bio | website | maths.dur.ac.uk/~lfvx79 |
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location | ||
age | ||
visits | member for | 4 years, 4 months |
seen | 2 hours ago | |
stats | profile views | 934 |
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 |
Oct 1 |
awarded | Caucus |
Sep 20 |
comment |
a very elementary question on the conjugated matrices
Could you specify what the square brackets in $B~[\pi^{d+1}]$ mean? |
Jun 26 |
awarded | Revival |