for the grassroots movements for my bread and butter. These topics are just to brush up my rusted studies. And in all candidness I must accept that it has rusted irreparably 🙂

.., and I’m not being sarcastic and asking that. I have a friend who’s a physicist she studies the earths atmosphere and the solar winds and the sign and stuff like that, but she knows very little about quantum physics and it’s kind of funny when I brought it up with her a couple times she knows very rudimentary things and I think I might even know conceptually more things about quantum physics and she even does which I think is kind of hilarious.