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26 votes
3 answers
4k views

How are these two ways of thinking about the cross product related?

I was always bothered by the definition of the cross product given in e.g. a calculus course because it's never made clear how one would go about defining the cross product in a coordinate-free manner....
Qiaochu Yuan's user avatar
16 votes
0 answers
755 views

Is there a "natural" proof of the equality $4^2=2^4$?

This question, or rather any answer that it might receive, would probably belong to the realm of Awfully sophisticated proof for simple facts. Still, I claim that I have quite serious motivation for ...
მამუკა ჯიბლაძე's user avatar
12 votes
4 answers
1k views

Real and quaternionic representations according to weights

According to this question, it is easy to know whether a (complex, finite-dimensional) representation is self-dual or not: just check if the weight distribution in space is symmetric about the origin. ...
Jjm's user avatar
  • 2,091
9 votes
1 answer
158 views

Eigenfunctions of the Laplace–Beltrami operator on the coadjoint orbit of $\mathfrak{su}(n)$

$\DeclareMathOperator\SU{SU}$For $\mathfrak{su}(2,\mathbb{C})$, the generic coadjoint orbit is $\mathbb{S}^2$, and the Laplace–Beltrami operator on it is given by $$ \Delta \equiv \frac{1}{\sin\theta} ...
SCh's user avatar
  • 195
8 votes
1 answer
799 views

higher Casimirs for $\mathfrak{sl}$

The Wikipedia universal enveloping algebra suggest a way to obtain higher Casimir operators (e.g. generators of the center of $\mathfrak{U(g)}$ for $\mathfrak{g}$ semisimple) by evaluating certain ...
Vít Tuček's user avatar
  • 8,597
6 votes
2 answers
832 views

Cyclic vectors in irreducible representations of simple Lie algebras

Is there a notion of "cyclic element" in a simple Lie algebra? In particular, is it independent of the irreducible representation chosen? Explanation. An endomorphism A is called cyclic if there is ...
AThomas's user avatar
  • 617
5 votes
3 answers
1k views

classifying space and cohomology of integer general linear group

I have obtained that the classifying space $$ BGL(\mathbb{R}^n)=BO(\mathbb{R}^n)=G_n(\mathbb{R}^\infty) $$ is the Grassmannian. I have also obtained that the mod 2 cohomology is the polynomial ...
Shiquan Ren's user avatar
  • 1,990
5 votes
1 answer
378 views

Plethysm of $S^3(S^2V)$ as $\mathfrak{sl}_3(\mathbb{C})$-module

I have asked this question in MSE before, but have not got any answer. So here I am asking it again with some more detail. I believe that the following sequence of $\mathfrak{sl}_3(\mathbb{C})$-...
Pritam Majumder's user avatar
5 votes
0 answers
231 views

Avoiding Cartan subalgebra in a Lie algebra

Let $G$ be a simple complex algebraic group acting on its Lie algebra $\mathfrak{g}$ via the adjoint representation. What is the largest integer $d$ such that every subspace $U \subseteq \mathfrak{g}$ ...
darko's user avatar
  • 309
5 votes
0 answers
428 views

Weyl's construction for symplectic groups--an exercise in Fulton and Harris's book

This is an exercise in section 17.3 in Fulton and Harris's book:Representation theory-a first course. Let $V=\mathbb{C}^{2n}$ and $Sp(2n)$ be the symplectic group w.r.t the nondegenerate bilinear ...
yang's user avatar
  • 181
4 votes
2 answers
360 views

Double centralizer in special linear algebra

It is well known that for a matrix $A$ in $\mathfrak{sl}_n(\mathbb{C})$, we have the following equivalence: $$\dim Z(A) \text{ is minimal} \leftrightarrow A \text{ is cyclic}$$ where $Z(A)$ is the ...
AThomas's user avatar
  • 617
4 votes
0 answers
104 views

Relationship between characteristic polynomials of a matrix and its adjoint representation

Let $A \in \mathrm{M}_n(F)$ be a matrix over a field $F$. Consider its adjoint representation $\mathrm{ad}_A \in \mathrm{End}(\mathrm{M}_n(F))$, defined by $$ \mathrm{ad}_A(X) = [A, X] = AX - XA. $$ I ...
darko's user avatar
  • 309
3 votes
1 answer
111 views

Generalization of a result of Kostant related to Gauss decomposition and Toda lattices

I found myself needing a generalization of a result of Kostant in his famous paper B. Kostant, The solution to a generalized Toda lattice and representation theory, Adv. in Math, Volume 34, 1979, ...
Three aggies's user avatar
3 votes
0 answers
236 views

Deligne-Simpson problem for classical groups

Additive Deligne-Simpson problem was partially prooved by Kostov. Also there is Crawley-Boevey's approach to the question. The problem is about existence of a solution of the equation $$ A_1 +...+A_n =...
quantum's user avatar
  • 181
2 votes
1 answer
168 views

Irreducible $G$-representations with unital algebra structure

Let us work over $\mathbb C$. Suppose that $G$ is a semisimple algebraic group and let $H \subset G$ be a maximal torus. Consider a dominant weight $\omega$, then one can associate a unique ...
Bobech's user avatar
  • 381
2 votes
0 answers
135 views

Set of Special Unitary Matrices that are dense in SU(4) and obey certain relations

I'm trying to find a finite set of 4x4 Unitary matricies $\{U_1,U_2,\ldots U_N\}$ such that the matrices are dense in SU(4), and obey the relations: $[U_i, U_j] = 0$ for $|i-j|>1$ $U_iU_{i+1}U_i=...
Ryan's user avatar
  • 71
1 vote
1 answer
298 views

Degenerate representation

Let $\rho : \mathbb{R}^n\to \mathfrak{so}(2m)$ be a faithful representation of the commutative Lie algebra $\mathbb{R}^n$ into the Lie algebra of skew-symmetric matrices. There is an orthonormal basis ...
user56980's user avatar
  • 442
1 vote
0 answers
58 views

Linear algebraic group, absolute root system, computing roots

Let $G(F)$ be a reductive linear algebraic group, where $F$ is a local field. Let $T(F)$ be a maximal anisotropic torus of $G$ that splits over a quadratic extension of $F$. Is there an efficient ...
user536406's user avatar
1 vote
0 answers
139 views

Integral convex polytopes formed from the weight diagrams of representations of $\mathfrak {sl}_4$($\mathbb{C}$)

I'm a student studying undergraduate abstract algebra and doing a summer research project in the mathematics department at my school. I'm barely familiar with the rudiments of representation theory; I ...
Caleb Williams's user avatar
1 vote
0 answers
71 views

What subspace of $\operatorname{SU}(4)$ group keeps an element of the $\mathfrak{su}(2)$ subalgebra within $\mathfrak{su}(2)$ upon adjoint action?

Consider the Lie group $G_4=\operatorname{SU}(4)$ with (15) generators $T^a$. A basis for the latter is $$\{\sigma^j \times 1_2, \quad \quad \sigma^i \times \sigma^j, \quad \quad 1_2 \times \sigma^j\},...
Rudyard's user avatar
  • 155
1 vote
0 answers
143 views

Why is this operator independent of the choice of basis

I asked this question in MSE but I received no answer https://math.stackexchange.com/questions/4009524/why-is-the-following-operator-independent-of-the-choice-of-basis/4013636#4013636 Let $G$ be a lie ...
Mira's user avatar
  • 139
1 vote
0 answers
84 views

Extra-Lorentzian Kac-Moody algebras

My question is about Kac-Moody (KM) algebras of finite rank with symmetrized Cartan matrices $B = C A$ ($A$ is Cartan matrix) of signatures $(-,-,+,...,+)$, $(-,-,-,+,...,+)$, etc. i.e. with $2$,...
Vladimir's user avatar
  • 371
0 votes
1 answer
775 views

Highest weights of irreducible components of tensor product of irreducible sl(3)-module [closed]

I am study the representation theory of $sl(3)$ and I have a question about the tensor representation of irreducible $sl(3)$-modules as follows: For each weight $\mu$, let $L(\mu)$ be the irreducible ...
L. Khako's user avatar
0 votes
1 answer
274 views

when $g^*$ is invariant under $Ad(G)$?

Let $G$ be a Lie Group and $\mathfrak{g}$ be its lie algebra. Let $\mathfrak{g}$ is semisimple or reductive lie algebra, then prove that $\mathfrak{g}^*$ (dual of $\mathfrak{g}$)is invariant under $...
user avatar
0 votes
0 answers
53 views

Largest Set of Special Unitary Matricies With Invariant Subspace For Adjoint Action

I am trying to solve the following. Given the special unitary group $SU(n)$ and its adjoint action $Ad_{U}: \mathfrak{su}(n) \rightarrow \mathfrak{su}(n)$, what is the largest subset of $SU(n)$ such ...
Benjamin's user avatar
  • 2,099