bio | website | maths.dur.ac.uk/~lfvx79 |
---|---|---|
location | ||
age | ||
visits | member for | 4 years, 9 months |
seen | 3 hours ago | |
stats | profile views | 1,164 |
Jul 24 |
comment |
Rational structures on the flag variety over a finite field
I do not understand why this question was put on hold as it is perfectly clear what is being asked in the context of common terminology used for algebraic groups over finite fields; see for instance Digne & Michel, Representations of Finite Groups of Lie type, or Malle & Testerman, Linear Algebraic Groups and Finite Groups of Lie Type. |
Jul 2 |
awarded | Curious |
May 7 |
awarded | Fanatic |
Mar 17 |
comment |
Weyl group action on continuous characters of the group of $\mathbf{Q}_p$-points of the torus
Isn't your remark valid for all reductive $G$ and all $\chi\in X(T)$? |
Feb 3 |
comment |
How can classifying irreducible representations be a “wild” problem?
By the way, doesn't the existence of a good character sheaf theory for $\mathbf{U}(\mathbb{F}_q)$ mean that the irreps are parametrised by objects related to the infinite algebraic group $\mathbf{U}$ in a way which is similar to the case of reductive groups? |
Feb 3 |
comment |
How can classifying irreducible representations be a “wild” problem?
While Lusztig's Jordan decomp of reps of reductive groups over finite fields is important, and the independence of the unipotent reps of $q$ is remarkable, it is not the lack of this which makes a classification problem wild in the technical sense of the original question. E.g., $\mathrm{GL}_2(\mathbb{Z}/p^r)$ has a Jordan decomp of reps (due to Hill) and the 'nilpotent' reps depend on $q$ and even on $r$, but classifying the reps of this group (over any finite local PIR) is a tame problem. |
Feb 2 |
revised |
How can classifying irreducible representations be a “wild” problem?
added 236 characters in body |
Feb 2 |
comment |
How can classifying irreducible representations be a “wild” problem?
I've added some more details about this. |
Feb 2 |
revised |
How can classifying irreducible representations be a “wild” problem?
added 584 characters in body |
Feb 1 |
answered | How can classifying irreducible representations be a “wild” problem? |
Feb 1 |
comment |
How can classifying irreducible representations be a “wild” problem?
I think the difference between the rep. theories of $U_n$ and $\mathrm{GL}_n$ over finite fields is that the irreps. of $U_n(\mathbb{F}_q)$ are inherently more difficult to parametrise (as per Geoff Robinson's comment above). Just knowing the dimensions and multiplicities of the irreps. does not imply that we have a reasonable classification or parametrisation. |
Dec 8 |
awarded | Yearling |
Oct 27 |
answered | Cayley-Hamilton revisited |
Oct 18 |
comment |
semisimple conjugacy classes over general bases
@Margaux: You are right, I should have said that the semisimple conjugacy classes in $G(R)$ are in bijection with those of $G(k)$. The semisimple elements in $T(R)K$ are the conjugates of elements in the reductive part. |
Oct 17 |
comment |
semisimple conjugacy classes over general bases
@Margaux: The kernel $K$ of $G(R)\rightarrow G(k)$ lies in the unipotent radical of $G(R)$. Hence any maximal torus of $G(R)$ lies inside $T(R)K$, where $T$ is a maximal torus of $G$. But the only semisimple elements in $T(R)K$ are the ones in the reductive part of $T(R)$, which is isomorphic to $T(k)$ (e.g. via the Teichmuller section $T(k)\rightarrow T(R)$). The group $G(R)$ has a canonical structure of lin. alg. gp. over $k$ so we have a natural notion of semisimple elements. If these are not appropriate for the present purposes perhaps one should not consider semisimple elements. |
Oct 16 |
comment |
semisimple conjugacy classes over general bases
To state the question precisely you need to define the Steinberg map over $R$. If this can be done the answer to the question should be analogous to the case over $k$. |
Oct 16 |
comment |
semisimple conjugacy classes over general bases
Suppose that $R$ is local Artinian with residue field $k$. Then the Greenberg functor identifies $G(R)$ with a linear algebraic group over $k$. We therefore have a natural notion of semisimple elements in $G(R)$. A maximal torus in $G(R)$ is then the reductive part of $T(R)$ where $T$ is a maximal torus in $G$, so the semisimple elements of $G(R)$ are in bijection with those of $G(k)$. |
Oct 7 |
revised |
Centralizers of elements in general linear group over Z mod prime power
added 580 characters in body |
Oct 7 |
revised |
Centralizers of elements in general linear group over Z mod prime power
added 4 characters in body |
Oct 7 |
answered | Centralizers of elements in general linear group over Z mod prime power |