Let $G$ be a simply-connected Lie group endowed with a biinvariant Riemannian metric. Can you please tell me, if the following is true: Is the Riemann curvature tensor $R(X, Y)Z$ of the above metric always parallel, i.e. do we have $\bigtriangledown R(X, Y)Z=0$? I wonder this because the situation looks so much like a symmetric space and in every Riemann symmetric space, the curvature tensor is parallel.



Question on simply-connected Lie group.
To elaborate a bit on Liviu's answwer: If a Riemannian metric $g$ on $G$ is bi-invariant, then it easily follows that $g$ is invariant under the inversion map $\iota: G\to G$, and then bi-invariance implies that, for any $a\in G$, the map $\iota\_a:G\to G$ defined by $\iota\_a(b) = a(b^{-1})a$ also leaves $g$ invariant and fixes $a$. Since $g$ allows an isometric inversion through each point $a$, it follows that $(G,g)$ is a symmetric space, by definition.


