Metric deformation:
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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 $R$.



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> **Q:** What is the explicit formula for $k'(0)=\frac{d}{dt}k(t)\big|\_{t=0}$?
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Possible (but different?) answers:
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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.



Behavior of sectional curvature under metric deformations
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. 


