Let $A:\mathbb{C}^*\to\mathrm{M}_n(\mathbb{C})$ be a holomorphic map. Consider the system of first order differential equations \begin{equation} \begin{cases} \frac{dY}{dz} = A Y\\ Y(1)= I, \end{cases} \end{equation} on $\mathbb{C}^*$ and assume that $z=0$ is an irregular singularity.

If $Y$ is a local solution of this system, it can be analytically continued along any path in $\mathbb{C}^*$, and the monodromy theorem implies that $\tilde{Y}=Y\circ\exp$ extends to a holomorphic function in $\tilde{\mathbb{C}^*}=\mathbb{C}$. Note that $\exp:\mathbb{C}\to\mathbb{C}^*$ is the universal cover. Take $z_0\in\mathbb{C}^*$ and $\tilde z_0\in\mathbb{C}$ such that $\exp(\tilde z_0)=z_0$, say $\tilde z_0=0$ and $z_0=1$.

Let me drop for a moment the usual notation "$\tilde{\cdot}$" for the universal cover. For any $n\in\mathbb{Z}$ the map $\tau_n:\mathbb{C}\to\mathbb{C}$ given by $z\mapsto z+2\pi in$ is a deck transformation. Then, one has that $$\tilde Y(\tau_n(\tilde z))=\tilde Y(\tilde z)M_{\tau_n},$$ where $M_{\tau_n}$ is called monodromy matrix.

I am unsure about two statements regarding this monodromy matrix:

  • Is it true that $M_{\tau_n}=\tilde{Y}(\tau_n(\tilde z_0))$ which is equal to $\tilde{Y}(2\pi in)$ in our case? From what I have read I would say this is the case if $z=0$ is a regular singularity, but I'm not sure if it also holds for irregular singularities.
  • Suppose that I continue the solution along a different path, this gives a different monodromy. In particular, suppose that I take now the deck transformation $\tau_{-n}$ that sends $z$ in the universal cover to $z-2\pi i n$. Is there any relation between the monodromy matrices $M_{\tau_n}$ and $M_{\tau_{-n}}$. I would expect something like $$M_{\tau_n}^{-1}=M_{\tau_{-n}},$$ but I don't know how to prove it.

Thanks in advance for any idea or help.

  • $\begingroup$ Did you mean $Y′(z)=A(z)Y(z)$ where $Y(z),A(z) \in M_n(\mathbb{C})$, if so then the point is that each column of the solution is linear in each column of the initial condition, so with $Y(z e^{2i \pi k})$ the continuation of $Y$ after $k$ loops around $z=0$ then $Y(e^{2i \pi})=MY(1) \implies Y(z e^{2i \pi})=MY(z)$ and $Y(z e^{2i \pi k})=M^k Y(z)$ $\endgroup$ – reuns Feb 22 at 23:23
  • $\begingroup$ @reuns Then I guess it is true that the continuation of $Y$ after one loop clockwise (assuming the others were counterclockwise) would be equal to $M^{-1}Y(z)$, right? $\endgroup$ – Edu Feb 22 at 23:58
  • $\begingroup$ I meant $Y(e^{2i \pi})=Y(1) M \implies Y(z e^{2i \pi})=Y(z)M,Y(z e^{2i \pi k})=Y(z)M^k$, which is valid for $k \in \mathbb{Z}$ since $M$ is inversible when $Y(1) = I$. What do you get when $A(z)$ is analytic on $\mathbb{C} - \{0,2\}$ for the monodromy on each loop $\in \pi_1(\mathbb{C} - \{0,2\})$ ? $\endgroup$ – reuns Feb 23 at 0:21

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