I have trouble understanding how the curvature and volume work in pseudo-Riemannian geometry. Something I will say should be wrong. Let $(\mathcal{M},g)$ be a pseudo Riemannian manifold of dimension $n$ and $B(\epsilon)\subset \mathcal{M}$ be the ball of radius $\epsilon$ centered on a point $p\in \mathcal{M}$. The wikipedia on scalar curvature gives
$$ \frac{vol(B(\epsilon))}{vol(B_{\mathbb{R}^n}(\epsilon))} = 1 - \frac{S}{6(n+2)}\epsilon^2+ O(\epsilon^4),$$
where $S=g^{ij}R_{ij}$, $R_{ij}$ being the Ricci tensor, is the scalar curvature at $p$. I am confused by the change of metric $g\rightarrow -g$.
-The Ricci tensor is defined from the curvature tensor, which can be defined from the Levi-Civita connection. $g\rightarrow -g$ does not change the connection so it does not change the Ricci tensor.
-Hence the sign of the scalar curvature should change.
-But on the other hand the volume of balls should not change, since in a coordinate system the volume form is defined as
$$\omega = \sqrt{|det(g)|}dx_1 \wedge ... \wedge dx_n.$$
This seems to be in contradiction with the above formula. Does anyone see what's wrong in what I said?
Thank you for your help.