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Suppose a group $G$ acts on a set $X$. I know what is meant by "an orbit of a point $x\in X$".

  1. For $Y\subset X$, what does "$Y$ is an orbit" mean? Does it mean "for some $x\in X$, $Y$ is an orbit of $x$" ?

  2. Consider a right action $(P, A) \mapsto {}^tPA\bar P$ of $\mathit{GL}_n(\mathbb{C})$ on $M_n(\mathbb{C})$. A textbook at hand says a subset $\{A\in M_n(\mathbb{C})\mid A\ \text{is a positive-definite Hermitian matrix}\}\subset M_n(\mathbb{C})$ is an orbit. Why is this true?

Edit: Since the word something seemed confusing, I replaced it with $Y$.

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An orbit is a set. For instance, the orbit of $1$ under the permutation $(124)$ is $\{1, 2, 4\}$ while the orbit of $3$ is $\{3\}$. So the "something" must be a set. –  user1729 Feb 29 '12 at 13:20
    
You may find this answer of mine, where I discuss a misconstrued definition helpful. –  user21436 Feb 29 '12 at 14:12

1 Answer 1

up vote 6 down vote accepted

For 1: It would be wrong to say "an orbit of a point $x\in X$", since a point has exactly one orbit (under a given action), so it's "the orbit of a point $x\in X$" (under the given action). Thus the indefinite article is free to be used in the case where $x$ isn't specified; so yes, "$O$ is an orbit" (under the given action) means "for some $x\in X$, $O$ is the [not an] orbit of $x$" (under the given action).

For 2: Positive-definite Hermitian matrices can be diagonalized. Let $A$ and $B$ be two such matrices, and let $A={^t}SC\bar S$ and $B={^t}TD\bar T$ be their diagonalizations, with ${^t}S\bar S=1$, ${^t}T\bar T=1$ and $C$ and $D$ positive real diagonal matrices. Then $B={^t}PA\bar P$ with $\bar P={^t}SC^{-1/2}D^{1/2}\bar T$. Thus all positive-definite Hermitian matrices are in the same orbit of this action.

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Since my native language (in which the textbook I'm reading is written) lacks articles, I wasn't sure which article should be used. –  Pteromys Feb 29 '12 at 13:37

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