Difference between revisions of "Orbit"

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An '''orbit''' is part of a [[set]] on which a [[group]] [[group action|acts]].
 
An '''orbit''' is part of a [[set]] on which a [[group]] [[group action|acts]].
  
Let <math>G</math> be a group, and let <math>S</math> be a <math>G</math>-set.  The '''orbit''' of an element <math>x\in S</math> is the set <math>Gx</math>, i.e., the set of [[conjugate (group theory) | conjugate]]s of <math>x</math>, or the set of elements <math>y</math> in <math>S</math> for which there exists <math>\alpha \in G</math> for which <math>\alpha x = y</math>.
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Let <math>G</math> be a group, and let <math>S</math> be a <math>G</math>-set.  The '''orbit''' of an element <math>x\in S</math> is the set <math>Gx</math>, i.e., the set of [[conjugate (group theory) | conjugate]]s of <math>x</math>, or the set of elements <math>y</math> in <math>S</math> for which there exists <math>\alpha \in G</math> for which <math>\alpha\cdot x = y</math>.
  
For <math>x\in S</math>, the mapping <math>\alpha \mapsto \alpha x</math> is sometimes known as the ''orbital mapping defined by <math>x</math>''; it is a homomorphism of the <math>G</math>-set <math>G</math> (with action on itself, by left translation) into <math>S</math>; the image of <math>G</math> is the orbit of <math>x</math>.  We say that <math>G</math> acts ''freely'' on <math>S</math> if the orbital mapping defined by <math>x</math> is [[injective]], for all <math>x \in S</math>.
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For <math>x\in S</math>, the mapping <math>\alpha \mapsto \alpha\cdot x</math> is sometimes known as the ''orbital mapping defined by <math>x</math>''; it is a homomorphism of the <math>G</math>-set <math>G</math> (with action on itself, by left translation) into <math>S</math>; the image of <math>G</math> is the orbit of <math>x</math>.  We say that <math>G</math> acts ''freely'' on <math>S</math> if the orbital mapping defined by <math>x</math> is [[injective]], for all <math>x \in S</math>.
  
 
The set of orbits of <math>S</math> is the [[quotient set]] of <math>S</math> under the relation of conjugation.  This set is denoted <math>G\backslash S</math>, or <math>S/G</math>.  (Sometimes the first notation is used when <math>G</math> acts on the left, and the second, when <math>G</math> acts on the right.)
 
The set of orbits of <math>S</math> is the [[quotient set]] of <math>S</math> under the relation of conjugation.  This set is denoted <math>G\backslash S</math>, or <math>S/G</math>.  (Sometimes the first notation is used when <math>G</math> acts on the left, and the second, when <math>G</math> acts on the right.)
  
Let <math>G</math> be a set acting on <math>S</math> from the right, and let <math>H</math> be a [[normal subgroup]] of <math>G</math>.  Then <math>G</math> acts on <math>S/H</math> from the right, under the action <math>g: xH \mapsto xHg = xgH</math>, for <math>x\in S</math>.  (<math>H</math> acts trivially on this set, so <math>(E/H)/G = (E/H)/(G/H)</math>.)  Consider the canonical mapping <math>\phi : E/H \to E/G</math>.  The inverse images of elements of <math>E/G</math> under <math>\phi</math> are the orbits of <math>E/H</math> under action of <math>G</math>; thus on passing to the quotient, <math>\phi</math> defines an isomorphism from <math>(E/H)/G</math> to <math>E/G</math>.
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Let <math>G</math> be a set acting on <math>S</math> from the right, and let <math>H</math> be a [[normal subgroup]] of <math>G</math>.  Then <math>G</math> acts on <math>S/H</math> from the right, under the action <math>xH\cdot g = xHg = xgH</math>, for <math>x\in S</math>.  (<math>H</math> acts trivially on this set, so <math>(E/H)/G = (E/H)/(G/H)</math>.)  Consider the canonical mapping <math>\phi : E/H \to E/G</math>.  The inverse images of elements of <math>E/G</math> under <math>\phi</math> are the orbits of <math>E/H</math> under action of <math>G</math>; thus on passing to the quotient, <math>\phi</math> defines an isomorphism from <math>(E/H)/G</math> to <math>E/G</math>.
  
 
Suppose <math>G</math> and <math>H</math> are groups, and <math>G</math> acts on <math>S</math> on the left, and <math>H</math> on the right; suppose furthermore that the operations of <math>G</math> and <math>H</math> commute, i.e., for all <math>g\in G</math>, <math>h\in H</math>, <math>x\in S</math>,
 
Suppose <math>G</math> and <math>H</math> are groups, and <math>G</math> acts on <math>S</math> on the left, and <math>H</math> on the right; suppose furthermore that the operations of <math>G</math> and <math>H</math> commute, i.e., for all <math>g\in G</math>, <math>h\in H</math>, <math>x\in S</math>,
<cmath> (gx)h = g(xh) .</cmath>
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<cmath> (g\cdot x)\cdot h = g\cdot (x\cdot h) .</cmath>
 
Let <math>H^0</math> be the opposite group of <math>G</math>; then the actions of <math>G</math> and <math>H</math> on <math>S</math> define a left action of <math>G \times H^0</math> on <math>S</math>.  The set <math>(G\times H^0)\backslash S</math> is denoted <math>G \backslash S /H</math>.  Since <math>G</math> and <math>H^0</math> are normal subgroups of <math>G \times H^0</math>, by the previous paragraph, the <math>G</math>-sets <math>(G \backslash S)/H</math>, <math>G \backslash ( S/H)</math>, <math>G \backslash S /H</math> are isomorphic and identitfied with each other.
 
Let <math>H^0</math> be the opposite group of <math>G</math>; then the actions of <math>G</math> and <math>H</math> on <math>S</math> define a left action of <math>G \times H^0</math> on <math>S</math>.  The set <math>(G\times H^0)\backslash S</math> is denoted <math>G \backslash S /H</math>.  Since <math>G</math> and <math>H^0</math> are normal subgroups of <math>G \times H^0</math>, by the previous paragraph, the <math>G</math>-sets <math>(G \backslash S)/H</math>, <math>G \backslash ( S/H)</math>, <math>G \backslash S /H</math> are isomorphic and identitfied with each other.
  

Latest revision as of 15:44, 7 September 2008

An orbit is part of a set on which a group acts.

Let $G$ be a group, and let $S$ be a $G$-set. The orbit of an element $x\in S$ is the set $Gx$, i.e., the set of conjugates of $x$, or the set of elements $y$ in $S$ for which there exists $\alpha \in G$ for which $\alpha\cdot x = y$.

For $x\in S$, the mapping $\alpha \mapsto \alpha\cdot x$ is sometimes known as the orbital mapping defined by $x$; it is a homomorphism of the $G$-set $G$ (with action on itself, by left translation) into $S$; the image of $G$ is the orbit of $x$. We say that $G$ acts freely on $S$ if the orbital mapping defined by $x$ is injective, for all $x \in S$.

The set of orbits of $S$ is the quotient set of $S$ under the relation of conjugation. This set is denoted $G\backslash S$, or $S/G$. (Sometimes the first notation is used when $G$ acts on the left, and the second, when $G$ acts on the right.)

Let $G$ be a set acting on $S$ from the right, and let $H$ be a normal subgroup of $G$. Then $G$ acts on $S/H$ from the right, under the action $xH\cdot g = xHg = xgH$, for $x\in S$. ($H$ acts trivially on this set, so $(E/H)/G = (E/H)/(G/H)$.) Consider the canonical mapping $\phi : E/H \to E/G$. The inverse images of elements of $E/G$ under $\phi$ are the orbits of $E/H$ under action of $G$; thus on passing to the quotient, $\phi$ defines an isomorphism from $(E/H)/G$ to $E/G$.

Suppose $G$ and $H$ are groups, and $G$ acts on $S$ on the left, and $H$ on the right; suppose furthermore that the operations of $G$ and $H$ commute, i.e., for all $g\in G$, $h\in H$, $x\in S$, \[(g\cdot x)\cdot h = g\cdot (x\cdot h) .\] Let $H^0$ be the opposite group of $G$; then the actions of $G$ and $H$ on $S$ define a left action of $G \times H^0$ on $S$. The set $(G\times H^0)\backslash S$ is denoted $G \backslash S /H$. Since $G$ and $H^0$ are normal subgroups of $G \times H^0$, by the previous paragraph, the $G$-sets $(G \backslash S)/H$, $G \backslash ( S/H)$, $G \backslash S /H$ are isomorphic and identitfied with each other.

Let $G$ be a group, and $H$ a subgroup of $G$; let it act on $G$ from the right. Then the set $G/H$ is the set of left cosets mod $H$.

If $G$ is a group and $H,K$ are subgroups of $G$, then the set $H \backslash G /K$ is called the set of double cosets mod $H$ and $K$.

See also

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