# Sectional curvature of Lie group

Good time of day. I try to prove that the sectional curvature for Lie group with bi-invariant metric is equal

$$K(\sigma)= \frac 14 |[X,Y]|^2$$

My attempt is the following I have proved that $$\nabla_X Y = \frac 12 [X,Y]$$

I know that $$K_{XY}(p)=\frac{R(X,Y,X,Y)}{||X||^2 ||Y||^2 - \langle X,Y\rangle^2}=R(X,Y,X,Y)=\langle R(X,Y)X,Y\rangle$$

If $$X,Y$$ are orthonormal fields

$$\langle R(X,Y)X,Y\rangle =\langle (\nabla_Y \nabla_X X - \nabla_X \nabla_Y X + \nabla_{[X,Y]} X),Y\rangle$$

1. first term is equal $$0$$ since $$\nabla _X X=0$$

2. second term $$-\langle\nabla_X \nabla_Y X,Y\rangle=-X\langle\nabla_Y X,Y\rangle+\langle\nabla_Y X,\nabla_X Y\rangle$$ $$\langle\nabla_Y X,Y\rangle=Y\langle X,Y\rangle-\langle X,\nabla_Y Y\rangle=0$$

And $$-\langle \nabla_X \nabla_Y X,Y\rangle =\langle \nabla_Y X,\nabla_X Y\rangle=\langle\frac12[Y,X],\frac12[X,Y]\rangle=-\frac 14 |[X,Y]|^2$$

1. third term is equal to $$0$$ since $$\langle\nabla_{[X,Y]} X,Y\rangle=\frac12[X,Y]\langle X,Y\rangle=0$$

And $$K(\sigma)= -\frac 14 |[X,Y]|^2$$

The third term should be $$\langle \nabla_{[X,Y]} X, Y \rangle = \langle \frac{1}{2} [ [X,Y], X], Y \rangle = \frac{1}{2} \langle [X,Y] , [X,Y] \rangle = \frac{1}{2} \lvert [X,Y] \rvert^2 .$$ If you combine this correction with the rest of your work you get the right formula.
It looks like your mistake was forgetting a term in the identity $$[X,Y] \langle X,Y \rangle = \langle \nabla_{[X,Y]} X,Y \rangle + \langle X, \nabla_{[X,Y]} Y \rangle .$$