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	<title>Comments on: Network Theory (Part 22)</title>
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	<link>http://johncarlosbaez.wordpress.com/2012/08/16/network-theory-part-22/</link>
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	<item>
		<title>By: John Baez</title>
		<link>http://johncarlosbaez.wordpress.com/2012/08/16/network-theory-part-22/#comment-18670</link>
		<dc:creator><![CDATA[John Baez]]></dc:creator>
		<pubDate>Wed, 22 Aug 2012 11:41:34 +0000</pubDate>
		<guid isPermaLink="false">http://johncarlosbaez.wordpress.com/?p=11586#comment-18670</guid>
		<description><![CDATA[Dan wrote:

&lt;blockquote&gt;
I’d love to see the rigorous proof showing the rate equation as the limit of the master equation...
&lt;/blockquote&gt;

Yes, that&#039;s something that&#039;s missing in these posts, which really should be included.  I haven&#039;t actually written up such a proof, and I haven&#039;t seen one anywhere though it seems like an obvious thing for people to have studied.  You sketched the basic idea quite nicely---and based on that idea, my &#039;analysis brain&#039; feels completely confident it can jump in and supply all the necessary epsilons and deltas, and hypotheses, that will lead to a nice theorem.  Getting the strongest possible theorem would be hard!  Getting a theorem that illustrates the idea should not be hard.  But I won&#039;t try to do it now... that&#039;s more like a blog post on its own, and one that will take about a week to write.

A crucial point is that the solution of the master equation will only be close to that of the rate equation if the &#039;spread&#039; of the probability distributions of the numbers of particles of each species is sufficiently small, as well as their means being large.   However, as time passes, the spread will tend to grow.  So, the rate equation will only be a good approximation to the master equation for a certain amount of time, not forever... except in certain very special cases.

For example, there are stochastic reaction networks whose rate equations have periodic solutions which never approach any sort of equilibrium.  However, the solutions of the corresponding master equation will probably &#039;smear out&#039; and approach an equilibrium, at least in many cases.

&lt;blockquote&gt;
I’m not sure what sort of convergence to try and whether I need to make additional assumptions, like possibly a condition on the variance of N?
&lt;/blockquote&gt;

You certainly need something like that, but probably more. I did a bunch of calculations at one point and it seems I needed not just bounds on the variance but also the higher moments.  Basically the trick is just to calculate the difference between what you want and what you&#039;ve got, and see what&#039;s required to make that be small!]]></description>
		<content:encoded><![CDATA[<p>Dan wrote:</p>
<blockquote><p>
I’d love to see the rigorous proof showing the rate equation as the limit of the master equation&#8230;
</p></blockquote>
<p>Yes, that&#8217;s something that&#8217;s missing in these posts, which really should be included.  I haven&#8217;t actually written up such a proof, and I haven&#8217;t seen one anywhere though it seems like an obvious thing for people to have studied.  You sketched the basic idea quite nicely&#8212;and based on that idea, my &#8216;analysis brain&#8217; feels completely confident it can jump in and supply all the necessary epsilons and deltas, and hypotheses, that will lead to a nice theorem.  Getting the strongest possible theorem would be hard!  Getting a theorem that illustrates the idea should not be hard.  But I won&#8217;t try to do it now&#8230; that&#8217;s more like a blog post on its own, and one that will take about a week to write.</p>
<p>A crucial point is that the solution of the master equation will only be close to that of the rate equation if the &#8216;spread&#8217; of the probability distributions of the numbers of particles of each species is sufficiently small, as well as their means being large.   However, as time passes, the spread will tend to grow.  So, the rate equation will only be a good approximation to the master equation for a certain amount of time, not forever&#8230; except in certain very special cases.</p>
<p>For example, there are stochastic reaction networks whose rate equations have periodic solutions which never approach any sort of equilibrium.  However, the solutions of the corresponding master equation will probably &#8216;smear out&#8217; and approach an equilibrium, at least in many cases.</p>
<blockquote><p>
I’m not sure what sort of convergence to try and whether I need to make additional assumptions, like possibly a condition on the variance of N?
</p></blockquote>
<p>You certainly need something like that, but probably more. I did a bunch of calculations at one point and it seems I needed not just bounds on the variance but also the higher moments.  Basically the trick is just to calculate the difference between what you want and what you&#8217;ve got, and see what&#8217;s required to make that be small!</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: Dan</title>
		<link>http://johncarlosbaez.wordpress.com/2012/08/16/network-theory-part-22/#comment-18668</link>
		<dc:creator><![CDATA[Dan]]></dc:creator>
		<pubDate>Wed, 22 Aug 2012 11:22:01 +0000</pubDate>
		<guid isPermaLink="false">http://johncarlosbaez.wordpress.com/?p=11586#comment-18668</guid>
		<description><![CDATA[Thank you for the detailed reply to a very vague question.  It hit exactly on what was confusing me, namely,  the $latex H$ you introduced above is (in retrospect, obviously) different from the hamiltonian that gives rise to the Master equation that the rate equation is a large number approximation for.  

But of course, the danger of a detailed reply is that it often raises other questions that can get you off topic.  So, rather than ask an off topic question, I&#039;ll just put in a request: I&#039;d love to see the rigorous proof showing the rate equation as the limit of the Master equation.  Defining $latex x=\sum N \Psi$, using some notation from a ways back, it&#039;s straightforward algebra to show that the Master equation implies something very much like the rate equation, only with falling powers in place of ordinary powers.  And intuitively, for large N those are the same.  But from there, I&#039;m not sure what sort of convergence to try and whether I need to make additional assumptions, like possibly a condition on the variance of $latex N$?]]></description>
		<content:encoded><![CDATA[<p>Thank you for the detailed reply to a very vague question.  It hit exactly on what was confusing me, namely,  the <img src='http://s0.wp.com/latex.php?latex=H&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='H' title='H' class='latex' /> you introduced above is (in retrospect, obviously) different from the hamiltonian that gives rise to the Master equation that the rate equation is a large number approximation for.  </p>
<p>But of course, the danger of a detailed reply is that it often raises other questions that can get you off topic.  So, rather than ask an off topic question, I&#8217;ll just put in a request: I&#8217;d love to see the rigorous proof showing the rate equation as the limit of the Master equation.  Defining <img src='http://s0.wp.com/latex.php?latex=x%3D%5Csum+N+%5CPsi&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='x=&#92;sum N &#92;Psi' title='x=&#92;sum N &#92;Psi' class='latex' />, using some notation from a ways back, it&#8217;s straightforward algebra to show that the Master equation implies something very much like the rate equation, only with falling powers in place of ordinary powers.  And intuitively, for large N those are the same.  But from there, I&#8217;m not sure what sort of convergence to try and whether I need to make additional assumptions, like possibly a condition on the variance of <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' />?</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: John Baez</title>
		<link>http://johncarlosbaez.wordpress.com/2012/08/16/network-theory-part-22/#comment-18630</link>
		<dc:creator><![CDATA[John Baez]]></dc:creator>
		<pubDate>Tue, 21 Aug 2012 06:06:39 +0000</pubDate>
		<guid isPermaLink="false">http://johncarlosbaez.wordpress.com/?p=11586#comment-18630</guid>
		<description><![CDATA[Ruggero wrote:

&lt;blockquote&gt;
This is like an eigenstate/eigenvector of the matrix power operator, and the whole story looks like a generalization of the usual ‘linear’ version of it, very common in linear algebra or quantum mechanics. 
&lt;/blockquote&gt;

Indeed this &#039;generalization&#039; is closely related to the usual linear story.  In the climactic moment of Part 24, I will suddenly pull out this formula:

$latex x^Y = \exp(Y^\dagger \ln(x)) $

and use it to reduce a problem involving $latex x^Y$ to a problem involving ordinary matrix multiplication.  

I&#039;ll explain this formula when I need it, but in fact I&#039;ve already begun to explain it &lt;a href=&quot;http://johncarlosbaez.wordpress.com/2012/08/16/network-theory-part-22/#comment-18597&quot; rel=&quot;nofollow&quot;&gt;down here&lt;/a&gt; in the comments.   The above formula would look nicer if I let myself multiply a vector by a matrix on the right:

$latex x^Y = \exp(\ln(x) Y) $

and it would look even nicer, perhaps, if I took the componentwise logarithm of each side:

$latex \ln(x^Y) = \ln(x) Y$

Anyway, this means that tools developed for matrix multiplication can be applied to matrix exponentiation.]]></description>
		<content:encoded><![CDATA[<p>Ruggero wrote:</p>
<blockquote><p>
This is like an eigenstate/eigenvector of the matrix power operator, and the whole story looks like a generalization of the usual ‘linear’ version of it, very common in linear algebra or quantum mechanics.
</p></blockquote>
<p>Indeed this &#8216;generalization&#8217; is closely related to the usual linear story.  In the climactic moment of Part 24, I will suddenly pull out this formula:</p>
<p><img src='http://s0.wp.com/latex.php?latex=x%5EY+%3D+%5Cexp%28Y%5E%5Cdagger+%5Cln%28x%29%29+&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='x^Y = &#92;exp(Y^&#92;dagger &#92;ln(x)) ' title='x^Y = &#92;exp(Y^&#92;dagger &#92;ln(x)) ' class='latex' /></p>
<p>and use it to reduce a problem involving <img src='http://s0.wp.com/latex.php?latex=x%5EY&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='x^Y' title='x^Y' class='latex' /> to a problem involving ordinary matrix multiplication.  </p>
<p>I&#8217;ll explain this formula when I need it, but in fact I&#8217;ve already begun to explain it <a href="http://johncarlosbaez.wordpress.com/2012/08/16/network-theory-part-22/#comment-18597" rel="nofollow">down here</a> in the comments.   The above formula would look nicer if I let myself multiply a vector by a matrix on the right:</p>
<p><img src='http://s0.wp.com/latex.php?latex=x%5EY+%3D+%5Cexp%28%5Cln%28x%29+Y%29+&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='x^Y = &#92;exp(&#92;ln(x) Y) ' title='x^Y = &#92;exp(&#92;ln(x) Y) ' class='latex' /></p>
<p>and it would look even nicer, perhaps, if I took the componentwise logarithm of each side:</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cln%28x%5EY%29+%3D+%5Cln%28x%29+Y&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='&#92;ln(x^Y) = &#92;ln(x) Y' title='&#92;ln(x^Y) = &#92;ln(x) Y' class='latex' /></p>
<p>Anyway, this means that tools developed for matrix multiplication can be applied to matrix exponentiation.</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: John Baez</title>
		<link>http://johncarlosbaez.wordpress.com/2012/08/16/network-theory-part-22/#comment-18629</link>
		<dc:creator><![CDATA[John Baez]]></dc:creator>
		<pubDate>Tue, 21 Aug 2012 06:00:14 +0000</pubDate>
		<guid isPermaLink="false">http://johncarlosbaez.wordpress.com/?p=11586#comment-18629</guid>
		<description><![CDATA[Okay.  If you have any questions, don&#039;t be shy.  The series is building up to a climax: in Part 24 I&#039;ll prove the deficiency zero theorem, or at least the part of it that I want to prove.  But as usual in math, this climax involves putting together lots of ideas that were discussed earlier... and as usual in math, most people will be left scratching their heads and wondering what just happened.  &lt;img src=&quot;http://math.ucr.edu/home/baez/emoticons/tongue2.gif&quot; alt=&quot;&quot; /&gt;   At least in a blog, unlike a textbook, one can ask questions or complain.]]></description>
		<content:encoded><![CDATA[<p>Okay.  If you have any questions, don&#8217;t be shy.  The series is building up to a climax: in Part 24 I&#8217;ll prove the deficiency zero theorem, or at least the part of it that I want to prove.  But as usual in math, this climax involves putting together lots of ideas that were discussed earlier&#8230; and as usual in math, most people will be left scratching their heads and wondering what just happened.  <img src="http://math.ucr.edu/home/baez/emoticons/tongue2.gif" alt="" />   At least in a blog, unlike a textbook, one can ask questions or complain.</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: John Baez</title>
		<link>http://johncarlosbaez.wordpress.com/2012/08/16/network-theory-part-22/#comment-18628</link>
		<dc:creator><![CDATA[John Baez]]></dc:creator>
		<pubDate>Tue, 21 Aug 2012 05:54:35 +0000</pubDate>
		<guid isPermaLink="false">http://johncarlosbaez.wordpress.com/?p=11586#comment-18628</guid>
		<description><![CDATA[Dan wrote:

&lt;blockquote&gt;
Where are the raising and lowering operators hiding in this formalism? 
&lt;/blockquote&gt;

If you watched the earlier episodes of this show, you know that raising and lowering operators, and Fock space, are extremely important in studying the &lt;a href=&quot;http://math.ucr.edu/home/baez/networks/networks_6.html&quot; rel=&quot;nofollow&quot;&gt;master equation&lt;/a&gt; for a stochastic reaction network (aka stochastic Petri net).    

But now we&#039;re talking about the rate equation.  This is like the &#039;classical limit&#039; of the master equation, or more precisely its &#039;large number limit&#039;.

The master equation is all about the probability that we have some number of things of each species.  These numbers are natural numbers, so they jump discretely... but we only know the &lt;i&gt;probability&lt;/i&gt; that these numbers are this or that, and these probabilities vary smoothly with the passage of time.

The rate equation is all about the number of things of each species.  Now we act like we know them for sure, and that they&#039;re positive real numbers, and that they vary smoothly with the passage of time.

That&#039;s all fun... and we can rigorously derive the rate equation from the master equation in a certain limit where there are large numbers of things.  

But in &lt;a href=&quot;http://johncarlosbaez.wordpress.com/2012/08/21/network-theory-part-23/&quot; rel=&quot;nofollow&quot;&gt;Part 23&lt;/a&gt; we&#039;ll see an amazing extra twist!  In our quest to prove the deficiency zero theorem, we&#039;ll introduce &lt;i&gt;another&lt;/i&gt; master equation, which describes a Markov process on the set of complexes (which are bunches of things of our various species).   The dynamics of this Markov process are in general different than that of the rate equation... but we&#039;ll get equilibrium solutions of the rate equation from equilibria of this Markov process!

Operators like $latex \partial$ and $latex s^\dagger$ play a huge role in constructing the Hamiltonian for this Markov process on complexes.  But they are not, as far as I can tell, directly related to operators on the Fock space we saw earlier. 

But of course, this is the sort of thing I&#039;m still busy trying to understand.

By the way, any Markov process gives rise to a stochastic Petri net of a special sort, where each transition has just one input and one output.   The space of states for the stochastic Petri net is the &#039;Fock space&#039; of that for the Markov process.   In other words, the space of states for the Markov process is the &#039;single-particle space&#039; of the space of states for the stochastic Petri net.   And here&#039;s what you&#039;ll like: time evolution for the Petri net is obtained by applying Reed and Simon&#039;s $latex \Gamma$, or at least its stochastic analogue, to time evolution for the Markov process!   

I&#039;ve been meaning to talk about this for a long time.  But this is &lt;i&gt;not&lt;/i&gt; one of the games I&#039;m playing right now.]]></description>
		<content:encoded><![CDATA[<p>Dan wrote:</p>
<blockquote><p>
Where are the raising and lowering operators hiding in this formalism?
</p></blockquote>
<p>If you watched the earlier episodes of this show, you know that raising and lowering operators, and Fock space, are extremely important in studying the <a href="http://math.ucr.edu/home/baez/networks/networks_6.html" rel="nofollow">master equation</a> for a stochastic reaction network (aka stochastic Petri net).    </p>
<p>But now we&#8217;re talking about the rate equation.  This is like the &#8216;classical limit&#8217; of the master equation, or more precisely its &#8216;large number limit&#8217;.</p>
<p>The master equation is all about the probability that we have some number of things of each species.  These numbers are natural numbers, so they jump discretely&#8230; but we only know the <i>probability</i> that these numbers are this or that, and these probabilities vary smoothly with the passage of time.</p>
<p>The rate equation is all about the number of things of each species.  Now we act like we know them for sure, and that they&#8217;re positive real numbers, and that they vary smoothly with the passage of time.</p>
<p>That&#8217;s all fun&#8230; and we can rigorously derive the rate equation from the master equation in a certain limit where there are large numbers of things.  </p>
<p>But in <a href="http://johncarlosbaez.wordpress.com/2012/08/21/network-theory-part-23/" rel="nofollow">Part 23</a> we&#8217;ll see an amazing extra twist!  In our quest to prove the deficiency zero theorem, we&#8217;ll introduce <i>another</i> master equation, which describes a Markov process on the set of complexes (which are bunches of things of our various species).   The dynamics of this Markov process are in general different than that of the rate equation&#8230; but we&#8217;ll get equilibrium solutions of the rate equation from equilibria of this Markov process!</p>
<p>Operators like <img src='http://s0.wp.com/latex.php?latex=%5Cpartial&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='&#92;partial' title='&#92;partial' class='latex' /> and <img src='http://s0.wp.com/latex.php?latex=s%5E%5Cdagger&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='s^&#92;dagger' title='s^&#92;dagger' class='latex' /> play a huge role in constructing the Hamiltonian for this Markov process on complexes.  But they are not, as far as I can tell, directly related to operators on the Fock space we saw earlier. </p>
<p>But of course, this is the sort of thing I&#8217;m still busy trying to understand.</p>
<p>By the way, any Markov process gives rise to a stochastic Petri net of a special sort, where each transition has just one input and one output.   The space of states for the stochastic Petri net is the &#8216;Fock space&#8217; of that for the Markov process.   In other words, the space of states for the Markov process is the &#8216;single-particle space&#8217; of the space of states for the stochastic Petri net.   And here&#8217;s what you&#8217;ll like: time evolution for the Petri net is obtained by applying Reed and Simon&#8217;s <img src='http://s0.wp.com/latex.php?latex=%5CGamma&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='&#92;Gamma' title='&#92;Gamma' class='latex' />, or at least its stochastic analogue, to time evolution for the Markov process!   </p>
<p>I&#8217;ve been meaning to talk about this for a long time.  But this is <i>not</i> one of the games I&#8217;m playing right now.</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: John Baez</title>
		<link>http://johncarlosbaez.wordpress.com/2012/08/16/network-theory-part-22/#comment-18598</link>
		<dc:creator><![CDATA[John Baez]]></dc:creator>
		<pubDate>Mon, 20 Aug 2012 09:32:56 +0000</pubDate>
		<guid isPermaLink="false">http://johncarlosbaez.wordpress.com/?p=11586#comment-18598</guid>
		<description><![CDATA[Dan wrote:

&lt;blockquote&gt;
Okay, well, I’ll take another stab at it then. If you think of vectors as column vectors (i.e., as $latex n\times 1$ matrices) then the two definitions coincide (assuming you identify $latex 1\times 1$ vectors with numbers), so that raising a vector to a vector is a special case of raising a vector to a matrix. Am I getting warmer or colder?
&lt;/blockquote&gt;

You&#039;re &lt;font color=&quot;red&quot;&gt;hot!&lt;/font&gt;

That&#039;s the answer I was fishing for: when you raise a vector to a vector power and get a number, it&#039;s a &lt;i&gt;special case&lt;/i&gt; of raising an $latex \ell \times n$ matrix to the power of an $latex n \times m$ matrix and getting an $latex \ell \times m$ matrix.  Just take $latex \ell = m = 1$.  

It just like how when you take the dot product of two vectors and get a number, you can see it as a special case of matrix multiplication.   Think of the first vector as a row vector: a $latex 1 \times n$ matrix.  Think of the second vector as a column vector: a $latex n \times 1$ matrix.  Multiply these matrices and presto, you get a $latex 1 \times 1$ matrix, or number.

Of course later, when we get sophisticated, we like to think of column vectors and row vectors as living in dual vector spaces, not the same vector space, and then this operation is called, not the &#039;dot product&#039;, but the pairing between a vector and a dual vector.  

But never mind: all that sophisticated stuff is designed to ultimately make things basis-independent, but matrix exponentiation definitely &lt;i&gt;does&lt;/i&gt; depend on a choice of basis.  That&#039;s okay, because our vectors here are really functions on a finite set: the set of complexes, say, or the set of species.  So, we have god-given bases, and we get to use them.]]></description>
		<content:encoded><![CDATA[<p>Dan wrote:</p>
<blockquote><p>
Okay, well, I’ll take another stab at it then. If you think of vectors as column vectors (i.e., as <img src='http://s0.wp.com/latex.php?latex=n%5Ctimes+1&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='n&#92;times 1' title='n&#92;times 1' class='latex' /> matrices) then the two definitions coincide (assuming you identify <img src='http://s0.wp.com/latex.php?latex=1%5Ctimes+1&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='1&#92;times 1' title='1&#92;times 1' class='latex' /> vectors with numbers), so that raising a vector to a vector is a special case of raising a vector to a matrix. Am I getting warmer or colder?
</p></blockquote>
<p>You&#8217;re <font color="red">hot!</font></p>
<p>That&#8217;s the answer I was fishing for: when you raise a vector to a vector power and get a number, it&#8217;s a <i>special case</i> of raising an <img src='http://s0.wp.com/latex.php?latex=%5Cell+%5Ctimes+n&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='&#92;ell &#92;times n' title='&#92;ell &#92;times n' class='latex' /> matrix to the power of an <img src='http://s0.wp.com/latex.php?latex=n+%5Ctimes+m&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='n &#92;times m' title='n &#92;times m' class='latex' /> matrix and getting an <img src='http://s0.wp.com/latex.php?latex=%5Cell+%5Ctimes+m&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='&#92;ell &#92;times m' title='&#92;ell &#92;times m' class='latex' /> matrix.  Just take <img src='http://s0.wp.com/latex.php?latex=%5Cell+%3D+m+%3D+1&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='&#92;ell = m = 1' title='&#92;ell = m = 1' class='latex' />.  </p>
<p>It just like how when you take the dot product of two vectors and get a number, you can see it as a special case of matrix multiplication.   Think of the first vector as a row vector: a <img src='http://s0.wp.com/latex.php?latex=1+%5Ctimes+n&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='1 &#92;times n' title='1 &#92;times n' class='latex' /> matrix.  Think of the second vector as a column vector: a <img src='http://s0.wp.com/latex.php?latex=n+%5Ctimes+1&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='n &#92;times 1' title='n &#92;times 1' class='latex' /> matrix.  Multiply these matrices and presto, you get a <img src='http://s0.wp.com/latex.php?latex=1+%5Ctimes+1&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='1 &#92;times 1' title='1 &#92;times 1' class='latex' /> matrix, or number.</p>
<p>Of course later, when we get sophisticated, we like to think of column vectors and row vectors as living in dual vector spaces, not the same vector space, and then this operation is called, not the &#8216;dot product&#8217;, but the pairing between a vector and a dual vector.  </p>
<p>But never mind: all that sophisticated stuff is designed to ultimately make things basis-independent, but matrix exponentiation definitely <i>does</i> depend on a choice of basis.  That&#8217;s okay, because our vectors here are really functions on a finite set: the set of complexes, say, or the set of species.  So, we have god-given bases, and we get to use them.</p>
]]></content:encoded>
	</item>
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		<title>By: John Baez</title>
		<link>http://johncarlosbaez.wordpress.com/2012/08/16/network-theory-part-22/#comment-18597</link>
		<dc:creator><![CDATA[John Baez]]></dc:creator>
		<pubDate>Mon, 20 Aug 2012 09:24:36 +0000</pubDate>
		<guid isPermaLink="false">http://johncarlosbaez.wordpress.com/?p=11586#comment-18597</guid>
		<description><![CDATA[Over on Google+, I admitted that with matrix exponentiation as defined here, $latex A^0$ needn&#039;t be an identity matrix.

Dayal D. Purohit replied:

&lt;blockquote&gt; In that case, in what sense will it be called &#039;matrix exponentiation&#039;? Does it satisfy any rule of exponentiation at all?﻿
&lt;/blockquote&gt;

I responded, a wee bit tetchily:

&lt;blockquote&gt;
I don&#039;t really care what this operation is called: what matters is its applications to chemical reaction networks. 

There&#039;s no operation on pairs of matrices that obeys all, or even most, of the usual rules of exponentiation.  The rules obeyed by this operation  are listed in Appendix A here:

• Jonathan M. Guberman, &lt;a href=&quot;http://www4.utsouthwestern.edu/altschulerwulab/theses/GubermanJ_Thesis.pdf&quot; rel=&quot;nofollow&quot;&gt;&lt;i&gt;Mass Action Reaction Networks and the Deficiency Zero Theorem&lt;/i&gt;&lt;/a&gt;, B.A. thesis, Department of Mathematics, Harvard University, 2003.  

One of the most important for applications is this.  There&#039;s a &#039;logarithm&#039; operation defined component-wise on vectors with positive components, and an &#039;exponential&#039; operation defined component-wise on all vectors, and then

$latex \ln(x^A) = \ln(x) A$

Here $latex x$ is a vector with $latex n$ components, which we can think of as a $latex 1 \times n$ matrix.  $latex A$ is an $latex n \times m$ matrix, $latex x^A$ is the $latex 1 \times m$ matrix given by matrix exponentiation as defined in this paper, and $latex \ln(x) A$ is given by ordinary matrix multiplication.

The reason this is important is that a list of concentrations of chemicals can be seen as a vector with nonnegative (and often positive) components.

But I also like this kind of matrix exponentiation because it arises from the usual concept of an &lt;a href=&quot;http://en.wikipedia.org/wiki/Exponential_object&quot; rel=&quot;nofollow&quot;&gt;&#039;exponential object&#039;&lt;/a&gt; in category theory.
&lt;/blockquote&gt;]]></description>
		<content:encoded><![CDATA[<p>Over on Google+, I admitted that with matrix exponentiation as defined here, <img src='http://s0.wp.com/latex.php?latex=A%5E0&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='A^0' title='A^0' class='latex' /> needn&#8217;t be an identity matrix.</p>
<p>Dayal D. Purohit replied:</p>
<blockquote><p> In that case, in what sense will it be called &#8216;matrix exponentiation&#8217;? Does it satisfy any rule of exponentiation at all?﻿
</p></blockquote>
<p>I responded, a wee bit tetchily:</p>
<blockquote><p>
I don&#8217;t really care what this operation is called: what matters is its applications to chemical reaction networks. </p>
<p>There&#8217;s no operation on pairs of matrices that obeys all, or even most, of the usual rules of exponentiation.  The rules obeyed by this operation  are listed in Appendix A here:</p>
<p>• Jonathan M. Guberman, <a href="http://www4.utsouthwestern.edu/altschulerwulab/theses/GubermanJ_Thesis.pdf" rel="nofollow"><i>Mass Action Reaction Networks and the Deficiency Zero Theorem</i></a>, B.A. thesis, Department of Mathematics, Harvard University, 2003.  </p>
<p>One of the most important for applications is this.  There&#8217;s a &#8216;logarithm&#8217; operation defined component-wise on vectors with positive components, and an &#8216;exponential&#8217; operation defined component-wise on all vectors, and then</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cln%28x%5EA%29+%3D+%5Cln%28x%29+A&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='&#92;ln(x^A) = &#92;ln(x) A' title='&#92;ln(x^A) = &#92;ln(x) A' class='latex' /></p>
<p>Here <img src='http://s0.wp.com/latex.php?latex=x&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='x' title='x' class='latex' /> is a vector 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' /> components, which we can think of as a <img src='http://s0.wp.com/latex.php?latex=1+%5Ctimes+n&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='1 &#92;times n' title='1 &#92;times n' class='latex' /> matrix.  <img src='http://s0.wp.com/latex.php?latex=A&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='A' title='A' class='latex' /> is an <img src='http://s0.wp.com/latex.php?latex=n+%5Ctimes+m&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='n &#92;times m' title='n &#92;times m' class='latex' /> matrix, <img src='http://s0.wp.com/latex.php?latex=x%5EA&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='x^A' title='x^A' class='latex' /> is the <img src='http://s0.wp.com/latex.php?latex=1+%5Ctimes+m&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='1 &#92;times m' title='1 &#92;times m' class='latex' /> matrix given by matrix exponentiation as defined in this paper, and <img src='http://s0.wp.com/latex.php?latex=%5Cln%28x%29+A&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='&#92;ln(x) A' title='&#92;ln(x) A' class='latex' /> is given by ordinary matrix multiplication.</p>
<p>The reason this is important is that a list of concentrations of chemicals can be seen as a vector with nonnegative (and often positive) components.</p>
<p>But I also like this kind of matrix exponentiation because it arises from the usual concept of an <a href="http://en.wikipedia.org/wiki/Exponential_object" rel="nofollow">&#8216;exponential object&#8217;</a> in category theory.
</p></blockquote>
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		<title>By: davidtweed</title>
		<link>http://johncarlosbaez.wordpress.com/2012/08/16/network-theory-part-22/#comment-18593</link>
		<dc:creator><![CDATA[davidtweed]]></dc:creator>
		<pubDate>Mon, 20 Aug 2012 06:58:36 +0000</pubDate>
		<guid isPermaLink="false">http://johncarlosbaez.wordpress.com/?p=11586#comment-18593</guid>
		<description><![CDATA[Continuing the thought in my (currently awaiting approval) comment, if we get an SVD-like decomposition $latex Y=\sum_i a^i {b^i}^T$ where we&#039;ve absorbed the diagonal term of the SVD into the $latex b$s (say), then then if $latex x^{a^i}=z_i$, then the contribution to first component is $latex z_i^{b^i_1}$, the contribution to the second component is $latex z_i^{b^i_2}$, etc. So &quot;having computed the $latex z_i$s&quot;, the $latex j$th component of $latex x^Y$ is $latex \prod_i z_i^{b^j_i}$. Not sure if this actually helps understand possible behaviours much better though...]]></description>
		<content:encoded><![CDATA[<p>Continuing the thought in my (currently awaiting approval) comment, if we get an SVD-like decomposition <img src='http://s0.wp.com/latex.php?latex=Y%3D%5Csum_i+a%5Ei+%7Bb%5Ei%7D%5ET&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='Y=&#92;sum_i a^i {b^i}^T' title='Y=&#92;sum_i a^i {b^i}^T' class='latex' /> where we&#8217;ve absorbed the diagonal term of the SVD into the <img src='http://s0.wp.com/latex.php?latex=b&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='b' title='b' class='latex' />s (say), then then if <img src='http://s0.wp.com/latex.php?latex=x%5E%7Ba%5Ei%7D%3Dz_i&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='x^{a^i}=z_i' title='x^{a^i}=z_i' class='latex' />, then the contribution to first component is <img src='http://s0.wp.com/latex.php?latex=z_i%5E%7Bb%5Ei_1%7D&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='z_i^{b^i_1}' title='z_i^{b^i_1}' class='latex' />, the contribution to the second component is <img src='http://s0.wp.com/latex.php?latex=z_i%5E%7Bb%5Ei_2%7D&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='z_i^{b^i_2}' title='z_i^{b^i_2}' class='latex' />, etc. So &#8220;having computed the <img src='http://s0.wp.com/latex.php?latex=z_i&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='z_i' title='z_i' class='latex' />s&#8221;, the <img src='http://s0.wp.com/latex.php?latex=j&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='j' title='j' class='latex' />th component of <img src='http://s0.wp.com/latex.php?latex=x%5EY&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='x^Y' title='x^Y' class='latex' /> is <img src='http://s0.wp.com/latex.php?latex=%5Cprod_i+z_i%5E%7Bb%5Ej_i%7D&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='&#92;prod_i z_i^{b^j_i}' title='&#92;prod_i z_i^{b^j_i}' class='latex' />. Not sure if this actually helps understand possible behaviours much better though&#8230;</p>
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		<title>By: Ruggero</title>
		<link>http://johncarlosbaez.wordpress.com/2012/08/16/network-theory-part-22/#comment-18590</link>
		<dc:creator><![CDATA[Ruggero]]></dc:creator>
		<pubDate>Mon, 20 Aug 2012 02:52:15 +0000</pubDate>
		<guid isPermaLink="false">http://johncarlosbaez.wordpress.com/?p=11586#comment-18590</guid>
		<description><![CDATA[Puzzle: if I see things correctly, as you say a vector raised to a vector power gives you a number, which is similar to a dot product, while a vector raised to a matrix power, gives you a vector, which is similar to a vector (external) product, since the dimensionality can be whatever you want. But there is something more I think that goes around here, I think. 

In fact, let&#039;s assume that you choose a square matrix so that the initial vector and the final vector both live on the same space, then I&#039;d guess you may find a bunch of matrices which leave the initial vector unchanged up to an overall constant. This is like an eigenstate/eigenvector of the matrix power operator, and the whole story looks like a generalization of the usual &#039;linear&#039; version of it, very common in linear algebra or quantum mechanics. You can find your projectors, orthogonal spaces, and so on, and even do a little group theory using matrices that will switch around components of the initial vector and so on.

I&#039;m too tired right now to actually try, but it seems very intriguing since the generators of the typical groups (rotations etc) must have the same algebra by definition but I think will have a different form, since the way the algebra acts on them is not the usual matrix multiplication but this power multiplications. It would be fun to see how this generators relate to the &#039;linear&#039; ones.

In some sense the S matrix in particle physics represents all the possible interactions between an initial and a final state, and now I&#039;m starting thinking whether this formulation could be useful in physics in general or at least in cosmology before the last scattering or so, when all the fundamental interactions may look more like chemistry. 

Maybe I&#039;m really to tired hahaha, anyhow I hope I&#039;m not seeing ghosts here, and thanks a lot for the very cool post!]]></description>
		<content:encoded><![CDATA[<p>Puzzle: if I see things correctly, as you say a vector raised to a vector power gives you a number, which is similar to a dot product, while a vector raised to a matrix power, gives you a vector, which is similar to a vector (external) product, since the dimensionality can be whatever you want. But there is something more I think that goes around here, I think. </p>
<p>In fact, let&#8217;s assume that you choose a square matrix so that the initial vector and the final vector both live on the same space, then I&#8217;d guess you may find a bunch of matrices which leave the initial vector unchanged up to an overall constant. This is like an eigenstate/eigenvector of the matrix power operator, and the whole story looks like a generalization of the usual &#8216;linear&#8217; version of it, very common in linear algebra or quantum mechanics. You can find your projectors, orthogonal spaces, and so on, and even do a little group theory using matrices that will switch around components of the initial vector and so on.</p>
<p>I&#8217;m too tired right now to actually try, but it seems very intriguing since the generators of the typical groups (rotations etc) must have the same algebra by definition but I think will have a different form, since the way the algebra acts on them is not the usual matrix multiplication but this power multiplications. It would be fun to see how this generators relate to the &#8216;linear&#8217; ones.</p>
<p>In some sense the S matrix in particle physics represents all the possible interactions between an initial and a final state, and now I&#8217;m starting thinking whether this formulation could be useful in physics in general or at least in cosmology before the last scattering or so, when all the fundamental interactions may look more like chemistry. </p>
<p>Maybe I&#8217;m really to tired hahaha, anyhow I hope I&#8217;m not seeing ghosts here, and thanks a lot for the very cool post!</p>
]]></content:encoded>
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		<title>By: Dan</title>
		<link>http://johncarlosbaez.wordpress.com/2012/08/16/network-theory-part-22/#comment-18583</link>
		<dc:creator><![CDATA[Dan]]></dc:creator>
		<pubDate>Sun, 19 Aug 2012 22:57:04 +0000</pubDate>
		<guid isPermaLink="false">http://johncarlosbaez.wordpress.com/?p=11586#comment-18583</guid>
		<description><![CDATA[On further reflection, that question is overly convoluted.  So, let me try a simpler version: Where are the raising and lowering operators hiding in this formalism?  Is it another map that isn&#039;t present here or do they arise from one of the maps you&#039;ve introduced (or perhaps a composition of some of the maps you&#039;ve introduced)?]]></description>
		<content:encoded><![CDATA[<p>On further reflection, that question is overly convoluted.  So, let me try a simpler version: Where are the raising and lowering operators hiding in this formalism?  Is it another map that isn&#8217;t present here or do they arise from one of the maps you&#8217;ve introduced (or perhaps a composition of some of the maps you&#8217;ve introduced)?</p>
]]></content:encoded>
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