Difference between revisions of "Circle"
(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== | ||
| − | A circle is usually defined by {{M|1=\mathcal{ | + | 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\} }} |
| + | |||
| + | ==Topological perspective== | ||
| + | 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}} | ||
| + | |||
| + | ===The circle as a quotient space=== | ||
| + | {{Begin Theorem}} | ||
| + | Theorem: The circle {{M|\mathbb{S}^1}} is [[Homeomorphism|homeomorphic]] to {{M|\frac{\mathbb{R} }{\mathbb{Z} } }} | ||
| + | {{Begin Proof}} | ||
| + | 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]]. | ||
| + | |||
| + | 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} } }} | ||
| + | |||
| + | |||
| + | 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)}} | ||
| + | |||
| + | |||
| + | [math] | ||
| + | \begin{CD} | ||
| + | R {{CD Hoz Eq}} R \\ | ||
| + | {{CD Down Arrow|q}} {{CD Down Arrow||f}} \\ | ||
| + | \frac{\mathbb{R}}{\mathbb{Z}} {{CD Right Arrow||\bar{f} }} \mathbb{S}^1 | ||
| + | \end{CD} | ||
| + | [/math] {{Triangle Wanted}} | ||
| + | |||
| + | (Where {{M|\bar{f}:\frac{\mathbb{R} }{\mathbb{Z} }\rightarrow{\mathbb{S}^1} }} is given by {{M|\bar{f}:[t]\rightarrow f(t)}}) | ||
| + | |||
| + | |||
| + | If {{M|\bar{f} }} is a homeomorphism the result is shown. | ||
| + | |||
| + | * {{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}} | ||
| + | * {{M|\mathbb{S}^1}} is [[Hausdorff]] since it is a [[Metric space|metric space]] and every metric space is Hausdorff. | ||
| + | * {{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. | ||
| + | ** Otherwise there'd be things {{M|f}} maps to that {{M|\bar{f}\circ q}} may not - contradicting the diagrams commute | ||
| + | * {{M|\bar{f} }} is [[Injection|injective]] | ||
| + | ** To be injective {{M|1=\bar{f}([t_1])=\bar{f}([t_2])\implies[t_1]=[t_2]}} | ||
| + | *** Showing that {{M|\bar{f} }} is well defined | ||
| + | ***: Given {{M|a,b\in [t]}} we know {{M|a\sim b}} as {{M|[t]}} is an [[Equivalence class]] | ||
| + | ***: So this means {{M|1=f(a)=f(b)}} because that's how we defined 'equivalent' | ||
| + | ***: Thus {{M|1=\forall a\in [t][\bar{f}([t])=f(a)]}} - so we can defined {{M|\bar{f} }} unambiguously! | ||
| + | *** 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 | ||
| + | ***: {{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}} | ||
| + | **** Now we know {{M|t_1\in[t_1]\cap[t_2] }} and {{M|t_2\in[t_1]cap[t_2] }} | ||
| + | ***: 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}}) | ||
| + | ***: so are equal - thus {{M|\bar{f} }} is injective. | ||
| + | * Thus {{M|\bar{f} }} is a [[Bijection|bijection]] | ||
| + | |||
| + | Using the [[Compact-to-Hausdorff theorem]] we conclude {{M|\bar{f} }} is a '''homeomorphism''' | ||
| + | {{End Proof}} | ||
| + | {{End Theorem}} | ||
==See also== | ==See also== | ||
Revision as of 18:53, 16 April 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{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.
- Showing that [ilmath]\bar{f} [/ilmath] is well defined
- To be injective [ilmath]\bar{f}([t_1])=\bar{f}([t_2])\implies[t_1]=[t_2][/ilmath]
- Thus [ilmath]\bar{f} [/ilmath] is a bijection
Using the Compact-to-Hausdorff theorem we conclude [ilmath]\bar{f} [/ilmath] is a homeomorphism
