# Five-dimensional non-Abelian supersymmetric Chern-Simons action in
Projective Superspace

Ariunzul Dagvadorj<EMAIL_ADDRESS>Department of Theoretical Physics and
Astrophysics, Faculty of Science, Masaryk University, Kotlářská 2, 611 37
Brno, Czech Republic

(August 28, 2024)

###### Abstract

In this paper, we derive the five-dimensional non-Abelian $\mathcal{N}=1$
Chern-Simons action from its established definition of variation with respect
to an infinitesimal deformation of the vector prepotential in a projective
superspace setting. It has been a long-standing open problem how to integrate
this variation. Now, thanks to the simplicity of a projective superspace
technique, we have been finally able to obtain this term. As a bonus, we also
give the supersymmetric Yang-Mills action with a half measure as well.

projective superspace, supersymmetric Chern-Simons, five-dimensions

###### Contents

  1. I Introduction
  2. II Yang-Mills action with a half measure
    1. II.1 5-dimensional superalgebra
    2. II.2 SYM action in terms of the field strength
    3. II.3 Reduction to 4-dimensions
    4. II.4 Discussion
  3. III Chern-Simons action
    1. III.1 Abelian Chern-Simons action
    2. III.2 Non-Abelian Chern-Simons action
    3. III.3 Discussion
  4. A Nonsymmetric splitting of the prepotential
  5. B Field strength in terms of nonsymmetric splittings

##  I Introduction

The most recognizable prototypical topological field theory is the Chern-
Simons theory in 3-dimensions . This theory is solvable exactly and Witten
showed that the expectation value of the topologically invariant observables,
e.g., Wilson loops, is equivalent to the Jones polynomial of knot homology
invariants emerging from the Chern-Simons theory on any 3-manifold Witten
(1989). Uplifting this approach (describing knot invariants via Chern-Simons
theory on the boundary of a higher-dimensional manifold) to five dimensions
has also been done, leading to the categorization of the Jones polynomial
Khovanov (2000); Witten (2011a); Gaiotto and Witten (2012); Witten (2011b).

Five-dimensional supersymmetric gauge theories have been discussed in
different contexts in the past, such as superconformal fixed points as an
infinite coupling limit of super Yang-Mills theory (SYM) with various gauge
groups, particularly with $U(N)$, $SU(N)$ groups, which allow the Chern-Simons
term in the action Seiberg (1996); Intriligator et al. (1997); Minahan et al.
(2013); Minahan and Nedelin (2014); or due to their relation to the six-
dimensional $\mathcal{N}=(2,0)$ superconformal field theories. In this case,
the UV complexion of the 5d effective field theory is a 6d theory, whose one
dimension is compactified on a circle Witten (1995); Douglas (2011); Lambert
et al. (2013). It was also argued that the Chern-Simons term can be obtained
from the super Yang-Mills theory with massive hypermultiplets by integrating
out the hypermultiplet Seiberg (1996); Intriligator et al. (1997).

In harmonic superspace settings Galperin et al. (1984, 2007), one-loop
effective action for a hypermultiplet with a background vector multiplet has
been obtained, and as a result the Chern-Simons action at the quantum level
has been generated. Kuzenko (2007); Buchbinder and Pletnev (2015). The leading
order of this quantum correction agrees with the action formula proposed in
Zupnik (1999). In that paper Zupnik defined the variation of the non-Abelian
Chern-Simons action with respect to the infinitesimal prepotential
deformation. However, the integrated form of this action has not been known.
The purpose of this paper is to finally find an answer to this question. In
the Abelian case, the action was obtained in Zupnik (1999); Kuzenko and Linch
(2006). The superform approach to the non-Abelian case was introduced in
Kuzenko and Novak (2014).

In this paper, we will employ the projective superspace Karlhede et al. (1984)
approach to this long-standing problem. Projective supespace formalism is a
manifestly supersymmetric method developed for extended supersymmetric
theories with a $R$-symmetry group $SU(2)$ or a product of it. In the case of
four-dimensions, the ordinary superspace $\mathbb{R}^{4|8}$ is appended by an
auxiliary $\mathbb{CP}^{1}=\frac{SU(2)}{U(1)}$ factor, on which all the fields
depend holomorphically. Contour integrals are then used on this extra
coordinate to construct invariant actions. A comparison between Projective and
Harmonic superspace can be found in Kuzenko (1999); Butter (2012). A hybrid
formalism between them was also developed in Jain and Siegel (2009, 2011,
2012a, 2013). Applications of the Projective superspace method, for instance,
to find hypermultiplet contributions to the effective action Gonzalez-Rey and
von Unge (1998); Gonzalez-Rey et al. (1998) or in the theory of Abelian super
Yang-Mills Lindstrom and Rocek (1990), in the quantization of the Abelian
vector multiplet Gonzalez-Rey (1997) and in the non-Abelian super Yang-Mills
theory Davgadorj and von Unge (2018) can be found. New techniques to treat,
more precisely, how to separate the field $e^{V(\zeta)}$ (constructed from the
unconstrained gauge prepotential $V(\zeta)$) into negative and positive powers
of $\mathbb{CP}^{1}$ holomorphic coordinate $\zeta$ was developed in Davgadorj
et al. (2023). Considering that all objects need to be expanded in powers of
this object $X=e^{V}-1$ at different $\zeta$-coordinates, particular
separation/projection methods make a difference, leading to significant
simplification of expressions for the field strength and Lagrangians. One of
the hallmarks of Projective superspace is its straightforwardness in reducing
the higher supersymmetric expression into lower supersymmetric components, the
new method enables this reduction efficiently. The power of this technique was
tested and applied to the $\mathcal{N}=3$ and $\mathcal{N}=4$ gauge theories
in three-dimensions as well as in four-dimensional, 8-supercharge Yang-Mills
action in that paper.

In this article, in section III, we will again utilize the power of this non-
symmetric separation technique of Davgadorj et al. (2023) in deriving the
five-dimensional $\mathcal{N}=1$ non-Abelian Chern-Simons term from its well-
defined variation formula. But before that, for the sake of completeness, in
Section II, we will describe the Yang-Mills action with a projective half-
measure. This formula will be proven to be correct, when we reduce it to the
4-dimensions. In order to be self-contained, we also include the non-symmetric
splitting techniques for the prepotential multiplet $e^{V}$ and subsequently
for the field strength in the appendices.

##  II Yang-Mills action with a half measure

###  II.1 5-dimensional superalgebra

Let us start with a comment on spinors in 5-dimensions. Here, one cannot have
Majorana spinors like in $d=2,3,4$, because the reality condition that is
$\psi=C(\Gamma^{0})^{T}\psi^{*}$ has no solution. Therefore, one constructs
pairs of Dirac spinors that transform to each other under charge conjugation:
$\psi_{1}=C(\Gamma^{0})^{T}\psi^{*}_{2}$ and
$\psi_{2}=-C(\Gamma^{0})^{T}\psi^{*}_{1}$. Concisely:
$\psi_{i}=\epsilon^{ij}C(\Gamma^{0})^{T}\psi^{*}_{j}$ This is called the
symplectic Majorana spinors / condition. The antisymmetric 5D $\Gamma$
matrices that satisfy a relation $C\Gamma^{a}=-(C\Gamma^{a})^{T}$ are:

$\displaystyle C\Gamma^{0}=\begin{pmatrix}0&\begin{pmatrix}0&1\\\
-1&0\end{pmatrix}\\\ \begin{pmatrix}0&1\\\
-1&0\end{pmatrix}&0\end{pmatrix}\;\;;\;\;C\Gamma^{1}=\begin{pmatrix}0&\begin{pmatrix}-1&0\\\
0&1\end{pmatrix}\\\ \begin{pmatrix}1&0\\\
0&-1\end{pmatrix}&0\end{pmatrix}\;\;;$ $\displaystyle
C\Gamma^{2}=\begin{pmatrix}0&\begin{pmatrix}i&0\\\ 0&i\end{pmatrix}\\\
\begin{pmatrix}-i&0\\\
0&-i\end{pmatrix}&0\end{pmatrix}\;\;;\;\;C\Gamma^{3}=\begin{pmatrix}0&\begin{pmatrix}0&-1\\\
-1&0\end{pmatrix}\\\ \begin{pmatrix}0&1\\\
1&0\end{pmatrix}&0\end{pmatrix}\;\;;$ $\displaystyle
C\Gamma^{5}=\begin{pmatrix}\begin{pmatrix}0&-i\\\ i&0\end{pmatrix}&0\\\
0&\begin{pmatrix}0&i\\\
-i&0\end{pmatrix}\end{pmatrix}\;=\;\begin{pmatrix}-i\epsilon^{\alpha\beta}&0\\\
0&i\epsilon^{\dot{\alpha}\dot{\beta}}\end{pmatrix}$ (1)

Here the antisymmetric
$C=\epsilon^{\hat{\alpha}\hat{\beta}}=\begin{pmatrix}\epsilon^{\alpha\beta}&0\\\
0&\epsilon^{\dot{\alpha}\dot{\beta}}\end{pmatrix}$ and the inverse
$C^{-1}=\epsilon_{\hat{\alpha}\hat{\beta}}$ serve as charge conjugation
matrices. $\epsilon$ matrices are used to raise and lower indices.
$\epsilon^{\hat{\alpha}\hat{\beta}}\epsilon_{\hat{\alpha}\hat{\beta}}=-4$.

Superfields in the superspace $\mathbb{R}^{5|8}$ depend on
$(x^{\hat{a}},\theta^{\hat{\alpha}}_{i})$ coordinates, where $\hat{a}$ and
$\hat{\alpha}$ are the 5-vector and 4-spinor indices of the Lorentz group
$SO(4,1)$, while $i$ is the index of the $SU(2)$ automorphism group (a pair of
Dirac spinors that we discussed above).

Gauged covariant derivatives satisfy the following algebra:

$\displaystyle\\{\mathcal{D}^{i}_{\hat{\alpha}},\mathcal{D}^{j}_{\hat{\beta}}\\}=i\epsilon^{ij}\left(\nabla_{\hat{\alpha}\hat{\beta}}+\epsilon_{\hat{\alpha}\hat{\beta}}\mathbb{W}\right)\;,\;\textit{where}\;\;\nabla_{\hat{\alpha}\hat{\beta}}=(C\Gamma^{\hat{a}})_{\hat{\alpha}\hat{\beta}}\partial_{\hat{a}}$
(2)

Now we apply this relation to write down the projective gauge covariantized
anticommutation rule for the $\nabla$ operator: $\nabla_{\hat{\alpha}}\equiv
D_{1\hat{\alpha}}+\zeta D_{2\hat{\alpha}}$

$\displaystyle\\{\boldsymbol{\nabla}_{\hat{\alpha}},\boldsymbol{\nabla}_{\hat{\beta}}\\}=0$
(3)

This does not tell us much except that this looks like the integrability
condition and it can be related with the ungauged $\nabla$ through the
projections of the prepotential $e^{V}$ just like in 4-dimensions. Therefore,
we take the $\zeta$’s at different points.

$\displaystyle\\{\boldsymbol{\nabla}_{\hat{\alpha}}(\zeta_{1}),\boldsymbol{\nabla}_{\hat{\beta}}(\zeta_{2})\\}=i(\zeta_{1}-\zeta_{2})\left(\nabla_{\hat{\alpha}\hat{\beta}}+\epsilon_{\hat{\alpha}\hat{\beta}}\mathbb{W}\right)$
(4)

We do the same trick as we did in 4D, take a $\zeta$-derivative with respect
to one of the $\zeta$’s and act with $e^{U}$ and $e^{-U}$ from both sides.
However, the difference from the 4-dimensional case is that now we have an
additional term $\nabla_{\hat{\alpha}\hat{\beta}}$, but this will not produce
anything problematic, since one can see that
$\epsilon^{\hat{\alpha}\hat{\beta}}(C\Gamma^{\hat{a}})_{\hat{\alpha}\hat{\beta}}=0$,
therefore:

$\displaystyle\epsilon^{\hat{\alpha}\hat{\beta}}\;\;|\;\;\\{\nabla_{\hat{\alpha}},[e^{-U}\partial_{\zeta}e^{U},\nabla_{\hat{\beta}}]\\}$
$\displaystyle=ie^{-U}(C\Gamma^{\hat{a}})_{\hat{\alpha}\hat{\beta}}\partial_{\hat{a}}e^{U}+i\epsilon_{\hat{\alpha}\hat{\beta}}e^{-U}\mathbb{W}e^{U}$
$\displaystyle-4i\mathcal{W}$
$\displaystyle=-\epsilon^{\hat{\alpha}\hat{\beta}}\nabla_{\hat{\alpha}}\nabla_{\hat{\beta}}A$
(5)

where, as usual, we define the arctic field strength
$\mathcal{W}=e^{-U}\mathbb{W}e^{U}$ and the connection
$A=e^{-U}(\partial_{\zeta}e^{U})$. Then the above relation can be written
compactly:

$\mathcal{W}=\frac{1}{2}i\nabla^{2}A$ (6)

###  II.2 SYM action in terms of the field strength

Zupnik has established long time ago in Zupnik (1999), that the 8-supercharge
SYM theories with full measure have an universal form in terms of the
prepotential. That is (in projective superspace description Jain and Siegel
(2012b); Davgadorj and von Unge (2018)):

$\displaystyle
S_{YM}=\frac{1}{g_{YM}^{2}}\sum_{n=2}^{\infty}\frac{(-1)^{n}}{n}\tr\int
d^{5}xd^{8}\theta\oint d\zeta_{1}\ldots d\zeta_{n}\frac{X_{1}X_{2}\ldots
X_{n}}{\zeta_{21}\ldots\zeta_{n,n-1}\zeta_{1n}}$ (7)

Then, one might want to construct the Yang-Mills action with a half-measure in
terms of the field strength just like in four-dimensions. However, in five-
dimensions a chiral description does not exist, as it is not Lorentz-
covariant; therefore, a projective half measure is an option. We also need to
construct a projective covariant superfield $\it{G}$ in terms of the field
strength Kuzenko and Linch (2006).

$G=\nabla^{4}\\{A,\mathcal{W}\\}$ (8)

The idea of applying Sohnius’s prescription to construct a supersymmetric
invariant action for an algebra with central charges Sohnius (1978) to a four-
dimensional $\mathcal{N}=2$ harmonic superspace in the context of vector-
tensor multiplet was developed in Dragon et al. (1998). Later, this was
applied to describe five-dimensional super Yang-Mills action in Buchbinder and
Pletnev (2015).

In our case, we take a similar route; first, parametrizing the Grassmann
coordinates with the $\mathbf{CP}^{1}$ parameter, such that our $\Delta$
operator that defines the projective measure is proportional to the partial
derivative with respect to this projective Grassmann variable.

Now let us redefine Grassmann coordinates into projective ones.

$\displaystyle\Theta_{\hat{\alpha}}=\theta_{\hat{\alpha}2}-\zeta\theta_{\hat{\alpha}1}\;\;\textit{such
that}\;\;\Delta^{\hat{\alpha}}=D^{\hat{\alpha}}_{2}-\frac{1}{\zeta}D^{\hat{\alpha}}_{1}\;\sim\;\frac{\partial}{\partial\Theta_{\hat{\alpha}}}$
(9)
$\displaystyle\Theta^{\hat{\alpha}}=\theta^{\hat{\alpha}}_{2}-\zeta\theta^{\hat{\alpha}}_{1}\;\;\textit{such
that}\;\;\Delta_{\hat{\alpha}}=D_{\hat{\alpha}2}-\frac{1}{\zeta}D_{\hat{\alpha}1}\;\sim\;\frac{\partial}{\partial\Theta^{\hat{\alpha}}}$
(10)

Then we define a projective, constant field $\tau$ that couples the projective
covariant multiplet $G$. The constant field $\tau$ is defined by:

$\displaystyle\Delta^{2}\tau=\frac{1}{g_{YM}^{2}};\hskip
28.45274pt\tau=\frac{1}{g_{YM}^{2}}\Theta^{2};\hskip 28.45274pt\nabla\tau=0$
(11)

Yang-Mills action then becomes:

$\displaystyle S_{YM}=\frac{1}{g_{YM}^{2}}\tr\int
d^{5}x\oint\frac{d\zeta}{\zeta}\Delta^{4}\Theta^{2}G,\;\;\textit{where}\;\;G=\nabla^{4}\\{A,\mathcal{W}\\}$
(12)

In the next part, we will show that this particular construction of action
reduces to the expected expression in four-dimensional Yang-Mills action
formula, proving that this is the correct form.

###  II.3 Reduction to 4-dimensions

One could want to see how this above-received construction works out if we
reduce the dimension, does it produce well-known 4-dimensional chiral actions?
For this purpose, let us write down the 5-dimensional symplectic Majorana
spinors as a pair of spinors that are constructed from 4-dimensional spinors.
The pairs are related to each other with the antisymmetric tensor
$\epsilon^{ij}$. Spinor relations between $\mathbb{R}^{5|8}$ and
$\mathbb{R}^{4|8}$ are:

$\displaystyle\theta^{\hat{\alpha}}_{1}=(\theta^{\alpha}_{1}\;,\;\mkern
1.5mu\overline{\mkern-1.5mu\theta\mkern-1.5mu}\mkern
1.5mu_{2\dot{\alpha}})\;\;,\;\;\theta^{\hat{\alpha}}_{2}=(\theta^{\alpha}_{2}\;,\;-\mkern
1.5mu\overline{\mkern-1.5mu\theta\mkern-1.5mu}\mkern 1.5mu_{1\dot{\alpha}})$
$\displaystyle\theta_{1\hat{\alpha}}=\binom{\theta_{1\alpha}}{\mkern
1.5mu\overline{\mkern-1.5mu\theta\mkern-1.5mu}\mkern
1.5mu^{\dot{\alpha}}_{2}}\;\;,\;\;\theta_{2\hat{\alpha}}=\binom{\theta_{2\alpha}}{-\mkern
1.5mu\overline{\mkern-1.5mu\theta\mkern-1.5mu}\mkern
1.5mu^{\dot{\alpha}}_{1}}$ (13)

The corresponding covariant derivatives will be:

$\displaystyle D^{\hat{\alpha}}_{1}=(D^{\alpha}\;,\;\mkern
1.5mu\overline{\mkern-1.5muQ\mkern-1.5mu}\mkern
1.5mu_{\dot{\alpha}})\;\;,\;\;D^{\hat{\alpha}}_{2}=(Q^{\alpha}\;,\;-\mkern
1.5mu\overline{\mkern-1.5muD\mkern-1.5mu}\mkern 1.5mu_{1\dot{\alpha}})$
$\displaystyle D_{1\hat{\alpha}}=\binom{D_{\alpha}}{\mkern
1.5mu\overline{\mkern-1.5muQ\mkern-1.5mu}\mkern
1.5mu^{\dot{\alpha}}}\;\;,\;\;D_{2\hat{\alpha}}=\binom{Q_{\alpha}}{-\mkern
1.5mu\overline{\mkern-1.5muD\mkern-1.5mu}\mkern 1.5mu^{\dot{\alpha}}}$ (14)

Then, as we defined in (9) the 5D projective Grassmann coordinates are in
terms of 4D variables:

$\displaystyle\Theta^{\hat{\alpha}}=(\theta^{\alpha}_{2}-\zeta\theta^{\alpha}_{1}\;,\;-\mkern
1.5mu\overline{\mkern-1.5mu\theta\mkern-1.5mu}\mkern
1.5mu_{\dot{\alpha}1}-\zeta\mkern
1.5mu\overline{\mkern-1.5mu\theta\mkern-1.5mu}\mkern
1.5mu_{\dot{\alpha}2})\;\;,\;\;\Theta_{\hat{\alpha}}=\binom{\theta_{2\alpha}-\zeta\theta_{1\alpha}}{-\mkern
1.5mu\overline{\mkern-1.5mu\theta\mkern-1.5mu}\mkern
1.5mu^{\dot{\alpha}}_{1}-\zeta\mkern
1.5mu\overline{\mkern-1.5mu\theta\mkern-1.5mu}\mkern
1.5mu^{\dot{\alpha}}_{2}}$
$\displaystyle\Theta^{2}_{5}=\frac{1}{2}\Theta^{\hat{\alpha}}\Theta_{\hat{\alpha}}=\theta^{2}_{4}+\mkern
1.5mu\overline{\mkern-1.5mu\theta\mkern-1.5mu}\mkern 1.5mu^{2}_{4}$ (15)

Above we also defined the 4-dimensional projective Grassmann variables.
In this description the $\nabla$ and $\Delta$ operators are expressed with 4D
pairs as:

$\displaystyle\nabla^{\hat{\alpha}}=D^{\hat{\alpha}}_{1}+\zeta
D^{\hat{\alpha}}_{2}=(\nabla^{\alpha}_{4}\;,\,\mkern
1.5mu\overline{\mkern-1.5mu\nabla\mkern-1.5mu}\mkern
1.5mu_{4\dot{\alpha}})\;\;,\;\;\nabla_{\hat{\alpha}}=D_{1\hat{\alpha}}+\zeta
D_{2\hat{\alpha}}=\binom{\nabla^{\alpha}_{4}}{\mkern
1.5mu\overline{\mkern-1.5mu\nabla\mkern-1.5mu}\mkern 1.5mu_{4\dot{\alpha}}}$
(16)
$\displaystyle\Delta^{\hat{\alpha}}=D^{\hat{\alpha}}_{2}-\frac{1}{\zeta}D^{\hat{\alpha}}_{1}=(\Delta^{\alpha}_{4}\;,\;-\mkern
1.5mu\overline{\mkern-1.5mu\Delta\mkern-1.5mu}\mkern
1.5mu_{4\dot{\alpha}})\;\;,\;\;\Delta_{\hat{\alpha}}=D_{\hat{\alpha}2}-\frac{1}{\zeta}D_{\hat{\alpha}1}=\binom{\Delta_{4\alpha}}{-\mkern
1.5mu\overline{\mkern-1.5mu\Delta\mkern-1.5mu}\mkern
1.5mu^{\dot{\alpha}}_{4}}$ (17)

Based on these relations, we now obtain the 4D chiral/antichiral SYM from the
action in (12). First, the field strength needs to be expressed by
$\mathcal{W}_{4}$ and $\mkern
1.5mu\overline{\mkern-1.5mu\mathcal{W}\mkern-1.5mu}\mkern 1.5mu_{4}$. To do
this, let us write the 5D superalgebra in terms of 2$\times$2 matrices.

$\displaystyle\\{\mathcal{D}^{i}_{\hat{\alpha}},\mathcal{D}^{j}_{\hat{\beta}}\\}=i\epsilon^{ij}\left[\begin{pmatrix}i\epsilon_{\alpha\beta}\mathcal{D}_{5}&\nabla_{\alpha\dot{\beta}}\\\
\nabla_{\dot{\alpha}\beta}&-i\epsilon_{\dot{\alpha}\dot{\beta}}\mathcal{D}_{5}\end{pmatrix}+\begin{pmatrix}\epsilon_{\alpha\beta}&0\\\
0&\epsilon_{\dot{\alpha}\dot{\beta}}\end{pmatrix}\mathbb{W}_{5}\right]$ (18)

After descending from 4-spinor to 2-spinor representations, the resulting
anticommutators are equivalent to the 4D superalgebra:

$\displaystyle\\{\mathcal{D}^{i}_{\alpha},\mathcal{D}^{j}_{\beta}\\}=i\epsilon^{ij}(\nabla_{5}+\epsilon_{\alpha\beta}\mathbb{W}_{5})=i\epsilon^{ij}\epsilon_{\alpha\beta}\mkern
1.5mu\overline{\mkern-1.5mu\mathbb{W}_{4}\mkern-1.5mu}\mkern 1.5mu\;;$ (19)
$\displaystyle\\{\mkern
1.5mu\overline{\mkern-1.5mu\mathcal{D}\mkern-1.5mu}\mkern
1.5mu^{i}_{\dot{\alpha}},\mkern
1.5mu\overline{\mkern-1.5mu\mathcal{D}\mkern-1.5mu}\mkern
1.5mu^{j}_{\dot{\beta}}\\}=i\epsilon^{ij}(-\nabla_{5}+\epsilon_{\dot{\alpha}\dot{\beta}}\mathbb{W}_{5})=i\epsilon^{ij}\epsilon_{\dot{\alpha}\dot{\beta}}\mathbb{W}_{4}$
(20)

Here we have expressed the gauge covariantized
$\nabla_{\hat{\alpha}\hat{\beta}}$ along the 5-th spacetime direction using
the 5D antisymmetric $\Gamma$ matrix:

$\displaystyle(\nabla_{5})_{\hat{\alpha}\hat{\beta}}=(\partial_{5}+A_{5})_{\hat{\alpha}\hat{\beta}}=(C\Gamma^{5})_{\hat{\alpha}\hat{\beta}}\mathcal{D}_{5}=\begin{pmatrix}i\epsilon_{\alpha\beta}&0\\\
0&-i\epsilon_{\dot{\alpha}\dot{\beta}}\end{pmatrix}\mathcal{D}_{5}$ (21)

Comparing the 5D and 4D algebras, one could find:

$\displaystyle i$ $\displaystyle\mathcal{D}_{5}+\mathbb{W}_{5}=\mkern
1.5mu\overline{\mkern-1.5mu\mathbb{W}\mkern-1.5mu}\mkern 1.5mu_{4}$
$\displaystyle-i$ $\displaystyle\mathcal{D}_{5}+\mathbb{W}_{5}=\mathbb{W}_{4}$
(22)

Then we have the 5D field strength in terms of 4D ones.

$\mathbb{W}_{5}=\frac{1}{2}(\mathbb{W}_{4}+\mkern
1.5mu\overline{\mkern-1.5mu\mathbb{W}\mkern-1.5mu}\mkern 1.5mu_{4})$ (23)

Now that we know 5D field strength is expressed with 4D chiral field strength
and its complex conjugation, we reduce the Yang-Mills action in (12).
Projective, descendent field $G$ in 4-dimensions:

$G|_{4D}=\nabla_{4}^{2}\mathcal{W}_{4}^{2}+\mkern
1.5mu\overline{\mkern-1.5mu\nabla\mkern-1.5mu}\mkern 1.5mu_{4}^{2}\mkern
1.5mu\overline{\mkern-1.5mu\mathcal{W}\mkern-1.5mu}\mkern 1.5mu_{4}^{2}$ (24)

The 4-th degree of $\Delta_{5}$ reduces to:

$\Delta_{5}^{4}=\left(\Delta_{4}^{2}+\mkern
1.5mu\overline{\mkern-1.5mu\Delta\mkern-1.5mu}\mkern
1.5mu_{4}^{2}\right)^{2}=2\Delta_{4}^{2}\mkern
1.5mu\overline{\mkern-1.5mu\Delta\mkern-1.5mu}\mkern 1.5mu_{4}^{2}$ (25)

There will be spacetime derivatives when swapping $\Delta$ with $\mkern
1.5mu\overline{\mkern-1.5mu\Delta\mkern-1.5mu}\mkern 1.5mu$, however, we can
ignore them, since they are integrated over the spacetime.

$\displaystyle S_{YM}=$ $\displaystyle\int d^{4}x\Delta^{2}\mkern
1.5mu\overline{\mkern-1.5mu\Delta\mkern-1.5mu}\mkern
1.5mu^{2}\left[(\theta^{2}+\mkern
1.5mu\overline{\mkern-1.5mu\theta\mkern-1.5mu}\mkern
1.5mu^{2})(\nabla^{2}\mathcal{W}^{2}+\mkern
1.5mu\overline{\mkern-1.5mu\nabla\mkern-1.5mu}\mkern 1.5mu^{2}\mkern
1.5mu\overline{\mkern-1.5mu\mathcal{W}\mkern-1.5mu}\mkern 1.5mu^{2})\right]$
$\displaystyle\sim$ $\displaystyle\int
d^{4}x\frac{\partial^{2}}{\partial\theta^{2}}\frac{\partial^{2}}{\partial\mkern
1.5mu\overline{\mkern-1.5mu\theta\mkern-1.5mu}\mkern
1.5mu^{2}}\left[(\theta^{2}+\mkern
1.5mu\overline{\mkern-1.5mu\theta\mkern-1.5mu}\mkern
1.5mu^{2})(\nabla^{2}\mathcal{W}^{2}+\mkern
1.5mu\overline{\mkern-1.5mu\nabla\mkern-1.5mu}\mkern 1.5mu^{2}\mkern
1.5mu\overline{\mkern-1.5mu\mathcal{W}\mkern-1.5mu}\mkern 1.5mu^{2})\right]$
(26)

Taking into account 4-dimensional chirality and the Bianchi identities, it is
easy to see the above action reduces to just the sum of chiral/antichiral
actions:

$S_{YM}\sim\int d^{4}x\Delta^{2}\nabla^{2}\mathcal{W}^{2}+\int d^{4}x\mkern
1.5mu\overline{\mkern-1.5mu\Delta\mkern-1.5mu}\mkern 1.5mu^{2}\mkern
1.5mu\overline{\mkern-1.5mu\nabla\mkern-1.5mu}\mkern 1.5mu^{2}\mkern
1.5mu\overline{\mkern-1.5mu\mathcal{W}\mkern-1.5mu}\mkern 1.5mu^{2}$ (27)

This shows that the proposed projective formula for the 5-dimensional YM
action (12) is correct.

###  II.4 Discussion

In this section, we see that 5-dimensional Yang-Mills action can be described
in terms of field strength in projective superspace. We coupled a constant
projective field depending on Grassmann coordinates with the projective
covariant field which is the descendent derivative of the field strength.
Because we defined the derivatives with respect to the projective Grassman
coordinates as proportional to the $\Delta$ derivatives, it was
straightforward to prove that this action reduces to the 4-dimensional chiral
action and its complex conjugate part.

##  III Chern-Simons action

We take a hint from the description of the variation of the 5-dimensional
Chern-Simons action proposed by Zupnik in Zupnik (1999) and later corrected in
the erratum Zupnik (2002). In projective superspace, the action should contain
$e^{-V}\delta e^{V}$ as the variation of the prepotential. This is because of
the way vector multiplets couple to polar/hypermultiplets. This form is also
consistent with the overall gauge invariance of the action, as we will see
below.

$\displaystyle\delta S_{CS}=k_{5}\tr\int d^{5}xd^{8}\theta\oint d\zeta
e^{-V}\delta e^{V}\\{A\;,\;\mathcal{W}\\}$ (28)

From the previous chapters, we know the following infinitesimal gauge
transformation rules and these apply in 5-dimensions as well. Remind you that
the gauge parameters $i\mkern
1.5mu\overline{\mkern-1.5mu\Lambda\mkern-1.5mu}\mkern 1.5mu$, $i\Lambda$ are
projective.

$\displaystyle\delta e^{V}=i\mkern
1.5mu\overline{\mkern-1.5mu\Lambda\mkern-1.5mu}\mkern
1.5mue^{V}-e^{V}i\Lambda\;\;,\;\;\delta
A=-i\partial_{\zeta}\Lambda+[i\Lambda,A]\;\;,\;\;\delta
W=[i\Lambda,\mathcal{W}]$ (29)

Additionally, useful to remember
$e^{V}Ae^{-V}=\widetilde{A}+\partial_{\zeta}e^{V}e^{-V}$ and none of the
$i\mkern 1.5mu\overline{\mkern-1.5mu\lambda\mkern-1.5mu}\mkern
1.5mu\widetilde{A}\widetilde{\mathcal{W}}$ and $i\Lambda A\mathcal{W}$
contains $\zeta^{-1}$ term, therefore vanishes after contour integrals.
Furthermore, one could easily check that $\nabla^{4}(i\mkern
1.5mu\overline{\mkern-1.5mu\Lambda\mkern-1.5mu}\mkern
1.5mu\widetilde{\mathcal{W}}\partial_{\zeta}e^{V}e^{-V})=0$. Therefore,

$\displaystyle\delta S_{\Lambda}=k_{5}\tr\int d^{5}xd^{8}\theta\oint
d\zeta\left(i\mkern 1.5mu\overline{\mkern-1.5mu\Lambda\mkern-1.5mu}\mkern
1.5mu\\{\widetilde{A}\;,\;\widetilde{\mathcal{W}}\\}+i\mkern
1.5mu\overline{\mkern-1.5mu\Lambda\mkern-1.5mu}\mkern
1.5mu\\{\partial_{\zeta}e^{V}e^{-V}\;,\;\widetilde{\mathcal{W}}\\}-i\Lambda\\{A\;,\;\mathcal{W}\\}\right)=0$
(30)

proven to be gauge-invariant.

###  III.1 Abelian Chern-Simons action

Now that we know this formula (28) for variation is gauge invariant, let us
first derive the Abelian action from its corresponding variation.

$\delta S_{CS}^{Ab}=k_{5}\tr\int d^{5}xd^{8}\theta\oint d\zeta\delta
VA\nabla^{2}A$ (31)

where the Abelian version of the gauge connection $A_{\zeta}$ is as we
established in the previous 4-dimensional chapter:

$A_{\zeta_{0}}=\oint d\zeta_{1}\frac{V_{1}}{\zeta_{10}^{2}}$ (32)

Then one can pull a partial derivative from the variation on (31).

$\displaystyle(\ref{AbC-S})=$ $\displaystyle\delta k_{5}\int
d^{5}xd^{8}\theta\oint d\zeta_{0}V_{0}\oint
d\zeta_{1}\frac{V_{1}}{\zeta_{10}^{2}}\nabla_{0}^{2}\oint
d\zeta_{2}\frac{V_{2}}{\zeta_{20}^{2}}+$ (33) $\displaystyle-2k_{5}\int
d^{5}xd^{8}\theta\oint d\zeta_{0}V_{0}\oint d\zeta_{1}\frac{\delta
V_{1}}{\zeta_{10}^{2}}\nabla_{0}^{2}\oint
d\zeta_{2}\frac{V_{2}}{\zeta_{20}^{2}}$ (34)

Here we can apply renaming $\zeta_{0}\longleftrightarrow\zeta_{1}$ to the
second line and there reorder the first and the second parts

$\displaystyle-2k_{5}\int d^{5}xd^{8}\theta\oint d\zeta_{0}\delta V_{0}\oint
d\zeta_{1}\frac{V_{1}}{\zeta_{01}^{2}}\nabla_{1}^{2}\oint
d\zeta_{2}\frac{V_{2}}{\zeta_{21}^{2}}$ (35)

Again, recalling the 4-dimensional chapter, we use the following relation:

$\oint d\zeta_{1}\frac{V_{1}}{\zeta_{01}^{2}}=-\partial_{\zeta_{0}}V_{0}+\oint
d\zeta_{1}\frac{V_{1}}{\zeta_{10}^{2}}$ (36)

Another known fact from the 4-dimensions for any projective function
$\mathcal{P}$:

$\nabla^{2}\partial_{\zeta}\mathcal{P}=0$ (37)

The same is here in 5-dimensions, leading to the conclusion, saying that
$\nabla^{2}$ acting on positive or negative projections of the
$\zeta$-derivative of any projective field is $\zeta$-independent.

$\mathcal{W}^{Ab}=\nabla_{1}^{2}\oint
d\zeta_{2}\frac{V_{2}}{\zeta_{21}^{2}}=\nabla_{1}^{2}\oint
d\zeta_{2}\frac{V_{2}}{\zeta_{12}^{2}}=D_{2}^{\hat{\alpha}}D_{2\hat{\alpha}}v_{-1}$
(38)

Therefore, the second line (34) is reduced to the same form as the original
variation itself (31). This allows us to find the Abelian action:

$\displaystyle 3\delta S_{CS}^{Ab}=$ $\displaystyle 3k_{5}\int
d^{5}xd^{8}\theta\oint d\zeta_{0}\delta V_{0}\oint
d\zeta_{1}\frac{V_{1}}{\zeta_{10}^{2}}\nabla_{0}^{2}\oint
d\zeta_{2}\frac{V_{2}}{\zeta_{20}^{2}}$ $\displaystyle=\delta k_{5}\int
d^{5}xd^{8}\theta\oint d\zeta_{0}V_{0}\oint
d\zeta_{1}\frac{V_{1}}{\zeta_{10}^{2}}\nabla_{0}^{2}\oint
d\zeta_{2}\frac{V_{2}}{\zeta_{20}^{2}}$ $\displaystyle S_{CS}^{Ab}=$
$\displaystyle\frac{1}{3}k_{5}\int d^{5}xd^{8}\theta\oint
d\zeta_{0}V_{0}A_{\zeta_{0}}\nabla_{0}^{2}A_{\zeta_{0}}$ (39)

In this derivation, the key simplification comes from the $\zeta$-independent
description of the Abelian field strength (38). Can we also describe the non-
Abelian field strength in a similar manner as a $\zeta$-dependent part
$\times$ $\zeta$-indepent term $\times$ $\zeta$-dependent part?

In fact, we used a similar description in 4-dimensions to prove the
equivalence of the chiral action and the full-measure action Davgadorj et al.
(2023). We splat the prepotential into nonsymmetric parts. In 5-dimensions
this splitting will lead us to find desirable expressions for the field
strength as we needed.

$\displaystyle\mathcal{W}=$ $\displaystyle
e^{-\check{U}}(D_{1}^{\hat{\alpha}}D_{1\hat{\alpha}}\mkern
1.5mu\overline{\mkern-1.5muF\mkern-1.5mu}\mkern 1.5mu)e^{\check{U}}$ (40)
$\displaystyle\mathcal{W}=$ $\displaystyle
e^{-\hat{U}}(D_{2}^{\hat{\alpha}}D_{2\hat{\alpha}}F)e^{\hat{U}}$ (41)

Here, $F$ and $\mkern 1.5mu\overline{\mkern-1.5muF\mkern-1.5mu}\mkern 1.5mu$
are related to each other through $e^{P}e^{\mkern
1.5mu\overline{\mkern-1.5muP\mkern-1.5mu}\mkern 1.5mu}$ function:

$D_{1}^{\hat{\alpha}}D_{1\hat{\alpha}}\mkern
1.5mu\overline{\mkern-1.5muF\mkern-1.5mu}\mkern 1.5mu=e^{-\mkern
1.5mu\overline{\mkern-1.5muP\mkern-1.5mu}\mkern
1.5mu}e^{-P}(D_{2}^{\hat{\alpha}}D_{2\hat{\alpha}}F)e^{P}e^{\mkern
1.5mu\overline{\mkern-1.5muP\mkern-1.5mu}\mkern 1.5mu}$ (42)

For the definitions of these functions and techniques how to do it one can
refer to the appendices.

###  III.2 Non-Abelian Chern-Simons action

Our task now is to derive the actual non-Abelian action from the proposed
variation form (28). Let us first notice that for this variation we can
equivalently write as

$\displaystyle\delta S_{CS}=-k_{5}\tr\int d^{5}xd^{8}\theta\oint d\zeta
e^{-V}\delta e^{V}\nabla^{\hat{\alpha}}A\nabla_{\hat{\alpha}}A$ (43)

We can define here the projective fields $G$ and $K$. In fact, the
construction (28) stems from first constructing the projective covariant
supefield $G$ and then coupling it to the vector $e^{-V}\delta e^{V}$;
finally, the whole coupling comes with the projective measure $\Delta^{4}$.

$\displaystyle
G=\nabla^{4}\\{A\;,\;\mathcal{W}\\}=2\left(\nabla^{2}\mathcal{W}^{2}+\nabla^{\hat{\alpha}}\mathcal{W}\nabla_{\hat{\alpha}}\mathcal{W}\right)$
(44) $\displaystyle
K=\nabla^{4}\left(\nabla^{\hat{\alpha}}A\nabla_{\hat{\alpha}}A\right)=-2\nabla^{2}\mathcal{W}^{2}+3\nabla^{\hat{\alpha}}\mathcal{W}\nabla_{\hat{\alpha}}\mathcal{W}$
(45)

The full measure of the action is
$\mathrm{d}^{8}\theta\thicksim\Delta^{4}\nabla^{4}$, then since the Lagrangian
$e^{-V}\delta e^{V}G$ is a projective, we can use the projective half measure:
$\frac{\left(D^{\hat{\alpha}}_{2}D_{1\hat{\alpha}}\right)^{2}}{\zeta^{2}}$
instead of $\Delta^{4}$. This comes from the fact that $\Delta$’s can be
written in terms of $\nabla$’s and the $D$’s as:

$\Delta^{\hat{\alpha}}=2D^{\hat{\alpha}}_{2}-\frac{1}{\zeta}\nabla^{\hat{\alpha}}\;\;,\;\;\Delta_{\hat{\alpha}}=-\frac{2}{\zeta}D_{1\hat{\alpha}}+\frac{1}{\zeta}\nabla_{\hat{\alpha}}$
(46)

Then
$\Delta^{4}\nabla^{4}=(\frac{1}{2}\Delta^{\hat{\alpha}}\Delta_{\hat{\alpha}})^{2}\nabla^{4}=\frac{4}{\zeta^{2}}(D^{\hat{\alpha}}_{2}D_{1\hat{\alpha}})^{2}\nabla^{4}$.
Hence, the variation is:

$\displaystyle\delta S_{CS}=4k_{5}\tr\int
d^{5}x\left(D^{\hat{\alpha}}_{2}D_{1\hat{\alpha}}\right)^{2}\oint\frac{d\zeta}{\zeta^{2}}\left(e^{-V}\delta
e^{V}G\right)$ (47)

Since (28) and (43) (notice the negative sign) are equivalent, one can imply:

$\displaystyle\tr\int d^{5}x\oint d\zeta\Delta^{4}\left(e^{-V}\delta
e^{V}\nabla^{\hat{\alpha}}\mathcal{W}\nabla_{\hat{\alpha}}\mathcal{W}\right)=0$
(48)

This means consequently we can write the variation further:

$\displaystyle\delta S_{CS}=8k_{5}\tr\int
d^{5}x\left(D^{\hat{\alpha}}_{2}D_{1\hat{\alpha}}\right)^{2}\oint\frac{\nabla^{2}}{\zeta^{2}}\left(e^{-V}\delta
e^{V}\mathcal{W}^{2}\right)$ (49)

Let us now rewrite the integrand using the decomposition $e^{V}=e^{\hat{\mkern
1.5mu\overline{\mkern-1.5muU\mkern-1.5mu}\mkern 1.5mu}}e^{\check{U}}$ and
remind you that in this representation (we will introduce this decomposition
in the appendices) the field strength is expressed as
$\mathcal{W}=e^{-\check{U}}\left(D^{\hat{\alpha}}_{1}D_{1\hat{\alpha}}\mkern
1.5mu\overline{\mkern-1.5muF\mkern-1.5mu}\mkern 1.5mu\right)e^{\check{U}}$.
Then we can write the following:

$\displaystyle\oint
d\zeta\frac{\left(D^{\hat{\alpha}}_{2}D_{1\hat{\alpha}}\right)^{2}}{\zeta^{2}}\nabla^{2}\left[e^{-\hat{\mkern
1.5mu\overline{\mkern-1.5muU\mkern-1.5mu}\mkern 1.5mu}}\delta e^{\hat{\mkern
1.5mu\overline{\mkern-1.5muU\mkern-1.5mu}\mkern
1.5mu}}\left(D^{\hat{\alpha}}_{1}D_{1\hat{\alpha}}\mkern
1.5mu\overline{\mkern-1.5muF\mkern-1.5mu}\mkern 1.5mu\right)^{2}+\delta
e^{\check{U}}e^{-\check{U}}\left(D^{\hat{\alpha}}_{1}D_{1\hat{\alpha}}\mkern
1.5mu\overline{\mkern-1.5muF\mkern-1.5mu}\mkern 1.5mu\right)^{2}\right]=$ (50)
$\displaystyle=$ $\displaystyle\oint
d\zeta\frac{\left(D^{\hat{\alpha}}_{2}D_{1\hat{\alpha}}\right)^{2}}{\zeta^{2}}\left\\{\zeta\left(D^{\hat{\alpha}}_{2}D_{1\hat{\alpha}}\right)\left[e^{-\mkern
1.5mu\overline{\mkern-1.5muP\mkern-1.5mu}\mkern
1.5mu}e^{-P}\delta(e^{P}e^{\mkern
1.5mu\overline{\mkern-1.5muP\mkern-1.5mu}\mkern
1.5mu})\left(D^{\hat{\alpha}}_{1}D_{1\hat{\alpha}}\mkern
1.5mu\overline{\mkern-1.5muF\mkern-1.5mu}\mkern
1.5mu\right)^{2}\right]+\right.$ $\displaystyle+$
$\displaystyle\frac{1}{2}\zeta^{2}\left(D^{\hat{\alpha}}_{2}D_{2\hat{\alpha}}\right)\left[\frac{N}{\zeta}\delta(e^{P}e^{\mkern
1.5mu\overline{\mkern-1.5muP\mkern-1.5mu}\mkern
1.5mu})\left(D^{\hat{\alpha}}_{1}D_{1\hat{\alpha}}\mkern
1.5mu\overline{\mkern-1.5muF\mkern-1.5mu}\mkern 1.5mu\right)^{2}+(e^{-\mkern
1.5mu\overline{\mkern-1.5muP\mkern-1.5mu}\mkern 1.5mu}e^{-P})\frac{\delta
M}{\zeta}\left(D^{\hat{\alpha}}_{1}D_{1\hat{\alpha}}\mkern
1.5mu\overline{\mkern-1.5muF\mkern-1.5mu}\mkern 1.5mu\right)^{2}\right]$
$\displaystyle+$
$\displaystyle\left.\frac{1}{2}\left(D^{\hat{\alpha}}_{1}D_{1\hat{\alpha}}\right)\left[\delta\mkern
1.5mu\overline{\mkern-1.5muF\mkern-1.5mu}\mkern
1.5mu\zeta\left(D^{\hat{\alpha}}_{1}D_{1\hat{\alpha}}\mkern
1.5mu\overline{\mkern-1.5muF\mkern-1.5mu}\mkern
1.5mu\right)^{2}\right]\right\\}$ (51)

Here, we have rewritten $\nabla^{2}$ in terms of $\zeta$’s:

$\displaystyle\nabla^{2}=\frac{1}{2}D^{\hat{\alpha}}_{1}D_{1\hat{\alpha}}+\zeta
D^{\hat{\alpha}}_{2}D_{1\hat{\alpha}}+\frac{1}{2}\zeta^{2}D^{\hat{\alpha}}_{2}D_{2\hat{\alpha}}$
(52)

$N$ and $M$ are so far not defined $\zeta$-independent functions. But we do
not need to worry about them, since we will replace this term with the
convenient $F$ function later. However, for the sake of consistency, let us
write them down. $F$, $\mkern 1.5mu\overline{\mkern-1.5muF\mkern-1.5mu}\mkern
1.5mu$ and $e^{P}e^{\mkern 1.5mu\overline{\mkern-1.5muP\mkern-1.5mu}\mkern
1.5mu}$ and its inverse are defined respectively in (114), (117) and (106),
(107).

$\displaystyle N=\sum^{\infty}_{n=1}(-1)^{n}\oint d\zeta_{1}\dots
d\zeta_{n}\frac{X_{1}\dots
X_{n}}{\zeta_{21}\dots\zeta_{n,n-1}}\frac{\zeta_{n}}{\zeta_{1}}$ (53)
$\displaystyle M=\sum^{\infty}_{n=1}(-1)^{n+1}\oint d\zeta_{1}\dots
d\zeta_{n}\frac{X_{1}\dots
X_{n}}{\zeta_{21}\dots\zeta_{n,n-1}}\frac{\zeta_{1}}{\zeta_{n}}$ (54)

Equivalently, we could also apply the decomposition $e^{V}=e^{\check{\mkern
1.5mu\overline{\mkern-1.5muU\mkern-1.5mu}\mkern 1.5mu}}e^{\hat{U}}$, and in
this splitting, the field strength is written as
$\mathcal{W}=e^{-\hat{U}}\left(D^{\hat{\alpha}}_{2}D_{2\hat{\alpha}}F\right)e^{\hat{U}}$.
Then:

$\displaystyle\oint
d\zeta\frac{\left(D^{\hat{\alpha}}_{2}D_{1\hat{\alpha}}\right)^{2}}{\zeta^{2}}\nabla^{2}\left[e^{-\check{\mkern
1.5mu\overline{\mkern-1.5muU\mkern-1.5mu}\mkern 1.5mu}}\delta e^{\check{\mkern
1.5mu\overline{\mkern-1.5muU\mkern-1.5mu}\mkern
1.5mu}}\left(D^{\hat{\alpha}}_{2}D_{2\hat{\alpha}}F\right)^{2}+\delta
e^{\hat{U}}e^{-\hat{U}}\left(D^{\hat{\alpha}}_{2}D_{2\hat{\alpha}}F\right)^{2}\right]=$
(55) $\displaystyle=$ $\displaystyle\oint
d\zeta\frac{\left(D^{\hat{\alpha}}_{2}D_{1\hat{\alpha}}\right)^{2}}{\zeta^{2}}\left\\{\zeta\left(D^{\hat{\alpha}}_{2}D_{1\hat{\alpha}}\right)\left[\delta(e^{P}e^{\mkern
1.5mu\overline{\mkern-1.5muP\mkern-1.5mu}\mkern 1.5mu})e^{-\mkern
1.5mu\overline{\mkern-1.5muP\mkern-1.5mu}\mkern
1.5mu}e^{-P}\left(D^{\hat{\alpha}}_{2}D_{2\hat{\alpha}}F\right)^{2}\right]+\right.$
$\displaystyle+$
$\displaystyle\frac{1}{2}\left(D^{\hat{\alpha}}_{1}D_{1\hat{\alpha}}\right)\left[\zeta\delta
S(e^{-\mkern 1.5mu\overline{\mkern-1.5muP\mkern-1.5mu}\mkern
1.5mu}e^{-P})\left(D^{\hat{\alpha}}_{2}D_{2\hat{\alpha}}F\right)^{2}+\zeta\delta(e^{P}e^{\mkern
1.5mu\overline{\mkern-1.5muP\mkern-1.5mu}\mkern
1.5mu})T\left(D^{\hat{\alpha}}_{2}D_{2\hat{\alpha}}F\right)^{2}\right]$
$\displaystyle+$
$\displaystyle\left.\frac{1}{2}\zeta^{2}\left(D^{\hat{\alpha}}_{2}D_{2\hat{\alpha}}\right)\left[\frac{\delta
F}{\zeta}\left(D^{\hat{\alpha}}_{2}D_{2\hat{\alpha}}F\right)^{2}\right]\right\\}$
(56)

$S$ and $T$ are not important here, but we give them away anyways.

$\displaystyle S=\sum^{\infty}_{n=1}(-1)^{n+1}\oint d\zeta_{1}\dots
d\zeta_{n}\frac{X_{1}\dots
X_{n}}{\zeta_{21}\dots\zeta_{n,n-1}}\frac{1}{\zeta_{n}^{2}}$ (57)
$\displaystyle T=\sum^{\infty}_{n=1}(-1)^{n}\oint d\zeta_{1}\dots
d\zeta_{n}\frac{X_{1}\dots
X_{n}}{\zeta_{21}\dots\zeta_{n,n-1}}\frac{1}{\zeta_{1}^{2}}$ (58)

Now, those two expressions (using different decompositions) for the variation
of the action are obviously equal, and therefore we can compare their terms
with the same measure. Using the relation

$\left(D^{\hat{\alpha}}_{2}D_{2\hat{\alpha}}F\right)=e^{P}e^{\mkern
1.5mu\overline{\mkern-1.5muP\mkern-1.5mu}\mkern
1.5mu}\left(D^{\hat{\alpha}}_{1}D_{1\hat{\alpha}}\mkern
1.5mu\overline{\mkern-1.5muF\mkern-1.5mu}\mkern 1.5mu\right)e^{-\mkern
1.5mu\overline{\mkern-1.5muP\mkern-1.5mu}\mkern 1.5mu}e^{-P}$ (59)

we can see that both the first lines with the
$\left(D^{\hat{\alpha}}_{2}D_{1\hat{\alpha}}\right)$ derivatives are indeed
the same expression. Comparing the other two lines, we realise the following:

$\displaystyle\int
d^{5}x\left(D^{\hat{\alpha}}_{1}D_{1\hat{\alpha}}\right)\left[\delta
S(e^{-\mkern 1.5mu\overline{\mkern-1.5muP\mkern-1.5mu}\mkern
1.5mu}e^{-P})\left(D^{\hat{\alpha}}_{2}D_{2\hat{\alpha}}F\right)^{2}+\delta(e^{P}e^{\mkern
1.5mu\overline{\mkern-1.5muP\mkern-1.5mu}\mkern
1.5mu})T\left(D^{\hat{\alpha}}_{2}D_{2\hat{\alpha}}F\right)^{2}\right]=$
$\displaystyle=$ $\displaystyle\int
d^{5}x\left(D^{\hat{\alpha}}_{1}D_{1\hat{\alpha}}\right)\left[\delta\mkern
1.5mu\overline{\mkern-1.5muF\mkern-1.5mu}\mkern
1.5mu\left(D^{\hat{\alpha}}_{1}D_{1\hat{\alpha}}\mkern
1.5mu\overline{\mkern-1.5muF\mkern-1.5mu}\mkern 1.5mu\right)^{2}\right]$ (60)
$\displaystyle\int
d^{5}x\left(D^{\hat{\alpha}}_{2}D_{2\hat{\alpha}}\right)\left[N\delta(e^{P}e^{\mkern
1.5mu\overline{\mkern-1.5muP\mkern-1.5mu}\mkern
1.5mu})\left(D^{\hat{\alpha}}_{1}D_{1\hat{\alpha}}\mkern
1.5mu\overline{\mkern-1.5muF\mkern-1.5mu}\mkern 1.5mu\right)^{2}+(e^{-\mkern
1.5mu\overline{\mkern-1.5muP\mkern-1.5mu}\mkern 1.5mu}e^{-P})\delta
M\left(D^{\hat{\alpha}}_{1}D_{1\hat{\alpha}}\mkern
1.5mu\overline{\mkern-1.5muF\mkern-1.5mu}\mkern 1.5mu\right)^{2}\right]=$
$\displaystyle=$ $\displaystyle\int
d^{5}x\left(D^{\hat{\alpha}}_{2}D_{2\hat{\alpha}}\right)\left[\delta
F\left(D^{\hat{\alpha}}_{2}D_{2\hat{\alpha}}F\right)^{2}\right]$ (61)

Therefore, we can use only the $F$ and $\mkern
1.5mu\overline{\mkern-1.5muF\mkern-1.5mu}\mkern 1.5mu$ expressions as:

$\displaystyle\delta S_{CS}=8k_{5}\tr\int
d^{5}\left(D^{\hat{\alpha}}_{2}D_{1\hat{\alpha}}\right)^{2}\left\\{\left(D^{\hat{\alpha}}_{2}D_{1\hat{\alpha}}\right)\left[\delta(e^{P}e^{\mkern
1.5mu\overline{\mkern-1.5muP\mkern-1.5mu}\mkern 1.5mu})e^{-\mkern
1.5mu\overline{\mkern-1.5muP\mkern-1.5mu}\mkern
1.5mu}e^{-P}\left(D^{\hat{\alpha}}_{2}D_{2\hat{\alpha}}F\right)^{2}\right]+\right.$
$\displaystyle+$
$\displaystyle\left.\frac{1}{2}\left(D^{\hat{\alpha}}_{1}D_{1\hat{\alpha}}\right)\left[\delta\mkern
1.5mu\overline{\mkern-1.5muF\mkern-1.5mu}\mkern
1.5mu\left(D^{\hat{\alpha}}_{1}D_{1\hat{\alpha}}\mkern
1.5mu\overline{\mkern-1.5muF\mkern-1.5mu}\mkern
1.5mu\right)^{2}\right]+\frac{1}{2}\left(D^{\hat{\alpha}}_{2}D_{2\hat{\alpha}}\right)\left[\delta
F\left(D^{\hat{\alpha}}_{2}D_{2\hat{\alpha}}F\right)^{2}\right]\right\\}$ (62)

One could have received the above three terms with different numerical factors
if they had started with the more general measure $\Delta^{4}$, instead of
switching to the projective measure
$\frac{(D^{\hat{\alpha}}_{2}D_{1\hat{\alpha}})^{2}}{\zeta^{2}}$. Remembering
that the projective function $G$ is reduced to the form
$\nabla^{2}\mathcal{W}^{2}$, we write the Chern-Simons variation in a general
form:

$\displaystyle\delta S_{CS}=k_{5}\tr\int
d^{5}x\oint\Delta^{4}\nabla^{2}\left(e^{-V}\delta e^{V}\mathcal{W}^{2}\right)$
(63)

Let us now find the component covariant derivative form of the measure
$\Delta^{4}\nabla^{2}$. Before that we establish that with the space-time
integration the following is true:

$\displaystyle\int
d^{5}x\left(D^{\hat{\alpha}}_{2}D_{1\hat{\alpha}}\right)\left(D^{\hat{\alpha}}_{2}D_{2\hat{\alpha}}\right)=\int
d^{5}x\left(D^{\hat{\alpha}}_{2}D_{2\hat{\alpha}}\right)\left(D^{\hat{\alpha}}_{2}D_{1\hat{\alpha}}\right)$
$\displaystyle\int
d^{5}x\left(D^{\hat{\alpha}}_{1}D_{1\hat{\alpha}}\right)\left(D^{\hat{\alpha}}_{2}D_{2\hat{\alpha}}\right)=\int
d^{5}x\left(D^{\hat{\alpha}}_{2}D_{2\hat{\alpha}}\right)\left(D^{\hat{\alpha}}_{1}D_{1\hat{\alpha}}\right)$
(64)

Taking those into account the measure becomes:

$\displaystyle\Delta^{4}\nabla^{2}=$
$\displaystyle\frac{3}{8}\left(D^{\hat{\alpha}}_{1}D_{1\hat{\alpha}}\right)\left(D^{\hat{\alpha}}_{2}D_{2\hat{\alpha}}\right)^{2}-\frac{1}{2}\left(D^{\hat{\alpha}}_{2}D_{2\hat{\alpha}}\right)\left(D^{\hat{\alpha}}_{2}D_{1\hat{\alpha}}\right)^{2}+$
$\displaystyle+\frac{1}{\zeta}\left[\left(D^{\hat{\alpha}}_{2}D_{1\hat{\alpha}}\right)^{3}-\frac{1}{2}\left(D^{\hat{\alpha}}_{2}D_{1\hat{\alpha}}\right)\left(D^{\hat{\alpha}}_{2}D_{2\hat{\alpha}}\right)\left(D^{\hat{\alpha}}_{1}D_{1\hat{\alpha}}\right)\right]+$
$\displaystyle+\frac{1}{\zeta^{2}}\left[\frac{3}{8}\left(D^{\hat{\alpha}}_{2}D_{2\hat{\alpha}}\right)\left(D^{\hat{\alpha}}_{1}D_{1\hat{\alpha}}\right)^{2}-\frac{1}{2}\left(D^{\hat{\alpha}}_{1}D_{1\hat{\alpha}}\right)\left(D^{\hat{\alpha}}_{2}D_{1\hat{\alpha}}\right)^{2}\right]$
(65)

Further, we can investigate the relations between these terms using the fact
that $\nabla^{6}=0$.

$\displaystyle 0=\int d^{5}x\oint\frac{d\zeta}{\zeta^{3}}\nabla^{6}=\int
d^{5}x\left[\left(D^{\hat{\alpha}}_{2}D_{1\hat{\alpha}}\right)^{3}+\frac{6}{4}\left(D^{\hat{\alpha}}_{2}D_{2\hat{\alpha}}\right)\left(D^{\hat{\alpha}}_{1}D_{1\hat{\alpha}}\right)\left(D^{\hat{\alpha}}_{2}D_{1\hat{\alpha}}\right)\right]$
$\displaystyle 0=\int d^{5}x\oint\frac{d\zeta}{\zeta^{4}}\nabla^{6}=\int
d^{5}x\left[\frac{3}{8}\left(D^{\hat{\alpha}}_{1}D_{1\hat{\alpha}}\right)\left(D^{\hat{\alpha}}_{2}D_{2\hat{\alpha}}\right)^{2}+\frac{3}{2}\left(D^{\hat{\alpha}}_{2}D_{2\hat{\alpha}}\right)\left(D^{\hat{\alpha}}_{2}D_{1\hat{\alpha}}\right)^{2}\right]$
$\displaystyle 0=\int d^{5}x\oint\frac{d\zeta}{\zeta^{2}}\nabla^{6}=\int
d^{5}x\left[\frac{3}{8}\left(D^{\hat{\alpha}}_{2}D_{2\hat{\alpha}}\right)\left(D^{\hat{\alpha}}_{1}D_{1\hat{\alpha}}\right)^{2}+\frac{3}{2}\left(D^{\hat{\alpha}}_{1}D_{1\hat{\alpha}}\right)\left(D^{\hat{\alpha}}_{2}D_{1\hat{\alpha}}\right)^{2}\right]$
(66)

Once we write the integrand $e^{-V}\delta e^{V}\mathcal{W}^{2}$ using non-
symmetric splitting and the whole expression becomes $\zeta$-independent, just
as we did earlier with the projective measure, we can then use the relations
(III.2) to simplify the measure in (III.2). This will give us the Chern-Simons
variation:

$\displaystyle\delta S_{CS}=2k_{5}\tr\int
d^{5}\left\\{\frac{4}{3}\left(D^{\hat{\alpha}}_{2}D_{1\hat{\alpha}}\right)^{3}\left[\delta(e^{P}e^{\mkern
1.5mu\overline{\mkern-1.5muP\mkern-1.5mu}\mkern 1.5mu})e^{-\mkern
1.5mu\overline{\mkern-1.5muP\mkern-1.5mu}\mkern
1.5mu}e^{-P}\left(D^{\hat{\alpha}}_{2}D_{2\hat{\alpha}}F\right)^{2}\right]\right.$
$\displaystyle\left.-2\left(D^{\hat{\alpha}}_{2}D_{1\hat{\alpha}}\right)^{2}\left(D^{\hat{\alpha}}_{1}D_{1\hat{\alpha}}\right)\left[\delta\mkern
1.5mu\overline{\mkern-1.5muF\mkern-1.5mu}\mkern
1.5mu\left(D^{\hat{\alpha}}_{1}D_{1\hat{\alpha}}\mkern
1.5mu\overline{\mkern-1.5muF\mkern-1.5mu}\mkern
1.5mu\right)^{2}\right]-2\left(D^{\hat{\alpha}}_{2}D_{1\hat{\alpha}}\right)^{2}\left(D^{\hat{\alpha}}_{2}D_{2\hat{\alpha}}\right)\left[\delta
F\left(D^{\hat{\alpha}}_{2}D_{2\hat{\alpha}}F\right)^{2}\right]\right\\}$ (67)

Two forms of variation (III.2) and (III.2) are equal. Therefore, we can find
the term with the $\left(D^{\hat{\alpha}}_{2}D_{1\hat{\alpha}}\right)^{3}$
measure from this equality.

$\displaystyle\tr\int
d^{5}x\left(D^{\hat{\alpha}}_{2}D_{1\hat{\alpha}}\right)^{3}\left[\delta(e^{P}e^{\mkern
1.5mu\overline{\mkern-1.5muP\mkern-1.5mu}\mkern 1.5mu})e^{-\mkern
1.5mu\overline{\mkern-1.5muP\mkern-1.5mu}\mkern
1.5mu}e^{-P}\left(D^{\hat{\alpha}}_{2}D_{2\hat{\alpha}}F\right)^{2}\right]=$
$\displaystyle=-\frac{3}{2}\tr\int
d^{5}x\left(D^{\hat{\alpha}}_{2}D_{1\hat{\alpha}}\right)^{2}\left\\{\left(D^{\hat{\alpha}}_{1}D_{1\hat{\alpha}}\right)\left[\delta\mkern
1.5mu\overline{\mkern-1.5muF\mkern-1.5mu}\mkern
1.5mu\left(D^{\hat{\alpha}}_{1}D_{1\hat{\alpha}}\mkern
1.5mu\overline{\mkern-1.5muF\mkern-1.5mu}\mkern
1.5mu\right)^{2}\right]+\left(D^{\hat{\alpha}}_{2}D_{2\hat{\alpha}}\right)\left[\delta
F\left(D^{\hat{\alpha}}_{2}D_{2\hat{\alpha}}F\right)^{2}\right]\right\\}$ (68)

Then, we replace it in the variation, so that we now have only two analogous
terms.

$\displaystyle\delta S_{CS}=-8k_{5}\tr\int
d^{5}x\left(D^{\hat{\alpha}}_{2}D_{1\hat{\alpha}}\right)^{2}\left\\{\left(D^{\hat{\alpha}}_{1}D_{1\hat{\alpha}}\right)\left[\delta\mkern
1.5mu\overline{\mkern-1.5muF\mkern-1.5mu}\mkern
1.5mu\left(D^{\hat{\alpha}}_{1}D_{1\hat{\alpha}}\mkern
1.5mu\overline{\mkern-1.5muF\mkern-1.5mu}\mkern
1.5mu\right)^{2}\right]+\left(D^{\hat{\alpha}}_{2}D_{2\hat{\alpha}}\right)\left[\delta
F\left(D^{\hat{\alpha}}_{2}D_{2\hat{\alpha}}F\right)^{2}\right]\right\\}$
$\displaystyle=$ $\displaystyle 2k_{5}\tr\int
d^{5}x\left(D^{\hat{\alpha}}_{2}D_{2\hat{\alpha}}\right)\left(D^{\hat{\alpha}}_{1}D_{1\hat{\alpha}}\right)\left\\{\left(D^{\hat{\alpha}}_{1}D_{1\hat{\alpha}}\right)\left[\delta\mkern
1.5mu\overline{\mkern-1.5muF\mkern-1.5mu}\mkern
1.5mu\left(D^{\hat{\alpha}}_{1}D_{1\hat{\alpha}}\mkern
1.5mu\overline{\mkern-1.5muF\mkern-1.5mu}\mkern
1.5mu\right)^{2}\right]+\left(D^{\hat{\alpha}}_{2}D_{2\hat{\alpha}}\right)\left[\delta
F\left(D^{\hat{\alpha}}_{2}D_{2\hat{\alpha}}F\right)^{2}\right]\right\\}$ (69)

Now, we treat these two terms separately, taking into account the fact that
their measures already contain four $D_{1}$’s or four $D_{2}$’s for the second
term. This fact leads to simple observations using partial integrations by
$D_{1}$ (or $D_{2}$):

$\displaystyle\left(D_{2}D_{2}\right)\left(D_{1}D_{1}\right)^{2}\left[D_{1}D_{1}\mkern
1.5mu\overline{\mkern-1.5muF\mkern-1.5mu}\mkern 1.5mu\cdot
D^{\hat{\alpha}}_{1}\bar{F}\cdot
D_{1\hat{\alpha}}\delta\bar{F}\right]=\left(D_{2}D_{2}\right)\left(D_{1}D_{1}\right)^{2}$
$\displaystyle\left[\frac{1}{4}D^{\hat{\alpha}}_{1}\bar{F}\cdot
D_{1}D_{1}\mkern 1.5mu\overline{\mkern-1.5muF\mkern-1.5mu}\mkern 1.5mu\cdot
D_{1\hat{\alpha}}\delta\bar{F}+\right.$
$\displaystyle\left.+\frac{1}{4}D^{\hat{\alpha}}_{1}\bar{F}\cdot
D_{1\hat{\alpha}}\bar{F}\cdot D_{1}D_{1}\delta\bar{F}\right]$ (70)
$\displaystyle\left(D_{2}D_{2}\right)\left(D_{1}D_{1}\right)^{2}\left[D^{\hat{\alpha}}_{1}\bar{F}\cdot
D_{1}D_{1}\mkern 1.5mu\overline{\mkern-1.5muF\mkern-1.5mu}\mkern 1.5mu\cdot
D_{1\hat{\alpha}}\delta\bar{F}\right]=\left(D_{2}D_{2}\right)\left(D_{1}D_{1}\right)^{2}$
$\displaystyle\left[\frac{1}{4}D_{1}D_{1}\mkern
1.5mu\overline{\mkern-1.5muF\mkern-1.5mu}\mkern 1.5mu\cdot
D^{\hat{\alpha}}_{1}\bar{F}\cdot D_{1\hat{\alpha}}\delta\bar{F}+\right.$
$\displaystyle\left.+\frac{1}{4}D^{\hat{\alpha}}_{1}\bar{F}\cdot
D_{1\hat{\alpha}}\bar{F}\cdot D_{1}D_{1}\delta\bar{F}\right]$ (71)

Those two relations are independent, therefore, substituting we find:

$\displaystyle\left(D_{2}D_{2}\right)\left(D_{1}D_{1}\right)^{2}\left[D_{1}D_{1}\mkern
1.5mu\overline{\mkern-1.5muF\mkern-1.5mu}\mkern 1.5mu\cdot
D^{\hat{\alpha}}_{1}\bar{F}\cdot
D_{1\hat{\alpha}}\delta\bar{F}\right]=\frac{1}{3}\left(D_{2}D_{2}\right)\left(D_{1}D_{1}\right)^{2}\left[D^{\hat{\alpha}}_{1}\bar{F}\cdot
D_{1\hat{\alpha}}\bar{F}\cdot D_{1}D_{1}\delta\bar{F}\right]$
$\displaystyle\left(D_{2}D_{2}\right)\left(D_{1}D_{1}\right)^{2}\left[D^{\hat{\alpha}}_{1}\bar{F}\cdot
D_{1}D_{1}\mkern 1.5mu\overline{\mkern-1.5muF\mkern-1.5mu}\mkern 1.5mu\cdot
D_{1\hat{\alpha}}\delta\bar{F}\right]=\frac{1}{3}\left(D_{2}D_{2}\right)\left(D_{1}D_{1}\right)^{2}\left[D^{\hat{\alpha}}_{1}\bar{F}\cdot
D_{1\hat{\alpha}}\bar{F}\cdot D_{1}D_{1}\delta\bar{F}\right]$ (72)

Further, we use the above 2 relations in the next relation repeating partial
integrations:

$\displaystyle\left(D_{2}D_{2}\right)\left(D_{1}D_{1}\right)^{2}\left[D^{\hat{\alpha}}_{1}\bar{F}\cdot
D_{1\hat{\alpha}}\bar{F}\cdot
D_{1}D_{1}\delta\bar{F}\right]=\left(D_{2}D_{2}\right)\left(D_{1}D_{1}\right)^{2}\left[D_{1}D_{1}\left(D^{\hat{\alpha}}_{1}\bar{F}\cdot
D_{1\hat{\alpha}}\bar{F}\right)\cdot\delta\bar{F}\right]=$
$\displaystyle\left(D_{2}D_{2}\right)\left(D_{1}D_{1}\right)^{2}\left[D^{\hat{\alpha}}_{1}D_{1}D_{1}\mkern
1.5mu\overline{\mkern-1.5muF\mkern-1.5mu}\mkern 1.5mu\cdot
D_{1\hat{\alpha}}\mkern 1.5mu\overline{\mkern-1.5muF\mkern-1.5mu}\mkern
1.5mu\cdot\delta\bar{F}+D^{\hat{\alpha}}_{1}\mkern
1.5mu\overline{\mkern-1.5muF\mkern-1.5mu}\mkern 1.5mu\cdot
D_{1\hat{\alpha}}D_{1}D_{1}\bar{F}\cdot\delta\bar{F}-2D^{\hat{\beta}}_{1}D^{\hat{\alpha}}_{1}\bar{F}\cdot
D_{1\hat{\beta}}D_{1\hat{\alpha}}\bar{F}\cdot\delta\bar{F}\right]=$
$\displaystyle\left(D_{2}D_{2}\right)\left(D_{1}D_{1}\right)^{2}\left[-\frac{5}{2}(D_{1}D_{1}\bar{F})^{2}\cdot\delta\bar{F}-D_{1}D_{1}\bar{F}\cdot
D^{\hat{\alpha}}_{1}\bar{F}\cdot
D_{1\hat{\alpha}}\delta\bar{F}-D^{\hat{\alpha}}_{1}\bar{F}\cdot
D_{1}D_{1}\bar{F}\cdot D_{1\hat{\alpha}}\delta\bar{F}\right]=$
$\displaystyle\left(D_{2}D_{2}\right)\left(D_{1}D_{1}\right)^{2}\left[-\frac{5}{2}(D_{1}D_{1}\bar{F})^{2}\cdot\delta\bar{F}-\frac{2}{3}D^{\hat{\alpha}}_{1}\bar{F}\cdot
D_{1\hat{\alpha}}\bar{F}\cdot D_{1}D_{1}\delta\bar{F}\right]$ (73)

This brings us to the next relation:

$\displaystyle\left(D_{2}D_{2}\right)\left(D_{1}D_{1}\right)^{2}\left[(D_{1}D_{1}\bar{F})^{2}\cdot\delta\bar{F}\right]=-\frac{2}{3}\left(D_{2}D_{2}\right)\left(D_{1}D_{1}\right)^{2}\left[D^{\hat{\alpha}}_{1}\bar{F}\cdot
D_{1\hat{\alpha}}\bar{F}\cdot D_{1}D_{1}\delta\bar{F}\right]$ (74)

On the other hand we can use the variation with respect to prepotentials
$X_{i}$.

$\displaystyle\left(D_{2}D_{2}\right)\left(D_{1}D_{1}\right)^{2}\left[D^{\hat{\alpha}}_{1}\bar{F}\cdot
D_{1\hat{\alpha}}\bar{F}\cdot
D_{1}D_{1}\delta\bar{F}\right]=\delta\left(D_{2}D_{2}\right)\left(D_{1}D_{1}\right)^{2}\left[D^{\hat{\alpha}}_{1}\bar{F}\cdot
D_{1\hat{\alpha}}\bar{F}\cdot D_{1}D_{1}\bar{F}\right]$
$\displaystyle-\left(D_{2}D_{2}\right)\left(D_{1}D_{1}\right)^{2}\left[D_{1}D_{1}\mkern
1.5mu\overline{\mkern-1.5muF\mkern-1.5mu}\mkern 1.5mu\cdot
D^{\hat{\alpha}}_{1}\bar{F}\cdot
D_{1\hat{\alpha}}\delta\bar{F}+D^{\hat{\alpha}}_{1}\bar{F}\cdot
D_{1}D_{1}\mkern 1.5mu\overline{\mkern-1.5muF\mkern-1.5mu}\mkern 1.5mu\cdot
D_{1\hat{\alpha}}\delta\bar{F}\right]$ (75)

And then apply here the relations in (III.2) to find:

$\displaystyle\left(D_{2}D_{2}\right)\left(D_{1}D_{1}\right)^{2}\left[D^{\hat{\alpha}}_{1}\bar{F}\cdot
D_{1\hat{\alpha}}\bar{F}\cdot
D_{1}D_{1}\delta\bar{F}\right]=\frac{3}{5}\delta\left(D_{2}D_{2}\right)\left(D_{1}D_{1}\right)^{2}\left[D^{\hat{\alpha}}_{1}\bar{F}\cdot
D_{1\hat{\alpha}}\bar{F}\cdot D_{1}D_{1}\bar{F}\right]$ (76)

This allows us to write:

$\displaystyle\left(D_{2}D_{2}\right)\left(D_{1}D_{1}\right)^{2}\left[(D_{1}D_{1}\bar{F})^{2}\cdot\delta\bar{F}\right]$
$\displaystyle=-\frac{2}{5}\delta\left(D_{2}D_{2}\right)\left(D_{1}D_{1}\right)^{2}\left[D^{\hat{\alpha}}_{1}\bar{F}\cdot
D_{1\hat{\alpha}}\bar{F}\cdot D_{1}D_{1}\bar{F}\right]=$ $\displaystyle=$
$\displaystyle\frac{2}{5}\delta\left(D_{2}D_{2}\right)\left(D_{1}D_{1}\right)^{2}\left[\bar{F}\cdot(D_{1}D_{1}\bar{F})^{2}-\frac{1}{2}\bar{F}^{2}\cdot(D_{1}D_{1})^{2}\bar{F}\right]$
(77)

Since we can exchange ($D_{1}\leftrightarrow D_{2},F\leftrightarrow\bar{F}$)
and employ the same steps as above, the non-Abelian Chern-Simons action
finally takes the form:

$\displaystyle S_{CS}=\frac{1}{5}k_{5}\tr\int
d^{5}x\left(D_{2}D_{2}\right)\left(D_{1}D_{1}\right)$
$\displaystyle\left\\{(D_{1}D_{1})\left[\bar{F}\cdot(D_{1}D_{1}\bar{F})^{2}-\frac{1}{2}\bar{F}^{2}\cdot(D_{1}D_{1})^{2}\bar{F}\right]+\right.$
$\displaystyle\left.+(D_{2}D_{2})\left[F\cdot(D_{2}D_{2}F)^{2}-\frac{1}{2}F^{2}\cdot(D_{2}D_{2})^{2}F\right]\right\\}$
(78)

###  III.3 Discussion

In this example, (however it is not a mere example, but it has been an
unsolved problem for a long time) we saw how useful it is to use the
nonsymmetric splitting techniques of the projective superspace to decompose
the prepotential product $e^{V}$ and subsequently express the field strength
in terms of the $e^{\check{U}}$ (or $e^{\hat{U}}$) projections and the
$\zeta$-independent term $D^{\hat{\alpha}}_{1}D_{\hat{\alpha}1}\mkern
1.5mu\overline{\mkern-1.5muF\mkern-1.5mu}\mkern 1.5mu$ (respectively
$D^{\hat{\alpha}}_{2}D_{\hat{\alpha}2}F$). The advantage of doing this was
that it is now easier to separate terms and the contour-integral eliminated
unwanted terms and left us with much simpler expressions. Furthermore,
analysis on the ordinary covariant derivatives in 5-dimensions and applying
partial integrations on them, we can finally pull out the variation and find
the long-awaited non-Abelian Chern-Simons action in terms of unconstrained
vector prepotentials $X=e^{V}-1$ (remind you, they are sitting in the
$\zeta$-independent $F$ and $\mkern
1.5mu\overline{\mkern-1.5muF\mkern-1.5mu}\mkern 1.5mu$ objects).

## Acknowledgements

The author would like to thank Rikard von Unge for fruitful discussions and
Silvia Penati for reading the earlier version of this article.

##  Appendix A Nonsymmetric splitting of the prepotential

Typically, one uses the arctic and antarctic prescription Davgadorj and von
Unge (2018) to describe the field strength, where the prepotential $e^{V}$ is
decomposed into the parts containing the negative and positive powers of
$\zeta$, respectively, $e^{\mkern
1.5mu\overline{\mkern-1.5muU\mkern-1.5mu}\mkern 1.5mu}$ and $e^{U}$. Those two
parts are projectively conjugate to each other and the half of the
$\zeta$-independent terms sit in the $e^{\mkern
1.5mu\overline{\mkern-1.5muU\mkern-1.5mu}\mkern 1.5mu}$ and the another half
in the $e^{U}$ part, therefore symmetric. Then we could consider what if we
let the zero-mode ($\zeta$ independent) terms to be contained in either the
negative or positive $\zeta$-power part.
In the symmetric splitting:

$e^{V}=e^{\mkern 1.5mu\overline{\mkern-1.5muU\mkern-1.5mu}\mkern 1.5mu}e^{U}$
(79)

if we expand $e^{U}$ and $e^{\mkern
1.5mu\overline{\mkern-1.5muU\mkern-1.5mu}\mkern 1.5mu}$ in powers of
$X=e^{V}-1$ as a sum:

$\displaystyle e^{U}=1+Y^{(1)}+Y^{(2)}+\dots$ (80) $\displaystyle e^{\mkern
1.5mu\overline{\mkern-1.5muU\mkern-1.5mu}\mkern 1.5mu}=1+\mkern
1.5mu\overline{\mkern-1.5muY\mkern-1.5mu}\mkern 1.5mu^{(1)}+\mkern
1.5mu\overline{\mkern-1.5muY\mkern-1.5mu}\mkern 1.5mu^{(2)}+\dots$ (81)

we get an infinite set of equations that can be solved recursively:

$\displaystyle\mkern 1.5mu\overline{\mkern-1.5muY\mkern-1.5mu}\mkern
1.5mu^{(1)}+Y^{(1)}$ $\displaystyle=X$ $\displaystyle\mkern
1.5mu\overline{\mkern-1.5muY\mkern-1.5mu}\mkern 1.5mu^{(2)}+Y^{(2)}$
$\displaystyle=-\mkern 1.5mu\overline{\mkern-1.5muY\mkern-1.5mu}\mkern
1.5mu^{(1)}Y^{(1)}$ $\displaystyle\vdots$ $\displaystyle\mkern
1.5mu\overline{\mkern-1.5muY\mkern-1.5mu}\mkern 1.5mu^{(n)}+Y^{(n)}$
$\displaystyle=-\mkern 1.5mu\overline{\mkern-1.5muY\mkern-1.5mu}\mkern
1.5mu^{(n-1)}Y^{(1)}-\dots-\mkern
1.5mu\overline{\mkern-1.5muY\mkern-1.5mu}\mkern 1.5mu^{(1)}Y^{(n-1)}$ (82)

In each stage of the solution, we need to project the expression on the right
hand side on positive or negative powers of $\zeta$ with a symmetric splitting
of the zero-mode terms so that the $\mkern
1.5mu\overline{\mkern-1.5muY\mkern-1.5mu}\mkern 1.5mu^{(n)}$ term is the
conjugate of the $Y^{(n)}$. As an example we can write down the first two
terms and see that this splitting is complicated.

$\displaystyle Y^{(1)}=X^{+}-\frac{1}{2}x_{0}\;\;,\;\;\mkern
1.5mu\overline{\mkern-1.5muY\mkern-1.5mu}\mkern
1.5mu^{(1)}=X^{-}+\frac{1}{2}x_{0}$ (83) $\displaystyle
Y^{(2)}=-[X^{-}X^{+}]^{+}+\frac{1}{2}[X^{-}X^{+}]^{0}-\frac{1}{2}x_{0}X^{+}+\frac{3}{8}x_{0}^{2}$
(84) $\displaystyle\mkern 1.5mu\overline{\mkern-1.5muY\mkern-1.5mu}\mkern
1.5mu^{(2)}=-[X^{-}X^{+}]^{-}-\frac{1}{2}[X^{-}X^{+}]^{0}+\frac{1}{2}X^{-}x_{0}-\frac{1}{8}x_{0}^{2}$
(85)

here, $X^{+}$ contains positive powers plus zero-modes of $\zeta$, $X^{-}$
contains only negative powers.
So it turns out that the projections by $\zeta$-integrals and the
$\epsilon$-prescription naturally favors nonsymmetric projections.

$\displaystyle\oint
d\zeta_{1}\frac{X_{1}}{\zeta_{10}}=\sum^{\infty}_{n=0}x_{n}\zeta_{0}^{n}$
$\displaystyle\oint
d\zeta_{1}\frac{X_{1}}{\zeta_{01}}=\sum^{-1}_{n=-\infty}x_{n}\zeta_{0}^{n}$
(86)

This inspires to split the prepotential as: $e^{V}=e^{\check{\mkern
1.5mu\overline{\mkern-1.5muU\mkern-1.5mu}\mkern 1.5mu}}e^{\hat{U}}$, where
$e^{\hat{U}}$ contains positive plus all zero-mode terms (at each powers of
X). Then the above recursion relations can be written in terms of:

$\displaystyle e^{\hat{U}}=1+\hat{Y}^{(1)}+\hat{Y}^{(2)}+\dots$ (87)
$\displaystyle e^{\check{\mkern
1.5mu\overline{\mkern-1.5muU\mkern-1.5mu}\mkern 1.5mu}}=1+\check{\mkern
1.5mu\overline{\mkern-1.5muY\mkern-1.5mu}\mkern 1.5mu}^{(1)}+\check{\mkern
1.5mu\overline{\mkern-1.5muY\mkern-1.5mu}\mkern 1.5mu}^{(2)}+\dots$ (88)

as sums of only negative terms and positive terms containing zero-mode terms.

$\displaystyle\check{\mkern 1.5mu\overline{\mkern-1.5muY\mkern-1.5mu}\mkern
1.5mu}^{(1)}+\hat{Y}^{(1)}$ $\displaystyle=X$ $\displaystyle\check{\mkern
1.5mu\overline{\mkern-1.5muY\mkern-1.5mu}\mkern 1.5mu}^{(2)}+\hat{Y}^{(2)}$
$\displaystyle=-\check{\mkern 1.5mu\overline{\mkern-1.5muY\mkern-1.5mu}\mkern
1.5mu}^{(1)}\hat{Y}^{(1)}$ $\displaystyle\vdots$ $\displaystyle\check{\mkern
1.5mu\overline{\mkern-1.5muY\mkern-1.5mu}\mkern 1.5mu}^{(n)}+\hat{Y}^{(n)}$
$\displaystyle=-\check{\mkern 1.5mu\overline{\mkern-1.5muY\mkern-1.5mu}\mkern
1.5mu}^{(n-1)}\hat{Y}^{(1)}-\dots-\check{\mkern
1.5mu\overline{\mkern-1.5muY\mkern-1.5mu}\mkern 1.5mu}^{(1)}\hat{Y}^{(n-1)}$
(89)

This allows us to write compactly using contour integral projections as:

$\displaystyle e^{\hat{U}}=$ $\displaystyle
1+\sum^{\infty}_{n=1}(-1)^{n+1}\oint d\zeta_{1}\dots
d\zeta_{n}\frac{X_{1}\dots
X_{n}}{\zeta_{21}\dots\zeta_{n,n-1}}\frac{1}{\zeta_{n0}}$ (90) $\displaystyle
e^{\check{\mkern 1.5mu\overline{\mkern-1.5muU\mkern-1.5mu}\mkern 1.5mu}}=$
$\displaystyle 1+\sum^{\infty}_{n=1}(-1)^{n+1}\oint d\zeta_{1}\dots
d\zeta_{n}\frac{1}{\zeta_{01}}\frac{X_{1}\dots
X_{n}}{\zeta_{21}\dots\zeta_{n,n-1}}$ (91)

In the language of simple projections (83) the above can be written:

$\displaystyle
e^{\hat{U}}=1+X^{+}-[X^{-}X]^{+}+\left[[X^{-}X]^{-}X\right]^{+}-\dots$
$\displaystyle e^{\check{\mkern
1.5mu\overline{\mkern-1.5muU\mkern-1.5mu}\mkern
1.5mu}}=1+X^{-}-[XX^{+}]^{-}+\left[X[XX^{+}]^{+}\right]^{-}-\dots$ (92)

One could also be interested in putting the zero-mode terms together with the
negative power terms as in:

$\displaystyle\oint
d\zeta_{1}\frac{X_{1}}{\zeta_{10}}\frac{\zeta_{0}}{\zeta_{1}}=\sum^{\infty}_{n=1}x_{n}\zeta_{0}^{n}$
$\displaystyle\oint
d\zeta_{1}\frac{X_{1}}{\zeta_{01}}\frac{\zeta_{0}}{\zeta_{1}}=\sum^{0}_{n=-\infty}x_{n}\zeta_{0}^{n}$
(93)

This kind of projection suggests to split the prepotential as
$e^{V}=e^{\hat{\mkern 1.5mu\overline{\mkern-1.5muU\mkern-1.5mu}\mkern
1.5mu}}e^{\check{U}}$, where $e^{\check{U}}$ does not contain zero-mode. Doing
the same recursion as above and reminding yourself the additional
$\frac{\zeta_{0}}{\zeta_{1}}$ projection, which will shift terms like this and
remove the zero-mode terms:

$\frac{1}{\zeta_{n,0}}\frac{\zeta_{0}}{\zeta_{1}}=\frac{1}{\zeta_{n}}\sum_{k=0}^{\infty}\left(\frac{\zeta_{0}}{\zeta_{n}}\right)^{k}\frac{\zeta_{0}}{\zeta_{1}}=\frac{1}{\zeta_{n}}\sum_{k=1}^{\infty}\left(\frac{\zeta_{0}}{\zeta_{n}}\right)^{k}\frac{\zeta_{n}}{\zeta_{1}}$
(94)

The contour integral projections are the following.

$\displaystyle e^{\check{U}}=$ $\displaystyle
1+\sum^{\infty}_{n=1}(-1)^{n+1}\oint d\zeta_{1}\dots
d\zeta_{n}\frac{\zeta_{0}}{\zeta_{1}}\frac{X_{1}\dots
X_{n}}{\zeta_{21}\dots\zeta_{n,n-1}}\frac{1}{\zeta_{n0}}$ (95) $\displaystyle
e^{\hat{\mkern 1.5mu\overline{\mkern-1.5muU\mkern-1.5mu}\mkern 1.5mu}}=$
$\displaystyle 1+\sum^{\infty}_{n=1}(-1)^{n+1}\oint d\zeta_{1}\dots
d\zeta_{n}\frac{1}{\zeta_{01}}\frac{X_{1}\dots
X_{n}}{\zeta_{21}\dots\zeta_{n,n-1}}\frac{\zeta_{0}}{\zeta_{n}}$ (96)

In the language of simple projections (83) the above are:

$\displaystyle
e^{\check{U}}-1=\left[\frac{X}{\zeta}\right]^{+}\zeta-\left[\left[\frac{X}{\zeta}\right]^{-}X\right]^{+}\zeta+\left[\left[\left[\frac{X}{\zeta}\right]^{-}X\right]^{-}X\right]^{+}\zeta-\dots$
$\displaystyle=\left(X^{+}-x_{0}-[X^{-}X^{+}]^{+}-x_{0}X^{+}+\left[[X^{-}X]^{-}X^{+}\right]^{+}+x_{0}[X^{-}X^{+}]^{+}\dots\right)\zeta$
$\displaystyle e^{\hat{\mkern 1.5mu\overline{\mkern-1.5muU\mkern-1.5mu}\mkern
1.5mu}}-1=\left[\frac{X}{\zeta}\right]^{-}\zeta-\left[X\left[\frac{X}{\zeta}\right]^{+}\right]^{-}\zeta+\left[X\left[X\left[\frac{X}{\zeta}\right]^{+}\right]^{+}\right]^{-}\zeta-\dots$
$\displaystyle=\left(X^{-}+x_{0}-[X^{-}X^{+}]^{-}+X^{-}x_{0}+\left[X^{-}[XX^{+}]^{+}\right]^{-}-[X^{-}X^{+}]^{-}x_{0}\dots\right)\zeta$
(97)

For completeness of the description and for the practical uses later we give
also the inverses.

$\displaystyle e^{-\hat{U}}=$ $\displaystyle
1+\sum^{\infty}_{n=1}(-1)^{n}\oint d\zeta_{1}\dots
d\zeta_{n}\frac{1}{\zeta_{10}}\frac{X_{1}\dots
X_{n}}{\zeta_{21}\dots\zeta_{n,n-1}}$ (98) $\displaystyle e^{-\check{\mkern
1.5mu\overline{\mkern-1.5muU\mkern-1.5mu}\mkern 1.5mu}}=$ $\displaystyle
1+\sum^{\infty}_{n=1}(-1)^{n}\oint d\zeta_{1}\dots d\zeta_{n}\frac{X_{1}\dots
X_{n}}{\zeta_{21}\dots\zeta_{n,n-1}}\frac{1}{\zeta_{0n}}$ (99) $\displaystyle
e^{-\check{U}}=$ $\displaystyle 1+\sum^{\infty}_{n=1}(-1)^{n}\oint
d\zeta_{1}\dots d\zeta_{n}\frac{1}{\zeta_{10}}\frac{X_{1}\dots
X_{n}}{\zeta_{21}\dots\zeta_{n,n-1}}\frac{\zeta_{0}}{\zeta_{n}}$ (100)
$\displaystyle e^{-\hat{\mkern 1.5mu\overline{\mkern-1.5muU\mkern-1.5mu}\mkern
1.5mu}}=$ $\displaystyle 1+\sum^{\infty}_{n=1}(-1)^{n}\oint d\zeta_{1}\dots
d\zeta_{n}\frac{\zeta_{0}}{\zeta_{1}}\frac{X_{1}\dots
X_{n}}{\zeta_{21}\dots\zeta_{n,n-1}}\frac{1}{\zeta_{0n}}$ (101)

To be clear let us write these different splittings together in one place.

$e^{V}=e^{\mkern 1.5mu\overline{\mkern-1.5muU\mkern-1.5mu}\mkern
1.5mu}e^{U}=e^{\check{\mkern 1.5mu\overline{\mkern-1.5muU\mkern-1.5mu}\mkern
1.5mu}}e^{\hat{U}}=e^{\hat{\mkern
1.5mu\overline{\mkern-1.5muU\mkern-1.5mu}\mkern 1.5mu}}e^{\check{U}}$ (102)

They differ by where the $\zeta$-independent terms sit, therefore one can
relate those splittings through a $\zeta$-independent object as in
$e^{\hat{U}}=e^{P}e^{U}$. From (102) we can write this $e^{P}$ and its
conjugate $e^{\mkern 1.5mu\overline{\mkern-1.5muP\mkern-1.5mu}\mkern 1.5mu}$
equivalently:

$\displaystyle e^{P}=e^{\hat{U}}e^{-U}=e^{-\check{\mkern
1.5mu\overline{\mkern-1.5muU\mkern-1.5mu}\mkern 1.5mu}}e^{\mkern
1.5mu\overline{\mkern-1.5muU\mkern-1.5mu}\mkern 1.5mu}$ (103) $\displaystyle
e^{\mkern 1.5mu\overline{\mkern-1.5muP\mkern-1.5mu}\mkern
1.5mu}=e^{U}e^{-\check{U}}$ (104) $\displaystyle e^{P}e^{\mkern
1.5mu\overline{\mkern-1.5muP\mkern-1.5mu}\mkern
1.5mu}=e^{\hat{U}}e^{-\check{U}}$ (105)

Employing again recursions we arrive at the concise expression for the
difference between the zero-mode containing and the no-zero-mode bridges.

$e^{P}e^{\mkern 1.5mu\overline{\mkern-1.5muP\mkern-1.5mu}\mkern
1.5mu}=1+\sum^{\infty}_{n=1}(-1)^{n+1}\oint d\zeta_{1}\dots
d\zeta_{n}\frac{X_{1}\dots
X_{n}}{\zeta_{21}\dots\zeta_{n,n-1}}\frac{1}{\zeta_{n}}$ (106)

its inverse is naturally as expected

$e^{-\mkern 1.5mu\overline{\mkern-1.5muP\mkern-1.5mu}\mkern
1.5mu}e^{-P}=1+\sum^{\infty}_{n=1}(-1)^{n+1}\oint d\zeta_{1}\dots
d\zeta_{n}\frac{1}{\zeta_{1}}\frac{X_{1}\dots
X_{n}}{\zeta_{21}\dots\zeta_{n,n-1}}$ (107)

Because this $e^{P}$ is $\zeta$-independent, consequently the gauge connection
$A_{\zeta}$ does not depend on which splitting we use.

$A_{\zeta}=e^{-U}(\partial_{\zeta}e^{U})=e^{-\hat{U}}(\partial_{\zeta}e^{\hat{U}})=e^{-\check{U}}(\partial_{\zeta}e^{\check{U}})$
(108)

After finding out the explicit form of the connection $A_{\zeta}$ we will
write down the field strength in terms of the nonsymmetric bridges
$e^{\hat{U}}$ and $e^{\check{U}}$.

##  Appendix B Field strength in terms of nonsymmetric splittings

Now that we are equipped with the explicit contour integral form of the gauge
connection and these new projecting devices (nonsymmetric splitting), we would
like to write down the polar field strength $i\mathcal{W}=\bar{\nabla}^{2}A$
in terms of this new language. In order to do that, first, we need to express
it in terms of $\mathcal{N}=1$ derivatives.

$\displaystyle\bar{\nabla}_{0}^{2}A_{0}=$
$\displaystyle\bar{\nabla}_{0}^{2}\sum_{n=1}^{\infty}\frac{(-1)^{n+1}X_{1}\dots
X_{n}}{\zeta_{10}\zeta_{21}\dots\zeta_{n,n-1}\zeta_{n0}}=$ $\displaystyle=$
$\displaystyle\sum_{n=1}^{\infty}(-1)^{n+1}\frac{1}{2}\sum_{k=1}^{n}\sum_{i=1}^{n}\frac{X_{1}\dots\bar{\nabla}_{0}^{\alpha}X_{k}\dots\bar{\nabla}_{0\alpha}X_{i}\dots
X_{n}}{\zeta_{10}\zeta_{21}\dots\zeta_{n,n-1}\zeta_{n0}}$ (109)

Projectivity of fields gives the relations:

$\displaystyle\bar{\nabla}_{0}X_{1}=(\zeta_{1}-\zeta_{0})\bar{D}X_{1}$
$\displaystyle\bar{\nabla}_{0}^{2}X_{1}=(\zeta_{1}-\zeta_{0})^{2}\bar{D}^{2}X_{1}$
(110)

Also, when one of $\nabla_{0}$’s hits the $X_{k}$, we decompose the
$\zeta_{k}-\zeta_{0}$ factor by shifting left or right depending on which side
the given $\nabla$ sits.

$\displaystyle\zeta_{k}-\zeta_{0}=(\zeta_{k}-\zeta_{k-1})+(\zeta_{k-1}-\zeta_{k-2})+\dots(\zeta_{2}-\zeta_{1})+(\zeta_{1}-\zeta_{0})$
$\displaystyle\zeta_{i}-\zeta_{0}=(\zeta_{i}-\zeta_{i+1})+(\zeta_{i+1}-\zeta_{i+2})+\dots(\zeta_{n-1}-\zeta_{n})+(\zeta_{n}-\zeta_{0})$
(111)

Each of these factors cancel one factor from the denominator, and this empty
spot in the string of factors in the denominator indicates where we separate
the part between the $\mkern 1.5mu\overline{\mkern-1.5muD\mkern-1.5mu}\mkern
1.5mu^{2}$ free left (right) side and the middle part with the $\mkern
1.5mu\overline{\mkern-1.5muD\mkern-1.5mu}\mkern 1.5mu^{2}$ derivatives.

$\displaystyle\bar{\nabla}_{0}^{2}A_{0}=\sum_{n=1}^{\infty}\sum_{a=0}^{b-1}\sum_{b=1}^{n}\oint$
$\displaystyle\frac{(-1)^{a}X_{1}\dots
X_{a}}{\zeta_{10}\dots\zeta_{a,a-1}}\cdot\frac{(-1)^{b-(a+1)}\bar{D}^{2}(X_{a+1}\dots
X_{b})}{\zeta_{a+2,a+1}\dots\zeta_{b,b-1}}\cdot$
$\displaystyle\cdot\frac{(-1)^{n+1-(b+1)}X_{b+1}\dots
X_{n}}{\zeta_{b+2,b+1}\dots\zeta_{n0}}$ (112)

On the left and right side of the $\bar{D}^{2}$ we recognize the
$e^{-\hat{U}}$ and $e^{\hat{U}}$ sums, therefore:

$\displaystyle\bar{\nabla}_{0}^{2}\mathcal{A}_{0}=e^{-\hat{U}_{0}}\bar{D}^{2}\sum_{b=1}^{\infty}\oint
d\zeta_{1}\dots d\zeta_{b}\frac{(-1)^{b-1}X_{1}\dots
X_{b}}{\zeta_{21}\dots\zeta_{b,b-1}}e^{\hat{U}_{0}}\equiv
e^{-\hat{U}_{0}}\bar{D}^{2}Fe^{\hat{U}_{0}}\;,$ (113)

where we defined a new $\zeta$ independent sum $F$:

$\displaystyle F=\sum^{\infty}_{n=1}(-1)^{n+1}\oint d\zeta_{1}\dots
d\zeta_{n}\frac{X_{1}\dots X_{n}}{\zeta_{21}\dots\zeta_{n,n-1}}$ (114)

We could have equivalently done the above in terms of $\mkern
1.5mu\overline{\mkern-1.5muQ\mkern-1.5mu}\mkern 1.5mu$ derivatives.

$\displaystyle\bar{\nabla}_{0}X_{1}=\frac{\zeta_{1}-\zeta_{0}}{\zeta_{1}}\bar{Q}X_{1}$
$\displaystyle\bar{\nabla}_{0}^{2}X_{1}=\frac{(\zeta_{1}-\zeta_{0})^{2}}{\zeta_{1}^{2}}\bar{Q}^{2}X_{1}$
(115)

By doing similar decompositions and separating terms, we will receive another
expression for the field strength.

$\displaystyle i\mathcal{W}=e^{-\check{U}}\bar{Q}^{2}\bar{F}e^{\check{U}}$
(116)

here the $\zeta$-independent term $\mkern
1.5mu\overline{\mkern-1.5muF\mkern-1.5mu}\mkern 1.5mu$ is found to be:

$\displaystyle\bar{F}=\sum^{\infty}_{n=1}(-1)^{n+1}\oint d\zeta_{1}\dots
d\zeta_{n}\frac{1}{\zeta_{1}}\frac{X_{1}\dots
X_{n}}{\zeta_{21}\dots\zeta_{n,n-1}}\frac{1}{\zeta_{n}}$ (117)

Summarizing them:

$\displaystyle
i\mathcal{W}=e^{-\hat{U}}\bar{D}^{2}Fe^{\hat{U}}=e^{-\check{U}}\bar{Q}^{2}\bar{F}e^{\check{U}}=e^{-U}e^{-P}\bar{D}^{2}Fe^{P}e^{U}=e^{-U}e^{\mkern
1.5mu\overline{\mkern-1.5muP\mkern-1.5mu}\mkern
1.5mu}\bar{Q}^{2}\bar{F}e^{-\mkern
1.5mu\overline{\mkern-1.5muP\mkern-1.5mu}\mkern 1.5mu}e^{U}$ $\displaystyle
i\bar{\mathcal{W}}=e^{-\hat{U}}Q^{2}Fe^{\hat{U}}=e^{-\check{U}}D^{2}\bar{F}e^{\check{U}}=e^{-U}e^{-P}Q^{2}Fe^{P}e^{U}=e^{-U}e^{\mkern
1.5mu\overline{\mkern-1.5muP\mkern-1.5mu}\mkern 1.5mu}D^{2}\bar{F}e^{-\mkern
1.5mu\overline{\mkern-1.5muP\mkern-1.5mu}\mkern 1.5mu}e^{U}$ (118)

Then from (B) it is clear that the field strength in the vector representation
can be written as:

$\displaystyle\mathbb{W}=$ $\displaystyle e^{-P}\bar{D}^{2}Fe^{P}=e^{\mkern
1.5mu\overline{\mkern-1.5muP\mkern-1.5mu}\mkern
1.5mu}\bar{Q}^{2}\bar{F}e^{-\mkern
1.5mu\overline{\mkern-1.5muP\mkern-1.5mu}\mkern 1.5mu}$
$\displaystyle\bar{\mathbb{W}}=$ $\displaystyle e^{\mkern
1.5mu\overline{\mkern-1.5muP\mkern-1.5mu}\mkern 1.5mu}D^{2}\bar{F}e^{-\mkern
1.5mu\overline{\mkern-1.5muP\mkern-1.5mu}\mkern 1.5mu}=e^{-P}Q^{2}Fe^{P}$
(119)

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