monodromy group

Consider an ordinary linear differential equation

$$\sum _{k=0}^n{c}_k(x)\frac{d^{n}y}{d{x}^n}=0$$

in which the coefficients are polynomials. If $c_n$is not constant, then it is possible that the solutionsof this equation will have branch points at the zerosof $c_n$. (To see if this actually happens, we need toexamine the indicial equation.)

By the persistence of differential equations, the analyticcontinuation of a solution of this equation will be anothersolution. Pick a neighborhood which does not contain anyzeros of $c_n$. Since the differential equation is of order$n$, there will be $n$ independent solutions $y=f_1(x),\mathrm{\dots },y=f_n(x)$. (For example, one may exhibit thesesolutions as power series about some point in the neighborhood.)

Upon analytic continuation back to the original neighborhood viaa chain of neghborhoods, supposethat the solution $y=f_i(x)$ is taken to a solution $y=g_i(x)$.Because the solutions are linearly independent, there will exist a matrix${\{m_{ij}\}}_{i,j=1}^n$ such that

$$g_i=\sum _{j=1}^n{m}_{ij}{f}_j.$$

Now consider the totality of all such matrices corresponding toall possible ways of making analytic continuations along chainswhich begin and end wit the original neighborhood. They form agroup known as the monodromy group of the differentialequation. The reason this set is a group is some basic factsabout analytic continuation. First, there is the trivial analyticcontinuation which simply takes a function element to itself. Thiswill correspond to the identity matrix. Second, we can reverse aprocess of analytic continuation. This will correspond to the inversematrix. Third, we can follow continuation along one chain ofneighborhoods by continuation along another chain. This willcorrespond to multiplying the matrices corresponding to the two chains.