1.1.21 Polygonal Rotations in $U(3)$

It is es­sen­tial to gain an in­tu­itive grasp of how in­fin­i­tes­i­mal ro­ta­tions of the equi­lat­er­al tri­an­gle might occur. Using the ex­am­ple for our ex­po­nen­ti­at­ed el­e­ment $X_{1}$, and re­call­ing that this per­mu­ta­tion re­flects the points $B$ and $C$ of the equi­lat­er­al tri­an­gle about an axis pass­ing through the point $A$, which splits the tri­an­gle into equal halves. The $U(3)$ gen­er­a­tor $G = e^{X_{1}}$ acts on the iden­ti­ty el­e­ment $\epsilon = (A,B,C)'$, and has the matrix-vector prod­uct rep­re­sen­ta­tion

where for clar­i­ty, the angle of ro­ta­tion is $\theta$. The re­sult­ing matrix iden­ti­ty-vector prod­uct ro­tates the vertex $A$ around the unit circle in the $X_{1}$-plane by $\angle\theta$, while the ver­tices $B,C$ are ro­tat­ing in the $X_{2}X_{3}$-planes by $\angle\theta$. The ro­ta­tion for the vertex $A$ around its fixed ge­o­des­ic is the same ro­ta­tion of a point around the com­plex unit circle like the $U(1)$ case, and, as such, the group $U(1)$ is em­bed­ded in the $G\left( X_{1} \right)$ gen­er­a­tor by de­fault. Since the ro­ta­tion of the $B,C$ ver­tices in the $X_{2}/X_{3}$-plane may be less in­tu­itive; we can con­sid­er it fur­ther in two dif­fer­ent ways. The first is to follow the real and imag­i­nary vector com­po­nents of the re­spec­tive $B,C$ ver­tices in­di­vid­u­al­ly by taking each $X_{2}$ and $X_{3}$-plane sep­a­rate­ly. To con­sid­er this geo­met­ri­cal­ly, we set $\angle\theta = \frac{\pi}{8}$, to then cal­cu­late our matrix-vector prod­uct as

and gath­er­ing like vertex com­po­nents by re­spec­tive planes, we have

In Figure 1.1.12, we graph these ${\widehat{x}}_{j}$ vec­tors, seeing that they main­tain the polyg­o­nal form of the equi­lat­er­al tri­an­gle, pic­tured in blue, with ro­tat­ing ver­tices on the sphere's sur­face. The re­la­tion­ship in Eq. 1.1.25 is the result of taking the iden­ti­ty $\epsilon$ to be equal to the fol­low­ing $3 \times 3$ matrix

and $G\left( X_{1}\left( \frac{i\pi}{8} \right) \right) \bullet \epsilon$ be­comes

where each column rep­re­sents the re­spec­tive ${\widehat{x}}_{j}$ axis vector. When con­sid­er­ing the matrix gen­er­a­tor $G\left( X_{1}\left( \frac{i\pi}{8} \right) \right)$ acting on a vector by left mul­ti­pli­ca­tion, per­form­ing a sin­gu­lar value de­com­po­si­tion of the matrix can give us a better idea of what steps happen to make this ro­ta­tion occur. The reader can refer to Sec­tion 1.1.8 to review the SVD of a matrix if needed. As $G\left( X_{1}\left( \frac{i\pi}{8} \right) \right)$ is di­ag­o­nal­iz­able, we take our $\Sigma$ to be a $3 \times 3$ matrix with the eigen­val­ues along the di­ag­o­nal, and our uni­tary ma­tri­ces, $V,V^{\dagger}$, are just the eigen­vec­tors arranged as columns in the $V$ matrix, co­in­cid­ing with the re­spec­tive $\sigma_{j}$ en­tries in $\Sigma$. The SVD can then be ex­pressed as

where $\Sigma,V,V^{\dagger}$ are

Above the eigen­vec­tors in $V$ are nor­mal­ized, and $V$ is equal to its con­ju­gate trans­pose $V^{\dagger}$. Set­ting our iden­ti­ty vector to $\widehat{\epsilon} = (A,B,C)^{T}$, for $A,B,C = 1$, choos­ing an angle value $\theta = \frac{\pi}{4}$, and then cal­cu­lat­ing the prod­uct $V^{\dagger}\widehat{\epsilon}$, we find

This shows that $B$ is taken to $B^{'} = \left( \frac{C}{\sqrt{2}\ } - \frac{B}{\sqrt{2}\ } \right) = 0$, the origin $C$ is taken to be $C^{'} = \left( \frac{C}{\sqrt{2}\ } + \frac{B}{\sqrt{2}\ } \right) = \sqrt{2}$, and that the $A$ vertex re­mains un­changed $A' = A$. Since $A,B,C$ are uni­tary in mag­ni­tude, the two-norm of the re­sult­ing vector $\left| {\widehat{\epsilon}}^{'} \right|$is the same as the iden­ti­ty vector's two-norm $\left| \widehat{\epsilon} \right|$, where both are equal to $\sqrt{3}$

For our case, the two-norm is the mag­ni­tude of a vector $\in \mathbb{C}^{3}$ with com­po­nents $A,B,C$, rep­re­sent­ing the co­or­di­nate values of the tri­an­gle's ver­tices in the $X_{1},X_{2},X_{3}$-planes re­spec­tive­ly. So, the result of $V^{\dagger}\widehat{\epsilon}$ is to rotate this tri­an­gle, whose ver­tices rep­re­sent a vector $\in \mathbb{C}^{3}$, with an ini­tial vector start­ing po­si­tion $\widehat{\epsilon} = (1,1,1)$, to the co­or­di­nates ${\widehat{\epsilon}}^{'} = \left( 1,0,\sqrt{2} \right)$. The ro­ta­tion of the tri­an­gle and its vector rep­re­sen­ta­tion re­sult­ing from $V^{\dagger}\widehat{\epsilon}$, is de­pict­ed in Figure 1.1.13, in which the start­ing tri­an­gle-$1$ with vertex co­or­di­nates $(1,1,1)$ has the vector rep­re­sen­ta­tion de­pict­ed as $(A,B,C)$, which both rotate as the result of the $V^{\dagger}\widehat{\epsilon}$ prod­uct to the tri­an­gle-$2$, de­pict­ed at co­or­di­nates $(1,0,\sqrt{2})$, along with the re­sult­ing vector rep­re­sen­ta­tion $(A',B',C').$This is con­sis­tent with what is the ex­pect­ed out­come from the $V^{\dagger}$ por­tion of the SVD, which acts to rotate the vector/tri­an­gle by ro­tat­ing the sphere.

The next step of the SVD is left mul­ti­pli­ca­tion by the $\Sigma$ matrix on the $\widehat{\epsilon}'$vector, which yields the${\ \widehat{\epsilon}}^{''}$ vector

Given that $\Sigma$ is a di­ag­o­nal­ized matrix of the eigen­val­ues of $G\left( X_{1}\left( \theta = \frac{\pi}{4} \right) \right)$, the $\Sigma{\widehat{\epsilon}}^{'}$ matrix-vector prod­uct scales ${\widehat{\epsilon}}^{'}$, and ac­cord­ing­ly, its geo­met­ric tri­an­gle/vector rep­re­sen­ta­tions. As the eigen­val­ues are com­plex and uni­tary, the vector $\widehat{\epsilon}'$is ro­tat­ed and scaled, as op­posed to solely the scal­ing seen for a $\Sigma$-matrix com­posed of real eigen­val­ues.

tri­an­gle-2, $\Sigma{\widehat{\epsilon}}^{'} = (A'',B'',C'')$: tri­an­gle-3, $\Sigma{\widehat{\epsilon}}^{''}(4) = (A''',B''',C''')^{T} = G\left( X_{1}(\theta) \right)\widehat{\epsilon}$: tri­an­gle-4.

In Figure 1.1.14, we track the path of the ini­tial vector points $A,B,C$, rep­re­sent­ing the tri­an­gle's ver­tices through the SVD de­com­po­si­tion $G\left( X_{1}\left( \theta = \frac{\pi}{4} \right) \right) = V\Sigma V^{\dagger}.$The matrix-vector prod­uct $V\widehat{\epsilon}''$brings us to the ul­ti­mate vector result, ${\widehat{\epsilon}}^{'''}$.

rep­re­sent­ing the linear trans­for­ma­tion of the $\widehat{\epsilon}$ vector, the result of the matrix-vector prod­uct $G\left( X_{1}\left( \theta = \frac{\pi}{4} \right) \right)\widehat{\epsilon}$.