Isomorphisms in Linear Mappings between Vector Spaces

Jul 22, 2017

Equipping a vector space with an inner product results in a natural isomorphism $\CV\to\CV^\ast$, where the metric tensor can be interpreted as the linear mapping $\Bg:\CV\to\CV^\ast$ and its inverse $\Bg\inv:\CV^\ast\to\CV$.

Notation: Given two real vector spaces $\CV$ and $\CW$, we denote their inner products as $$\dabrn{\cdot,\cdot}_{\CV}$$ and $$\dabrn{\cdot,\cdot}_{\CW}$$ respectively. Given vectors $\Bv\in\CV$ and $\Bw\in\CW$, we define their lengths as

$$$\Norm{\Bv}_\CV = \sqrt{\dabrn{\Bv,\Bv}_\CV} \eqand \Norm{\Bw}_\CW = \sqrt{\dabrn{\Bw,\Bw}_\CW}.$$$

Regarding $\CV$ and $\CW$,

1. their bases are denoted $\cbrn{\BE_A}$ and $\cbrn{\Be_a}$,
2. their dual bases are denoted $\cbrn{\BE^A}$ and $\cbrn{\Be^a}$,
3. their metrics are denoted $\BG$ and $\Bg$ with the components $$G_{AB}=\dabrn{\BE_A,\BE_B}_\CV$$ and $$g_{ab}=\dabrn{\Be_a,\Be_b}_\CW$$,

respectively. Here, the indices pertaining to $\CV$ are uppercase $(ABC\dots)$ and the indices pertaining to $\CW$ are lowercase $(abc\dots)$.

Definition: Let $\BP:\CV\to\CW$ be a linear mapping. Then the transpose, or adjoint of $\BP$, written $\BP\tra$, is the linear mapping

$$$\boxed{ \BP\tra: \CW\to\CV \quad\text{such that}\quad \dabrn{\Bv,\BP\tra\Bw}_\CV = \dabrn{\BP\Bv,\Bw}_\CW }$$$

for all $\Bv\in\CV$ and $\Bw\in\CW$. Carrying out the products,

$$$G_{BA} v^B (P\tra){}^{A}{}_{d} w^d = g_{ab} P{}^{b}{}_{C}v^Cw^a.$$$

For arbitrary $\Bv$ and $\Bw$,

$$$G_{BA} (P\tra){}^{A}{}_{a} = g_{ab} P{}^{b}{}_{A}$$$

from which we can obtain the components of the transpose as

$$$\boxed{ (P\tra){}^{A}{}_{a} = g_{ab} P{}^{b}{}_{B} G^{AB} \eqwith \BP\tra = (P\tra){}^{A}{}_{a} \BE_A\dyd\Be^a . }$$$

If $\BB:\CV\to\CV$ is a linear mapping, it is called symmetric if $\BB=\BB\tra$.

Definition: Let $\BP:\CV\to\CW$ be a linear mapping. Then the dual of $\BP$ is a metric independent mapping

$$$\boxed{ \BP^\ast: \CW^\ast\to\CV^\ast \quad\text{such that}\quad \abrn{\Bv,\BP^\ast\Bbeta}_\CV = \abrn{\BP\Bv,\Bbeta}_\CW }$$$

defined through natural pairings for all $\Bv\in\CV$ and $\Bbeta\in\CW^\ast$. Carrying out the products,

$$$v^A (P^\ast){}_{A}{}^{a} \beta_a = P{}^{b}{}_{B} v^B \beta_b.$$$

For arbitrary $\Bv$ and $\Bbeta$, we obtain the components of the dual mapping as

$$$\boxed{ (P^\ast){}_{A}{}^{a} = P{}^{a}{}_{A} \eqwith \BP^\ast = (P^\ast){}_{A}{}^{a} \BE^A\dyd\Be_a = P{}^{a}{}_{A} \BE^A\dyd\Be_a . }$$$

To fully appreciate the symmetry that originates from the duality, we can think of not just the mappings between $\CV$ and $\CW$, but also between their dual spaces. To this end we can enumerate four mappings corresponding to $\cbr{\CV,\CV^\ast}\to\cbr{\CW,\CW^\ast}$ and their duals, corresponding to $\cbr{\CW,\CW^\ast}\to\cbr{\CV,\CV^\ast}$. Their definitions can be found in the table below.

 $\BP\in\CW \dyd\CV^\ast$ $P^a_{\idxsep A}=\BP(\Be^a,\BE_A)$ $\BP = P^{a}_{\idxsep A}\, \Be_a \dyd \BE^A$ Mappings Duals

The commutative diagrams pertaining to these mappings can be found in the figure below