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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 $K_{XY}(p)=\frac{R(X,Y,X,Y)}{||X||^2 ||Y||^2 - <X,Y>^2}=R(X,Y,X,Y)=<R(X,Y)X,Y>$

If $X,Y$ are orthonormal fields

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

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

  2. second term $-<\nabla_X \nabla_Y X,Y>=-X<\nabla_Y X,Y>+<\nabla_Y X,\nabla_X Y>$ $<\nabla_Y X,Y>=Y<X,Y>-<X,\nabla_Y Y>=0$

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

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

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

Please help me. I don't know where I have made mistake. Thank you

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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 $$ [X,Y] \langle X,Y \rangle = \langle \nabla_{[X,Y]} X,Y \rangle + \langle X, \nabla_{[X,Y]} Y \rangle .$$

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    $\begingroup$ thank you for your answer $\endgroup$ – UserE Jun 14 at 16:03

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