Masteries · Jason GlowneyArithmoí Book 1

Masteries

1.6.7 Transformations and Representation Theory

The Fouri­er trans­form rep­re­sents the means to de­ter­mine the an­gu­lar fre­quen­cy com­po­nents and or wavenum­ber com­po­nents of a wave­func­tion/signal. The Fouri­er trans­form in­volves mul­ti­ply­ing a func­tion by an ex­po­nen­tial series and then in­te­grat­ing the result, making it pos­si­ble to trans­form from the time and space do­mains to that of the fre­quen­cy domain and wave domain re­spec­tive­ly. We in­di­cate trans­formed func­tions with a hat above the func­tion and use T,X,T,X,for time-based and space-based func­tions. For a Fouri­er trans­form of a time-based wave­func­tion to the fre­quen­cy domain, the math­e­mat­i­cal rep­re­sen­ta­tion is

Tˆ(f)=∫−∞∞T(t)e−i(2πft)dt,\begin{array}{r} \widehat{T}(f) = \int_{- \infty}^{\infty}{T(t)e^{- i(2\pi ft)}dt}, \end{array}
(1.6.213)

and the trans­form from the space-domain to the wave domain,

Xˆ(k)=∫−∞∞X(x)e−i(2πkx)dx.\begin{array}{r} \widehat{X}(k) = \int_{- \infty}^{\infty}{X(x)e^{- i(2\pi kx)}dx}. \end{array}
(1.6.214)

The reader's at­ten­tion should be di­rect­ed to the change in vari­able above from time to fre­quen­cy (t→f)\ (t \rightarrow f), from po­si­tion to wavenum­ber (x→k)(x \rightarrow k), and that the Fouri­er trans­form takes the in­verse of the wave­form units used for the re­sult­ing trans­form. Recall that fre­quen­cy is the in­verse of a wave's period, f=T−1f = T^{- 1}, and sim­i­lar­ly for the wavenum­ber, there is an in­verse re­la­tion­ship to the wave­length, k=2πλ.k = \frac{2\pi}{\lambda}.

Just as we can trans­form a wave­form from the time and space do­mains to the Fouri­er domain, an in­verse trans­form will bring us from Fouri­er domain back into the time and space do­mains. Using sim­i­lar no­ta­tion to the above Fouri­er trans­forms, the in­verse trans­forms for time/space-do­mains are

T(t)=∫−∞∞Tˆ(f)ei(2πft) df,\begin{array}{r} T(t) = \int_{- \infty}^{\infty}{\widehat{T}(f)e^{i(2\pi ft)}\ df,} \end{array}
(1.6.215)

and

X(x)=∫−∞∞Xˆ(k)ei(2πkx) dk.\begin{array}{r} X(x) = \int_{- \infty}^{\infty}{\widehat{X}(k)e^{i(2\pi kx)}\ dk.} \end{array}
(1.6.216)

For these Fouri­er trans­forms to be an­a­lyt­ic, we re­quire that the in­te­grals are con­ver­gent and that the trans­formed wave­forms are con­tin­u­ous. To demon­strate, we take the ex­am­ple for the Fouri­er trans­form of the wave­form X(x)=cos(2πk0x)X(x) = cos(2\pi k_{0}x), where

Xˆ(k)=∫−∞∞cos(2πk0x)e−i(2πkx)dx=∫−∞∞(ei2πk0x+e−i2πk0x2)e−i(2πkx)dx\widehat{X}(k) = \int_{- \infty}^{\infty}{cos({2\pi k}_{0}x)}e^{- i(2\pi kx)}dx = \int_{- \infty}^{\infty}\left( \frac{e^{i{2\pi k}_{0}x} + e^{- i2\pi k_{0}x}}{2} \right)e^{- i(2\pi kx)}dx
=12∫−∞∞(e2π(k+k0)x+e2π(−k−k0)x)dx=12(δ(k+k0)+δ(−k−k0)),\begin{array}{r} = \frac{1}{2}\int_{- \infty}^{\infty}\left( e^{{2\pi(k + k}_{0})x} + e^{{2\pi( - k - k}_{0})x} \right)dx = \frac{1}{2}\left( \delta\left( k + k_{0} \right) + \delta\left( - k - k_{0} \right) \right), \end{array}
(1.6.217)

con­tin­u­ing, we take the Fouri­er trans­form X(x)=sin(2πk0x)X(x) = sin(2\pi k_{0}x), find­ing

Xˆ(k)=∫−∞∞sin⁡(2πk0x)e−i(2πkx)dx=12(δ(k+k0)−δ(−k−k0)).\begin{array}{r} \widehat{X}(k) = \int_{- \infty}^{\infty}{\sin\left( 2\pi k_{0}x \right)}e^{- i(2\pi kx)}dx = \frac{1}{2}\left( \delta\left( k + k_{0} \right) - \delta\left( - k - k_{0} \right) \right). \end{array}
(1.6.218)

Adding the two cal­cu­lat­ed trans­forms from above, we find

12(δ(k+k0)+δ(−k−k0))+12(δ(k+k0)−δ(−k−k0))=δ(k+k0),\begin{array}{r} \frac{1}{2}\left( \delta\left( k + k_{0} \right) + \delta\left( - k - k_{0} \right) \right) + \frac{1}{2}\left( \delta\left( k + k_{0} \right) - \delta\left( - k - k_{0} \right) \right) = \delta\left( k + k_{0} \right), \end{array}
(1.6.219)

for δ(k+k0)\delta\left( k + k_{0} \right), the Dirac delta func­tion, and by taking its in­verse trans­form, Euler's for­mu­la is the ex­pect­ed result

eik0x=∫−∞∞δ(k+k0)ei(2πkx) dk.\begin{array}{r} e^{ik_{0}x} = \int_{- \infty}^{\infty}{\delta\left( k + k_{0} \right)e^{i(2\pi kx)}\ dk.} \end{array}
(1.6.220)

This gives us a Fouri­er domain rep­re­sen­ta­tion com­posed of Dirac delta func­tions, which war­rants fur­ther dis­cus­sion. The Dirac delta func­tion is a rep­re­sen­ta­tion of a unit im­pulse, which equals zero for all values except k=0k = 0, and this func­tion is typ­i­cal­ly taken to rep­re­sent the unit im­pulse of a point charge in physics. The func­tion at zero is in­fi­nite­ly narrow and in­fi­nite­ly tall, where the in­te­gral of the func­tion is equal to one, and math­e­mat­i­cal­ly, it is de­fined, as

δ(k)={+∞ for k=00 for k ≠0, \begin{array}{r} \delta(k) = \left\{ \begin{array}{r} + \infty\ for\ k = 0 \\ 0\ for\ k\ \neq 0 \end{array},\ \right. \end{array}
(1.6.221)

with an in­te­gral equal to unity,

∫−∞∞δ(k) dk=1.\begin{array}{r} \int_{- \infty}^{\infty}{\delta(k)\ dk = 1.} \end{array}
(1.6.222)

A plot of the Dirac delta func­tion δ(k)\delta(k), is shown in Figure 1.6.19, where it should be clear to see that the func­tion is equal to zero every­where except for a line at k=0k = 0, which rises to in­fin­i­ty.

Figure
Figure 1.6.19– The Dirac delta func­tion δ(k)\delta(k)

Re­turn­ing to the Fouri­er trans­form from the pre­vi­ous ex­am­ple, Xˆ(k)=cos(2πk0x)=12(δ(k−k0)+δ(k+k0))\widehat{X}(k) = cos\left( 2\pi k_{0}x \right) = \frac{1}{2}\left( \delta\left( k - k_{0} \right) + \delta\left( k + k_{0} \right) \right), we see that it is com­prised of two im­puls­es in the Fouri­er domain that are non-zero only for k=±k0.k = \pm k_{0}.The fact that an in­fi­nite trigono­met­ric series can be sim­pli­fied into im­pulse spikes in the Fouri­er domain is pro­found. This spike is anal­o­gous to the im­pulse cre­at­ed by a pebble hit­ting the sur­face of an undis­turbed body of water, and the cir­cu­lar waves em­a­nat­ing away from the im­pulse would rep­re­sent its in­verse Fouri­er trans­form, de­pict­ed as a sur­face plot for the sinc func­tion, in Figure 1.6.20.

Figure
Figure 1.6.20 – The sinc func­tion as the unit im­pulse and its effect in the time domain.

The sinc func­tion, also known as the sine car­di­nal func­tion, is fre­quent­ly used in signal pro­cess­ing and Fouri­er trans­form theory to rep­re­sent these unit im­puls­es, rep­re­sent­ed as

sinc(x)={      1     for x=0sin⁡(x)x for x≠0 ,\begin{array}{r} sinc(x) = \left\{ \begin{array}{r} \ \ \ \ \ \ 1\ \ \ \ \ for\ x = 0 \\ \frac{\sin(x)}{x}\ for\ x \neq 0 \end{array} \right.\ , \end{array}
(1.6.223)

which is close­ly re­lat­ed to the spher­i­cal Bessel func­tion of the first kind,

sinc(z)=j0(z)=∑k=0∞(−1)kz2k(2k+1)!,   z∈ℂ,\begin{array}{r} sinc(z) = j_{0}(z) = \sum_{k = 0}^{\infty}\frac{( - 1)^{k}z^{2k}}{(2k + 1)!},\ \ \ z\in \mathbb{C}, \end{array}
(1.6.224)

The spher­i­cal Bessel func­tion of the first kind is re­lat­ed to the Bessel func­tion of the first kind, by

jv(z)=π2z J(v+12)(z),\begin{array}{r} j_{v}(z) = \sqrt{\frac{\pi}{2z}}\ J_{\left( v + \frac{1}{2} \right)}(z), \end{array}
(1.6.225)

for v,z∈ℂv,z\in \mathbb{C}, and J(v+12)J_{\left( v + \frac{1}{2} \right)}, the Bessel func­tion of the first kind. Eval­u­at­ing the fol­low­ing in­te­gral

∫−a2a2ei2πftdt= ei2πfti2πf|−a2a2=e(i2πfa2)−e(−i2πfa2)i2πf=sin⁡(πfa)πf=asinc(πfa),\begin{array}{r} {\int_{- \frac{a}{2}}^{\frac{a}{2}}{e^{i2\pi ft}dt = \left. \ \frac{e^{i2\pi ft}}{i2\pi f} \right|}}_{- \frac{a}{2}}^{\frac{a}{2}} = \frac{e^{\left( \frac{i2\pi fa}{2} \right)} - e^{\left( - \frac{i2\pi fa}{2} \right)}}{i2\pi f} = \frac{\sin(\pi fa)}{\pi f} = asinc(\pi fa), \end{array}
(1.6.226)

re­sults in the sinc func­tion, which in its nor­mal­ized form can be rep­re­sent­ed, as

sinc(πx)=sin⁡(πx)πx,\begin{array}{r} sinc(\pi x) = \frac{\sin(\pi x)}{\pi x}, \end{array}
(1.6.227)

and by taking its limit as x→∞x \rightarrow \infty, it equals unity

limx→0⁡(sin⁡(πx)πx)=1.\begin{array}{r} \lim_{x \rightarrow 0}{\left( \frac{\sin(\pi x)}{\pi x} \right) = 1}. \end{array}
(1.6.228)

For the func­tion asinc(fa)asinc(fa), when taking the limit, as a→∞,a \rightarrow \infty,it be­comes in­fin­i­ty at f=0f = 0, making it clear that the ini­tial in­te­gral equals the Dirac delta func­tion with in­te­gra­tion bounds of neg­a­tive to pos­i­tive in­fin­i­ty

∫−∞∞ei2πftdt=lima→∞⁡(asinc(fa))=δ(f).\begin{array}{r} \int_{- \infty}^{\infty}{e^{i2\pi ft}dt} = \lim_{a \rightarrow \infty}{\left( asinc(fa) \right) = \delta(f).} \end{array}
(1.6.229)

At­ten­tion should be di­rect­ed to the re­la­tion­ship be­tween the sinc and cosine func­tions, by plot­ting the two func­tions in Figure 1.6.21 below, where both equal 11 at x=0x = 0, and the peaks and troughs of the sinc func­tion, rep­re­sent­ing its local ex­trema, cor­re­spond with in­ter­sec­tion points on the cosine func­tion's wave­form.

Figure
Figure 1.6.21 – Plots of sinc(x)sinc(x) and cos⁡(x)\cos(x)

From the in­tu­itive stand­point, the in­te­gral of a si­nu­soid over in­fi­nite time in both di­rec­tions is equal to zero, except for f=0f = 0, when sin⁡(0)=0\sin(0) = 0, but at the same time, cos⁡(0)=1\cos(0) = 1, yield­ing an in­fi­nite in­te­gral at this point.

In Lie theory we typ­i­cal­ly con­sid­er the ex­po­nen­ti­at­ed group forms as a trans­for­ma­tion of the tan­gent space by way of the ex­po­nen­tial map­ping of its Lie al­ge­bra. When we con­sid­er in­fi­nite in­te­gra­tion as a con­tin­u­um in the coun­ter­clock­wise and clock­wise di­rec­tions, pre­vi­ous­ly stated as

…+e3i𝒶mr33!+e2i𝒶mr22!+ei𝒶mr11!+e0×i𝒶mr00!+ei𝒶m(−r)11!+e2i𝒶m(−r)22!+e3i𝒶m(−r)33!+…,\begin{array}{r} \ldots + \frac{e^{3i\mathcal{a}_{m}}r^{3}}{3!} + \frac{e^{2i\mathcal{a}_{m}}r^{2}}{2!} + \frac{e^{i\mathcal{a}_{m}}r^{1}}{1!} + \frac{e^{0 \times i\mathcal{a}_{m}}r^{0}}{0!} + \frac{e^{i\mathcal{a}_{m}}({- r)}^{1}}{1!} + \frac{e^{2i\mathcal{a}_{m}}({- r)}^{2}}{2!} + \frac{e^{3i\mathcal{a}_{m}}({- r)}^{3}}{3!} + \ldots, \end{array}
(1.6.230)

we see in re­la­tion to Fouri­er trans­forms, that the in­fi­nite spike em­bod­ied in the Dirac delta and sinc func­tions is re­lat­ed to the iden­ti­ty el­e­ment sit­u­at­ed at (e0×i𝒶mr00!)\left( \frac{e^{0 \times i\mathcal{a}_{m}}r^{0}}{0!} \right). Taking the de­riv­a­tive of a Lie group rep­re­sen­ta­tion with its vari­able equal to zero yields that group's Lie al­ge­bra, and this process is re­liant on the even trigono­met­ric func­tions of the group's de­riv­a­tive. Much the same occurs when taking the in­verse Fouri­er trans­form of a wave func­tion, which has its root in the in­fi­nite spikes co­in­cid­ing with the even cosine func­tion. We use an ex­am­ple to il­lus­trate a Dirac delta rep­re­sen­ta­tion for the prod­uct of two matrix groups from the in­fi­nite sym­me­try group S3ℂ∞S_{3\mathbb{C}}^{\infty}

eZ=eX×eY=(ei2πf1t1000cos⁡(2πf1t1)isin(2πf1t1)0isin(2πf1t1)cos⁡(2πf1t1))(cos⁡(2πf2t2)0isin(2πf2t2)0ei2πf2t20isin(2πf2t2)0cos⁡(2πf2t2))=e^{Z} = e^{X} \times e^{Y} = \begin{pmatrix} e^{i2\pi f_{1}t_{1}} & 0 & 0 \\ 0 & \cos\left( 2\pi f_{1}t_{1} \right) & isin\left( 2\pi f_{1}t_{1} \right) \\ 0 & isin\left( 2\pi f_{1}t_{1} \right) & \cos\left( 2\pi f_{1}t_{1} \right) \end{pmatrix}\begin{pmatrix} \cos\left( 2\pi f_{2}t_{2} \right) & 0 & isin\left( 2\pi f_{2}t_{2} \right) \\ 0 & e^{i2\pi f_{2}t_{2}} & 0 \\ isin\left( 2\pi f_{2}t_{2} \right) & 0 & \cos\left( 2\pi f_{2}t_{2} \right) \end{pmatrix} =
(ei2πf1t1cos⁡(2πf2t2)0ei2πf1t1isin(2πf2t2)(isin(2πf1t1))(isin(2πf2t2))cos⁡(2πf1t1)ei2πf2t2isin(2πf1t1)cos⁡(2πf2t2)cos⁡(2πf1t1)sin⁡(2πf2t2)isin(2πf1t1)ei2πf2t2cos⁡(2πf1t1)cos⁡(2πf2t2)).\begin{array}{r} \begin{pmatrix} e^{i2\pi f_{1}t_{1}}\cos\left( 2\pi f_{2}t_{2} \right) & 0 & e^{i2\pi f_{1}t_{1}}isin\left( 2\pi f_{2}t_{2} \right) \\ \left( isin\left( 2\pi f_{1}t_{1} \right) \right)\left( isin\left( 2\pi f_{2}t_{2} \right) \right) & \cos\left( 2\pi f_{1}t_{1} \right)e^{i2\pi f_{2}t_{2}} & isin\left( 2\pi f_{1}t_{1} \right)\cos\left( 2\pi f_{2}t_{2} \right) \\ \cos\left( 2\pi f_{1}t_{1} \right)\sin\left( 2\pi f_{2}t_{2} \right) & {isin}\left( 2\pi f_{1}t_{1} \right)e^{i2\pi f_{2}t_{2}} & \cos\left( 2\pi f_{1}t_{1} \right)\cos\left( 2\pi f_{2}t_{2} \right) \end{pmatrix}. \end{array}
(1.6.231)

Mul­ti­ply­ing the left and right sides by the X,YX,Y in­te­gra­tion op­er­a­tors, we find the fol­low­ing matrix that rep­re­sents the prod­uct of sinc func­tions

(∫−a2a2dt10000∫−a2a2dt10∫−a2a2dt10)×eZ×(00∫−b2b2dt20∫−b2b2dt20∫−b2b2dt200)=\begin{pmatrix} \int_{- \frac{a}{2}}^{\frac{a}{2}}{dt_{1}} & 0 & 0 \\ 0 & 0 & \int_{- \frac{a}{2}}^{\frac{a}{2}}{dt_{1}} \\ 0 & \int_{- \frac{a}{2}}^{\frac{a}{2}}{dt_{1}} & 0 \end{pmatrix} \times e^{Z} \times \begin{pmatrix} 0 & 0 & \int_{- \frac{b}{2}}^{\frac{b}{2}}{dt_{2}} \\ 0 & \int_{- \frac{b}{2}}^{\frac{b}{2}}{dt_{2}} & 0 \\ \int_{- \frac{b}{2}}^{\frac{b}{2}}{dt_{2}} & 0 & 0 \end{pmatrix} =
(∫−a2a2∫−b2b2ei2πf1t1isin(2πf2t2) dt1dt20∫−a2a2∫−b2b2ei2πf1t1cos⁡(2πf2t2)dt1dt2∫−a2a2∫−b2b2cos⁡(2πf1t1)cos⁡(2πf2t2) dt1dt2∫−a2a2∫−b2b2isin(2πf1t1)ei2πf2t2dt1dt2∫−a2a2∫−b2b2cos⁡(2πf1t1)sin⁡(2πf2t2)dt1dt2∫−a2a2∫−b2b2isin(2πf1t1)cos⁡(2πf2t2) dt1dt2∫−a2a2∫−b2b2cos⁡(2πf1t1)ei2πf2t2 dt1dt2∫−a2a2∫−b2b2(isin(2πf1t1))(isin(2πf2t2))dt1dt2)\begin{pmatrix} \int_{- \frac{a}{2}}^{\frac{a}{2}}{\int_{- \frac{b}{2}}^{\frac{b}{2}}{e^{i2\pi f_{1}t_{1}}isin\left( 2\pi f_{2}t_{2} \right)\ dt_{1}dt_{2}}} & 0 & \int_{- \frac{a}{2}}^{\frac{a}{2}}{\int_{- \frac{b}{2}}^{\frac{b}{2}}{e^{i2\pi f_{1}t_{1}}\cos\left( 2\pi f_{2}t_{2} \right)dt_{1}dt_{2}}} \\ \int_{- \frac{a}{2}}^{\frac{a}{2}}{\int_{- \frac{b}{2}}^{\frac{b}{2}}{\cos\left( 2\pi f_{1}t_{1} \right)\cos{\left( 2\pi f_{2}t_{2} \right)\ dt_{1}dt_{2}}}} & \int_{- \frac{a}{2}}^{\frac{a}{2}}{\int_{- \frac{b}{2}}^{\frac{b}{2}}{{isin}\left( 2\pi f_{1}t_{1} \right)e^{i2\pi f_{2}t_{2}}dt_{1}dt_{2}}} & \int_{- \frac{a}{2}}^{\frac{a}{2}}{\int_{- \frac{b}{2}}^{\frac{b}{2}}{\cos\left( 2\pi f_{1}t_{1} \right)\sin\left( 2\pi f_{2}t_{2} \right)dt_{1}dt_{2}}} \\ \int_{- \frac{a}{2}}^{\frac{a}{2}}{\int_{- \frac{b}{2}}^{\frac{b}{2}}{isin\left( 2\pi f_{1}t_{1} \right)\cos{\left( 2\pi f_{2}t_{2} \right)\ dt_{1}dt_{2}}}} & \int_{- \frac{a}{2}}^{\frac{a}{2}}{\int_{- \frac{b}{2}}^{\frac{b}{2}}{\cos\left( 2\pi f_{1}t_{1} \right)e^{i2\pi f_{2}t_{2}}\ dt_{1}dt_{2}}} & \int_{- \frac{a}{2}}^{\frac{a}{2}}{\int_{- \frac{b}{2}}^{\frac{b}{2}}{(isin\left( 2\pi f_{1}t_{1} \right))(isin\left( 2\pi f_{2}t_{2}) \right)dt_{1}dt_{2}}} \end{pmatrix}
=(00sin⁡(πaf1)sin⁡(πbf2)(πf1)(πf2)sin⁡(πaf1)sin⁡(πbf2)(πf1)(πf2)000sin⁡(πaf1)sin⁡(πbf2)(πf1)(πf2)0).\begin{array}{r} = \begin{pmatrix} 0 & 0 & \frac{\sin\left( \pi af_{1} \right)\sin\left( \pi bf_{2} \right)}{\left( \pi f_{1} \right)\left( \pi f_{2} \right)} \\ \frac{\sin\left( \pi af_{1} \right)\sin\left( \pi bf_{2} \right)}{\left( \pi f_{1} \right)\left( \pi f_{2} \right)} & 0 & 0 \\ 0 & \frac{\sin\left( \pi af_{1} \right)\sin\left( \pi bf_{2} \right)}{\left( \pi f_{1} \right)\left( \pi f_{2} \right)} & 0 \end{pmatrix}. \end{array}
(1.6.232)

Taking the limit, as a,b→∞a,b \rightarrow \infty for the re­spec­tive sinc func­tions in the matrix above re­sults in a matrix with the fol­low­ing prod­uct of Dirac delta func­tions

(00δ(f1)δ(f2)δ(f1)δ(f2)000δ(f1)δ(f2)0).\begin{array}{r} \begin{pmatrix} 0 & 0 & \delta\left( f_{1} \right)\delta\left( f_{2} \right) \\ \delta\left( f_{1} \right)\delta\left( f_{2} \right) & 0 & 0 \\ 0 & \delta\left( f_{1} \right)\delta\left( f_{2} \right) & 0 \end{pmatrix}. \end{array}
(1.6.233)

In our ex­am­ple, when taking double in­te­grals over time, the matrix en­tries with sine func­tions will become zero, while those with cosine func­tions will become sinc func­tions since cos⁡(0)=1\cos{(0) = 1}. The re­sult­ing delta matrix shows us that the trans­formed matrix group is the prod­uct com­po­si­tion of two matrix groups, X1,X2X_{1},X_{2}

X×Y=(δ(f1)0000δ(f1)0δ(f1)0)×(00δ(f2)0δ(f2)0δ(f2)00)=(00δ(f1)δ(f2)δ(f1)δ(f2)000δ(f1)δ(f2)0),\begin{array}{r} X \times Y = \begin{pmatrix} \delta\left( f_{1} \right) & 0 & 0 \\ 0 & 0 & \delta\left( f_{1} \right) \\ 0 & \delta\left( f_{1} \right) & 0 \end{pmatrix} \times \begin{pmatrix} 0 & 0 & \delta\left( f_{2} \right) \\ 0 & \delta\left( f_{2} \right) & 0 \\ \delta\left( f_{2} \right) & 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 & \delta\left( f_{1} \right)\delta\left( f_{2} \right) \\ \delta\left( f_{1} \right)\delta\left( f_{2} \right) & 0 & 0 \\ 0 & \delta\left( f_{1} \right)\delta\left( f_{2} \right) & 0 \end{pmatrix}, \end{array}
(1.6.234)

and much like in Fouri­er analy­sis, with its com­po­si­tion of si­nu­soids, the matrix group wave gen­er­a­tors of Lie theory and the S3ℂ∞S_{3\mathbb{C}}^{\infty} sin­gu­lar­i­ty can be com­po­si­tions of matrix groups.

We have seen the link that in­fi­nite in­te­gra­tion has with ex­po­nen­ti­a­tion, and when con­sid­er­ing delta im­puls­es being in­te­grat­ed over in­fini­ties with Fouri­er trans­for­ma­tions, it is anal­o­gous to how the ex­po­nen­tial map­ping trans­forms a Lie al­ge­bra. To demon­strate this con­cept, a Lie al­ge­bra can be stated in terms of the Dirac delta dis­tri­b­u­tion func­tion, as

𝔰𝔲(X1)=(ik0x10000ik0x10ik0x10)→δ(δ(k+k0)0000δ(k+k0)0δ(k+k0)0),\begin{array}{r} \mathfrak{su}\left( X_{1} \right) = \begin{pmatrix} ik_{0}x_{1} & 0 & 0 \\ 0 & 0 & ik_{0}x_{1} \\ 0 & ik_{0}x_{1} & 0 \end{pmatrix}\overset{\delta}{\rightarrow}\begin{pmatrix} \delta\left( k + k_{0} \right) & 0 & 0 \\ 0 & 0 & \delta\left( k + k_{0} \right) \\ 0 & \delta\left( k + k_{0} \right) & 0 \end{pmatrix}, \end{array}
(1.6.235)

where (k+k0)\left( k + k_{0} \right), de­notes the im­puls­es which are lo­cat­ed a dis­tance k0k_{0} from the origin, and we typ­i­cal­ly set k0=1k_{0} = 1, to co­in­cide with our uni­tary study ap­proach. Mind­ful of this, for the Fouri­er trans­form, we have that

eik0x=∫−∞∞δ(k+k0)ei(2πkx) dk,\begin{array}{r} e^{ik_{0}x} = \int_{- \infty}^{\infty}{\delta\left( k + k_{0} \right)e^{i(2\pi kx)}\ dk,} \end{array}
(1.6.236)

but the matrix rep­re­sen­ta­tion would differ as a higher-di­men­sion­al form ∈ℂ3\in \mathbb{C}^{3}, rep­re­sent­ing a ro­tat­ing equi­lat­er­al tri­an­gle in the X2,X3X_{2},X_{3}-plane with an X1X_{1}-axis el­e­ment. The in­te­gra­tion op­er­a­tor that we pre­sent­ed ear­li­er acts to parse the delta func­tions from the lower right 2×22 \times 2 matrix ro­ta­tion subset into cosine and sine func­tions

(∫ei(2πkx)dk0000∫ei(2πkx)dk0∫ei(2πkx)dk0)(δ(k+k0)0000δ(k+k0)0δ(k+k0)0)\begin{pmatrix} \int_{}^{}{e^{i(2\pi kx)}dk} & 0 & 0 \\ 0 & 0 & \int_{}^{}{e^{i(2\pi kx)}dk} \\ 0 & \int_{}^{}{e^{i(2\pi kx)}dk} & 0 \end{pmatrix}\begin{pmatrix} \delta\left( k + k_{0} \right) & 0 & 0 \\ 0 & 0 & \delta\left( k + k_{0} \right) \\ 0 & \delta\left( k + k_{0} \right) & 0 \end{pmatrix}
=(δ(k+k0)000δ(k+k0)+δ(k−k0)2δ(k+k0)−δ(k−k0)20δ(k+k0)−δ(k−k0)2δ(k+k0)+δ(k−k0)2),\begin{array}{r} = \begin{pmatrix} \delta\left( k + k_{0} \right) & 0 & 0 \\ 0 & \frac{\delta\left( k + k_{0} \right) + \delta\left( k - k_{0} \right)}{2} & \frac{\delta\left( k + k_{0} \right) - \delta\left( k - k_{0} \right)}{2} \\ 0 & \frac{\delta\left( k + k_{0} \right) - \delta\left( k - k_{0} \right)}{2} & \frac{\delta\left( k + k_{0} \right) + \delta\left( k - k_{0} \right)}{2} \end{pmatrix}, \end{array}
(1.6.237)

which yields the in­fi­nite in­te­gra­tion fℂ−n(X1)f_{\mathbb{C}}^{- n}\left( X_{1} \right), ex­pressed using the in­fi­nite in­te­gra­tion op­er­a­tor ψ(X1)−n(−∞,∞){\psi\left( X_{1} \right)}^{- n( - \infty,\infty)}

ψ(X1)−n(−∞,∞)×(δ(k+k0)0000δ(k+k0)0δ(k+k0)0)=(eik0x000cos⁡(k0x)isin(k0x)0isin(k0x)cos⁡(k0x)).\begin{array}{r} {\psi\left( X_{1} \right)}^{- n( - \infty,\infty)} \times \begin{pmatrix} \delta\left( k + k_{0} \right) & 0 & 0 \\ 0 & 0 & \delta\left( k + k_{0} \right) \\ 0 & \delta\left( k + k_{0} \right) & 0 \end{pmatrix} = \begin{pmatrix} e^{ik_{0}x} & 0 & 0 \\ 0 & \cos\left( k_{0}x \right) & isin\left( k_{0}x \right) \\ 0 & isin\left( k_{0}x \right) & \cos\left( k_{0}x \right) \end{pmatrix}. \end{array}
(1.6.238)

The cen­tral tenet what we are con­sid­er­ing is that a deep con­nec­tion exists be­tween rep­re­sen­ta­tion theory and the Fouri­er trans­form. The im­puls­es in the tan­gent and sub­tan­gent spaces of the sin­gu­lar­i­ty trans­form into wave­forms in the ex­tend­ed space. The Baker-Camp­bell-Haus­dorff for­mu­la tells us that the prod­uct of two matrix groups is con­nect­ed to a Lie al­ge­bra that is part of the tan­gent space and com­posed of Lie brack­ets. These Lie al­ge­bras are rep­re­sen­ta­tion theory's ana­logue to the im­puls­es of Dirac delta dis­tri­b­u­tion for­mu­la.

When con­sid­er­ing the sin­gu­lar­i­ty, we will need to expand our notion of the trans­for­ma­tion of the delta func­tion from an ori­en­ta­tion along the real-axis, to one that ori­ents not only on the com­plex unit circle but any­where on the com­plex unit sphere ∈ℂ3\in \mathbb{C}^{3} space. This means that for any delta rep­re­sen­ta­tion δ(k+k0)\delta\left( k + k_{0} \right), k0k_{0} is not ∈ℝ\in \mathbb{R}, but in­stead k0∈ℂk_{0}\in \mathbb{C}, and fur­ther, k0k_{0} is ori­ent­ed some­where in the higher di­men­sion­al­i­ty of the three com­plex planes X1,X2,X3X_{1},X_{2},X_{3}. The typ­i­cal treat­ment of the Dirac delta func­tion for k0∈ℝk_{0}\in \mathbb{R}, is shown in Figure 1.6.22, where an in­fi­nite­ly narrow line, rooted at k0=1,k_{0} = 1,goes off to in­fin­i­ty for δ(k+k0)\delta\left( k + k_{0} \right).

Figure
Figure 1.6.22 – δ(k+k0)\delta\left( k + k_{0} \right), for k0=1∈ℝk_{0} = 1 \in \mathbb{R}

To demon­strate delta im­puls­es on the sin­gu­lar­i­ty in ℂ3\mathbb{C}^{3} space, we con­sid­er the equi­lat­er­al tri­an­gle sit­u­at­ed with ver­tices on the sin­gu­lar­i­ty's bound­ary, de­not­ing the delta func­tion ∈ℂ3\in \mathbb{C}^{3}, by δ(k+(k1,k2,k3))\delta\left( k + \left( k_{1},k_{2},k_{3} \right) \right), for kjk_{j}, the lo­ca­tions of each vertex in the X1,X2,X3X_{1},X_{2},X_{3}-planes re­spec­tive­ly. Chos­ing k1=i, k2=i, k3=−ik_{1} = i,\ k_{2} = i,\ k_{3} = - i, we plot the com­plex spher­i­cal delta func­tion in Figure 1.6.23, for which, the reader's at­ten­tion should be drawn to the ori­en­ta­tion of the delta im­pulse lines from var­i­ous points on the sin­gu­lar­i­ty.

Figure
Figure 1.6.23 – The com­plex spher­i­cal delta func­tion δ(k+(ik1,ik2,−ik3))\delta\left( k + \left( {ik}_{1},{ik}_{2},{- ik}_{3} \right) \right)

De­f­i­n­i­tion (1.6.239) – The Spher­i­cal Delta Func­tion is the rep­re­sen­ta­tion of the Dirac delta func­tion in mm-di­men­sion­al com­plex space ℂm\mathbb{C}^{m}, de­not­ed as δz(μ+μj)\delta^{z}\left( \mu + \mu_{j} \right), where there are mm vari­ables for μj\mu_{j}, which is the dis­tance from the origin for each polyg­o­nal vertex in com­plex space, and μj\mu_{j} can rep­re­sent space, time, wavenum­ber, or a fre­quen­cy rep­re­sen­ta­tion. For m=1m = 1 and μ1∈ℝ\mu_{1}\in \mathbb{R}, the spher­i­cal delta func­tion re­duces to the Dirac delta func­tion δ(μ+μ1).\delta\left( \mu + \mu_{1} \right).

The im­puls­es of the com­plex spher­i­cal delta func­tion em­anate tan­gen­tial­ly from the sin­gu­lar­i­ty's bound­ary from each of the ver­tices and this cor­re­sponds well with the treat­ment of the tan­gent space in Lie theory. The trans­la­tion of the spher­i­cal delta func­tion into its Lie al­ge­bra rep­re­sen­ta­tion is straight­for­ward when we con­sid­er that the delta func­tion δ(k+1)\delta(k + 1) is equal to the iden­ti­ty for δ(k+k0)\delta\left( k + k_{0} \right), when k0=1k_{0} = 1. In­te­grat­ing the delta func­tion over neg­a­tive to pos­i­tive in­fin­i­ty

∫−∞∞δ(k+1) dk=1,\begin{array}{r} \int_{- \infty}^{\infty}{\delta(k + 1)\ dk} = 1, \end{array}
(1.6.240)

we know, based on the delta func­tion, is equal to zero every­where except at k0=1k_{0} = 1, where it is pos­i­tive in­fin­i­ty in value

δ(k+1)={+∞ for k0=10     for k0≠1 .\begin{array}{r} \delta(k + 1) = \left\{ \begin{matrix} {+ \infty\ for\ k}_{0} = 1 \\ 0\ \ \ \ \ for\ k_{0} \neq 1 \end{matrix} \right.\ . \end{array}
(1.6.241)

An ex­pres­sion for the re­la­tion­ship of the spher­i­cal delta func­tion to its Lie al­ge­bra in the in­fi­nite sym­me­try group 𝔰3ℂ∞\mathfrak{s}_{3\mathbb{C}}^{\infty}, is stated as

∫−∞∞δ(k+k1)dk1+∫−∞∞δ(k+k2)dk2+∫−∞∞δ(k+k3)dk3=𝔰3ℂ∞,\begin{array}{r} \int_{- \infty}^{\infty}{\delta\left( k + k_{1} \right)dk_{1} + \int_{- \infty}^{\infty}{\delta\left( k + k_{2} \right)dk_{2} +}\int_{- \infty}^{\infty}{\delta\left( k + k_{3} \right)dk_{3}}} = \mathfrak{s}_{3\mathbb{C}}^{\infty}, \end{array}
(1.6.242)

which can be de­not­ed using our in­te­gra­tion op­er­a­tor no­ta­tion

ψδ−n(−∞,∞)×δ(k+(k1,k2,k3))=𝔰3ℂ∞.\begin{array}{r} \psi_{\delta}^{- n( - \infty,\infty)} \times \delta\left( k + \left( k_{1},k_{2},k_{3} \right) \right) = \mathfrak{s}_{3\mathbb{C}}^{\infty}. \end{array}
(1.6.243)

The Lie al­ge­bra 𝔰3ℂ∞\mathfrak{s}_{3\mathbb{C}}^{\infty} of the in­fi­nite sym­me­try group takes on com­plex-valued ex­po­nents, where real-valued ex­po­nen­tials will not nec­es­sar­i­ly have pe­ri­od­ic­i­ty akin to wave­forms, and this is based on the ec­cen­tric­i­ty of the geom­e­try we are con­sid­er­ing. To ad­dress this, in the up­com­ing book Phe­nom­e­nol­o­gy and the Geom­e­try of Light, we will study the use of the Laplace trans­form to ensure math­e­mat­i­cal con­sis­ten­cy with Lie theory. The Laplace trans­form is stated as

ℒ{F(t)}=f(s)=∫0∞F(t)e−stdt,\begin{array}{r} \mathcal{L}\left\{ F(t) \right\} = f(s) = \int_{0}^{\infty}{F(t)e^{- st}dt,} \end{array}
(1.6.244)

with a trans­form vari­able s∈ℂs\in \mathbb{C}, mean­ing it can take both real and imag­i­nary number values, en­com­pass­ing the math­e­mat­i­cal frame­work nec­es­sary for the trans­for­ma­tions of the com­plex spher­i­cal delta func­tion. The reader should be mind­ful that when taking the Laplace trans­form of a wave func­tion, it will simply reduce to the Fouri­er trans­form.