Differential of flow related to gradient Ricci soliton on $S^2$ I'm confused by a claim made in Brendle's Ricci Flow and the Sphere Theorem on page 37. 
Let $(S^2,g)$ be a gradient Ricci soliton with soliton function $f:S^2\to\Bbb R$. Since the sphere is $2$-dimensional, we have 
$$\nabla^2f=(\rho-\tfrac{1}{2}s)g,$$
where $\nabla$ is the LC connection, $s$ is the scalar curvature, and $\rho$ is a constant. Let $J$ be an almost complex structure compatible with $g$, i.e. $(x,y)\mapsto g(x,Jy)$ is a symplectic form, $g(Jx,Jy)=g(x,y)$, and $\nabla J=0$. Letting $\xi$ be the gradient of $f$ wrt. $g$, we have that $J\xi$ is a Killing field and its flow $\varphi_t$ is a one-parameter group of isometries. Let $p\in S^2$ be a critical point of $f$ and let $a=\rho-\tfrac{1}{2}s(p)$, so that 
$$\tag{1}(\nabla^2 f)_p(v,v)=a|v|^2.$$
The author claims this implies $$\tag{2}(\mathrm d\varphi_t)_p(v)=\cos(at)v+\sin(at)Jv,$$
for all $t\in\Bbb R$ and $v\in T_pS^2$, which I am having a very hard time seeing. 
Since $p$ is a critical point $J\xi_p=0$, and $\varphi_t(p)=p$. We also have $g(v,Jv)=\omega(v,v)=0$, i.e. $v\bot Jv$ and are thus linearly independent. So we may write $$(\mathrm d\varphi_t)_pv=\alpha(t)v+\beta(t)Jv,$$
for smooth functions $\alpha,\beta:\Bbb R\to\Bbb R$. Furthermore, since $\varphi_t$ is an isometry, we have 
\begin{align*}g(v,v)&=g((\mathrm d\varphi_t)_p(v),(\mathrm d\varphi_t)_p(v))\\
&=g(\alpha v+\beta Jv,\alpha v+\beta Jv)\\
&=\alpha^2 g(v,v)+2\alpha\beta g(v,Jv)+\beta^2g(Jv,Jv)\\
&=(\alpha^2+\beta^2)g(v,v)\implies \alpha^2(t)+\beta^2(t)=1.\end{align*}
Clearly $\alpha(0)=1$ and $\beta(0)=0$. To get sine and cosine, it seems like one should derive the ODEs $\alpha''=-a^2\alpha$ and $\beta''=-a^2\beta$ with the correct initial values from $(1)$, but I don't know how to do that.
Taking the derivative of $\alpha^2+\beta^2=1$ gives $\alpha\alpha'+\beta\beta'=0$, whence $\alpha'(0)=0$. Let $c(t)=\alpha(t)v+\beta(t)Jv$. Then $t\mapsto g(c(t),c(t))$ is a constant function, so $g(c',c)=0$. This fixes $c'=-k\beta v+k\alpha Jv$, where $k$ is a constant. Thus $
\alpha'=-k\beta$ and $\beta'=k\alpha$. Thus $\beta'(0)=k$ and we get $c''=-k^2c$. Together with the initial values, this gives $c(t)=\cos(kt)v+\sin(kt)Jv$. How does one see that $a=k$?
It's not clear to me from this why $\alpha,\beta$ should be the same for all vectors, either. Assuming $(\mathrm d\varphi_t)_p(w)=\gamma(t)w+\delta(t)Jw$ and computing $g(v,w)$ gives a horrible mess. 
How does one obtain $(2)$?
 A: The main idea of this proof is due to a professor of mine.
Because the differential flows $\{(\mathrm{d}\varphi_t)_p\}_{t\in\Bbb R}$ are isometries of the $2$-dimensional inner product space $(T_pS^2,g_p)$, we have $(\mathrm{d}\varphi_t)_p\in \mathrm{SO}(2)$ for each $t$. Since $T_pS^2$ has an orthogonal basis $\{v,Jv\}$, we can write $(\mathrm{d}\varphi_t)_p(v)=\cos(\alpha)v+\sin(\alpha)Jv$ for each $t$. The group property of flows means that $t\mapsto (\mathrm{d}\varphi_t)_p$ is a (smooth) homomorphism $\Bbb R\to\mathrm{SO}(2)$, and we may write 
$$(\mathrm{d}\varphi_t)_p(v)=\cos(\theta t)v+\sin(\theta t)Jv,$$
for $\theta\in\Bbb R/2\pi$.
Note that $$\nabla_Y\xi^\flat=\nabla_Y\nabla f=(\rho-\tfrac{1}{2}s)Y^\flat\implies \nabla_Y\xi=(\rho-\tfrac{1}{2}s)Y,$$
for any vector field $Y$. Then, at $p$, since $J$ is parallel:
$$\nabla_YJ\xi=aJY.$$
From do Carmo Riemannian Geometry, on page 28, 
$$[X,Y](p)=\lim_{t\to 0}\frac{Y(\varphi_t(p))-(\mathrm{d}\varphi_t)_pY(p)}{t},$$
where $X,Y$ are vector fields and $\varphi_t$ is the flow of $X$. Letting $X=J\xi$ and $Y=$ arbitrary extension of $v\in T_pS^2$, we obtain 
$$[J\xi,Y](p)=-\left.\frac{\mathrm d}{\mathrm dt}\right|_{t=0}(\mathrm d\varphi_t)_p(v)=-\theta Jv.$$
Now, since $J\xi(p)=0$, the zero torsion condition gives 
$$[J\xi,Y](p)=\nabla_{J\xi}Y(p)-\nabla_YJ\xi(p)=-aJv,$$
so that $a=\theta$, as desired. 
