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	<title>Comments on: Norm of traceless Ricci tensor</title>
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	<link>http://anhngq.wordpress.com/2011/05/30/norm-of-traceless-ricci-tensor/</link>
	<description>Học học nữa học mãi</description>
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		<title>By: Ngô Quốc Anh</title>
		<link>http://anhngq.wordpress.com/2011/05/30/norm-of-traceless-ricci-tensor/#comment-764</link>
		<dc:creator><![CDATA[Ngô Quốc Anh]]></dc:creator>
		<pubDate>Wed, 01 Jun 2011 11:19:43 +0000</pubDate>
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		<description><![CDATA[Ah yes, that&#039;s interesting. I was thinking in a traditional way, by calculus only. Thank you for your beautiful approach.]]></description>
		<content:encoded><![CDATA[<p>Ah yes, that&#8217;s interesting. I was thinking in a traditional way, by calculus only. Thank you for your beautiful approach.</p>
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		<title>By: K</title>
		<link>http://anhngq.wordpress.com/2011/05/30/norm-of-traceless-ricci-tensor/#comment-763</link>
		<dc:creator><![CDATA[K]]></dc:creator>
		<pubDate>Wed, 01 Jun 2011 11:16:52 +0000</pubDate>
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		<description><![CDATA[I think a slightly tidier way to prove it, is to see that 

$latex \displaystyle\mbox{Ric} -f g=\stackrel{\circ}{\mbox{Ric}}+(R/n-f)g$ 

and 

$latex \displaystyle g\perp \stackrel{\circ}{\mbox{Ric}},$

so the identity follows from Pythagoras theorem.]]></description>
		<content:encoded><![CDATA[<p>I think a slightly tidier way to prove it, is to see that </p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle%5Cmbox%7BRic%7D+-f+g%3D%5Cstackrel%7B%5Ccirc%7D%7B%5Cmbox%7BRic%7D%7D%2B%28R%2Fn-f%29g&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='&#92;displaystyle&#92;mbox{Ric} -f g=&#92;stackrel{&#92;circ}{&#92;mbox{Ric}}+(R/n-f)g' title='&#92;displaystyle&#92;mbox{Ric} -f g=&#92;stackrel{&#92;circ}{&#92;mbox{Ric}}+(R/n-f)g' class='latex' /> </p>
<p>and </p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+g%5Cperp+%5Cstackrel%7B%5Ccirc%7D%7B%5Cmbox%7BRic%7D%7D%2C&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='&#92;displaystyle g&#92;perp &#92;stackrel{&#92;circ}{&#92;mbox{Ric}},' title='&#92;displaystyle g&#92;perp &#92;stackrel{&#92;circ}{&#92;mbox{Ric}},' class='latex' /></p>
<p>so the identity follows from Pythagoras theorem.</p>
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