# Monodromies of complex differential equation

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

• 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 $Y(z e^{2i \pi k})=M^k Y(z)$ – reuns Feb 22 at 23:23
• @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? – Edu Feb 22 at 23:58
• 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 $\in \pi_1(\mathbb{C} - \{0,2\})$ ? – reuns Feb 23 at 0:21