Geometric intuition about Jacobian of unit normal vector to surface and orthogonal projections Background:
Let $M=f^{-1}(\{0\})$ be a surface in $\mathbb{R}^3$ defined by an implicit equation for a smooth $f:\mathbb{R}^3\to \mathbb{R}$ with non-zero gradient. We can define the unit normal vector
$$n(x)=\frac{\nabla f(x)}{\|\nabla f(x)\|},$$
everywhere in a neighborhood of $M$ in $\mathbb{R}^3$.
Then we can define the orthogonal projection onto the tangent plane of $M$ at a point $x$ by
$$P(x)=I-n(x)n(x)^T,$$
where matrix-multiplication is being used. So $P^2=P$.
My question:
Let $J_n(x)$ be the Jacobian of the normal vector $n$ at $x$ and let $a(x):=(J_n(x))n(x)$. Is it always true that
$$P(x)a(x)=a(x)?$$
Some thoughts:
I have verified this for some examples, like the sphere (in which case $a(x)=\vec{0}$), and the ellipsoid (in which case $a(x)\neq \vec{0}$). I am not sure how to prove it in general, but it seems there is some good geometric intuition that I am missing here that if someone could comment on, it would be very helpful. I have also noted that, since $P=I-nn^T$, the claim is equivalent to saying
$$(n(x)n(x)^T) a(x) =0,$$
i.e.
$$a(x)\in \ker n(x)n(x)^T,$$
for all $x\in M$. Writing out the matrices and trying to solve the equation in general does not seem feasible, so I think I am missing out on some linear algebra or differential geometry facts. Perhaps it is something like saying $Jn\cdot n$ is in the tangent space of $M$ at $x$ so that $P$ leaves it as it is. Like I said, my geometric intuition is a bit lacking in high-dimensions, so it would be great if someone could clarify this not just analytically but with some geometric flare.
 A: If I understand it correctly, it should be as follows. The mentioned Jacobian is the derivative of the map (vector field)
\begin{align}
n:&\mathbb R^3 \to T\mathbb R^3 \cong \mathbb R^3 & x &\mapsto (\nabla f(x))/\|\nabla f(x)\|
\end{align}
The integral curves (aka filed lines) of the gradient and those of $n$ are the same lines, (both are tangent to these curves), but $n(x)$ are by definition unit tangent, so its derivative along an integral curve $\gamma: \mathbb R \supset I \to \mathbb R^3$ is orthogonal to it, i.e. (the crucial point is)
$$
n(\gamma(t)) \perp \frac{d}{dt}n(\gamma(t)) \equiv n'(\gamma(t)) = J_n(\gamma(t))(\gamma'(t)).
$$
where the chain rule was used. The jacobian is the linear-map-valued field
\begin{align}
J_n:& \mathbb R^3 \to L(\mathbb R^3;\mathbb R^3) & \left[J_n(x)\right]_{ij} = \partial_j n_i(x)  
\end{align}
Now acting the Jacobian at a point $J_n(x)$ on a small vector $v \in \mathbb R^3$ gives us the small change in $n(x)$, i.e $(J_n(x))(v) \approx n(x+v) - n(x)$. If $v = \gamma'(t)dt$, i.e. is a small tangent vector to an integral curve $\gamma$ of $n(x)$, we get that corresponding change $$
n(\gamma'(t)dt +x) - n(x)\approx (J_n(x))(\gamma'(t))dt
$$
which is orthogonal to $n(x)$. So your defined map $a: x \mapsto (J_n(x))(n(x))$ has its images orthogonal to the integral curve $\gamma$, and so orthogonal to $n(x)$, and so its projection is
$$
P(a(x)) = a(x).
$$
This $a$ is the acceleration of the integral curves of $n$, and it is normal to these curves.
A: $
\newcommand\PD[2]{\frac{\partial#1}{\partial#2}}
\newcommand\R{\mathbb R}
$
Edit:
I decided I should maybe explain why $a(x) = (n(x)\cdot\nabla)n(x)$.
Rather than as a matrix, for fixed $x$ we may view the Jacobian of $f : \R^m \to \R^n$ at $x$ as a linear function $Jf(x) : \R^m \to \R^n$ which may be expressed as
$$
  [Jf(x)](y) = (y\cdot\nabla)f(x),
$$
where $\nabla$ acts on $f$ and the resulting derivative is evaluated at $x$; perhaps more clearly
$$
    [Jf(x)](y) = \bigl[(y\cdot\nabla_z)f(z)\bigr]_{z=x}.
$$
To verify this, consider $y = e_i$ of the standard basis:
$$
  [Jf(x)](e_i) = (e_i\cdot\nabla)f(x) = \PD f{x_i}(x),
$$
for $x = \sum_{i=1}^mx_ie_i$, which is exactly the $i^{\mathrm{th}}$ column of the Jacobian matrix.

We may write $a(x) = (n(x)\cdot\nabla)n(x)$. Note that $\nabla$ is only acting on the $n(x)$ on the right. The projection $P(x)a(x) = a(x)$ if and only if $a(x)$ is orthogonal to $n(x)$; so we look at $a(x)\cdot n(x)$, abbreviating $\PD{}r := n(x)\cdot\nabla$ and suppressing $x$ dependence:
$$
  a\cdot n
  = \PD nr\cdot n
  = \frac12\PD{}r(n\cdot n)
  = \frac12\PD{}r1
  = 0,
$$
so $a(x)$ and $n(x)$ are orthogonal.
