User matthew dodelson - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T10:10:47Z http://mathoverflow.net/feeds/user/3372 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/118432/lax-pairs-for-linear-pdes Lax Pairs for Linear PDEs Matthew Dodelson 2013-01-09T09:43:05Z 2013-01-09T21:34:46Z <p>I'm trying to understand the discussion around equation (2.1) of the paper <a href="http://www.jstor.org/stable/53053" rel="nofollow">http://www.jstor.org/stable/53053</a>. It says that the linear PDE $M(\partial_x,\partial_y)q=0$ with constant coefficients has the Lax pair $\mu_x+ik\mu=q$ and $M(\partial_x,\partial_y)\mu=0$, where k is any complex number and $\mu$ is a function.</p> <p>The way I'm used to thinking of Lax pairs is as operators $L$ and $B$ such that $\dot{L}+[L,B]=0$ is equivalent to the original PDE. This is equivalent to requiring that the equations $L\phi=\lambda\phi$ and $\dot{\phi}=B\phi$ are compatible, where $\lambda$ is a fixed eigenvalue and $\phi$ is any function. Can anyone explain how this connects with the discussion in the paper? What are $L$ and $B$ in the above case?</p> <p>Thanks!</p> http://mathoverflow.net/questions/12248/gauge-connections-and-lie-algebras Gauge connections and Lie algebras? Matthew Dodelson 2010-01-18T22:55:20Z 2010-11-08T01:06:15Z <p>I'm probably missing something obvious, but I've been wondering what the motivation is for requiring the components <code>$A_\mu$</code> in a local trivialization of a gauge connection on a smooth principal <code>$G$</code>-bundle to lie in <code>$\mathfrak{g}$</code>, the Lie algebra of <code>$G$</code>. I can see that this gives a couple of nice properties; for example, in a local trivialization it ensures that under a gauge transformation <code>$A'_\mu=gA_\mu g^{-1}+g\partial_\mu g$</code> lies in <code>$\mathfrak{g}$</code>, and that the curvature form <code>$F=dA+A\wedge A$</code> lies in <code>$\mathfrak{g}$</code> (since <code>$\mathfrak{g}$</code> is closed under the Lie bracket). But is there a more intrinsic or geometric reason that <code>$A_\mu$</code> must be in <code>$\mathfrak{g}$</code>? Thanks.</p> http://mathoverflow.net/questions/12248/gauge-connections-and-lie-algebras/12257#12257 Comment by Matthew Dodelson Matthew Dodelson 2010-01-18T23:37:24Z 2010-01-18T23:37:24Z Exactly what I was looking for. Thanks! http://mathoverflow.net/questions/12248/gauge-connections-and-lie-algebras Comment by Matthew Dodelson Matthew Dodelson 2010-01-18T23:08:59Z 2010-01-18T23:08:59Z Yes, but I guess I'm asking what the reason/motivation is for including it in the definition.