Floquet Theory $\phi(t)=P(t)e^{tR}$, can be $R$ Hermitian?

Question

Let ${\displaystyle {\dot {x}}=A(t)x}$ be a linear first order differential equation, where ${\displaystyle x(t)}$ is a column vector of length ${\displaystyle n}$ and ${\displaystyle A(t)}$ an ${\displaystyle n\times n}$ periodic matrix with period ${\displaystyle T}$ (that is ${\displaystyle A(t+T)=A(t)}$ for all real values of ${\displaystyle t}$. Let ${\displaystyle \phi(t)}$ be a fundamental matrix solution of this differential equation. The Floquet Theory tells us that there is a constant matrix $R$ (possibly complex) and a T-periodic matrix valued function ${\displaystyle t\mapsto P(t)}$ such that $$\phi(t)=P(t)e^{tR}.$$ This gives rise to a time-dependent change of coordinates ${\displaystyle y=P^{-1}(t)x}$, under which our original system becomes a linear system with constant coefficients ${\displaystyle {\dot {y}}=Ry}$.

My question: Does $R$ follow the algebraic properties of $A(t)$. For example if $A(t)$ is Hermitian, can we chose $R$ to be also Hermitian? The same question when $A(t)$ is skew-Hermitian, unitary,...

Answer 1

Let $A : \mathbb{R} \to M_n(\mathbb{C})$ be continuous and periodic with period $T>0$, so that the principal fundamental solution of the vector-valued ODE $\dot{x} = Ax$ is the unique continuously differentiable function $\Phi : \mathbb{R} \to \operatorname{GL}(n,\mathbb{C})$, such that $$ \dot{\Phi} = A\Phi, \quad \Phi(0) = I_n. $$ Something remarkable happens when $A$ is valued in a matrix Lie algebra $\mathfrak{g} \leq M_n(\mathbb{C})$, i.e., an $\mathbb{R}$-linear subspace of $M_n(\mathbb{C})$, such that $$ \forall X,Y \in \mathfrak{g}, \quad XY-YX \in \mathfrak{g}. $$ In this case, by standard theory of ODE on manifolds, the principal fundamental solution $\Phi$ is necessarily valued in the connected matrix Lie group $$ G := \{\exp(X) : X \in \mathfrak{g}\} \leq \operatorname{GL}(n,\mathbb{C}) $$ corresponding to the matrix Lie algebra $\mathfrak{g}$. Hence, there exists $R \in \mathfrak{g}$, such that $\Phi(T) = \exp(TR)$. Since $\mathfrak{g} = \{X \in M_n(\mathbb{C}) : \forall t \in \mathbb{R}, \, \exp(tX) \in G\}$, we can safely define $T$-periodic $P : \mathbb{R} \to G$ by $$ \forall t \in \mathbb{R}, \quad P(t) := \Phi(t) \exp(-tR). $$ In turn, given $g_0 \in G$, the unique global solution $g : \mathbb{R} \to G$ of $$ \dot{g} = Ag, \quad g(0) = g_0, $$ i.e., the fundamental solution with initial value $g_0$, is given by $$ \forall t \in \mathbb{R}, \quad g(t) := (P(t)g_0)\exp(t g_0^{-1}Xg_0), $$ where $g_0^{-1}Xg_0 \in \mathfrak{g}$ by $\mathrm{Ad}$-invariance of $\mathfrak{g}$ and where $(t \mapsto P(t)g_0) : \mathbb{R} \to G$ since $G$ is a subgroup of $\operatorname{GL}(n,\mathbb{C})$ and since $P$ is $G$-valued.

Let me close by mentioning some concrete special cases:

1. if $\mathfrak{g} = \mathfrak{u}(n) := \{X \in M_n(\mathbb{C}) : X^\ast + X=0\}$, then $G = \operatorname{U}(n) := \{g \in M_n(\mathbb{C}) : g^\ast g = I_n\}$;
2. if $\mathfrak{g} = \mathfrak{su}(n) := \{X \in \mathfrak{su}(n) : \operatorname{tr}(X) = 0\}$, then $G = \operatorname{SU}(n) := \{g \in \operatorname{U}(n) : \det g = 1\}$;
3. if $\mathfrak{g} = \mathfrak{so}(n) := \{X \in M_n(\mathbb{R}) : X^T + X = 0\}$, then $G = \operatorname{SO}(n) := \{g \in M_n(\mathbb{R}) : g^T g = I_n, \, \det g = 1\}$.

Note, however, that the $n \times n$ Hermitian matrices do not define a matrix Lie algebra, so that this general method is inapplicable to that case.