Is the Scalar \( R^{ji} R_{ij} - R^2 \) Recognized in Mathematical Physics?

[joschu]

In my calculations, I come across the scalar
R^{ji} R_{ij} - R^2
(R_{ij} is the Ricci tensor, R is the Ricci scalar)
More specifically, I come across the integral of this scalar over a compact manifold.
Has anyone seen it before, and does it have any nice properties?

John Schulman

 

[robphy]

It seems to me...
Up to a constant factor, that scalar is the second principal-invariant of the Ricci tensor. It is proportional to the second elementary symmetric function of its eigenvalues. How does it arise? (I've been interested in this invariant [not necessarily for Ricci] and have been searching for a geometrical interpretation [and physical interpretation] for it.)

 

[pmb_phy]

In my calculations, I come across the scalar
R^{ji} R_{ij} - R^2
(R_{ij} is the Ricci tensor, R is the Ricci scalar)
More specifically, I come across the integral of this scalar over a compact manifold.
Has anyone seen it before, and does it have any nice properties?

John Schulman

Since

R^{ji} R_{ij} = R

it follows that

R^{ji} R_{ij} - R^2 = R - R^2 = R(1 - R)

That's about all I can see about it.

Pete

 

[Stingray]

Since

R^{ji} R_{ij} = R

it follows that

R^{ji} R_{ij} - R^2 = R - R^2 = R(1 - R)

That's about all I can see about it.

Pete

No. R=R_{ij} g^{ij}. The Ricci scalar has units, by the way...

I have no idea if the original poster is looking for a purely mathematical answer or not, but if Einstein's equations hold, and you have a perfect fluid with density \rho and pressure p, that scalar is proportional to p(\rho-p). Normally, \rho \gg p, so the square root of your scalar is basically a geometric average of the density and pressure.

 

[robphy]

In my earlier post, I was too lazy to write
R^{ji} R_{ij} - R^2=2R^i{}_{[j}R^j{}_{i]},
where I've used the metric to raise and lower indices.

 

[joschu]

Thanks for the responses, especially robphy. I'll do a little research about the eigenvalues of R^{i}_j.

Here's how I came across this quantity:
I'm working in a rather specific area: I'm studying the relationship between the curvature of a Kahler supermanifold with the curvature of the underlying complex manifold.
I found that if a supermanifold with two "fermionic" dimensions satisfies R=0 then there's a scalar differential equation with some curvature tensors that the underlying complex "bosonic" manifold must satisfy. This equation has a bunch of terms in it, but if we take the integral of the expression over a compact manifold, we get
\int (R^{ji} R_{ij} - R^2) dV=0

Here's the consequence of this calculation that interests me:
If I can find a complex manifold where the above integral CAN'T equal zero regardless of metric, it is likely that I will be able to disprove a certain conjecture concerning Yau's theorem generalized to supermanifolds. Unfortunately, I don't know if such a manifold exists.