Timeline for Representation theory of p-groups in particular upper tringular matrices over F_p
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Sep 10, 2012 at 7:16 | comment | added | Alexander Chervov | Okay. Thank you very much ! Might be interesting: groupprops.subwiki.org/wiki/… | |
| Sep 9, 2012 at 19:24 | comment | added | Frieder Ladisch | Yes, in this special family of groups every irrep of dim $q$ comes from $G/Z(G)$, that is, $Z(G)$ is contained in the kernel. I already mentioned this in my second comment to this answer: First, every irred. char. of a $p$-group is induced from a linear char. If $\chi\in \operatorname{Irr}(G)$ has $\chi(1)=q$, then $\chi=\lambda^G$ and $\lambda\in \operatorname{Lin}(N)$, where $|G:N|=\chi(1)=q$. So $|N|=q^5$ and $N$ is non-abelian and one can show that $Z(G)\leq [N,N]$, so $Z(G)\leq \ker(\lambda)$. But then also $Z(G)\leq \ker(\chi)$. | |
| Sep 9, 2012 at 13:31 | comment | added | Alexander Chervov | Do you rely on the fact that all q-dim irreps of G comes from G/Z(G) ? If yes how to justify it? If not I do not quite understand: how do you get " one can compute the number of irreps of dim q " ? | |
| Sep 8, 2012 at 23:09 | comment | added | Frieder Ladisch | It is a general fact that $\chi(1)^2\leq |G:Z(G)|$ for any irred. character $\chi$ of a group $G$ (see Isaacs' book on character theory, Corollary 2.30). I apply this to $G/Z(G)$. Since $|(G/Z(G)):Z(G/Z(G))|=q^3$, irreps of $G/Z(G)$ have dim at most $q$. Since $|G/Z(G)|=$ sum of squares of the dims of the irreps, and we have $q^3=|G/[G,G]|$ irreps of dim 1, one can compute the number of irreps of dim $q$. A monomial group is a group where every character is induced from a linear character, and nilpotent groups are known to be monomial (see, for example, Corollary 6.14 in Isaacs' book). | |
| Sep 8, 2012 at 19:48 | comment | added | Alexander Chervov | By the way I've found page about UT(4,2) groupprops.subwiki.org/wiki/Unitriangular_matrix_group:UT(4,2) its count of conj. classes 2q^3+q^2-2q agrees with yours count of irreps number | |
| Sep 8, 2012 at 17:35 | vote | accept | Alexander Chervov | ||
| Sep 8, 2012 at 17:35 | vote | accept | Alexander Chervov | ||
| Sep 8, 2012 at 17:35 | |||||
| Sep 8, 2012 at 17:32 | comment | added | Alexander Chervov | Thank you very much ! How to you see "so that the irreps of G/Z(G) have dim at most q" and "So G/Z(G) must have q3−q irreps of dim q." ? Also this seems to appeal to some background which is not known for me "If such an irrep had dim q, then it would have to be induced from a linear character of a subgroup of order q5 --||WHY?||-- (since p-groups are monomial)--||?||--, but Z(G) is contained in the derived subgroup of every subgroup of order q5, contradiction. " | |
| Sep 8, 2012 at 14:25 | comment | added | Frieder Ladisch | $G=U_4(F_q)$ has center of order $q$ and $G/Z(G)$ has center of order $q^2$ and index $q^3$, so that the irreps of $G/Z(G)$ have dim at most $q$. So $G/Z(G)$ must have $q^3-q$ irreps of dim $q$. Finally, one needs to see that irreps where $Z(G)$ is not in the kernel have dimension $q^2$. If such an irrep had dim $q$, then it would have to be induced from a linear character of a subgroup of order $q^5$ (since $p$-groups are monomial), but $Z(G)$ is contained in the derived subgroup of every subgroup of order $q^5$, contradiction. | |
| Sep 8, 2012 at 13:04 | comment | added | Alexander Chervov | Ooops, too much thinking ))... Thank you ! How did you get "q3−q dim q's and q2−q dim q2's" ? | |
| Sep 8, 2012 at 12:33 | comment | added | Frieder Ladisch | @Alexander Chervov: for $n=4$, you get $\lfloor 9/4 \rfloor = 2$, correct, but this means that the dims of the irreps are $q^0$, $q^1$ and $q^2$, so for $q=2$ they are $1$, $2$ and $4$. There are $q^3$ dim 1's, $q^3-q$ dim $q$'s and $q^2-q$ dim $q^2$'s. | |
| Sep 8, 2012 at 11:15 | comment | added | Alexander Chervov | Thank you very much for yours kind answer ! Let me ask something. You write that dim(irreps)<= floor[ (n-1)^2/4], is it correct, I mean "floor" not "ceil" rounding ? So in particular if n=4, p= 2 we get [9/4] = 2, so this means that dims of irreps are 1,2. So 4 is forbidden. If i understand correctly G/[G,G] in this case contain 8 elements. So we see that there should be 8*dim1 + 16*dim2. So in total there should be 24 irreps and same number of conjugacy classes. Does it sounds reasonable ? Or I miscalculated ? | |
| Sep 7, 2012 at 13:37 | history | answered | Frieder Ladisch | CC BY-SA 3.0 |