Higher order expansion of hypersurface about a point (beyond second fundamental form/extrinsic curvature) Consider a smooth, compact $(d-1)$-dimensional hypersurface $S$ without boundary embedded in $\mathbb{R}^d$. The surface $S$ can be described as the graph of a function $f(x_1,x_2,\cdots,x_{d-1})$. Using the tangent plane to $S$ at some point $\mathbf{x}$, endowed with an orthonormal basis $\{{\bf e}_1,\cdots,{\bf e}_{d-1}\}$, one can Taylor expand the function $f$ around ${\bf x} = 0$ as
\begin{aligned}\label{graph}
f({\bf x })  =  \frac{1}{2} K_{ij} x^{i} x^{j} + \frac{1}{3!} A_{ijk}x^{i} x^{j} x^k +  \frac{1}{4!} B_{ijkl}x^{i} x^{j} x^k x^l+ O \left(| {\bf x} |^5 \right).
\end{aligned}
The quadratic term is the extrinsic curvature of $S$ as embedded in $\mathbb{R}^d$, also called the second fundamental form. However, the higher order terms in the expansion are less known and I was wondering what the tensors $A$ and $B$ correspond to? I would guess they depend also on the extrinsic curvature (and its derivatives), but what is their exact form? I cannot find anything beyond second order.
Disclaimer: physicist notations
 A: Let me start by summarizing what you're doing: Fix a point $p$ on the hypersurface $S$. Move the hypersurface rigidly so that the point is at the origin and the hypersurface is the graph of a function $f$ that has a critical point at the origin. Then the $k$-th order term of the Taylor series of $f$ at the origin defines a symmetric $k$-tensor on $\mathbb{R}^{d-1}$. If you do this for each point in the hypersurface, you get a symmetric $k$-tensor field along the entire hypersurface.
We can calculate the second and third order tensors at the origin as follows: $\newcommand\R{\mathbb{R}}$
Consider a function $f: O \rightarrow \R$, where $O\subset \R^{d-1}$ is open. Assume $$f(0) = \partial_1f(0) = \dots = \partial_{d-1}f(0) = 0.$$ Let $\phi: O \rightarrow \R^d$ be the graph map
$$
  \phi(x^1, \dots, x^{d-1}) = (x^1, \dots, x^{d-1},f(x^1, \dots, x^{d-1})).
$$
Let $\nu: O \rightarrow S^{d-1} \subset \R^d$ be the Gauss map composed with $\phi$. Observe that for $k \ge 2$, the $k$-order tensor $A$ at $0$ is given by
$$
  A_{i_1\dots i_k}(0) = \nu(0)\cdot \partial^k_{i_1\cdots i_k}\phi(0).
$$
On the other hand, if $k=2$,
\begin{align*}
  \nu\cdot\partial^2_{ij}\phi &= \partial_j(\nu\cdot\partial_i\phi) - \partial_j\nu\cdot\partial_i\phi\\
                              &= -\partial_i\phi\cdot S(\partial_j\phi)\\
                              &= -K_{ij},
\end{align*}
where $S$ is the shape operator and $K$ is the second fundamental form. Evaluating this at $0$, we get $A_{ij} = -K_{ij}$.
Next,
\begin{align*}
  \nu\cdot\partial^3_{ijk}\phi &= \partial_k(\nu\cdot\partial^2_{ij}\phi) - \partial_k\nu\cdot\partial^2_{ij}\phi\\
                               &= -\partial_kK_{ij} - S(\partial_k\phi)\cdot\partial^2_{ij}\phi.
\end{align*}
Evaluating this at $0$,
\begin{align*}
  A_{ijk} &=   \nu\cdot\partial^3_{ijk}\phi\\
  &= -\partial_kK_{ij}\\
  &= -\nabla_kK_{ij}.
\end{align*}
Observe that a consequence of this is the Codazzi-Mainardi equations.
I do not know what happens for $k=4$.
