Let $X$ be a Killing field on a Riemannian manifold $M$. Define a mapping $A_X : \mathcal X(M) \to \mathcal X(M)$ by $A_X(Z)=\nabla_Z X$, $Z \in \mathcal X(M)$. Consider the function $f : M \to \mathbb R$ given by $f(q) = \langle X,X \rangle_q$, $q \in M$. Let $p \in M$ be a critical point of $f$ (that is, $df_p=0$). Prove that for any $Z \in \mathcal X(M)$, at $p$, $\langle A_X(Z),A_X(Z) \rangle(p)=\frac 12Z_p(Z\langle X,X\rangle)+\langle R(X,Z)X,Z\rangle$.

Hint: Put $S=\frac 12Z_p(Z\langle X,X\rangle)+\langle R(X,Z)X,Z\rangle(p)$. Using the Killing equation $\langle \nabla_Z X,X\rangle+\langle \nabla_X X,Z \rangle=0$, we obtain $$ S=\langle \nabla_{[X,Z]}X,Z\rangle-\langle \nabla_X X,\nabla_Z Z \rangle-\langle \nabla_X \nabla_Z X, Z \rangle. $$ Using the Killing equation again, we obtain \begin{align} S&=-\langle \nabla_Z X,\nabla_X Z\rangle + \langle \nabla_Z X, \nabla_Z X \rangle + \langle \nabla_Z X,\nabla_X Z \rangle - \langle \nabla_X X,\nabla_Z Z \rangle \\ &= \langle \nabla_Z X, \nabla_Z X \rangle - \langle \nabla_X X,\nabla_Z Z \rangle. \end{align} Because of the Killing equation at $p$, $\nabla_X X(p)=0$, and we conclude the assertion.

How can we get from the first expression of $S$ to the second expression of $S$? To answer that question, it suffices to show that $$ \langle \nabla_{[X,Z]}X,Z\rangle-\langle \nabla_X \nabla_Z X, Z \rangle = -\langle \nabla_Z X,\nabla_X Z\rangle + \langle \nabla_Z X, \nabla_Z X \rangle + \langle \nabla_Z X,\nabla_X Z \rangle. $$ So how can we establish this equality?

The only thing I could do so far was that I algebraically rearranged some terms (so that they are all positive) as follows: \begin{align} \langle \nabla_{[X,Z]}X,Z\rangle + \langle \nabla_Z X, \nabla_X Z \rangle &= \langle \nabla_X \nabla_Z X, Z \rangle + \langle \nabla_Z X,\nabla_X Z \rangle + \langle \nabla_Z X,\nabla_Z X \rangle \\ &= X\langle \nabla_Z X,Z \rangle + \langle \nabla_Z X,\nabla_Z X \rangle \end{align} I'm hoping that this is progress.

What can I do now? (Particularly, I have a hard time working with the connection $\nabla_{[X,Z]}X$, with the Lie bracket...)


The Killing equations is

$$ \langle\nabla_BA,C\rangle+\langle B,\nabla_CA\rangle=0 $$

which gives you

$$ \langle\nabla_{[X,Y]}X,Z\rangle+\langle [X,Y],\nabla_ZX\rangle=0 $$

then you use

$$ [X,Y]=\nabla_XY-\nabla_YX $$

  • $\begingroup$ Thanks, I kept trying to implement the Killing equation given in the exercise's hint, but that was actually a special case so that did not help. I should have gone back to the original Killing equation which is what you pointed out implicitly in your answer. $\endgroup$ – New day rising May 21 '17 at 20:18

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