# Projective-invariant differential operator

This question has been cross-posted to MathOverflow.

Suppose we want a differential operator $$T$$ acting on functions $$\mathbb{R}^n \rightarrow \mathbb{R}^n$$ such that

\begin{align*} &T(g) = 0 \Longleftrightarrow g \in G \\ &g \in G \Longrightarrow T(g \circ f) = T(f) \end{align*}

where $$G = \text{Aff}(n, \mathbb{R})$$ is the affine group. Consider the operator

$$T(f) = (\nabla f)^{-1} \cdot \nabla \nabla f$$

where $$\nabla f$$ is the gradient of $$f$$ and $$\nabla \nabla f$$ is its Hessian. This seems to satisfy the criteria since

$$\nabla \nabla f = 0 \Longleftrightarrow f(x) = A \cdot x + b$$

and

\begin{align*} T(A \cdot f + b) &= (\nabla (A \cdot f + b))^{-1} \cdot \nabla \nabla (A \cdot f + b) \\ &= (\nabla A \cdot f)^{-1} \cdot \nabla \nabla A \cdot f \\ &= (A \cdot \nabla f)^{-1} \cdot \nabla A \cdot \nabla f \\ &= (\nabla f)^{-1} \cdot A^{-1} \cdot A \cdot \nabla \nabla f \\ &= (\nabla f)^{-1} \cdot \nabla \nabla f \\ &= T(f) \end{align*}

My question is this: Is there a similar operator that is invariant under the projective group $$G = \text{PGL}(n, \mathbb{R})$$? For $$G = \text{PGL}(1,\mathbb{R})$$, an example is the Schwarzian derivative

$$S(f) = \frac{f'''}{f'} - \frac{3}{2} \left(\frac{f''}{f'}\right)^2$$

Projective differential geometry old and new by Ovsienko and Tabachnikov states in chapter 1.3 page 10 that $$S(g) = 0$$ iff $$g$$ is a projective transformation and $$S(g \circ f) = S(f)$$ if $$g$$ is a projective transformation. They also give a multidimensional generalization of the Schwarzian derivative in equation 7.1.6 page 191:

$$L(f)_{ij}^k = \sum_\ell \frac{\partial^2 f^\ell}{\partial x^i \partial x^j} \frac{\partial x^k}{\partial f^\ell} - \frac{1}{n+1} \left(\delta_j^k \frac{\partial}{\partial x^i} + \delta_i^k \frac{\partial}{\partial x^j}\right) \log J_f$$

where $$J_f = \det \frac{\partial f^i}{\partial x^j}$$ is the Jacobian. However, Schwarps by Pizarro et al. states in section 3.3 page 97 that this "cannot be used to ensure infinitesimally homographic warps as it also vanishes for other functions than homographies" (are there examples?). Instead, they give a system of 2D Schwarzian equations that "vanish if and only if the warp is a homography" (page 94). These are given in section 4.2 equation 29 page 98:

\begin{align*} S_1[\eta] &= \eta^x_{uu} \eta^y_u - \eta^y_{uu} \eta^x_u \\ S_2[\eta] &= \eta^x_{vv} \eta^y_v - \eta^y_{vv} \eta^x_v \\ S_3[\eta] &= (\eta^x_{uu} \eta^y_v - \eta^y_{uu} \eta^x_v) + 2(\eta^x_{uv} \eta^y_u - \eta^y_{uv} \eta^x_u) \\ S_4[\eta] &= (\eta^x_{vv} \eta^y_u - \eta^y_{vv} \eta^x_u) + 2(\eta^x_{uv} \eta^y_v - \eta^y_{uv} \eta^x_v) \end{align*}

What is the geometric intuition behind these equations? Can they be stated more compactly/concisely? Also, how can we normalize them so that $$S_i[\eta]$$ is actually invariant under projective transformations of $$\eta$$?

Finally, is there a compact expression for the $$n$$-dimensional generalization of this derivative?

• In your first equations, you state that $g\in G\Longrightarrow T(g\circ f)=T(f)$. May I ask what this $f$ stands for, an element in $G$ or an arbitrary $\mathbb{R}^n$-valued function? Thanks. – hypernova Jan 8 at 23:53
• In that expression, $f$ is an arbitrary function $\mathbb{R}^n \rightarrow \mathbb{R}^n$. – user76284 Jan 9 at 0:04
• Thank you for your clarification. – hypernova Jan 9 at 0:09