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


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.


1 Answer 1


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 $V(g) = v$.


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