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Metric deformation:

Let $(M,g_0)$ be a Riemannian manifold and consider a (sufficiently smooth) deformation of $g_0$, $$g_t=g_0+th+O(t^2), \quad 0< t<\varepsilon $$ where $h$ is some symmetric (0,2)-tensor. A natural (and important) question is how the sectional curvatures of $g_0$ change under this deformation, e.g., what is the infinitesimal change in terms of $h$. More precisely, given two $g_0$-orthonormal vectors $X$ and $Y$ in $T_pM$, define the (unnormalized) sectional curvature $$k(t)=g_t(R_t(X,Y)Y,X),$$ where we are using the appropriate sign convention on $R$.

Q: What is the explicit formula for $k'(0)=\frac{d}{dt}k(t)\big|_{t=0}$?


Possible (but different?) answers:

I have found a few papers with an answer, but (understandably) none provide the complete argument. Unfortunately, it seems like some of them are really different, and it would be very helpful if someone could point out if they coincide for some (possibly silly) reason I am not seeing.

  1. Berger'66 (Trois remarques sur les variétés riemanniennes à courbure positive)/Bourguignon, Deschamps, Sentenac'72 (Conjecture de H. Hopf sur les produits de variétés): $$k'(0)=\nabla_X\nabla_Y h(X,Y)-\tfrac12\nabla_X\nabla_X h(Y,Y)-\tfrac12\nabla_Y\nabla_Y h(X,X)$$

  2. Strake'87 (Curvature increasing metric variations): $$k'(0)=\nabla_X\nabla_Y h(X,Y)-\tfrac12\nabla_X\nabla_X h(Y,Y)-\tfrac12\nabla_Y\nabla_Y h(X,X)+h(R_0(X,Y)Y,X)$$

$$-k(0)(h(X,X)+h(Y,Y))$$

3. Topping'06 (Lectures on Ricci Flow): $$k'(0)=\nabla_X\nabla_Y h(X,Y)-\tfrac12\nabla_X\nabla_X h(Y,Y)-\tfrac12\nabla_Y\nabla_Y h(X,X)+\tfrac12h(R_0(X,Y)Y,X)$$

$$-\tfrac{1}{2}h(R_0(X,Y)X,Y)$$

EDIT: I should point out that, although formula 3) as I wrote above is NOT a correct expression for $k'(0)$, I (embarrassingly) misinterpreted it from Prop. 2.3.5 in Topping's lecture notes -- which actually contains the correct formula (matching Vitali's answer below). He gives a general expression for $\frac{d}{dt}g_t(R^t(X,Y)Z,W)\big|_{t=0}$, and by using the Ricci identity it becomes clear that his formula is indeed the same as Vitali's. I sincerely apologize for the confusion.

Note that all answers above coincide if $(M,g_0)$ has non-negative sectional curvature and $X$ and $Y$ span a plane of zero $g_0$-curvature. As Strake remarks, the self-adjoint endomorphism $A_X Z=R(Z,X)X$ is positive-semidefinite and $g(A_X Y,Y)=0$. (Nevertheless, to the best of my understanding, THESE HYPOTHESES ARE NOT ASSUMED in the references in 1). Also, it seems to me that answers 2 and 3 DO NOT COINCIDE.

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Check what happens with the vectors $X$ and $Y$ in these papers. One can keep them fixed, but there is a smarter way which produce a one parameter of vectors $X_t$ and $Y_t$ so that $g_t(X_t,Y_t)=\mathrm{const}$. (So all the answers might be correct.) –  Anton Petrunin Jan 15 '12 at 1:40

2 Answers 2

up vote 11 down vote accepted

Formula 2) is the correct one in general except it's the derivative of the sectional curvature i.e of $\frac{k_t(X,Y)}{|X\wedge Y|^2_t}$ (and not just of $k_t(X,Y)$) for an orthonormal frame $X,Y$ with respect to the original metric. This accounts for the last term in formula 2. For $k'(0)$ itself the correct formula is $$ k'(0)=\nabla_X\nabla_Y h(X,Y)-\tfrac12\nabla_X\nabla_X h(Y,Y)-\tfrac12\nabla_Y\nabla_Y h(X,X)+h(R(X,Y)Y,X) $$

As Deane said, this is a straightforward calculation. But since the OP seems to be struggling with it I'll supply some details. Let $\nabla^t$ be the Levi-Civita connection for $g_t$. Then $\nabla^t_XY=\nabla_XY+tS(X,Y)+O(t^2)$ where $S$ is a (2,1)-tensor and $\nabla=\nabla^0$. Let's first compute $k'(0)$ in terms of $S$. As usual let's work in normal coordinates around $p$ with $X,Y$ coordinate fields.

We have $$\langle R^t(X,Y)Y,X\rangle_t=\langle R^t(X,Y)Y,X\rangle_0+t\cdot h(R(X,Y)Y,X)+O(t^2)=$$

$$=\langle \nabla^t_X\nabla^t_YY,X\rangle_0-\langle \nabla^t_Y\nabla^t_XY,X\rangle_0+t\cdot h(R(X,Y)Y,X)+O(t^2)$$ Next we expand the first term

$$\langle \nabla^t_X\nabla^t_YY,X\rangle_0=\langle \nabla^t_X(\nabla_YY+tS(Y,Y)),X\rangle_0+O(t^2)=$$

$$\langle \nabla_X\nabla_YY,X\rangle_0+t\langle S(X,\nabla_YY)+\nabla_XS(Y,Y), X\rangle_0+O(t^2)=$$

$$ \langle \nabla_X\nabla_YY,X\rangle_0+t\langle \nabla_XS(Y,Y), X\rangle_0+O(t^2)$$

where in the last equality we used that $\nabla_YY(p)=0$. After a similar computation for $\langle \nabla^t_Y\nabla^t_XY,X\rangle_0$ we get that

$$k'(0)=\langle\nabla_XS(Y,Y)-\nabla_YS(X,Y), X\rangle_0+h(R(X,Y)Y,X)$$

$$=\nabla_X S(Y,Y,X)-\nabla_YS(X,Y,X)+h(R(X,Y)Y,X)$$

where we lowered the index and turned $S$ into a $(3,0)$-tensor $S(X,Y,Z)=\langle S(X,Y),Z\rangle_0$

Lastly, recall that

$\langle \nabla_XY,Z\rangle=\frac{1}{2}[X\langle Y,Z\rangle+Y\langle X,Z\rangle -Z\langle X, Y\rangle]$ for coordinate fileds. This easily gives

$S(Y,Y,X)=\frac{1}{2}[Yh(X,Y)+Yh(X,Y)-Xh(Y,Y)]=\nabla_Yh(X,Y)-\frac{1}{2}\nabla_Xh(Y,Y)$ and $S(X,Y,X)=\frac{1}{2}\nabla_Yh(X,X)$ written invariantly as tensors.

Altogether this gives $$ k'(0)=\nabla_X\nabla_Y h(X,Y)-\tfrac12\nabla_X\nabla_X h(Y,Y)-\tfrac12\nabla_Y\nabla_Y h(X,X)+h(R(X,Y)Y,X) $$ as promised.

Formulas 1) and 3) are not true in general but might be true in the specific circumstances where they are applied in the papers in question.

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@Vitali: Thank you for your time and patience to explain so many details. This morning, after finally obtaining the first 3 terms that are common to all three answers, I figured out that the last two terms in (2) where coming from the normalization, as you mentioned. But for some reason I was still missing the last term, $h(R^0(X,Y)Y,X)$, in the formula for $k'(0)$, that now is very clear. I think I was using a poor notation for $\langle,\rangle_t$ and $\langle,\rangle_0$ that eventually got both mixed up... Thank you again for confirming and clarifying everything. –  Renato G Bettiol Jan 15 '12 at 22:25
    
I realized that I made a mistake when interpreting Topping's formula (consequently when writing the expression 3) in my original post). Topping's original formula (Prop 2.3.5) is perfectly correct, one just needs to use the Ricci identity to get it to look exactly like Vitali's expression for $k'(0)$. I edited my post to mention this fact, but I'm not sure whether formula 3) should also be updated to be consistent with what actually is in Topping's notes. Regarding expression 1), it seems like the authors actually forgot to mention they are considering $X$ and $Y$ so that $h(R_0(X,Y)Y,X)=0$.. –  Renato G Bettiol Jan 16 '12 at 5:05
    
@Renato: Glad you have worked this out and there is no mistake in Topping's paper. BTW, you might want to work out the general formula for the derivative of the curvature that he gives. It is easily obtained using the same method I used above. –  Vitali Kapovitch Jan 16 '12 at 14:41
    
@Vitali: If I may, there is one final detail that I just noticed that is not so clear to me. When you use the Koszul formula for $g_t$ to obtain an expression for $S(X,Y,Z)=g_0(S(X,Y),Z)$, shouldn't there also be a term $h(\nabla_X Y,Z)$? More precisely, I get $S(X,Y,Z)=\tfrac12(Xh(Y,Z)+Yh(X,Z)-Zh(X,Y))-h(\nabla_X Y,Z)$, and it seems your formula doesn't have this last term. For the computation of $S(Y,Y,X)$ this doesn't matter since $\nabla_Y Y=0$ at $p$, but I don't see why this extra term $-h(\nabla_X Y,X)$ does not appear on $S(X,Y,X)$. Sorry to bother again with something possibly silly.. –  Renato G Bettiol Jan 17 '12 at 17:02
    
@Rentao: the last term is not there because we are computing at $p$ in normal coordinates so that $\nabla_XY(p)=0$. –  Vitali Kapovitch Jan 17 '12 at 20:29

I guess I say this a lot, but this is something you really should work out yourself. You might get lost in the calculations the first few or many times you do it, but after a while you should get the hang of it.

It's probably best to do it first in local co-ordinates. It's relatively straightforward to compute the variation of the Christoffel symbol and the Riemann curvature tensor written with respect to the co-ordinates. It is therefore relatively easy to compute the first variation of $R(X,Y)Y\cdot X$, where $X$ and $Y$ are assumed to be fixed, independent of the variation.

Finally, it is important to note that even if you assume $X$ and $Y$ are orthonormal for the original metric, they do not remain so under the variation. So you have to normalize $R_t(X,Y)Y\cdot X$ to get the sectional curvature. So you have to compute the variation of the normalization, too.

If you do this enough times, you will know which of the formulas above are right. That's a lot better than taking anyone else's word for it. And you will develop a much greater facility for doing such calculations.

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Very well said, Deanne. –  Spiro Karigiannis Jan 15 '12 at 0:03
    
@Deane: I agree with you, and indeed I've been trying to compute it myself (but invariantly, no coordinates, since it gets messy with too many indices). I still did not manage to conclude, but hopefully I will. The purpose of posting the question was mainly that some specialists might have already noticed the differences in those formulas in the literature; perhaps it is a well-known typo in one of the papers, the kind of thing that "everybody in the area knows" (but not me, given that I'm fairly new to it). It seems like this is quite common..I apologize if my post conveys a different message –  Renato G Bettiol Jan 15 '12 at 0:28
    
I recommend trying more than one approach to the calculation. But it is difficult to do it "invariantly" the first time, because too much is hidden or implied by the notation. Do it in co-ordinates first (and get used to working with all those indices) and then use that to help figure out how to do the invariant calculation. The key point is to make sure you really are differentiating every instance of the metric and connection in your formula. –  Deane Yang Jan 15 '12 at 10:36
    
I also advise reading the three papers more carefully. It appears to me that all three formulas are probably what the respective authors intended. Given what I wrote, it's rather easy to identify which of the three is the variation of sectional curvature of the 2-plane spanned by a fixed pair of vectors $X$ and $Y$. The other two differ only slightly and are variations of two different forms of the Riemann curvature tensor evaluated using fixed vectors $X$ and $Y$. –  Deane Yang Jan 15 '12 at 14:13

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