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0 votes

Bounding the projective dimension of modules by the number of points and arrows

A counterexample is the algebra $k\langle x,y| xyxy^2, xy^2xy^3, \dots ,xy^{k-1} xy^{k}\rangle $. It is connected ($n=1$) with $m=2$ generators and have global dimension $k$. So, the trivial module $...
1 vote

Dual Clifford module

$ \newcommand\Cl{\mathrm{Cl} \newcommand\tr{\mathop{\mathrm{tr}}}} \newcommand\Ext{{\textstyle\bigwedge}} \newcommand\form[1]{\langle#1\rangle} \newcommand\Hom{\mathop{\mathrm{Hom}}} \newcommand\rev[1]...
4 votes
Accepted

Length of $\mathbb{C}^\infty$ as an $S_\infty$-representation

To make sure I have understood correctly, I have repeated part of what you wrote in the next paragraph. The aim is to to compare the lengths of the representation $(\mathbb{C}^{\infty})^{\otimes k}$ ...
2 votes
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Reference request: decomposability of $\mathbb{G}$-Hilbert modules

Let's assume that $G$ is a reduced compact quantum group, that is, the Haar state on $C(G)$ is faithful. (1): A direct reference is [1, Lemma 4.2]. You can also get this from a careful study of [2, ...
5 votes

learning Deligne-Lusztig theory

Cédric Bonnafé is currently giving a course entitled "Introduction to Deligne-Lusztig theory" at YMSC (Tsinghua University, Beijing, China), from Nov.9 to Dec.16, 2022. All the slides, ...
2 votes
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Basic algebra of $\mathcal{O}_0(\mathfrak{sl}_n(\mathbb{C}))$ — Reference request

You can find quiver and relations (not sure if they are admissible always) here: http://www.math.uni-bonn.de/ag/stroppel/Quivers.pdf In particular the explicit algebra is only fully understoof for $...
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5 votes

A class function defined using Frobenius-Schur indicators

The group $𝐺=\mathtt{PerfectGroup}(43008,10)≅𝐶_2^7\rtimes 𝑆𝐿(2,7)$ has an element $g$ such that $𝑏(𝑔)<0$. Hence, $b$ cannot count something. The first such example was found by Gabriel ...
4 votes

How do you construct elements in $\operatorname{Ind}_P^G\pi$?

To have it written explicitly, $\pi$ is a smooth representation of $M$, extended trivially across $N$ to $P$. Choose an open subgroup $K_M$ of $M$ that is so small that $\pi^{K_M} \ne 0$, a $K_M$-...
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3 votes

Structure of cuspidal Bernstein components—do non-commutative endomorphism rings ever really show up?

A lot of time has passed, so perhaps this question is moot for you [I mean, I saw you give the Plücker lectures last week on something very different], or perhaps you now know the answer. But since ...
3 votes

On a theorem of Bernstein-Zelevinsky regarding supercuspidal resentations

Higher multiplicities can occur. See Keys, L-indistinguishability and R-groups for quasisplit groups: unitary groups in even dimension, Ann. Sci. ENS, 4th series, vol 20, no. 1, 1987, pp. 31-64. ...
12 votes
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Applications of equivariant homotopy theory to representation theory

There are decades and decades of algebraic results that use techniques from equivariant homotopy theory. Some examples ... (1) Quillen's work on ring theoretic aspects of the cohomology of finite ...
2 votes

Homogeneous components of Cox RIngs

The answer to all these questions is no. Let $X=F_1$, the first Hirzebruch surface. Let $D_0$ be a fibre of the projection to ${\mathbb P}^1$. Let $D_{-1}$ be the "negative" section of this ...
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14 votes
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When the Littlewood-Richardson rule gives only irreducibles?

The answer is Yes, but this requires some elaboration. Knutson-Tao-Woodward prove Fulton's conjecture in $\S$6.1. In principle, you can follow the approach by De Loera-McAllister or Mulmuley-...
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0 votes

Galois representations with semisimple residue representation

As pointed out in the comments, the answer to the modified question for the case of $m=1$ is Yes, and I try to give my own proof as follows. But I do not know if there is also an affirmative answer ...
2 votes
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Representation of $\mathrm{AGL}(V)$ on the homology of the poset of affine subspaces of $V$

I suspect that Solomon, Louis The affine group. I. Bruhat decomposition. proves what you are looking for. Let $A_n(q)$ denote the poset of proper affine subspaces of $\mathbf{F}_q^n$. The only non-...
1 vote

Counting adjoints in the symmetric or antisymmetric square of a Lie group representation

I don't have a proof but I noticed some things while having a quick look for some counterexamples that I thought were worth sharing. First things first what you are looking for is not special to the ...
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0 votes

Dual Clifford module

$\DeclareMathOperator\Cl{Cl}\DeclareMathOperator\char{char}$Not sure about i), but at least ii) is true, so the isomorphism is canonical (at least over $\mathbb{R}$ or $\mathbb{C}$). To see this it is ...

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