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	<title>Comments on: Symmetry and the Fourth Dimension (Part 4)</title>
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	<link>http://johncarlosbaez.wordpress.com/2012/07/26/symmetry-and-the-fourth-dimension-part-4/</link>
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		<title>By: Weekly links for July 30 &#171; God plays dice</title>
		<link>http://johncarlosbaez.wordpress.com/2012/07/26/symmetry-and-the-fourth-dimension-part-4/#comment-17357</link>
		<dc:creator><![CDATA[Weekly links for July 30 &#171; God plays dice]]></dc:creator>
		<pubDate>Mon, 30 Jul 2012 17:51:50 +0000</pubDate>
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		<description><![CDATA[[...] couple series of posts from John Baez: Symmetry and the four dimension one, two, three, four. The Mathematics of Biodiversity eight-part [...]]]></description>
		<content:encoded><![CDATA[<p>[...] couple series of posts from John Baez: Symmetry and the four dimension one, two, three, four. The Mathematics of Biodiversity eight-part [...]</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: John Baez</title>
		<link>http://johncarlosbaez.wordpress.com/2012/07/26/symmetry-and-the-fourth-dimension-part-4/#comment-17309</link>
		<dc:creator><![CDATA[John Baez]]></dc:creator>
		<pubDate>Sun, 29 Jul 2012 01:39:22 +0000</pubDate>
		<guid isPermaLink="false">http://johncarlosbaez.wordpress.com/?p=10900#comment-17309</guid>
		<description><![CDATA[Greg Egan solved today&#039;s puzzle &lt;a href=&quot;http://johncarlosbaez.wordpress.com/2012/07/26/symmetry-and-the-fourth-dimension-part-4/#comment-17186&quot; rel=&quot;nofollow&quot;&gt;here&lt;/a&gt;, but there&#039;s another slightly different solution nobody has mentioned yet.  This involves an &lt;i&gt;equation&lt;/i&gt; instead of an inequality.  

Suppose are trying to classify all the Platonic solids. We&#039;re looking for ways to tile the surface of a sphere with regular $latex m$-gons, with $latex n$ meeting at each vertex. Suppose there are a total of $latex V$ vertices, $latex E$ edges, and $latex F$ faces. Since the Euler characteristic of the sphere is 2, we have

$latex V - E + F = 2$

Since each face has $latex m$ edges but 2 faces meet along each edge, we have

$latex mF = 2E$

Since each vertex has $latex m$ edges meeting it but each edge meets 2 vertices, we also have

$latex nV = 2E$

Putting these equations together we get

$latex \displaystyle{ 2E\left( \frac{1}{m} + \frac{1}{n} - \frac{1}{2} \right) = 2 }$

or

$latex \displaystyle{  \frac{1}{m} + \frac{1}{n} = \frac{1}{2} + \frac{1}{E} } $

Of course this implies the inequality we&#039;ve already seen:

$latex \displaystyle{  \frac{1}{m} + \frac{1}{n} &gt; \frac{1}{2} } $

&lt;i&gt;A priori&lt;/i&gt; the equation is stronger than the inequality, but it just happens to be equivalent, at least when 

$latex m, n \ge 2$

Whenever a sum of two reciprocals of two numbers like this exceeds 1/2, it exceeds 1/2 by a reciprocal of a number like this! And this number is the number of &lt;i&gt;edges&lt;/i&gt; of your Platonic solid...

... or hosohedron, or dihedron.

For example

$latex \displaystyle{  \frac{1}{4} + \frac{1}{3} &gt; \frac{1}{2} } $

gives the cube, but in fact

$latex \displaystyle{  \frac{1}{4} + \frac{1}{3} = \frac{7}{12} = \frac{1}{2} + \frac{1}{12} }  $

so the cube has 12 edges!  Simple but pretty stuff.]]></description>
		<content:encoded><![CDATA[<p>Greg Egan solved today&#8217;s puzzle <a href="http://johncarlosbaez.wordpress.com/2012/07/26/symmetry-and-the-fourth-dimension-part-4/#comment-17186" rel="nofollow">here</a>, but there&#8217;s another slightly different solution nobody has mentioned yet.  This involves an <i>equation</i> instead of an inequality.  </p>
<p>Suppose are trying to classify all the Platonic solids. We&#8217;re looking for ways to tile the surface of a sphere with regular <img src='http://s0.wp.com/latex.php?latex=m&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='m' title='m' class='latex' />-gons, with <img src='http://s0.wp.com/latex.php?latex=n&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='n' title='n' class='latex' /> meeting at each vertex. Suppose there are a total of <img src='http://s0.wp.com/latex.php?latex=V&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='V' title='V' class='latex' /> vertices, <img src='http://s0.wp.com/latex.php?latex=E&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='E' title='E' class='latex' /> edges, and <img src='http://s0.wp.com/latex.php?latex=F&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='F' title='F' class='latex' /> faces. Since the Euler characteristic of the sphere is 2, we have</p>
<p><img src='http://s0.wp.com/latex.php?latex=V+-+E+%2B+F+%3D+2&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='V - E + F = 2' title='V - E + F = 2' class='latex' /></p>
<p>Since each face has <img src='http://s0.wp.com/latex.php?latex=m&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='m' title='m' class='latex' /> edges but 2 faces meet along each edge, we have</p>
<p><img src='http://s0.wp.com/latex.php?latex=mF+%3D+2E&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='mF = 2E' title='mF = 2E' class='latex' /></p>
<p>Since each vertex has <img src='http://s0.wp.com/latex.php?latex=m&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='m' title='m' class='latex' /> edges meeting it but each edge meets 2 vertices, we also have</p>
<p><img src='http://s0.wp.com/latex.php?latex=nV+%3D+2E&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='nV = 2E' title='nV = 2E' class='latex' /></p>
<p>Putting these equations together we get</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle%7B+2E%5Cleft%28+%5Cfrac%7B1%7D%7Bm%7D+%2B+%5Cfrac%7B1%7D%7Bn%7D+-+%5Cfrac%7B1%7D%7B2%7D+%5Cright%29+%3D+2+%7D&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='&#92;displaystyle{ 2E&#92;left( &#92;frac{1}{m} + &#92;frac{1}{n} - &#92;frac{1}{2} &#92;right) = 2 }' title='&#92;displaystyle{ 2E&#92;left( &#92;frac{1}{m} + &#92;frac{1}{n} - &#92;frac{1}{2} &#92;right) = 2 }' class='latex' /></p>
<p>or</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle%7B++%5Cfrac%7B1%7D%7Bm%7D+%2B+%5Cfrac%7B1%7D%7Bn%7D+%3D+%5Cfrac%7B1%7D%7B2%7D+%2B+%5Cfrac%7B1%7D%7BE%7D+%7D+&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='&#92;displaystyle{  &#92;frac{1}{m} + &#92;frac{1}{n} = &#92;frac{1}{2} + &#92;frac{1}{E} } ' title='&#92;displaystyle{  &#92;frac{1}{m} + &#92;frac{1}{n} = &#92;frac{1}{2} + &#92;frac{1}{E} } ' class='latex' /></p>
<p>Of course this implies the inequality we&#8217;ve already seen:</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle%7B++%5Cfrac%7B1%7D%7Bm%7D+%2B+%5Cfrac%7B1%7D%7Bn%7D+%3E+%5Cfrac%7B1%7D%7B2%7D+%7D+&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='&#92;displaystyle{  &#92;frac{1}{m} + &#92;frac{1}{n} &gt; &#92;frac{1}{2} } ' title='&#92;displaystyle{  &#92;frac{1}{m} + &#92;frac{1}{n} &gt; &#92;frac{1}{2} } ' class='latex' /></p>
<p><i>A priori</i> the equation is stronger than the inequality, but it just happens to be equivalent, at least when </p>
<p><img src='http://s0.wp.com/latex.php?latex=m%2C+n+%5Cge+2&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='m, n &#92;ge 2' title='m, n &#92;ge 2' class='latex' /></p>
<p>Whenever a sum of two reciprocals of two numbers like this exceeds 1/2, it exceeds 1/2 by a reciprocal of a number like this! And this number is the number of <i>edges</i> of your Platonic solid&#8230;</p>
<p>&#8230; or hosohedron, or dihedron.</p>
<p>For example</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle%7B++%5Cfrac%7B1%7D%7B4%7D+%2B+%5Cfrac%7B1%7D%7B3%7D+%3E+%5Cfrac%7B1%7D%7B2%7D+%7D+&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='&#92;displaystyle{  &#92;frac{1}{4} + &#92;frac{1}{3} &gt; &#92;frac{1}{2} } ' title='&#92;displaystyle{  &#92;frac{1}{4} + &#92;frac{1}{3} &gt; &#92;frac{1}{2} } ' class='latex' /></p>
<p>gives the cube, but in fact</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle%7B++%5Cfrac%7B1%7D%7B4%7D+%2B+%5Cfrac%7B1%7D%7B3%7D+%3D+%5Cfrac%7B7%7D%7B12%7D+%3D+%5Cfrac%7B1%7D%7B2%7D+%2B+%5Cfrac%7B1%7D%7B12%7D+%7D++&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='&#92;displaystyle{  &#92;frac{1}{4} + &#92;frac{1}{3} = &#92;frac{7}{12} = &#92;frac{1}{2} + &#92;frac{1}{12} }  ' title='&#92;displaystyle{  &#92;frac{1}{4} + &#92;frac{1}{3} = &#92;frac{7}{12} = &#92;frac{1}{2} + &#92;frac{1}{12} }  ' class='latex' /></p>
<p>so the cube has 12 edges!  Simple but pretty stuff.</p>
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		<title>By: John Baez</title>
		<link>http://johncarlosbaez.wordpress.com/2012/07/26/symmetry-and-the-fourth-dimension-part-4/#comment-17308</link>
		<dc:creator><![CDATA[John Baez]]></dc:creator>
		<pubDate>Sun, 29 Jul 2012 01:14:08 +0000</pubDate>
		<guid isPermaLink="false">http://johncarlosbaez.wordpress.com/?p=10900#comment-17308</guid>
		<description><![CDATA[In response to a puzzle last time, &lt;a href=&quot;http://johncarlosbaez.wordpress.com/2012/07/22/symmetry-and-the-fourth-dimension-part-3/#comment-17303&quot; rel=&quot;nofollow&quot;&gt;Chris Namaste&lt;/a&gt; has worked out the Coxeter diagrams of all the regular tilings of the plane:

&lt;div align=&quot;center&quot;&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Square_tiling&quot; rel=&quot;nofollow&quot;&gt;square tiling&lt;/a&gt;:&lt;br /&gt; &lt;b&gt;V---4---E---4---F&lt;/b&gt;&lt;br /&gt;&lt;br /&gt;&lt;/div&gt;

&lt;div align=&quot;center&quot;&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Square_tiling&quot; rel=&quot;nofollow&quot;&gt;&lt;img src=&quot;http://math.ucr.edu/home/baez/mathematical/square_tiling.png&quot; /&gt;&lt;/a&gt;&lt;/div&gt;
&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;

&lt;div align=&quot;center&quot;&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Triangular_tiling&quot; rel=&quot;nofollow&quot;&gt;triangular tiling&lt;/a&gt;:&lt;br /&gt; &lt;b&gt;V---3---E---6---F&lt;/b&gt;&lt;br /&gt;&lt;br /&gt;&lt;/div&gt;

&lt;div align=&quot;center&quot;&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Triangular_tiling&quot; rel=&quot;nofollow&quot;&gt;&lt;img src=&quot;http://math.ucr.edu/home/baez/mathematical/triangular_tiling.png&quot; /&gt;&lt;/a&gt;&lt;/div&gt;
&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;

&lt;div align=&quot;center&quot;&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Hexagonal_tiling&quot; rel=&quot;nofollow&quot;&gt;hexagonal tiling&lt;/a&gt;:&lt;br /&gt; &lt;b&gt;V---6---E---3---F&lt;/b&gt;&lt;br /&gt;&lt;br /&gt;&lt;/div&gt;

&lt;div align=&quot;center&quot;&gt;&lt;a href=&quot;http://en.wikipedia.org/wiki/Hexagonal_tiling&quot; rel=&quot;nofollow&quot;&gt;&lt;img src=&quot;http://math.ucr.edu/home/baez/mathematical/hexagonal_tiling.png&quot; /&gt;&lt;/a&gt;&lt;/div&gt;
&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;

The square tiling is self-dual; the other two are dual to each other.

These tilings correspond to the solutions of

$latex \displaystyle{ \frac{1}{m} + \frac{1}{n} = \frac{1}{2}}$

where $latex m, n$ are positive integers.]]></description>
		<content:encoded><![CDATA[<p>In response to a puzzle last time, <a href="http://johncarlosbaez.wordpress.com/2012/07/22/symmetry-and-the-fourth-dimension-part-3/#comment-17303" rel="nofollow">Chris Namaste</a> has worked out the Coxeter diagrams of all the regular tilings of the plane:</p>
<div align="center"><a href="http://en.wikipedia.org/wiki/Square_tiling" rel="nofollow">square tiling</a>:<br /> <b>V&#8212;4&#8212;E&#8212;4&#8212;F</b></p>
</div>
<div align="center"><a href="http://en.wikipedia.org/wiki/Square_tiling" rel="nofollow"><img src="http://math.ucr.edu/home/baez/mathematical/square_tiling.png" /></a></div>
<p></p>
<div align="center"><a href="http://en.wikipedia.org/wiki/Triangular_tiling" rel="nofollow">triangular tiling</a>:<br /> <b>V&#8212;3&#8212;E&#8212;6&#8212;F</b></p>
</div>
<div align="center"><a href="http://en.wikipedia.org/wiki/Triangular_tiling" rel="nofollow"><img src="http://math.ucr.edu/home/baez/mathematical/triangular_tiling.png" /></a></div>
<p></p>
<div align="center"><a href="http://en.wikipedia.org/wiki/Hexagonal_tiling" rel="nofollow">hexagonal tiling</a>:<br /> <b>V&#8212;6&#8212;E&#8212;3&#8212;F</b></p>
</div>
<div align="center"><a href="http://en.wikipedia.org/wiki/Hexagonal_tiling" rel="nofollow"><img src="http://math.ucr.edu/home/baez/mathematical/hexagonal_tiling.png" /></a></div>
<p></p>
<p>The square tiling is self-dual; the other two are dual to each other.</p>
<p>These tilings correspond to the solutions of</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle%7B+%5Cfrac%7B1%7D%7Bm%7D+%2B+%5Cfrac%7B1%7D%7Bn%7D+%3D+%5Cfrac%7B1%7D%7B2%7D%7D&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='&#92;displaystyle{ &#92;frac{1}{m} + &#92;frac{1}{n} = &#92;frac{1}{2}}' title='&#92;displaystyle{ &#92;frac{1}{m} + &#92;frac{1}{n} = &#92;frac{1}{2}}' class='latex' /></p>
<p>where <img src='http://s0.wp.com/latex.php?latex=m%2C+n&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='m, n' title='m, n' class='latex' /> are positive integers.</p>
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		<title>By: John Baez</title>
		<link>http://johncarlosbaez.wordpress.com/2012/07/26/symmetry-and-the-fourth-dimension-part-4/#comment-17307</link>
		<dc:creator><![CDATA[John Baez]]></dc:creator>
		<pubDate>Sun, 29 Jul 2012 01:05:01 +0000</pubDate>
		<guid isPermaLink="false">http://johncarlosbaez.wordpress.com/?p=10900#comment-17307</guid>
		<description><![CDATA[Those notes by Qi Phillip are very nice---thanks!  I wish they&#039;d been around when I was just learning this stuff!  And his handwriting, if that&#039;s what is, looks like it&#039;s &lt;i&gt;typed&lt;/i&gt; in a special font.]]></description>
		<content:encoded><![CDATA[<p>Those notes by Qi Phillip are very nice&#8212;thanks!  I wish they&#8217;d been around when I was just learning this stuff!  And his handwriting, if that&#8217;s what is, looks like it&#8217;s <i>typed</i> in a special font.</p>
]]></content:encoded>
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		<title>By: John Baez</title>
		<link>http://johncarlosbaez.wordpress.com/2012/07/26/symmetry-and-the-fourth-dimension-part-4/#comment-17288</link>
		<dc:creator><![CDATA[John Baez]]></dc:creator>
		<pubDate>Sat, 28 Jul 2012 11:30:39 +0000</pubDate>
		<guid isPermaLink="false">http://johncarlosbaez.wordpress.com/?p=10900#comment-17288</guid>
		<description><![CDATA[Tobias wrote:

&lt;blockquote&gt;
That&#039;s weird.
&lt;/blockquote&gt;

Yes, the modularity theorem is weirdly self-referential.   I&#039;ve always been puzzled why everyone explaining this stuff doesn&#039;t mention that.

It&#039;s even more amusing that this self-referential result is what it took to prove something seemingly &#039;concrete&#039; like Fermat&#039;s Last Theorem.  As it happened, &lt;a href=&quot;http://en.wikipedia.org/wiki/Frey_curve&quot; rel=&quot;nofollow&quot;&gt;Gerard Frey&lt;/a&gt; suggested that a counterexample to Fermat&#039;s Last Theorem 

$latex a^n + b^n = c^n$

would give an elliptic curve

$latex y^2 = x(x - a^n)(x + b^n)$

 that couldn&#039;t be covered by a modular one.   This was later proved by Jean-Pierre Serre and Kenneth Ribet.  And that made it clear what had do be done to prove Fermat&#039;s Last Theorem: prove the modularity theorem, or at least enough of it to cover this case.]]></description>
		<content:encoded><![CDATA[<p>Tobias wrote:</p>
<blockquote><p>
That&#8217;s weird.
</p></blockquote>
<p>Yes, the modularity theorem is weirdly self-referential.   I&#8217;ve always been puzzled why everyone explaining this stuff doesn&#8217;t mention that.</p>
<p>It&#8217;s even more amusing that this self-referential result is what it took to prove something seemingly &#8216;concrete&#8217; like Fermat&#8217;s Last Theorem.  As it happened, <a href="http://en.wikipedia.org/wiki/Frey_curve" rel="nofollow">Gerard Frey</a> suggested that a counterexample to Fermat&#8217;s Last Theorem </p>
<p><img src='http://s0.wp.com/latex.php?latex=a%5En+%2B+b%5En+%3D+c%5En&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='a^n + b^n = c^n' title='a^n + b^n = c^n' class='latex' /></p>
<p>would give an elliptic curve</p>
<p><img src='http://s0.wp.com/latex.php?latex=y%5E2+%3D+x%28x+-+a%5En%29%28x+%2B+b%5En%29&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='y^2 = x(x - a^n)(x + b^n)' title='y^2 = x(x - a^n)(x + b^n)' class='latex' /></p>
<p> that couldn&#8217;t be covered by a modular one.   This was later proved by Jean-Pierre Serre and Kenneth Ribet.  And that made it clear what had do be done to prove Fermat&#8217;s Last Theorem: prove the modularity theorem, or at least enough of it to cover this case.</p>
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		<title>By: Tobias Fritz</title>
		<link>http://johncarlosbaez.wordpress.com/2012/07/26/symmetry-and-the-fourth-dimension-part-4/#comment-17286</link>
		<dc:creator><![CDATA[Tobias Fritz]]></dc:creator>
		<pubDate>Sat, 28 Jul 2012 11:00:49 +0000</pubDate>
		<guid isPermaLink="false">http://johncarlosbaez.wordpress.com/?p=10900#comment-17286</guid>
		<description><![CDATA[John explained:
&lt;blockquote&gt;Modular curves can be seen as parametrizing families of elliptic curves. [..] Sometimes an elliptic curve can be covered by a modular curve using a so-called ‘branched cover’. There’s a big theorem called the modularity theorem which says (very roughly) that all elliptic curves defined by equations using just rational numbers can be covered in this way by modular curves.&lt;/blockquote&gt;
That&#039;s weird: a modular curve, equipped with some additional structure defining the &#039;branched quotient&#039; and this torsion subgroup iso, becomes itself a *point* in a modular curve! (In general, I suppose that this second curve is in general different from the first.)]]></description>
		<content:encoded><![CDATA[<p>John explained:</p>
<blockquote><p>Modular curves can be seen as parametrizing families of elliptic curves. [..] Sometimes an elliptic curve can be covered by a modular curve using a so-called ‘branched cover’. There’s a big theorem called the modularity theorem which says (very roughly) that all elliptic curves defined by equations using just rational numbers can be covered in this way by modular curves.</p></blockquote>
<p>That&#8217;s weird: a modular curve, equipped with some additional structure defining the &#8216;branched quotient&#8217; and this torsion subgroup iso, becomes itself a *point* in a modular curve! (In general, I suppose that this second curve is in general different from the first.)</p>
]]></content:encoded>
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	<item>
		<title>By: John Baez</title>
		<link>http://johncarlosbaez.wordpress.com/2012/07/26/symmetry-and-the-fourth-dimension-part-4/#comment-17270</link>
		<dc:creator><![CDATA[John Baez]]></dc:creator>
		<pubDate>Sat, 28 Jul 2012 06:25:22 +0000</pubDate>
		<guid isPermaLink="false">http://johncarlosbaez.wordpress.com/?p=10900#comment-17270</guid>
		<description><![CDATA[Oh, I was talking about the essence of ADE theory, not all its ramifications and applications...   

I guess you&#039;ve seen John McKay&#039;s &lt;a href=&quot;http://math.ucr.edu/home/baez/ADE.html&quot; rel=&quot;nofollow&quot;&gt;terse blast of information&lt;/a&gt; on this subject.  It mainly serves to make you want to learn more.]]></description>
		<content:encoded><![CDATA[<p>Oh, I was talking about the essence of ADE theory, not all its ramifications and applications&#8230;   </p>
<p>I guess you&#8217;ve seen John McKay&#8217;s <a href="http://math.ucr.edu/home/baez/ADE.html" rel="nofollow">terse blast of information</a> on this subject.  It mainly serves to make you want to learn more.</p>
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		<title>By: John Baez</title>
		<link>http://johncarlosbaez.wordpress.com/2012/07/26/symmetry-and-the-fourth-dimension-part-4/#comment-17264</link>
		<dc:creator><![CDATA[John Baez]]></dc:creator>
		<pubDate>Sat, 28 Jul 2012 03:06:54 +0000</pubDate>
		<guid isPermaLink="false">http://johncarlosbaez.wordpress.com/?p=10900#comment-17264</guid>
		<description><![CDATA[Well, I sort of changed course midstream, switching from my intended goal to something I find easier to understand.  If I&#039;d kept marching boldly ahead, I would have said something like this (but longer, and maybe more helpful, though maybe less):

Since Klein&#039;s quartic curve is a curled-up piece of the hyperbolic plane, it&#039;s called a &lt;a href=&quot;http://en.wikipedia.org/wiki/Modular_curve&quot; rel=&quot;nofollow&quot;&gt;&#039;modular curve&#039;&lt;/a&gt;.  Modular curves can be seen as parametrizing families of elliptic curves (roughly speaking, &lt;a href=&quot;http://en.wikipedia.org/wiki/Modular_curve&quot; rel=&quot;nofollow&quot;&gt;tori&lt;/a&gt;) equipped with extra structure.  

This is a nice big story already.  However, it then takes a weird turn and gets much more intense.

There&#039;s another weirder relation between modular curves and elliptic curves.  Sometimes an elliptic curve can be &lt;i&gt;covered&lt;/i&gt; by a modular curve using a so-called &#039;branched cover&#039;.   There&#039;s a big theorem called the &lt;a href=&quot;http://en.wikipedia.org/wiki/Modularity_theorem&quot; rel=&quot;nofollow&quot;&gt;modularity theorem&lt;/a&gt; which says (very roughly) that all elliptic curves defined by equations using just rational numbers can be covered in this way by modular curves. 

And this theorem implies Fermat&#039;s Last Theorem!

So, as usual in number theory, the flashy easy-to-explain result follows as a corollary from something that&#039;s harder to explain, but ultimately more interesting and more connected to &lt;i&gt;geometry&lt;/i&gt; and &lt;i&gt;symmetry&lt;/i&gt;. 

The proof of the modularity theorem is not easy: Andrew Wiles and a grad student of his proved enough of it to get Fermat&#039;s Last Theorem, and other good mathematicians finished off the job.   

Instead of reading about that, it&#039;s a lot less stressful to start with:

&#8226; Tim Silverman, &lt;a href=&quot;http://golem.ph.utexas.edu/category/2010/10/test.html&quot; rel=&quot;nofollow&quot;&gt;Pictures of Modular Curves (Part I)&lt;/a&gt;, &lt;i&gt;n&lt;/i&gt;-Category Caf&amp;eeacute;, 10 October 2006.

and read &lt;a href=&quot;https://www.google.com/search?q=silverman+pictures+of+modular+curves&quot; rel=&quot;nofollow&quot;&gt;all 11 parts&lt;/a&gt;.   You&#039;ll get a deeper understanding of creatures like

&lt;div align=&quot;center&quot;&gt;&lt;b&gt; V---8---F---3---E&lt;/b&gt;&lt;/div&gt;

as shown here:

&lt;div align=&quot;center&quot;&gt;&lt;a href=&quot;http://golem.ph.utexas.edu/category/2010/11/pictures_of_modular_curves_iii.html&quot; rel=&quot;nofollow&quot;&gt;&lt;img width=&quot;500&quot; src=&quot;http://www.tisloc.me.uk/Farey%20Pictures/N=8.png&quot; /&gt;&lt;/a&gt;&lt;/div&gt;

]]></description>
		<content:encoded><![CDATA[<p>Well, I sort of changed course midstream, switching from my intended goal to something I find easier to understand.  If I&#8217;d kept marching boldly ahead, I would have said something like this (but longer, and maybe more helpful, though maybe less):</p>
<p>Since Klein&#8217;s quartic curve is a curled-up piece of the hyperbolic plane, it&#8217;s called a <a href="http://en.wikipedia.org/wiki/Modular_curve" rel="nofollow">&#8216;modular curve&#8217;</a>.  Modular curves can be seen as parametrizing families of elliptic curves (roughly speaking, <a href="http://en.wikipedia.org/wiki/Modular_curve" rel="nofollow">tori</a>) equipped with extra structure.  </p>
<p>This is a nice big story already.  However, it then takes a weird turn and gets much more intense.</p>
<p>There&#8217;s another weirder relation between modular curves and elliptic curves.  Sometimes an elliptic curve can be <i>covered</i> by a modular curve using a so-called &#8216;branched cover&#8217;.   There&#8217;s a big theorem called the <a href="http://en.wikipedia.org/wiki/Modularity_theorem" rel="nofollow">modularity theorem</a> which says (very roughly) that all elliptic curves defined by equations using just rational numbers can be covered in this way by modular curves. </p>
<p>And this theorem implies Fermat&#8217;s Last Theorem!</p>
<p>So, as usual in number theory, the flashy easy-to-explain result follows as a corollary from something that&#8217;s harder to explain, but ultimately more interesting and more connected to <i>geometry</i> and <i>symmetry</i>. </p>
<p>The proof of the modularity theorem is not easy: Andrew Wiles and a grad student of his proved enough of it to get Fermat&#8217;s Last Theorem, and other good mathematicians finished off the job.   </p>
<p>Instead of reading about that, it&#8217;s a lot less stressful to start with:</p>
<p>&bull; Tim Silverman, <a href="http://golem.ph.utexas.edu/category/2010/10/test.html" rel="nofollow">Pictures of Modular Curves (Part I)</a>, <i>n</i>-Category Caf&amp;eeacute;, 10 October 2006.</p>
<p>and read <a href="https://www.google.com/search?q=silverman+pictures+of+modular+curves" rel="nofollow">all 11 parts</a>.   You&#8217;ll get a deeper understanding of creatures like</p>
<div align="center"><b> V&#8212;8&#8212;F&#8212;3&#8212;E</b></div>
<p>as shown here:</p>
<div align="center"><a href="http://golem.ph.utexas.edu/category/2010/11/pictures_of_modular_curves_iii.html" rel="nofollow"><img width="500" src="http://www.tisloc.me.uk/Farey%20Pictures/N=8.png" /></a></div>
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		<title>By: John Baez</title>
		<link>http://johncarlosbaez.wordpress.com/2012/07/26/symmetry-and-the-fourth-dimension-part-4/#comment-17263</link>
		<dc:creator><![CDATA[John Baez]]></dc:creator>
		<pubDate>Sat, 28 Jul 2012 02:53:51 +0000</pubDate>
		<guid isPermaLink="false">http://johncarlosbaez.wordpress.com/?p=10900#comment-17263</guid>
		<description><![CDATA[Later in this series we&#039;ll certainly be using Coxeter diagrams to help classify Platonic solids in 4 dimensions.   They&#039;re also good for classifying &#039;tilings&#039; of 3d Euclidean space and 3d hyperbolic space by polyhedra.  In case that mixture of 4&#039;s and 3&#039;s annoys you: these examples are actually all living in the same dimension, since a 4d Platonic solid is a special tiling of a 3-sphere by polyhedra.  Spherical, planar and hyperbolic geometry always go hand in hand in mathematics.

But, understanding which Coxeter diagrams give which things isn&#039;t simply a matter of taking sums of reciprocals.  We&#039;ll see, I guess!]]></description>
		<content:encoded><![CDATA[<p>Later in this series we&#8217;ll certainly be using Coxeter diagrams to help classify Platonic solids in 4 dimensions.   They&#8217;re also good for classifying &#8217;tilings&#8217; of 3d Euclidean space and 3d hyperbolic space by polyhedra.  In case that mixture of 4&#8242;s and 3&#8242;s annoys you: these examples are actually all living in the same dimension, since a 4d Platonic solid is a special tiling of a 3-sphere by polyhedra.  Spherical, planar and hyperbolic geometry always go hand in hand in mathematics.</p>
<p>But, understanding which Coxeter diagrams give which things isn&#8217;t simply a matter of taking sums of reciprocals.  We&#8217;ll see, I guess!</p>
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		<title>By: Robert</title>
		<link>http://johncarlosbaez.wordpress.com/2012/07/26/symmetry-and-the-fourth-dimension-part-4/#comment-17261</link>
		<dc:creator><![CDATA[Robert]]></dc:creator>
		<pubDate>Fri, 27 Jul 2012 23:24:56 +0000</pubDate>
		<guid isPermaLink="false">http://johncarlosbaez.wordpress.com/?p=10900#comment-17261</guid>
		<description><![CDATA[Your example doesn&#039;t look very scary to me. That&#039;s nice. Next time i&#039;m lost in the jargon i&#039;ll try and imagine klein&#039;s quartic rolling over the hyperbolic plane. I can now see how spheres are limited and why a cylinder is a bit too plain for this kind of fun. Thanks for the insight, oh and for the link to the Congruence Groups.]]></description>
		<content:encoded><![CDATA[<p>Your example doesn&#8217;t look very scary to me. That&#8217;s nice. Next time i&#8217;m lost in the jargon i&#8217;ll try and imagine klein&#8217;s quartic rolling over the hyperbolic plane. I can now see how spheres are limited and why a cylinder is a bit too plain for this kind of fun. Thanks for the insight, oh and for the link to the Congruence Groups.</p>
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