F}_{13\,C^{\prime},C}^{(0)}=iV^{-1/4}\sum_{p=0}^{\scriptsize\frac{r}{\mbox{gcd}(k,r)}-1}\left\\{-{\cal
C}_{4}^{[C^{\prime}+pk]_{r}}{\cal
G}^{(p)}_{1,C^{\prime},C}(x,\hat{\phi})+{\cal
C}_{2}^{[C^{\prime}+pk]_{r}}{\cal
G}^{(p)}_{3,C^{\prime},C}(x,\hat{\phi})\right\\}\,,$ $\displaystyle{\cal
F}_{12\,C^{\prime},C}^{(0)}$ $\displaystyle=$ $\displaystyle{\cal
F}_{34\,C^{\prime},C}^{(0)}=0\,,$ (189)

where the functions ${\cal G}^{(p)}_{1,C^{\prime},C}(x,\hat{\phi})$ and ${\cal
G}^{(p)}_{3,C^{\prime},C}(x,\hat{\phi})$ are defined as

$\displaystyle{\cal G}^{(p)}_{1,C^{\prime},C}(x,\hat{\phi})$ $\displaystyle=$
$\displaystyle\hat{D}_{1}\Phi^{(p)}_{C^{\prime},C}(x,\hat{\phi})$ (191)
$\displaystyle=$ $\displaystyle-\frac{2\pi r}{kL_{1}L_{2}}\ \sum_{\scriptsize
m=p+\frac{rm^{\prime}}{\mbox{gcd}(k,r)},\,m^{\prime}\in\mathbb{Z}}\sum_{n^{\prime}\in\mathbb{Z}}e^{\frac{i2\pi
x_{2}}{L_{2}}(m+\frac{2C^{\prime}-1-k}{2k})}e^{\frac{i2\pi
x_{4}}{L_{4}}(n^{\prime}-\frac{2C-1-\ell}{2\ell})}$
$\displaystyle\times\left(x_{1}-\frac{kL_{1}L_{2}\hat{\phi}_{2}^{[C^{\prime}]_{r}}}{2\pi
r}-\frac{L_{1}}{r}\left(km+\frac{2C^{\prime}-1-k}{2}\right)\right)$
$\displaystyle\times e^{-\frac{\pi
r}{kL_{1}L_{2}}\left[x_{1}-\frac{kL_{1}L_{2}}{2\pi
r}(\hat{\phi}_{2}^{[C^{\prime}]_{r}}-i\hat{\phi}_{1}^{[C^{\prime}]_{r}})-\frac{L_{1}}{r}\left(km+\frac{2C^{\prime}-1-k}{2}\right)\right]^{2}}$
$\displaystyle\times e^{-\frac{\pi}{\ell L_{3}L_{4}}\left[x_{3}-\frac{\ell
L_{3}L_{4}}{2\pi}(\hat{\phi}_{4}^{[C^{\prime}]_{r}}-i\hat{\phi}_{3}^{[C^{\prime}]_{r}})-L_{3}\left(\ell
n^{\prime}-\frac{2C-1-\ell}{2}\right)\right]^{2}}\,,$

and

$\displaystyle{\cal G}^{(p)}_{3,C^{\prime},C}(x,\hat{\phi})$ $\displaystyle=$
$\displaystyle\hat{D}_{3}\Phi^{(p)}_{C^{\prime},C}(x,\hat{\phi})$ (193)
$\displaystyle=$ $\displaystyle-\frac{2\pi}{\ell L_{3}L_{4}}\
\sum_{\scriptsize
m=p+\frac{rm^{\prime}}{\mbox{gcd}(k,r)},\,m^{\prime}\in\mathbb{Z}}\sum_{n^{\prime}\in\mathbb{Z}}e^{\frac{i2\pi
x_{2}}{L_{2}}(m+\frac{2C^{\prime}-1-k}{2k})}e^{\frac{i2\pi
x_{4}}{L_{4}}(n^{\prime}-\frac{2C-1-\ell}{2\ell})}$
$\displaystyle\times\left(x_{3}-\frac{\ell
L_{3}L_{4}\hat{\phi}_{4}^{[C^{\prime}]_{r}}}{2\pi}-L_{3}\left(\ell
n^{\prime}-\frac{2C-1-\ell}{2}\right)\right)$ $\displaystyle\times
e^{-\frac{\pi r}{kL_{1}L_{2}}\left[x_{1}-\frac{kL_{1}L_{2}}{2\pi
r}(\hat{\phi}_{2}^{[C^{\prime}]_{r}}-i\hat{\phi}_{1}^{[C^{\prime}]_{r}})-\frac{L_{1}}{r}\left(km+\frac{2C^{\prime}-1-k}{2}\right)\right]^{2}}$
$\displaystyle\times e^{-\frac{\pi}{\ell L_{3}L_{4}}\left[x_{3}-\frac{\ell
L_{3}L_{4}}{2\pi}(\hat{\phi}_{4}^{[C^{\prime}]_{r}}-i\hat{\phi}_{3}^{[C^{\prime}]_{r}})-L_{3}\left(\ell
n^{\prime}-\frac{2C-1-\ell}{2}\right)\right]^{2}}\,.$

Owing to the self-duality of the solution, we also have:

$\displaystyle{\cal F}_{23\,C^{\prime},C}^{(0)}={\cal
F}_{14\,C^{\prime},C}^{(0)}\,,\quad{\cal F}_{24\,C^{\prime},C}^{(0)}$
$\displaystyle=$ $\displaystyle-{\cal F}_{13\,C^{\prime},C}^{(0)}\,.$ (194)

In the following, we calculate the action density
$\mbox{tr}\left[F_{\mu\nu}F_{\mu\nu}\right]$ of the twisted solution. Using
(63), the square of the field strength is

$\displaystyle F_{\mu\nu}F_{\mu\nu}$ $\displaystyle=$
$\displaystyle\omega^{2}\left(\hat{F}_{\mu\nu}^{\omega}+F_{\mu\nu}^{s}\right)^{2}+4\pi\left(\hat{F}_{\mu\nu}^{\omega}+F_{\mu\nu}^{s}\right)\left[\begin{array}[]{cc}\ell
F_{\mu\nu}^{k}&{\cal F}_{\mu\nu}^{k\times\ell}\\\ {\cal
F}_{\mu\nu}^{\dagger\ell\times k}&-kF_{\mu\nu}^{\ell}\end{array}\right]$ (200)
$\displaystyle+\left[\begin{array}[]{cc}F_{\mu\nu}^{k}F_{\mu\nu}^{k}+{\cal
F}_{\mu\nu}^{k\times\ell}{\cal F}_{\mu\nu}^{\dagger\ell\times
k}&F^{k}_{\mu\nu}{\cal F}_{\mu\nu}^{k\times\ell}+{\cal
F}_{\mu\nu}^{k\times\ell}F_{\mu\nu}^{\ell}\\\ {\cal
F}_{\mu\nu}^{\dagger\ell\times k}F_{\mu\nu}^{k}+F_{\mu\nu}^{\ell}{\cal
F}_{\mu\nu}^{\dagger\ell\times k}&F_{\mu\nu}^{\ell}F_{\mu\nu}^{\ell}+{\cal
F}_{\mu\nu}^{\dagger\ell\times k}{\cal
F}_{\mu\nu}^{k\times\ell}\end{array}\right]\,.$

Then, the action density is given by the trace

$\displaystyle\mbox{tr}\left[F_{\mu\nu}F_{\mu\nu}\right]$ $\displaystyle=$
$\displaystyle\mbox{tr}\left[\omega^{2}\right]\left(\hat{F}_{\mu\nu}^{\omega}+F_{\mu\nu}^{s}\right)^{2}$
$\displaystyle+4\pi\ell\left(\hat{F}_{\mu\nu}^{\omega}+F_{\mu\nu}^{s}\right)\mbox{tr}_{k}\left[F_{\mu\nu}^{k}\right]-4\pi
k\left(\hat{F}_{\mu\nu}^{\omega}+F_{\mu\nu}^{s}\right)\mbox{tr}_{\ell}\left[F_{\mu\nu}^{\ell}\right]$
$\displaystyle+\mbox{tr}_{k}\left[F_{\mu\nu}^{k}F_{\mu\nu}^{k}+{\cal
F}_{\mu\nu}^{k\times\ell}{\cal F}_{\mu\nu}^{\dagger\ell\times
k}\right]+\mbox{tr}_{\ell}\left[F_{\mu\nu}^{\ell}F_{\mu\nu}^{\ell}+{\cal
F}_{\mu\nu}^{\dagger\ell\times k}{\cal F}_{\mu\nu}^{k\times\ell}\right]\,.$

To leading order in $\Delta$ we have:

$\displaystyle F_{\mu\nu}^{s}$ $\displaystyle=$
$\displaystyle\Delta\left(\partial_{\mu}S_{\nu}^{\omega(0)}-\partial_{\nu}S_{\mu}^{\omega(0)}\right)\,,$
$\displaystyle F_{\mu\nu}^{k}$ $\displaystyle=$
$\displaystyle\Delta\left(\partial_{\mu}S_{\nu}^{k(0)}-\partial_{\nu}S_{\mu}^{k(0)}+i{\cal
W}_{\mu}^{(0)k\times\ell}{\cal W}_{\nu}^{\dagger(0)\ell\times k}-i{\cal
W}_{\nu}^{(0)k\times\ell}{\cal W}_{\mu}^{\dagger(0)\ell\times k}\right)\,,$
$\displaystyle F_{\mu\nu}^{\ell}$ $\displaystyle=$
$\displaystyle\Delta\left(\partial_{\mu}S_{\nu}^{\ell(0)}-\partial_{\nu}S_{\mu}^{\ell(0)}+i{\cal
W}_{\mu}^{\dagger(0)\ell\times k}{\cal W}_{\nu}^{(0)k\times\ell}-i{\cal
W}_{\nu}^{\dagger(0)\ell\times k}{\cal W}_{\mu}^{(0)k\times\ell}\right)\,,$
$\displaystyle{\cal F}_{\mu\nu}^{k\times\ell}$ $\displaystyle=$
$\displaystyle\sqrt{\Delta}{\cal
F}_{\mu\nu}^{(0)k\times\ell}=\sqrt{\Delta}\left(\hat{D}_{\mu}{\cal
W}_{\nu}^{(0)k\times\ell}-\hat{D}_{\nu}{\cal
W}_{\mu}^{(0)k\times\ell}\right)\,.$ (202)

Substituting (202) into (LABEL:action_density_in_full), we find to ${\cal
O}(\Delta)$:

$\displaystyle\mbox{tr}\left[F_{\mu\nu}F_{\mu\nu}\right]$ $\displaystyle=$
$\displaystyle\mbox{tr}\left[\omega^{2}\right]\left(\hat{F}_{\mu\nu}^{\omega}\hat{F}_{\mu\nu}^{\omega}+2\Delta(\partial_{\mu}{\cal
S}_{\nu}^{\omega(0)}-\partial_{\nu}{\cal
S}_{\mu}^{\omega(0)})\hat{F}_{\mu\nu}^{\omega}\right)$ (203)
$\displaystyle+4\pi\ell\Delta\hat{F}_{\mu\nu}^{\omega}\mbox{tr}_{k}\left[\partial_{\mu}{\cal
S}_{\nu}^{k(0)}-\partial_{\nu}{\cal S}_{\mu}^{k(0)}\right]-4\pi
k\Delta\hat{F}_{\mu\nu}^{\omega}\mbox{tr}_{\ell}\left[\partial_{\mu}{\cal
S}_{\nu}^{\ell(0)}-\partial_{\nu}{\cal S}_{\mu}^{\ell(0)}\right]$
$\displaystyle+i4\pi\ell\Delta\hat{F}_{\mu\nu}^{\omega}\mbox{tr}_{k}\left[{\cal
W}_{\mu}^{(0)k\times\ell}{\cal W}_{\nu}^{\dagger(0)\ell\times k}-{\cal
W}_{\nu}^{(0)k\times\ell}{\cal W}_{\mu}^{\dagger(0)\ell\times k}\right]$
$\displaystyle-i4\pi
k\Delta\hat{F}_{\mu\nu}^{\omega}\mbox{tr}_{\ell}\left[{\cal
W}_{\mu}^{\dagger(0)\ell\times k}{\cal W}_{\nu}^{(0)k\times\ell}-{\cal
W}_{\nu}^{\dagger(0)\ell\times k}{\cal W}_{\mu}^{(0)k\times\ell}\right]$
$\displaystyle+\Delta\mbox{tr}_{k}\left[{\cal
F}_{\mu\nu}^{(0)k\times\ell}{\cal F}_{\mu\nu}^{\dagger(0)\ell\times
k}\right]+\Delta\mbox{tr}_{\ell}\left[{\cal F}_{\mu\nu}^{\dagger(0)\ell\times
k}{\cal F}_{\mu\nu}^{(0)k\times\ell}\right]\,.$

Then, using the trace properties $\mbox{tr}_{k}[{\cal
S}_{\mu}^{(0)k}]=\mbox{tr}_{\ell}[{\cal S}_{\mu}^{(0)\ell}]=0$, along with

$\displaystyle\mbox{tr}_{k}\left[{\cal F}_{\mu\nu}^{(0)k\times\ell}{\cal
F}_{\mu\nu}^{\dagger(0)\ell\times k}\right]$ $\displaystyle=$
$\displaystyle\mbox{tr}_{\ell}\left[{\cal F}_{\mu\nu}^{\dagger(0)\ell\times
k}{\cal F}_{\mu\nu}^{(0)k\times\ell}\right]\,,$
$\displaystyle\mbox{tr}_{k}\left[{\cal W}_{\mu}^{(0)k\times\ell}{\cal
W}_{\nu}^{\dagger(0)\ell\times k}-{\cal W}_{\nu}^{(0)k\times\ell}{\cal
W}_{\mu}^{\dagger(0)\ell\times k}\right]$ $\displaystyle=$
$\displaystyle-\mbox{tr}_{\ell}\left[{\cal W}_{\mu}^{\dagger(0)\ell\times
k}{\cal W}_{\nu}^{(0)k\times\ell}-{\cal W}_{\nu}^{\dagger(0)\ell\times k}{\cal
W}_{\mu}^{(0)k\times\ell}\right]\,,$

we find to ${\cal O}(\Delta)$

$\displaystyle\mbox{tr}\left[F_{\mu\nu}F_{\mu\nu}\right]$ $\displaystyle=$
$\displaystyle\mbox{tr}\left[\omega^{2}\right]\left(\hat{F}_{\mu\nu}^{\omega}\hat{F}_{\mu\nu}^{\omega}+2\Delta(\partial_{\mu}{\cal
S}_{\nu}^{\omega(0)}-\partial_{\nu}{\cal
S}_{\mu}^{\omega(0)})\hat{F}_{\mu\nu}^{\omega}\right)$ (205)
$\displaystyle+i4\pi N\Delta\hat{F}_{\mu\nu}^{\omega}\mbox{tr}_{k}\left[{\cal
W}_{\mu}^{(0)k\times\ell}{\cal W}_{\nu}^{\dagger(0)\ell\times k}-{\cal
W}_{\nu}^{(0)k\times\ell}{\cal W}_{\mu}^{\dagger(0)\ell\times k}\right]$
$\displaystyle+2\Delta\mbox{tr}_{k}\left[{\cal
F}_{\mu\nu}^{(0)k\times\ell}{\cal F}_{\mu\nu}^{\dagger(0)\ell\times
k}\right]\,.$

In the following, we perform the calculation of the action setting ${\cal
C}_{4}^{[C^{\prime}]_{r}}=0$. Thus, recalling (98), we are particularly
interested in the cases $r=1$ and $r=k,k>1$. However, the conclusion should
hold in the general case. Keeping only the non-zero entries and using $-i{\cal
F}_{14}^{(0)\bm{\beta}}={\cal F}_{13}^{(0)\bm{\beta}}$ along with the self-
duality property, we arrive at

$\displaystyle\mbox{tr}_{k}\left[{\cal F}_{\mu\nu}^{(0)k\times\ell}{\cal
F}_{\mu\nu}^{\dagger(0)\ell\times k}\right]$ $\displaystyle=$ $\displaystyle
2\mbox{tr}_{k}\left[{\cal F}_{13}^{(0)k\times\ell}{\cal
F}_{13}^{\dagger(0)\ell\times k}\right]+2\mbox{tr}_{k}\left[{\cal
F}_{14}^{(0)k\times\ell}{\cal F}_{14}^{\dagger(0)\ell\times k}\right]$ (206)
$\displaystyle+$ $\displaystyle 2\mbox{tr}_{k}\left[{\cal
F}_{23}^{(0)k\times\ell}{\cal F}_{23}^{\dagger(0)\ell\times
k}\right]+2\mbox{tr}_{k}\left[{\cal F}_{24}^{(0)k\times\ell}{\cal
F}_{24}^{\dagger(0)\ell\times k}\right]$ $\displaystyle=$ $\displaystyle
8\mbox{tr}_{k}\left[{\cal F}_{13}^{(0)k\times\ell}{\cal
F}_{13}^{\dagger(0)\ell\times k}\right]\,.$

Likewise:

$\displaystyle\hat{F}_{\mu\nu}^{\omega}\mbox{tr}_{k}\left[{\cal
W}_{\mu}^{(0)k\times\ell}{\cal W}_{\nu}^{\dagger(0)\ell\times k}-{\cal
W}_{\nu}^{(0)k\times\ell}{\cal W}_{\mu}^{\dagger(0)\ell\times
k}\right]=-4i\hat{F}_{12}^{\omega}\mbox{tr}_{k}\left[{\cal
W}_{2}^{(0)k\times\ell}{\cal W}_{2}^{\dagger(0)\ell\times k}\right]\,.$ (207)

Thus, the action density is given by the expression

$\displaystyle\mbox{tr}\left[F_{\mu\nu}F_{\mu\nu}\right]$ $\displaystyle=$
$\displaystyle\mbox{tr}\left[\omega^{2}\right]\left(\hat{F}_{\mu\nu}^{\omega}\hat{F}_{\mu\nu}^{\omega}+2\Delta(\partial_{\mu}{\cal
S}_{\nu}^{\omega}-\partial_{\nu}{\cal
S}_{\mu}^{\omega})\hat{F}_{\mu\nu}^{\omega}\right)$ (208) $\displaystyle+16\pi
N\Delta\hat{F}_{12}^{\omega}\mbox{tr}_{k}\left[{\cal
W}_{2}^{(0)k\times\ell}{\cal W}_{2}^{\dagger(0)\ell\times k}\right]$
$\displaystyle+16\Delta\mbox{tr}_{k}\left[{\cal F}_{13}^{(0)k\times\ell}{\cal
F}_{13}^{\dagger(0)\ell\times k}\right]\,.$

The action is

$\displaystyle
S=\frac{2}{g^{2}}\int_{\mathbb{T}^{4}}\mbox{tr}\left[F_{\mu\nu}F_{\mu\nu}\right]\,,$
(209)

and upon integrating, the term $\partial_{\mu}{\cal S}_{\nu}^{(0)\omega}$
drops out because ${\cal S}_{\nu}^{(0)\omega}$ satisfies periodic boundary
conditions. Thus, we finally have to ${\cal O}(\Delta)$:

$\displaystyle S$ $\displaystyle=$ $\displaystyle
S_{0}+\frac{\Delta}{2g^{2}}\int_{\mathbb{T}^{4}}16\pi
N\hat{F}_{12}^{\omega}\mbox{tr}_{k}\left[{\cal W}_{2}^{(0)k\times\ell}{\cal
W}_{2}^{\dagger(0)\ell\times
k}\right]+\frac{\Delta}{2g^{2}}\int_{\mathbb{T}^{4}}16\mbox{tr}_{k}\left[{\cal
F}_{13}^{(0)k\times\ell}{\cal F}_{13}^{\dagger(0)\ell\times k}\right]\,,$

where

$\displaystyle S_{0}$ $\displaystyle=$
$\displaystyle\frac{1}{2g^{2}}\int_{\mathbb{T}^{4}}\mbox{tr}\left[\omega^{2}\right]\left(\hat{F}_{\mu\nu}^{\omega}\hat{F}_{\mu\nu}^{\omega}\right)=\frac{1}{2g^{2}}\int_{\mathbb{T}^{4}}\mbox{tr}\left[\omega^{2}\right]\left\\{2\left(\hat{F}_{12}^{\omega}\hat{F}_{12}^{\omega}+\hat{F}_{34}^{\omega}\hat{F}_{34}^{\omega}\right)\right\\}$
(210) $\displaystyle=$
$\displaystyle(4\pi^{2}Nk\ell)\frac{1}{g^{2}N^{2}}\left(\frac{r^{2}}{k^{2}}\frac{L_{3}L_{4}}{L_{1}L_{2}}+\frac{1}{\ell^{2}}\frac{L_{1}L_{2}}{L_{3}L_{4}}\right)\,.$

Using the definition of $\Delta$ (67) we readily find

$\displaystyle S_{0}=\frac{8\pi^{2}r}{Ng^{2}}+{\cal O}(\Delta^{2})\,.$ (211)

Then, using $\hat{F}_{12}^{\omega}=-\frac{r}{kNL_{1}L_{2}}$, we have

$\displaystyle S=S_{0}+\frac{\Delta}{g^{2}}\left(-\frac{8\pi
r}{kL_{1}L_{2}}\int_{\mathbb{T}^{4}}\mbox{tr}_{k}\left[{\cal
W}_{2}^{(0)k\times\ell}{\cal W}_{2}^{\dagger(0)\ell\times
k}\right]+8\int_{\mathbb{T}^{4}}\mbox{tr}_{k}\left[{\cal
F}_{13}^{(0)k\times\ell}{\cal F}_{13}^{\dagger(0)\ell\times
k}\right]\right)\,.$

Finally, the remaining integrals are given by (we set all holonomies to $0$,
as the final answer will not depend on them):

$\displaystyle\int_{\mathbb{T}^{4}}\mbox{tr}_{k}\left[{\cal
W}_{2}^{(0)k\times\ell}{\cal W}_{2}^{\dagger(0)\ell\times k}\right]$
$\displaystyle=$
$\displaystyle\sqrt{L_{1}L_{2}L_{3}L_{4}}\sum_{C=1}^{\ell}\sum_{C^{\prime}=1}^{k}|{\cal
C}_{2}^{[C^{\prime}]_{r}}|^{2}$ (213) $\displaystyle\sum_{\scriptsize
m=p+\frac{rm^{\prime}}{\mbox{gcd}(k,r)},\,m^{\prime}\in\mathbb{Z}}\int_{0}^{1}d\tilde{x}_{1}e^{-\frac{2\pi
rL_{1}}{kL_{2}}\left(\tilde{x}_{1}-\frac{2mk+2(j+nr)-1-k}{2r}\right)^{2}}$
$\displaystyle\times\sum_{n^{\prime}\in\mathbb{Z}}\int_{0}^{1}d\tilde{x}_{3}e^{-\frac{2\pi
L_{3}}{\ell
L_{4}}\left(\tilde{x}_{3}-\frac{(2n^{\prime}+1)\ell-(2C-1)}{2}\right)^{2}}\,,$

and

$\displaystyle\int_{\mathbb{T}^{4}}\mbox{tr}_{k}\left[{\cal
F}_{13}^{(0)k\times\ell}{\cal F}_{13}^{\dagger(0)\ell\times k}\right]$
$\displaystyle=$
$\displaystyle\frac{4\pi^{2}}{\ell^{2}}\sqrt{\frac{L_{1}L_{2}L_{3}}{L_{4}^{3}}}\sum_{C=1}^{\ell}\sum_{C^{\prime}=1}^{k}|{\cal
C}_{2}^{[C^{\prime}]_{r}}|^{2}$ $\displaystyle\sum_{\scriptsize
m=p+\frac{rm^{\prime}}{\mbox{gcd}(k,r)},\,m^{\prime}\in\mathbb{Z}}\int_{0}^{1}d\tilde{x}_{1}e^{-\frac{2\pi
rL_{1}}{kL_{2}}\left(\tilde{x}_{1}-\frac{2mk+2(j+nr)-1-k}{2r}\right)^{2}}$
$\displaystyle\times\sum_{n^{\prime}\in\mathbb{Z}}\int_{0}^{1}d\tilde{x}_{3}\left(\tilde{x}_{3}-\frac{\left((2n^{\prime}+1)\ell-(2C-1)\right)}{2}\right)^{2}e^{-\frac{2\pi
L_{3}}{\ell
L_{4}}\left(\tilde{x}_{3}-\frac{(2n^{\prime}+1)\ell-(2C-1)}{2}\right)^{2}}\,.$

Now, collecting terms of ${\cal O}(\Delta)$ and using $r\ell
L_{3}L_{4}=kL_{1}L_{2}$, thus ignoring corrections ${\cal O}(\Delta^{2})$, we
find:

$\displaystyle S$ $\displaystyle=$ $\displaystyle
S_{0}+8\pi\sqrt{\frac{r}{\ell
k}}\frac{\Delta}{g^{2}}\sum_{C=1}^{\ell}\sum_{C^{\prime}=1}^{k}|{\cal
C}_{2}^{[C^{\prime}]_{r}}|^{2}$ $\displaystyle\sum_{\scriptsize
m=p+\frac{rm^{\prime}}{\mbox{gcd}(k,r)},\,m^{\prime}\in\mathbb{Z}}\int_{0}^{1}d\tilde{x}_{1}e^{-\frac{2\pi
rL_{1}}{kL_{2}}\left(\tilde{x}_{1}-\frac{2mk+2(j+nr)-1-k}{2r}\right)^{2}}$
$\displaystyle\times\sum_{n^{\prime}}\int_{0}^{1}d\tilde{x}_{3}\left\\{-1+\frac{4\pi}{\ell}\frac{L_{3}}{L_{4}}\left(\tilde{x}_{3}-\frac{\left((2n^{\prime}+1)\ell-(2C-1)\right)}{2}\right)^{2}\right\\}e^{-\frac{2\pi
L_{3}}{\ell
L_{4}}\left(\tilde{x}_{3}-\frac{(2n^{\prime}+1)\ell-(2C-1)}{2}\right)^{2}}\,.$

One can check (using Mathematica) that262626One can show that (216) is true by
converting the combined infinite sum and the integral over the unit interval
to an infinite integral.:

$\displaystyle\sum_{C=1}^{\ell}\sum_{n}\int_{0}^{1}d\tilde{x}_{3}\left\\{-1+\frac{4\pi}{\ell}\frac{L_{3}}{L_{4}}\left(\tilde{x}_{3}-\frac{\left((2n+1)\ell-(2C-1)\right)}{2}\right)^{2}\right\\}e^{-\frac{2\pi
L_{3}}{\ell
L_{4}}\left(\tilde{x}_{3}-\frac{(2n+1)\ell-(2C-1)}{2}\right)^{2}}=0\,,$ (216)

and thus, we conclude that, as expected

$\displaystyle S$ $\displaystyle=$ $\displaystyle S_{0}+{\cal
O}(\Delta^{2})=\frac{r}{N}\frac{8\pi^{2}}{g^{2}}+{\cal O}(\Delta^{2})\,,$
(217)

i.e. the action of the multifractional instanton is, to the order in $\Delta$
we are working on, equal to ${r\over N}$ times the BPST instanton action.

##  Appendix D Blow up of the gauge invariant local densities along the
noncompact moduli of the $k\neq r$ solution

To determine the gauge invariant density (110), we need to solve for ${\cal
S}^{(0)\omega}_{\nu}$. To this end, we use (LABEL:Sequations) (or the
equivalent forms (84, 88)). Acting on these equations with
$\partial=\sigma^{\nu}\partial_{\nu}$ and using the identity
$\sigma^{\nu}\bar{\sigma}^{\mu}+\sigma^{\mu}\bar{\sigma}^{\nu}=2\delta_{\mu\nu}$,
we find the expression

$\displaystyle\square{\cal S}^{(0)\omega}=-\frac{i}{\pi\ell
k}\sigma^{\nu}\partial_{\nu}{\cal Y}\,,$ (218)

where (once more, for brevity, we omit the $k\times\ell$ and $\ell\times k$
superscripts)

$\displaystyle{\cal Y}=\left[\begin{array}[]{cc}\mbox{tr}_{k}\left[{\cal
W}_{2}^{(0)}{\cal W}_{2}^{\dagger(0)}-{\cal W}_{4}^{(0)}{\cal
W}_{4}^{\dagger(0)}\right]&-2\mbox{tr}_{k}\left[{\cal W}_{2}^{(0)}{\cal
W}_{4}^{\dagger(0)}\right]\\\ -2\mbox{tr}_{k}\left[{\cal W}_{4}^{(0)}{\cal
W}_{2}^{\dagger(0)}\right]&-\mbox{tr}_{k}\left[{\cal W}_{2}^{(0)}{\cal
W}_{2}^{\dagger(0)}-{\cal W}_{4}^{(0)}{\cal
W}_{4}^{\dagger(0)}\right]\end{array}\right]\,.$ (221)

Equating the components of (218), we arrive at the following set of equations:

$\displaystyle i\pi\ell k\square\left({\cal S}^{(0)\omega}_{4}+i{\cal
S}^{(0)\omega}_{3}\right)=$
$\displaystyle\left(\partial_{4}+i\partial_{3}\right)\mbox{tr}_{k}\left[{\cal
W}_{2}^{(0)}{\cal W}_{2}^{\dagger(0)}-{\cal W}_{4}^{(0)}{\cal
W}_{4}^{\dagger(0)}\right]-2\left(i\partial_{1}+\partial_{2}\right)\mbox{tr}_{k}\left[{\cal
W}_{4}^{(0)}{\cal W}_{2}^{\dagger(0)}\right]\,,$ $\displaystyle i\pi\ell
k\square\left({\cal S}^{(0)\omega}_{4}-i{\cal S}^{(0)\omega}_{3}\right)=$
$\displaystyle-\left(\partial_{4}-i\partial_{3}\right)\mbox{tr}_{k}\left[{\cal
W}_{2}^{(0)}{\cal W}_{2}^{\dagger(0)}-{\cal W}_{4}^{(0)}{\cal
W}_{4}^{\dagger(0)}\right]-2\left(i\partial_{1}-\partial_{2}\right)\mbox{tr}_{k}\left[{\cal
W}_{2}^{(0)}{\cal W}_{4}^{\dagger(0)}\right]\,,$ $\displaystyle i\pi\ell
k\square\left(i{\cal S}^{(0)\omega}_{1}+{\cal S}^{(0)\omega}_{2}\right)=$
$\displaystyle-2\left(\partial_{4}+i\partial_{3}\right)\mbox{tr}_{k}\left[{\cal
W}_{2}^{(0)}{\cal
W}_{4}^{\dagger(0)}\right]-\left(i\partial_{1}+\partial_{2}\right)\mbox{tr}_{k}\left[{\cal
W}_{2}^{(0)}{\cal W}_{2}^{\dagger(0)}-{\cal W}_{4}^{(0)}{\cal
W}_{4}^{\dagger(0)}\right]\,,$ $\displaystyle i\pi\ell k\square\left(i{\cal
S}^{(0)\omega}_{1}-{\cal S}^{(0)\omega}_{2}\right)=$
$\displaystyle-2\left(\partial_{4}-i\partial_{3}\right)\mbox{tr}_{k}\left[{\cal
W}_{4}^{(0)}{\cal
W}_{2}^{\dagger(0)}\right]+\left(i\partial_{1}-\partial_{2}\right)\mbox{tr}_{k}\left[{\cal
W}_{2}^{(0)}{\cal W}_{2}^{\dagger(0)}-{\cal W}_{4}^{(0)}{\cal
W}_{4}^{\dagger(0)}\right]\,.$ (222)

Thus, we find

$\displaystyle\pi\ell k\Box{\cal S}_{4}^{(0)\omega}=$
$\displaystyle\partial_{3}\mbox{tr}_{k}\left[{\cal W}_{2}^{(0)}{\cal
W}_{2}^{\dagger(0)}-{\cal W}_{4}^{(0)}{\cal
W}_{4}^{\dagger(0)}\right]-(\partial_{1}-i\partial_{2})\mbox{tr}_{k}\left[{\cal
W}_{4}^{(0)}{\cal
W}_{2}^{\dagger(0)}\right]-(\partial_{1}+i\partial_{2})\mbox{tr}_{k}\left[{\cal
W}_{2}^{(0)}{\cal W}_{4}^{\dagger(0)}\right]\,,$ $\displaystyle-\pi\ell
k\Box{\cal S}_{3}^{(0)\omega}=$
$\displaystyle\partial_{4}\mbox{tr}_{k}\left[{\cal W}_{2}^{(0)}{\cal
W}_{2}^{\dagger(0)}-{\cal W}_{4}^{(0)}{\cal
W}_{4}^{\dagger(0)}\right]-(i\partial_{1}+\partial_{2})\mbox{tr}_{k}\left[{\cal
W}_{4}^{(0)}{\cal
W}_{2}^{\dagger(0)}\right]+(i\partial_{1}-\partial_{2})\mbox{tr}_{k}\left[{\cal
W}_{2}^{(0)}{\cal W}_{4}^{\dagger(0)}\right]\,.$

and

$\displaystyle\left(\partial_{3}{\cal S}_{4}^{(0)\omega}-\partial_{4}{\cal
S}_{3}^{(0)\omega}\right)$ $\displaystyle=$ $\displaystyle\left(\pi\ell
k\Box\right)^{-1}\left\\{\left(\partial_{3}^{2}+\partial_{4}^{2}\right)\mbox{tr}_{k}\left[{\cal
W}_{2}^{(0)}{\cal W}_{2}^{\dagger(0)}-{\cal W}_{4}^{(0)}{\cal
W}_{4}^{\dagger(0)}\right]\right.$ $\displaystyle+$
$\displaystyle\left.\left(-\partial_{1}\partial_{3}-\partial_{2}\partial_{4}+i\partial_{2}\partial_{3}-i\partial_{1}\partial_{4}\right)\mbox{tr}_{k}\left[{\cal
W}_{4}^{(0)}{\cal W}_{2}^{\dagger(0)}\right]\right.$ $\displaystyle+$
$\displaystyle\left.\left(-\partial_{1}\partial_{3}-\partial_{2}\partial_{4}-i\partial_{2}\partial_{3}+i\partial_{1}\partial_{4}\right)\mbox{tr}_{k}\left[{\cal
W}_{2}^{(0)}{\cal W}_{4}^{\dagger(0)}\right]\right\\}\,.$

We are interested in the case $r>1$ and $\mbox{gcd}(k,r)=1$. Let us consider
the example $r=2,k=3$. Then, using the parameterization of (103), taking the
upper sign for definiteness,

$\displaystyle{\cal C}_{2}^{0}=u\,,\quad{\cal C}_{2}^{1}=u\,,\quad{\cal
C}_{4}^{0}=-iu\,,\quad{\cal C}_{4}^{1}=iu\,.$ (225)

we find272727The sums over $C^{\prime}$ and $C$ should be really thought of as
being over $0,...,k-1$ and $0,...,\ell-1$, respectively, to be consistent with
the main body of the paper. We apologize to the reader for this slight
mismatch.

$\displaystyle\mbox{tr}_{k}\left[{\cal W}_{2}^{(0)}{\cal
W}_{2}^{\dagger(0)}\right]$ $\displaystyle=$ $\displaystyle
u^{2}\sum_{C=1}^{\ell}\sum_{C^{\prime}=1}^{k}\left[\Phi_{C^{\prime},C}^{0}+\Phi_{C^{\prime},C}^{1}\right]\left[\Phi_{C^{\prime},C}^{*0}+\Phi_{C^{\prime},C}^{*1}\right]$
$\displaystyle=$ $\displaystyle
u^{2}\sum_{C=1}^{\ell}\sum_{C^{\prime}=1}^{k}|\Phi_{C^{\prime},C}^{0}|^{2}+|\Phi_{C^{\prime},C}^{1}|+\Phi_{C^{\prime},C}^{0}\Phi_{C^{\prime},C}^{*1}+\Phi_{C^{\prime},C}^{*0}\Phi_{C^{\prime},C}^{1}\,,$
$\displaystyle\mbox{tr}_{k}\left[{\cal W}_{4}^{(0)}{\cal
W}_{4}^{\dagger(0)}\right]$ $\displaystyle=$ $\displaystyle
u^{2}\sum_{C=1}^{\ell}\sum_{C^{\prime}=1}^{k}\left[\Phi_{C^{\prime},C}^{0}-\Phi_{C^{\prime},C}^{1}\right]\left[\Phi_{C^{\prime},C}^{*0}-\Phi_{C^{\prime},C}^{*1}\right]$
$\displaystyle=$ $\displaystyle
u^{2}\sum_{C=1}^{\ell}\sum_{C^{\prime}=1}^{k}|\Phi_{C^{\prime},C}^{0}|^{2}+|\Phi_{C^{\prime},C}^{1}|-\Phi_{C^{\prime},C}^{0}\Phi_{C^{\prime},C}^{*1}-\Phi_{C^{\prime},C}^{*0}\Phi_{C^{\prime},C}^{1}\,,$

and

$\displaystyle\mbox{tr}_{k}\left[{\cal W}_{4}^{(0)}{\cal
W}_{2}^{\dagger(0)}\right]$ $\displaystyle=$ $\displaystyle
iu^{2}\sum_{C=1}^{\ell}\left[\left(-\Phi_{1,C}^{0}+\Phi_{1,C}^{1}\right)\left(\Phi_{1,C}^{*0}+\Phi_{1,C}^{*1}\right)\right.$
(228)
$\displaystyle+\left.\left(\Phi_{2,C}^{0}-\Phi_{2,C}^{1}\right)\left(\Phi_{2,C}^{*0}+\Phi_{2,C}^{*1}\right)\right.$
$\displaystyle+\left.\left(-\Phi_{3,C}^{0}+\Phi_{3,C}^{1}\right)\left(\Phi_{3,C}^{*0}+\Phi_{C^{\prime},C}^{*1}\right)\right]\,,$
$\displaystyle\mbox{tr}_{k}\left[{\cal W}_{2}^{(0)}{\cal
W}_{4}^{\dagger(0)}\right]$ $\displaystyle=$ $\displaystyle-
iu^{2}\sum_{C=1}^{\ell}\left[\left(\Phi_{1,C}^{0}+\Phi_{1,C}^{1}\right)\left(-\Phi_{1,C}^{*0}+\Phi_{1,C}^{*1}\right)\right.$
(229)
$\displaystyle+\left.\left(\Phi_{2,C}^{0}+\Phi_{2,C}^{1}\right)\left(\Phi_{2,C}^{*0}-\Phi_{2,C}^{*1}\right)\right.$
$\displaystyle+\left.\left(\Phi_{3,C}^{0}+\Phi_{3,C}^{1}\right)\left(-\Phi_{3,C}^{*0}+\Phi_{3,C}^{*1}\right)\right]\,.$

It is not hard, using (154), to check that the combinations

$\displaystyle|\Phi_{C^{\prime},C}^{0}(x)|^{2},|\Phi_{C^{\prime},C}^{1}(x)|^{2},\left(-\Phi_{1,C}^{0}+\Phi_{1,C}^{1}\right)(x)\left(\Phi_{1,C}^{*0}+\Phi_{1,C}^{*1}\right)(x),$
$\displaystyle\left(\Phi_{2,C}^{0}-\Phi_{2,C}^{1}\right)(x)\left(\Phi_{2,C}^{*0}+\Phi_{2,C}^{*1}\right)(x),\left(-\Phi_{3,C}^{0}+\Phi_{3,C}^{1}\right)(x)\left(\Phi_{3,C}^{*0}+\Phi_{C^{\prime},C}^{*1}\right)(x)$

satisfy periodic boundary conditions.282828However, the component that carries
the subscript $C^{\prime},C$ is sent to one with subscript $C^{\prime}-r,C+1$.
Nevertheless, the combinations that give the gauge invariant density are
periodic. Also, from the linearity of the Fourier analysis of the Fourier-
transformed components below, the superposition of the various terms makes
sense. The difficulty in the analysis below is that numerical convergence is
hard to achieve. Then, we use the Fourier transform of these combinations,
namely,

$\displaystyle|\Phi_{C^{\prime},C}^{0}(x)|^{2}$ $\displaystyle=$
$\displaystyle\sum_{p_{\mu}\in\mathbb{Z}}e^{i\frac{2\pi
p_{\mu}x)\mu}{L_{\mu}}}{\cal X}_{0;C^{\prime},C}(p)\,,$
$\displaystyle|\Phi_{C^{\prime},C}^{1}(x)|^{2}$ $\displaystyle=$
$\displaystyle\sum_{p_{\mu}}e^{i\frac{2\pi p_{\mu}x_{\mu}}{L_{\mu}}}{\cal
X}_{1;C^{\prime},C}(p)\,,$
$\displaystyle\Phi_{C^{\prime},C}^{0}(x)\Phi_{C^{\prime},C}^{*1}(x)$
$\displaystyle=$ $\displaystyle\sum_{p_{\mu}}e^{i\frac{2\pi
p_{\mu}x_{\mu}}{L_{\mu}}}{\cal X}_{2;C^{\prime},C}(p)\,,$
$\displaystyle\left(-\Phi_{1,C}^{0}+\Phi_{1,C}^{1}\right)(x)\left(\Phi_{1,C}^{*0}+\Phi_{1,C}^{*1}\right)(x)$
$\displaystyle=$ $\displaystyle\sum_{p_{\mu}}e^{i\frac{2\pi
p_{\mu}x_{\mu}}{L_{\mu}}}{\cal X}_{3;C}(p)\,,$
$\displaystyle\left(\Phi_{2,C}^{0}-\Phi_{2,C}^{1}\right)(x)\left(\Phi_{2,C}^{*0}+\Phi_{2,C}^{*1}\right)(x)$
$\displaystyle=$ $\displaystyle\sum_{p_{\mu}}e^{i\frac{2\pi
p_{\mu}x_{\mu}}{L_{\mu}}}{\cal X}_{4;C}(p)\,,$
$\displaystyle\left(-\Phi_{3,C}^{0}+\Phi_{3,C}^{1}\right)(x)\left(\Phi_{3,C}^{*0}+\Phi_{3,C}^{*1}\right)(x)$
$\displaystyle=$ $\displaystyle\sum_{p_{\mu}}e^{i\frac{2\pi
p_{\mu}x_{\mu}}{L_{\mu}}}{\cal X}_{5;C}(p)\,.$ (231)

to find

$\displaystyle\mbox{tr}_{k}\left[{\cal W}_{4}^{(0)}{\cal
W}_{4}^{\dagger(0)}\right]$ $\displaystyle=$ $\displaystyle
u^{2}\sum_{p_{\mu}}e^{i\frac{2\pi p_{\mu}x_{\mu}}{L_{\mu}}}{\cal H}(p)$
$\displaystyle\equiv u^{2}\sum_{p_{\mu},C,C^{\prime}}e^{i\frac{2\pi
p_{\mu}x_{\mu}}{L_{\mu}}}\left({\cal X}_{0;C^{\prime},C}(p)+{\cal
X}_{1;C^{\prime},C}(p)-{\cal X}_{2;C^{\prime},C}(p)-{\cal
X}_{2;C^{\prime},C}^{*}(p)\right)\,.$

The function ${\cal H}(p)$, the Fourier transform of $\mbox{tr}_{k}\left[{\cal
W}_{4}^{(0)}{\cal W}_{4}^{\dagger(0)}\right]$ modulo $u^{2}$, will play an
important role below. In addition, we find

$\displaystyle\left(\pi\ell
k\Box\right)^{-1}\left\\{\left(\partial_{3}^{2}+\partial_{4}^{2}\right)\mbox{tr}_{k}\left[{\cal
W}_{2}^{(0)}{\cal W}_{2}^{\dagger(0)}-{\cal W}_{4}^{(0)}{\cal
W}_{4}^{\dagger(0)}\right]\right\\}$ $\displaystyle=\frac{2u^{2}}{\pi\ell
k}\sum_{p_{\mu},C,C^{\prime}}\frac{\left[\frac{p_{3}^{2}}{L_{3}^{2}}+\frac{p_{4}^{2}}{L_{4}^{2}}\right]e^{i\frac{2\pi
p_{\mu}x_{\mu}}{L_{\mu}}}}{\frac{p_{1}^{2}}{L_{1}^{2}}+\frac{p_{2}^{2}}{L_{2}^{2}}+\frac{p_{3}^{2}}{L_{3}^{2}}+\frac{p_{4}^{2}}{L_{4}^{2}}}\left({\cal
X}_{2;C^{\prime},C}(p)+{\cal X}_{2;C^{\prime},C}^{*}(p)\right)\,,$

and

$\displaystyle\left(\pi\ell
k\Box\right)^{-1}\left\\{\left(-\partial_{1}\partial_{3}-\partial_{2}\partial_{4}+i\partial_{2}\partial_{3}-i\partial_{1}\partial_{4}\right)\mbox{tr}_{k}\left[{\cal
W}_{4}^{(0)}{\cal W}_{2}^{\dagger(0)}\right]\right.$ (234)
$\displaystyle\left.+\left(-\partial_{1}\partial_{3}-\partial_{2}\partial_{4}-i\partial_{2}\partial_{3}+i\partial_{1}\partial_{4}\right)\mbox{tr}_{k}\left[{\cal
W}_{2}^{(0)}{\cal W}_{4}^{\dagger(0)}\right]\right\\}$ $\displaystyle=$
$\displaystyle\frac{u^{2}}{\pi\ell k}\sum_{p_{\mu},C}\frac{e^{i\frac{2\pi
p_{\mu}x_{\mu}}{L_{\mu}}}}{\frac{p_{1}^{2}}{L_{1}^{2}}+\frac{p_{2}^{2}}{L_{2}^{2}}+\frac{p_{3}^{2}}{L_{3}^{2}}+\frac{p_{4}^{2}}{L_{4}^{2}}}\left\\{\left(-i\frac{p_{1}p_{3}}{L_{1}L_{3}}-i\frac{p_{2}p_{4}}{L_{2}L_{4}}-\frac{p_{2}p_{3}}{L_{2}L_{3}}+\frac{p_{1}p_{4}}{L_{1}L_{4}}\right)\right.$
$\displaystyle\left.\times({\cal X}_{3;C}(p)+{\cal X}_{4;C}(p)+{\cal
X}_{5;C}(p))\right.$
$\displaystyle\left.+\left(i\frac{p_{1}p_{3}}{L_{1}L_{3}}+i\frac{p_{2}p_{4}}{L_{2}L_{4}}-\frac{p_{2}p_{3}}{L_{2}L_{3}}+\frac{p_{1}p_{4}}{L_{1}L_{4}}\right)\right.$
$\displaystyle\left.\times({\cal X}_{3;C}^{*}(p)+{\cal X}_{4;C}^{*}(p)+{\cal
X}_{5;C}^{*}(p))\right\\}\,.$

Finally, one can also define the Fourier components of
$\mbox{tr}[F_{34}F_{34}](x)$:

$\displaystyle\mbox{tr}[F_{34}F_{34}](x)=\sum_{p_{\mu}\in\mathbb{Z}}e^{i\frac{2\pi
p_{\mu}x_{\mu}}{L_{\mu}}}{\cal Q}(p)\,.$ (235)

Using (110), we find, apart from an additive constant:

$\displaystyle\frac{{\cal Q}(p)}{u^{2}\hat{F}_{34}^{\omega}\Delta}$
$\displaystyle=$ $\displaystyle 8\pi N{\cal H}(p)+\frac{4}{\pi\ell
k}\sum_{C,C^{\prime}}\frac{\left[\frac{p_{3}^{2}}{L_{3}^{2}}+\frac{p_{4}^{2}}{L_{4}^{2}}\right]}{\frac{p_{1}^{2}}{L_{1}^{2}}+\frac{p_{2}^{2}}{L_{2}^{2}}+\frac{p_{3}^{2}}{L_{3}^{2}}+\frac{p_{4}^{2}}{L_{4}^{2}}}\left({\cal
X}_{2;C^{\prime},C}(p)+{\cal X}_{2;C^{\prime},C}^{*}(p)\right)$ (236)
$\displaystyle+\frac{2}{\pi\ell
k}\sum_{C}\frac{1}{\frac{p_{1}^{2}}{L_{1}^{2}}+\frac{p_{2}^{2}}{L_{2}^{2}}+\frac{p_{3}^{2}}{L_{3}^{2}}+\frac{p_{4}^{2}}{L_{4}^{2}}}\left\\{\left(-i\frac{p_{1}p_{3}}{L_{1}L_{3}}-i\frac{p_{2}p_{4}}{L_{2}L_{4}}-\frac{p_{2}p_{3}}{L_{2}L_{3}}+\frac{p_{1}p_{4}}{L_{1}L_{4}}\right)\right.$
$\displaystyle\left.\times({\cal X}_{3;C}(p)+{\cal X}_{4;C}(p)+{\cal
X}_{5;C}(p))\right.$
$\displaystyle\left.+\left(i\frac{p_{1}p_{3}}{L_{1}L_{3}}+i\frac{p_{2}p_{4}}{L_{2}L_{4}}-\frac{p_{2}p_{3}}{L_{2}L_{3}}+\frac{p_{1}p_{4}}{L_{1}L_{4}}\right)\right.$
$\displaystyle\left.\times({\cal X}_{3;C}^{*}(p)+{\cal X}_{4;C}^{*}(p)+{\cal
X}_{5;C}^{*}(p))\right\\}\,.$

We need to check whether the expression on the R.H.S. vanishes for all values
of $p_{\mu}$. The easiest check to perform is to choose
$p_{\mu}=(0,p_{2},0,0)$. With this choice, all terms vanish except ${\cal
H}(p)$, the Fourier transform of $\mbox{tr}_{k}\left[{\cal W}_{4}^{(0)}{\cal
W}_{4}^{\dagger(0)}\right]$ modulo $u^{2}$. One can check numerically that
${\cal H}(p)$ is non-vanishing, indicating that the gauge-invariant density
$\mbox{tr}[F_{34}F_{34}](x)$ increases indefinitely as $u\rightarrow\infty$.

##  Appendix E Fermion zero modes on the deformed-$\mathbb{T}^{4}$, for $k=r$

In this Appendix, we solve for the fermion zero modes in the background (53),
which we rewrite in the familiar $k$/$\ell$ block matrix form, using the
notation of (34):

$\displaystyle
A_{\mu}=\left(\begin{array}[]{cc}||\left(2\pi\ell\;(A_{\mu}^{\omega}-{z_{\mu}\over
L_{\mu}})+\phi_{\mu}^{C^{\prime}}\right)\delta_{C^{\prime}B^{\prime}}+\epsilon^{2}\;{\cal
S}_{\mu\;C^{\prime}B^{\prime}}||&||\epsilon\;{\cal
W}_{\mu\;C^{\prime}B}||\cr||\epsilon\;({\cal
W}_{\mu}^{\dagger})_{CB^{\prime}}||&||-2\pi k\;(A_{\mu}^{\omega}-{z_{\mu}\over
L_{\mu}})\delta_{CB}+\epsilon^{2}\;{\cal S}_{\mu\;CB}||\end{array}\right)~{}.$
(239)

Here we consider exclusively the $k=r$ case, where:

  1. 1.

$A_{\mu}^{\omega}$ is the constant flux background
$A_{1}^{\omega}=A_{3}^{\omega}=0$, $A_{2}^{\omega}=-{x_{1}\over NL_{1}L_{2}}$,
$A_{4}^{\omega}=-{x^{3}\over N\ell L_{3}L_{4}}$.

  2. 2.

$\phi^{C^{\prime}}_{\mu}$ are the $r-1$ allowed holonomies in $SU(k=r)$ (thus
obeying $\sum_{C^{\prime}}\phi_{\mu}^{C^{\prime}}=0$) from (34) and $z_{\mu}$
are the holonomies in the $U(1)$-direction $\omega$, eqn. (11).

We also recall that these are, after computing the commutator in the Weyl
equation, combined into the $r$ independent $\hat{\phi}_{\mu}^{C^{\prime}}$ of
eqns. (50, 51) with no constraint on the trace.

  3. 3.

${\cal W}_{\mu}$ is leading order $k=r$ solution. Thus, ${\cal W}_{3}={\cal
W}_{4}=0$ and ${\cal W}_{1}=-i{\cal W}_{2}$, and with ${\cal W}_{2}$ given by
(83), with the $r$ coefficients ${\cal C}_{2}^{A}$ fixed by solving eqn. (98).

  4. 4.

The components of ${\cal S}_{\mu}$ are obtained by solving (84, 88) (recall
that they obey the tracelessness condition
$\sum_{C^{\prime}}S_{\mu\;C^{\prime}C^{\prime}}+\sum_{C}S_{\mu\;CC}=0$).

  5. 5.

Finally, to remind us of the powers of $\sqrt{\Delta}$ appearing in the
leading order solution for ${\cal W}_{\mu}$ and ${\cal S}_{\mu}$, we
introduced a parameter $\epsilon\;(\equiv 1)$.

Our goal is to solve the Weyl equation
$\partial_{\mu}\bar{\sigma}^{\mu}\lambda+i\bar{\sigma}^{\mu}[A_{\mu},\lambda]=0$
in the $k=r$ background (239), using a series expansion in $\epsilon$, to
leading order. We take $\lambda$ also in the block-diagonal form (39), obeying
(40):

$\displaystyle\lambda$ $\displaystyle=$
$\displaystyle\left[\begin{array}[]{cc}||\lambda_{C^{\prime}B^{\prime}}||&||\lambda_{C^{\prime}B}||\\\
||\lambda_{CB^{\prime}}||&||\lambda_{CB}||\end{array}\right]~{},~{}C^{\prime},B^{\prime}\in\\{0,...k-1\\},~{}C,B\in\\{0,...\ell-1\\}~{}.$
(243)

Newt, write the Weyl equation, using the quaternionic notation of Section 4:
$\bar{\partial}=\partial_{\mu}\bar{\sigma}_{\mu}$ and
$\bar{A}=\bar{\sigma}_{\mu}A_{\mu}$ (and similar for all other vectors in
(239), with $\bar{\sigma}_{\mu}$ defined in Footnote 5) and obtain the
following equations for the components of $\lambda$ of (243), with a sum over
repeated indices $B,B^{\prime}$ implied:

$\displaystyle\bar{\partial}\lambda_{C^{\prime}D^{\prime}}$ $\displaystyle=$
$\displaystyle-i\epsilon\bar{\sigma}_{\mu}({\cal
W}_{\mu\;C^{\prime}B}\lambda_{BD^{\prime}}-\lambda_{C^{\prime}B}({\cal
W}^{\dagger})_{\mu\;BD^{\prime}})-i\epsilon^{2}\bar{\sigma}_{\mu}({\cal
S}_{\mu\;C^{\prime}B^{\prime}}\lambda_{B^{\prime}D^{\prime}}-\lambda_{C^{\prime}B^{\prime}}{\cal
S}_{\mu\;B^{\prime}D^{\prime}}),$ $\displaystyle\bar{\partial}\lambda_{CD}$
$\displaystyle=$ $\displaystyle-i\epsilon\bar{\sigma}_{\mu}(({\cal
W}^{\dagger})_{\mu\;CB^{\prime}}\lambda_{B^{\prime}D}-\lambda_{CB^{\prime}}{\cal
W}_{\mu\;B^{\prime}D})-i\epsilon^{2}\bar{\sigma}_{\mu}({\cal
S}_{\mu\;CB}\lambda_{BD}-\lambda_{CB}{\cal S}_{\mu\;BD}),$
$\displaystyle\bar{\partial}\lambda_{C^{\prime}D}$ $\displaystyle=$
$\displaystyle-i(2\pi
N\bar{A}^{\omega}+\bar{\hat{\phi}}^{C^{\prime}})\lambda_{C^{\prime}D}-i\epsilon\bar{\sigma}_{\mu}({\cal{W}}_{\mu\;C^{\prime}B}\lambda_{BD}-\lambda_{C^{\prime}B^{\prime}}{\cal{W}}_{\mu\;B^{\prime}D})$
$\displaystyle-i\epsilon^{2}\bar{\sigma}_{\mu}({\cal
S}_{\mu\;C^{\prime}B^{\prime}}\lambda_{B^{\prime}D}-\lambda_{C^{\prime}B}{\cal{S}}_{\mu\;BD}),$
$\displaystyle\bar{\partial}\lambda_{CD^{\prime}}$ $\displaystyle=$
$\displaystyle i(2\pi
N\bar{A}^{\omega}+\bar{\hat{\phi}}^{C^{\prime}})\lambda_{CD^{\prime}}-i\epsilon\bar{\sigma}_{\mu}(({\cal
W}^{\dagger})_{\mu\;CB^{\prime}}\lambda_{B^{\prime}D^{\prime}}-\lambda_{CB}({\cal
W}^{\dagger})_{\mu\;BD^{\prime}})$ (244)
$\displaystyle-i\epsilon^{2}\bar{\sigma}_{\mu}({\cal
S}_{\mu\;CB}\lambda_{BD^{\prime}}-\lambda_{CB^{\prime}}{\cal{S}}_{\mu\;B^{\prime}D^{\prime}})\,.$

We now observe that we can consistently solve (E) in a series expansion in
$\epsilon$, assigning the following (leading-order only shown)
$\epsilon$-scaling of the various components of $\lambda$:

$\displaystyle\lambda_{C^{\prime}D^{\prime}}$ $\displaystyle=$
$\displaystyle\epsilon^{0}\lambda_{C^{\prime}D^{\prime}}+{\cal
O}(\epsilon)\,,$ $\displaystyle\lambda_{CD}$ $\displaystyle=$
$\displaystyle\epsilon^{0}\lambda_{C^{\prime}D^{\prime}}+{\cal
O}(\epsilon)\,,$ $\displaystyle\lambda_{C^{\prime}D}$ $\displaystyle=$
$\displaystyle\epsilon\;\lambda_{C^{\prime}D}+{\cal O}(\epsilon^{2})\,,$
$\displaystyle\lambda_{CD^{\prime}}$ $\displaystyle=$
$\displaystyle\epsilon\;\lambda_{CD^{\prime}}+{\cal O}(\epsilon^{2})\,.$ (245)

Substituting into (E) and keeping only the leading terms in $\epsilon$ in each
equation, we find the following equations for the leading order (in
$\sqrt{\Delta}$) fermion zero modes in the background (239):

$\displaystyle\bar{\partial}\lambda_{C^{\prime}D^{\prime}}$ $\displaystyle=$
$\displaystyle 0,$ $\displaystyle\bar{\partial}\lambda_{CD}$ $\displaystyle=$
$\displaystyle 0,$ $\displaystyle(\bar{\partial}+i(2\pi
N\bar{A}^{\omega}+\bar{\hat{\phi}}^{C^{\prime}}))\lambda_{C^{\prime}D}$
$\displaystyle=$
$\displaystyle-i({\cal{W}}_{\mu\;C^{\prime}B}\;\bar{\sigma}_{\mu}\lambda_{BD}-\bar{\sigma}_{\mu}\lambda_{C^{\prime}B^{\prime}}{\cal{W}}_{\mu\;B^{\prime}D}),$
$\displaystyle(\bar{\partial}-i(2\pi
N\bar{A}^{\omega}+\bar{\hat{\phi}}^{C^{\prime}}))\lambda_{CD^{\prime}}$
$\displaystyle=$ $\displaystyle-i({\cal
W}^{*}_{\mu\;B^{\prime}C}\;\bar{\sigma}_{\mu}\lambda_{B^{\prime}D^{\prime}}-\bar{\sigma}_{\mu}\lambda_{CB}{\cal
W}^{*}_{\mu\;D^{\prime}B}).$ (246)

Now, we recall that the first two equations were already solved in Section
3.4.1. From eqn. (48), taken with $k=r$, we have the diagonal zero mode
solutions

$\displaystyle\lambda_{\alpha\;B^{\prime}C^{\prime}}$ $\displaystyle=$
$\displaystyle\delta_{B^{\prime}C^{\prime}}\;\theta_{\alpha}^{C^{\prime}},$
$\displaystyle\lambda_{\alpha\;BC}$ $\displaystyle=$
$\displaystyle-\delta_{BC}\;{1\over\ell}\sum_{C^{\prime}}\theta_{\alpha}^{C^{\prime}},$
(247)

where we momentarily restored the spinor index $\alpha$. We first define the
spinor

$\eta^{C^{\prime}}\equiv\theta^{C^{\prime}}+{1\over\ell}\sum\limits_{B^{\prime}=0}^{k-1}\theta^{B^{\prime}},$
(248)

and then plug (E) into the last two equations in (E) to obtain:

$\displaystyle(\bar{\partial}+i(2\pi
N\bar{A}^{\omega}+\bar{\hat{\phi}}^{C^{\prime}}))\lambda_{C^{\prime}D}$
$\displaystyle=$ $\displaystyle
i{\cal{W}}_{\mu\;C^{\prime}D}\;\bar{\sigma}_{\mu}\;\eta^{C^{\prime}},$
$\displaystyle(\bar{\partial}-i(2\pi
N\bar{A}^{\omega}+\bar{\hat{\phi}}^{C^{\prime}}))\lambda_{CD^{\prime}}$
$\displaystyle=$ $\displaystyle-i{\cal
W}^{*}_{\mu\;D^{\prime}C}\;\bar{\sigma}_{\mu}\;\eta^{D^{\prime}}.$ (249)

We now recall that for the $k=r$ solution, ${\cal W}_{3}={\cal W}_{4}=0$ and
${\cal W}_{1}=-i{\cal W}_{2}$, hence

$\displaystyle{\cal{W}}_{\mu\;C^{\prime}D}\;\bar{\sigma}_{\mu}$
$\displaystyle=$ $\displaystyle(-i\bar{\sigma}_{1}+\bar{\sigma}_{2}){\cal
W}_{2\;C^{\prime}D}={\cal W}_{2\;C^{\prime}D}\left(\begin{array}[]{cc}0&-2\cr
0&0\end{array}\right)\,,$ (252) $\displaystyle{\cal
W}^{*}_{\mu\;D^{\prime}C}\;\bar{\sigma}_{\mu}$ $\displaystyle=$
$\displaystyle(i\bar{\sigma}_{1}+\bar{\sigma}_{2}){\cal
W}_{2\;D^{\prime}C}^{*}={\cal
W}_{2\;D^{\prime}C}^{*}\left(\begin{array}[]{cc}0&0\cr
2&0\end{array}\right)\,,$ (255)

and that ${\cal W}_{2\,C^{\prime},C}=V^{-1/4}{\cal
C}_{2}^{C^{\prime}}\Phi^{(0)}_{C^{\prime},C}(x,\hat{\phi})$, where ${\cal
C}_{2}^{C^{\prime}}$ is as determined in Section 5.

The equation for $\lambda_{C^{\prime}D}$ then takes the form, using the
derivatives from (187) and noting that the equations for each
$C^{\prime}=1,...,r$ decouple:

$\displaystyle(\hat{D}_{4}-i\hat{D}_{3})\lambda_{1\;C^{\prime}D}-(i\hat{D}_{1}+\hat{D}_{2})\lambda_{2\;C^{\prime}D}$
$\displaystyle=$
$\displaystyle\bar{\eta}_{2}^{C^{\prime}}\;\Phi^{(0)}_{2\;C^{\prime}D}\,,$
$\displaystyle(-i\hat{D}_{1}+\hat{D}_{2})\lambda_{1\;C^{\prime}D}+(\hat{D}_{4}+i\hat{D}_{3})\lambda_{2\;C^{\prime}D}$
$\displaystyle=$ $\displaystyle 0~{},$ (256)

where we absorb various inessential constants in the redefined
$\bar{\eta}_{2}^{C^{\prime}}$ coefficient. The solution of these equations is
given by the function ${\cal G}_{3\;C^{\prime}D}^{(0)}$ defined in (193),
explicitly

$\displaystyle\lambda_{1\;C^{\prime}D}$ $\displaystyle=$
$\displaystyle\bar{\eta}_{2}^{C^{\prime}}{\cal G}_{3\;C^{\prime}D}^{(0)},$
$\displaystyle\lambda_{2\;C^{\prime}D}$ $\displaystyle=$ $\displaystyle 0~{}.$
(257)

Similarly, one finds that the other zero mode is

$\displaystyle\lambda_{1\;CD^{\prime}}$ $\displaystyle=$ $\displaystyle 0,$
$\displaystyle\lambda_{2\;CD^{\prime}}$ $\displaystyle=$
$\displaystyle\bar{\eta}_{1}^{D^{\prime}}{\cal G}_{3\;D^{\prime}C}^{*\;(0)}.$
(258)

Thus, there are in total $2r$ zero modes labeled by
$\bar{\eta}_{1,2}^{C^{\prime}}$, with $C^{\prime}=1,...,r$. The $x$-dependence
of the zero mode labeled by a given $C^{\prime}$ is governed only by the
holonomies $\hat{\phi}^{C^{\prime}}_{\mu}$, similar to the bosonic case
discussed earlier.

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