Difference between revisions of "Circle"

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(Created page with " ==Definition== A circle is usually defined by {{M|1=\mathcal{C}=\{ {{!}} \} }} ==See also== * Sphere {{Definition|Manifolds|Topology}}")
 
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{{Extra Maths}}
 
==Definition==
 
==Definition==
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A circle is usually defined by {{M|1=\mathcal{C}=\{ {{!}} \} }}
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A circle is usually defined by {{M|1=\mathcal{S}^1=\Big\{(x,y)\in\mathbb{R}^2{{!}}d\Big((0,0),(x,y)\Big)=1 \Big\} }}
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==Topological perspective==
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The map {{M|f:\mathbb{R}\rightarrow\mathbb{S}^1}} given by {{M|f:t\mapsto e^{2\pi jt} }} is significant. As it makes {{M|\mathbb{R} }} a [[Covering space|covering space]] of {{M|\mathbb{S}^1}}
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===The circle as a quotient space===
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{{Begin Theorem}}
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Theorem: The circle {{M|\mathbb{S}^1}} is [[Homeomorphism|homeomorphic]] to {{M|\frac{\mathbb{R} }{\mathbb{Z} } }}
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{{Begin Proof}}
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Using the map above, we see that this just wraps the real line around the circle over and over again, specifically {{M|1=f(t_1)=f(t_2)\iff t_1-t_2\in\mathbb{S} }}, this suggests an [[Equivalence relation]].
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Using a bit of abstract algebra it is not hard to see that the [[Equivalence class|equivalence classes]] are exactly the [[Coset|cosets]] of {{M|\mathbb{Z} }} in {{M|\mathbb{R} }}. So it is no problem to write {{M|1=\tfrac{\mathbb{R} }{\sim}=\tfrac{\mathbb{R} }{\mathbb{Z} } }}
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Using [[Passing to the quotient]] we see that {{M|\exists\bar{f} }} that makes the diagram below commute '''if and only if''' {{M|1=t_1\sim t_2\implies f(t_1)=f(t_2)}}
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<math>
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\begin{xy}
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\xymatrix{
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{\mathbb{R}} \ar[d] \ar[dr] &\\
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{\frac{\mathbb{R}}{\mathbb{Z}}} \ar[r]_{\bar{f}} & {\mathbb{S}^1}
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}
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\end{xy}
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</math>
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[math]
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\begin{CD}
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R {{CD Hoz Eq}} R \\
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{{CD Down Arrow|q}} {{CD Down Arrow||f}} \\
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\frac{\mathbb{R}}{\mathbb{Z}} {{CD Right Arrow||\bar{f} }} \mathbb{S}^1
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\end{CD}
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[/math] {{Triangle Wanted}}
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(Where {{M|\bar{f}:\frac{\mathbb{R} }{\mathbb{Z} }\rightarrow{\mathbb{S}^1} }} is given by {{M|\bar{f}:[t]\rightarrow f(t)}})
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If {{M|\bar{f} }} is a homeomorphism the result is shown.
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* {{M|\frac{\mathbb{R} }{\mathbb{Z} } }} is compact as it is the [[Image of a compact set is compact|image of a compact set]], namely {{M|[0,1]}} under {{M|q}}
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* {{M|\mathbb{S}^1}} is [[Hausdorff]] since it is a [[Metric space|metric space]] and every metric space is Hausdorff.
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* {{M|f}} is [[Surjection|surjective]], so as {{M|1=f=\bar{f}\circ q}} and {{M|q}} is surjective, {{M|\bar{f} }} must be too.
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** Otherwise there'd be things {{M|f}} maps to that {{M|\bar{f}\circ q}} may not - contradicting the diagrams commute
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* {{M|\bar{f} }} is [[Injection|injective]]
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** To be injective {{M|1=\bar{f}([t_1])=\bar{f}([t_2])\implies[t_1]=[t_2]}}
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*** Showing that {{M|\bar{f} }} is well defined
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***: Given {{M|a,b\in [t]}} we know {{M|a\sim b}} as {{M|[t]}} is an [[Equivalence class]]
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***: So this means {{M|1=f(a)=f(b)}} because that's how we defined 'equivalent'
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***: Thus {{M|1=\forall a\in [t][\bar{f}([t])=f(a)]}} - so we can defined {{M|\bar{f} }} unambiguously!
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*** Using this we see {{M|1=\bar{f}([t_1])=f(t_1)}} (choosing {{M|t_1}} as the representative of {{M|[t_1]}}) and
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***: {{M|1=\bar{f}([t_2])=f(t_2)}}, so we have {{M|1=f(t_1)=f(t_2)}} so {{M|t_1\sim t_2}}
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**** Now we know {{M|t_1\in[t_1]\cap[t_2] }} and {{M|t_2\in[t_1]cap[t_2] }}
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***: As [[Coset#Cosets are either disjoint or equal|cosets are either disjoint or equal]], and they're not disjoint! (we know {{M|t_1}} is in the intersection even if {{M|1=t_1=t_2}})
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***: so are equal - thus {{M|\bar{f} }} is injective.
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* Thus {{M|\bar{f} }} is a [[Bijection|bijection]]
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Using the [[Compact-to-Hausdorff theorem]] we conclude {{M|\bar{f} }} is a '''homeomorphism'''
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{{End Proof}}
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{{End Theorem}}
  
 
==See also==
 
==See also==

Latest revision as of 16:57, 11 May 2015

[math]\newcommand{\bigudot}{ \mathchoice{\mathop{\bigcup\mkern-15mu\cdot\mkern8mu}}{\mathop{\bigcup\mkern-13mu\cdot\mkern5mu}}{\mathop{\bigcup\mkern-13mu\cdot\mkern5mu}}{\mathop{\bigcup\mkern-13mu\cdot\mkern5mu}} }[/math][math]\newcommand{\udot}{\cup\mkern-12.5mu\cdot\mkern6.25mu\!}[/math][math]\require{AMScd}\newcommand{\d}[1][]{\mathrm{d}^{#1} }[/math]

Definition

A circle is usually defined by [ilmath]\mathcal{S}^1=\Big\{(x,y)\in\mathbb{R}^2|d\Big((0,0),(x,y)\Big)=1 \Big\}[/ilmath]

Topological perspective

The map [ilmath]f:\mathbb{R}\rightarrow\mathbb{S}^1[/ilmath] given by [ilmath]f:t\mapsto e^{2\pi jt} [/ilmath] is significant. As it makes [ilmath]\mathbb{R} [/ilmath] a covering space of [ilmath]\mathbb{S}^1[/ilmath]

The circle as a quotient space

Theorem: The circle [ilmath]\mathbb{S}^1[/ilmath] is homeomorphic to [ilmath]\frac{\mathbb{R} }{\mathbb{Z} } [/ilmath]

Using the map above, we see that this just wraps the real line around the circle over and over again, specifically [ilmath]f(t_1)=f(t_2)\iff t_1-t_2\in\mathbb{S}[/ilmath], this suggests an Equivalence relation.

Using a bit of abstract algebra it is not hard to see that the equivalence classes are exactly the cosets of [ilmath]\mathbb{Z} [/ilmath] in [ilmath]\mathbb{R} [/ilmath]. So it is no problem to write [ilmath]\tfrac{\mathbb{R} }{\sim}=\tfrac{\mathbb{R} }{\mathbb{Z} }[/ilmath]


Using Passing to the quotient we see that [ilmath]\exists\bar{f} [/ilmath] that makes the diagram below commute if and only if [ilmath]t_1\sim t_2\implies f(t_1)=f(t_2)[/ilmath]

[math] \begin{xy} \xymatrix{ {\mathbb{R}} \ar[d] \ar[dr] &\\ {\frac{\mathbb{R}}{\mathbb{Z}}} \ar[r]_{\bar{f}} & {\mathbb{S}^1} } \end{xy} [/math]


[math] \begin{CD} R @= R \\ @V q V V @V Vf V \\ \frac{\mathbb{R}}{\mathbb{Z}} @> >\bar{f} > \mathbb{S}^1 \end{CD} [/math] (Triangle diagram wanted)

(Where [ilmath]\bar{f}:\frac{\mathbb{R} }{\mathbb{Z} }\rightarrow{\mathbb{S}^1} [/ilmath] is given by [ilmath]\bar{f}:[t]\rightarrow f(t)[/ilmath])


If [ilmath]\bar{f} [/ilmath] is a homeomorphism the result is shown.

  • [ilmath]\frac{\mathbb{R} }{\mathbb{Z} } [/ilmath] is compact as it is the image of a compact set, namely [ilmath][0,1][/ilmath] under [ilmath]q[/ilmath]
  • [ilmath]\mathbb{S}^1[/ilmath] is Hausdorff since it is a metric space and every metric space is Hausdorff.
  • [ilmath]f[/ilmath] is surjective, so as [ilmath]f=\bar{f}\circ q[/ilmath] and [ilmath]q[/ilmath] is surjective, [ilmath]\bar{f} [/ilmath] must be too.
    • Otherwise there'd be things [ilmath]f[/ilmath] maps to that [ilmath]\bar{f}\circ q[/ilmath] may not - contradicting the diagrams commute
  • [ilmath]\bar{f} [/ilmath] is injective
    • To be injective [ilmath]\bar{f}([t_1])=\bar{f}([t_2])\implies[t_1]=[t_2][/ilmath]
      • Showing that [ilmath]\bar{f} [/ilmath] is well defined
        Given [ilmath]a,b\in [t][/ilmath] we know [ilmath]a\sim b[/ilmath] as [ilmath][t][/ilmath] is an Equivalence class
        So this means [ilmath]f(a)=f(b)[/ilmath] because that's how we defined 'equivalent'
        Thus [ilmath]\forall a\in [t][\bar{f}([t])=f(a)][/ilmath] - so we can defined [ilmath]\bar{f} [/ilmath] unambiguously!
      • Using this we see [ilmath]\bar{f}([t_1])=f(t_1)[/ilmath] (choosing [ilmath]t_1[/ilmath] as the representative of [ilmath][t_1][/ilmath]) and
        [ilmath]\bar{f}([t_2])=f(t_2)[/ilmath], so we have [ilmath]f(t_1)=f(t_2)[/ilmath] so [ilmath]t_1\sim t_2[/ilmath]
        • Now we know [ilmath]t_1\in[t_1]\cap[t_2] [/ilmath] and [ilmath]t_2\in[t_1]cap[t_2] [/ilmath]
        As cosets are either disjoint or equal, and they're not disjoint! (we know [ilmath]t_1[/ilmath] is in the intersection even if [ilmath]t_1=t_2[/ilmath])
        so are equal - thus [ilmath]\bar{f} [/ilmath] is injective.
  • Thus [ilmath]\bar{f} [/ilmath] is a bijection

Using the Compact-to-Hausdorff theorem we conclude [ilmath]\bar{f} [/ilmath] is a homeomorphism


See also

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