Seminar Sophus Lie 3 (1993) 89–105

Generating Functions of π_{2n−1}(SU_{n})

Christian Gross

In many questions concerning homotopy groups of Lie groups G, especially of the unitary groups U_{m} and SU_{m}, it suffices to know the mere group structure of π_{n}(G). For this purpose one can consult tables. For example, it is known that

(1) π_{2n}(SU_{m}) = π_{2n}(U_{m}) = 0, π_{2n+1}(SU_{m}) ≅ π_{2n+1}(U_{m}) ≅ ℤ, m > n ∈ ℕ

by Bott’s periodicy theorem [1]. However, often we have to know representatives U: S^{n} → G for the generators of these homotopy groups.

One example for this situation is the Skyrme model [5] in theoretical nuclear physics, a chiral invariant effective field theory describing the low energy limit of the quantum chromodynamics (QCD). By compactification of euclidian space ℝ^{3} , resp., of space-time ℝ^{4} , the meson fields are differentiable functions Û: ℝ_{(t)} × S^{3} → SU_{N_{F}} , resp., U: S^{4} → SU_{N_{F}} , N_{F} being the number of flavors in the QCD (N_{F} = 2, resp., N_{F} = 3). In this model nucleons appear as topological soliton solutions of these field configurations. The number of nucleons described by a certain meson field Û can be computed by integration of the pullback

(2)  \frac{1}{24π^{2}} Û^{⋆}ω_{3} = \frac{1}{24π^{2}} Tr (L ∧ L ∧ L)

over the space manifold, with L: = Û^{†} dÛ, where ∧ is the wedge product of differential forms and ω_{3} is the generator of the de-Rham cohomology H_{3}(SU_{m}) ≅ H_{3}(U_{m}) ≅ ℝ for m ≥ 2.

The meson fields obey the field equations derived as Euler-Lagrange equations from a lagrangian L(U, dU) by variation of the action integral ∫_{S^{4}} L dV . Let ϵ^{μνρσ} denote the totally antisymmetrical Levi-Civita symbol, L_{μ} = U^{†} ∂_{μ}U and λ a coupling constant. Then for N_{F} = 3 the field equations involve an additional term

(3)  λ ϵ^{μνρσ} L_{μ}L_{ν}L_{ρ}L_{σ},

that describes anomalous processes of the QCD. (In (3) we have used the Einstein summation convention.) Unfortunately, it is impossible to build up the global corresponding term in the lagrangian from which (3) could be derived by variation. Instead by using π_{4}(SU_{3}) = 0 from (1) one argues that U can be extended to a differentiable function U′: D^{5} → SU_{3} from a five-dimensional disc D^{5} whose boundary ∂D^{5} is space-time S^{4} [7]. Now the corresponding term for (3), the so-called Wess-Zumino term [6], is λ ∫_{D^{5}}(U′)^{⋆}ω_{5} , with ω_{5} being the generator of H_{5}(SU_{m}) ≅ ℝ for m ≥ 3. Using Stokes’ theorem we can perform the integration along space-time which leads — at least locally — to (3).

For any possible extension U′ the result has to be unique. This is equivalent to the requirement that

λ ∫_{S^{5}}(Ũ)^{⋆}ω_{5} = 2πz, z ∈ ℤ,

where S^{5} is the 5-sphere which one obtains by gluing any two 5-cells D_{(1)}^{5} and D_{(2)}^{5} at space-time S^{4} = ∂D_{(1)}^{5} = ∂D_{(2)}^{5} together, and where we have defined Ũ = U′_{(1)} ∪ U′_{(2)}: S^{5} → SU_{3} as the corresponding extension to this 5-sphere. This forces λ to be set equal to \frac{1}{240π^{2}} by the following index theorem (cf. Bott, Seeley [2]). The factor \frac{1}{24π^{2}} in (2) can also be deduced from this conclusion. Recall L_{μ} = U^{†} ∂_{μ}U.

Theorem 1. For every map U: S^{2n−1} → U_{m} the integral

\begin{matrix}∫_{S^{2n−1}}(\frac{i}{2π} )^{n}\frac{(n − 1)!}{(2n − 1)!} U^{⋆}ω_{2n−1} = \\ ∫_{S^{2n−1}}(\frac{i}{2π} )^{n}\frac{(n − 1)!}{(2n − 1)!} ϵ^{μ_{1}μ_{2}···μ_{2n−1}} Tr (L_{μ_{1}}L_{μ_{2}}⋯ L_{μ_{2n−1}}) dx_{1} ∧ dx_{2} ∧ ⋯ ∧ dx_{2n−1} \end{matrix}

is an integer n(U). The assignment [U] ↦ n(U): π_{2n−1}(U_{m}) → ℤ is an isomorphism for m ≥ n.

We have seen that in the case of the Skyrme model, explicit representatives Û(t, ·): S^{3} → SU_{N_{F}} and Ũ: S^{3} → SU_{N_{F}} for the generators of π_{3}(SU_{N_{F}}) and π_{5}(SU_{N_{F}}) have physical significance. Thus it is worthwhile to look for such explicit representatives. This is the task of the following article.

For π_{3}(SU_{2}) there is the so-called Hedgehog Ansatz [5] where the field equations can be transformed into a differential equation for the radial part of this ansatz. Unfortunately, this is not transferable to π_{5}(SU_{m}), let alone π_{2n−1}(SU_{m}). In order to achieve such an extension we take the more mathematical point of view and do not demand our representatives to obey certain physical field equations. A first result is the following: having found a generator U of π_{2n−1}(SU_{n}) one also has a generator j ∘ U of π_{2n−1}(SU_{m}) for m > n through the inclusion

j: SU_{n} → SU_{m}, U ↦ ( \begin{matrix}U & 0 \\ 0 & 1l_{m−n} \end{matrix} ) ,

because of U^{⋆}ω_{k} = (j ∘ U)^{⋆}ω_{k}. On the other hand one obtains a generator i ∘ U of π_{2n−1}(U_{n}) (and thereby of π_{2n−1}(U_{m}) for m ≥ n) via the inclusion i: SU_{n} → U_{n}.

So the main problem is to find representatives for π_{2n−1}(SU_{n}). By looking at the Lie algebra of U_{n} and the use of the exponential map we make the following ansatz for a function of a (2n − 1)-dimensional disc D_{(1)}^{2n−1} into U_{n}: let H denote the hermitian operator

(4)  ( \begin{matrix}x_{0} & z_{1} & 0 \\ \overline{z_{1}} & −x_{0} & z_{2} & 0 \\ 0 & \overline{z_{2}} & x_{0} & z_{3} & ... \\ & 0 & \overline{z_{3}} & −x_{0} & ... & 0 \\ & & ... & ... & ... & z_{n−1} \\ & & & 0 & \overline{z_{n−1}} & ±x_{0} \end{matrix} ) ,

where z_{j} := x_{2j−1} + ix_{2j} for j = 1, …, n − 1. Let x = (x_{0}, x_{1}, …, x_{2n−2}) and define U_{1}: D_{(1)}^{2n−1} → U_{n} by U_{1}(x) = exp(iπH(x)). To obtain a representative of a generator of π_{2n−1}(U_{n}), resp., of π_{2n−1}(SU_{n}) one has to construct a second function Ũ_{1} of a second disc D_{(2)}^{2n−1} (northern and southern hemisphere) so that

U_{1} ∪ Ũ_{1} is a continous function of D_{(1)}^{2n−1} ∪ D_{(2)}^{2n−1} ≅ S^{2n−1} , well defined on the equator ∂D_{(1)}^{2n−1} = ∂D_{(2)}^{2n−1} ≅ S^{2n−2} . In order to get a generator we must make

sure that this “gluing process” is not trivial: if we were so careless as to choose Ũ_{1} so that U_{1}∪Ũ_{1} is symmetric about the equator, then we would obtain a candidate for the zero element of π_{2n−1}(U_{n}) instead of a generator. In this paper we shall carry out this program for n = 1, 2, 3.

In [3] Lundell has proven an iteration for the construction of representatives for generators of π_{2n−1}(SU_{n}). This iteration even leads to functions U: S^{2n−1} → SU_{n} directly, one doesn’t have to look for fitting second functions on the northern hemisperes. But unfortunately, as he himself admits, “the actual formulae are too complicated for reasonable calculation”. They do not inherit any symmetries between the matrix elements — like the ones built up by (4) — that allow for the calculation of the integral in Theorem 1. So this iteration is of more theoretical interest, whereas the representatives presented here could be of practical use whenever the problem of finding functions for the northern hemispheres is solved for n ≥ 4.

Using Clifford algebras Lundell and Tosa constructed representatives for generators of the stabe homotopy groups of SO, SU and Sp [4]. In the case of SU their formalism leads to functions U: S^{2n+1} → SU_{2^{n}}, so π_{2n−1}(SU_{n}), n ≥ 3 isn’t covered either.

A Generator of π_{1}(U_{1})

For the sake of illustration and completeness we begin by discussing the simplest case. The isomorphism U_{1} → S^{1} yields a representative for the generator of π_{1}(U_{1}). We also obtain this representative by using our scheme in (4). In this case we set:

H = (x_{0}), U_{1}(x) = U_{1}(x_{0}) = exp(iπx_{0}).

Here we have U_{1}(−1) = U_{1}(1) = −1. Therefore we can map D^{1} onto S^{1} by identifying 1 and −1 (and so we define our second function from D_{(2)}^{1} to S^{1} by

Ũ_{1}(x) = −1 = const). The mapping U: S^{1} → U_{1} we obtain is a homeomorphism and thus generates π(U_{1}). This is confirmed by our invoking Theorem 1: Because of (U_{1})^{⋆}ω_{1} = Tr [exp(−iπx_{0})iπ exp(iπx_{0})] dx_{0} = iπ dx_{0} , integration gives

∫_{−1}^{+1} (\frac{i}{2π} )^{1} \frac{0!}{1!} iπ dx_{0} = −1.

Note. Representatives for the other elements of π_{1}(U_{1}) are obtained by expanding the domain for U_{1} to be n · D^{1} = [−n, n]. Because of U_{1}(−n) = U_{1}(n) = (−1)^{n}, we can again identify n and −n and thereby transform n · D^{1} into S^{1} . Integration leads to

∫_{−n}^{+n} (\frac{i}{2π} )^{1} \frac{0!}{1!} iπ dx_{0} = −n.

If we keep D^{1} = [−1, 1] as domain, U_{n} = exp(inπx_{0}), resp., U_{−n} = exp(−inπx_{0}) is a representative for the n-th element of π_{1}(U_{1}).

A Generator of π_{3}(SU_{2})

Here we have SU_{2} ≅ S^{3} . Under this identification the identity on S^{3} is again a representative for the generator of π_{3}(SU_{2}). We are led to it through our scheme defined by (4) (remember z_{1} = x_{1} + ix_{2} ):

H = ( \begin{matrix}x_{0} & z_{1} \\ \overline{z_{1}} & −x_{0} \end{matrix} ) , U_{1}(x) = U_{1}(x_{0}, x_{1}, x_{2}) = exp(iπH(x)).

Evaluating the exponential map and using R^{2} := x_{0}^{2} + x_{1}^{2} + x_{2}^{2} , we obtain

U_{1}(x) = ( \begin{matrix}cos πR + i\frac{x_{0}}{R} sin πR & i\frac{z_{1}}{R} sin πR \\ i\frac{\overline{z_{1}}}{R} sin πR & cos πR − i\frac{x_{0}}{R} sin πR \end{matrix} ) .

Setting y_{0} := cos πR, y_{1} := \frac{x_{1}}{R} sin πR, y_{2} := \frac{x_{2}}{R} sin πR, and y_{3} := \frac{x_{0}}{R} sin πR, we get ∑_{j=0}^{3} y_{j}^{2} = 1 and realize the isomorphism χ: SU(2) → S^{3} as follows:

( \begin{matrix}y_{0} + iy_{3} & −y_{2} + iy_{1} \\ y_{2} + iy_{1} & y_{0} − iy_{3} \end{matrix} ) ↦ (y_{0}, y_{1}, y_{2}, y_{3}) ∈ S^{3}.

In particular, R = 1 yields

U_{1}(x) = ( \begin{matrix}−1 & 0 \\ 0 & −1 \end{matrix} ) ,

and therefore, similarly to the previous case, we can transform our function U_{1}: D^{3} → SU_{2} into a continous mapping U: S^{3} → SU_{2} by collapsing all points x ∈ D^{3} with R = 1 into one single point ∞, the “North Pole”.

For the evaluation of the integral in Theorem 1 we use the three independent parameters x_{0}, r_{1}, ϕ_{1} defined by z_{1} = r_{1}e^{iϕ_{1}} . We thus rather compute ψ^{⋆}(U_{1})^{⋆}ω_{3} instead of U^{⋆}ω_{3} , where ψ: ℝ × [0, 1] × [−π, +π] → D^{3} is defined by ψ(x_{0}, r_{1}, ϕ_{1}) = (x_{0}, r_{1} cos ϕ_{1}, r_{1} sin ϕ_{1}). By cyclic permutation under the trace we get

ψ^{⋆}(U_{1})^{⋆}ω_{3} = −3 · Tr{L_{x_{0}}[L_{ϕ_{1}}, L_{r_{1}}]} dx_{0} ∧ dr_{1} ∧ dϕ_{1}.

Next we compute the L_{μ} ’s, using the abbrevations c := cos πR, s := sin πR:

L_{x_{0}} = i ( \begin{matrix}\frac{πx_{0}^{2}}{R^{2}} + \frac{r_{1}^{2}sc}{R^{3}} & z_{1} [\frac{πx_{0}}{R^{2}} − \frac{x_{0}sc}{R^{3}} + i\frac{s^{2}}{R^{2}} ] \\ \overline{z_{1}} [\frac{πx_{0}}{R^{2}} − \frac{x_{0}sc}{R^{3}} − i\frac{s^{2}}{R^{2}} ] & −\frac{πx_{0}^{2}}{R^{2}} − \frac{r_{1}^{2}sc}{R^{3}} \end{matrix} ) ,  (5)

L_{ϕ_{1}} = i ( \begin{matrix}−\frac{r_{1}^{2}s^{2}}{R^{2}} & z_{1} [\frac{x_{0}s^{2}}{R^{2}} + i\frac{sc}{R}] \\ \overline{z_{1}} [\frac{x_{0}s^{2}}{R^{2}} − i\frac{sc}{R}] & \frac{r_{1}^{2}s^{2}}{R^{2}} \end{matrix} ) ,  (6)

L_{r_{1}} = ir_{1} ( \begin{matrix}\frac{πx_{0}}{R^{2}} − \frac{x_{0}sc}{R^{3}} & z_{1} [\frac{π}{R^{2}} + \frac{x_{0}^{2}sc}{r_{1}^{2}R^{3}} − i\frac{x_{0}s^{2}}{r_{1}^{2}R^{2}} ] \\ \overline{z_{1}} [\frac{π}{R^{2}} + \frac{x_{0}^{2}sc}{r_{1}^{2}R^{3}} + i\frac{x_{0}s^{2}}{r_{1}^{2}R^{2}} ] & −\frac{πx_{0}}{R^{2}} + \frac{x_{0}sc}{R^{3}} \end{matrix} ) .  (7)

This yields

[L_{ϕ_{1}}, L_{r_{1}}] = \frac{2ir_{1}}{R^{2}} ( \begin{matrix}−\frac{πr_{1}^{2}sc}{R} − \frac{x_{0}^{2}s^{2}}{R^{2}} & z_{1} [\frac{πx_{0}sc}{R} − \frac{x_{0}s^{2}}{R^{2}} − iπs^{2}] \\ \overline{z_{1}} [\frac{πx_{0}sc}{R} − \frac{x_{0}s^{2}}{R^{2}} + iπs^{2}] & \frac{πr_{1}^{2}sc}{R} + \frac{x_{0}^{2}s^{2}}{R^{2}} \end{matrix} )

and Tr\{L_{x_{0}}[L_{ϕ_{1}}, L_{r_{1}}]\} = \frac{4πr_{1}}{R^{2}} sin^{2}πR, from which we deduce

ψ^{⋆}(U_{1})^{⋆}ω_{3} = −\frac{12πr_{1}}{R^{2}} sin^{2}πR dx_{0} ∧ dr_{1} ∧ dϕ_{1},
resp.,  U^{⋆}ω_{3} = −\frac{12π}{R^{2}} sin^{2}πR dx_{0} ∧ dx_{1} ∧ dx_{2}.

By the transformation rule for integrals we obtain for the integral in theorem 1

I_{1} =∫_{S^{3}} −\frac{1}{24π^{2}} U^{⋆}ω_{3} =∫_{−1}^{+1}dx_{0} ∫_{0}^{\sqrt{1−x_{0}^{2}}} dr_{1}∫_{0}^{2π}dϕ_{1} \frac{r_{1}}{2πR^{2}} sin^{2}πR =∫_{0}^{+1}dx_{0} ∫_{0}^{\sqrt{1−x_{0}^{2}}} dr_{1} \frac{2r_{1}}{R^{2}} sin^{2}πR

(the integrand is even in x_{0} ). We choose new variables R, r_{1}^{2} , observe dR∧d(r_{1}^{2}) = \frac{2x_{0}r_{1}}{R} dx_{0} ∧ dr_{1} , and finally get

I_{1} = ∫_{0}^{+1}dR ∫_{0}^{R^{2}}d(r_{1}^{2})\frac{sin^{2}πR}{R\sqrt{R^{2} − r_{1}^{2}}} = ∫_{0}^{+1}2 sin^{2}πR dR = 1.

This confirmes that U is a representative for the generator of π_{3}(SU_{2}).

Note. As for π_{1}(U_{1}), we obtain representatives for all other elements of π_{3}(SU_{2}) by expanding our domain to the ball of radius R = n. For x with ∥x∥ = R = n we have

U_{1}(x) = ( \begin{matrix}(−1)^{n} & 0 \\ 0 & (−1)^{n} \end{matrix} ) .

For this reason even a mapping from n · D^{3} can be transformed into a continous mapping from S^{3} , resp., n · S^{3} , into S^{3} , which yields

I_{n} = ∫_{0}^{n} 2 sin^{2}πR dR = n.

If we want to keep our domain D^{3} , we simply replace x by nx and obtain U_{n}: D^{3} → SU_{2} :

U_{n}(x) = ( \begin{matrix}cos nπR + i\frac{x_{0}}{R} sin nπR & i\frac{z_{1}}{R} sin nπR \\ i\frac{\overline{z_{1}}}{R} sin nπR & cos nπR − i\frac{x_{0}}{R} sin nπR \end{matrix} )

as representative for the n-th element of π_{3}(SU_{2}). In order to get the inverse elements we replace x_{0} by −x_{0} , then L_{x_{0}} changes into −L_{x_{0}} and U^{⋆}ω_{3} changes into −U^{⋆}ω_{3} . We obtain U_{−n}: D^{3} → SU_{2} :

U_{−n}(x) = ( \begin{matrix}cos nπR − i\frac{x_{0}}{R} sin nπR & i\frac{z_{1}}{R} sin nπR \\ i\frac{\overline{z_{1}}}{R} sin nπR & cos nπR + i\frac{x_{0}}{R} sin nπR \end{matrix} ) .

A Generator of π_{5}(SU_{3})

There is no isomorphism between SU_{3} and a sphere and for the first time we will have to make use of the gluing process described in the introduction. Two mappings U_{1}: D_{(1)}^{5} → SU_{3} and Ũ_{1}: D_{(2)}^{5} → SU_{3} that coincide on the boundaries ∂D_{(1)}^{5} = ∂D_{(2)}^{5} = S^{4} , are transformed into a well defined continous

function U = U_{1} ∪ Ũ_{1}: S^{5} → SU_{3} . In analogy with (4) we have

H = ( \begin{matrix}x_{0} & z_{1} & 0 \\ \overline{z_{1}} & −x_{0} & z_{2} \\ 0 & \overline{z_{2}} & x_{0} \end{matrix} ) , \{ \begin{matrix}z_{1} = x_{1} + ix_{2} = r_{1}e^{iϕ_{1}} \\ z_{2} = x_{3} + ix_{4} = r_{2}e^{iϕ_{2}} \end{matrix} \} , ∑_{i=0}^{4} x_{i}^{2} = x_{0}^{2} + r^{2} = R^{2},

and a mapping U′_{1}: D^{5} → U_{3} defined by U′_{1}(x) = exp(iπH)(x).

det U′_{1} = exp(iπ Tr H) = exp(iπx_{0}),

so we have U′_{1}(D^{5}) ⊈ SU_{3} . Using the diagonalisation of H we compute

U′_{1}(x) = ( \begin{matrix}\frac{r_{1}^{2}}{r^{2}} (c + i\frac{x_{0}}{R} s) + \frac{r_{2}^{2}}{r^{2}} e^{iπx_{0}} & i\frac{z_{1}}{R} s & \frac{z_{1}z_{2}}{r^{2}} (c + i\frac{x_{0}}{R} s − e^{iπx_{0}}) \\ i\frac{\overline{z_{1}}}{R} s & c − i\frac{x_{0}}{R} s & i\frac{z_{2}}{R} s \\ \frac{\overline{z_{1}z_{2}}}{r^{2}} (c + i\frac{x_{0}}{R} s − e^{iπx_{0}}) & i\frac{\overline{z_{2}}}{R} s & \frac{r_{2}^{2}}{r^{2}} (c + i\frac{x_{0}}{R} s) + \frac{r_{1}^{2}}{r^{2}} e^{iπx_{0}} \end{matrix} ) ,

where we again used c = cos πR and s = sin πR for convenience. In order to obtain U_{1}: D^{5} → SU_{3} , we multiply every matrix U′_{1}(x) by a matrix T (x) of determinant det T (x) = exp(−iπx_{0}), preserving a convenient degree of symmetry between its elements. Thus we choose

(8) U_{1}(x) = T (x) · U′_{1}(x)  with  T (x) = ( \begin{matrix}e^{−i \frac{π}{2} x_{0}} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & e^{−i \frac{π}{2} x_{0}} \end{matrix} ) .

Using π^{±} := c ± i\frac{x_{0}}{R} s for further convenience we obtain

U_{1}(x) = ( \begin{matrix}\frac{r_{1}^{2}}{r^{2}} π^{+}e^{−i \frac{π}{2} x_{0}} + \frac{r_{2}^{2}}{r^{2}} e^{+i \frac{π}{2} x_{0}} & i\frac{z_{1}}{R} se^{−i \frac{π}{2} x_{0}} & \frac{z_{1}z_{2}}{r^{2}} (π^{+}e^{−i \frac{π}{2} x_{0}} − e^{+i \frac{π}{2} x_{0}}) \\ i\frac{\overline{z_{1}}}{R} s & π^{−} & i\frac{z_{2}}{R} s \\ \frac{\overline{z_{1}z_{2}}}{r^{2}} (π^{+}e^{−i \frac{π}{2} x_{0}} − e^{+i \frac{π}{2} x_{0}}) & i\frac{\overline{z_{2}}}{R} se^{−i \frac{π}{2} x_{0}} & \frac{r_{2}^{2}}{r^{2}} π^{+}e^{−i \frac{π}{2} x_{0}} + \frac{r_{1}^{2}}{r^{2}} e^{+i \frac{π}{2} x_{0}} \end{matrix} )

as the function on the southern hemisphere of S^{5} . On the equator (R = 1) it turns out to be

U_{1}(x) = ( \begin{matrix}−\frac{r_{1}^{2}}{r^{2}} e^{−i \frac{π}{2} x_{0}} + \frac{r_{2}^{2}}{r^{2}} e^{+i \frac{π}{2} x_{0}} & 0 & −\frac{z_{1}z_{2}}{r^{2}} (e^{+i \frac{π}{2} x_{0}} + e^{−i \frac{π}{2} x_{0}}) \\ 0 & −1 & 0 \\ \frac{\overline{z_{1}z_{2}}}{r^{2}} (e^{+i \frac{π}{2} x_{0}} + e^{−i \frac{π}{2} x_{0}}) & 0 & −\frac{r_{2}^{2}}{r^{2}} e^{−i \frac{π}{2} x_{0}} + \frac{r_{1}^{2}}{r^{2}} e^{+i \frac{π}{2} x_{0}} \end{matrix} ) .

Obviously this is not constant like in the previous cases, so it is impossible to contract the boundary into one single point, but we have to look for a nontrivial mapping Ũ_{1} on the northern hemisphere, that coincides with U_{1} on the equator. There are two possibilities:

Ũ_{1}(x)=(\begin{matrix}−\frac{r_{2}^{2}}{r^{2}} π^{−}e^{+i \frac{π}{2} x_{0}} − \frac{r_{1}^{2}}{r^{2}} e^{−i \frac{π}{2} x_{0}} & ±i\frac{z_{2}}{R} se^{+i \frac{π}{2} x_{0}} & \frac{z_{1}z_{2}}{r^{2}} (π^{−}e^{+i \frac{π}{2} x_{0}} − e^{−i \frac{π}{2} x_{0}}) \\ ∓i\frac{\overline{z_{2}}}{R} s & π^{+} & ±i\frac{z_{1}}{R} s \\ \frac{\overline{z_{1}z_{2}}}{r^{2}} (π^{−}e^{+i \frac{π}{2} x_{0}} − e^{−i \frac{π}{2} x_{0}}) & ∓i\frac{\overline{z_{1}}}{R} se^{+i \frac{π}{2} x_{0}} & −\frac{r_{1}^{2}}{r^{2}} π^{−}e^{+i \frac{π}{2} x_{0}} − \frac{r_{2}^{2}}{r^{2}} e^{−i \frac{π}{2} x_{0}} \end{matrix}).

To secure the property of being unitary we have to choose either the upper or the lower signs. Once the choice has been made, it propagates to all products, its derivations and inverses, and so — by forming the trace at the end of the computation of (Ũ_{1})^{⋆}ω_{5} — does not influence the value of this pullback. In the following we choose the upper signs.

Again we will use polar coordinates for the evaluation of our 5-form: let K = ℝ × ℝ_{0}^{+} × ℝ_{0}^{+} × [−π, +π] × [−π, +π] and define ψ: K → ℝ^{5} by ψ(x_{0}, r_{1}, r_{2}, ϕ_{1}, ϕ_{2}) = (x_{0}, r_{1} cos ϕ_{1}, r_{1} sin ϕ_{1}, r_{2} cos ϕ_{2}, r_{2} sin ϕ_{2}), resp., restrict K to ψ^{−1}(D^{5}). By cyclic permutation under the trace we then obtain

(9)  ψ^{⋆}(U_{1})^{⋆}ω_{5} = 5 · Tr{L · L_{x_{0}}} dx_{0} ∧ dr_{1} ∧ dr_{2} ∧ dϕ_{1} ∧ dϕ_{2}

with the hermitian matrix

(10) \begin{aligned}L & = +[L_{ϕ_{1}}, L_{ϕ_{2}}][L_{r_{1}}, L_{r_{2}}] − [L_{ϕ_{1}}, L_{r_{1}}][L_{ϕ_{2}}, L_{r_{2}}] + [L_{ϕ_{1}}, L_{r_{2}}][L_{ϕ_{2}}, L_{r_{1}}] \\ & +[L_{r_{1}}, L_{r_{2}}][L_{ϕ_{1}}, L_{ϕ_{2}}] − [L_{ϕ_{2}}, L_{r_{2}}][L_{ϕ_{1}}, L_{r_{1}}] + [L_{ϕ_{2}}, L_{r_{1}}][L_{ϕ_{1}}, L_{r_{2}}] \end{aligned}

(L^{†} = L is a consequence of [L_{μ}, L_{ν}]^{†} = −[L_{μ}, L_{ν}], which itself follows from L_{μ}^{†} = −L_{μ}, cf. (5) to (7)).

The computation of ψ^{⋆}(U_{1})^{⋆}ω_{5} is straightforward but long and tedious. We have collected the main steps in the appendix. We end up with (14):

ψ^{⋆}(U_{1})^{⋆}ω_{5} =30iπ \frac{r_{1}r_{2}}{R^{3}} [−π sin^{2}πR(sinπR cosπx_{0} − \frac{x_{0}}{R} cosπR sinπx_{0})

+ 2π sinπR(cosπx_{0}−cosπR) + (2\frac{R^{2}}{r^{2}} +2)\frac{sin^{2}πR}{R}(1−cosπR cosπx_{0})

− (2\frac{R^{2}}{r^{2}} +1)\frac{x_{0}}{R^{2}} sin^{3}πR sinπx_{0}]dx_{0}∧dr_{1}∧dr_{2}∧dϕ_{1}∧dϕ_{2}.

For the mapping on the northern hemisphere, it turns out that — cf. (15) —

(Ũ_{1})^{⋆}ω_{5} = \overline{(U_{1})^{⋆}ω_{5}} = −(U_{1})^{⋆}ω_{5}.

Fortunately, the negative sign compensates the factor (−1), that arises as a consequence of the opposite orientation of the northern hemisphere. So both integrals yield the same value:

I_{1} = ∫_{S^{5},“south”}−\frac{i}{480π^{3}} (U_{1})^{⋆}ω_{5} + ∫_{S^{5},“north”}−\frac{i}{480π^{3}} (Ũ_{1})^{⋆}ω_{5}

= 2 ∫_{S^{5},“south”}−\frac{i}{480π^{3}} (U_{1})^{⋆}ω_{5} = 2 ∫_{ψ^{−1}(D^{5})} −\frac{i}{480π^{3}} ψ^{⋆}(U_{1})^{⋆}ω_{5}.

Because ψ^{⋆}(U_{1})^{⋆}ω_{5} is even in x_{0} , we integrate twice over positive values of x_{0} , the integration over ϕ_{1} and ϕ_{2} just yields the factor 4π^{2} . Using new variables R, x_{0} and r_{1}^{2} and observing dR ∧ dx_{0} ∧ dr_{1}^{2} = 2\frac{r_{1}r_{2}}{R} dx_{0} ∧ dr_{1} ∧ dr_{2} , we obtain

I_{1} =∫_{0}^{1} \frac{dR}{R^{2}} ∫_{0}^{R} dx_{0} ∫_{0}^{R^{2}−x_{0}^{2}}[−π sin^{2}πR(sinπR cosπx_{0} − \frac{x_{0}}{R} cosπR sinπx_{0})

+ 2π sinπR(cosπx_{0} − cosπR) − (2\frac{R^{2}}{r^{2}} + 1)\frac{x_{0}}{R^{2}} sin^{3}πR sinπx_{0}

+ (2\frac{R^{2}}{r^{2}} + 2)\frac{sin^{2}πR}{R}(1 − cosπR cosπx_{0})]dr_{1}^{2}

=∫_{0}^{1} \frac{dR}{R^{2}} ∫_{0}^{R}[−π(R^{2}−x_{0}^{2}) sin^{2}πR(sinπR cosπx_{0} − \frac{x_{0}}{R} cosπR sinπx_{0})

+ 2π(R^{2}−x_{0}^{2}) sinπR(cosπx_{0} − cosπR) − (3R^{2}−x_{0}^{2})\frac{x_{0}}{R^{2}} sin^{3}πR sinπx_{0}

+ (4R^{2}−2x_{0}^{2})\frac{sin^{2}πR}{R}(1 − cosπR cosπx_{0})]dx_{0}.

Partial integration yields:

I_{1} = ∫_{0}^{1} [+2 sin^{2}πR − \frac{2}{3}πR sinπR cosπR − \frac{1}{3} sin^{2}πR

−2\frac{sinπR cosπR}{πR} + \frac{sin^{2}πR}{π^{2}R^{2}} + 4\frac{sin^{3}πR cosπR}{π^{3}R^{3}} − 3\frac{sin^{4}πR}{π^{4}R^{4}}] dR

= ∫_{0}^{1} 2 sin^{2}πR dR + [−\frac{1}{3}R sin^{2}πR − \frac{sin^{2}πR}{π^{2}R} + \frac{sin^{4}πR}{π^{4}R^{3}}]_{R=0}^{R=1}

= 1 + 0 = 1.

This finally proves that our mapping constructed from U_{1} and Ũ_{1} represents the generator of π_{5}(SU_{3}).

Representatives for further Elements of π_{5}(SU_{3})

Having found a representative U for the generator [U] of π_{5}(SU_{3}), we could use standard techniques to construct representatives for the powers [U]^{n}, notably, since SU_{3} is a group. But neither of these is practical for an explicit numerical representation of a V_{n} with [V_{n}] = [U]^{n}. Fortunately, there is a simple technique due to the fact that we can expand the domains for U_{1} and Ũ_{1} . They not only can be glued together at R = 1, but as well at R = 2n + 1, yet not at R = 2n (n ∈ N_{0}).

I_{2n+1} = ∫_{0}^{2n+1} 2 sin^{2}πR dR = 2n + 1,

we thus easily obtain further representatives for all odd products of [U]. For even products U_{1} has to be combined with another function Û_{1}: 2n · D^{5} → SU_{3} . Choosing

Û_{1}(x)=(\begin{matrix}\frac{r_{2}^{2}}{r^{2}} π^{−}e^{+i \frac{π}{2} x_{0}} + \frac{r_{1}^{2}}{r^{2}} e^{−i \frac{π}{2} x_{0}} & i\frac{z_{2}}{R} se^{+i \frac{π}{2} x_{0}} & −\frac{z_{1}z_{2}}{r^{2}} (π^{−}e^{+i \frac{π}{2} x_{0}} −e^{−i \frac{π}{2} x_{0}}) \\ i\frac{\overline{z_{2}}}{R} s & π^{+} & i\frac{z_{1}}{R} s \\ −\frac{\overline{z_{1}z_{2}}}{r^{2}} (π^{−}e^{+i \frac{π}{2} x_{0}} −e^{−i \frac{π}{2} x_{0}}) & i\frac{\overline{z_{1}}}{R} se^{+i \frac{π}{2} x_{0}} & \frac{r_{1}^{2}}{r^{2}} π^{−}e^{+i \frac{π}{2} x_{0}} + \frac{r_{2}^{2}}{r^{2}} e^{−i \frac{π}{2} x_{0}} \end{matrix})

for points on the northern hemisphere, we recognize that Û_{1} can be glued together with U_{1} at R = 2n, since

Û_{1}(x) =(\begin{matrix}\frac{r_{1}^{2}}{r^{2}} e^{−i \frac{π}{2} x_{0}} + \frac{r_{2}^{2}}{r^{2}} e^{+i \frac{π}{2} x_{0}} & 0 & −\frac{z_{1}z_{2}}{r^{2}} (e^{+i \frac{π}{2} x_{0}} −e^{−i \frac{π}{2} x_{0}}) \\ 0 & 1 & 0 \\ \frac{\overline{z_{1}z_{2}}}{r^{2}} (e^{+i \frac{π}{2} x_{0}} −e^{−i \frac{π}{2} x_{0}}) & 0 & \frac{r_{2}^{2}}{r^{2}} e^{−i \frac{π}{2} x_{0}} + \frac{r_{1}^{2}}{r^{2}} e^{+i \frac{π}{2} x_{0}} \end{matrix})= U_{1}(x)

for all points x with ∥x∥ = R = 2n. Because of Û_{1}(x = 0) = 1l_{3} = U_{1}(x = 0), both North Pole and South Pole of S^{5} are mapped onto the base point of SU_{3} . Using the fact that the L_{μ} are invariant under left multiplications, we have

Û_{1}(x) = ( \begin{matrix}−1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & −1 \end{matrix} ) · Ũ_{1}(x) ⟹ L̂_{μ}(x) = L̃_{μ}(x).

This yields (Û_{1})^{⋆}ω_{5} = (Ũ_{1})^{⋆}ω_{5} and thus:

I_{2n} =  ∫_{2n·S^{5},“south”}−\frac{i}{480π^{3}} (U_{1})^{⋆}ω_{5} +  ∫_{2n·S^{5},“north”}−\frac{i}{480π^{3}} (Û_{1})^{⋆}ω_{5}

= 2 ∫_{2n·S^{5},“south”}−\frac{i}{480π^{3}} (U_{1})^{⋆}ω_{5} = ∫_{0}^{2n} 2 sin^{2}πR dR = 2n.

In order to obtain representatives for the corresponding inverse elements of π_{5}(SU_{3}) we replace x_{0} by −x_{0} , or define U_{1} to be the mapping of the northern hemisphere and Ũ_{1} , resp., Û_{1} to be the mapping of the southern hemisphere of S^{5} . If we replace x by 2nx , resp., by (2n + 1)x , we can keep D^{5} instead of 2n · D^{5} , resp., (2n + 1) · D^{5} as domain.

Representatives for Elements of π_{5}(U_{3})

As already mentioned in the introduction, U: S^{5} → SU_{3} constructed above also is a representative for the generator of π_{5}(U_{3}) via the inclusion i: SU_{3} → U_{3} . Alternatively, we can also use the function U′_{1}: D^{5} → U_{3} that we had obtained by (4) directly, to build up representatives for all the elements of π_{5}(U_{3}), as a short computation will show.

For the mappings on the northern hemisphere we define:

\begin{aligned}Ũ′_{1}(x) & = ( \begin{matrix}−\frac{r_{2}^{2}}{r^{2}} π^{−}e^{iπx_{0}} − \frac{r_{1}^{2}}{r^{2}} & i\frac{z_{2}}{R} se^{i \frac{π}{2} x_{0}} & \frac{z_{1}z_{2}}{r^{2}} (π^{−}e^{iπx_{0}} − 1) \\ −i\frac{\overline{z_{2}}}{R} se^{i \frac{π}{2} x_{0}} & π^{+} & i\frac{z_{1}}{R} se^{i \frac{π}{2} x_{0}} \\ \frac{\overline{z_{1}z_{2}}}{r^{2}} (π^{−}e^{iπx_{0}} − 1) & −i\frac{\overline{z_{1}}}{R} se^{i \frac{π}{2} x_{0}} & −\frac{r_{1}^{2}}{r^{2}} π^{−}e^{iπx_{0}} − \frac{r_{2}^{2}}{r^{2}} \end{matrix} ) , resp., \\ Û′_{1}(x) & = ( \begin{matrix}\frac{r_{2}^{2}}{r^{2}} π^{−}e^{iπx_{0}} + \frac{r_{1}^{2}}{r^{2}} & i\frac{z_{2}}{R} se^{i \frac{π}{2} x_{0}} & \frac{z_{1}z_{2}}{r^{2}} (−π^{−}e^{iπx_{0}} + 1) \\ i\frac{\overline{z_{2}}}{R} se^{i \frac{π}{2} x_{0}} & π^{+} & −i\frac{z_{1}}{R} se^{i \frac{π}{2} x_{0}} \\ \frac{\overline{z_{1}z_{2}}}{r^{2}} (−π^{−}e^{iπx_{0}} + 1) & −i\frac{\overline{z_{1}}}{R} se^{i \frac{π}{2} x_{0}} & \frac{r_{1}^{2}}{r^{2}} π^{−}e^{iπx_{0}} + \frac{r_{2}^{2}}{r^{2}} \end{matrix} ); \end{aligned}

these can be glued together with U′_{1} at R = 2n + 1, resp., R = 2n, because

Ũ′_{1}(x) = ( \begin{matrix}−\frac{r_{1}^{2}}{r^{2}} + \frac{r_{2}^{2}}{r^{2}} e^{iπx_{0}} & 0 & −\frac{z_{1}z_{2}}{r^{2}} (1 + e^{iπx_{0}}) \\ 0 & −1 & 0 \\ \frac{\overline{z_{1}z_{2}}}{r^{2}} (1 + e^{iπx_{0}}) & 0 & −\frac{r_{2}^{2}}{r^{2}} + \frac{r_{1}^{2}}{r^{2}} e^{iπx_{0}} \end{matrix} ) = U′_{1}(x)

for all x with ∥x∥ = R = 2n + 1 and

Û′_{1}(x) = ( \begin{matrix}\frac{r_{1}^{2}}{r^{2}} + \frac{r_{2}^{2}}{r^{2}} e^{iπx_{0}} & 0 & \frac{z_{1}z_{2}}{r^{2}} (1 − e^{iπx_{0}}) \\ 0 & 1 & 0 \\ \frac{\overline{z_{1}z_{2}}}{r^{2}} (1 − e^{iπx_{0}}) & 0 & \frac{r_{2}^{2}}{r^{2}} + \frac{r_{1}^{2}}{r^{2}} e^{iπx_{0}} \end{matrix} ) = U′_{1}(x)

for all x with ∥x∥ = 2n. Recalling T (x) from (8) we get

U′_{1}(x) = T^{−1}(x)·U_{1}(x), Ũ′_{1}(x) = Ũ_{1}(x)·T^{−1}(x), Û′_{1}(x) = Û_{1}(x)·T^{−1}(x).

T (x) only depends on x_{0} , so the matrices that occur in our calculation of (U_{1})^{⋆}ω_{5} (conf. (9)), only change in the following manner (we omit the argument x for convenience):

\begin{aligned}L′_{x_{0}} & = L_{x_{0}} + i\frac{π}{2} U_{1}^{†}EU_{1}, \\ L̃′_{x_{0}} & = T · L̃_{x_{0}} · T^{−1} + i\frac{π}{2} E, \\ L̂′_{x_{0}} & = T · L̂_{x_{0}} · T^{−1} + i\frac{π}{2} E, \\ L′ & = L,  L̃′ = T · L̃ · T^{−1},  L̂′ = T · L̂ · T^{−1}, \end{aligned}

where we have defined E := ( \begin{matrix}1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{matrix} ). We easily deduce

Tr\{L′ · L′_{x_{0}}\} = Tr\{L · L_{x_{0}}\} + i\frac{π}{2} Tr\{L · U_{1}^{†}EU_{1}\} and
Tr\{L̃′ · L̃′_{x_{0}}\} = Tr\{L̂′ · L̂′_{x_{0}}\} = − Tr\{L · L_{x_{0}}\} + i\frac{π}{2} Tr\{L · E\}.

For the total integral we get I′_{n} = I_{n} + Δ with

Δ = ∫_{0}^{n} \frac{dR}{R^{2}} ∫_{−R}^{R} dx_{0} ∫_{0}^{R^{2}−x_{0}^{2}}dr_{1}^{2} \frac{1}{48} \frac{R^{3}}{r_{1}r_{2}} Tr\{L · (U_{1}^{†}EU_{1} − E)\},

and for M := U_{1}^{†}EU_{1} − E we compute

M = ( \begin{matrix}−\frac{r_{1}^{2}s^{2}}{R^{2}} & z_{1}(x_{0} \frac{s^{2}}{R^{2}} + i\frac{sc}{R}) & −z_{1}z_{2} \frac{s^{2}}{R^{2}} \\ \overline{z_{1}}(x_{0} \frac{s^{2}}{R^{2}} − i\frac{sc}{R}) & \frac{r^{2}s^{2}}{R^{2}} & z_{2}(x_{0} \frac{s^{2}}{R^{2}} − i\frac{sc}{R}) \\ −\overline{z_{1}z_{2}} \frac{s^{2}}{R^{2}} & \overline{z_{2}}(x_{0} \frac{s^{2}}{R^{2}} + i\frac{sc}{R}) & −\frac{r_{2}^{2}s^{2}}{R^{2}} \end{matrix} ) .

Using (11) to (13) from the appendix we obtain

\begin{aligned}Tr\{L · M\} & = \frac{s^{2}}{R^{2}} S + 2\frac{sc}{R}(r_{1}^{2} Im\{\frac{L_{12}}{z_{1}} \} − r_{2}^{2} Im\{\frac{L_{23}}{z_{2}} \}) \\ & = \frac{r_{1}r_{2}}{R^{3}} [−24πs^{3}c(1−cc_{x}− \frac{x_{0}}{R} ss_{x}) + 24πs^{3}c(1−cc_{x}− \frac{x_{0}}{R} ss_{x})] = 0. \end{aligned}

This yields Δ = 0, and thus — as expected —

I′_{n} = n.

This result once again confirms that our ansatz (4) directly leads to representatives for generators of π_{2n−1}(SU_{n}), resp., π_{2n−1}(U_{n}), depending on n being even or odd — at least for the lower dimensions examinated here.

Appendix

In order to compute ψ^{⋆}(U_{1})^{⋆}ω_{5} we first calculate the antihermitian L_{μ} ’s. Throughout all computations we will use the following abbrevations for convenience and clarity:

c = cos πR,
π^{+} = cos πR + i\frac{x_{0}}{R} sin πR,
c_{x} = cos πx_{0},
e^{+} = exp(+iπx_{0}),

s = sin πR,
π^{−} = cos πR − i\frac{x_{0}}{R} sin πR,
s_{x} = sin πx_{0},
e^{−} = exp(−iπx_{0}).

Remember K = ℝ × ℝ_{0}^{+} × ℝ_{0}^{+} × [−π, +π] × [−π, +π] as domain for the polar coordinate function ψ and let v := (x_{0}, r_{1}, r_{2}, ϕ_{1}, ϕ_{2}) ∈ K. Define the linear involution Λ: K → K by w = Λ(v) = (x_{0}, r_{2}, r_{1}, −ϕ_{2}, −ϕ_{1}). So U_{1}(ψ(w)) is the matrix we obtain from U_{1}(ψ(v)) by replacing (z_{1}, z_{2}) by (\overline{z_{2}}, \overline{z_{1}}), resp., (r_{1}, ϕ_{1}) by (r_{2}, −ϕ_{2}), and vice versa.

Let A^{P} denote the matrix A “rotated by 180^{∘} ”, so that A_{11} becomes A_{33} , A_{12} becomes A_{32} , A_{13} becomes A_{31} , and so on. Obviously this operation commutes with the hermitian conjugation and the derivation of A. We have (AB)^{P} = A^{P} B^{P} and Tr{A^{P} } = Tr{A}. Because of U_{1}^{P} ∘ψ = U_{1}∘ψ∘Λ we obtain \frac{∂}{∂x_{0}} (U_{1}^{P} ∘ψ) = \frac{∂}{∂x_{0}} (U_{1} ∘ψ ∘Λ) = \frac{∂}{∂x_{0}} (U_{1} ∘ψ)∘Λ, \frac{∂}{∂r_{1}} (U_{1}^{P} ∘ψ) = \frac{∂}{∂r_{1}} (U_{1} ∘ψ ∘Λ) = \frac{∂U_{1}}{∂r_{2}} (U_{1} ∘ ψ) ∘ Λ and \frac{∂}{∂ϕ_{1}} (U_{1}^{P} ∘ ψ) = \frac{∂}{∂ϕ_{1}} (U_{1} ∘ ψ ∘ Λ) = −\frac{∂U_{1}}{∂ϕ_{2}} (U_{1} ∘ ψ) ∘ Λ. We thus have an additional symmetry between the elements of the antihermitian L_{μ} ’s (here L_{μ} = (U_{1} ∘ ψ)^{†} \frac{∂}{∂μ}(U_{1} ∘ ψ): K → M_{3}(ℂ) for μ = x_{0}, r_{1}, r_{2}, ϕ_{1}, ϕ_{2} ):

L_{x_{0}}(v) = L_{x_{0}}^{P}(w),
L_{ϕ_{1}}(v) = −L_{ϕ_{2}}^{P}(w), L_{ϕ_{2}}(v) = −L_{ϕ_{1}}^{P}(w),
L_{r_{1}}(v) = +L_{r_{2}}^{P}(w), L_{r_{2}}(v) = +L_{r_{1}}^{P}(w),

which makes life a bit easier. We obtain

L_{x_{0}}(v) = \frac{iπ}{2R^{2}} ·

(\begin{matrix}R^{2}+r_{1}^{2}( \frac{2sc}{πR} −1−c^{2}) & z_{1}[−x_{0}(\frac{2sc}{πR} −1−c^{2})+i(\frac{2s^{2}}{π} −Rsc)] & z_{1}z_{2}( \frac{2sc}{πR} −1−c^{2}) \\ \overline{z_{1}}[−x_{0}(\frac{2sc}{πR} −1−c^{2})−i(\frac{2s^{2}}{π} −Rsc)] & −2R^{2}−r^{2}( \frac{2sc}{πR} −1−c^{2}) & z_{2}[−x_{0}(\frac{2sc}{πR} −1−c^{2})−i(\frac{2s^{2}}{π} −Rsc)] \\ \overline{z_{1}z_{2}}( \frac{2sc}{πR} −1−c^{2}) & \overline{z_{2}}[−x_{0}(\frac{2sc}{πR} −1−c^{2})+i(\frac{2s^{2}}{π} −Rsc)] & R^{2}+r_{2}^{2}( \frac{2sc}{πR} −1−c^{2}) \end{matrix} ),

L_{ϕ_{1}}(v) = i ·

( \begin{matrix}− \frac{r^{2}_{1}s^{2}}{R^{2}} − \frac{r^{2}_{1}r^{2}_{2}}{r^{4}} |π^{+}−e^{+}|^{2} & +iz_{1} \frac{s}{R}(\frac{r^{2}_{1}}{r^{2}} π^{−}+ \frac{r^{2}_{2}}{r^{2}} e^{−}) & + \frac{z_{1}z_{2}}{r^{4}} (r_{1}^{2}π^{−}+r_{2}^{2}e^{−})(π^{+}−e^{+}) \\ −i\overline{z_{1}} \frac{s}{R}(\frac{r^{2}_{1}}{r^{2}} π^{+}+ \frac{r^{2}_{2}}{r^{2}} e^{+}) & + \frac{r^{2}_{1}s^{2}}{R^{2}} & −iz_{2} \frac{s}{R} \frac{r^{2}_{1}}{r^{2}} (π^{+}−e^{+}) \\ + \frac{\overline{z_{1}z_{2}}}{r^{4}} (r_{1}^{2}π^{+}+r_{2}^{2}e^{+})(π^{−}−e^{−}) & +i\overline{z_{2}} \frac{s}{R} \frac{r^{2}_{1}}{r^{2}} (π^{−}−e^{−}) & + \frac{r^{2}_{1}r^{2}_{2}}{r^{4}} |π^{+}−e^{+}|^{2} \end{matrix} ),

L_{ϕ_{2}}(v) = i ·

( \begin{matrix}− \frac{r^{2}_{1}r^{2}_{2}}{r^{4}} |π^{+}−e^{+}|^{2} & −iz_{1} \frac{s}{R} \frac{r^{2}_{2}}{r^{2}} (π^{−}−e^{−}) & − \frac{z_{1}z_{2}}{r^{4}} (r_{2}^{2}π^{+}+r_{1}^{2}e^{+})(π^{−}−e^{−}) \\ +i\overline{z_{1}} \frac{s}{R} \frac{r^{2}_{2}}{r^{2}} (π^{+}−e^{+}) & − \frac{r^{2}_{2}s^{2}}{R^{2}} & +iz_{2} \frac{s}{R}(\frac{r^{2}_{2}}{r^{2}} π^{+}+ \frac{r^{2}_{1}}{r^{2}} e^{+}) \\ − \frac{\overline{z_{1}z_{2}}}{r^{4}} (r_{2}^{2}π^{−}+r_{1}^{2}e^{−})(π^{+}−e^{+}) & −i\overline{z_{2}} \frac{s}{R}(\frac{r^{2}_{2}}{r^{2}} π^{−}+ \frac{r^{2}_{1}}{r^{2}} e^{−}) & + \frac{r^{2}_{2}s^{2}}{R^{2}} + \frac{r^{2}_{1}r^{2}_{2}}{r^{4}} |π^{+}−e^{+}|^{2} \end{matrix} ),

L_{r_{1}}(v) = ir_{1} ·

(\begin{matrix}\frac{x_{0}r^{2}_{1}}{r^{2}} (\frac{π}{R^{2}} − \frac{sc}{R^{3}})+i \frac{r^{2}_{2}}{r^{4}} (π^{−}e^{+}−π^{+}e^{−}) & z_{1}[(\frac{π}{R^{2}}−\frac{sc}{R^{3}})+\frac{s(r^{2}_{1}π^{−}+r^{2}_{2}e^{−})}{r^{2}_{1}r^{2}R}] & z_{1}z_{2}[\frac{x_{0}}{r^{2}} (\frac{π}{R^{2}}−\frac{sc}{R^{3}})−\frac{r^{2}−r^{2}_{1}π^{−}e^{+}−r^{2}_{2}π^{+}e^{−}}{ir^{2}_{1}r^{4}}] \\ \overline{z_{1}}[( \frac{π}{R^{2}} − \frac{sc}{R^{3}} )+ \frac{s(r^{2}_{1}π^{+}+r^{2}_{2}e^{+})}{r^{2}_{1}r^{2}R}] & −x_{0}( \frac{π}{R^{2}} − \frac{sc}{R^{3}} ) & z_{2}[( \frac{π}{R^{2}} − \frac{sc}{R^{3}} )+ \frac{s(π^{+}−e^{+})}{r^{2}R}] \\ \overline{z_{1}z_{2}}[\frac{x_{0}}{r^{2}} (\frac{π}{R^{2}}−\frac{sc}{R^{3}})+\frac{r^{2}−r^{2}_{1}π^{−}e^{+}−r^{2}_{2}π^{+}e^{−}}{ir^{2}_{1}r^{4}}] & \overline{z_{2}}[(\frac{π}{R^{2}} − \frac{sc}{R^{3}})+ \frac{s(π^{−}−e^{−})}{r^{2}R}] & \frac{x_{0}r^{2}_{2}}{r^{2}} (\frac{π}{R^{2}} − \frac{sc}{R^{3}})−i \frac{r^{2}_{2}}{r^{4}} (π^{−}e^{+}−π^{+}e^{−}) \end{matrix}),

L_{r_{2}}(v) = ir_{2} ·

(\begin{matrix}\frac{x_{0}r^{2}_{1}}{r^{2}} (\frac{π}{R^{2}} − \frac{sc}{R^{3}})−i \frac{r^{2}_{1}}{r^{4}} (π^{−}e^{+}−π^{+}e^{−}) & z_{1}[(\frac{π}{R^{2}} − \frac{sc}{R^{3}})+ \frac{s(π^{−}−e^{−})}{r^{2}R}] & z_{1}z_{2}[\frac{x_{0}}{r^{2}} (\frac{π}{R^{2}}−\frac{sc}{R^{3}})+\frac{r^{2}−r^{2}_{1}π^{−}e^{+}−r^{2}_{2}π^{+}e^{−}}{ir^{2}_{2}r^{4}}] \\ \overline{z_{1}}[( \frac{π}{R^{2}} − \frac{sc}{R^{3}} )+ \frac{s(π^{+}−e^{+})}{r^{2}R}] & −x_{0}( \frac{π}{R^{2}} − \frac{sc}{R^{3}} ) & z_{2}[( \frac{π}{R^{2}} − \frac{sc}{R^{3}} )+ \frac{s(r^{2}_{2}π^{+}+r^{2}_{1}e^{+})}{r^{2}_{2}r^{2}R}] \\ \overline{z_{1}z_{2}}[\frac{x_{0}}{r^{2}} (\frac{π}{R^{2}}−\frac{sc}{R^{3}})−\frac{r^{2}−r^{2}_{1}π^{−}e^{+}−r^{2}_{2}π^{+}e^{−}}{ir^{2}_{2}r^{4}}] & \overline{z_{2}}[(\frac{π}{R^{2}}−\frac{sc}{R^{3}})+\frac{s(r^{2}_{2}π^{−}+r^{2}_{1}e^{−})}{r^{2}_{2}r^{2}R}] & \frac{x_{0}r^{2}_{2}}{r^{2}} (\frac{π}{R^{2}} − \frac{sc}{R^{3}})+i \frac{r^{2}_{1}}{r^{4}} (π^{−}e^{+}−π^{+}e^{−}) \end{matrix}).

For the antihermitian [L_{μ}, L_{ν}] we have the following additional symmetries:

[L_{ϕ_{1}}, L_{ϕ_{2}}](v) = −[L_{ϕ_{1}}, L_{ϕ_{2}}]^{P} (w), [L_{r_{1}}, L_{r_{2}}](v) = −[L_{r_{1}}, L_{r_{2}}]^{P} (w),
[L_{ϕ_{1}}, L_{r_{1}}](v) = −[L_{ϕ_{2}}, L_{r_{2}}]^{P} (w), [L_{ϕ_{2}}, L_{r_{1}}](v) = −[L_{ϕ_{1}}, L_{r_{2}}]^{P} (w),
[L_{ϕ_{1}}, L_{r_{2}}](v) = −[L_{ϕ_{2}}, L_{r_{1}}]^{P} (w), [L_{ϕ_{2}}, L_{r_{2}}](v) = −[L_{ϕ_{1}}, L_{r_{1}}]^{P} (w),

so that [L_{ϕ_{1}}, L_{r_{2}}](v) and [L_{ϕ_{2}}, L_{r_{2}}](v) do not need to be computed. For the others we obtain

[L_{ϕ_{1}}, L_{ϕ_{2}}](v) = ( \begin{matrix}0 & −iz_{1} \frac{s}{R} \frac{r^{2}_{2}}{r^{2}} (π^{−}−e^{−}) & z_{1}z_{2} \frac{s^{2}}{R^{2}} \\ −i\overline{z_{1}} \frac{s}{R} \frac{r^{2}_{2}}{r^{2}} (π^{+}−e^{+}) & 0 & iz_{2} \frac{s}{R} \frac{r^{2}_{1}}{r^{2}} (π^{+}−e^{+}) \\ −\overline{z_{1}z_{2}} \frac{s^{2}}{R^{2}} & i\overline{z_{2}} \frac{s}{R} \frac{r^{2}_{1}}{r^{2}} (π^{−}−e^{−}) & 0 \end{matrix} ),

[L_{r_{1}}, L_{r_{2}}](v) = \frac{r_{1}r_{2}}{r^{2}R^{2}} ·

(\begin{matrix}\begin{matrix}πRs(e^{+}−e^{−}) \\ +iπx_{0}(ce^{+}+ce^{−}−2) \\ +ix_{0} \frac{s}{R} (2c−e^{+}−e^{−}) \end{matrix} & \frac{iz_{1}}{r_{1}^{2}} \{ \begin{matrix}πr^{2}(ce^{−}−1) \\ +x_{0}^{2} \frac{s}{R} (e^{−}−c) \\ +ix_{0}s^{2} \end{matrix} \} & \frac{z_{1}z_{2}}{r_{1}^{2}r_{2}^{2}} \{ \begin{matrix}−πRs(r_{1}^{2}e^{+}+r_{2}^{2}e^{−}) \\ +iπx_{0}r_{2}^{2}(ce^{−}−1) \\ −iπx_{0}r_{1}^{2}(ce^{+}−1) \\ +ix_{0}r_{1}^{2} \frac{s}{R} (e^{+}−c) \\ −ix_{0}r_{2}^{2} \frac{s}{R} (e^{−}−c) \end{matrix} \} \\ \frac{i\overline{z_{1}}}{r_{1}^{2}} \{ \begin{matrix}πr^{2}(ce^{+}−1) \\ +x_{0}^{2} \frac{s}{R} (e^{+}−c) \\ −ix_{0}s^{2} \end{matrix} \} & 0 & − \frac{iz_{2}}{r_{2}^{2}} \{ \begin{matrix}πr^{2}(ce^{+}−1) \\ +x_{0}^{2} \frac{s}{R} (e^{+}−c) \\ −ix_{0}s^{2} \end{matrix} \} \\ \frac{\overline{z_{1}z_{2}}}{r_{1}^{2}r_{2}^{2}} \{ \begin{matrix}πRs(r_{1}^{2}e^{−}+r_{2}^{2}e^{+}) \\ +iπx_{0}r_{2}^{2}(ce^{+}−1) \\ −iπx_{0}r_{1}^{2}(ce^{−}−1) \\ +ix_{0}r_{1}^{2} \frac{s}{R} (e^{−}−c) \\ −ix_{0}r_{2}^{2} \frac{s}{R} (e^{+}−c) \end{matrix} \} & − \frac{i\overline{z_{2}}}{r_{2}^{2}} \{ \begin{matrix}πr^{2}(ce^{−}−1) \\ +x_{0}^{2} \frac{s}{R} (e^{−}−c) \\ +ix_{0}s^{2} \end{matrix} \} & \begin{matrix}−πRs(e^{+}−e^{−}) \\ −iπx_{0}(ce^{+}+ce^{−}−2) \\ −ix_{0} \frac{s}{R} (2c−e^{+}−e^{−}) \end{matrix} \end{matrix}),

[L_{ϕ_{1}}, L_{r_{1}}](v) = r_{1} ·

(\begin{matrix}\begin{matrix}\frac{2iπx_{0}r^{2}_{1}r^{2}_{2}csx}{r^{4}R^{2}} − \frac{2iπr^{4}_{1}sc}{r^{2}R^{3}} \\ − \frac{2iπr^{2}_{1}r^{2}_{2}scx}{r^{4}R} − \frac{2ix_{0}r^{2}_{1}r^{2}_{2}ssx}{r^{4}R^{3}} \\ − \frac{2ir^{2}_{2}(r^{2}_{2}−r^{2}_{1})}{r^{6}} |π^{+}−e^{+}|^{2} \\ + \frac{2ir^{2}_{1}r^{2}_{2}s^{2}}{r^{4}R^{2}} + \frac{2ir^{4}_{1}s^{2}}{r^{2}R^{4}} − \frac{2is^{2}}{R^{2}} \end{matrix} & z_{1}\begin{cases}\frac{2πr^{2}_{1}s^{2}}{r^{2}R^{2}} + \frac{r^{2}_{2}s}{r^{2}R^{3}} (e^{−}−c) \\ − \frac{πr^{2}_{2}}{r^{2}R^{2}} (ce^{−}−1) \\ + \frac{2ix_{0}r^{2}_{2}s}{r^{2}R^{4}} (πRc−s) \\ + \frac{2r^{2}_{2}s}{r^{4}R} (e^{−}−π^{−}) \end{cases}\} & z_{1}z_{2}\begin{cases}−\frac{x_{0}}{r^{4}R^{3}} (r_{1}^{2}e^{+}+r_{2}^{2}e^{−})(πRc−s) \\ + \frac{iπs}{r^{4}R} (r_{1}^{2}e^{+}−r_{2}^{2}e^{−}) \\ + \frac{πx_{0}}{r^{2}R^{2}} − \frac{2iπr^{2}_{1}sc}{r^{2}R^{3}} \\ + \frac{is^{2}}{R^{4}} + \frac{4ir^{2}_{2}}{r^{6}} |π^{+}−e^{+}|^{2} \\ − \frac{x_{0}sc}{r^{2}R^{3}} − \frac{ix^{2}_{0}(r^{2}_{1}−r^{2}_{2})s^{2}}{r^{4}R^{4}} \end{cases}\} \\ \overline{z_{1}}\begin{cases}− \frac{2πr^{2}_{1}s^{2}}{r^{2}R^{2}} − \frac{r^{2}_{2}s}{r^{2}R^{3}} (e^{+}−c) \\ + \frac{πr^{2}_{2}}{r^{2}R^{2}} (ce^{+}−1) \\ + \frac{2ix_{0}r^{2}_{2}s}{r^{2}R^{4}} (πRc−s) \\ − \frac{2r^{2}_{2}s}{r^{4}R} (e^{+}−π^{+}) \end{cases}\} & \frac{2is}{R^{4}} [r_{1}^{2}(πRc−s)+R^{2}s] & z_{2}\{ \begin{matrix}− \frac{2πr^{2}_{1}s^{2}}{r^{2}R^{2}} + \frac{r^{2}_{1}s}{r^{2}R^{3}} (e^{+}−c) \\ − \frac{πr^{2}_{1}}{r^{2}R^{2}} (ce^{+}−1) \\ + \frac{2ix_{0}r^{2}_{2}s}{r^{2}R^{4}} (πRc−s) \\ − \frac{2r^{2}_{2}s}{r^{4}R} (e^{+}−π^{+}) \end{matrix} \} \\ \overline{z_{1}z_{2}}\begin{cases}\frac{x_{0}}{r^{4}R^{3}} (r_{1}^{2}e^{−}+r_{2}^{2}e^{+})(πRc−s) \\ + \frac{iπs}{r^{4}R} (r_{1}^{2}e^{−}−r_{2}^{2}e^{+}) \\ − \frac{πx_{0}}{r^{2}R^{2}} − \frac{2iπr^{2}_{1}sc}{r^{2}R^{3}} \\ + \frac{is^{2}}{R^{4}} + \frac{4ir^{2}_{2}}{r^{6}} |π^{+}−e^{+}|^{2} \\ + \frac{x_{0}sc}{r^{2}R^{3}} − \frac{ix^{2}_{0}(r^{2}_{1}−r^{2}_{2})s^{2}}{r^{4}R^{4}} \end{cases}\} & \overline{z_{2}}\begin{cases}\frac{2πr^{2}_{1}s^{2}}{r^{2}R^{2}} − \frac{r^{2}_{1}s}{r^{2}R^{3}} (e^{−}−c) \\ + \frac{πr^{2}_{1}}{r^{2}R^{2}} (ce^{−}−1) \\ + \frac{2ix_{0}r^{2}_{2}s}{r^{2}R^{4}} (πRc−s) \\ + \frac{2r^{2}_{2}s}{r^{4}R} (e^{−}−π^{−}) \end{cases}\} & \begin{matrix}− \frac{2iπx_{0}r^{2}_{1}r^{2}_{2}csx}{r^{4}R^{2}} − \frac{2iπr^{2}_{1}r^{2}_{2}sc}{r^{2}R^{3}} \\ + \frac{2iπr^{2}_{1}r^{2}_{2}scx}{r^{4}R} + \frac{2ix_{0}r^{2}_{1}r^{2}_{2}ssx}{r^{4}R^{3}} \\ + \frac{2ir^{2}_{2}(r^{2}_{2}−r^{2}_{1})}{r^{6}} |π^{+}−e^{+}|^{2} \\ −2ir_{1}^{2}r_{2}^{2}s^{2}r^{4}R^{2}+ \frac{2ir^{2}_{1}r^{2}_{2}s^{2}}{r^{2}R^{4}} \end{matrix} \end{matrix}),

\begin{matrix}[L_{ϕ_{2}}, L_{r_{1}}](v) = r_{1} · \\ (\begin{matrix}\begin{matrix}\frac{2iπx_{0}r^{2}_{1}r^{2}_{2}csx}{r^{4}R^{2}} + \frac{2iπr^{2}_{1}r^{2}_{2}sc}{r^{2}R^{3}} \\ − \frac{2iπr^{2}_{1}r^{2}_{2}scx}{r^{4}R} − \frac{2ix_{0}r^{2}_{1}r^{2}_{2}ssx}{r^{4}R^{3}} \\ − \frac{2ir^{2}_{2}(r^{2}_{2}−r^{2}_{1})}{r^{6}} |π^{+}−e^{+}|^{2} \\ + \frac{2ir^{2}_{1}r^{2}_{2}s^{2}}{r^{4}R^{2}} − \frac{2ir^{2}_{1}r^{2}_{2}s^{2}}{r^{2}R^{4}} \end{matrix} & z_{1}\begin{cases}− \frac{2πr^{2}_{2}s^{2}}{r^{2}R^{2}} + \frac{r^{2}_{2}s}{r^{2}R^{3}} (e^{−}−c) \\ − \frac{πr^{2}_{2}}{r^{2}R^{2}} (ce^{−}−1) \\ − \frac{2ix_{0}r^{2}_{2}s}{r^{2}R^{4}} (πRc−s) \\ + \frac{r^{2}_{2}(r^{2}_{1}−r^{2}_{2})s}{r^{2}_{1}r^{4}R} (e^{−}−π^{−}) \end{cases}\} & z_{1}z_{2}\begin{cases}−\frac{x_{0}}{r^{4}R^{3}} (r_{1}^{2}e^{+}+r_{2}^{2}e^{−})(πRc−s) \\ + \frac{iπs}{r^{4}R} (r_{1}^{2}e^{+}−r_{2}^{2}e^{−}) \\ + \frac{πx_{0}}{r^{2}R^{2}} − \frac{2iπr^{2}_{1}sc}{r^{2}R^{3}} \\ +\frac{ir^{2}_{2}s^{2}}{r^{2}_{1}r^{2}R^{2}} + \frac{4ir^{2}_{2}}{r^{6}} |π^{+}−e^{+}|^{2} \\ − \frac{x_{0}sc}{r^{2}R^{3}} + \frac{2ix^{2}_{0}r^{2}_{2}s^{2}}{r^{4}R^{4}} \end{cases}\} \\ \overline{z_{1}}\begin{cases}\frac{2πr^{2}_{2}s^{2}}{r^{2}R^{2}} − \frac{r^{2}_{2}s}{r^{2}R^{3}} (e^{+}−c) \\ + \frac{πr^{2}_{2}}{r^{2}R^{2}} (ce^{+}−1) \\ − \frac{2ix_{0}r^{2}_{2}s}{r^{2}R^{4}} (πRc−s) \\ − \frac{r^{2}_{2}(r^{2}_{1}−r^{2}_{2})s}{r^{2}_{1}r^{4}R} (e^{+}−π^{+}) \end{cases}\} & − \frac{2is}{R^{4}} [r_{2}^{2}(πRc−s)] & z_{2}\begin{cases}\frac{2πr^{2}_{2}s^{2}}{r^{2}R^{2}} + \frac{r^{2}_{1}s}{r^{2}R^{3}} (e^{+}−c) \\ − \frac{πr^{2}_{1}}{r^{2}R^{2}} (ce^{+}−1) \\ − \frac{2ix_{0}r^{2}_{2}s}{r^{2}R^{4}} (πRc−s) \\ − \frac{(r^{2}_{1}+3r^{2}_{2})s}{r^{4}R} (e^{+}−π^{+}) \end{cases}\} \\ \overline{z_{1}z_{2}}\begin{cases}\frac{x_{0}}{r^{4}R^{3}} (r_{1}^{2}e^{−}+r_{2}^{2}e^{+})(πRc−s) \\ + \frac{iπs}{r^{4}R} (r_{1}^{2}e^{−}−r_{2}^{2}e^{+}) \\ − \frac{πx_{0}}{r^{2}R^{2}} + \frac{2iπr^{2}_{2}sc}{r^{2}R^{3}} \\ +\frac{ir^{2}_{2}s^{2}}{r^{2}_{1}r^{2}R^{2}} + \frac{4ir^{2}_{2}}{r^{6}} |π^{+}−e^{+}|^{2} \\ + \frac{x_{0}sc}{r^{2}R^{3}} + \frac{2ix^{2}_{0}r^{2}_{2}s^{2}}{r^{4}R^{4}} \end{cases}\} & \overline{z_{2}}\begin{cases}− \frac{2πr^{2}_{2}s^{2}}{r^{2}R^{2}} − \frac{r^{2}_{1}s}{r^{2}R^{3}} (e^{−}−c) \\ + \frac{πr^{2}_{1}}{r^{2}R^{2}} (ce^{−}−1) \\ − \frac{2ix_{0}r^{2}_{2}s}{r^{2}R^{4}} (πRc−s) \\ + \frac{(r^{2}_{1}+3r^{2}_{2})s}{r^{4}R} (e^{−}−π^{−}) \end{cases}\} & \begin{matrix}− \frac{2iπx_{0}r^{2}_{1}r^{2}_{2}csx}{r^{4}R^{2}} + \frac{2iπr^{4}_{2}sc}{r^{2}R^{3}} \\ + \frac{2iπr^{2}_{1}r^{2}_{2}scx}{r^{4}R} + \frac{2ix_{0}r^{2}_{1}r^{2}_{2}ssx}{r^{4}R^{3}} \\ + \frac{2ir^{2}_{2}(r^{2}_{2}−r^{2}_{1})}{r^{6}} |π^{+}−e^{+}|^{2} \\ − \frac{2ir^{2}_{1}r^{2}_{2}s^{2}}{r^{4}R^{2}} + \frac{2ir^{2}_{1}r^{2}_{2}s^{2}}{r^{2}R^{4}} − \frac{2ir^{2}_{2}s^{2}}{R^{4}} \end{matrix} \end{matrix} ). \end{matrix}

We further have

[L_{ϕ_{1}}, L_{ϕ_{2}}][L_{r_{1}}, L_{r_{2}}](v) = ([L_{ϕ_{1}}, L_{ϕ_{2}}][L_{r_{1}}, L_{r_{2}}])^{P} (w),
[L_{r_{1}}, L_{r_{2}}][L_{ϕ_{1}}, L_{ϕ_{2}}](v) = ([L_{r_{1}}, L_{r_{2}}][L_{ϕ_{1}}, L_{ϕ_{2}}])^{P} (w),
[L_{ϕ_{1}}, L_{r_{1}}][L_{ϕ_{2}}, L_{r_{2}}](v) = ([L_{ϕ_{1}}, L_{r_{1}}][L_{ϕ_{2}}, L_{r_{2}}])^{P} (w),
[L_{ϕ_{2}}, L_{r_{2}}][L_{ϕ_{1}}, L_{r_{1}}](v) = ([L_{ϕ_{2}}, L_{r_{2}}][L_{ϕ_{1}}, L_{r_{1}}])^{P} (w),
[L_{ϕ_{2}}, L_{r_{1}}][L_{ϕ_{1}}, L_{r_{2}}](v) = ([L_{ϕ_{2}}, L_{r_{1}}][L_{ϕ_{1}}, L_{r_{2}}])^{P} (w),
[L_{ϕ_{1}}, L_{r_{2}}][L_{ϕ_{2}}, L_{r_{1}}](v) = ([L_{ϕ_{1}}, L_{r_{2}}][L_{ϕ_{2}}, L_{r_{1}}])^{P} (w),

from which we deduce for L defined by (10)

L(v) = L^{P} (w), L · L_{x_{0}}(v) = (L · L_{x_{0}})^{P} (w) and Tr{L · L_{x_{0}}}(v) = Tr{L · L_{x_{0}}}(w).

Since dx_{0} ∧ dr_{1} ∧ dr_{2} ∧ dϕ_{1} ∧ dϕ_{2} = dx_{0} ∧ dr_{2} ∧ dr_{1} ∧ d(−ϕ_{2}) ∧ d(−ϕ_{1}), we have ψ^{⋆}(U_{1})^{⋆}ω_{5}(v) = ψ^{⋆}(U_{1})^{⋆}ω_{5}(w) by (9). Even if we make good use of these symmetries there is still some work left over to compute L. We finally obtain

\begin{matrix}L(v) = \frac{r_{1}r_{2}}{R^{3}} · \\ (\begin{matrix}\begin{matrix}\frac{12r^{2}_{1}s^{2}}{r^{2}} (πsc_{x}−π \frac{x_{0}}{R} cs_{x}+\frac{x_{0}}{R^{2}} ss_{x}) \\ +8s(2−3 \frac{r^{2}_{2}}{r^{2}} ) [π(c−c_{x}) \\ + \frac{x^{2}_{0}s}{r^{2}R} (1−cc_{x})− \frac{x_{0}}{r^{2}} s^{2}s_{x}] \end{matrix} & z_{1}\begin{cases}\frac{12s}{r^{2}} (1−cc_{x}− \frac{x_{0}}{R} ss_{x})· \\ ·(−πx_{0}c+iπRs+ \frac{x_{0}}{R} s) \end{cases}\} & z_{1}z_{2}\begin{cases}\frac{12s^{2}}{r^{2}} (πsc_{x}−π \frac{x_{0}}{R} cs_{x}+\frac{x_{0}}{R^{2}} ss_{x}) \\ +24 \frac{s}{r^{2}} [π(c−c_{x}) \\ + \frac{x^{2}_{0}s}{r^{2}R} (1−cc_{x}− \frac{x_{0}}{R} ss_{x})] \end{cases}\} \\ \overline{z_{1}}\begin{cases}\frac{12s}{r^{2}} (1−cc_{x}− \frac{x_{0}}{R} ss_{x})· \\ ·(−πx_{0}c−iπRs+ \frac{x_{0}}{R} s) \end{cases}\} & \begin{matrix}−12s^{2}(πsc_{x}−π \frac{x_{0}}{R} cs_{x}+\frac{x_{0}}{R^{2}} ss_{x}) \\ −8s [π(c−c_{x}) \\ + \frac{x^{2}_{0}s}{r^{2}R} (1−cc_{x})− \frac{x_{0}}{r^{2}} s^{2}s_{x}] \end{matrix} & z_{2}\begin{cases}\frac{12s}{r^{2}} (1−cc_{x}− \frac{x_{0}}{R} ss_{x})· \\ ·(−πx_{0}c−iπRs+ \frac{x_{0}}{R} s) \end{cases}\} \\ \overline{z_{1}z_{2}}\begin{cases}\frac{12s^{2}}{r^{2}} (πsc_{x}−π \frac{x_{0}}{R} cs_{x}+\frac{x_{0}}{R^{2}} ss_{x}) \\ +24 \frac{s}{r^{2}} [π(c−c_{x}) \\ + \frac{x^{2}_{0}s}{r^{2}R} (1−cc_{x}− \frac{x_{0}}{R} ss_{x})] \end{cases}\} & \overline{z_{2}}\begin{cases}\frac{12s}{r^{2}} (1−cc_{x}− \frac{x_{0}}{R} ss_{x})· \\ ·(−πx_{0}c+iπRs+ \frac{x_{0}}{R} s) \end{cases}\} & \begin{matrix}\frac{12r^{2}_{2}s^{2}}{r^{2}} (πsc_{x}−π \frac{x_{0}}{R} cs_{x}+\frac{x_{0}}{R^{2}} ss_{x}) \\ +8s(2−3 \frac{r^{2}_{1}}{r^{2}} ) [π(c−c_{x}) \\ + \frac{x^{2}_{0}s}{r^{2}R} (1−cc_{x})− \frac{x_{0}}{r^{2}} s^{2}s_{x}] \end{matrix} \end{matrix}). \end{matrix}

Gross 103

Using

Tr{L · L_{x_{0}}} = ∑_{ij} L_{ij}(L_{x_{0}})_{ji} = ∑_{i} L_{ii}(L_{x_{0}})_{ii} + 2i ∑_{i<j} Im{L_{ij}(L_{x_{0}})_{ji}}

we have (omitting the argument v )

Tr\{L · L_{x_{0}}\} =i(\frac{2s^{2}}{R^{2}} − \frac{πsc}{R})(r_{1}^{2} Im\{\frac{L_{12}}{z_{1}} \} − r_{2}^{2} Im\{\frac{L_{23}}{z_{2}} \})

+ i\frac{π}{2} (L_{11} − 2L_{22} + L_{33}) + i(\frac{π(1 + c^{2})}{2R^{2}} − \frac{sc}{R^{3}} )S

with  S =[−r_{1}^{2}L_{11} + r^{2}L_{22} − r_{2}^{2}L_{33} − 2r_{1}^{2}r_{2}^{2} Re\{\frac{L_{13}}{z_{1}z_{2}} \}
+ 2x_{0}(r_{1}^{2} Re\{\frac{L_{12}}{z_{1}} \} + r_{2}^{2} Re\{\frac{L_{23}}{z_{2}} \})].  (11)

We obtain

r_{1}^{2} Im\{\frac{L_{12}}{z_{1}} \} − r_{2}^{2} Im\{\frac{L_{23}}{z_{2}} \} =\frac{r_{1}r_{2}}{R^{3}} 12πRs^{2}(1 − cc_{x} − \frac{x_{0}}{R} ss_{x}),  (12)
r_{1}^{2} Re\{\frac{L_{12}}{z_{1}} \} + r_{2}^{2} Re\{\frac{L_{23}}{z_{2}} \} =\frac{r_{1}r_{2}}{R^{3}} 12x_{0}s(\frac{s}{R} − πc)(1 − cc_{x} − \frac{x_{0}}{R} ss_{x}),
L_{11} − 2L_{22} + L_{33} =\frac{r_{1}r_{2}}{R^{3}} \{36s^{2}[π(sc_{x} − \frac{x_{0}}{R} cs_{x}) + \frac{x_{0}}{R^{2}} ss_{x}]

+ 24s[π(c − c_{x}) + \frac{x_{0}^{2}s}{r^{2}R}(1 − cc_{x}) − \frac{x_{0}}{r^{2}} s^{2}s_{x}]\},

−2r_{1}^{2}r_{2}^{2} Re\{\frac{L_{13}}{z_{1}z_{2}} \} =\frac{r_{1}r_{2}}{R^{3}} \frac{r_{1}^{2}r_{2}^{2}}{r^{2}} \{−24s^{2}[π(sc_{x} − \frac{x_{0}}{R} cs_{x}) − \frac{x_{0}}{R^{2}} ss_{x}]

− 48s[π(c − c_{x}) + \frac{x_{0}^{2}s}{r^{2}R}(1 − cc_{x} − \frac{x_{0}}{R} ss_{x})]\},

−r_{1}^{2}L_{11}+r^{2}L_{22}−r_{2}^{2}L_{33} =\frac{r_{1}r_{2}}{R^{3}} r^{2}\{−24s^{2}(1− \frac{r_{1}^{2}r_{2}^{2}}{r^{4}} )[π(sc_{x} − \frac{x_{0}}{R} cs_{x}) + \frac{x_{0}}{R^{2}} ss_{x}]

− 24s\frac{r_{1}^{4}+r_{2}^{4}}{r^{4}} [π(c−c_{x}) + \frac{x_{0}^{2}s}{r^{2}R}(1−cc_{x}) − \frac{x_{0}}{r^{2}} s^{2}s_{x}]\},

thus: S = − \frac{r_{1}r_{2}}{R^{3}} 24πR^{2}sc(1 − cc_{x} − \frac{x_{0}}{R} ss_{x}) (13)
and  Tr\{L · L_{x_{0}}\} =i\frac{r_{1}r_{2}}{R^{3}} \{12π^{2}s(c_{x}−c) − 6π^{2}s^{2}(sc_{x} − \frac{x_{0}}{R} cs_{x})

− 6π(2\frac{R^{2}}{r^{2}} +1)\frac{x_{0}}{R^{2}} s^{3}s_{x} + 6π(2\frac{R^{2}}{r^{2}} +2)\frac{s^{2}}{R} (1−cc_{x})\}.

It is finally done. For the desired 5-form on the southern hemisphere of the S^{5} we end up with

ψ^{⋆}(U_{1})^{⋆}ω_{5} =30iπ \frac{r_{1}r_{2}}{R^{3}} [−π sin^{2}πR(sinπR cosπx_{0} − \frac{x_{0}}{R} cosπR sinπx_{0})

+ 2π sinπR(cosπx_{0}−cosπR) + (2\frac{R^{2}}{r^{2}} +2)\frac{sin^{2}πR}{R}(1−cosπR cosπx_{0})

− (2\frac{R^{2}}{r^{2}} +1)\frac{x_{0}}{R^{2}} sin^{3}πR sinπx_{0}]dx_{0}∧dr_{1}∧dr_{2}∧dϕ_{1}∧dϕ_{2}.  (14)

After this preliminary work it is quite easy now to compute (Ũ_{1})^{⋆}ω_{5} on the northern hemisphere. We rewrite Ũ_{1} as

Ũ_{1} = F \overline{U_{1}}^{P} F^{P} = (F^{P} \overline{U_{1}}F )^{P} ,

where we have defined F := ( \begin{matrix}−1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} ). This yields

Ũ_{1}^{†} = (F \overline{U_{1}}^{†}F^{P} )^{P} , \frac{∂Ũ_{1}}{∂x_{i}} = (F^{P} \overline{\frac{∂U_{1}}{∂x_{i}}} F )^{P} ,

because complex conjugation and differentiation along real variables commute. We thus have

L̃_{x_{i}} = (F \overline{L_{x_{i}}}F)^{P} ,
Tr\{L̃ · L̃_{x_{0}}\} = Tr\{(F \overline{L · L_{x_{0}}}F)^{P} \} = Tr\{F \overline{L · L_{x_{0}}}F \}
= Tr\{\overline{L · L_{x_{0}}}\} = \overline{Tr\{L · L_{x_{0}}\}}

and immediately obtain

(15)  (Ũ_{1})^{⋆}ω_{5} = \overline{(U_{1})^{⋆}ω_{5}} = −(U_{1})^{⋆}ω_{5}.

Acknowledgement

The author is very grateful to Prof. Dr. K. H. Hofmann for numerous fruitful discussions and his great help in revising this work.

References

[1] Bott, R., The stable homotopy of the classical groups, Ann. Math. 70 (1959), 313–337.

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Fachbereich Mathematik Technische Hochschule Schlossgartenstr. 7 D-6100 Darmstadt, Germany gross@mathematik.th-darmstadt.de

Received April 15, 1993