6
$\begingroup$

I'm trying to understand Lie groups method for solving differential equations and I have some basic questions.

For instance, Let $\Sigma = 0$ be a differential equation of order $n$ and let $X^{(n)}$ be the n-th prolongation of the vector field $X$. Then $e^{\epsilon X}$ is a symmetry if $X^{(n)} \Sigma = 0$ whenever $\Sigma = 0$. These two simultaneous problems gives rise to determining equations which are usually solved by means of a polynomial ansatz.

So here are my questions:

  1. How can one know the degree of the polynomial one should use?

  2. Does one always find more symmetry generators by increasing the degree of the polynomial? Or maybe there is a maximum bound?

  3. All symmetry generators found in this way necessarily close a Lie algebra?

Thanks very much in advance for any insights!

$\endgroup$

1 Answer 1

3
$\begingroup$

All of the questions are answered immediately by looking at the symmetries of the Cauchy--Riemann equations in complex analysis of one complex variable, or the Laplace equation on the 3-sphere.

  1. You can't know in advance what degrees to try: all complex polynomial vector fields prolong to preserve the Cauchy--Riemann equations.
  2. You do not always get more symmetry generators by increasing the degree. Typically, most differential equations one encounters have only a finite dimensional symmetry Lie algebra. The Laplace equation on the 3-sphere has symmetry group the conformal group, a finite dimensional Lie group, and the symmetry vector fields are polynomial in the standard Ptolemaic projection coordinates.
  3. The Lie bracket of any two symmetry vector fields is also a symmetry vector field, but might not be one you have already encountered, because it might be of a higher degree of polynomial than you have checked for. For example, with the Cauchy--Riemann equations, two generic complex polynomial vector fields in one complex variable, of degree 3 or more have higher degree bracket. Of course, there is no guarantee that the symmetries must be polynomial, or even that there must be a choice of coordinates in which all symmetry vector fields are polynomial; but Lie bracket of polynomial vector fields yields a polynomial vector field. If the symmetry Lie algebra (the Lie algebra of the symmetry pseudogroup, not necessarily of the global symmetry group) is infinite dimensional (as, for example, with the Cauchy--Riemann equations on a complex manifold), then you will never close up to find a Lie algebra, unless you happen by accident to have all of your symmetry vector fields land inside some finite dimensional subalgebra. For example, you might find the subalgebra of polynomial vector fields of degree at most 2 inside the Lie algebra of holomorphic vector fields, and this might incorrectly lead you to believe that you have found all of the symmetries.
$\endgroup$
2
  • $\begingroup$ By "Ptolemaic projection coordinates", do you mean the stereographic projection? $\endgroup$ Commented Sep 24, 2018 at 8:27
  • 1
    $\begingroup$ @GiuseppeNegro: yes. $\endgroup$
    – Ben McKay
    Commented Sep 24, 2018 at 10:26

You must log in to answer this question.

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