Timeline for Orthogonal Groups over finite fields
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Jul 16, 2012 at 17:34 | comment | added | Dima Pasechnik | Geoff, thanks, I stand corrected. I've removed the wrong comment. I'll dig the stuff on spinor norm up when/if I get to teach graduate group theory :–) | |
| Jul 15, 2012 at 13:18 | comment | added | Geoff Robinson | @M.B. I do not think you are making a mistake. I think the Spinor norm homomorphism has different kernel from the determinant in odd characteristic. | |
| Jul 15, 2012 at 8:57 | history | edited | Dima Pasechnik | CC BY-SA 3.0 |
an example added
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| Jul 14, 2012 at 20:46 | comment | added | Natalie | you should also have a look at Kleidman and Liebeck's book The Subgroup Structure of the Finite Classical Groups". $\Omega$ is the group you are looking at and which is explained there in detail. | |
| Jul 14, 2012 at 18:39 | comment | added | Dima Pasechnik | en.wikipedia.org/wiki/Orthogonal_group#The_Dickson_invariant says "Dickson invariant can be defined as D(f)= rank (I-f) modulo 2, where I is the identity (Taylor 1992, Theorem 11.43). Over fields that are not of characteristic 2 it is equivalent to the determinant: the determinant is −1 to the power of the Dickson invariant." I don't know. At least I am glad that I wrote "(sometimes you might have to take the commutator subgroup)" in my answer. :–) | |
| Jul 14, 2012 at 18:10 | comment | added | M.B | Dima: If the spinor norm is not trivial then $SO^+_{2n}(\mathbb{F}_p)$ is not perfect. I am looking at the O'Meara's book on "Introduction to Quadratic Forms" page 139 lemma 55:2a, which shows that spinor norm is not trivial. Where am I making mistake? | |
| Jul 14, 2012 at 18:01 | comment | added | Dima Pasechnik | according to Wikipedia, Dickson invariant only matters in even characteristic. In odd characteristic it cuts out the same normal subgroup as the determinant. | |
| Jul 14, 2012 at 17:57 | comment | added | Dima Pasechnik | Geoff, are you saying that for $p$ even one actually would have an index 2 subgroup regardless, as $SO^+_{2n}(p)=O^+{2n}(p)=D_n(p)$ in this case, but one still has nontrivial Dickson invariant? Then, oops, my last comment is only true for $p$ odd. As a lame excuse I can only say that my copy of the Atlas is in the office :–) | |
| Jul 14, 2012 at 17:51 | comment | added | M.B | Geoff: Here I am considering the special orthogonal group, then the determinant map is a trivial map. Let me mention that, by $SO^+_{2n}(\mathbb{F}_p)$ I mean, the special orthogonal group of the identity matrix $I_{2n}$. But it is not clear to me that why the spinor norm should be a trivial map. If the spinor norm is not trivial then, indeed this implies that $SO^+_{2n}(\mathbb{F}_p)$ is not perfect. Would you please tell me why the spinor norm is not trivial? | |
| Jul 14, 2012 at 17:36 | comment | added | Geoff Robinson | I find the "standard" notation for orthogonal groups confusing. When $p$ is odd, the usual orthogonal group in even dimension has more than one normal subgroup of index $2$. As well as the kernel of the determinant, there is the kernel of the spinor norm. | |
| Jul 14, 2012 at 16:59 | comment | added | M.B | Let just consider $SO_{2n}^+(\mathbb{F}_p)$, for $n\geq 3$. Is it true now that, this group is perfect? | |
| Jul 14, 2012 at 16:11 | history | answered | Dima Pasechnik | CC BY-SA 3.0 |