This seems to be a straightforward result and it is intuitively true, but I some steps of this elude me. I will summarize it below.
Let $f$ be a smooth function on a Riemannian manifold $(M, g)$ s.t.
$$
\text{Hess}(f)(\nabla f, X) = \frac 12 (\mathcal L_{\nabla f} g)(\nabla f, X) = \lambda g(\nabla f, X)
$$
for any smooth vector field $X$, where $\nabla f$ is the gradient of $f$, characterized by
$$
g(\nabla f, X) = \text d f(X).
$$
For brevity, write $Y = \nabla f$, $b = g(Y, Y)$, and $\theta_Y (X) = g(X, Y)$. Note that, by the definition of the gradient, $\theta_Y(X) = \text d f(X)$, so $\text d \theta_Y = 0$. By Koszul's formula for the Riemannian connection:
\begin{align*}
Xb = X g(Y, Y) &= 2 g(\nabla_X Y, Y) = (\mathcal L_Yg)(X, Y) + \text d \theta_Y(X, Y)
\\
&= 2 \lambda g(X, Y) = 2 \lambda g(X, \nabla f) = 2 \lambda \text d f(X),
\end{align*}
in particular, $Yb = 2\lambda g(Y, Y) = 2\lambda b$. Additionally, we can calculate
\begin{align*}
(\mathcal L_Y \text d f)(X) &= Y (\text d f(X)) – \text d f([Y, X])
\\
&= Y(Xf) – Y(X f) + X(Y f) = X (\text d f(Y))
\\
&= X(g(\nabla f, Y)) = X g(Y, Y),
\end{align*}
so by our previous computation, we see that
$$
\mathcal L_Y \text d f = 2 \lambda \text d f.
$$
By expanding in local coordinates, it is easy to see that $Y b ^{-1} = -b ^{-2} Yb = – 2 \lambda b ^{-1} $. Thus, applying the product rule for the Lie derivative, we arrive at
\begin{align*}
\mathcal L_Y (b^{-1} \text d f \otimes \text d f) &= (Y b^{-1}) \text d f \otimes \text d f + b^{-1} (\mathcal L_Y \text d f) \otimes \text d f + b^{-1} \text d f \otimes (\mathcal L_Y \text d f)
\\
&= 2 \lambda b ^{-1} \text d f \otimes \text d f.
\end{align*}
Now, by assumption,
$$
\mathcal L_Y g = 2 \lambda g,
$$
and the author of the notes I linked above states that we can define a tensor $h$ such that $b(c) h$ is
… the restriction of $g$ to the level set $f = c$. In particular, we have that
$$
g|_p = \left(\frac 1b \text d f \otimes \text d f + b h \right)|_p, \ \text{when } f(p) = c.
$$
so that
\begin{align*}
\mathcal L_{\nabla f} \left(\frac 1b \text d f \otimes \text d f + b h \right) &= 2 \lambda \left(\frac 1b \text d f \otimes \text d f + b h \right)
\end{align*}
implies $g$ and $\frac 1b \text d f \otimes \text d f + b h$ satisfy the same differential equation with equal initial conditions, so they must be equal.
What I am failing to understand is what exactly is $h$, and how do we compute its Lie derivative?
Best Answer
We can prove a more general result:
If $X$ is a conformal Killing vector field with an integrable orthogonal distribution on a Riemannian manifold $(M, g)$, then, around every point where $X \neq 0$ we can find a chart $(V, (y ^i))$ s.t. $$ g(y) = \kappa(y)f_{00}(y') \text{d} y ^{0} \otimes \text{d} y ^0 + \kappa(y)\sum_{i, j = 1} ^{n-1} f_{ij}(y') \text{d} y ^i \otimes \text{d} y ^j $$ and, additionally, $$ X = g(X, X) \nabla y ^0 = \kappa(y) g(y')(X(y'), X(y'))\nabla y ^0. $$ Furthermore, if $\mathcal L_X g = 0$, then $\kappa = 1$.
Proof The conformality of $X$ can be restated as $$ (\phi_t ^* g)_{p} = \kappa_p(t) g_p. $$ Because $\phi_t$ is a diffeomorphism $\phi_{t*}(Y_i(y'))$ remains a frame over $V$. In particular, the coefficients of the metric can be computed as: $$ g(y)(Y_i(y), Y_j(y)) = \kappa(y ^0, y') g(y')(Y_i(y'), Y_j(y')) = \kappa(y)f_{ij}(y'). $$ Additionally, $X(y')$ is orthogonal to $ Y_i(c')$ by virtue of the $Y_i$'s being tangent to the integral manifold of the orthogonal distribution to $X$, so we can see that: $$ g(y) (X(y), Y_i(y)) = 0 \quad i = 1, \dots, n -1. $$ Abbreviate $\Omega(y) = g(y)(X(y), X(y))$ so that the above orthogonality relation yields: \begin{align} g(y)(X(y), a_i(y)Y_i(y)) &= a_0(y)\Omega(y) \\ &= \Omega(y) \sum_{i = 0} ^n a_i(y) \frac{\partial }{\partial y ^{i}} \Big|_{\psi ^{-1}(y)} y ^0 \\ &= \Omega(y)\sum_{i = 0} ^n a_i(y) Y_i(y)y ^0, \end{align} whence $g(X, v) = \Omega(y) v(y ^0) = \Omega(y)\text{d} y ^0(v) = g(\Omega(y)\nabla y ^0, v)$, but by definition of the gradient, this means $$ X = \Omega(y) \nabla y ^0 = \kappa(y) g(y')(X(y'), X(y')). $$ Note that $\kappa_p(t) = e ^{\int_0 t \lambda(\phi(s, p)) \text{d}s} $, and $\phi_t$ sends leaves of the foliation to other leaves, so if $\lambda$ is constant on the foliation, $\kappa_p$ does not depend on $p$. This is how Petersen manages to eliminate the dependence on the first parameter for the first component of the metric.