User matthew dodelson - MathOverflow

Lax Pairs for Linear PDEs

2013-01-09T09:43:05Z

I'm trying to understand the discussion around equation (2.1) of the paper http://www.jstor.org/stable/53053. 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.

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?

Thanks!

Gauge connections and Lie algebras?

2010-01-18T22:55:20Z

I'm probably missing something obvious, but I've been wondering what the motivation is for requiring the components $A_\mu$ in a local trivialization of a gauge connection on a smooth principal $G$ -bundle to lie in $\mathfrak{g}$ , the Lie algebra of $G$ . I can see that this gives a couple of nice properties; for example, in a local trivialization it ensures that under a gauge transformation $A'_\mu=gA_\mu g^{-1}+g\partial_\mu g$ lies in $\mathfrak{g}$ , and that the curvature form $F=dA+A\wedge A$ lies in $\mathfrak{g}$ (since $\mathfrak{g}$ is closed under the Lie bracket). But is there a more intrinsic or geometric reason that $A_\mu$ must be in $\mathfrak{g}$ ? Thanks.

Comment by Matthew Dodelson

2010-01-18T23:37:24Z

Exactly what I was looking for. Thanks!

Comment by Matthew Dodelson

2010-01-18T23:08:59Z

Yes, but I guess I'm asking what the reason/motivation is for including it in the definition.