# Tag Info

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### 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 ...
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### 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 ...
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### 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 ...
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### 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 ...
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### 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....
1 vote
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### 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.
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1 vote
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1 vote
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1 vote
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### 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 ...
1 vote

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 $\Sigma_{\log}... 1 vote ### Let$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.

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