0
$\begingroup$

Let $(x, y, f(x,y))$ be a surface in $\mathbb{R^3}$. It is written in a book without proof that all rotation invariant (rotating around $z$-axis) of $f$ are combinations of the following four quantities:

  • $\Delta f = f_{xx}+f_{yy}$
  • $||\nabla f||^2=f_x^2 + f_y ^2$
  • $\det{H} = f_{xx} f_{yy} - f_{xy}^2$
  • $(\nabla f, H \nabla f) = f_{xx}f_x^2 + 2f_{xy}f_xf_y+f_{yy}f_y^2$

How do I prove this?

Definition of rotation invariant

My book doesn't provide a precise definition of rotation invariant but I think this is defined as follows:

Let $R[F]={C^{\infty}(\mathbb{R}^2)}[F_{x}, F_{y}, F_{xx}, F_{xy}, F_{yy}, ...]$ be a polynomial ring of infinite variables and $R(F)$ be its field of fractions. Elements in $R(F)$ can be seen as functionals. For a element $H(F)\in R(F)$, we define that $H(F)$ is rotation invariant iff for any rotation $T:\mathbb{R}^2 \rightarrow \mathbb{R}^2$around the origin and for any $f\in C^{\infty}(\mathbb{R}^2)$, $H(f\circ T)=H(f)\circ T$ holds.

Definition of combinations

Let $S(F)$ be a field of fractions of the ring $C^{\infty}(\mathbb{R}^2)[\Delta F, ||\nabla F||, \det{H},(\nabla F, H \nabla F)]$. A combination of the four quantities means an element of $S(F)$.

Improve this question
$\endgroup$
9
  • $\begingroup$ Are we rotating around the line $0=x=y$? $\endgroup$
    – Ben McKay
    Commented Jul 20, 2018 at 9:24
  • $\begingroup$ If $f(x,y)$ is not rotation invariant, and not linear, then surely $\Delta f$ is also not rotation invariant. What do you mean? $\endgroup$
    – Ben McKay
    Commented Jul 20, 2018 at 9:26
  • $\begingroup$ @BenMcKay If $T:(u, v)\mapsto (x,y)$ is a rotation, $(f_{xx}+f_{yy})(T(u,v)) = ((f\circ T)_{uu}+(f\circ T)_{vv})(u,v)$. $\endgroup$
    – marimo
    Commented Jul 20, 2018 at 9:28
  • $\begingroup$ Consider the invariant $I=1$ if $f>0$ everywhere and $I=0$ otherwise. This is a rotation invariant, not given as above. Please make a more precise definition of what you mean by a rotation invariant. Clearly you mean something like a nonlinear differential operator, but what is a combination, and do you require $f$ to belong to some particular function space? $\endgroup$
    – Ben McKay
    Commented Jul 20, 2018 at 9:33
  • $\begingroup$ @BenMcKay: Could the OP be asking for all the rotation-invariant differential operators in $\mathbb R^3$ (excluding the multiplication by functions)? Or maybe only those of order $2$? $\endgroup$
    – Alex M.
    Commented Jul 20, 2018 at 9:41

1 Answer 1

Reset to default
2
$\begingroup$

The result is not correct. Rotations around the $z$-axis preserve the radial vector field $X=\partial_r$, i.e. $X(x,y)=(x,y)$. Such rotations hence preserve $df(X)=xf_x+yf_y$. This is not expressible in terms of the above invariants, as they do not have any first order linear invariants.

Maybe you want to consider the group generated by rotations around all vertical axes, i.e. the group of orientation preserving rigid motions of the plane. Your differential operators are invariant under that group, and that group is probably what the author of the book intended.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .