# Definition Consider a [[vector space]] $V$ equipped with a [[Non-Degeneracy (Quadratic Forms)|non-degenerate]] [[Hermitian form]] $\braket{.|.}$. The set of *isometries* (or *congruences*) on this vector space, denoted $\text{Isom}(V)$, is the set of operators which "preserve" $\braket{.|.}$, i.e., $ \braket{T\pmb{v}|T\pmb{w}} = \braket{\pmb{v}|\pmb{w}} \quad \forall \, \pmb{v}, \pmb{w}, \in V. $ It is a [[subgroup]] of the [[general linear group]] on $V$, $GL(V)$. The elements of the isometry group are sometimes called the motions of the space. # Examples ## Example 1: $O(n)$ Let $V$ be an $n$-dimensional *real* [[inner product space]]. Then the isometries of $V$ preserve both lengths and angles, since angles are defined purely in terms of lengths in [[Inner Product Space|inner product spaces]]. Let $T \in \text{Isom}(V)$, and pick an [[Orthogonality|orthonormal]] basis for $V$, then $ \begin{align} \braket{\pmb{v}|\pmb{w}} &= [\pmb{v}]^T[\pmb{w}] = \\ \braket{T\pmb{v}|T\pmb{w}} &= [T\pmb{v}]^T[T\pmb{w}]= [\pmb{v}]^T[T]^T [T][\pmb{w}]\ \end{align} $ which implies that $ [T]^T T = I \iff [T]^T = T^{-1} $ This is merely the [[Orthogonal Operator|orthogonality]] condition, and the set of all orthogonal matrices with dimension $n\times n$ is just the [[Orthogonal Group]] $O(n)$. ## Example 2: $U(n)$ If $V$ is equipped with a positive-definite [[Non-Degeneracy (Quadratic Forms)|non-degenerate]] [[Hermitian form]], then for a $T \in \text{Isom(V)}$ we can write: $ \braket{T \pmb{v} | T \pmb{w}} = \braket{T T^\dagger \pmb{v}|\pmb{w}} = \braket{\pmb{v}|\pmb{w}}, $ i.e., $ T T^\dagger = I \iff T^\dagger = T^{-1} $ This implies for the matrices $ [T][T]^\dagger = I \iff [T]^\dagger [T]^{-1} $ which is the familiar [[Unitary Operator|unitarity]] condition. These matrices form the [[Unitary Group|unitary group]] $U(n)$. ## Example 3: $O(n-1, n)$ Let $V$ be a real vector space with a [[Minkowski Metric|Minkowski metric]] $\eta$. We can consider its group of isometries. Let $T \in \text{Isom}(V)$ and work in an arbitrary basis, then: $ \begin{align} \eta(\pmb{v}, \pmb{w}) &= [\pmb{v}]^T [\eta] [\pmb{w}] = ,\\ \eta(T\pmb{v}, T\pmb{w}) &= [\pmb{v}]^T [T]^T [\eta] [T] [\pmb{w}] \end{align} $ so we get: $ \begin{align} [\eta] &= [T]^T [\eta] [T],\\ \eta_{\mu\nu} &= T_{\mu}^{\,\,\rho} T_{\nu}^{\,\,\sigma} \eta_{\rho\sigma} \end{align} $ which is just the definition of [[Lorentz Transformation|Lorentz transformations]]. The set of all Lorentz transformations forms the [[Lorentz Group|Lorentz group]], $O(n-1, 1)$ # Isomorphisms As shown [[General Linear Group#The Isomorphism|here]], $GL(V)$ and $GL(n,\mathbb{F})$ are isomorphic. If we restrict this isomorphism to $\text{Isom}(V) \subset GL(V)$ and equip $V$ with a [[Non-Degeneracy (Quadratic Forms)|non-degenerate]] [[Hermitian form]] $\braket{\cdot|\cdot}$, we end up with the following: 1. $\text{Isom(V)} \cong U(n)$ when $V$ is complex and $\braket{\cdot|\cdot}$ is positive definite. 2. $\text{Isom(V)} \cong O(n)$ when $V$ is real and $\braket{\cdot|\cdot}$ is positive definite. 3. $\text{Isom(V)} \cong O(n-1,1)$ when $V$ is real and $\braket{\cdot|\cdot}$ is a [[Minkowski metric]]. # Lie Algebra Using the identification between finite [[dimensiom|dimensional]] [[linear operator|linear operators]] and [[matrix|matrices]], we may think of $\text{Isom}(V)$ as a [[Matrix Lie Group]] and find out the properties of its [[Lie Algebra]] via the usual methods. Let $X$ be an element of the Lie algebra of $\text{Isom}(V)$, then: $ \braket{e^{tX} \pmb{v}|e^{tX}\pmb{w}} = \braket{\pmb{v}|\pmb{w}} \quad \forall \pmb{v},\pmb{w} \in V, t \in \mathbb{R} $ now differentiating with respect to $t$ and evaluating at $t = 0$, we get: $ \braket{X \pmb{v} | \pmb{w}} = - \braket{\pmb{v}|X\pmb{w}} \quad \forall \pmb{v}, \pmb{w} \in V $ Thus, the elements of the Lie algebra of *any* Isometry group obeys the equation above. # Lie Algebras Consider a vector space $V$ of arbitrary (possibly infinite) dimension, and the vector space of [[linear operator|linear operators]] on that space, $\mathcal{L}(V)$. It is shown [[General Linear Group#Lie Algebra|here]] that this vector space, when equipped with the [[commutator]], forms a [[Lie algebra]] known as $\mathfrak{gl}(V)$. Now define $\mathfrak{isom}(V) \subset \mathfrak{gl}(V)$ as the set of all *anti*-Hermitian operators. ==(Physicists would likely define this as Hermitian)== It is clear that $\mathfrak{isom}(V)$ is a [[Lie Subalgebra|Lie subalgebra]] of $\mathfrak{gl}(V)$. Note that, if $V$ is finite [[dimension|dimensional]] ($\dim(V) = n$) over a [[field]] $\mathbb{F} = \mathbb{C}$ *and* we pick an [[Orthogonality|orthonormal]] basis, then $\mathfrak{isom}(V) \cong \mathfrak{u}(n)$, where $\cong$ denotes a [[Lie Algebra Isomorphism]]. See the Lie algebra of the [[special unitary group#Lie Algebra|special unitary group]]. Similarly, if the field is $\mathbb{F} = \mathbb{R}$, then we get $\mathfrak{isom}(V) \cong \mathfrak{o}(n)$ $\mathfrak{isom}(V)$ is different from $\mathfrak{u}(n)$ in that it works for infinite dimensional spaces as well. It is a generalization in some sense, and it is *not* associated with a [[matrix Lie group]].