📂Matrix Algebra

One-Parameter Subgroup

Definition¹

If the function $A : \mathbb{R} \to \operatorname{GL}(n, \mathbb{C})$ satisfies the following, $A$ is called a one-parameter subgroup of $\operatorname{GL}(n, \mathbb{C})$.

1. $A$ is a continuous function.

2. $A(0) = I$. ($I$ is $n \times n$ identity matrix.)

3. $A(s + t) = A(s) A(t)$, $\forall s, t \in \mathbb{R}$.

Theorem

If $A$ is a one-parameter subgroup of $\operatorname{GL}(n, \mathbb{C})$, then there exists a unique $X$, a complex matrix $n \times n$, satisfying the following.

$$ A(t) = e^{tX} $$

Here $e^{tX}$ is the matrix exponential.

Explanation

In the three conditions of the definition, $A(t + s) = A(t) A(s)$ means that $A$ is a homomorphism from $\mathbb{R}$ to $\operatorname{GL}(n, \mathbb{C})$. In other words, $t \mapsto A(t)$ is a continuous path in $\operatorname{GL}(n, \mathbb{C})$ driven by a single real parameter $t$, and it preserves the group structure.

The above theorem has important implications. Although the definition of a one-parameter subgroup gives an algebraic condition (3.), the theorem shows not only that the analytic form of $A$ is fixed, but that its form is the exponential function $A(t) = e^{tX}$. In particular, the entire trajectory of a one-parameter subgroup is determined by a single matrix $X=A^{\prime}(0)$.

Proof

$X = A^{\prime}(0)$ will be the matrix sought in the theorem. To this end, we first show that $A$ is differentiable. The integral of a matrix-valued function is defined componentwise.

Step 1. $A$ is differentiable.

Since the integral of a matrix-valued function is defined componentwise, consider the auxiliary function $H$ below.

$$ H(s) := \int_{0}^{s} A(u) du, \qquad H_{ij}(s) = \int_{0}^{s} A_{ij}(u) du $$

Each component $A_{ij}$ of $A$ is continuous, so by the Fundamental Theorem of Calculus each $H_{ij}$ is differentiable.

$$ H_{ij}^{\prime}(s) = A_{ij}(s), \qquad H^{\prime}(s) = A(s) \tag{1} $$

We now show two facts.

(i) Invertibility of $B$.

$\frac{1}{\epsilon} \int_{0}^{\epsilon} A(t) dt$ is the derivative at $s = 0$ of $H$. By $(1)$

$$ \lim\limits_{\epsilon \to 0}\frac{1}{\epsilon} \int_{0}^{\epsilon} A(t) dt = \lim\limits_{\epsilon \to 0}\frac{H(\epsilon) - H(0)}{\epsilon} = H^{\prime}(0) = A(0) = I $$

The determinant $\det$ is a continuous function and since $\det I = 1 \ne 0$, by choosing sufficiently small $\epsilon > 0$ we have $\det \left( \dfrac{1}{\epsilon} \int_{0}^{\epsilon} A(t) dt \right) \ne 0$. Therefore $B := \int_{0}^{\epsilon} A(t) dt$ is invertible.

(ii) Differentiability of $A$.

Using condition $A(s)A(t) = A(s+t)$ and the substitution $u = s + t$ we obtain

$$ A(s) B = \int_{0}^{\epsilon} A(s) A(t) dt = \int_{0}^{\epsilon} A(s+t) dt = \int_{s}^{s+\epsilon} A(u) du $$

and, since $B$ is invertible,

$$ A(s) = \left( \int_{s}^{s+\epsilon} A(u) du \right) B^{-1} $$

Rewriting the integral on the right-hand side using $H$ yields $\int_{s}^{s+\epsilon} A(u) du = H(s+\epsilon) - H(s)$, and because $H$ is differentiable by $(1)$, this function is also differentiable with respect to $s$ and the following holds.

$$ \frac{d}{ds} \left( \int_{s}^{s+\epsilon} A(u) du \right) = H^{\prime}(s+\epsilon) - H^{\prime}(s) = A(s+\epsilon) - A(s) $$

The integral inside the parentheses is differentiable and $B^{-1}$ is a constant matrix independent of $s$, so their product is differentiable. Also, since its derivative $\big( A(s+\epsilon) - A(s) \big) B^{-1}$ is continuous, $A$ is $C^{1}$.

Step 2. $A^{\prime} (t) = A(t) X$.

Since $A$ is differentiable, from condition $A(t+h) = A(t)A(h)$ and $A(0) = I$ we obtain the following.

$$ \begin{align*} A^{\prime}(t) &= \lim_{h \to 0} \dfrac{A(t+h) - A(t)}{h} \\ &= \lim_{h \to 0} A(t) \dfrac{A(h) - A(0)}{h} \\ &= A(t) A^{\prime}(0) = A(t) X \end{align*} $$

Step 3. $A(t) = e^{tX}$.

Let us set $C(t) := A(t) e^{-tX}$. By the properties of the matrix exponential we have $\dfrac{d}{dt} e^{-tX} = -X e^{-tX}$, and by the differentiation rules for matrix-valued functions the following holds.

$$ C^{\prime}(t) = A^{\prime}(t) e^{-tX} + A(t) \left( -X e^{-tX} \right) = A(t) X e^{-tX} - A(t) X e^{-tX} = 0 $$

Therefore $C$ is constant and $C(t) = C(0) = A(0) e^{O} = I^{2} = I$. That is $A(t) e^{-tX} = I$, and multiplying both sides by $e^{tX}$ yields $A(t) = e^{tX}$.

Step 4. Uniqueness.

For every $t$ let $A(t) = e^{tX} = e^{tY}$. Differentiating both sides with respect to $t$ and substituting ▷eq71◯ yields ▷eq72◯.

■