For questions on the adjoint action of a Lie group $G$, which represents the elements of the group as linear transformations of the group's Lie algebra, considered as a vector space.
Questions tagged [adjoint-action]
14 questions
2
votes
1 answer
Understanding $\text{Ad}\circ \exp = \exp \circ \text{ad}$, domains and ranges
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I am…
Jagerber48
- 1,631
2
votes
0 answers
Equivalent representations of $SO(3)$
Definition. Let $\rho_V$ and $\rho_W$ be representations of a Lie group $G$ on vector spaces $V$ and $W$. Then a isomorphism of the representations if a linear bijective map $f:V\rightarrow W$ so that
$$ f((\rho_V)_g v)=(\rho_W)_g f(v)…
Powder
- 975
2
votes
1 answer
How is the coadjoint action of $\mathfrak {g}^{\ast}$ on $\mathfrak {g}$ defined?
Let $\mathfrak {g}$ be a Lie bialgebra. Then $\mathfrak {g}^{\ast}$ is also a Lie bialgebra which is dual to $\mathfrak {g}.$ Let the brackets on $\mathfrak {g}$ and $\mathfrak {g}^{\ast}$ be denoted by $b$ and $b'$ respectively. Then how to define…
ACB
- 3,068
1
vote
1 answer
Derivative of group adjoint at the Lagrangian flow is composition of Lie algebra adjoint with group adjoint
Let $G$ be a (not necessarily finite-dimensional) Lie group with Lie algebra $\mathfrak{g}.$ Define the group adjoint $\operatorname{Ad}_\eta:\mathfrak{g}\to\mathfrak{g}$ by $\operatorname{Ad}_\eta v=dL_{\eta}dR_{\eta^{-1}}v$ and the Lie algebra…
Stuck
- 1,754
1
vote
0 answers
Path of connections pastes to a connection on cylinder
Suppose that $G$ is a Lie group and $\pi:P\rightarrow X$ is a principal $G$-bundle, where $X$ is taken to be a closed manifold. Denote the affine space of connections for this principal bundle as $\mathcal{A}_P$, to which we endow a differential…
tsch_
- 41
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vote
0 answers
Compute the derivative of adjoint action on Heisenberg group
Let $(V,\Omega)$ be a symplectic vector space, and consider $H:=V\times \mathbb{R}=\{(v,t)\}$. This space $H$ with the multiplication
$$(v_1,t_1)\cdot(v_2,t_2)=\left( v_1+v_2,\frac{1}{2}\Omega(v_1,v_2)+t_1+t_2 \right) $$
is a Lie group called the…
danimalabares
- 153
1
vote
2 answers
What are the eigenvalues and eigenvectors of $\operatorname{ad}x$ for non-diagonalizable $x$?
We know the following proposition is true. The proof together with the specification of the eigenvectors of $\operatorname{ad}x$ is here.
Let $x\in \operatorname{gl}(n,F)$ be diagonalizable with $n$ eigenvalues $a_1,\ldots,a_n$ in $F$. The…
Hans
- 10,484
0
votes
0 answers
Can the closure of an arbitrary adjoint orbit be written in terms of the closure of a nilpotent orbit?
For any $x \in \mathfrak{g}$, is it true that the closure of its $G$-orbit is given by $$\overline{G \cdot x} = G \cdot \left(x_s + \overline{{C_Gx_s}^{\circ} \cdot x_n}\right)?$$
Notation:
Let $G$ be a connected reductive algebraic group, defined…
0
votes
1 answer
Two questions about a proof of $\text{ad}$ action
I don't understand the first paragraph of the proof of the Theorem 20.27 from Lee's Introduction to Smooth Manifolds.
First, I don't understand why he uses $t=0$ in the derivate, I think that the map
$t \mapsto \text{exp}\ tX$ satisfies $t=t_0$ and…
0
votes
1 answer
On the adjoint representation of $GL_n(\mathbb{C})$
I'm studying the adjoint representation of $GL_n(\mathbb{C})$, and I'm encountering some confusion about the denominator that appears in the matrix entries of the adjoint action.
The adjoint representation of $GL_n(\mathbb{C})$ is given by the map…
user969954
0
votes
1 answer
Adjoint representation on basis vectors of $T_e G$
Lets consider the adjoint action
$$\mathrm{Ad}: G\rightarrow G$$
$$\mathrm{Ad}_g(g')=gg'g^{-1}$$
This can equally be written as
$$(l_g\circ r_{g^{-1}})g'=\mathrm{Ad}_g(g')$$
Lets consider a Lie group, for example the $SL(n)$ Lie group of $n\times n$…
0
votes
2 answers
Is the Lie algebra of a Lie subgroup Ad-invariant?
Let $G$ be a Lie group with Lie algebra $\mathfrak{g}$, and $H \subset G$ be a Lie subgroup of $G$ with Lie algebra $\mathfrak{h}$. Suppose that there is some inner-product $\left<\cdot,\cdot\right>$ on $\mathfrak{g}$ and define $\mathfrak{m} =…
0
votes
0 answers
Any element of a semisimple Lie group is conjugate (via the adjoint action) to an element in the Cartan subalgebra?
The statement:
If $G$ is a semisimple Lie group and $\mathfrak{g}$ the corresponding Lie algebra, then the general statement is that every element of $\mathfrak{g}$ is conjugate, via the adjoint action of $G$, to an element in any fixed Cartan…
Fabian
- 24,230