1.2.6 The Lie Algebra and Differential Forms

We return to the $\left( {x_{j}X}_{j,k}^{\mathbb{c}} \right)^{'}$ de­riv­a­tive al­ge­bras to high­light how the matrix de­riv­a­tive op­er­a­tion on a group $G\left( x_{j}{\alpha_{k}X}_{j} \right)$ is close­ly re­lat­ed to its Lie al­ge­bra $X_{j}$. For an ex­po­nen­tial func­tion de­fined as $e^{\alpha xt}\in \mathbb{C}$, its de­riv­a­tive with re­spect to the vari­able $t$ is

and we would like to show that this holds for the de­riv­a­tives of our groups $G\left( x_{j}{\alpha_{k}X}_{j} \right) = e^{{x_{j}\alpha}_{k}X_{j}} \in \mathbb{C}^{3}$, which has the dif­fer­en­tial rep­re­sen­ta­tion

for $x_{j}\in \mathbb{R}$ and $X_{j} \in \mathbb{R}^{3}$. So, the vari­able $\alpha_{k}$ is now equal to the scalar value$\ \in \mathbb{C}$, which forms a prod­uct with the $x_{j}$ vari­ables to rep­re­sent the matrix en­tries for any Lie al­ge­bra set $\left\{ {x_{j}X}_{j}^{\mathbb{R}},\ {x_{j}X}_{j}^{\mathbb{C}},{x_{j}X}_{j,k}^{\mathbb{c}} \right\} \in A.$To max­i­mize the in­tu­ition un­der­ly­ing matrix-by-scalar de­riv­a­tives of the group $G\left( x_{j}{\alpha_{k}X}_{j} \right)$, we use the di­ag­o­nal­iza­tion de­com­po­si­tion tech­nique. Taking the $x_{1}\alpha_{k}X_{1}$ Lie al­ge­bra, as an ex­am­ple, a gen­er­al­ized rep­re­sen­ta­tion of the eigen­vec­tor/eigen­val­ue de­com­po­si­tion is the ex­pres­sion

for which the ex­po­nen­tial map­ping of its di­ag­o­nal matrix of eigen­val­ues is

and its de­riv­a­tive, $\left( e^{D} \right)^{'}$, eval­u­ates to

Cal­cu­lat­ing the matrix log­a­rithm of $\left( e^{D} \right)^{'}$ is straight­for­ward in this di­ag­o­nal­ized form

with which the $\left( x_{1}\alpha_{k}X_{1} \right)^{'}$ de­riv­a­tive al­ge­bra can now be de­ter­mined by the prod­uct $VD^{'}V^{- 1}$

which sim­pli­fies to

Like our other $\left( {x_{j}X}_{j,k} \right)^{'}$ de­riv­a­tive al­ge­bras, $\left( x_{1}\alpha_{k}X_{1} \right)^{'}$ is de­fined by its anti-de­riv­a­tive Lie al­ge­bra in ad­di­tion to the de­riv­a­tive-in­duced phase-shifts. The ex­po­nen­tial map­ping of this de­riv­a­tive al­ge­bra yields the fol­low­ing matrix group

which, after fac­tor­ing out the $\alpha_{k}$ terms, $G\left( \left( x_{1}\alpha_{k}X_{1} \right)^{'} \right)$ be­comes

and it is im­por­tant to keep in mind the equiv­a­lences: $G\left( \left( x_{1}\alpha_{k}X_{1} \right)^{'} \right) = G^{'}\left( x_{1}\alpha_{k}X_{1} \right) = G^{'}\left( x_{1}X_{1,k} \right)$. We pro­ceed by solv­ing the left-hand side of the fol­low­ing matrix-prod­uct equa­tion

find­ing that

and since $\alpha_{k}$ is a com­plex-valued scalar term, the left-hand side is equal to $\alpha_{k}X_{1}$

Next, we mul­ti­ply both sides of the ex­pres­sion by the group $G\left( x_{1}\alpha_{k}X_{1} \right)$

and by 1.2.98, we have that

which shows the gen­er­al­ized group dif­fer­en­tial ex­pres­sion $G^{'}\left( {x_{j}\alpha}_{k}X_{j} \right)$

holds for the $X_{1}$-type al­ge­bras. It is straight­for­ward to show that this holds for all $X_{j}$-type al­ge­bras, seen by uti­liz­ing the de­f­i­n­i­tion of the matrix-by-scalar de­riv­a­tive for $G^{'}\left( {x_{2}\alpha}_{k}X_{2} \right)$

and for $G^{'}\left( {x_{3}\alpha}_{k}X_{3} \right)$

Given that $G^{'}\left( {x_{j}\alpha}_{k}X_{j} \right) = \alpha_{k}X_{j}e^{{x_{j}\beta}_{k}X_{j}}$, the higher-order de­riv­a­tives for the $X_{j}$-type al­ge­bras will then have the form

where the $n$ in the su­per­script of $G^{'\lbrack n\rbrack}\left( {x_{j}\alpha}_{k}X_{j} \right)$ rep­re­sents the de­riv­a­tive's order. It turns out that de­riv­a­tives of the $Y_{j}$-type al­ge­bras $G^{'}\left( {y_{j}\alpha}_{k}Y_{j} \right)$will all behave sim­i­lar­ly. Using the ex­am­ple of a $Y_{2}$-type al­ge­bra, the gen­er­al­ized group $G\left( y_{2}\alpha_{k}Y_{2} \right)$ has the matrix rep­re­sen­ta­tion

where the cyclic trigono­met­ric func­tions $\mathfrak{O}_{l}^{3}\left( \beta_{k} \right)$ in their ex­pand­ed forms are

Cal­cu­lat­ing the matrix group de­riv­a­tive $G^{'}\left( y_{j}\alpha_{k}Y_{j} \right)$ first re­quires eval­u­at­ing the in­di­vid­ual de­riv­a­tives of the cyclic trigono­met­ric func­tions. Start­ing with the $\mathfrak{O}_{0}^{3}\left( \alpha_{k} \right)$ func­tion, the de­riv­a­tive is found to be

which is equal to the prod­uct of $\alpha_{k}$ with the left modulo $\mathfrak{O}_{2}^{3}\left( \alpha_{k} \right)$ func­tion, seen in the group rep­re­sen­ta­tion from Eq. 1.2.108. As ex­pect­ed, the de­riv­a­tives for the other two func­tions are sim­i­lar

After fac­tor­ing out the uni­form scalar mul­ti­pli­er $\alpha_{k}$, $G^{'}\left( y_{2}\alpha_{k}Y_{2} \right)$ can be rep­re­sent­ed as

which rep­re­sents a scaled by $\alpha_{k}$ cyclic right­ward shift of the $G\left( y_{2}\alpha_{k}Y_{2} \right)$ group's el­e­ments. This cyclic shift is simply due to the $Y_{2}$ al­ge­bra's action on $G\left( y_{2}\alpha_{k}Y_{2} \right)$, seen by the $Y_{2}G\left( y_{2}\alpha_{k}Y_{2} \right)$ prod­uct

and we have shown that the fol­low­ing re­la­tion­ship holds for the $Y_{2}$-type al­ge­bras

The $G\left( y_{3}\alpha_{k}Y_{3} \right)$ group dif­fers from the $G\left( y_{2}\alpha_{k}Y_{2} \right)$ by vari­able choice and by the swap­ping of the $\mathfrak{O}_{2}^{3}\left( \alpha_{k} \right)$ and $\mathfrak{O}_{1}^{3}\left( \alpha_{k} \right)$ func­tions in its matrix rep­re­sen­ta­tion

The de­riv­a­tive of $G\left( y_{3}\alpha_{k}Y_{3} \right)$ has the form

which rep­re­sents a matrix scaled by $\alpha_{k}$, with a cyclic left­ward shift of the ini­tial $G\left( y_{3}\alpha_{k}Y_{3} \right)$ el­e­ments. This ori­ent­ed shift is the result of the $Y_{3}$-al­ge­bra's action on $G\left( y_{3}\alpha_{k}Y_{3} \right)$, which to­geth­er with the $\alpha_{j}$ scalar rep­re­sents the prod­uct

show­ing that

also holds for de­riv­a­tives of $G\left( y_{3}\alpha_{k}Y_{3} \right)$. De­riv­a­tives of the $G\left( {y_{1}\alpha}_{k}Y_{1} \right)$ group are triv­ial, given that the $Y_{1}$ al­ge­bra is the iden­ti­ty, where these reduce to

which is equiv­a­lent to $G^{'}\left( {y_{1}\alpha}_{k}Y_{1} \right) = {\alpha_{k}Y}_{1}G\left( y_{1}\alpha_{k}1 \right)$. Given that $G^{'}\left( {y_{1}\alpha}_{k}Y_{1} \right) = {\alpha_{k}Y}_{1}G\left( y_{1}\alpha_{k}1 \right)$, the higher-order de­riv­a­tives for the $Y_{j}$-type al­ge­bras also adhere to the form

where the $n$ in the su­per­script of $G^{'\lbrack n\rbrack}\left( {y_{j}\alpha}_{k}Y_{j} \right)$ rep­re­sents the de­riv­a­tive-order. It needs to be clar­i­fied that the prod­uct ${\alpha_{k}Y}_{j}G\left( y_{j}\alpha_{k}Y_{j} \right)$ turns out to be a com­mu­ta­tive one, mean­ing $G\left( y_{j}\alpha_{k}Y_{j} \right){\alpha_{k}Y}_{j} = {\alpha_{k}Y}_{j}G\left( y_{j}\alpha_{k}Y_{j} \right),$which can be rep­re­sent­ed as the left and right-sided prod­ucts of $Y_{3}$ with $M$, a cyclic-struc­tured matrix $M$

and the com­mu­ta­tiv­i­ty in this ex­am­ple is due to the cyclic nature of both ma­tri­ces. We should not assume com­mu­ta­tiv­i­ty holds be­tween $Y_{j}$ and non-cyclic matrix rep­re­sen­ta­tions, seen by left and right-sided prod­ucts of $Y_{3}$ with the $xX_{1}$ al­ge­bra

Upon ending this sec­tion, we call at­ten­tion to a sig­nif­i­cant result that is present in the $\left( x_{j}\alpha_{k}X_{j} \right)^{'}$ de­riv­a­tive al­ge­bras, which is that for $\alpha_{k} = (a + ib)$, phase-shifts will be present for $a \neq 0$, and for $b \neq \pm 1$ in the non-axis di­ag­o­nal en­tries of the $\left( x_{j}\alpha_{k}X_{j} \right)^{'}$ de­riv­a­tive al­ge­bras, except when $\alpha_{k}$ is purely $\pm$ uni­tary-imag­i­nary valued. Using the $\left( x_{1}\alpha_{k}X_{1} \right)^{'}$ ex­am­ple

when $\alpha_{k} = i$, we find that the non-axis di­ag­o­nal el­e­ments eval­u­ate to

Eval­u­at­ing the gen­er­al case for $\alpha_{k}= \mathbb{C}$, this will sim­pli­fy by first con­vert­ing $\alpha_{k}$ into polar co­or­di­nates

which, de­pend­ing on the values of $a,b$, the $atan2(b,a)$ func­tion eval­u­ates per the fol­low­ing

So, when $a = 0$ and $b$ is non-zero, $\alpha_{k}$ be­comes

and eval­u­at­ing the log­a­rithm term

in­di­cates the entry is zero-valued only when $b$ is uni­tary in mag­ni­tude. When $b$ is non-uni­tary in mag­ni­tude and $a$ is non-zero, the log­a­rithm term be­comes the fol­low­ing

and using the two polar form pos­si­bil­i­ties, $\alpha_{k} = re^{i\theta}$ and $re^{i\theta \pm i\pi}$, there are two pos­si­ble re­sults for $\ln\left( \alpha_{k}^{2} \right)$

The log­a­rithm term will then eval­u­ate to one of the two fol­low­ing pos­si­bil­i­ties

both of which will be solely imag­i­nary-valued, except when the radial mod­u­lus $r$ in the $\ln(r)$ term is non-uni­tary. This result shows that de­riv­a­tives of matrix groups have de­riv­a­tive Lie al­ge­bras that exist in com­plex space, and with our knowl­edge about the de­riv­a­tive op­er­a­tions on the $D_{3\mathbb{R}},\ D_{3\mathbb{C}},\ D_{3\mathbb{c}}$ matrix groups, we are now pre­pared to con­sid­er tan­gent spaces. The reader should con­sid­er the Lie al­ge­bra to be the kernel of the group it be­comes through the ex­po­nen­tial map­ping. The term kernel is ap­pro­pri­ate be­cause it con­tains the group's prop­er­ties and given the role it plays in the group de­riv­a­tive, it guides the said group's dy­nam­ics.