📂Representation Theory

Euclidean group

Definition¹

The set of all compositions of translations and orthogonal transformations defined on the real space $\mathbb{R}^{n}$ is called the Euclidean group and is denoted by $\operatorname{E}(n)$. For given $\mathbf{x} \in \mathbb{R}^{n}$ and $R \in \operatorname{O}(n)$, it is the set of maps $( \mathbf{x}, R )$ as follows.

$$ \operatorname{E}(n) := \left\{ ( \mathbf{x}, R ) : \forall \mathbf{x} \in \mathbb{R}^{n},\quad \forall R \in \operatorname{O}(n) \right\} \\[1em] \begin{align*} \text{where } ( \mathbf{x}, R ) : \mathbb{R}^{n} &\to \mathbb{R}^{n} \\ \mathbf{y} &\mapsto R\mathbf{y} + \mathbf{x} \end{align*} $$

Here $\operatorname{O}(n)$ is the orthogonal group.

Explanation

Intuitively, the Euclidean group is the set of all compositions of translations, reflections, and rotations.

Group

It is a group with respect to composition of functions. For the composition of two transformations the following holds.

$$ (\mathbf{x}_{1}, R_{1})(\mathbf{x}_{2}, R_{2})\mathbf{y} = (\mathbf{x}_{1}, R_{1})(R_{2}\mathbf{y} + \mathbf{x}_{2}) = R_{1}R_{2}\mathbf{y} + R_{1}\mathbf{x}_{2} + \mathbf{x}_{1} $$

$$ \implies (\mathbf{x}_{1}, R_{1})(\mathbf{x}_{2}, R_{2}) = (\mathbf{x}_{1} + R_{1}\mathbf{x}_{2}, R_{1}R_{2}) \tag{1} $$

• Associativity $$ \begin{align*} \Big[ (\mathbf{x}_{1}, R_{1})(\mathbf{x}_{2}, R_{2}) \Big] (\mathbf{x}_{3}, R_{3}) &= (\mathbf{x}_{1} + R_{1}\mathbf{x}_{2}, R_{1}R_{2}) (\mathbf{x}_{3}, R_{3}) \\ &= (\mathbf{x}_{1} + R_{1}\mathbf{x}_{2} + R_{1}R_{2}\mathbf{x}_{3}, R_{1}R_{2}R_{3}) \\ &= (\mathbf{x}_{1} + R_{1}(\mathbf{x}_{2} + R_{2}\mathbf{x}_{3}), R_{1}(R_{2}R_{3})) \\ &= (\mathbf{x}_{1}, R_{1})(\mathbf{x}_{2} + R_{2}\mathbf{x}_{3}, R_{2}R_{3}) \\ &= (\mathbf{x}_{1}, R_{1}) \Big[ (\mathbf{x}_{2}, R_{2})(\mathbf{x}_{3}, R_{3}) \Big] \\ \end{align*} $$

• Identity element

  For the zero vector $\mathbf{0}$ and the identity matrix $I$, $(\mathbf{0}, I)$ is the identity element of $\operatorname{E}(n)$.

• Inverse

  The inverse of $(\mathbf{x}, R)$ is $(-R^{-1}\mathbf{x}, R^{-1})$. $$ (\mathbf{x}, R)(-R^{-1}\mathbf{x}, R^{-1}) = (\mathbf{x} - RR^{-1}\mathbf{x}, RR^{-1}) = (\mathbf{0}, I) $$

Subgroup

Since translations are not linear, $\operatorname{E}(n)$ is not a subgroup of $\operatorname{GL}(n, \mathbb{R})$. However, it is a subgroup of $\operatorname{GL}(n+1, \mathbb{R})$ and is isomorphic to the following set.

$$ \left\{ \begin{bmatrix} R & \mathbf{x} \\ \mathbf{0}^{\mathsf{T}} & 1 \end{bmatrix} : R \in \operatorname{O}(n), \mathbf{x} \in \mathbb{R}^{n} \right\} $$

Thus an element of $\operatorname{E}(n)$ can be represented as a $n+1$-dimensional affine transformation. This transformation is a mapping of the form $[\mathbf{y}, 1]^{\mathsf{T}} \mapsto [R\mathbf{y} + \mathbf{x}, 1]^{\mathsf{T}}$ and follows the multiplication rule of $(1)$.

$$ \begin{bmatrix} R_{1} & \mathbf{x}_{1} \\ \mathbf{0}^{\mathsf{T}} & 1 \end{bmatrix} \begin{bmatrix} R_{2} & \mathbf{x}_{2} \\ \mathbf{0}^{\mathsf{T}} & 1 \end{bmatrix} = \begin{bmatrix} R_{1}R_{2} & R_{1}\mathbf{x}_{2} + \mathbf{x}_{1} \\ \mathbf{0}^{\mathsf{T}} & 1 \end{bmatrix} $$

$$ \begin{bmatrix} R & \mathbf{x} \\ \mathbf{0}^{\mathsf{T}} & 1 \end{bmatrix} \begin{bmatrix} R^{-1} & -R^{-1}\mathbf{x} \\ \mathbf{0}^{\mathsf{T}} & 1 \end{bmatrix} = \begin{bmatrix} I & \mathbf{0} \\ \mathbf{0}^{\mathsf{T}} & 1 \end{bmatrix} $$

Let $\begin{bmatrix} R_{1} & \mathbf{x}_{1} \\ \mathbf{0}^{\mathsf{T}} & 1 \end{bmatrix}, \begin{bmatrix} R_{2} & \mathbf{x}_{2} \\ \mathbf{0}^{\mathsf{T}} & 1 \end{bmatrix} \in \operatorname{E}(n)$. Then, since the following holds, by the subgroup test, $\operatorname{E}(n)$ is a subgroup of $\operatorname{GL}(n+1, \mathbb{R})$.

$$ \begin{bmatrix} R_{1} & \mathbf{x}_{1} \\[0.5em] \mathbf{0}^{\mathsf{T}} & 1 \end{bmatrix} \begin{bmatrix} R_{2}^{-1} & -R_{2}^{-1}\mathbf{x}_{2} \\[0.5em] \mathbf{0}^{\mathsf{T}} & 1 \end{bmatrix} = \begin{bmatrix} R_{1}R_{2}^{-1} & R_{1}(-R_{2}^{-1}\mathbf{x}_{2}) + \mathbf{x}_{1} \\[0.5em] \mathbf{0}^{\mathsf{T}} & 1 \end{bmatrix} \in \operatorname{E}(n) $$

Matrix Lie group

$\operatorname{E}(n)$ is a closed subgroup of $\operatorname{GL}(n+1, \mathbb{R})$, hence a matrix Lie group. Define the function $f : \operatorname{GL}(n+1, \mathbb{R}) \to M_{n \times n}(\mathbb{R})$ as follows. It is the function that takes only the upper-left $n \times n$ block of a given matrix.

$$ f (A) = \begin{bmatrix} A_{11} & \cdots & A_{1n} \\ \vdots & \ddots & \vdots \\ A_{n1} & \cdots & A_{nn} \end{bmatrix}, \qquad A \in M_{(n+1) \times (n+1)}(\mathbb{R}) $$

Then $f$ is continuous. Recall that the preimage of a closed set under a continuous function is closed. Since $\operatorname{O}(n)$ is a closed set in $M_{n \times n}$, the following preimage is also a closed set.

$$ f^{-1}(\left\{ \operatorname{O}(n) \right\}) = \left\{ \begin{bmatrix} R & \mathbf{x} \\ \mathbf{y}^{\mathsf{T}} & z \end{bmatrix} : R \in \operatorname{O}(n), \mathbf{x}, \mathbf{y} \in \mathbb{R}^{n}, z \in \mathbb{R} \right\} $$

And define the function $g$ as follows. It takes only the last row of a given matrix.

$$ g(A) = \begin{bmatrix} A_{n+1, 1} & A_{n+1, 2} & \cdots & A_{n+1, n+1} \end{bmatrix}, \qquad A \in M_{(n+1) \times (n+1)}(\mathbb{R}) $$

Then $g$ is also continuous, and the preimage of the closed set $\left\{ \begin{bmatrix} 0 & \cdots & 0 & 1 \end{bmatrix} \right\}$ is likewise closed.

$$ g^{-1}(\left\{ \begin{bmatrix} 0 & \cdots & 0 & 1 \end{bmatrix} \right\}) = \left\{ \begin{bmatrix} A & \mathbf{x} \\ \mathbf{0}^{\mathsf{T}} & 1 \end{bmatrix} : A \in M_{n \times n}(\mathbb{R}), \mathbf{x} \in \mathbb{R}^{n} \right\} $$

Since the intersection of closed sets is closed, $\operatorname{Sp}(n, \mathbb{R})$ is a closed subgroup and thus a matrix Lie group.

$$ \begin{align*} \operatorname{Sp}(n, \mathbb{R}) &= \left\{ \begin{bmatrix} R & \mathbf{x} \\ \mathbf{0}^{\mathsf{T}} & 1 \end{bmatrix} : R \in \operatorname{O}(n), \mathbf{x} \in \mathbb{R}^{n} \right\} \\ &= f^{-1}(\left\{ \operatorname{O}(n) \right\}) \cap g^{-1}(\left\{ \begin{bmatrix} 0 & \cdots & 0 & 1 \end{bmatrix} \right\}) \end{align*} $$