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I have some question about detail on Chapter 9 Riemannian submersion in the book [Be].

[Be] A. L. Besse, Einstein Manifolds, Springer-Verlag

In the above book, in the proof of 9.24 Proposition (See p. 240) it is described that $[X,U]$ is vertical.

Let me explain details : On $(M = B\times_f F, g=\hat{g} + f\check{g})$ with a
submersion $\pi : M \rightarrow B$, let $U$ be a vertical vector field and $X$ be a horizontal vector field.

The proof used the argument that $[X,U]$ is vertical, which is not proved. I can not understand why $[X,U]$ is vertical.

In pp. 239-240, 9.22 also says that $[X,U]$ is vertical, but no explanation. And I know that $d\pi[X,Y] = [d\pi X, d\pi Y]$. From this can we derive that $[X,U]$ is vertical ?

Please give me an explanation

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I don no have a copy of Besse at hand, but I am pretty sure that X needs to be a horizontal lift of a vectorfield on the base. If not, the commutator is in general not vertical, as you can see for example in the Hopf fibration on $S^3:$ Let $I,J,K$ be the left invariant fields correspondding to $i,j,k\in im \mathbb{H}.$ Then $I$ is the generator of the Hopf fibers, $J,k$ are horizontal, but $[I,J]=2K$ is also horizontal. –  Sebastian Apr 30 '12 at 7:25
    
You are right. In the above, $X$ is a horizontal lift of a vector field $d\pi X =X_0$. Also $d\pi U$ is a 0-vector field on $B$. Thank you. –  Hee Kwon Lee Apr 30 '12 at 7:51
    
After consideration about definition of Lie derivative I have the following: If $f: M \rightarrow N$ is a smooth onto map and $df X$ is a vector field on $N$, then $df [X,Y] = [df X, df Y]$ for any vector field $Y$ on $M$ From this the above question is solved. –  Hee Kwon Lee Apr 30 '12 at 8:01
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1 Answer 1

up vote 2 down vote accepted

Fix arbitrary $\phi:B\to \mathbb R$ and let $f=\phi\circ\pi$. It is sufficient to show that $[X,U]f=0$.

Since $U$ is vertical, $Uf=0$. SInce $X$ is a horizontal lift, $Xf$ is constant on each fiber and therefore $U(Xf)=0$. Hence $$[X,U]f=X(Uf)-U(Xf)=0.$$

(In case if $X$ is only horizontal, the function $Xf$ is arbitrary, in this case $[X,U]f=-U(Xf)\ne 0$.)

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Thank you. Your explanation is helpful for me. –  Hee Kwon Lee May 1 '12 at 0:41
    
You are very welcome :) –  Anton Petrunin May 1 '12 at 3:42
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