New answers tagged

1

It seems that any module $M$ whose double $A$-dual $M^{**}$ is not reflexive gives a counterexample. In this case $G(M) = M^{**}$ is a summand of $G^2(M) = (M^{**})^{**}$ with non-trivial complement, and $G^2(M)$ can't be reflexive since this property is inherited by summands. Repeating this argument with $G(M)$ instead of $M$, we see that $G^l(M)$ is a ...


5

I would have preferred to not answer my own question, but here it goes. Yes, the two groups have the same number of conjugacy classes and in fact, the groups $\mathrm{SL}_{n}(W_{2}(\mathbb{F}_{q}))$ and $\mathrm{SL}_{n}(\mathbb{F}_{q}[t]/t^{2})$, for $q$ a power of a prime $p$ dividing $n$, have the same number of irreducible characters of dimension $d$ for ...


4

I am also very far from an expert here, but I think there's a case to be made that the "correct notion" involves actions on categories of sheaves, as Donu says in the comments. Consider the following toy model: if $A$ is, say, a finite abelian group then its Pontryagin dual $\widehat{A}$ can be defined as the group $\text{Hom}(A, \mathbb{G}_m)$ of ...


2

The quickest way to display the character table for a fixed prime-power $q$ is: Display(CharacterTable( "PSL", 2, q) The generic character table is in fact available on GAP4 (as pointed out in private by Frank Lübeck) with the following commands: For $q$ even: gap> Print(CharacterTableFromLibrary("SL2even")); For $q \equiv 1 \mod 4$: ...


10

Here is a counterexample of size 38. Let $m=8$ and let $$P=\{0,a_1,\dots,a_m,1\}\cup\{b_{ij}: 1\le i<j\le m\}$$ where $0<a_i<b_{jk}<1$ whenever $i$ is distinct from $j$ and $k$. The cardinality of $P$ is $n=1+m+\binom{m}{2}+1=38$. The join-irreducible elements are only the $a_i$ and $0$, since $b_{ij}=\bigvee\{a_k:k\ne i, k\ne j\}$. The $a_i$ ...


3

I don't see a way at the moment to a full answer, but I mention the following in case someone can make use of it : the class function Sqr defined by Sqr(g) = the number of square roots of G in G is always a generalzied character: we have ${\rm Sqr}(g) = \sum_{ \chi \in {\rm Irr}(G)} \nu_{2}(\chi) \chi(g)$ for each $g \in G$, and clearly Sqr contains the ...


2

From the source code of GAP: ############################################################################# ## #M IndicatorOp( <ordtbl>, <characters>, <n> ) #M IndicatorOp( <modtbl>, <characters>, 2 ) ## InstallMethod( IndicatorOp, "for an ord. character table, a hom. list, and a pos. integer", [ ...


2

A partial answer in a very special case, but still possibly useful to record. The search for idempotents in a group algebra can be simplified by using GAP's MeatAxe: https://www.gap-system.org/Manuals/doc/ref/chap69.html For example, if $G$ is the dihedral group with eight elements, and $F$ the field with eight elements, the following code verifies that $kG$ ...


2

As you say you have a given system of equations, you can do this by brute-force in GAP. The following constructs a group algebra $kG$ in GAP -- this one constructs $kG$ when $k = GF(4)$ and $G = C_2$. G := CyclicGroup(2); k := FiniteField(4); kG := GroupRing(k,G); The output is then gap> kG; <algebra-with-one of dimension 2 over GF(2^2)> and, as ...


3

Integrality doesn't really seem to play a role in the statement of Beilinson-Bernstein as far as I can tell. For a general weight $\lambda \in \mathfrak h^\ast$ it makes sense to talk about the notions of dominant and regular. If $\lambda$ is regular, then there is a derived localization: $D(D_{G/B}^\lambda-mod) \simeq D(U_\lambda -mod)$ If $\lambda$ is both ...


2

This is not an answer, more an extended comment. Though I am far from being an expert, I would have rather thought of an abelian variety (say $X$) to be analoguous to a vector (or projective) space (say $V$) and $\textrm{End}(X)$, the ring of isogenies of $X$, to be analogous to $\mathrm{GL}(V)$. Note that there is a notion of dual abelian variety which is ...


2

Heintze, Ernst; Groß, Christian. Finite order automorphisms and real forms of affine Kac-Moody algebras in the smooth and algebraic category. (English summary) Mem. Amer. Math. Soc. 219 (2012), no. 1030, viii+66 pp. ISBN: 978-0-8218-6918-5 Edit: I recommend the above reference for the following reasons. Through the work of F. Levstein, G. Rousseau, their ...


12

I get a slightly different formula: $$\sum_{V} \frac{d(V)^{2g-2-n}}{|G|^{2g-2}} \prod_{i=1}^n |k_i|\chi_V(k_i)$$ Here $|k_i|$ denotes the size of the conjugacy class. I prefer to express this in a slightly different way. For a conjugacy class $k$ and irreducible representation $V$ of $G$, let $f^V_k$ denote the scalar by which the indicator function $k \in Z(...


3

Since you tagged this with "symplectic geometry", I'm going to give an answer from a symplectic geometry perspective, which may not be what you're looking for, but (as a symplectic person) I find it is a helpful point of view. This will use the language of $A_\infty$-categories as well as dg-categories. Given a manifold $X$, take its cotangent ...


5

One way to discuss $\infty$-local systems over a space $X$ is in terms of the fundamental $\infty$-groupoid of $X$. To motivate this, recall that for classical local systems, one has the equivalence of categories $$LocSys(X) \simeq Rep(\pi_{1}(X)) $$ between the local systems on $X$ and representations of the fundamental group. It's easily verified that ...


8

Nate's suggestion on math.SE works. We'll show that if $A = k[x, \partial_x]$ and $B = k[y, \partial_y]$ are both taken to be the Weyl algebra, then the module over $A_2 = A \otimes B \cong k[x, \partial_x, y, \partial_y]$ generated by $e^{xy}$ is 1) simple and 2) not a tensor product of simple modules of $A$ and $B$. Explicitly this module $M$ consists of ...


1

First of all, you're right that by "the Casimirs" the authors mean the eigenvalues of the quadratic Casimir operator on the irreps in question — this is a common phrasing in the physics literature. For $\mathfrak{su}(n)$, the Young diagram with $m$ rows of lengths $n_i$ corresponds to the highest weight $\mu=\sum_i n_i\lambda_i$ (cf. e.g. these ...


11

For every group $G$, the left action $g\cdot f(x)=f(g^{-1}x)$ yields an action on the space $\mathbf{F}_p^{(G)}$ of finitely supported functions $G\to \mathbf{F}_p$. If $G$ is infinite, this action has no nonzero fixed point. If $G=\mathbf{F}_p^{(\alpha)}$ with $\alpha$ infinite, $G$ itself has cardinal $\alpha$ and hence $\mathbf{F}_p^{(G)}$ is isomorphic ...


1

Thanks to the outstanding help of LSpice I present a version of more detailed proof of the two parts above. Do not hesitate to point out mistakes. "$(1) \Rightarrow (2)$": Fix $\alpha \in I$ and $\mu \in \Pi(M)$. Observe that for $\mu(h_\alpha)=0$, we have $s_\alpha\mu=\mu-\langle \mu, \alpha^{\vee}\rangle \alpha = \mu - \mu(h_\alpha)\alpha=\mu$ ...


3

This is (part of) Theorem 6.13 in Serre's Finite Groups: An Introduction, which says the following. Say that an action of a group $H$ on another group $N$ is almost free if the action on $N \setminus \{ e \}$ is free. Then TFAE: $H$ is a Frobenius complement; equivalently, $H$ acts almost freely on another finite group $N$. $H$ acts almost freely on a ...


2

Given a prime $p$ and a connected reductive algebraic group $G$ over $\mathbb{F}_p^{\mathrm{alg}}$ with a Frobenius map $F$, the fixed points $G^F$ are a finite group 'of Lie type'. The finite groups of Lie type are the main case in the Classification Theorem of Finite Simple Groups. They can all be obtained in a uniform way by this construction, except for ...


6

An immediate application is the existence of the Langlands complex dual group. If $G$ is a connected reductive group over a field $F$ and $\overline{F}/F$ is an algebraic closure, then $G^\vee$ is the connected reductive group over $\mathbb{C}$ whose root datum is dual to that of $G_{\overline{F}}$, i.e. is obtained by interchanging roots with coroots and ...


2

Come that far, one can begin to exploit the subgroups, turn one's attention to symmetric spaces, of which there are the compact and noncompact ones, the riemannian and the hermitian ones. And one can begin to exploit the arithmetic subgroups and so delve into locally symmetric varieties with various geometric and arithmetic properties. Come that far, whole ...


1

check work of David Vogan, and a book of Trappa Vogan ...


6

In principle one can develop (1) using the coproduct in the ring of symmetric functions. By the Littlewood–Richardson rule, $\Delta(s_\nu) = \sum_{\alpha}\sum_\beta c^\nu_{\alpha\beta} s_\alpha \otimes s_\beta$ where $c^\nu_{\alpha\beta}$ is a Littlewood–Richardson coefficient, and correspondingly $$s_\nu[s_\lambda + s_\mu] = \sum_{\alpha}\sum_\beta c^\nu_{\...


8

I can't give you a very deep reason for why root data appear in this context (because, let's face it, root systems spring out everywhere), but there are some very elementary reasons to why the action in question is very natural with regard to the classification. Let me start with the following two considerations: When one tries to distinguish between the ...


4

For the sake of completeness, here is the answer to Question 1, part of which is missing from the other answers: Proposition 1. Let $G$ be a finite group. Consider the representations of $G$ over $\mathbb{C}$. Let $\det G$ denote the determinant of the character table of $G$. (Note that this is only defined up to sign, since the order of the rows and of the ...


7

look at C. De Concini, C. Procesi, A characteristic free approach to invariant theory, Adv. Math. 21 (1976), 330–354.


5

In this commutative situation, a Frobenius algebra is the same as an artinian Gorenstein ring. In general, if $\theta_1,\dots,\theta_n$ are homogeneous elements of positive degree of $A=K[x_1,\dots,x_n]$ and if $K[x_1,\dots,x_n]/(\theta_1,\dots,\theta_n)$ is artinian (i.e., a finite-dimensional vector space in this situation), then $A/(\theta_1,\dots,\...


11

As for the second question, the equivalence relation on the class of countable ICC groups given by $G \sim \Gamma$ if and only if $L(G) \cong L(\Gamma)$ is very interesting and usually called $W^*$-equivalence of $G$ and $\Gamma$. On page 45 of [Con82], Connes conjectures that the $W^*$-equivalence class of an ICC property (T) group is a singleton (up to ...


13

The set of countable ICC groups doesn't exist, so I think you're asking about isomorphism classes. There are continuum many non-isomorphic locally finite fields (e.g., take, for $S$ any set of primes, the invariants by $\prod_{p\in S}\mathbf{Z}_p$, which has Galois group $\prod_{p\notin S}\mathbf{Z}_p)$ in the algebraic closure of $\mathbf{F}_p$. It's known ...


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