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I am currently working through the proof of the following equation for the curvature of a Riemannian submersion:

\begin{align} \langle \overline R(\overline X,\overline Y)\overline Z,\overline W \rangle &= \langle R(X,Y)Z,W \rangle - \frac 14 \left\langle [\overline X,\overline Z]^v,[\overline Y,\overline W]^v \right\rangle \\ &\qquad + \frac 14 \left\langle [\overline Y,\overline Z]^v,[\overline X,\overline W]^v \right\rangle - \frac 12 \left\langle [\overline Z,\overline W]^v,[\overline X,\overline Y]^v \right\rangle. \end{align}

The full context of this question, including do Carmo's notation reproduced here, comes from Problems 8.8-8.10 of do Carmo's Riemannian Geometry (PDF file of those problems). But my question focuses only on Problem 8.10(a).

I am trying to fill in the details of do Carmo's hint leading up to the formula. I am reproducing the hint below but I am also attempting to fill in the details of the hint. My questions below center around filling in such details.

Hint: Observe that $\overline X \langle \overline \nabla_{\overline Y}\overline Z,\overline W \rangle = X\langle \nabla_Y Z,W\rangle$. Therefore, \begin{align} \tag{1} \langle \overline \nabla_{\overline X} \overline \nabla_{\overline Y} \overline Z,\overline W\rangle &= \overline X\langle \overline \nabla_{\overline Y}\overline Z,\overline W\rangle - \langle \overline \nabla_{\overline Y}\overline Z,\overline \nabla_{\overline X}\overline W\rangle \\ &= \langle \nabla_X \nabla_Y Z,W\rangle - \frac 14 \langle [\overline Y,\overline Z]^v,[\overline X,\overline W]^v \rangle. \end{align} On the other hand, if $T \in \mathcal X(\overline M)$ is vertical, \begin{align} \tag{2} \langle \overline \nabla_T \overline X,\overline Y \rangle = \langle \overline \nabla_\overline X T,\overline Y \rangle + \langle [T,\overline X],\overline Y\rangle = -\langle T,\overline \nabla_{\overline X} \overline Y \rangle. \end{align} Therefore, \begin{align} \tag{3} \langle \overline \nabla_{[\overline X,\overline Y]} \overline Z,\overline W\rangle &= \langle \overline \nabla_{[\overline X,\overline Y]^h} \overline Z,\overline W \rangle + \langle \overline \nabla_{[\overline X,\overline Y]^v} \overline Z,\overline W \rangle \\ &= \langle \overline \nabla_{[X,Y]} Z,W \rangle - \frac 12 \langle [\overline X,\overline Y]^v,[\overline Z,\overline W]^v \rangle. \end{align} Putting the above together, we obtain the desired equation.

My thoughts so far:

Once the above three identities in the hint are established, I would certainly know how to apply them to the definition of the curvture of $\overline R$ to establish the desired equation.

But I cannot thoroughly understand how those above identities are established.


In $(1)$, I worked from a term on the RHS of $(1)$---and tried to apply the formula from Exercise 8.9(b) (can be viewed in PDF file linked above)---as follows: \begin{align} &\frac 14 \langle [\overline Y,\overline Z]^v,[\overline X,\overline W]^v \rangle \\ &\qquad = \frac 14 \langle 2(\overline \nabla_{\overline Y}\overline Z-\overline{(\nabla_Y Z)}),2(\overline \nabla_{\overline X}\overline W-\overline{(\nabla_X W)})\rangle \\ &\qquad = \langle \overline \nabla_{\overline Y}\overline Z-\overline{(\nabla_Y Z)},\overline \nabla_{\overline X}\overline W-\overline{(\nabla_X W)}\rangle \\ &\qquad = \langle \overline \nabla_{\overline Y}\overline Z,\overline \nabla_{\overline X}\overline W \rangle - \langle \nabla_{\overline Y}\overline Z, \overline{(\nabla_X W)} \rangle - \langle \overline{(\nabla_Y Z)} , \overline \nabla_{\overline X} \overline W \rangle + \langle \overline{(\nabla_Y Z)}, \overline{(\nabla_X W)} \rangle \end{align} but I'm not sure how to proceed (though I see a term on the LHS of $(1)$, which is a good sign).


In $(2)$, the first equality is simple. But it is the second equality of $(2)$ that is giving me problems. I tried working from the RHS again (like I did in $(1)$): \begin{align} -\langle T, \overline \nabla_{\overline X}\overline Y \rangle = -\frac 12 \langle T,[\overline X,\overline Y] \rangle = -\frac 12 \langle T,\overline \nabla_{\overline X}\overline Y \rangle + \frac 12 \langle T,\overline \nabla_{\overline Y} \overline X \rangle, \end{align} which would imply algebraically that $-\langle T, \overline \nabla_{\overline X} \overline Y \rangle = \langle T, \overline \nabla_{\overline Y} \overline X \rangle$, or $\overline \nabla_{\overline Y}\overline X = -\overline \nabla_{\overline X}\overline Y$ if it holds for all $T$. I'm not sure how this would help me get to establishing \begin{align} -\langle T,\overline \nabla_{\overline X} \overline Y \rangle=\langle \overline \nabla_\overline X T,\overline Y \rangle + \langle [T,\overline X],\overline Y\rangle, \end{align} though.


In $(3)$, the first equality is easy because one need only recognize that $[\overline X,\overline Y] = [\overline X,\overline Y]^h+[\overline X,\overline Y]^v$. However, the second equality of $(3)$ is not easy for me to figure out. But I am guessing that I use something similar to the work I did in $(1)$.


This is a long post, so my work can be at times sloppy (though I hope not). Please ask me if there is something that you read that you need me to clarify better.

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A direct computation shows for $(1)$: \begin{align*} &\left\langle\overline\nabla_{\overline{X}}\overline\nabla_{\overline{Y}}\overline{Z},\overline{W}\right\rangle \\ &\qquad = \overline{X}\left\langle\overline{\nabla}_{\overline{Y}}\overline{Z} ,\overline{W}\right\rangle - \left\langle\overline\nabla_{\overline{Y}}\overline{Z},\overline\nabla_{\overline{X}}\overline{W}\right\rangle \quad \text{metric compatibility} \\ &\qquad = X\left\langle\nabla_{{Y}}{Z} ,{W}\right\rangle - \left\langle\overline\nabla_{\overline{Y}}\overline{Z},\overline\nabla_{\overline{X}}\overline{W}\right\rangle \quad \text{8.9(c)} \\ &\qquad = X\left\langle\nabla_{{Y}}{Z} ,{W}\right\rangle - \left\langle\overline{\nabla_YZ} + \frac{1}{2}[\overline{Y},\overline{Z}]^v,\overline{\nabla_XW} + \frac{1}{2}[\overline{X},\overline{W}]^v\right\rangle \quad \text{8.9(b)}\\ &\qquad = X\left\langle\nabla_{{Y}}{Z} ,{W}\right\rangle - \left\langle\overline{\nabla_YZ},\overline{\nabla_XW}\right\rangle - \left\langle\overline{\nabla_YZ}, \frac{1}{2}[\overline{X},\overline{W}]^v\right\rangle \\ &\qquad -\left\langle\frac{1}{2}[\overline{Y},\overline{Z}]^v,\overline{\nabla_XW} \right\rangle - \frac{1}{4}\left\langle[\overline{Y},\overline{Z}]^v,[\overline{X},\overline{W}]^v\right\rangle \quad \text{bilinearity of $\langle\cdot,\cdot\rangle$} \\ &\qquad = X\left\langle\nabla_{{Y}}{Z} ,{W}\right\rangle - \left\langle\overline{\nabla_YZ},\overline{\nabla_XW}\right\rangle - \frac{1}{4}\left\langle[\overline{Y},\overline{Z}]^v,[\overline{X},\overline{W}]^v\right\rangle \quad \text{$\langle\overline{X},T\rangle=0$ in 8.9(c)} \\ &\qquad = X\left\langle\nabla_{{Y}}{Z} ,{W}\right\rangle - \left\langle\overline{\nabla_YZ},\overline{\nabla_XW}\right\rangle - \frac{1}{4}\left\langle[\overline{Y},\overline{Z}]^v,[\overline{X},\overline{W}]^v\right\rangle \\ &\qquad = X\left\langle\nabla_{{Y}}{Z} ,{W}\right\rangle - \left\langle{\nabla_YZ},{\nabla_XW}\right\rangle - \frac{1}{4}\left\langle[\overline{Y},\overline{Z}]^v,[\overline{X},\overline{W}]^v\right\rangle \quad \text{definition of horizontal lift}\\ &\qquad = \langle{\nabla_X\nabla_YZ,W}\rangle - \frac{1}{4}\left\langle[\overline{Y},\overline{Z}]^v,[\overline{X},\overline{W}]^v\right\rangle \quad \text{metric compatibility} \end{align*}

The only confusing thing in the above is that there are two different Riemannian metrics in play, and it would have been more clear to explitictly refer to them as $g$ and $\overline{g}.$

Now for $(2)$: \begin{align*} \left\langle\overline{\nabla}_T\overline{X},\overline{Y}\right\rangle& =\left\langle\overline{\nabla}_{\overline{X}}T,\overline{Y}\right\rangle + \left\langle[T,\overline{X}],\overline{Y}\right\rangle \quad \text{torsion free}\\ & = -\left\langle T,\overline{\nabla}_{\overline{X}}\overline{Y} \right\rangle + \overline{X}\left\langle T,\overline{Y} \right\rangle + \left\langle[T,\overline{X}],\overline{Y}\right\rangle \quad \text{metric compatibility}\\ & = -\left\langle T,\overline{\nabla}_{\overline{X}}\overline{Y} \right\rangle + \left\langle[T,\overline{X}],\overline{Y}\right\rangle \\ & = -\left\langle T,\overline{\nabla}_{\overline{X}}\overline{Y} \right\rangle \quad \text{8.9(c)} \end{align*}

Finally, for $(3)$: \begin{align*} \left\langle{\overline\nabla_{[\overline{X},\overline{Y}]}\overline{Z},\overline{W}}\right\rangle & = \left\langle\overline\nabla_{[\overline{X},\overline{Y}]^h}\overline{Z},\overline{W}\right\rangle + \left\langle\overline\nabla_{[\overline{X},\overline{Y}]^v}\overline{Z},\overline{W}\right\rangle \\ & = \left\langle\overline\nabla_{[\overline{X},\overline{Y}]^h}\overline{Z},\overline{W}\right\rangle - \left\langle[\overline{X},\overline{Y}]^v,\overline{\nabla}_{\overline{Z}}\overline{W}\right\rangle \quad \text{by (2)}\\ & = \left\langle\overline\nabla_{[\overline{X},\overline{Y}]^h}\overline{Z},\overline{W}\right\rangle - \left\langle[\overline{X},\overline{Y}]^v,\overline{\nabla_ZW}+\frac{1}{2}[\overline{Z},\overline{W}]^v\right\rangle \\ & = \left\langle\overline\nabla_{[\overline{X},\overline{Y}]^h}\overline{Z},\overline{W}\right\rangle - \frac{1}{2}\left\langle[\overline{X},\overline{Y}]^v,[\overline{Z},\overline{W}]^v\right\rangle \end{align*} Now note that $[\overline{X},\overline{Y}]-\overline{[X,Y]}$ is vertical, and so $([\overline{X},\overline{Y}]-\overline{[X,Y]})^h=0$. In particular, we see that $[\overline{X},\overline{Y}]^h=\overline{[X,Y]}$. We finish the computation for $(3)$: \begin{align*} \left\langle\overline\nabla_{[\overline{X},\overline{Y}]^h}\overline{Z},\overline{W}\right\rangle - \frac{1}{2}\left\langle[\overline{X},\overline{Y}]^v,[\overline{Z},\overline{W}]^v\right\rangle & = \left\langle\overline\nabla_{\overline{[X,Y]}}\overline{Z},\overline{W}\right\rangle - \frac{1}{2}\left\langle[\overline{X},\overline{Y}]^v,[\overline{Z},\overline{W}]^v\right\rangle \\ & = \left\langle\nabla_{{[X,Y]}}{Z},{W}\right\rangle - \frac{1}{2}\left\langle[\overline{X},\overline{Y}]^v,[\overline{Z},\overline{W}]^v\right\rangle. \quad \text{8.9(c)} \end{align*} Note that this final formula in $(3)$ has $\nabla$ instead of $\overline\nabla$, but I think that is how it should be since you can now derive the desired curvature relationship.

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  • $\begingroup$ In your work of $(1)$, can I ask how does the definition of horizontal lift $\overline X$---the horizontal field defined by $df_{\overline p} (\overline X(\overline p))=X(f(p))$---imply that $$\langle \overline{(\nabla_Y Z)}, \overline{(\nabla_X W)} \rangle=\langle \nabla_Y Z, \nabla_X W \rangle$$ $\endgroup$ – New day rising Jun 25 '17 at 4:03
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    $\begingroup$ All that I'm using is the following: Consider $X,Y\in\mathfrak{X}(M)$. Then $\bar{g}(\bar X,\bar Y)=g(X,Y)\circ f$, where $f:\bar M\to M$ is the Riemannian submersion. To see that this holds note that since $f$ is a Riemannian submersion that $df$ preserves the length of vectors orthogonal to the fibers. So in particular, $g(X,Y)\circ f = g(df\bar X,df\bar Y)=\bar{g}(\bar X,\bar Y)$. $\endgroup$ – yousuf soliman Jun 25 '17 at 4:37
  • $\begingroup$ The thing that generally confuses me about this notation is that there are plenty of implicit assumptions regarding which inner product is being used. In particular, we have functions living on different manifolds and we don't explicitly say whether we are postcomposing with the Riemannian submersion or not. I hope this post at least helps a little! $\endgroup$ – yousuf soliman Jun 25 '17 at 4:39
  • $\begingroup$ I have now a question with $(2)$, the last step. Why is $$\langle [T,\overline X],Y \rangle = 0$$ for all $X,Y \in \mathcal X(M)$ and $T \in \mathcal X(\overline M)$? You cited Exercise 8.9(c), but all that asserts is that $[\overline X,\overline Y]^v(\overline p)$ depends only on $\overline X(\overline p)$ and $\overline Y(\overline p)$. I currently do not see why that statement is helpful. I thought I should use instead identities like $\langle \overline X,T\rangle=0$ or $T \langle \overline X,\overline Y \rangle = 0$, along with torsion-free and compatibility properties of the metric. $\endgroup$ – New day rising Jun 25 '17 at 7:27
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    $\begingroup$ My bad with the citations... I made the mistake of citing 8.9(c) incorrectly multiple times. I meant to cite the hint for 8.9(b), since I assumed you had worked through that already. The last step in (2) follows since $T$ is vertical and $\bar X$ and $\bar Y$ are horizontal. (Note this doesn't hold for all $T\in\mathfrak X(\bar M)$! Just the vertical vector fields.) $\endgroup$ – yousuf soliman Jun 25 '17 at 7:32

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