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For the differential operator $$ D := i I \frac{d}{dx} + A(x) \colon C^\infty_T([0,\beta]),\mathbb{C}^m) \to C^\infty_T ([0,\beta],\mathbb{C}^m) $$ where $A(x)$ is Hermitian and $C^\infty_T [0,\beta]$ denotes the linear subspace in $C^\infty [0,\beta]$ consisting of all functions that satisfy $ f(0) = Tf(\beta)$ for some $T \in GL(n,\mathbb{C})$ ( I will denote this bvp by $D_T$ ):

I think I am ok to find the kernel of the inverse $D_T^{-1}$. For $p \colon [0,\beta] \to GL(n,\mathbb{C})$ a differentiable map that satisfies $Dp = 0$, $p(0) = I$ I get

\begin{equation} k(x,y) = \begin{cases} -ip(x) (I-Th)^{-1}Th p(y)^{-1} & x<y \\ -ip(x) (I-Th)^{-1} p(y)^{-1}& x>y \end{cases} \end{equation} where $h = p(\beta)$.

My question is how the kernel would be affected if I now conjugate the operator with a unitary gauge transformation, i.e. if I consider $ g^{-1}D_Tg $ for $g:[0,\beta] \to U(n)$.

I have next to no experience working with gauge transformations, so any hints or references to get me started would be highly appreciated. In particular is it the case that I can simply conjugate the respective expressions in the kernel above to get the new kernel for the conjugated operator ?

share|cite|improve this question
what is $h$ in your kernel ? – harlekin Apr 6 '14 at 12:28
@harlekin thanks, question edited accordingly. – Beltrame Apr 6 '14 at 14:58

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