note-math

Prop Generally, for $𝐴 \in SL (2, ℂ)$ , $𝐴, 𝐴^{*}$ are not equivalent in the sense of changing coordinates: There does not exist $𝐵 \in GL (2, ℂ)$ , $𝐴^{*} = 𝐵 \cdot 𝐴 \cdot 𝐵^{- 1}$

Proof

Eigenvalues remain unchanged under coordinate transformation

Generally, $𝐴, 𝐴^{*}$ have different eigenvalues

Example $𝐴 = (\begin{matrix} 2 i \\ - \frac{1}{2} i \end{matrix}), 𝐴^{*} = (\begin{matrix} - 2 i \\ \frac{1}{2} i \end{matrix})$

Prop $𝐴, {(𝐴^{⊺})}^{- 1}$ are equivalent, $𝐴^{*}, 𝐴^{†}$ are equivalent

use $j = (\begin{matrix} - 1 \\ 1 \end{matrix}), j^{2} = - 𝟙 ⟹ j^{- 1} = - j$

j \cdot 𝐴 \cdot j^{- 1} = {(𝐴^{⊺})}^{- 1}

Its complex conjugate conjugate-representation_(tag)

j \cdot 𝐴^{*} \cdot j^{- 1} = {(𝐴^{†})}^{- 1}

The above only works for $dim = 2$

Second-order tensor, with one undergoing complex conjugation

\begin{matrix} (ℂ^{2}, ℂ^{2}) & ⟶ & ⨂^{* 2} ℂ^{2} \\ (𝑣, 𝑤) & ⟿ & ⨂^{* 2} (𝑣, 𝑤) & = & 𝑣 \otimes 𝑤^{*} \end{matrix}

can be decompose to

Hermitian-tensor_(tag)

⨀^{* 2} (𝑣, 𝑤) = \frac{1}{2} (𝑣 \otimes 𝑤^{*} + 𝑤 \otimes 𝑣^{*})

anti-Hermitian-tensor_(tag)

\overset{* 2}{⋀} (𝑣, 𝑤) = \frac{1}{2} (𝑣 \otimes 𝑤^{*} - 𝑤 \otimes 𝑣^{*})

The conjugate modification in the other direction $𝑣^{*} \otimes 𝑤$ is similar

Hermitian-tensor-induced-linear-map_(tag) The induced action of $𝐴 \in GL (2, ℂ)$ on $⨂^{* 2} ℂ^{2}$ :=

𝐴^{\otimes * 2} : ⨂^{* 2} (𝑣, 𝑤) ⇝ ⨂^{* 2} (𝐴 𝑣, 𝐴 𝑤)

${(𝐴 𝑣)}^{*} = 𝐴^{*} 𝑣^{*}$

$⨂^{* 2} (𝜆 𝑣, 𝜆 𝑤) = | 𝜆 |^{2} ⨂^{* 2} (𝑣, 𝑤)$

matrix-description-of-Hermitian-tensor_(tag)

Use tensor base

𝑣 \otimes 𝑤^{*} = (\sum_{𝑖} 𝑣_{𝑖} 𝑒_{𝑖}) \otimes (\sum_{𝑗} 𝑤_{𝑗}^{*} 𝑒_{𝑗}) = \sum_{𝑖, 𝑗} 𝑣_{𝑖} 𝑤_{𝑗}^{*} (𝑒_{𝑖} \otimes 𝑒_{𝑗})

Corresponding to the matrix representation

(\begin{matrix} 𝑣_{1} 𝑤_{1}^{*} & 𝑣_{1} 𝑤_{2}^{*} \\ 𝑣_{2} 𝑤_{1}^{*} & 𝑣_{2} 𝑤_{2}^{*} \end{matrix}) = (\begin{matrix} 𝑣_{1} \\ 𝑣_{2} \end{matrix}) (\begin{matrix} 𝑤_{1}^{*} & 𝑤_{2}^{*} \end{matrix}) = 𝑣 \cdot 𝑤^{†}

Or written in Dirac notation

| 𝑣 ⟩ ⟨ 𝑤 |

The version without complex conjugation $𝑣 \otimes 𝑤 ⟷ 𝑣 \cdot 𝑤^{⊺}$

notation-overload: The space of matrix representation is also denoted as $⨂^{* 2} (𝑣, 𝑤)$

Hermitian matrix

\begin{matrix} ⨀^{* 2} (𝑣, 𝑤) & = & \frac{1}{2} (𝑣 𝑤^{†} + 𝑤 𝑣^{†}) = Re (𝑣 \cdot 𝑤^{†}) \\ ⨀^{* 2} (𝑣, 𝑤) & ≕ & Re (⨂^{* 2} (𝑣, 𝑤)) \end{matrix}

anti-Hermitian matrix

\begin{matrix} \overset{* 2}{⋀} (𝑣, 𝑤) & = & \frac{1}{2} (𝑣 𝑤^{†} - 𝑤 𝑣^{†}) = Im (𝑣 \cdot 𝑤^{†}) \\ \overset{* 2}{⋀} (𝑣, 𝑤) & ≕ & Im (⨂^{* 2} (𝑣, 𝑤)) \end{matrix}

For $ℍ, 𝕆$ , since $dim Im (ℍ), dim Im (𝕆) > 1$ , the dimension of anti-Hermitian is higher than Hermitian

Hermitian-tensor-induced-linear-map-matrix_(tag) The matrix representation of $𝐴^{\otimes * 2}$

𝑣 \cdot 𝑤^{†} ⇝ (𝐴 𝑣) \cdot {(𝐴 𝑤)}^{†} = 𝐴 (𝑣 \cdot 𝑤^{†}) 𝐴^{†}

$𝐴^{\otimes * 2}$ preserves the decomposition into Hermitian and anti-Hermitian

⨂^{* 2} ℂ^{2} = (⨀^{* 2} ℂ^{2}) \oplus (\overset{* 2}{⋀} ℂ^{2})

For general $𝑃 \in ⨂^{* 2} ℂ^{2}$ , there is also

𝐴^{\otimes * 2} : 𝑃 ⇝ 𝐴 𝑃 𝐴^{†}

The "matrix" representation of $𝕆$ needs to be handled separately, the composition of $Lin (2, 𝕆)$ cannot be represented as usual matrix multiplication. It can still make $𝐴 𝑃 𝐴^{†}$ well-defined

spacetime-momentum-spinor-representation_(tag)

( $𝑝$ represents "momentum" or "velocity" or tangent vector)

Bijection

\begin{matrix} ⨀^{* 2} ℂ^{2} & ⟶ & ℝ^{1, 3} \\ (\begin{matrix} 𝑝_{0} + 𝑝_{1} & 𝑝_{2} + 𝑝_{3} i \\ 𝑝_{2} - 𝑝_{3} i & 𝑝_{0} - 𝑝_{1} \end{matrix}) & ⟿ & (\begin{matrix} 𝑝_{0} \\ 𝑝_{1} \\ 𝑝_{2} \\ 𝑝_{3} \end{matrix}) \end{matrix}

metric

\begin{matrix} det (\begin{matrix} 𝑝_{0} + 𝑝_{1} & 𝑝_{2} + 𝑝_{3} i \\ 𝑝_{2} - 𝑝_{3} i & 𝑝_{0} - 𝑝_{1} \end{matrix}) & = & (𝑝_{0}^{2} - 𝑝_{1}^{2}) - (𝑝_{2}^{2} + 𝑝_{3}^{2}) \\ = & | 𝑝 |^{2} \end{matrix}

let $𝑝_{spin} ≔ (\begin{matrix} 𝑝_{0} + 𝑝_{1} & 𝑝_{2} + 𝑝_{3} i \\ 𝑝_{2} - 𝑝_{3} i & 𝑝_{0} - 𝑝_{1} \end{matrix})$ and $𝐴 \in SL (2, ℂ)$ , action $𝑝_{spin} ⇝ 𝐴 𝑝_{spin} 𝐴^{†}$

det 𝐴^{⊙ * 2} (𝑝_{spin}) = det 𝐴 𝑝_{spin} 𝐴^{†} = det 𝑝_{spin}

Because multiplication is non-commutative, the general definition of $det$ for $ℍ, 𝕆$ is problematic. However, the definition of $det 𝑝_{spin}$ does not require general multiplicative commutativity. At this time, $SL (2, 𝕂)$ is defined as $𝐴 : det 𝐴 𝑝_{spin} 𝐴^{†} = det 𝑝_{spin}$ . This is not a good notation because it may not be generalized to $dim > 3$

$SL (2, ℍ), SL (2, 𝕆)$ are also the spinor lifts of $SO (1, 5), SO (1, 9)$ . Similarly, $SU (2, 𝕂)$ is also the spinor lift of $SO (5), SO (9)$

Example Pauli-matrix_(tag) alias sigma-matrix_(tag)

for $(\begin{matrix} 𝑝_{0} + 𝑝_{1} & 𝑝_{2} + 𝑝_{3} i \\ 𝑝_{2} - 𝑝_{3} i & 𝑝_{0} - 𝑝_{1} \end{matrix})$

time-like $𝑝_{0} = 1 ⟷ (\begin{matrix} 1 \\ 1 \end{matrix}) ≕ 𝜎_{0}$
light-like $𝑝_{0} = 𝑝_{1} = 1 ⟷ (\begin{matrix} 2 \\ 0 \end{matrix}) = 𝜎_{0} + 𝜎_{1}$
space-like

$𝑝_{1} = 1 ⟷ (\begin{matrix} 1 \\ - 1 \end{matrix}) ≕ 𝜎_{1}$

$𝑝_{2} = 1 ⟷ (\begin{matrix} 1 \\ 1 \end{matrix}) ≕ 𝜎_{2}$

$𝑝_{3} = 1 ⟷ (\begin{matrix} i \\ - i \end{matrix}) ≕ 𝜎_{3}$ (when generalized to $ℍ, 𝕆$ , it corresponds to all imaginary elements)

$𝑝_{spin} = \sum 𝑝_{𝜇} 𝜎_{𝜇}$
$det 𝜎^{0} = 1$
$det 𝜎^{𝑖} = - 1, 𝑖 = 1, 2, 3$
$𝜎^{0}, 𝜎^{1}, 𝜎^{2}, 𝜎^{3}$ is orthonormal base
$𝑝_{0} + 𝑝_{1} i_{split} \in ℂ_{split} ≃ ℝ^{1, 1}$
$𝑝_{2} + 𝑝_{3} i \in ℂ ≃ ℝ^{2}$

Question What is the cognitive motivation for these constructions of $⨀^{* 2} ℂ^{2}, (\begin{matrix} 𝑝_{0} + 𝑝_{1} & 𝑝_{2} + 𝑝_{3} i \\ 𝑝_{2} - 𝑝_{3} i & 𝑝_{0} - 𝑝_{1} \end{matrix}), det 𝑝_{spin} = | 𝑝 |^{2}$ ?

$SO (1, 3) ≃ \frac{SL (2, ℂ)}{ℤ_{2}}$ #link(<Lorentz-group-spinor-representation>)[acts on] ${ℂ ℙ}^{1}$ lifts to $SL (2, ℂ)$ act on $ℂ^{2}$

$𝐴$ action, denoted as $(\frac{1}{2}, 0)$
$𝐴^{*}$ action, denoted as $(0, \frac{1}{2})$

square-root-of-Lorentz-group_(tag)

$SO (1, 3)$ act on $ℝ^{1, 3}$ is some kind of "square" of $𝐴, 𝐴^{*}$ , i.e. $(\frac{1}{2}, 0) \otimes (0, \frac{1}{2})$ or $(\frac{1}{2}, \frac{1}{2})$ represents the "real part" or "symmetric part"

Re (⨂^{* 2} (𝑣, 𝑤)) ⨀^{* 2} ℂ^{2} ⟷ ℝ^{1, 3}

Thus $𝐴, 𝐴^{*}$ are some kind of "square root" of $SO (1, 3)$ act on $ℝ^{1, 3}$

square-root-of-spacetime-metric-1_(tag) (inspired by ref-14, ch.11)

$det \in ⋀^{2} (ℂ^{2} \to ℂ), det (𝑣_{1}, 𝑣_{2}) = 𝑤^{⊺} j 𝑣, j = (\begin{matrix} - 1 \\ 1 \end{matrix})$

${det}^{\otimes 2} (𝑣_{1} \otimes 𝑤_{1}, 𝑣_{2} \otimes 𝑤_{2}) ≔ det (𝑣_{1}, 𝑣_{2}) det (𝑤_{1}, 𝑤_{2})$ . Note that it is not alternating for $𝑣_{1} \otimes 𝑤_{1}, 𝑣_{2} \otimes 𝑤_{2}$

metric $𝑔 \in (⨀^{2} ℝ^{1, 3} \to ℝ)$ with $ℝ^{1, 3} ≃ ⨀^{* 2} ℂ^{2}$ is some kind of "square" of $det$ , i.e. $\frac{1}{2} 𝑔 ≃ {det}^{⊙ * 2}$

\begin{matrix} {det}^{⊙ * 2} (⨀^{* 2} (𝑣_{1}, 𝑤_{1}), ⨀^{* 2} (𝑣_{2}, 𝑤_{2})) \\ = & \frac{1}{2^{4}} (det (𝑣_{1}, 𝑣_{2}) det (𝑤_{1}^{*}, 𝑤_{2}^{*}) + det (𝑤_{1}, 𝑣_{2}) det (𝑣_{1}^{*}, 𝑤_{2}^{*}) \\ + det (𝑣_{1}, 𝑤_{2}) det (𝑤_{1}^{*}, 𝑣_{2}^{*}) + det (𝑤_{1}, 𝑤_{2}) det (𝑣_{1}^{*}, 𝑣_{2}^{*})) \end{matrix}

quadratic-form is

{det}^{⊙ * 2} (⨀^{* 2} (𝑣, 𝑤), ⨀^{* 2} (𝑣, 𝑤)) = - \frac{1}{2^{3}} | det (𝑣, 𝑤) |^{2}

cf. #link(<Pauli-matrix>)[]

\begin{matrix} 𝑣 & 𝑤 & ⨀^{* 2} (𝑣, 𝑤) \\ (\begin{matrix} 1 \\ 0 \end{matrix}) & (\begin{matrix} 0 \\ 1 \end{matrix}) & \frac{1}{2} 𝜎_{2} \\ (\begin{matrix} 1 \\ 0 \end{matrix}) & (\begin{matrix} 0 \\ i \end{matrix}) & \frac{1}{2} 𝜎_{3} \\ (\begin{matrix} 1 \\ 1 \end{matrix}) & (\begin{matrix} 1 \\ - 1 \end{matrix}) & 𝜎_{1} \\ (\begin{matrix} 1 \\ 0 \end{matrix}) & (\begin{matrix} 1 \\ 0 \end{matrix}) & (\begin{matrix} 1 \\ 0 \end{matrix}) \\ (\begin{matrix} 0 \\ 1 \end{matrix}) & (\begin{matrix} 0 \\ 1 \end{matrix}) & (\begin{matrix} 0 \\ 1 \end{matrix}) \end{matrix}

The calculation shows that $\frac{1}{2} 𝑔 = {det}^{⊙ * 2}$ is correct for $𝜎^{1, 2, 3}$ . For $𝜎^{0}$ , use sum $⨀^{* 2} ((\begin{matrix} 1 \\ 0 \end{matrix}), (\begin{matrix} 1 \\ 0 \end{matrix})) + ⨀^{* 2} ((\begin{matrix} 0 \\ 1 \end{matrix}), (\begin{matrix} 0 \\ 1 \end{matrix}))$

\begin{matrix} 0 & = & {det}^{⊙ * 2} (⨀^{* 2} ((\begin{matrix} 1 \\ 0 \end{matrix}), (\begin{matrix} 1 \\ 0 \end{matrix})), ⨀^{* 2} ((\begin{matrix} 1 \\ 0 \end{matrix}), (\begin{matrix} 1 \\ 0 \end{matrix}))) \\ 0 & = & {det}^{⊙ * 2} (⨀^{* 2} ((\begin{matrix} 0 \\ 1 \end{matrix}), (\begin{matrix} 0 \\ 1 \end{matrix})), ⨀^{* 2} ((\begin{matrix} 0 \\ 1 \end{matrix}), (\begin{matrix} 0 \\ 1 \end{matrix}))) \frac{1}{4} & = & {det}^{⊙ * 2} (⨀^{* 2} ((\begin{matrix} 1 \\ 0 \end{matrix}), (\begin{matrix} 1 \\ 0 \end{matrix})), ⨀^{* 2} ((\begin{matrix} 0 \\ 1 \end{matrix}), (\begin{matrix} 0 \\ 1 \end{matrix}))) \end{matrix}

orthogonal of sigma matrix can also be obtained through calculation, thus ${det}^{⊙ * 2} = \frac{1}{2} 𝑔$

Thus $det$ is some kind of "square root" of metric $𝑔$

Question Still no direct intuitive calculation equation ${det}^{⊙ * 2} (\sum 𝑝_{𝜇} 𝜎_{𝜇}, \sum 𝑝_{𝜇} 𝜎_{𝜇}) = \frac{1}{2} det (\sum 𝑝_{𝜇} 𝜎_{𝜇})$ …

spacetime-momentum-aciton-spinor-representation_(tag)

let $𝑓 : SL (2, ℂ), ⨀^{* 2} ℂ^{2} ↠ SO (1, 3), ℝ^{1, 3}$ .

where $𝑓 (𝐴)$ is #link(<Lorentz-group-spinor-representation>)[]

$𝑓 (𝑝_{spin}) = 𝑝$ is #link(<spacetime-momentum-spinor-representation>)[]

Then there is a homomorphism

\begin{matrix} 𝑓 (𝐴^{⊙ * 2} 𝑝_{spin}) \\ = & 𝑓 (𝐴) 𝑓 (𝑝_{spin}) \\ = & 𝑓 (𝐴) 𝑝 \end{matrix}

Proof Use the $SL (2, ℂ) ↠ SO (1, 3)$ correspondence of 3 rotations, 3 boosts

spinor-representation-adjoint_(tag) $𝑓 (𝐴^{†}) = {𝑓 (𝐴)}^{⊺}$

Proof

use 3 boost, 3 rotation

use ${(𝐴 𝐵)}^{†} = 𝐵^{†} 𝐴^{†}, {(𝐴 𝐵)}^{⊺} = 𝐵^{⊺} 𝐴^{⊺}$

$𝐴 \in SO (1, 3) ⟹ 𝐴^{⊺} 𝜂 𝐴 = 𝜂 = (\begin{matrix} 1 \\ - 1 \\ - 1 \\ - 1 \end{matrix})$ , $𝑣^{⊺} 𝜂 𝑤 = 𝑔 (𝑣, 𝑤)$

$𝐴^{⊺} = 𝜂 𝐴^{- 1} 𝜂$

$𝐴^{- 1} = 𝜂 𝐴^{⊺} 𝜂$

\begin{matrix} {𝑓 (𝐴)}^{⊺} 𝑓 (𝑝_{spin}) & = & 𝑓 (𝐴^{†}) 𝑓 (𝑝_{spin}) \\ = & 𝑓 (𝐴^{†} 𝑝_{spin} 𝐴) \end{matrix}

Prop Use #link(<spacetime-momentum-spinor-representation>)[] for $𝑣 \cdot 𝑤^{†}$ , $𝑣 = 𝑤$ + $ℂ$ projection $𝜆 𝑣$ gives projective-lightcone

(\exists 𝑣 \in ℂ^{2} ∖ 0, 𝑝_{spin} = 𝑣 𝑣^{†}) ⟺ (𝑝_{0} > 0, det (𝑝_{spin}) = 0)

Therefore the following are equivalent

$SL (2, ℂ)$ act on ${ℂ ℙ}^{1}$ via $ℂ^{2}$
$SL (2, ℂ)$ act on $Cone- ℙ (1, 3)$ via $⨀^{* 2} ℂ^{2} ≃ ℝ^{1, 3}$

Proof

$⟹$

$𝑣 𝑣^{†} = (\begin{matrix} 𝑣_{1} \\ 𝑣_{2} \end{matrix}) (\begin{matrix} 𝑣_{1}^{*} & 𝑣_{2}^{*} \end{matrix}) = (\begin{matrix} 𝑣_{1} 𝑣_{1}^{*} & 𝑣_{1} 𝑣_{2}^{*} \\ 𝑣_{2} 𝑣_{1}^{*} & 𝑣_{2} 𝑣_{2}^{*} \end{matrix})$ with $det (𝑣 𝑣^{†}) = 𝑣_{1} 𝑣_{1}^{*} \cdot 𝑣_{2} 𝑣_{2}^{*} - 𝑣_{1} 𝑣_{2}^{*} \cdot 𝑣_{2} 𝑣_{1}^{*} = 0$ (requires $ℂ$ associative law of multiplication?)

$𝑝_{0} = \frac{1}{2} tr (𝑣 𝑣^{†}) = \frac{1}{2} | 𝑣 |^{2} > 0$

$⟸$

Given $𝑝_{spin} = (\begin{matrix} 𝑝_{0} + 𝑝_{1} & 𝑝_{2} + 𝑝_{3} i \\ 𝑝_{2} - 𝑝_{3} i & 𝑝_{0} - 𝑝_{1} \end{matrix}) ≃ (\begin{matrix} 𝑝_{0} \\ 𝑝_{1} \\ 𝑝_{2} \\ 𝑝_{3} \end{matrix})$

in $ℂ$ , $𝑣_{𝑖} = | 𝑣_{𝑖} | 𝑒^{𝜃_{𝑖} i}$

let
$| 𝑣_{1} |^{2} = 𝑝_{0} + 𝑝_{1}$
$| 𝑣_{2} |^{2} = 𝑝_{0} - 𝑝_{1}$

Also need to calculate $𝜃_{1}, 𝜃_{2}$

In order to get $𝑝_{2} + 𝑝_{3} i = 𝑣_{1} 𝑣_{2}^{*}$ , compare norm, phase

\begin{matrix} 𝑝_{2} + 𝑝_{3} i & = & | 𝑝_{2} + 𝑝_{3} i | \frac{𝑝_{2} + 𝑝_{3} i}{| 𝑝_{2} + 𝑝_{3} i |} \\ 𝑣_{1} 𝑣_{2}^{*} & = & | 𝑣_{1} | | 𝑣_{2} | 𝑒^{i (𝜃_{1} - 𝜃_{2})} \end{matrix}

norm

\begin{matrix} | 𝑝_{2} + 𝑝_{3} i |^{2} & = & 𝑝_{2}^{2} + 𝑝_{3}^{2} \\ = & 𝑝_{0}^{2} - 𝑝_{1}^{2} & (use | 𝑥 |^{2} = 0) \\ = & | 𝑣_{1} |^{2} | 𝑣_{2} |^{2} \\ = & | 𝑣_{1} 𝑣_{2}^{*} |^{2} \end{matrix}

phase

$arg (\frac{𝑝_{2} + 𝑝_{3} i}{| 𝑝_{2} + 𝑝_{3} i |}) \in ℝ$

so let $𝜃_{1}, 𝜃_{2} \in ℝ$ with $𝜃_{1} - 𝜃_{2} = arg (\frac{𝑝_{2} + 𝑝_{3} i}{| 𝑝_{2} + 𝑝_{3} i |})$

Generally $⨀^{* 2} (𝑣, 𝑤) = (\begin{matrix} Re (𝑣_{1} 𝑤_{1}^{*}) & \frac{1}{2} (𝑣_{1} 𝑤_{2}^{*} + 𝑤_{1} 𝑣_{2}^{*}) \\ {(\dots)}^{*} & Re (𝑣_{2} 𝑤_{2}^{*}) \end{matrix})$ . Compare $(\begin{matrix} 𝑝_{0} + 𝑝_{1} & 𝑝_{2} + 𝑝_{3} i \\ 𝑝_{2} - 𝑝_{3} i & 𝑝_{0} - 𝑝_{1} \end{matrix})$ to get

\begin{matrix} 𝑝_{0} = \frac{1}{2} Re (𝑣_{1} 𝑤_{1}^{*} + 𝑣_{2} 𝑤_{2}^{*}) \\ 𝑝_{1} = \frac{1}{2} Re (𝑣_{1} 𝑤_{1}^{*} - 𝑣_{2} 𝑤_{2}^{*}) \\ 𝑝_{2} = \frac{1}{2} Re (𝑣_{1} 𝑤_{2}^{*} + 𝑣_{2} 𝑤_{1}^{*}) \\ 𝑝_{3} = \frac{1}{2} Im (𝑣_{1} 𝑤_{2}^{*} - 𝑣_{2} 𝑤_{1}^{*}) \end{matrix}

parity_(tag)

parity corresponds to $(\frac{1}{2}, 0)$ vs $(0, \frac{1}{2})$ representation, or $𝐴$ vs $𝐴^{*}, {(𝐴^{†})}^{- 1}$ , cf. #link(<conjugate-representation>)[]

let $𝑃 \in ⨀^{* 2} ℂ^{2}$ . $𝑃^{*} = 𝑃^{⊺}$

𝑃^{◊} ≔ j \cdot 𝑃^{*} \cdot j^{- 1} with j = (\begin{matrix} - 1 \\ 1 \end{matrix})

parity corresponds to space inversion

${(\begin{matrix} 𝑝_{0} + 𝑝_{1} & 𝑝_{2} + 𝑝_{3} i \\ 𝑝_{2} - 𝑝_{3} i & 𝑝_{0} - 𝑝_{1} \end{matrix})}^{◊} = (\begin{matrix} 𝑝_{0} - 𝑝_{1} & - (𝑝_{2} + 𝑝_{3} i) \\ - (𝑝_{2} - 𝑝_{3} i) & 𝑝_{0} + 𝑝_{1} \end{matrix}) ⟷ (\begin{matrix} 𝑝_{0} \\ - 𝑝_{2} \\ - 𝑝_{3} \\ - 𝑝_{1} \end{matrix})$

$- 𝑃^{◊}$ corresponds to time inversion

parity corresponds to trace or determinant reversal

determinant-reversal_(tag)

let $𝑃 = (\begin{matrix} 𝑎 & 𝑏 \\ 𝑐 & 𝑑 \end{matrix}) \in Matrix (2, ℂ)$

determinant reversal $𝑃^{◊} ≔ (\begin{matrix} 𝑑 & - 𝑏 \\ - 𝑐 & 𝑎 \end{matrix})$ with

$𝑃 𝑃^{◊} = 𝑃^{◊} 𝑃 = det (𝑃) \cdot 𝟙$

$det 𝑃^{◊} = det 𝑃$

$𝑃 \in GL ⟹ 𝑃^{◊} = (det 𝑃) 𝑃^{- 1}$

trace-reversal_(tag) := $𝑃 + 𝑃^{◊} = tr (𝑃) \cdot 𝟙$ . or $𝑃^{◊} = (\begin{matrix} 𝑑 & - 𝑏 \\ - 𝑐 & 𝑎 \end{matrix})$ . $tr 𝑃^{◊} = tr 𝑃$

$dim = 2$ ==> determinant reversal is the same as trace reversal

square-root-of-spacetime-metric-2_(tag) a "square root" of $1, 3$ metric

let $𝑝_{spin} \in ⨀^{* 2} ℂ^{2} ≃ ℝ^{1, 3}$ . $det (𝑝_{spin}) = 𝑔 (𝑝, 𝑝) = | 𝑝 |^{2}$

\begin{matrix} | 𝑝 |^{2} 𝟙 & = & det (𝑝_{spin}) 𝟙 \\ = & {𝑝_{spin}}^{◊} 𝑝_{spin} \\ = & 𝑝_{spin} {𝑝_{spin}}^{◊} \end{matrix}

$2 𝑔 (𝑝, 𝑝^{'}) = | 𝑝 + 𝑝^{'} |^{2} - (| 𝑝 |^{2} + | 𝑝^{'} |^{2})$ give

\begin{matrix} 𝑔 (𝑝, 𝑝^{'}) 𝟙 & = & \frac{1}{2} ({𝑝_{spin}}^{◊} {𝑝_{spin}}^{'} + {𝑝_{spin}}^{' ◊} 𝑝_{spin}) \\ = & \frac{1}{2} (𝑝_{spin} {𝑝_{spin}}^{' ◊} + {𝑝_{spin}}^{'} {𝑝_{spin}}^{◊}) \end{matrix}

Also have $𝑔 (𝑝, 𝑝^{'}) = \frac{1}{2} Re (tr ({𝑝_{spin}}^{◊} {𝑝_{spin}}^{'})) = \frac{1}{2} Re (tr (𝑝_{spin} {𝑝_{spin}}^{' ◊}))$

for #link(<Pauli-matrix>)[]

$𝜎_{𝜇}^{◊} 𝜎_{𝜈} + 𝜎_{𝜈}^{◊} 𝜎_{𝜇} = 2 𝑔_{𝜇 𝜈} 𝟙$ or ${𝜎_{𝜇}, 𝜎_{𝜈}}_{◊} = 2 𝑔_{𝜇 𝜈} 𝟙$
$𝜎_{0}^{◊} = 𝜎_{0}$ , $𝜎_{𝑖}^{◊} = - 𝜎_{𝑖}$ for $𝑖 = 1, 2, 3$ (because parity is spatial inversion)

A better explanation of this "square root"?

Direct matrix multiplication without parity will give the square root of the $ℝ^{4}$ metric, with $𝜎_{𝜇}^{2} = 𝟙$ , $𝜎_{𝜇}^{- 1} = 𝜎_{𝜇}$

This makes it possible for the spacetime momentum spin representation to be connected to the concept of classical fermions. Spinors belong to the light cone projection ${ℂ ℙ}^{1}$ . If $𝑝_{spin}$ is on the light cone, then its square $\frac{1}{2} ({𝑝_{spin}}^{◊} 𝑝_{spin} + 𝑝_{spin} {𝑝_{spin}}^{◊}) = 𝑔 (𝑝_{spin}, 𝑝_{spin}) 𝟙 = 0$ . This seems to be related to the Pauli exclusion principle. But note that, in general, $𝑔 (𝑝_{spin}, {𝑝_{spin}}^{'}) \neq 0$ unless $𝑝_{spin}, {𝑝_{spin}}^{'}$ are collinear (#link(<signature-of-2d-subspace-of-spacetime>)[]). Therefore, the result of this multiplication, $𝑔 (𝑝_{spin}, {𝑝_{spin}}^{'}) 𝟙 ≃ (\begin{matrix} 𝑔 (𝑝_{spin}, {𝑝_{spin}}^{'}) \\ 0 \\ 0 \\ 0 \end{matrix})$ , will not be on the light cone. If you want you can extend it to Clifford algebra

square-root-of-Lorentz-Lie-algebra_(tag) "square root" of spacetime Lie-algebra

{[\frac{1}{2} 𝜎_{𝜇}, \frac{1}{2} 𝜎_{𝜈}]}_{◊} ≔ \frac{1}{4} (𝜎_{𝜇}^{◊} 𝜎_{𝜈} - 𝜎_{𝜈}^{◊} 𝜎_{𝜇}) ≃ 𝐿_{𝜇 𝜈}

where $𝐿_{𝜇 𝜈}$ is #link(<rotation-boost-spinor-representation>)[Lorentz-Lie-algebra]

Proof

${[\frac{1}{2} 𝜎_{𝑖}, \frac{1}{2} 𝜎_{𝑖^{'}}]}_{◊} = \frac{1}{2} i 𝜎_{𝑖^{″}} ≃ 𝐿_{𝑖 𝑖^{'}}$ is δ rotation in $𝑝_{𝑖^{″}}$ where $𝑖, 𝑖^{'}, 𝑖^{″}$ is any cyclic $123$
${[\frac{1}{2} 𝜎_{0}, \frac{1}{2} 𝜎_{𝑖}]}_{◊} = \frac{1}{2} 𝜎_{𝑖} ≃ 𝐿_{0 𝑖}$ where $𝑖 = 1, 2, 3$

Question A better explanation? Representation?

property-of-parity_(tag)

$\forall 𝑎, 𝑏 \in ℂ, {(𝑎 𝐴 + 𝑏 𝐵)}^{◊} = 𝑎 𝐴^{◊} + 𝑏 𝐵^{◊}$
${(𝐴 𝐵)}^{◊} = 𝐵^{◊} 𝐴^{◊}$
$𝟙^{◊} = 𝟙$
${(𝐴^{†})}^{◊} = {(𝐴^{◊})}^{†}$
$◊ : ⨀^{* 2} ℂ^{2} \to self$ i.e. parity preserve Hermitian
$𝐴 \in GL (2, ℂ) ⟹ 𝐴^{◊} = det (𝐴) \cdot 𝐴^{- 1}$
$𝐴 \in SL (2, ℂ) ⟹ 𝐴^{◊} = 𝐴^{- 1}, 𝐴 𝐴^{◊} = 𝟙, {(𝐴^{◊})}^{◊} = 𝐴$

parity-Euclidean-invariant_(tag) parity commutes with spatial action $SU (2)$ . In $ℝ^{3}$ , it manifests as $- 𝟙$ and commutes with $SO (3)$ . let $𝑝 \in ℝ^{3}, 𝐴 \in SU (2)$

$𝐴 \in SU (2) ⟹ 𝐴^{†} = 𝐴^{- 1} = 𝐴^{◊} ⟹ {(𝐴^{⊙ * 2} (𝑝_{spin}))}^{◊} = 𝐴^{⊙ * 2} ({𝑝_{spin}}^{◊})$

Generally, they do not commute, for example, $𝟙_{ℝ^{3}}$ certainly does not commute with the time-changing part in $SO (1, 3)$

let $𝑝_{spin} = 𝜎_{0} = (\begin{matrix} 1 \\ 1 \end{matrix}) = 𝟙, 𝐴 = (\begin{matrix} 𝑒^{\frac{𝜑}{2}} \\ 𝑒^{- \frac{𝜑}{2}} \end{matrix}), 𝐴^{†} = 𝐴$

${𝑝_{spin}}^{◊} = 𝑝_{spin}$

$𝐴 𝑝_{spin} 𝐴^{†} = (\begin{matrix} 𝑒^{𝜑} \\ 𝑒^{- 𝜑} \end{matrix})$ or ${\begin{matrix} 𝑝_{0} = cosh 𝜑 \\ 𝑝_{1} = sinh 𝜑 \\ 𝑝_{2} = 𝑝_{3} = 0 \end{matrix}$

${(𝐴 𝑝_{spin} 𝐴^{†})}^{◊} = (\begin{matrix} 𝑒^{- 𝜑} \\ 𝑒^{𝜑} \end{matrix})$ or ${\begin{matrix} 𝑝_{0} = cosh 𝜑 = cosh (- 𝜑) \\ 𝑝_{1} = - sinh 𝜑 = sinh (- 𝜑) \end{matrix}$

${(𝐴 𝑝_{spin} 𝐴^{†})}^{◊} \neq 𝐴 𝑝_{spin} 𝐴^{†} = 𝐴 {𝑝_{spin}}^{◊} 𝐴^{†}$

parity-reverse-boost_(tag) The effect of parity on the Lie-algebra is that it does not change δ rotation, but multiplies δ boost by $- 1$

Euclidean-spinor_(tag)

replace lightcone $Cone- ℙ (1, 3)$ with just sphere $𝕊^{2} = {ℂ ℙ}^{1}$ acted by $SO (3)$ and $SU (2)$

replace $SL (2, ℂ) ↠ SO (1, 3)$ with $SU (2) ↠ SO (3)$ , $(\begin{matrix} 𝑎 & 𝑏 \\ - 𝑏^{*} & 𝑎^{*} \end{matrix}) \in SU (2)$ with $| 𝑎 |^{2} + | 𝑏 |^{2} = 1$

use trace-free Hermitian $(\begin{matrix} 𝑝_{3} & 𝑝_{1} + 𝑝_{2} i \\ 𝑝_{1} - 𝑝_{2} i & - 𝑝_{3} \end{matrix}) ⟷ (\begin{matrix} 𝑝_{1} \\ 𝑝_{2} \\ 𝑝_{3} \end{matrix})$