17
$\begingroup$

Let $G$ be a finite group acting linearly on a finite dimensional vector space $V$ over a finite field. By Burnside's lemma, $$ |V/G| = \frac 1{|G|} \sum_{g\in G} q^{\dim(ker(g - I))}. $$ Since $g-I$ and its dual map $g^*-I$ have kernels of the same dimension, it follows that $|V/G|=|V^*/G|$.

The above argument shows that a vector space and its dual have the same number of orbits under the action of linear group. Can it happen that the cardinalities of the orbits are different?

$\endgroup$
  • 2
    $\begingroup$ One can also prove that the number of orbits are the same using the Fourier transform. It gives a canonical isomorphism $Fun(V) \to Fun(V^*)$ and hence $Fun_G(V) \to Fun_G(V^*)$. The dimensions of both sides is the number of $G$-orbits. $\endgroup$ – Geordie Williamson Apr 15 '14 at 13:25
21
$\begingroup$

Yes. In particular, it can happen that $V$ has non-zero fixed points, but $V^*$ doesn't.

For example, let $G$ be the symmetric group of degree 3 acting in the obvious way on the set $\{e_1,e_2,e_3\}$, and let $W$ be the corresponding permutation module over the field of 3 elements. Let $V$ be the submodule spanned by $e_1-e_2$ and $e_1-e_3$. Then $V$ has orbits of lengths $6,1,1,1$, but $V^*$ has orbits of lengths $3,3,2,1$.

$\endgroup$

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.