# Bi-invariant Metrics on Lie-Groups

In the Book "Morse Theory" by J. Milnor in chapter 21 they study the structure of Riemannian Lie Groups with bi-invariant metrics.

For Left-Invariant vector fields $$X,Y,Z,W$$ and a bi-invariant riemannian metric $$g$$ it is quite straightforward to show that

$$g(R(X,Y)Z,W)=1/4g([X,Y],[Z,W])$$.

Now he uses this relation to prove that the sectional curvature

$$K(X,Y)=g([X,Y],[X,Y])=1/4g([X,Y],[X,Y]) \geq 0$$.

What i don't understand at this point, is why it is sufficient to look only at left-invariant vectorfields X,Y to prove that the sectional curvature is generally positive.

My guess is that is has to do with the bi-invariance of the metric and the fact that the space of left-invariant vector fields is isomorphic to the tangent Space at the identity. But i don't see how exactly i can make sense of that.

For any $$g\in G$$ and any two planes $$\Pi$$ in $$T_gG$$, let $$v, w$$ be an orthonormal basis for $$\Pi$$. Let $$V, W$$ be two left-invariant vector fields on $$G$$ so that $$V(g) = v$$ and $$W(g) = w$$. Then the sectional curvature with respect to $$\Pi$$ is given by
$$S_\Pi = Rm(v, w, w, v) = Rm (V, W, W, V) = \frac 14 g ([V, W], [V, W]) \ge 0.$$
We are using that $$Rm$$ is a tensor, so it suffices to use any vector fields $$\tilde V, \tilde W$$ with $$\tilde V (g) = v, \tilde W(g) = w$$ to calculate $$Rm(v, w, w, v)$$. We also use that for all $$v\in T_gG$$, there is a left-invariant vector fields $$V$$ with $$V(g) = v$$.