6
$\begingroup$

While studying the particle interchange symmetry of the Bosons in physics, I have arrived at the notion of $\mathbb{R}^n$ with coordinate interchange symmetry.

That is, I take the quotient of $\mathbb{R}^n$ modulo the permutation group $S_n$. For example, the point $(1,0)$ is identified with the point $(0,1)$ in the case $n=2$.

Upon searching on the Google, this looks like an example of "orbifold". However, I cannot find a particular reference specializing on detailed properties of these objects $\mathbb{R}^n/S^n$.

Could anyone please suggest anything? Especially, I am thinking of the following:

  1. What would be the direct limit of $\{\mathbb{R}^n/S^n\}$ as $n \to \infty$? Since the "symmetry" $S^n$ becomes "larger" as $n \to \infty$, will the direct limit be something "trivial"?

  2. Provided that some direct limit exists, does there exist any notion of convergnce for the collection of canonical projections $\{ \pi_n : \mathbb{R}^n \to \mathbb{R}^n/S^n \}$ as $n \to \infty$?

I am quite curious about the above two issues.

Improve this question
$\endgroup$
6
  • 4
    $\begingroup$ You might be interested in "homological stability of configuration spaces." $\endgroup$
    – Sam Hopkins
    Commented Apr 23, 2023 at 21:33
  • 2
    $\begingroup$ The direct limit should still contain a copy of the real line, no? It seems to me it should be something like the space of finite but arbitrarily long non-increasing sequences, padded with an infinite tail of zeros. $\endgroup$
    – David Roberts ♦
    Commented Apr 23, 2023 at 22:08
  • $\begingroup$ @DavidRoberts could you tell me what you mean by non-increasing? $\endgroup$
    – Isaac
    Commented Apr 23, 2023 at 22:33
  • 3
    $\begingroup$ Your construction $\mathbb{R}^n/S^n$ is an example of a symmetric product and the direct limit is known as the infinite symmetric product. See en.m.wikipedia.org/wiki/Symmetric_product_(topology) for instance. $\endgroup$
    – David.D
    Commented Apr 24, 2023 at 5:30
  • 2
    $\begingroup$ @Isaac $x_1\geq x_2\geq x_3\geq\ldots\geq x_n$ and then zeros forever after that, where here $n$ is not fixed. $\endgroup$
    – David Roberts ♦
    Commented Apr 24, 2023 at 8:05

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .