New answers tagged

1

Under these assumptions, note that $\mathrm{Rad}(kG)$ coincides with the augmentation ideal $\omega(kG)$ of $kG$. Now, as remarked by Jeremy Rickard, the question is equivalent to whether the elements $(g-1)^s$ with $g\in G$ generate $\mathrm{Rad}^s(kG)$ as an ideal of $kG$. However, according to Theorem 3.7 in Section 5.3 of the book [D.S. Passman: The ...


15

When $m$ is prime there is a simpler proof. The Smith normal form of the character table of $S_n$ is computed at Problem 14 here (solution here). From this it follows that the rank of the character table mod $m$ equals the number of partitions of $n$ for which every part has multiplicity less than $m$. By a simple generating function argument (generalizing ...


17

This is true when $m$ is prime and false in general. Counterexample. Take $S_8$ with $m=6$. Computer calculations show that the $\mathbb{Z}$-rank of the character table of $S_8$ with entries taken modulo $6$ to lie in $\{0,1,2,3,4,5\}$ is $22$. Thus this matrix has full rank, equal to its number of columns. (Its rank as a matrix with entries in $\mathbb{Z}/6\...


1

The answer of your question is true if the orbit $O_v$ is simply connected. Montgomery, D. Zippin, L. Topological transformations groups p. 226


19

Chevalley's theorem (see Theorem 4.19 in Milne) is very close to this. It says Let $G$ be a linear algebraic group and let $H$ be a Zariski closed subgroup. Then there is a representation $V$ of $G$ and a one dimensional subspace $L$ of $V$ such that $H$ is the stabilizer of $\mathbb{P}(L)$ in the action of $G$ on $\mathbb{P}(V)$. So this is close to what ...


6

I think, the statement "the set of Borel subalgebras of a semi-simple Lie algebra $\mathfrak g$ forms a Zariski-closed subset of a Grassmannian" can be understood in more than one way and then the question becomes less trivial than one thinks. Everything is over $\mathbb C$, by the way, otherwise things become even more complicated. First ...


3

It is not a tannakian category. The issue is the tensor product. Let $V$ and $W$ be automorphic representations. The algebraic tensor product $V\otimes W$ is no longer automorphic. You need some kind of completion $V\widehat{\otimes}W$ but, according to David Loeffler, the completion is too big to be automorphic.


0

I found the answer in theorem 8.7 in the survey article on periodic algebras by Erdmann and Skowronski. They are certain (twisted) mesh algebras of Dynkin type.


3

Yes, of course. You still have a natural homomorphism $A\rightarrow END(U_A)$. Since ${}_AA$ is a free $A$-module, an endomorphism $x\in END(U_A)$ is determined by its value $x_A$ on ${}_AA$. This proves that the natural homomorphism is an isomorphism: $$x_A \in End (A_{End_{{}_AA}})= End (A_{{A}})=A.$$


3

There are many such splittings. Any complementary (aka opposite) parabolic subalgebra (i.e. $\mathfrak{q}$ such that $\mathfrak{p} \oplus \mathfrak{q}^\perp = \mathfrak{g}$ and so on) provides a unique splitting $\mathfrak{p} = \mathfrak{p} \cap \mathfrak{q} \oplus \mathfrak{p}^\perp$. The space of complementary parabolic subalgebras is a $\exp \mathfrak{p}^...


2

$\DeclareMathOperator\Spec{Spec}$Let $k$ be a field. Let $S= \mathbb{A}^1_k = \Spec(k[x])$ and $X= \mathbb{A}^2_k = \Spec(k[x,y])$.The constant additive group $\mathbb{G}_a\times S$ over $S$ acts on $X$ by the equation $$ t \cdot (x,y) = (x, y+xt).$$ The orbit of the $0$ section $\phi: S \to X$ defined by $$ \phi(x) = (x,0)$$ consists of the the union $V \...


3

There's another answer which doesn't work in full generality, but is often easier to work with, especially if you know the A-type theta functions really well. When the exchange matrix has full rank, X-type theta functions are certain A-type theta functions after a change of variables Specifically, let $B$ be an $d\times r$ extended exchange matrix whose rank ...


8

Probably the shortest answer, especially in finite-type: X-type theta functions are monomials times F-polynomials Making this precise is a bit fiddly, because there are multiple conventions and (more importantly) two different notions of `the' dual cluster algebra to an A-type cluster algebra: The "Fock-Goncharov dual" (sometimes called the ...


4

It turns out that this is indeed true. The relevant proofs are given in Chapter 3 of this extended abstract.


5

It is equal. Since the ground field is of characteristic zero and $v$ is symmetric one may as well compute the stabilizer of the polynomial $f:=\sum_{i=1}^nx_i^m$. Assume $g\in\mathrm{GL}_n(\mathbb C)$ stabilizes $f$. Then it also stabilizes the Hessian $$\det\nolimits_{ij}(\partial_i\partial_jf)\in\mathbb C^*(x_1\cdots x_n)^{m-2}$$ up to a factor. Hence it ...


4

(Comment converted to answer per request:) The “surprising result” about simply connected homogeneous spaces in your (currently) last paragraph is Montgomery's Theorem (1950): generally (in your notation) if $G/H$ is compact and $H$ closed connected, then $K$ is transitive on $G/H$. The theorem is also discussed in Samelson (1952, p. 17).


4

The following was my original answer, dealing with the case where $X$ is unitary. It is a nontrivial fact that the orthogonal complement of an invariant subspace is again an invariant subspace. Thus, the projection $p$ in the question will automatically satisfy the stronger property $(p \otimes 1)X = X(p \otimes 1)$. This was part of Woronowicz' original ...


2

The book by Skowronski and Simson: "Elements of the Representation Theory of Associative Algebras: Volume 2, Tubes and Concealed Algebras of Euclidean type" tread the representation theory of tame algebras in detail and in section 13 you can find a classifcation of indecomposable modules for tame KQ with a detailed example of the four subspace ...


3

This conjecture has now been proved by Valentin Bonzom, Guillaume Chapuy and Maciej Dołęga in their paper $b$-monotone Hurwitz numbers: Virasoro constraints, BKP hierarchy, and O(N)-BGW integral.


8

$\DeclareMathOperator\GL{GL}$Let me try to clarify. The formula for the Whittaker functional in Theorem 4.6.5 of "Automorphic forms and representations" of Bump states that $$W\left(\pi\begin{pmatrix}p^k& \\ &1\end{pmatrix}v_0\right) = W(v_0)\frac{\alpha_1^{k+1}-\alpha_2^{k+1}}{\alpha_1-\alpha_2},$$ where $v_0$ is a $\GL_2(\mathbb{Z}_p)$-...


3

I consider the situation over $\mathbb{C}$. We can replace $G$ by the subgroup generated by the $g_j$'s, so assume that $G$ is generated by the $g_j$'s. That $G$ contains a spanning set of the space of all matrices yields that the center of $G$ only contains scalar matrices, and also that the inclusion $G \hookrightarrow \operatorname{GL}(n,\mathbb{C}) $ is ...


8

I am not that strong on the representation-theory side, but know more about the combinatorics side. If you want to get an overview of the symmetric functions and the associated combinatorics (crystal bases, RSK etc), then one starting point (with references!) is www.symmetricfunctions.com. I am the admin for this site, so all errors and issues are completely ...


11

M. Haiman "Notes on Macdonald polynomials and the geometry of the Hilbert scheme of points on $\mathbb{P}^2$". By one of the greatest specialists of interactions between combinatorics and algebraic geometry.


0

Maybe here there is some usefull information, I'm reaserching these same field but still learning https://arxiv.org/pdf/0805.2531.pdf


1

The quadratic form whose matrix is $\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & I_n \\ 0 & I_n & 0 \end{pmatrix}$ gives an embedding of $\operatorname{SO}(2n + 1, \mathbb C)$ in $\operatorname{GL}(2n + 1, \mathbb C)$ whose derivative is your specified embedding $\mathfrak{so}(2n + 1, \mathbb C) \to \mathfrak{gl}(2n + 1, \mathbb C)$. Under ...


0

I found it confusing, and hope that this is correct. We work with right modules, and let $K$ be the base field. Representing $F$ by a complex of injective modules, $\operatorname{RHom}_{\Pi_G}(S_i,F)$ becomes a complex of vector spaces; call it $V$. Since $S_i \cong \Pi_G/I_i$ is a $\Pi_G$-$\Pi_G$-bimodule, $\operatorname{RHom}_{\Pi_G}(S_i,F)$ is also a ...


8

The answer is no in general: for example, if $\phi$ is a cuspidal automorphic form in a cuspidal automorphic representation $\pi$ of $\mathrm{GL}_n(\mathbb{A}_F)$ and $\Phi$ is an Eisenstein series in the noncuspidal automorphic representation $\Pi = \widetilde{\pi} \boxplus \omega$ of $\mathrm{GL}_n(\mathbb{A}_F)$, where $\omega$ is a Hecke character, then ...


3

If I'm not mistaken, $\pi_{K,\chi}$ and $\pi_{K',\chi}$ can be different. Consider a bipartite bi-regular tree, with degrees $q_1+1<q_2+1$. There are $p$-adic unitary groups that their building in such a tree with $q_2=q_1^3$, see for example https://arxiv.org/abs/1005.3504. Then there are two maximal compact subgroups, corresponding to the stabilizers of ...


8

$\def\Spin{\text{Spin}}$The cover $\Spin(n) \to SO(n)$ is $2$ to $1$, the nontrivial element kernel is a central element of $\Spin(n)$ which I'll call $z$. Since $z$ is central, it acts on any irrep of $\Spin(n)$ by $\pm 1$; if $z$ acts by $1$, then the irrep factors through $SO(n)$, if $z$ acts by $-1$, then it does not. Irreps of Lie groups are usually ...


1

It could be that the first reference uses a different terminology. There clearly are Majorana spinors for Spin(6)=SU(4). There are two ways of seeing this. One way is to note that there exists an anti-linear map (commuting with all transformations from Spin(6)), sending the two types of semi-spinors $S_+, S_-$ into each other and squaring to the identity. So,...


10

This isn't true. Let $V$ be the standard two dimensional representation of $U(2)$ and let $W = (\det V)^{-1} \otimes \mathrm{Sym}^2(V)$. If the eigenvalues of $g$ acting on $V$ are $z_1$ and $z_2$, then the eigenvalues of $g$ acting on $W$ are $z_1 z_2^{-1}$, $1$ and $z_1^{-1} z_2$. So $\det (\mathrm{Id} - \pi_W(g))=0$ for all $g$. Okay, general nonsense ...


2

As mentioned, the present question is related to Question on characterising CQGs. Since the questions are related, the answers are also. I will work in the setting of Theorem 3.2.12 from Timmerman's book "An invitation to quantum groups and duality". Let $(A,\Delta)$ be the given Hopf $*$-algebra. As in (v) in Theorem 3.2.12, let $(\delta_\alpha)_{...


18

$\begin{pmatrix} {\mathbb Z} & {\mathbb Q}\\ 0 & {\mathbb Q} \end{pmatrix}$ is a canonical example for such things. Let us list all simple right and left modules: $$R_p=\begin{pmatrix} {\mathbb Z}/(p) & 0\end{pmatrix}, \ R_0 =\begin{pmatrix} 0 & {\mathbb Q} \end{pmatrix}, \ L_p=\begin{pmatrix} {\mathbb Z}/(p) \\ 0 \end{pmatrix}, \ L_0=\...


17

Let $Q$ be a finite acyclic quiver and $K$ a field. Let $Q^{op}$ be the opposite quiver (where all the arrows are reversed). Let $KQ$ be the path algebra. Then the category of left $KQ$-modules is equivalent to the category of right $KQ^{op}$-modules. It is well known that if $Q,Q'$ are acyclic quivers, then the categories of right $KQ$- and $KQ'$-modules ...


1

See Appendix A in M. Borovoi and D. A. Timashev, Galois cohomology and component group of a real reductive group.


1

$\newcommand{\mk}{\mathfrak}$Let $\mk{h}_{2n+1}$ be the $(2n+1)$-dimensional Lie algebra (basis $x_1,\dotsc,x_n,y_1,\dotsc,y_n,z$, nonzero brackets $[x_i,y_i]=z$). The $n$-dimensional abelian Lie algebra $\mk{a}_n$ (basis $a_1,\dotsc,a_n$) acts on it by $a_i\cdot x_i=x_i$, $a_i\cdot y_i=-y_i$, rest of the action being zero. Through the quotient map $\mk{h}_{...


5

The set in question is isomorphic to the fiber product $F\times^SH$ where $F=SU(2)$, $H=S(U(3)\times U(3))$ and $S=F\cap H=U(1)$. In particular, it is a connected manifold. This can be seen as follows: Let $\xi=1_3\otimes\sigma^3$ und $U\in\mathfrak{su}(6)$ with $U\xi U^{-1}\in\mathfrak{su}(2)$. Since $\xi$ and $U\xi U^{−1}$ are elements of $\mathfrak{su}(2)$...


2

This was too big to fit as a comment. Here is a cute, completely trivial, but incredibly useful fact. Suppose $(f^*, f_*)$ is an adjoint pair of functors between additive categories (not necessary abelian or anything of that nature). If $X$ is an object such that $f^*X \neq 0$, then the unit map $\eta\colon X \to f_*f^*X$ is not zero. This is because this ...


3

This is correct. The complex span of $\mathfrak{su}(2)$ extended by the identity matrix (which commutes with everything) is the full matrix algebra $M_2(\mathbb{C})$, so the centralizer, $C$ say, of the matrix subalgebra $\langle I_3 \rangle \otimes \mathfrak{su}(2)$ of $M_6(\mathbb{C})$ is the centralizer of $\langle I_3 \rangle \otimes M_2(\mathbb{C})$ in ...


Top 50 recent answers are included