All Questions
15 questions
15
votes
3
answers
1k
views
Group of matrices in which every matrix is similar to unitary
$\DeclareMathOperator\GL{GL}$Let $G$ be a subgroup of $\GL_n(\mathbb{C})$ such that for every $g \in G$ there exists $c \in \GL_n(\mathbb{C})$ for which $cgc^{-1}$ is unitary (or, which is the same, $...
0
votes
0
answers
99
views
Unimodular matrices fixing $(1, 1, \cdots, 1)$
What is known about the subgroup of $GL(n, \mathbb Z)$ fixing (under left multiplication) the vector ${(1, 1, \cdots, 1)}^T$ ('T' denotes transposition). I'm particularly interested in the case $n = 5$...
8
votes
1
answer
305
views
Groups that act transitively on $\mathrm{Gr}(k,\Bbb R^n)$ but not transitively on $\mathrm{Gr}(k+1,\Bbb R^n)$
Is it known for which $n, k\in\Bbb N$ there exists a matrix group $\Gamma\subseteq\mathrm{GL}(\Bbb R^n)$ that
acts transitively on $\mathrm{Gr}(k,n)$, i.e., on the $k$-dimensional subspaces of $\Bbb ...
2
votes
1
answer
270
views
Explicit Normalizer of SU(3) Cartan subalgebra
The normalizer $N(\mathfrak{h})$ of the Cartan subalgebra $\mathfrak{h}$ of $\mathfrak{su}(3)$ is defined as
$$N=\left\{ x \in SU(3)\;|\; x^\dagger\mathfrak{h}x \in \mathfrak{h}\right\}$$
I would like ...
4
votes
1
answer
441
views
Smallest subgroup of unitary group, containing diagonal matrices and a fixed unitary matrix is the whole group
Suppose that $U_n(\mathbb{C})$ is the group of unitary matrices of dimension $n$ over complex numbers. Fix a unitary matrix $A \in U_n(\mathbb{C})$ and consider the smallest closed subgroup $K \...
3
votes
3
answers
643
views
A question on linear groups
Asking this question I have made a mistake joining my main question with two simple ones, so it hasn't received enough attention, however there was a partial answer, which was not elaborated, and now ...
1
vote
2
answers
289
views
Any analysis on phase of eigenvalue of unitary matrix?
I understand that there are invariant Haar measure for eigenvalues of unitary matrix. I further understand that absolute value of eigenvalues of unitary matrix is 1. But, I could not find any analysis ...
5
votes
1
answer
481
views
Centralizer of hermitian matrices with zero trace
In Quantum Physics one often has to deal with commutators.
Here I want to denote by $H_0$ the set of all hermitian matrices with trace equal to zero!
One can easily relate it to $\mathfrak{su}(N)=...
1
vote
1
answer
308
views
Suppose that $G$ is a subgroup of $GL_n(\mathbb C)$ with finite exponent. Then is $G$ a finite group? [closed]
As title. the exponent of $G$ is the least number $n$ (if exists) such that $g^n=e$ holds for all $g\in G$ or $+\infty$.
5
votes
0
answers
148
views
Groups of operators between local unitaries and full unitaries
Consider the group $U(d_1) \otimes U(d_2)$ of "local unitary" operators acting on the complex space $\mathbb{C}^{d_1} \otimes \mathbb{C}^{d_2}$ (i.e., $U(d_1) \otimes U(d_2)$ is the group of unitary ...
2
votes
0
answers
212
views
Compute the discriminant for reductive groups
Consider $G=GL_{2}$ and $F=k((\pi))$, and a diagonal matrix $t=\left(\begin{array}{cc}a&0\\0&b\end{array}\right)$.
The characteristic polynomial of $t$ is $X^{2}-(a+b)X+ab$, and the ...
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 $...
2
votes
1
answer
297
views
equations over (some) lie groups
To be concrete, let $G=SL(n, \mathbb{C}),$ $\phi$ an automorphism of $G.$ Is there a characterization of those $x$ for which there exists a $y$ such that $x = y \phi(y)?$ In the special case, the ...
5
votes
4
answers
3k
views
Parametrization of O(3)
Is there a simple way to parametrize the orthogonal group O(3) of 3 by 3 orthogonal matrices?
7
votes
2
answers
1k
views
Abelianization of Lie groups
If G is a group, its abelianization is the abelian group A and the map G → A such that any map G → B with B abelian factors through A. Abelianization is a functor, and in general a very ...