2
$\begingroup$

Looking for nicer ways to work this out than having to check all permutations.

If we have the polynomial:

$p=x_1^2+x_2^2+x_3^2+x_4$

Then In order for $S_4$ to stabilize it it must leave $x_4$ uuntouched and we can move around the other variables so the stabilizer is just the number of permutations which leave $x_4$ which is quite straightforward. Is there a similar approach with the polynomial

$x_1x_2^2x_3^3+x_3x_4^2x_1^3+x_2x_3^2x_4^3+x_4x_1^2x_2^3$

When I say similar I just mean what's the smart way to solve this :-)

$\endgroup$
  • 1
    $\begingroup$ Write variables to the 0th power too, and sort them. The answer should be clear then. $\endgroup$ – Jack Schmidt Jun 21 '12 at 17:33
1
$\begingroup$

Let $f=x_1x_2^2x_3^3x_4^0+x_4x_1^2x_2^3x_3^0+x_3x_4^2x_1^3x_2^0+x_2x_3^2x_4^3x_1^0$ be the given symmetric polynomial i.e. it is invariant under the cyclic permutation $\sigma=(1\; 2\; 3\; 4)$ so is under the powers of $\sigma$ which has order $4.$ Now, $S_4$ is generated by $\tau=(1 \; 2)$ and $\sigma,$ but $f$ is not invariant under the action of $\tau$ i.e. $\tau f \neq f$ therefore, is not invariant under the elements of the form $\tau \sigma^i$ and $\sigma^i \tau.$ Since the stabilizer of $f$ in $S_4$ is the subgroup of $S_4,$ it must contain $\{1, \sigma, \sigma^2, \sigma^3\}$ and thus nothing more.

$\endgroup$
  • $\begingroup$ The $\tau$ argument isn't completely convincing. There is a subgroup strictly between $\langle \sigma\rangle$ and $\langle \sigma,\tau\rangle$, and I don't think you mention why it cannot be this group. You have the right answer, and I still think it is clear. (if anyone cares, it's a holomorph!) $\endgroup$ – Jack Schmidt Jun 21 '12 at 17:58
  • $\begingroup$ Dear @Jack Schmidt: but I mentioned that $\tau \sigma^i f \neq f$ and $\sigma^i \tau f\neq f$ so the stabilizer cannot contain an element involving $\tau.$ Isn't it enough? $\endgroup$ – Ehsan M. Kermani Jun 21 '12 at 18:06
  • 1
    $\begingroup$ Oddly, no. $\langle \tau \sigma^2 \tau, \sigma \rangle$ is a proper subgroup, and you didn't check whether $\tau \sigma^2 \tau f = f$ (or equivalently, that $\sigma^2 \tau f = \tau f$). $\endgroup$ – Jack Schmidt Jun 21 '12 at 18:12
  • $\begingroup$ indeed you're right! this is the only thing that must be examined by hand which I'm quite reluctant to do so. $\endgroup$ – Ehsan M. Kermani Jun 21 '12 at 21:29
  • $\begingroup$ Thanks I follow most of this. But how is it possible to know $S_4$ is generated by the above permutations? $\endgroup$ – Ben Davidson Jun 22 '12 at 13:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.