Curvature of Lie group endowed with bi-invariant Riemmanian metric.

Question

Let $G$ be a Lie group endowed by a biinvariant Riemmanian metric $\langle\cdot,\cdot\rangle$ and $\mathfrak{g}$ its Lie algebra.
It is known that the metric is fully determined by its behaviour at the identity element, i.e. the scalar product $\langle\cdot,\cdot\rangle_e$ on $T_eG\cong\mathfrak{g}$.

Now, the following properties hold for the curvature $R$, Ricci tensor $Ric$ the Killing form $B$ , at the identity element, i. $\forall x,y,z\in\mathfrak{g}$:

1. $R(x,y)z=\frac{1}{4}[[x,y],z]$

2. $R(x,y,x,y)=\frac{1}{4}||[x,y]||^2$,

3. $\operatorname{Ric}(x,y)=-\frac{1}{4}B(x,y)$

Now, I agree that the same relations will hold when evaluating the corresponding tensors on the unique left invariant vector fields $X,Y,Z$ determined, respectively by $x,y,z$.

But the sources presenting these results then use them to say that, for instance the sectional curvature is nonnegative or, if $G$ is compact and semisimple $(G,-B)$ is Einstein. I mean this would certainly be true if we could extend this relations to hold for any vector field, and not only for left invariants one. And I don't see why this should be true.

What am I missing?