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