# What is Ricci trace-free tensor?

according to wikipedia (https://en.m.wikipedia.org/wiki/Ricci_curvature) you can derive a new 2-tensor from the Ricci tensor, it would be the ricci trace-free. I would like a light to understand some things about this tensor:

(1) To define it, i imagine that i must already have constant scalar curvature on manifold, correct?

(2) What does the minus sign in its definition represent? After all, I'm taking from $$Ric$$ the symmetric bilinear form $$- (S/ n) g$$, where $$S$$ is the scalar curvature, $$n$$ os the dimension and $$g$$ is the metric of ambient.

(3) Does having lower ricci limitation help you, for example, to compare volume of a manifold with a form space, does having lower limitation for Ricci trace free help you study volume, or area of ​​manifold?

Your reference defines the trace-free Ricci tensor as $$Z = \text{Ric}- \frac{S}{n}g$$, where $$S$$ is the scalar curvature, $$g$$ the metric and $$n$$ the dimension of your manifold. Two things to realize at this point are that

1. $$Tr(Ric) = S$$ (this is just the definition of $$S$$)
2. $$Tr(g)=n$$ (this is easily checked in a coordinate chart)

From this it follows immediately that $$Z$$ is trace-free, as

$$Tr(Z) = Tr(\text{Ric}) - \frac{S}{n}Tr(g) = S - S = 0.$$

Concerning your first question, $$Z$$ is again a tensor field because it is the sum of two tensor fields; this is independent of whether your manifold has constant scalar curvature or not. You can always define this tensor.

As to your second question, the short calculation above shows why we need a minus sign to make $$Z$$ trace-free.