4
votes
Accepted
Group laws of the flow of time-dependent vector field
You're mixing the notations. With $\psi_t$ you'd be fine to add them. However this is $\psi_{t_0,t_1}$ which is actually the curve $\beta_{t_1-t_0}$ as explained on page 238. Note that $$\psi_{t_0,t_1}...
- 10.5k
3
votes
Accepted
A question concerning group action and stabilizer.
You can't prove that if $g \circ x = k \circ x$ then $k \in K$, because it doesn't have to be in $K$. However, you are given that $K$ acts transitively on $X$. Can you use that to find $k \in K$ such ...
- 52
3
votes
G abelian Lie group, $(\pi, V)$ a finite-dim unitary rep. $\exists $ mutually orthog. 1-dim invariant linear subspaces s.t $V= \bigoplus_i V_i$
Let $\{\pi(g):g\in G\}\subset GL(V)$. These are unitary operatots that also commute ($\pi(g_1)\pi(g_2)=\pi(g_1g_2)=\pi(g_2g_1)=\pi(g_2)\pi(g_1))$.
Moreover, as they are unitary, they are also ...
- 4,386
3
votes
Accepted
The difference in the definition of the action of the symmetric group on abstract tensor spaces and concrete tensor spaces
This is a place that is particularly susceptible to typos, mistakes and misinterpretations. In what follows, the composition of permutations is defined as $(\sigma\tau)(i) = \sigma(\tau(i))$ (as it ...
- 468
2
votes
Accepted
Can any arbitrary group act on any arbitrary set?
Let $G$ be a group and $X$ be a set.
All that is required for a map $\cdot:G\times X\to X$ to be an action is:
for all $g,h\in G, x\in X$, $$g\cdot(h\cdot x)=(gh)\cdot x,$$
for all $x\in X$, $$e\...
- 43.6k
2
votes
Accepted
Action of symmetry group on homology?
A problem with this construction is that the vertex permutation operation does not necessarily map cycles to cycles. Therefore, it cannot possibly define a map on homology groups.
For example, let's ...
- 31.8k
2
votes
Accepted
details in the proof: G a commutative Lie group and $(\pi, V)$ a finite-dim unitary representation of G. Then $\pi$ is irreducible iff dim $V = 1$
A representation assumes that the map $\pi:G\times V\to V$ is continuous. So the composition $V\to \{g\}\times V\subseteq G\times V \xrightarrow{\pi} V$ is continuous. The topologies taken are the ...
- 2,474
2
votes
Accepted
Why is $ V \longrightarrow V: v \mapsto \pi(g)(v)$ an endomorphism of $G$-modules?
Lets write down explicitly:
$$f:V\to V$$
$$f(v)=\pi(g)(v)$$
Or simply $f=\pi(g)$. It is linear because $\pi(g)$ is. It is a bijection because $\pi(g^{-1})$ is inverse of $\pi(g)$.
So the only question ...
- 39.8k
2
votes
Accepted
Closed G-invariant set in a metric space contained in an open G-invariant set?
Here is an actual proof. We have two disjoint closed $G$-invariant subsets of $X$: $V$ and $U^c=X\setminus U$. The functions $f, h: X\to {\mathbb R}$, $f(x)=d(x, V)$, $h(x)=d(x, U^c)$ are continuous ...
- 93.8k
2
votes
Accepted
A question related to frame bundle of a vector bundle
I would probably reformulate the text you copied by saying first that a trivialization of $E$ over $U$ induces a trivialization of $\text{Hom}(\mathbb R^k_M,E)$ over $U$. This has values in $U\times\...
- 20k
2
votes
Accepted
A question related to the action of a group on the real projective space
The action $G \times \mathbb{P} \to \mathbb{P}$ is what we might expect: $(g, \mathbb{R}v) \mapsto \mathbb{R}gv$, but we have to verify that it's well-defined. For this consider any nonzero multiple $...
- 23.4k
2
votes
Accepted
Finite group acting on projective line effectively and holomorphically
The groups are the isometry groups of a tetrahedron, a cube, and a dodecahedron respectively. They are contained in $SO(3)$ which operates isometrically on the sphere i.e. the complex projective line....
- 7,363
1
vote
Accepted
Ping Pong Lemma
This was answered in the comments section. Briefly, either $w$ or $w^{-1}$ may be conjugated so that its final letter is in $S$, and then the argument applies.
- 10.7k
1
vote
Let $G$ act on $X$, and let $x\in X$ be given Show there is a unique action of $G$ on $Orb_G (x)$ such that the inclusion $Orb_G (x)$ is G-equivariant
You had it, but you got confused. The best way to not get confused is to get back to the definition of group actions: It is a morphism $\rho : G \to \mathfrak{S} (X)$. (I denote by $[x]$ the orbit of $...
- 2,935
1
vote
Understanding Representation theory
We know the group $G=\mathrm{SL}(2,\mathbb{R})$ acts on $\mathbb{R}^2$. Thus, it acts on the real projective line $\mathbb{RP}^1$. With the congruence $\big[\begin{smallmatrix} x \\ y \end{smallmatrix}...
- 2,447
1
vote
Accepted
Conjecture: For transitive imprimitive permutation groups, the intersection of the stabilisers of any pair of blocks is not trivial
This does not hold. Small counterexample:
$$
g:=\langle (1,3,5)(2,4,6), (1,4)(2,3)(5,6) \rangle\cong S_3
$$
Then $\{\{1,4\},\{2,5\},\{3,6\}\}$ is a block system, but the stabilizers of $\{1,4\}$ and $\...
- 17.6k
1
vote
$H$ acts on $G/H$. Show that $\mathrm{Fix}(H) = G/H$
The orbit equation reads:
$$|G/H|=\left|\operatorname{Fix}(H)\right|+\sum_i[H:\operatorname{Stab}(g_iH)]$$
where the $g_iH$'s are representatives of the non-sigleton orbits, necessarily each of size $...
- 2,585
1
vote
Accepted
$H$ acts on $G/H$. Show that $\mathrm{Fix}(H) = G/H$
My answer builds up from the hint that user297024 gave, which is crucial in my opinion. The "previous part of the question" that you proved $H \in \...
1
vote
Exchangeable strictly vs. almost surely
Assumption 1. Let $p$ be invariant under the measurable transformation/action/dynamical system $\sigma$, i.e., $p(\sigma^{-1}(A))=p(A)$ for all $A\in\mathcal{B}(\mathcal{X})$.
Assumption 2. Let $\...
- 1,887
1
vote
Accepted
Exchangeable strictly vs. almost surely
The key observation is that strict exchangeability means $1_A(\sigma(X)) = 1_A(X)$ for all $X$ while almost sure exchangeability means the equality only holds almost surely.
Suppose $A$ is almost ...
- 10.2k
1
vote
Hopf curve isomorphic to an elliptic curve $\mathbb{C}/\Gamma$.
I shall first claim that, the lattice $\Lambda$ is generated by $\log\lambda$ and $2\pi \sqrt{-1}$.
Recall that, the logarithm function $\log$ is only well-defined on the Riemann surface $\Sigma_{\log}...
- 448
1
vote
Let $G$ a simple group. Suppose exists a conjugacy class $C$ in $G$ such that $C \neq\{e\}$ is finite. Prove $G$ is finite.
$G$ acts transitively by conjugation on $C$. Since $G$ is simple, the action is faithful. Therefore, $G$ embeds into $S_C$, which is finite because $C$ is finite. So, $G$ is finite.
- 2,585
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