text:
\section{Introduction}\label{sec:intro}

In 3D supersymmetric gauge theories, there has been a remarkable
technical breakthrough based on exact computation of partition functions
of Euclidean theories on three-sphere $S^3$ using localization principle.
It was first developed for superconformal theories by \cite{KWY1, KWY2},
and recently extended to non-conformal theories by \cite{Jafferis, HHL}.
The result is expected to provide us with a new quantitative tool to
study strongly coupled IR physics of 3D supersymmetric theories. In
particular, in \cite{Jafferis} it has  been argued that the 3D partition
function can be used to determine the superconformal R-symmetry at the
infrared fixed points, in a similar spirit to $a$-maximization in 4D
gauge theories. It has also been attracting wide attention from various
other fields, for example Wilson loops \cite{Suy,RS}, 3D dualities
\cite{KWY2}, large-$N$ duality of topological string \cite{MPP} and the
ABJM theory at strong coupling \cite{DT,MP,DMP}. The localization
technique can also be applied to the theory on $S^2\times S^1$ and
allows one to compute the superconformal index. Using this idea, the
superconformal index of ${\cal N}=6$ Chern-Simons matter theories
was computed by \cite{Seok}. It has recently been generalized
to arbitrary ${\cal N}=2$ supersymmetric gauge theories by \cite{IY}.
See also \cite{IYY,CGKP}.

Another aspect, which we will focus on in this paper, is the application
to the AGT relation. In \cite{Pe} the exact partition function
of 4D ${\cal N}=2$ gauge theories on $S^4$ has been worked out,
and it has led to a discovery of a remarkable correspondence
between partition functions of 4D gauge theories and correlation
functions of 2D Liouville or Toda conformal field theories \cite{AGT,Wy}.
In this context, certain 3D supersymmetric gauge theories on $S^3$ play
the role of domain walls \cite{DGG, HLP}, and their partition functions
were shown to correspond to transformation coefficients of conformal
blocks under S-duality (or Moore-Seiberg groupoid operation).

In \cite{HLP} it was observed that the building blocks of 3D partition
functions on round $S^3$, namely the measure over Coulomb branch and the
one-loop determinants, are precisely the building blocks for the structure
constants in Liouville or Toda CFTs with $b=1$. Here $b$ is the coupling
which determines the central charge of Liouville or Toda theories, for
example for Liouville theory one has $c=1+6(b+b^{-1})^2$. In particular,
the one-loop determinant of a charged matter multiplet is given by a
double sine function $s_b(x)$ at $b=1$,
\begin{equation}
 s_b(x)~\equiv~\prod_{m,n\in\mathbb Z_{\ge0}}
 \frac{mb+nb^{-1}+\frac Q2-ix}{mb+nb^{-1}+\frac Q2+ix}\,,
\quad \big(Q\equiv b+\tfrac1b\big)
\end{equation}
and $x$ is related to a coordinate on the Coulomb branch. See
\cite{KLS,BT} for more details on this function. This is plausible if we
recall that 4D ${\cal N}=2$ gauge theories on round $S^4$ are in
correspondence with Liouville or Toda theories with $b=1$.
It is not yet known how to obtain theories with more general $b$ though,
as suggested in \cite{HLP}, a reasonable guess would be that the
background sphere should admit a continuous deformation to account for
this. In this paper we present the answer to this question for 3D theories.

\vskip2mm

We begin by presenting two kinds of squashed $S^3$ in Section
\ref{sec:three-sphere}. The first one appears frequently in the
literature; the metric is written in terms of left-invariant
one-forms and preserves $SU(2)\times U(1)$ symmetry.
The second one is less familiar and preserves only $U(1)\times U(1)$
symmetry, but it has a simple definition as a hyper-ellipsoid embedded
in flat $\mathbb R^4$. For both squashings we show that, if a suitable
background $U(1)$ gauge field is turned on, one can find a pair of
Killing spinors with the $U(1)$ charges $\pm1$ which is necessary for
defining ${\cal N}=2$ supersymmetric gauge theories. We give the general
construction of supersymmetric gauge theories in Section \ref{sec:SUSY}.

We then turn to the computation of partition functions. Section
\ref{sec:partition1} discusses the case of $SU(2)\times U(1)$ symmetric
squashing, where we use the spherical harmonics to work out all the
eigenmodes of the relevant Laplace and Dirac operators. We then use them
to compute the one-loop determinant at each saddle point. Disappointingly,
the partition function turns out to be essentially the same as that for
round sphere, essentially due to $SU(2)$ isometry.
Next we discuss the less symmetric squashing in Section
\ref{sec:partition2}. Rather than working out the full spectrum of the
Laplace and Dirac operators, we look closely into the structures in
which the bosonic and fermionic modes are paired, and exhaust the modes
unpaired which give nontrivial contributions to the one-loop
determinant. The final expression for the integration measure and
determinant are found to be precisely the building blocks for structure
constants in Liouville or Toda theories with general $b$.

\vskip2mm

We end the introduction by summarizing our conventions for bilinear
products of spinors.
\begin{equation}
 \bar\epsilon\lambda~=~ \bar\epsilon^\alpha C_{\alpha\beta}\lambda^\beta,\quad
 \bar\epsilon\gamma^a\lambda~=~
 \bar\epsilon^\alpha(C\gamma^a)_{\alpha\beta}\lambda^\beta,\quad
 \text{etc}.
\end{equation}
Here $C$ is the charge conjugation matrix with nonzero elements
$C_{12}=-C_{21}=1$, and $\gamma^a$ are Pauli's matrices.
Noticing that $C$ is antisymmetric and $C\gamma^a$ are symmetric,
one finds
\begin{equation}
 \bar\epsilon\lambda=\lambda\bar\epsilon,\quad
 \bar\epsilon\gamma^a\lambda=-\lambda\gamma^a\bar\epsilon
\end{equation}
for all spinors $\bar\epsilon,\lambda$ which we assume to be Grassmann
odd.

\section{Three-Spheres, Round and Squashed}\label{sec:three-sphere}

Here we summarize our notations for various geometric quantities on
round or squashed $S^3$. By squashing we mean certain deformations to
its round metric. We will restrict to those preserving at least
$U(1)\times U(1)$ isometry. The reason is that, when our $S^3$ is
embedded as a domain wall in a 4D space on which an ${\cal N}=2$
supersymmetric gauge theory is defined, the $U(1)\times U(1)$ isometry
is necessary for the omega deformation. There are many metrics
preserving this isometry, of which we take only two simple examples.
After presenting the metrics for these two squashings we show that, with
a suitable background $U(1)$ gauge field turned on, the two squashed
spheres both admit a pair of charged Killing spinors.

The three-sphere is parametrized by an element $g$ of the Lie group
$SU(2)$, and two copies of $SU(2)$ symmetry act on $g$ from the left
and the right. We introduce the left-invariant (LI) and right-invariant
(RI) one-forms
$\mu^a=\mu^a_\nu d\xi^\nu$ and $\tilde\mu^a=\tilde\mu^a_\nu d\xi^\nu$,
\begin{equation}
 g^{-1}dg ~=~ i\mu^a\gamma^a,\quad
 dg g^{-1}~=~ i\tilde\mu^a\gamma^a,
\end{equation}
where $\gamma^a$ are Pauli matrices.
These one-forms satisfy
\begin{equation}
 d\mu^a=\epsilon^{abc}\mu^b\mu^c,\quad
 d\tilde\mu^a=-\epsilon^{abc}\tilde\mu^b\tilde\mu^c.
\end{equation}

\paragraph{Round sphere.}

The left-right invariant metric on the round sphere with radius $\ell$ is
\begin{equation}
 ds^2 ~=~ \tfrac12\ell^2\text{tr}(dgdg^{-1})
 ~=~ \ell^2\mu^a\mu^a~=~\ell^2\tilde\mu^a\tilde\mu^a.
\end{equation}
We define the vielbein in the ``LI frame'' as
$e^a = e^a_\mu d\xi^\mu = \ell\mu^a$.
The spin connection in this frame is
$\omega^{ab}= \varepsilon^{abc}\mu^c$ and satisfies the torsion-free
condition $de^a+\omega^{ab}e^b=0$.
If we define the vielbein from $\tilde\mu^a$ (``RI frame''),
the spin connection is $\tilde\omega^{ab}=-\varepsilon^{abc}\tilde\mu^c$.

Killing spinor $\epsilon$ satisfies the following equation
\begin{equation}
 D\epsilon
 ~\equiv~ d\epsilon+\tfrac14\gamma^{ab}\omega^{ab}\epsilon
 ~=~ e^a\gamma^a\tilde\epsilon,
\end{equation}
for a certain $\tilde\epsilon$.
Here we used $\gamma^{ab}=i\varepsilon^{abc}\gamma^c$.
There are two types of Killing spinors. The first one is constant in the
LI frame,
\begin{equation}
 \epsilon=\epsilon_0~(\text{constant}), \quad
 \tilde\epsilon=+\tfrac i{2\ell}\epsilon.
\end{equation}
The second one reads
\begin{equation}
 \epsilon=g^{-1}\epsilon_0, \quad
 \tilde\epsilon=-\tfrac i{2\ell}\epsilon,
\end{equation}
and is constant in the RI frame.

Let us next introduce the Killing vector fields
$\mathscr{L}^a=\mathscr{L}^{a\mu}\frac\partial{\partial\xi^\mu}$ and
$\mathscr{R}^a=\mathscr{R}^{a\mu}\frac\partial{\partial\xi^\mu}$
which generate the left and the right actions of $SU(2)$.
They can be determined from
\begin{equation}
 \mathscr{L}^ag= i\gamma^ag,\quad
 \mathscr{R}^ag= ig\gamma^a.
\end{equation}
The vector fields $\tfrac i2\mathscr{L}^a$ and $-\tfrac i2\mathscr{R}^a$
satisfy the standard commutation relations of $SU(2)$ Lie algebra.
It is also easy to find
$\mathscr{R}^{a\nu}\mu_\nu^b=\mathscr{L}^{a\nu}\tilde\mu_\nu^b=\delta^{ab}$,
in other words $\mathscr{R}^{a\nu}$ and $\mathscr{L}^{a\nu}$ are proportional
to the inverse-vielbeins in LI or RI frames.
The action of these Killing vector fields on the LI and RI one-forms
is given by
\begin{equation}
 \mathscr{L}^a\tilde\mu^b=2\varepsilon^{abc}\tilde\mu^c,\quad
 \mathscr{R}^a\mu^b=-2\varepsilon^{abc}\mu^c,\quad
 \mathscr{L}^a\mu^b=\mathscr{R}^a\tilde\mu^b=0.
\end{equation}
It therefore follows that $\mu^1\mu^2\mu^3= d^3\xi\text{det}(\mu^a_\nu)$
can be used to define the invariant volume form.

\paragraph{Squashed sphere (familiar one).}

Squashed spheres are defined by the metric or vielbein one-forms
\begin{eqnarray}
&& ds^2 ~=~ \ell^2(\mu^1\mu^1+\mu^2\mu^2)+\tilde\ell^2\mu^3\mu^3,
\nonumber \\
&& (e^1,e^2,e^3)~=~(\ell\mu^1,\ell\mu^2,\tilde\ell\mu^3).
\label{fssgmn}
\end{eqnarray}
The squashed metric preserves the
$SU(2)_{\mathscr{L}}\times U(1)_{\mathscr R^3}$ symmetry.
The torsion-free condition determines the spin connection $\omega^{ab}$
as follows,
\begin{eqnarray}
\omega^{12}&=& (2\tilde\ell^{-1}-f^{-1})e^3,\nonumber \\
\omega^{23}&=& f^{-1}e^1,\nonumber \\
\omega^{31}&=& f^{-1}e^2,
\end{eqnarray}
where $f\equiv \ell^2\tilde\ell^{-1}$ is a constant. Then one can show
that any constant spinor $\psi$ satisfies
\begin{equation}
 d\psi+
 \frac14\gamma^{ab}\omega^{ab}\psi
 ~=~\frac i{2f}\gamma^ae^a\psi
 +i\gamma^3 V\psi,\quad
 V~\equiv~e^3\Big(\frac1{\tilde\ell}-\frac1f\Big).
\label{nks1}
\end{equation}
This is the Killing spinor equation were it not for the last term
in the right hand side.

Now, let us think of the one-form $V$ as a background gauge field
for a certain $U(1)$ symmetry. Let $\epsilon$ be a complex spinor
field with charge $+1$ under this $U(1)$, so that its covariant
derivative becomes
\begin{equation}
 D\epsilon~\equiv~ d\epsilon+\frac14\gamma^{ab}\omega^{ab}\epsilon
 -iV\epsilon.
\end{equation}
Then the constant spinor $\epsilon=(1,0)^t$ satisfies the
($U(1)$-covariant version of) Killing spinor equation,
\begin{equation}
 D\epsilon~=~ \frac{i}{2f}\gamma^ae^a\epsilon.
\end{equation}
Likewise, a spinor $\bar\epsilon$ carrying the $U(1)$ charge $-1$ is
a Killing spinor if $\bar\epsilon=(0,1)^t$.
One also finds
\begin{equation}
\bar\epsilon\epsilon=-1,\quad
\bar\epsilon\gamma^a\epsilon=-\delta^{a3}.
\end{equation}
The latter represents nothing but the Killing vector $\mathcal{R}^3$.
The squashed $S^3$ can be regarded as an $S^1$ fibration over $S^2$,
where each fiber is an orbit of the action of $\mathcal{R}^3$.

\paragraph{Squashed sphere (less familiar one).}

Next we consider the less familiar version of squashed sphere which
preserve only $U(1)_{\mathscr{L}^3}\times U(1)_{\mathscr{R}^3}$ symmetry.
We take one of the simplest metrics with this property,
\begin{eqnarray}
 ds^2 &=& \ell^2(dx_0^2+dx_1^2)+ \tilde\ell^2(dx_2^2+dx_3^2)\,.
 \nonumber \\&&
 (x_0^2+x_1^2+x_2^2+x_3^2=1)
\end{eqnarray}
Inserting $(x_0,x_1,x_2,x_3)=
(\cos\theta\cos\varphi,\cos\theta\sin\varphi,
 \sin\theta\cos\chi,\sin\theta\sin\chi)$
one obtains
\begin{eqnarray}
ds^2 &=&  f(\theta)^2d\theta^2
 +\ell^2\cos^2\theta d\varphi^2+\tilde\ell^2\sin^2\theta d\chi^2,
 \nonumber \\
 && f(\theta) ~\equiv~
 \sqrt{\ell^2\sin^2\theta+\tilde\ell^2\cos^2\theta}.
\end{eqnarray}
We set the vielbein as follows,
\begin{equation}
 e^1=\ell\cos\theta d\varphi,\quad
 e^2=\tilde\ell\sin\theta d\chi,\quad
 e^3=f(\theta)d\theta.
\label{ss2vb}
\end{equation}
The spin connection then becomes
\begin{equation}
 \omega^{12}=0, \quad
 \omega^{13}=-\frac\ell f\sin\theta d\varphi,\quad
 \omega^{23}= \frac{\tilde\ell}f\cos\theta d\chi.
\label{ss2sc}
\end{equation}

The Killing spinor equation for a spinor field $\psi$ can be reduced to
\begin{eqnarray}
 if\partial_\varphi\psi &=&
 \ell\gamma^2
 (\tfrac12\sin\theta+\cos\theta\partial_\theta)\psi,\nonumber \\
 if\partial_\chi\psi &=&
 \tilde\ell\gamma^1
 (\tfrac12\cos\theta-\sin\theta\partial_\theta)\psi.
\end{eqnarray}
For round sphere one has $\ell=\tilde\ell=f$, and the above equation is
easily shown to have four independent solutions,
\begin{equation}
 \psi_{st}=\left( \begin{array}{r}
 e^{\frac i2(s\chi+t\varphi-st\theta)}\\
-s e^{\frac i2(s\chi+t\varphi+st\theta)}
 \end{array} \right),\qquad
 (s,t=\pm)
\end{equation}
satisfying
\begin{equation}
D_\mu\psi_{st}~=~ -\frac{ist}{2\ell}\gamma_\mu\psi_{st}.
\end{equation}
For squashed spheres with $\ell\neq\tilde\ell$, the same spinor fields
$\psi_{st}$ fail to satisfy the Killing spinor equations.
\begin{equation}
 -\frac{ist}{2f}\gamma_\mu\psi_{st}
 ~=~ D_\mu\psi_{st}-iV_\mu^{(st)}\psi_{st},
\label{ss2ks}
\end{equation}
where
\begin{equation}
 V^{(st)}~=~ \frac t2\Big(1-\frac\ell f\Big)d\varphi
     +\frac s2\Big(1-\frac{\tilde\ell}f\Big)d\chi.
\label{ss2v}
\end{equation}
Again, one can reinterpret the unwanted term in the right hand side
of (\ref{ss2ks}) as the coupling to a background $U(1)$ gauge field.
For the discussions in later sections, we choose to turn on the gauge
field $V=V^{(+-)}$, so that the spinors $\epsilon=\psi_{+-}$ and
$\bar\epsilon=\psi_{-+}$ satisfy the Killing spinor equation with $U(1)$
chatges $\pm1$.

\section{SUSY Theories on Squashed $\bf S^3$}\label{sec:SUSY}

Here we present the Euclidean 3D ${\cal N}=2$ gauge theories on manifolds
with generalized Killing spinors. The formulae for the action and
supersymmery rules are almost the same as those for round $S^3$ (see
e.g. \cite{HHL}), but now contain the background $U(1)$ gauge field we
have turned on. Since the $U(1)$ symmetry rotates the Killing spinors,
it is the R-symmetry which has been gauged. In the following we show
that the closure of supersymmetry algebra and the invariance of
Lagrangian are achieved if the background gauge field is coupled to the
fields according to their R-charge.

\paragraph{Vector multiplets.}

Vector multiplet fields obey the following transformation laws,
\begin{eqnarray}
 \delta A_\mu &=&
 -\tfrac i2(\bar\epsilon\gamma_\mu\lambda-\bar\lambda\gamma_\mu\epsilon),
 \nonumber \\
 \delta\sigma &=&
 \tfrac12(\bar\epsilon\lambda-\bar\lambda\epsilon),
 \nonumber \\
 \delta\lambda &=&
 \tfrac12\gamma^{\mu\nu}\epsilon F_{\mu\nu}-D\epsilon
      +i\gamma^\mu\epsilon D_\mu\sigma
 +\tfrac{2i}3\sigma\gamma^\mu D_\mu\epsilon,
 \nonumber \\
 \delta\bar\lambda &=&
 \tfrac12\gamma^{\mu\nu}\bar\epsilon F_{\mu\nu}+D\bar\epsilon
                 -i\gamma^\mu\bar\epsilon D_\mu\sigma
 -\tfrac{2i}3\sigma\gamma^\mu D_\mu\bar\epsilon,
 \nonumber \\
 \delta D &=&
 -\tfrac i2\bar\epsilon\gamma^\mu D_\mu\lambda
 -\tfrac i2D_\mu\bar\lambda\gamma^\mu\epsilon
 +\tfrac i2[\bar\epsilon\lambda,\sigma]
 +\tfrac i2[\bar\lambda\epsilon,\sigma]
 -\tfrac i6(D_\mu\bar\epsilon\gamma^\mu\lambda
         +\bar\lambda\gamma^\mu D_\mu\epsilon).
\label{trvec}
\end{eqnarray}
Here and throughout this paper, $D_\mu$ denotes the covariant derivative
with respect to gauge, local Lorentz and background gauged R-symmetries
as well as general covariance. $\gamma^\mu=e^\mu_a\gamma^a$ is the Dirac
matrix with curved index.
The spinors $\epsilon,\bar\epsilon$ are assumed to satisfy
Killing spinor equation. Namely there are spinors
$\tilde\epsilon,\tilde{\bar\epsilon}$ which satisfy
\begin{eqnarray}
 D_\mu\epsilon &\equiv&
 (\partial_\mu+\tfrac14\omega_\mu^{ab}\gamma^{ab}-iV_\mu)\epsilon
 ~=~ \gamma_\mu\tilde\epsilon,\nonumber \\
 D_\mu\bar\epsilon &\equiv&
 (\partial_\mu+\tfrac14\omega_\mu^{ab}\gamma^{ab}+iV_\mu)\bar\epsilon
 ~=~ \gamma_\mu\tilde{\bar\epsilon}.
\label{KSeq1}
\end{eqnarray}
Denoting $\delta$ as the sum of unbarred and barred parts,
$\delta=\delta_\epsilon+\delta_{\bar\epsilon}$, one can show that
two unbarred or two unbarred supersymmetries commute. Also, on most of
the fields the commutator $[\delta_\epsilon,\delta_{\bar\epsilon}]$ is a
sum of translation by $iv^\mu$, gauge transformation by $\Lambda$,
Lorentz rotation by $\Theta^{\mu\nu}$, dilation by $\rho$ and R-rotation
by $\alpha$, where
\begin{eqnarray}
 v^\mu &=& \bar\epsilon\gamma^\mu\epsilon,
 \nonumber \\
 \Theta^{\mu\nu} &=& iD^{[\mu}v^{\nu]}+iv^\lambda\omega_\lambda^{\mu\nu},
 \nonumber \\
 \Lambda &=& -iA_\mu v^\mu+\sigma\bar\epsilon\epsilon,
 \nonumber \\
 \rho &=& \tfrac i3
 (\bar\epsilon\gamma^\mu D_\mu\epsilon
 +D_\mu\bar\epsilon\gamma^\mu\epsilon),
 \nonumber \\
 \alpha &=& \tfrac
 i3(D_\mu\bar\epsilon\gamma^\mu\epsilon-\bar\epsilon\gamma^\mu D_\mu\epsilon)
 +v^\mu V_\mu.
\label{sypar}
\end{eqnarray}
The only exception is that
\begin{eqnarray}
 [\delta_\epsilon,\delta_{\bar\epsilon}]D &=&
 iv^\mu\partial_\mu D + i[\Lambda,D]+2\rho D
 \nonumber \\&&
 +\tfrac13\sigma
 (\bar\epsilon\gamma^\mu\gamma^\nu D_\mu D_\nu\epsilon
 -\epsilon\gamma^\mu\gamma^\nu D_\mu D_\nu\bar\epsilon).
\end{eqnarray}
The last term in the right hand side vanishes provided that
$\epsilon$ and $\bar\epsilon$ satisfy, in addition to (\ref{KSeq1}), the
following equations
\begin{eqnarray}
 \gamma^\mu\gamma^\nu D_\mu D_\nu\epsilon &=&
 -\tfrac38(R+2iV_{\mu\nu}\gamma^{\mu\nu})\epsilon,
 \nonumber \\
 \gamma^\mu\gamma^\nu D_\mu D_\nu\bar\epsilon &=&
 -\tfrac38(R-2iV_{\mu\nu}\gamma^{\mu\nu})\bar\epsilon
\label{kseq2}
\end{eqnarray}
for a certain set of functions $(R,V_{\mu\nu})$. By arguing in a similar
way to \cite{HHL} one finds that $R$ is the scalar curvature of the 3D
manifold and $V_{\mu\nu}\equiv\partial_\mu V_\nu-\partial_\nu V_\mu$ is
the field strength of the background gauge field. Note that the Killing
spinors on the squashed $S^3$ of our interest actually all satisfy a
stronger condition (\ref{nscd}), so the supersymmetry is not reduced
by the above conition (\ref{kseq2}). For later convenience,
we give here some additional formulae which are related to (\ref{kseq2}).
\begin{eqnarray}
 \gamma^\mu\gamma^\nu D_\mu D_\nu\epsilon &=& 3D_\mu D^\mu\epsilon,\nonumber\\
 \gamma^\mu\gamma^\nu D_\mu D_\nu\bar\epsilon &=& 3D_\mu D^\mu\bar\epsilon,
\end{eqnarray}

\paragraph{Matter multiplets.}

The fields in a chiral multiplet coupled to a gauge symmetry
transform as follows,
\begin{eqnarray}
 \delta\phi &=& \bar\epsilon\psi, \nonumber \\
 \delta\bar\phi &=& \epsilon\bar\psi, \nonumber \\
 \delta\psi &=& i\gamma^\mu\epsilon D_\mu\phi +i\epsilon\sigma\phi
 +\tfrac{2qi}3\gamma^\mu D_\mu\epsilon\phi+\bar\epsilon F,
 \nonumber \\
 \delta\bar\psi &=& i\gamma^\mu\bar\epsilon D_\mu\bar\phi
 +i\bar\phi\sigma\bar\epsilon+\tfrac{2qi}3\bar\phi\gamma^\mu D_\mu\bar\epsilon
 +\bar F\epsilon,
 \nonumber \\
 \delta F &=&
 \epsilon(i\gamma^\mu D_\mu\psi-i\sigma\psi-i\lambda\phi)
 +\tfrac i3(2q-1)D_\mu\epsilon\gamma^\mu\psi,
 \nonumber \\
 \delta\bar F &=&
 \bar\epsilon(i\gamma^\mu D_\mu\bar\psi-i\bar\psi\sigma+i\bar\phi\bar\lambda)
 +\tfrac i3(2q-1)D_\mu\bar\epsilon\gamma^\mu\bar\psi.
\label{dncm}
\end{eqnarray}
Here we assumed the fields $\phi,\psi,F$ ($\bar\phi,\bar\psi,\bar F$) to be
column vectors (resp. row vectors) on which the vector multiplet fields
act as matrices from the left (right).
The lowest components $(\phi,\bar\phi)$ are assigned the dimension $q$
and R-charge $(-q,+q)$, as one can obtain from the supersymmetry
algebra realized on these fields.
The supersymmetry algebra can be easily shown to close off-shell,
except that two unbarred supersymmetries do not simply commute on $F$,
\begin{equation}
 [\delta_\epsilon,\delta_{\epsilon'}]F ~=~
 \epsilon\gamma^{\mu\nu}\epsilon'(2D_\mu D_\nu\phi+iF_{\mu\nu}\phi)
+\tfrac{2q}3\phi\cdot
 (\epsilon \gamma^\mu\gamma^\nu D_\mu D_\nu\epsilon'
 -\epsilon'\gamma^\mu\gamma^\nu D_\mu D_\nu\epsilon).
\end{equation}
The right hand side does not vanish for general Killing spinors
$\epsilon,\epsilon'$. With an additional condition (\ref{kseq2})
one finds
\begin{equation}
 [\delta_\epsilon,\delta_{\epsilon'}]F ~=~
 \epsilon\gamma^{\mu\nu}\epsilon'
 (2D_\mu D_\nu\phi+iF_{\mu\nu}\phi-iq\phi V_{\mu\nu}).
\end{equation}
The right hand side vanishes if $\phi$ couples to $V_\mu$ according
to its R-charge $-q$, namely,
\begin{equation}
 D_\mu\phi\equiv (\partial_\mu-iA_\mu+iqV_\mu)\phi.
\end{equation}
Likewise, two barred supersymmetry commute on $\bar F$ provided that
the Killing spinors satisfy (\ref{kseq2}) and $\bar\phi$ couples
with $V_\mu$ according to its charge $+q$.

\paragraph{Supersymmetric Lagrangians.}

The Chern-Simons Lagrangian for ${\cal N}=2$ vector multiplet is invariant
under supersymmetry.
\begin{equation}
 {\cal L}_{\rm CS} ~=~ {\rm Tr}\left[
   \tfrac1{\sqrt g}\varepsilon^{\mu\nu\lambda}
  (A_\mu\partial_\nu A_\lambda-\tfrac{2i}3A_\mu A_\nu A_\lambda)
   -\bar\lambda\lambda+2D\sigma  \right].
\end{equation}
The $F$-term of gauge-invariant chiral multiplets of R-charge $q=2$
is invariant under supersymmetry up to total derivatives.
\begin{equation}
 \delta F ~=~ iD_\mu(\epsilon\gamma^\mu\psi),\quad
 \delta\bar F ~=~ iD_\mu(\bar\epsilon\gamma^\mu\bar\psi).
\end{equation}
These terms are invariant under $\delta$ for any Killing spinors
$\epsilon,\bar\epsilon$.
In addition, chiral matter multiplets with canonical dimensions have
the kinetic Lagrangian,
\begin{eqnarray}
 {\cal L} &=& D_\mu\bar\phi D^\mu\phi -i\bar\psi\gamma^\mu D_\mu\psi
 +\tfrac R8\bar\phi\phi+i\bar\psi\sigma\psi
 \nonumber \\ &&
 +i\bar\psi\lambda\phi-i\bar\phi\bar\lambda\psi
 +i\bar\phi D\phi+\bar\phi\sigma^2\phi+\bar FF,
\label{Lcm}
\end{eqnarray}
which is invariant under supersymmetry if the Killing spinors
$\epsilon,\bar\epsilon$ satisfy (\ref{kseq2}).

There are Lagrangians which are not superconformal but are
still invariant under some supersymmetry. In the following we look for
the quantities which are invariant if the parameters
$\epsilon,\bar\epsilon$ satisfy
\begin{equation}
 D_\mu\epsilon=\tfrac{i}{2f}\gamma_\mu\epsilon,\quad
 D_\mu\bar\epsilon=\tfrac{i}{2f}\gamma_\mu\bar\epsilon
\label{nscd}
\end{equation}
for some function $f$.
Note that by combining these with (\ref{kseq2}) one finds
\begin{eqnarray}
 (R+2i\gamma^{\mu\nu}V_{\mu\nu})\epsilon &=&
 (6f^{-2}-4i\gamma^\mu\partial_\mu f^{-1})\epsilon,\nonumber\\
 (R-2i\gamma^{\mu\nu}V_{\mu\nu})\bar\epsilon &=&
 (6f^{-2}-4i\gamma^\mu\partial_\mu f^{-1})\bar\epsilon.
\end{eqnarray}
One example for such Lagrangian is the kinetic Lagrangian
for matter fields with non-canonical R-charges.
\begin{eqnarray}
 {\cal L}_\text{mat} &=&
  D_\mu\bar\phi D^\mu\phi
 +\bar\phi\sigma^2\phi
 +\tfrac{i(2q-1)}f\bar\phi\sigma\phi
 -\tfrac{q(2q-1)}{2f^2}\bar\phi\phi
 +\tfrac q4R\bar\phi\phi
 +i\bar\phi D\phi
 +\bar FF
 \nonumber \\ &&
 -i\bar\psi\gamma^\mu D_\mu\psi
 +i\bar\psi\sigma\psi
 -\tfrac{(2q-1)}{2f}\bar\psi\psi
 +i\bar\psi\lambda\phi
 -i\bar\phi\bar\lambda\psi.
\label{Lncm}
\end{eqnarray}
Another example is the Yang-Mills Lagrangian for vector multiplet.
\begin{eqnarray}
 {\cal L}_\text{YM} &=& \text{Tr}\Big(
  \tfrac14F_{\mu\nu}F^{\mu\nu}+\tfrac12D_\mu\sigma D^\mu\sigma
 +\tfrac12(D+\tfrac\sigma f)^2
 \nonumber \\ && ~~~
 +\tfrac i2\bar\lambda\gamma^\mu D_\mu\lambda
 +\tfrac i2\bar\lambda[\sigma,\lambda]
 -\tfrac1{4f}\bar\lambda\lambda
 \Big).
\end{eqnarray}
Note that ${\cal L}_\text{mat}$ and ${\cal L}_\text{YM}$ can be
expressed as total-superderivatives,
\begin{eqnarray}
 \bar\epsilon\epsilon\cdot{\cal L}_\text{mat} &=&
 \delta_{\bar\epsilon}\delta_\epsilon\Big(
  \bar\psi\psi-2i\bar\phi\sigma\phi+\tfrac{2(q-1)}{f}\bar\phi\phi
 \Big),
 \nonumber \\
 \bar\epsilon\epsilon\cdot{\cal L}_\text{YM} &=&
 \delta_{\bar\epsilon}\delta_\epsilon\text{Tr}\Big(
  \tfrac12\bar\lambda\lambda-2D\sigma \Big).
\label{Ltot}
\end{eqnarray}
Finally, there is an analogue of FI D-term for abelian vector multiplet.
The auxiliary field $D$ in abelian vector multiplet is also invariant
up to total derivative,
\begin{equation}
 {\cal L}_\text{FI}\equiv D-\tfrac\sigma f,\quad
 \delta {\cal L}_\text{FI} =
 -\tfrac i2D_\mu(\bar\epsilon\gamma^\mu\lambda
                       +\bar\lambda\gamma^\mu\epsilon).
\end{equation}

\section{Partition function on Squashed $\bf S^3$ (Familiar One)}
\label{sec:partition1}

Here we compute partition functions of supersymmetric gauge theories
on squashed $S^3$ based on localization principle. As has been explained
in \cite{KWY1,KWY2}, the path integral localizes onto the saddle points
where the supersymmetry variation of all the fermions vanish.
They are characterized by
\begin{equation}
 A_\mu = \phi= 0,\quad
 \sigma=-f D= \text{constant}.
\end{equation}
The integration over all the modes transverse to the locus of saddle
points can be made finite by introducing arbitrary weight $e^{-S}$,
where the regulator action $S$ should be supersymmery exact. Our
Lagrangians ${\cal L}_\text{mat}$ and ${\cal L}_\text{YM}$ are both
supersymmetry exact so they can be included in $S$ with arbitrary
coefficient. When this coefficient is taken larger and larger, then the
saddle-point (Gaussian) approximation for the path integral becomes more
and more accurate. The partition function can therefore be computed
by truncating the regulator action up to quadratic order at each saddle
point, evaluating the Gaussian path integral (one-loop determinant) and
then integrating over the space of saddle points parametrized by $\sigma$.

In the following we calculate these determinants for chiral and vector
multiplets on two different versions of squashed $S^3$.
In this section we focus on the familiar squashing preserving
$SU(2)_\mathscr{L}\times U(1)_{\mathscr{R}^3}$ symmetry, for which the
metric is given by (\ref{fssgmn}) and the background $U(1)$ gauge field
$V$ is given in (\ref{nks1}).

\paragraph{Matter multiplets.}

Let us evaluate the determinant from matter chiral multiplets first. For
simplicity, we focus on the simplest example of a single chiral
multiplet of R-charge $q$ coupled to an abelian vector multiplet. The
matter kinetic term on the saddle points is ${\cal L}_\phi+{\cal L}_\psi$, with
\begin{eqnarray}
 {\cal L}_\phi &=&
 g^{\mu\nu}D_\mu\bar\phi D_\nu\phi +\bar\phi\sigma^2\phi
 +\tfrac{2i(q-1)}f\bar\phi\sigma\phi
 +\Big\{\tfrac{q(1-2q)}{2f^2}+\tfrac{qR}{4}\Big\}\bar\phi\phi,
 \nonumber \\
 {\cal L}_\psi &=&
 -i\bar\psi\gamma^\mu D_\mu\psi
 +i\bar\psi\sigma\psi
 -\tfrac{2q-1}{2f}\bar\psi\psi.
\end{eqnarray}
We rewrite them using the differential operators $\mathscr{R}^a$
as well as $R=\frac8{f\tilde\ell}-\frac2{f^2}$ and get
\begin{eqnarray}
 {\cal L}_\phi &=&
 \tfrac1{\ell^2}
 \Big( \mathscr{R}^1\bar\phi\mathscr{R}^1\phi
      +\mathscr{R}^2\bar\phi\mathscr{R}^2\phi \Big)
+\tfrac1{\tilde\ell^2}
 \Big\{\mathscr{R}^3\bar\phi-iq(1-\tfrac{\tilde\ell}f)\bar\phi\Big\}
 \Big\{\mathscr{R}^3\phi+iq(1-\tfrac{\tilde\ell}f)\phi\Big\}
 \nonumber \\&&
 +\bar\phi\Big(\sigma^2+\tfrac{2i(q-1)\sigma}f-\tfrac{q^2}{f^2}
 +\tfrac{2q}{f\tilde\ell}\Big)\phi,
 \nonumber \\
 {\cal L}_\psi &=&
 \bar\psi\left\{
  -\tfrac i\ell(\gamma^1\mathscr R^1+\gamma^2\mathscr R^2)
  -\tfrac i{\tilde\ell}\gamma^3\mathscr R^3
  +\tfrac{1-q}{f}+\tfrac1{\tilde\ell}
  +(q-1)(\tfrac1{\tilde\ell}-\tfrac1f)\gamma^3
  +i\sigma
 \right\}\psi.
\end{eqnarray}
We further rewrite them in terms of $J^a=\frac1{2i}\mathscr{R}^a$ and
$S^a=\frac12\gamma^a$ satisfying standard $SU(2)$ commutation relations,
and obtain the Laplace operator $\Delta_\phi$ for scalar field and
the Dirac operator $\Delta_\psi$ for spinor field,
\begin{eqnarray}
 \Delta_\phi
 &=&
  \frac4{\ell^2}(J^1J^1+J^2J^2)
 +\frac4{\tilde\ell^2}\Big\{J^3+\frac q2(1-\frac{\tilde\ell}f)\Big\}^2
 +\sigma^2+\frac{2i(q-1)\sigma}f-\frac{q^2}{f^2}
 +\frac{2q}{f\tilde\ell},
 \nonumber \\
 \Delta_\psi &=&
 \frac4\ell(S^1J^1+S^2J^2)+\frac4{\tilde\ell}S^3J^3
 +\frac1{\tilde\ell}+\frac{1-q}f+2(q-1)\Big(\frac1{\tilde\ell}-\frac1f\Big)S^3.
\end{eqnarray}
The one-loop determinant can thus be computed from the spectrum of these
operators.

The Laplace operator is diagonalized by scalar spherical harmonics which
belong to the representations $(j,j)$ of
$SU(2)_\mathscr{R}\times SU(2)_\mathscr{L}$. There are therefore $2j+1$
scalar wave functions with angular momentum $j$ and $J^3=m$ corresponding to
the eigenvalue
\begin{eqnarray}
 \Delta_\phi
 &=&
  \frac{4j(j+1)-4m^2}{\ell^2}
 +\frac{\{2m+q(1-\frac{\tilde\ell}f)\}^2}{\tilde\ell^2}
 +\sigma^2+\frac{2i(q-1)\sigma}f-\frac{q^2}{f^2}
 +\frac{2q}{f\tilde\ell},
\label{Dphi1}
\end{eqnarray}
Note that for $m=j$ this can be factorized as follows,
\begin{equation}
\Delta_\phi =
  \left(\frac{2j+q}{\tilde\ell}+\frac{2-2q}{f}+i\sigma\right)
  \left(\frac{2j+q}{\tilde\ell}-i\sigma\right).
\end{equation}

Next we turn to the spectrum of Dirac operator $\Delta_\psi$.
Its generic eigenstates are suitable linear combinations of two states
$\ket{j;m,\frac12}$ and $\ket{j;m+1,-\frac12}$, where $j$ labels the
orbital angular momentum and
\[
 \begin{array}{rcr}
 J^3\ket{j;m,s}&=& m\ket{j;m,s}, \\
 S^3\ket{j;m,s}&=& s\ket{j;m,s},
 \end{array}\quad
 \begin{array}{lcl}
 (J^1\pm iJ^2)\ket{j;m,s}&=& (j\mp m)\ket{j;m\pm1,s}, \\
 (S^1\pm iS^2)\ket{j;m,\mp\tfrac12}&=& \ket{j;m,\pm\tfrac12}.
 \end{array}
\]
The action of $\Delta_\psi$ on the states of the form
\begin{equation}
 x_+\ket{j;m,\tfrac12}+x_-\ket{j,m+1,-\tfrac12},
\end{equation}
can be translated into the following $2\times2$ matrix acting on $(x_+,x_-)^t$.
\[
 \left(\begin{array}{cc}
  \frac{2m+1}{\tilde\ell}+(q-1)(\frac1{\tilde\ell}-\frac1f)
 +\frac{1-q}f+i\sigma &
 \frac2\ell(j+m+1) \\ \frac2\ell(j-m) &
 -\frac{2m+1}{\tilde\ell}-(q-1)(\frac1{\tilde\ell}-\frac1f)
 +\frac{1-q}f+i\sigma
 \end{array}\right).
\]
Its determinant is precisely $(-1)$ times the expression (\ref{Dphi1})
for the eigenvalue of $\Delta_\phi$. Note that $m$ can take values
between $-j$ to $j-1$. In addition to these generic eigenstates, the
states $\ket{j,\pm j,\pm\frac12}$ are themselves the eigenstates of
$\Delta_\psi$ for the eigenvalues
\begin{equation}
 \Delta_\psi~=~
 \frac{2j+q}{\tilde\ell}+\frac{2-2q}f+i\sigma,\quad
 \frac{2j+2-q}{\tilde\ell}+i\sigma.
\end{equation}
Since $SU(2)_\mathscr{L}$ is unbroken, all these spectra aquire
the multiplicity $(2j+1)$.

Combining everything together, we obtain the one-loop determinant,
\begin{eqnarray}
 \frac{\text{det}\Delta_\psi}{\text{det}\Delta_\phi} &=&
 \prod_{2j\in\mathbb{Z}_{\ge0}}
 \left[
 (-1)^{2j}
 \frac{
  (\frac{2j+q}{\tilde\ell}+\frac{2-2q}f+i\sigma)
  (\frac{2j+2-q}{\tilde\ell}+i\sigma)}{
  (\frac{2j+q}{\tilde\ell}+\frac{2-2q}f+i\sigma)
  (\frac{2j+q}{\tilde\ell}-i\sigma) }
 \right]^{2j+1}
 \nonumber \\ &=&
 \prod_{n>0}
 \left(\frac{n+1-q+i\tilde\ell\sigma}{n-1+q-i\tilde\ell\sigma}\right)^n
 ~=~ s_{b=1}(i-iq-\tilde\ell\sigma)\,.
\label{detmat1}
\end{eqnarray}
This is essentially the same as the result for round $S^3$ (see, e.g.
\cite{HHL}) except that the radius of round $S^3$ is replaced by
$\tilde\ell$.

\paragraph{Vector multiplets.}

Next we study vector multiplets. We denote by $\varphi$ the fluctuation
mode of the scalar field away from its classical value $\sigma$, and
consider the path integral with linearized Lagrangian
${\cal L}={\cal L}_B+{\cal L}_F$,
\begin{eqnarray}
 {\cal L}_B &=&
 \text{Tr}\Big(\tfrac14
 \hat F_{\mu\nu}\hat F^{\mu\nu}
 +\tfrac12\partial_\mu\varphi\partial^\mu\varphi
 -\tfrac12[A_\mu,\sigma][A^\mu,\sigma]
 -i[A_\mu,\sigma]\partial^\mu\varphi\Big),
 \nonumber \\
 {\cal L}_F &=&
 \text{Tr}\Big(
  \tfrac i2\bar\lambda\gamma^\mu D_\mu\lambda
 +\tfrac i2\bar\lambda[\sigma,\lambda]
 -\tfrac1{4f}\bar\lambda\lambda  \Big),
\label{Llin}
\end{eqnarray}
where $\hat F_{\mu\nu}\equiv\partial_\mu A_\nu-\partial_\nu A_\mu$.
We decompose all the adjoint-valued fields with respect to the
Cartan-Weyl basis $(H_i,E_\alpha,E_{-\alpha})$ satisfying the
commutation relations,
\[
 [H_i,H_j]=0,\quad
 [H_i,E_\alpha]=\alpha_i E_\alpha,\quad
 [E_\alpha,E_{-\alpha}]=\tfrac2{|\alpha|^2}\alpha_iH_i.
\]
We may assume that $\sigma$ takes values in the Cartan subalgebra, i.e.
$\sigma=\sigma_iH_i$. Then the Lagrangian for fermions can be rewritten
into the form
\begin{equation}
 {\cal L}_F~=~
 \sum_i\bar\lambda_i(i\gamma^\mu D_\mu-\tfrac1{2f})\lambda_i
+\sum_{\alpha\in\Delta}
 \bar\lambda_{-\alpha}(i\gamma^\mu D_\mu+i\sigma\alpha-\tfrac1{2f})
 \lambda_\alpha,
\end{equation}
where $\sigma\alpha\equiv\sigma_i\alpha_i$. Note that $D_\mu\lambda$
contains the couplings to spin connection as well as the background
$U(1)$ vector field according to the R-charge of $\lambda$. The Dirac
operator for these fermions is thus the same as that of matter fermions
in a chiral multiplet with $q=0$. The determinant of Dirac operator for
the fields $\lambda_{\pm\alpha},\bar\lambda_{\pm\alpha}$ is thus given by
\begin{eqnarray}
&&
 \prod_{j\ge0}
\left[
 \Big(\tfrac{2j}{\tilde\ell}+\tfrac2f-i\sigma\alpha\Big)
 \Big(\tfrac{2j}{\tilde\ell}+\tfrac2f+i\sigma\alpha\Big)
 \Big(\tfrac{2j+2}{\tilde\ell}-i\sigma\alpha\Big)
 \Big(\tfrac{2j+2}{\tilde\ell}+i\sigma\alpha\Big)
\right]^{2j+1}
\nonumber \\ && \times
 \prod_{j\ge1/2}
 \prod_{m=-j}^{j-1}
 \left[ \Delta_0(j,m)+(\sigma\alpha)^2+\tfrac{2i}f\sigma\alpha\right]^{2j+1}
 \left[ \Delta_0(j,m)+(\sigma\alpha)^2-\tfrac{2i}f\sigma\alpha\right]^{2j+1},
\end{eqnarray}
where
\begin{equation}
\Delta_0(j,m)\equiv
  \frac{4j(j+1)-4m^2}{\ell^2}
 +\frac{4m^2}{\tilde\ell^2}.
\label{Delta0}
\end{equation}

The bosonic Lagrangian is easier to handle once written in terms
of differential forms,
\begin{equation}
 d^3\xi\sqrt{g}{\cal L}_B~=~
 \tfrac12\text{Tr}\Big(dA\wedge\ast dA+d\varphi\wedge\ast d\varphi
 -[\sigma,A]\wedge[\sigma,\ast A]+i[\sigma,A]\wedge\ast d\varphi\Big).
\end{equation}
This leads to the Laplace operator acting on a pair $(A,\varphi)$
as follows,
\begin{equation}
 \Delta_B~:~\left(\begin{array}c A\\ \varphi\end{array}\right)~\mapsto
 \left(\begin{array}{l}
 \ast d\ast dA+[\sigma,[\sigma,A]]-i[\sigma,d\varphi]\\
 -i[\sigma,\ast d\ast A]-\ast d\ast d\varphi
 \end{array}\right).
\end{equation}
Here the Hodge star $\ast$ is a linear map from $d$-forms
to $(3-d)$-forms satisfying $\ast\ast=1$.
It is defined by the equations
\begin{equation}
\ast 1=e^1e^2e^3,\quad
\ast e^1=e^2e^3,\quad
\ast e^2=e^3e^1,\quad
\ast e^3=e^1e^2.
\end{equation}
Let us hereafter focus on the components of $(A,\varphi)$ proportional to
the Lie algebra element $E_\alpha$, so that the commutator with $\sigma$
becomes just the multiplication by $\sigma\alpha$.

Generic eigenmodes of $\Delta_B$ take the form
\begin{equation}
 A_\alpha = x^+Y_{j,m+1}e^- + x^-Y_{j,m-1}e^+ + x^3Y_{j,m}e^3,\quad
 \varphi_\alpha = x Y_{j,m},
\end{equation}
where $e^\pm\equiv\frac12(e^1\pm ie^2)$, and the spherical
harmonics $Y_{j,m}$ is normalized to satisfy
\[
 J^3Y_{j,m}=mY_{j,m},\quad
 (J^1\pm iJ^2)Y_{j,m}=(j\mp m)Y_{j,m\pm1}.
\]
The Laplace operator $\Delta_B$ then translates into a
matrix $X$ acting on $(x^+,x^3,x^-,x)^t$.
It takes the $(3+1)\times(3+1)$ block decomposed form
\begin{equation}
 X~=~
 \left(\begin{array}{cc}U^2+(\sigma\alpha)^2 &
 ~~(\sigma\alpha)\vec v \\
   (\sigma\alpha)\vec w & \Delta_0
\end{array}\right),
\end{equation}
with
\begin{eqnarray}
 U&=& \left(\begin{array}{ccc}
 \displaystyle -\frac{2m}{\tilde\ell} &
 \displaystyle \frac{2(j-m)}\ell & 0 \\
 \displaystyle \frac{j+m+1}\ell &
 \displaystyle \frac2f &
 \displaystyle -\frac{j-m+1}\ell \\
 0 &
 \displaystyle -\frac{2(j+m)}\ell &
 \displaystyle \frac{2m}{\tilde\ell}
 \end{array}\right),
 \nonumber \\
 \vec v &=&
 \left(\frac{2(j-m)}\ell,\frac {2m}{\tilde\ell},\frac{2(j+m)}\ell\right)^t,
 \nonumber \\
 \vec w &=&
 \left(\frac{j+m+1}\ell,\frac{2m}{\tilde\ell},\frac{j-m+1}\ell\right),
\end{eqnarray}
and $\Delta_0$ given in (\ref{Delta0}).
Noticing that $U\vec v=\vec wU=0$ and $\Delta_0=\vec w\cdot\vec v$,
one can easily find two eigenvectors of $X$,
\begin{eqnarray}
 \big(\vec v,-(\sigma\alpha)\big)^t&~:~&X=0,\nonumber \\
 \big((\sigma\alpha)\vec v,\Delta_0\big)^t&~:~&X=\Delta_0+(\sigma\alpha)^2.
\end{eqnarray}
These two modes are longitudinal in the sense that $A$ is a total derivative.
The above two eigenvalues do not enter the formula for one-loop
determinant, as we explain later. The other two eigenvalues of $X$
correspond to transverse modes, for which $\varphi\equiv0$ and $A$ is
divergenceless. They have eigenvalues  $X=\zeta_1^2+(\sigma\alpha)^2$
and $\zeta_2^2+(\sigma\alpha)^2$, where $\zeta_1,\zeta_2$ are the
two nonzero eigenvalues of $U$. They can therefore be found as two
nonzero roots of the characteristic equation
\begin{equation}
 0~=~ \text{Det}(\zeta-U)~=~\zeta^3-\frac2f\zeta^2-\Delta_0\zeta.
\end{equation}
The product of the eigenvalues of $X$ from the two transverse modes is
\begin{equation}
 \big\{\zeta_1^2+(\sigma\alpha)^2\big\}\big\{\zeta_2^2+(\sigma\alpha)^2\big\}
 ~=~\big\{\Delta_0-\tfrac{2i}f(\sigma\alpha)+(\sigma\alpha)^2\big\}
    \big\{\Delta_0+\tfrac{2i}f(\sigma\alpha)+(\sigma\alpha)^2\big\}.
\end{equation}

Up to now it was assumed that $|m|\le j-1$ and therefore $j\ge1$.
When $m=j$ and $j\ge1/2$, one is interested in the eigenmodes for which
$A$ has no component proportional to $e^-$, so that now $X$
becomes a $3\times 3$ matrix. Its three eigenvalues are
\begin{equation}
 0,\quad\Delta_0+(\sigma\alpha)^2,\quad
 \Big(\tfrac2f+\tfrac{2j}{\tilde\ell}\Big)^2+(\sigma\alpha)^2.
\end{equation}
Similarly, when $m=-j$ and $j\ge1/2$ the matrix $X$ becomes $3\times 3$,
and obtains the same three eigenvalues as above.
When $m=\pm(j+1)$ and $j\ge1/2$ one finds that $X$ becomes one-dimensional,
\begin{equation}
 X~=~\Big(\tfrac{2j+2}{\tilde\ell}\Big)^2+(\sigma\alpha)^2.
\end{equation}
Finally, for $j=0$ the four eigenmodes and eigenvalues of $\Delta_B$
are given by
\begin{eqnarray}
 (A=e^\pm,~\varphi=0) && \Delta_B=\tfrac4{\tilde\ell^2}+(\sigma\alpha)^2,
 \nonumber \\
 (A=e^3,~\varphi=0) && \Delta_B=\tfrac4{f^2}+(\sigma\alpha)^2,
 \nonumber \\
 (A=0,~\varphi=1) && \Delta_B=0.
\end{eqnarray}

Almost all the zero eigenvalues of $\Delta_B$ correspond to gauge
symmetry, so they should be excluded from the physical determinant. The
only exception is the constant mode of $\varphi$, which correspond to
shifting the saddle point and therefore should also be excluded. To
evaluate the partition function correctly, one also has to take proper
account of Faddeev-Popov determinant. Instead of introducing ghost
fields and modifying the supersymmetry by BRST transformation, we choose
to take a quicker route which takes advantage of saddle point approximation.

Now that we worked out the spectrum and eigenmode decomposition of
the operator $\Delta_B$, one can regard the path integral (before gauge
fixing) as an integral over the mode variables $x_i$ and $\tilde x_j$
corresponding to zero and nonzero eigenmodes of $\Delta_B$,
\begin{equation}
 DAD\varphi~=~
  \prod_{(\Delta_B=0)}\!\!dx_i~\times
  \prod_{(\Delta_B\ne0)}\!\!d\tilde x_j.
\end{equation}
We assume the mode variables are normalized to satisfy
\begin{equation}
\frac12\int\text{Tr}\big(
 A\wedge\ast A+\varphi\wedge\ast\varphi\big)~=~
 \pi\sum_ix_i^2
+\pi\sum_j\tilde x_j^2,
\label{normmode}
\end{equation}
so that the following equality can be reproduced using variables
$x_i,\tilde x_j$,
\begin{equation}
 \int DAD\varphi\exp\left(-\frac12\int\text{Tr}\big(
 A\wedge\ast A+\varphi\wedge\ast\varphi\big)\right)~=~1.
\label{mes}
\end{equation}
Excluding the zero eigenvalues from the determinant of $\Delta_B$
corresponds to the insertion of $\prod\delta(x_i)$. But the correct gauge
fixing is given by $\prod\delta(\omega_i)={\cal J}\prod\delta(x_i)$, where
$\omega_i$ are the mode variables of gauge transformation which acts on
the fields $A$ and $\varphi$ as
\begin{equation}
 \delta_\omega A ~=~ d\omega,
 \quad
 \delta_\omega\varphi ~=~i[\omega,\sigma]
\label{domega}
\end{equation}
on the saddle point labelled by $\sigma$. The variables $\omega_i$ also
satisfy the normalization condition similar to (\ref{normmode}). The
Jacobian ${\cal J}$ for the change of variables is called Faddeev-Popov
determinant. Noticing $\prod dx_i={\cal J}D\omega$, one can determine
${\cal J}$ by inserting (\ref{domega}) into (\ref{mes}),
\begin{equation}
 1~=~ {\cal J}\cdot
 \int D'\omega\exp\left(-\frac12\int d^3\xi\sqrt{g}\;\text{Tr}\big(
 \partial_\mu\omega\partial^\mu\omega-[\sigma,\omega]^2\big)\right).
\end{equation}
Note that the constant mode should be excluded from the measure
$D'\omega$ as explained in the previous paragraph. This path integral
can be easily worked out using spherical harmonics. The mode $\omega\sim
Y_{j,m}E_\alpha$ is an eigenmode of the Laplace operator with the eigenvalue
\begin{equation}
 \Delta_0(j,m)+(\sigma\alpha)^2.
\end{equation}
This precisely cancels with the contribution of longitudinal vector
eigenmode to the determinant. The path integral over bosons
$A_{\pm\alpha},\varphi_{\pm\alpha}$ modulo gauge equivalence finally becomes
\begin{eqnarray}
&&
 \Big(\tfrac4{f^2}+(\sigma\alpha)^2\Big)^{-1}
 \prod_{j\ge1/2}
 \Big[\big(\tfrac2f+\tfrac{2j}{\tilde\ell}\big)^2+(\sigma\alpha)^2\Big]^{-2(2j+1)}
 \prod_{j\ge0}
 \Big[\big(\tfrac{2j+2}{\tilde\ell}\big)^2+(\sigma\alpha)^2\Big]^{-2(2j+1)}
 \nonumber \\ &&\times
 \prod_{j\ge1}\prod_{m=1-j}^{j-1}
 \Big(\Delta_0(j,m)-\tfrac{2i}f\sigma\alpha+(\sigma\alpha)^2\Big)^{-2j-1}
 \Big(\Delta_0(j,m)+\tfrac{2i}f\sigma\alpha+(\sigma\alpha)^2\Big)^{-2j-1}.
\end{eqnarray}
where $j$ runs over half-integers.

Combining the bosonic and fermionic contributions together, we obtain
the one-loop determinant of vector multiplet.
\begin{eqnarray}
 \prod_{\alpha\in\Delta_+}\prod_{n>0}
 \Big[\frac{n^2}{\tilde\ell^2}+(\sigma\alpha)^2\Big]^2
 ~\sim~  \prod_{\alpha\in\Delta_+}
 \Big(\frac{\sinh(\pi\tilde\ell\sigma\alpha)}{\pi\tilde\ell\sigma\alpha}\Big)^2.
\end{eqnarray}
Here the product runs over all positive roots $\alpha$. The factors in the
denominator cancel against the Vandermonde determinant which arises when
the integration over the Lie algebra is reduced to that on
Cartan subalgebra. Again, the result is essentially the same as for
round $S^3$ \cite{KWY1} except now $\tilde\ell$ is playing the role of the
radius of round $S^3$.

\vskip2mm

Thus, after all these tedious computations, we found a rather
disappointing result that the familiar squashing of $S^3$ with
$SU(2)_{\mathscr{L}}\times U(1)_{\mathscr{R}^3}$ gives nothing new.
We notice here that there is a simple reason for this. Since our
squashed $S^3$ has an unbroken $SU(2)$ symmetry, the eigenmodes of
Laplace or Dirac operators naturally form multiplets with the same
eigenvalues. The exponent $n$ in the formula (\ref{detmat1}) simply
reflects the fact that the multiplicity becomes larger as the angular
momentum $j$ gets larger. This is tied to the degeneration of zeroes and
poles of $s_b(x)$ for $b=1$. Therefore, in order to find generalization
to $b\neq1$, we need to look for a less symmetric squashing of $S^3$.

\section{Partition function on Squashed $\bf S^3$ (Less Familiar One)}
\label{sec:partition2}

In this section we study the partition function on less familiar
version of squashed $S^3$ which preserves only
$U(1)_{\mathscr{L}^3}\times U(1)_{\mathscr{R}^3}$ symmetry.
The vielbein, spin connection, Killing spinors and the background
$U(1)$ gauge field $V$ are summarized in section \ref{sec:three-sphere}.
In the following discussion we regard the Killing spinors $\epsilon,
\bar\epsilon$ to be Grassmann-even. For convenience, we also renormalize
them by constants so that they satisfy
\begin{equation}
 \bar\epsilon\epsilon=1,\quad
 v^\mu v_\mu=
 \bar\epsilon\gamma^\mu\epsilon\cdot
 \bar\epsilon\gamma_\mu\epsilon=1.
\end{equation}
More explicitly, we choose
\begin{equation}
 \epsilon=\frac1{\sqrt2}\left(\begin{array}{r}
  -e^{\frac i2(\chi-\varphi+\theta)}\\
   e^{\frac i2(\chi-\varphi-\theta)}\end{array} \right),\quad
 \bar\epsilon=\frac1{\sqrt2}\left(\begin{array}{r}
   e^{\frac i2(-\chi+\varphi+\theta)}\\
   e^{\frac i2(-\chi+\varphi-\theta)}\end{array} \right),
\end{equation}
so that
\begin{eqnarray}
 \bar\epsilon\gamma^a\epsilon &=&
 (-\cos\theta,\sin\theta,0), \nonumber \\
 \epsilon\gamma^a\epsilon &=&
 (i\sin\theta,i\cos\theta,+1)e^{i(\chi-\varphi)}, \nonumber \\
 \bar\epsilon\gamma^a\bar\epsilon &=&
 (i\sin\theta,i\cos\theta,-1)e^{-i(\chi-\varphi)}.
\end{eqnarray}
Note also the equalities
\begin{equation}
 v_\mu\bar\epsilon\gamma^\mu~=~\bar\epsilon,\quad
 \gamma^\mu\epsilon v_\mu=\epsilon,
\end{equation}
which will be frequently used in what follows.

Since the squashed $S^3$ of our interest here has a reduced symmetry and
the metric cannot be written in terms of LI or RI forms, it will be a
tedious task to work out all the eigenmodes of the bosonic and fermionic
kinetic operators. In fact, the precise form of most of the eigenmodes
and eivenvalues is irrelevant since, as we have seen in the previous
section, they give cancelling contributions to the determinant due to
the pairing of bosonic and fermionic modes by supersymmetry. Therefore,
to compute the one-loop determinant, it will be useful to understand how
this cancellation happens.

\paragraph{Matter multiplets.}

We study the matter fields first, focusing on the case with
a single chiral multiplet coupled to an abelian vector multiplet.
What we need is the spectrum of the kinetic operators $\Delta_\phi$ and
$\Delta_\psi$ for bosons and fermions. In this section we define them
from the regulator Lagrangian
\begin{eqnarray}
 {\cal L}_\text{reg}&=& \delta_{\bar\epsilon}\delta_\epsilon
 \Big(\bar\psi\psi-2i\bar\phi\sigma\phi\Big)
 \nonumber \\ &=&
  D_\mu\bar\phi D^\mu\phi
 +\tfrac{2i(q-1)}f v^\mu D_\mu\bar\phi\phi
 +\bar\phi\sigma^2\phi
 +i\bar\phi\big(\tfrac{\sigma}f+D\big)\phi
 +\tfrac{2q^2-3q}{2f^2}\bar\phi\phi
 +\tfrac q4R\bar\phi\phi
 \nonumber \\ &&
 -i\bar\psi\gamma^\mu D_\mu\psi
 +i\bar\psi\sigma\psi
 -\tfrac{1}{2f}\bar\psi\psi
 +\tfrac{q-1}f\bar\psi\gamma^\mu v_\mu\psi
 +i\bar\psi\lambda\phi
 -i\bar\phi\bar\lambda\psi
 +\bar FF.
\label{Lreg}
\end{eqnarray}
This leads to the kinetic operators
\begin{eqnarray}
 \Delta_\phi &=& -D_\mu D^\mu-\tfrac{2i(q-1)}fv^\mu D_\mu+\sigma^2
 +\tfrac{2q^2-3q}{2f^2}+\tfrac{qR}4,
\nonumber \\
 \Delta_\psi &=& -i\gamma^\mu D_\mu+i\sigma-\tfrac{1}{2f}
 +\tfrac{q-1}f\gamma^\mu v_\mu,
\end{eqnarray}
acting on scalars and spinors of R-charges $-q, 1-q$ respectively.
Note that in deriving $\Delta_\phi$ we used
\begin{equation}
D_\mu v^\mu=0,\quad
 v^\mu \partial_\mu f=0,
\end{equation}
which can be shown using Killing spinor equation.

Now, it is a tedious but straightforward computation to show the following.
First, if $\Psi$ be a spinor eigenmode for $\Delta_\psi=M$, then
$\bar\epsilon\Psi$ is a scalar eigenmode for $\Delta_\phi=M(M-2i\sigma)$.
Second, let $\Phi$ be a scalar eigenmode for $\Delta_\phi=M(M-2i\sigma)$.
Then if we define a pair of spinor wave functions as
\begin{equation}
 \Psi_1 = \epsilon\Phi,\quad
 \Psi_2 = i\gamma^\mu\epsilon D_\mu\Phi
 +i\epsilon\sigma\Phi-\tfrac qf\epsilon\Phi,
\label{btof}
\end{equation}
then $\Delta_\psi$ acts on them as follows.
\begin{equation}
 \left(\begin{array}c \Delta_\psi\Psi_1 \\ \Delta_\psi\Psi_2 \end{array}\right)
~=~\left(\begin{array}{cr}
 2i\sigma & -1 \\ -M(M-2i\sigma) & 0 \end{array}\right)
\left(\begin{array}c \Psi_1 \\ \Psi_2 \end{array}\right).
\end{equation}
The $2\times 2$ matrix on the right hand side has eigenvalues $M$
and $2i\sigma-M$. We thus found a pairing between a scalar eigenmode
with $\Delta_\phi=M(M-2i\sigma)$ and two spinor eigenmodes with
$\Delta_\psi=M,\;2i\sigma-M$. Any modes which take part in this pairing
can be neglected when computing the one-loop determinant.
Note that part of this pairing, namely $\Phi=\bar\epsilon\Psi$ and the
construction (\ref{btof}) of $\Psi_2$ from $\Phi$, could be guessed from
the supersymmetry transformation rules (\ref{dncm}).

Had we chosen the Lagrangian (\ref{Lncm}) for the regulator, we would
find that the relation between scalar and spinor eigenmodes contains
$f$ and therefore becomes coordinate-dependent. This can be traced
to the third term in the right hand side of the first equation in
(\ref{Ltot}). Since one may use arbitrary regulator Lagrangian
as long as it regulates the integral over all the modes transverse to
the saddle point locus, we may choose it so that the relation between
bosonic and fermionic eigenmodes becomes as simple as possible. This is
why we chose the Lagrangian (\ref{Lreg}).

Nontrivial contributions to one-loop determinant come from the modes
which do not fall into the multiplet structure explained in the last
paragraph. There are two types of such modes.

The first is unpaired spinor eigenmodes $\Psi$, which vanish
when contracted with $\bar\epsilon$ and therefore do not have scalar
partners. Such modes can be expressed as $\Psi=\bar\epsilon F$, where
$F$ is a scalar with R-charge $2-q$. The eigenmode equation
$\Delta_\psi\Psi=M\Psi$ can be rewritten into
\begin{eqnarray}
 \big(M+\tfrac{q-2}f-i\sigma\big)F &=&
 -i\cos\theta D_1F+i\sin\theta D_2F,
 \nonumber \\
 D_3F &=& +i\sin\theta D_1F+i\cos\theta D_2F,
\label{detF}
\end{eqnarray}
where we have used the explicit form of $\bar\epsilon$, and
\begin{eqnarray}
 D_1F &=& \frac1{\ell\cos\theta}
 \Big\{\partial_\varphi-\tfrac i2(q-2)(1-\tfrac\ell f)\Big\}F,
 \nonumber \\
 D_2F &=& \frac1{\tilde\ell\cos\theta}
 \Big\{\partial_\chi+\tfrac i2(q-2)(1-\tfrac{\tilde\ell}f)\Big\}F,
 \nonumber \\
 D_3F &=& \frac1f\partial_\theta F.
\end{eqnarray}
If we assume $F\sim e^{im\varphi-in\chi}$, then the first equation in
(\ref{detF}) determines the eigenvalue $M$,
\begin{equation}
 M~=~ i\sigma+\frac m\ell +\frac n{\tilde\ell}
 -\frac{q-2}2\Big(\frac1\ell+\frac1{\tilde\ell}\Big),
\end{equation}
while the second equation determines the $\theta$-dependence of $F$.
\begin{equation}
 \frac1f\partial_\theta F ~=~
 -\frac{\sin\theta}{\ell\cos\theta}
  \Big\{m-\tfrac{q-2}2(1-\tfrac\ell f)\Big\}F
 +\frac{\cos\theta}{\tilde\ell\sin\theta}
  \Big\{n-\tfrac{q-2}2(1-\tfrac{\tilde\ell}f)\Big\}F.
\label{detF2}
\end{equation}
The precise form of the solutions to this differential equation is not
important, but we need to know what values of $(m,n)$ leads to normalizable
eigenmodes.
We recall here that the squared norm of $F$ is defined by an integral
over $\theta\in[0,\pi/2]$ with a measure which is proportional to
$d\theta f(\theta)\sin\theta\cos\theta$. The solution to (\ref{detF2})
may develop singularities at the two ends of the integration domain.
The behavior of $F$ there is easily found to be
\begin{equation}
 F \sim \cos^m\theta\sin^n\theta.
 \quad(\theta\sim 0~~\text{or}~~\pi/2)
\end{equation}
Therefore the normalizability requires $m,n$ to be nonnegative.

The second is the missing spinor eigenmodes. This is the case where
the map (\ref{btof}) does not give two independent spinor eigenmodes
from one scalar eigenmode. Namely $\Psi_1,\Psi_2$ are proportional
to each other. Let us put $\Psi_2=M\Psi_1$, i.e.
\begin{equation}
 i\gamma^\mu\epsilon D_\mu\Phi+i\epsilon\sigma\Phi-\tfrac qf\epsilon\Phi
 ~=~ M\epsilon\Phi.
\label{detPhi}
\end{equation}
Then one can show that $\Psi_2=M\Psi_1$ is a spinor eigenmode for
$\Delta_\psi=2i\sigma-M$, and moreover $\Phi$ has the eigenvalue
$\Delta_\phi=M(M-2i\sigma)$. That is to say, a spinor eigenmode for
$\Delta_\psi=M$ is missing.
To work out the spectrum of missing eigenvalues, we rewrite
(\ref{detPhi}) into the following form,
\begin{eqnarray}
 \big(M+\tfrac qf-i\sigma\big)\Phi &=&
 -i\cos\theta D_1\Phi+i\sin\theta D_2\Phi,
 \nonumber \\
 D_3\Phi &=& -i\sin\theta D_1\Phi-i\cos\theta D_2\Phi.
\end{eqnarray}
The rest of the computation is the same as for unpaired spinor eigenmodes.
The missing spinor eigenvalues are thus given by
\begin{equation}
 M~=~ i\sigma-\frac m\ell -\frac n{\tilde\ell}
 -\frac q2\Big(\frac1\ell+\frac1{\tilde\ell}\Big),\quad
 m,n\ge0.
\end{equation}

The one-loop determinant is the product of all the unpaired spinor
eigenvalues divided by the product of all the missing spinor
eigenvalues. Ignoring the sign factors we find
\begin{equation}
 \frac{\text{det}\Delta_\psi}
      {\text{det}\Delta_\phi}
  ~=~ \prod_{m,n\ge0}
 \frac{\frac m\ell+\frac n{\tilde\ell}+i\sigma
       -\frac{q-2}2(\frac1\ell+\frac1{\tilde\ell})}
      {\frac m\ell+\frac n{\tilde\ell}-i\sigma
       +\frac q2(\frac1\ell+\frac1{\tilde\ell})}.
\end{equation}
Introducing
\begin{equation}
 b\equiv(\tilde\ell/\ell)^{\frac12},\quad
 Q\equiv b+b^{-1},\quad
 \hat\sigma\equiv (\ell\tilde\ell)^{\frac12}\sigma,
\label{Lipar}
\end{equation}
one can rewrite this using the double sine function $s_b(x)$
given in the Introduction,
\begin{equation}
 \frac{\text{det}\Delta_\psi}
      {\text{det}\Delta_\phi}
  ~=~ \prod_{m,n\ge0}
 \frac{mb+nb^{-1}+\frac Q2+i\hat\sigma+\frac Q2(1-q)}
      {mb+nb^{-1}+\frac Q2-i\hat\sigma-\frac Q2(1-q)}
 ~=~ s_b\big(\tfrac{iQ}2(1-q)-\hat\sigma\big).
\label{nicedet1}
\end{equation}
Now $b$ is determined from the shape of the background squashed sphere
and it can take any values, which is precisely what we have been seeking for!

A comment on ${\cal N}=4$ extended supersymmetry is in order. Recall a
${\cal N}=4$ hypermultiplet is a pair of ${\cal N}=2$ chiral multiplets
$(q,\tilde q)$ with R-charge $1/2$ and opposite gauge charges. Its
one-loop determinant becomes very simple on round sphere ($b=1$),
\begin{equation}
 s_b(\tfrac{iQ}4-\hat\sigma)s_b(\tfrac{iQ}4+\hat\sigma)\Big|_{b=1}
~=~
 s_1(\tfrac i2-\hat\sigma)s_1(\tfrac i2+\hat\sigma)
~=~\frac1{2\cosh\pi\hat\sigma}.
\end{equation}
This follows from an identity of double sine function,
\begin{equation}
 s_b(\tfrac{ib}2-x)s_b(\tfrac{ib}2+x)~=~\frac{1}{2\cosh\pi bx}.
\end{equation}
However, this simplification does not happen on squashed sphere, unless
one turns on a mass deformation by gauging the global symmetry under
which $q,\tilde q$ have the same charge. Therefore, matrix models on
the squashed $S^3$ will be much more complicated than on round $S^3$.

\paragraph{Vector multiplets.}

Next we study the vector multiplets. As in the previous section we denote
by $\varphi$ the fluctuation mode of the scalar field away from its
classical value $\sigma$, and use the original Lagrangian for the
regulator which becomes (\ref{Llin}) after Gaussian approximation.
We also decompose all the adjoint fields into Cartan-Weyl basis.
The modes proportional to the root $\alpha$ get a mass $\sigma\alpha$,
and we denote them with suffix $\alpha$. We thus need to find out what
kind of multiplet structure is formed by the bosonic eigenmode of
$(A_\alpha,\varphi_\alpha)$ and the fermionic eigenmodes of
$\lambda_\alpha$.

We saw in the previous section that, for generic quantum number of
$SU(2)_\mathscr{L}\times SU(2)_{\mathcal{R}}$, the four bosonic
eigenmodes for $(A_\alpha,\varphi_\alpha)$ split into two transverse
modes with $dA_\alpha=0$ and two longitudinal modes with
$A_\alpha \sim d\varphi_\alpha$. After fixing a gauge and combining
with the volume of the gauge group, the longitudinal modes were shown
to yield no net contribution to the one-loop determinant. This property
can be shown to be independent of the 3D metric. Our problem is therefore how
the remaining two transverse modes are paired with the spinor eigenmodes.

The eigenmodes are generically paired in the following manner.
Let $\Lambda$ and $\mathscr{A}$ be spinor and transverse vector
eigenmodes for the eigenvalue $M$ satisfying
\begin{eqnarray}
 M\Lambda &=& \big(i\gamma^\mu D_\mu+i\sigma\alpha-\tfrac1{2f}\big)\Lambda,
\label{MLambda}
 \\
 M\mathscr{A} &=&i\sigma\alpha\mathscr{A}-\ast d\mathscr{A}.
\label{MA}
\end{eqnarray}
Then, up to constant multiplication, they are mapped to each other by
\begin{equation}
 \mathscr{A}\equiv
 d(\bar\epsilon\Lambda)+(iM+\sigma\alpha)
 \bar\epsilon\gamma_\mu\Lambda d\xi^\mu,
 \quad
 \Lambda\equiv \gamma^\mu\epsilon \mathscr{A}_\mu.
\label{ALMap}
\end{equation}
This relation could again be guessed from supersymmetry transformation
rules (\ref{trvec}). First, inserting a transverse vector eigenmode
${\cal A}$ satisfying (\ref{MA}) and a constant $\sigma$ into the right
hand side of $\delta\lambda$ in (\ref{trvec}) gives
\begin{equation}
 \delta\lambda~=~\tfrac12\gamma^{\mu\nu}\epsilon
 (\partial_\mu{\cal A}_\nu-\partial_\nu{\cal A}_\mu)
 -(\sigma\alpha)\gamma^\mu\epsilon{\cal A}_\mu,
\end{equation}
which is proportional to $\Lambda=\gamma^\mu\epsilon{\cal A}_\mu$.
On the other hand, the naive map from spinor to vector eigenmodes is
${\cal A}\simeq \bar\epsilon\gamma_\mu\Lambda d\xi^\mu$, but it can be
shown to satisfy
\begin{equation}
 -i\ast d(\bar\epsilon\gamma_\mu\Lambda d\xi^\mu)
 ~=~
 d(\bar\epsilon\Lambda)
 +(iM+\sigma\alpha)\bar\epsilon\gamma_\mu\Lambda d\xi^\mu,
\end{equation}
namely it fails to satisfy (\ref{MA}) due to a nonzero divergence term.
The right hand side of this equality is by construction divergenceless,
and can be used to define the map from $\Lambda$ to ${\cal A}$.

The modes which have a superpartner under the relation (\ref{ALMap}) are
irrelevant in computing the one-loop determinant. Unpaired spinor eigenmodes
satisfy (\ref{MLambda}) and are annihilated by (\ref{ALMap}), and
contribute the eigenvalue $M$ to the enumerator of the determinant.
Missing spinor eigenmodes correspond to $\mathscr A$ satisfying
(\ref{MA}) and are annihilated by (\ref{ALMap}), and contribute $M$
to the denominator.

We solve the equation for unpaired spinor eigenmodes by putting
\begin{equation}
 \Lambda ~=~ \epsilon\Phi_0+\bar\epsilon\Phi_2,
\end{equation}
where $\Phi_0$ and $\Phi_2$ are scalars with R-charges $0$ and $2$.
The equations for $\Phi_0$ and $\Phi_2$ at first look overdetermined
but turn out to have solutions.
If we put the ansatz
\begin{equation}
 \Phi_0=\varphi_0(\theta)e^{im\varphi-in\chi},\quad
 \Phi_2=\varphi_2(\theta)e^{i(m-1)\varphi-i(n-1)\chi},
\end{equation}
then the solution is determined by
\begin{equation}
 M=\frac m\ell+\frac n{\tilde\ell}+i\sigma\alpha,\quad
 \partial_\theta\varphi_0=
 if\cdot\Big(\frac m\ell+\frac n{\tilde\ell}\Big)\varphi_2,
\nonumber \\
\end{equation}
\begin{equation}
 \Big(\frac1f\partial_\theta+\frac{m\sin\theta}{\ell\cos\theta}
 -\frac{n\cos\theta}{\tilde\ell\sin\theta}\Big)\varphi_0~=~0.
\end{equation}
Normalizability of the solution requiers $m,n$ to be nonnegative.

Also, note that for $m=n=0$ or $M=i\sigma\alpha$ the second
term in the first equation of (\ref{ALMap}) vanishes, so that it does
not give a map from spinor eigenmode to transverse vector eigenmode.
By a direct check one finds for $m=n=0$ there is no normalizable
unpaired spinor eigenmode.

To find the transverse vector eigenmodes with missing spinor partners,
we begin by solving $\mathscr{A}_a\gamma^a\epsilon=0$ by
\begin{equation}
 \mathscr A_1=iY\sin\theta,\quad
 \mathscr A_2=iY\cos\theta,\quad
 \mathscr A_3=Y.
\end{equation}
It is straightforward to solve the eigenmode equation (\ref{MA}) in
components. Putting the ansatz $Y=y(\theta)e^{im\varphi-in\chi}$
one finds
\begin{eqnarray}
 M &=& \frac m\ell+\frac n{\tilde\ell}+i\sigma\alpha,\nonumber \\
 0 &=& \Big(\frac1f\partial_\theta
 -\frac{\sin\theta}{\cos\theta}\big(\frac m\ell+\frac1f\big)
 +\frac{\cos\theta}{\sin\theta}\big(\frac n{\tilde\ell}+\frac1f\big)\Big)y.
\end{eqnarray}
Normalizability of the solution requires $m,n\le-1$.

Now we combine all the contributions to compute the one-loop
determinant.
\begin{eqnarray}
\lefteqn{
 \prod_{\alpha\in\Delta}
 \Bigg[
 \frac1{i\sigma\alpha}
 \prod_{m,n\ge0}
 \frac{\frac m\ell+\frac n{\tilde\ell}+i\sigma\alpha}
      {\frac {-m-1}\ell+\frac {-n-1}{\tilde\ell}+i\sigma\alpha}
 \Bigg]
} \nonumber \\ &=&
 \prod_{\alpha\in\Delta_+}
 \prod_{n>0}
 \Big(\tfrac{n^2}{\ell^2}+(\sigma\alpha)^2\Big)
 \Big(\tfrac{n^2}{\tilde\ell^2}+(\sigma\alpha)^2\Big)
 ~=~ \prod_{\alpha\in\Delta_+}
 \frac{\sinh(\pi b\hat\sigma\alpha)\sinh(\pi b^{-1}\hat\sigma\alpha)}
      {(\pi\hat\sigma\alpha)^2}.
\label{nicedet2}
\end{eqnarray}
Here the product is over all the positive roots $\alpha$, and we used the
parameters of (\ref{Lipar}). The factors in the denominator in the right
hand side cancel against the Vandermonde determinant when reducing the
integral from Lie algebra to its Cartan subalgebra.

\vskip2mm

We thus found that, by putting the 3D theory on the $U(1)\times U(1)$
symmetric squashed $S^3$ (or the hyper-ellipsoid in $\mathbb{R}^4$ with
the four axis-length parameters $\ell,\ell,\tilde\ell,\tilde\ell$), the
partition function is expressed in terms of the double sine function with
$b=(\tilde\ell/\ell)^{\frac12}$, namely the formula (\ref{nicedet1}).
Also, the integration measure over the Coulomb branch (\ref{nicedet2})
is given by products of pairs of $\sinh$ functions. These are both
identified with the building blocks of structure constants in Liouville
or Toda CFTs.

\section{Concluding Remarks}

As was studied in \cite{DGG,HLP}, certain 3D supersymmetric gauge
theories arise on domain walls in 4D ${\cal N}=2$ gauge theories.
When the fields in the 4D theories on the two sides are connected on the
wall via S-duality, the AGT relation relates the 3D partition
functions on the wall with the kernels of the corresponding S-duality
transformation acting on conformal blocks. By replacing the round metric
of the wall $S^3$ by a squashed one, one obtains the kernels for general $b$.
Also, the AGT relation was interpreted in \cite{NW} as the equivalence
between the space of supersymmetric ground states of 4D ${\cal N}=2$ gauge
theories on $\mathbb R\times S^3$ (with omega deformation) and the space
of conformal blocks. Our results should give a generalization of the AGT
relation to $b\ne 1$ along this line, too.

We conclude with one immediate conjecture: there should be a similar
squashing of $S^4$, with some background gauge field turned on, which
gives the generalization of AGT relation to $b\ne 1$. The metric
and gauge field should be such that, when we view $S^4$ as a bundle of
$S^3$ fibred over a line segment, they take precisely the forms given in
this paper when restricted to each fiber.

\vskip1cm

redpajama_set_name: RedPajamaArXiv


text:
By Glenn Walker Posted on February 27, 2014 February 24, 2014
February Faves: Early Performances You May Have Missed
For the month of February, we've been doing something special here at Biff Bam Pop! Each week our writers and correspondents will be compiling lists of their favorite things. Leiki is back to take a look at her five favorite early performances that you may have missed, and well worth re-watching. Got it? Quite a mouthful, but a very cool list of February Faves, check it out after the jump.
1. Paul Bettany in After The Rain (1999). Two years before his Hollywood breakthrough performance in A Beautiful Mind Paul stole the film in this intense drama about relationships and apartheid that you would've likely only caught at a film festival. The film overall didn't impress critics but his performance was magnetic and ensured his stardom was just a matter of time.
2. Natalie Portman in Léon: The Professional (1994). In her first feature film, Natalie exceeded expectations by holding her own against seasoned Jean Reno who had nearly twenty years under his belt at the time. Given her Oscar cred and impressive filmography since, it's easy to forget, but this film reminds everyone of her innate and raw talent from her very first moments on screen.
3. Mariel Hemingway in Manhattan (1979). Director Woody Allen has been known to choose wisely in his female leads. Diane Keaton is the most obvious example. Mariel was another great casting choice. Despite her young age she effortlessly expressed the nuances of Allen's dialogue to give the viewer the voyeuristic feeling scenes were occurring in real time.
4. Nicole Kidman in Dead Calm (1989) was unpolished, unknown and fabulously frizzy-haired. Way before she became daily tabloid fodder as Mrs. Cruise, she conveyed fear and moxy simultaneously with her steely stare keeping viewers invested and on the edge of their seats. Nicole proved in this film and from that point forward that she is a force to be reckoned with.
5. TIE: Judd Nelson in The Breakfast Club (1985) and Charlie Sheen in Ferris Bueller's Day Off (1986). Both of these former feature film teen heartthrobs had commanding and charismatic performances in their respective films. Unfortunately, their film career decline since then means their strength onscreen is practically a vague memory. Remember and relish them in these timeless classics from the eighties.
Categories: February Faves, Film, Leiki VeskimetsTags: Charlie Sheen, Diane Keaton, February Faves, jean reno, judd nelson, leiki veskimets, mariel hemingway, natalie portman, Nicole Kidman, oscars, paul bettany, Woody Allen
Published by Glenn Walker
Glenn Walker is a professional writer, and editor-in-chief and contributing writer at Biff Bam Pop!. A blogger, podcaster, and reviewer of pop culture in all its forms, he's done stints in radio, journalism and video retail. Ask him anything about movies, television, music, or especially comics or French fries, and you'll be hard pressed to stump him or shut him up.	View all posts by Glenn Walker
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2 Replies to "February Faves: Early Performances You May Have Missed"
morrighansmuse says: 7 years ago Reply
I remember crushing on Charlie Sheen majorly when FBDO came out, and Leon: The Professional is one of my favorite films ever. Natalie Portman did hold her own in this one, and of course, there was also Gary Oldman 🙂
leocravenauthor says: 7 years ago Reply
Dead Calm was such a great movie! Kidman, Neil & Zane all did a fantastic job!
RT @Schwarzenegger: I'm rooting for you @JoeBiden. Your success is the country's success. https://t.co/BTvwsGN5Ih 1 hour ago

redpajama_set_name: RedPajamaCommonCrawl


Arizona Theatre Company and Chef Janos Wilder are thrilled to announce a new collaboration that pairs Arizona's preeminent producer of world-class theatrical productions with one of Tucson's most acclaimed and beloved chefs and restaurateurs. Janos at the Temple premiered in tandem with the opening of ATC's 2018/2019 season and will offer a unique dining experience complementing each of Arizona Theatre Company's productions at the Temple of Music and Art.
Janos at the Temple will be open starting 90 minutes before every matinee and evening ATC performance creating lunch and dinner menus that tease culinary themes from the plot lines of each play. The menus will be tasty, quick and price conscious at $14.50 for lunch and $24.50 for dinner. The lunch and dinner will be complete meals offering a variety of salads, starches, main dishes and desserts.
Janos Wilder's culinary flare, artistry, execution and good taste that has become his signature at DOWNTOWN KITCHEN AND COCKTAIL and THE CARRIAGE HOUSE, and the virtuosic theatrical talent and creativity displayed by Arizona Theatre Company productions, including last season's sell-out performances of OUTSIDE MULLINGAR and MAN OF LA MANCHA, are a perfect match to give patrons an unforgettable experience for all the senses.


If you are Reba, please join here and update your profile!
Updates to Reba's profile since your last visit are highlighted in yellow.
Reba Shumway Carr changed "Then" picture.
Reba Shumway Carr added a photo to profile gallery.


This Inside Out Crewneck Sweatshirt revisits one of the season's most eye-catching visual effects. Tailored in a slightly loose fit from pure cotton, it features visible stitching on the sleeves to create the impression that the piece is being worn inside out. Signatures include a signed palladium staple and woven Louis Vuitton Staples Edition label stitched on the back.


text:
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const path = require('path');

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redpajama_set_name: RedPajamaGithub


text:
Q: SQL - Select two different values based on the same condition Is it possible in SQL to have a case statement that selects two different values based on the same condition?
For example, if I had a table called Person with the following columns:

*

*Name

*Age

*EyeColor

So if the person is 18 years old or older, I want to select their name. But if they are less than 18, I want to select their eye color. How can I do this using a case statement?
Sample Data:
NAME    AGE    EyeColor
------  -----  --------
John    17     Brown
Mike    21     Blue
Peter   16     Green
Alan    18     Brown

Expected result:
NameOrEyeColor
--------------
Brown
Mike
Green
Alan

A: If both columns are the same data type, then:
select
  case
    when age >= 18 then name
    else eyecolor
  end as columnname
from person

although in a Mysql database this would work for columns with different data types because of the implicit conversion done (check this link).

A: This will do it
SELECT CASE WHEN age = 18 THEN name
WHEN age < 18 THEN eyeColor END AS mySelection
FROM my_table

A: One way is to simply use a CASE STATEMENTfor each condition:
SELECT (CASE WHEN Age = 18 THEN Name END) AS Name, (CASE WHEN Age < 18 THEN EyeColour END) AS EyeColour
FROM Person

Or, perhaps you're looking for:
SELECT (CASE WHEN Age = 18 THEN Name WHEN Age < 18 THEN EyeColour END) as Condition
FROM Person

Although that last option would give you results that are not clear.  What if the person's name is Blue ?

A: You can use IF statement in select
For example
SELECT
IF(Age > 18, Name, EyeColor) as val
FROM table

https://dev.mysql.com/doc/refman/8.0/en/if.html

redpajama_set_name: RedPajamaStackExchange


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Somsara Rielly is a Los Angeles Artist & Designer. Her name means "the eternal cycle of birth, suffering, death and rebirth" in Sanskrit – a pretty heavy name to be born with for someone who just wants to make the world a more beautiful place. Always driven by conceptual thinking, Somsara's blend of classical and modern styles results in a timeless and unique aesthetic. Her collage and mixed-media work weaves in layers of interpretation and explores the boundaries between art and design. A lifelong love affair with paper, typography, and printing techniques continue to inspire new methods and connections resulting in a unique style and voice. Be it design, illustration or fine art, there's always a dash of wit and a wink. Somsara worked with Scout Books to create her beautiful notebooks for collecting all your goals for the new year and was kind enough to answer a few questions about the project.
What inspired you to create the GOAL GETTER notebooks?
It was such a combination of things! I've been making a concerted effort to blend my collage work and design work, and had been wanting to create something as a self-promotion piece and holiday gift. It's important to me to make thoughtful things with a purpose – something that is actually useful and maybe even dares to be inspirational.
I am also SUCH a new year's "clean slate" kind of gal – I love the new year rituals. I enjoy the process of reflection, and setting new goals and intentions every year. This year in particular was so difficult on so many levels, and many people understandably seemed to be feeling "Ugh let this year be over already!" That made me kind of sad because I also saw so many great things happen this year – so I wanted to do something to focus on the positive and try and change some of that narrative. It takes real effort to keep the good things we accomplish front and center in our lives and not focus on the bad, and I personally have to remember and record that I actually DID make progress this year. Writing things down really helps you see that you are taking positive steps big and small to make your life – and hopefully the world – a better and more beautiful place.
When I read that statistic about how writing down goals makes them 42% more likely to happen, I just thought "well that's some concrete evidence to back up a process I already believe in!" So, creating a cool, inspiring, easy-to-reach place to record your goals and any other ideas or thoughts for the new year (or any time of year really) seemed to hit a lot of sweet spots for me.
We'd love to learn more about your collage process. Do you work with a blade and paper, digitally, or between mediums? Where do you find your favorite source material?
I create everything by hand, nothing digitally. I make a point to stay away from my computer and just commune with all the random stuff I collect and see what happens. It's become such a window into where my head is at, and sometimes it still surprises me (which is what I love about it!). I frequent flea markets, thrift stores, Goodwill, people's garages (legally of course) and have been at this for so long now that people gift me plenty of paper treasures. I love the hunt! I've also been known to pick up "trash" off the street that looks interesting. I try not to plan too much ahead of time, I start sifting through all my images and suddenly an idea comes. Or I'll read a phrase in a 1940s Woman's Day magazine that sparks something, and then I start cutting, layering, painting – it just goes from there. I do also have a more formal process for a larger new project I'm working on that involves a series of smaller collaged cards I've made that informs larger pieces, but that project is still taking shape.
What appealed to you about designing in the Scout Books format?
I always carry a small notebook with me, so first and foremost it's a useful and practical object! What I loved about Scout were all the design details and page options – that you could have all that with such a small print run, which is always the trick when you want to produce something great on a small budget. I was also interested in the challenge of producing a one-color photographic image on your cover stock to see if I could get it to a place where I was really happy with the production of it. The Scout design team was so great in helping me achieve that and answered so many (many!) questions – the entire process was really easy and pleasurable.
Can you tell us a bit more about the beautiful foiled belly bands? They're amazing!
Thanks! My good friends at Paperleaf Press in Downtown Los Angeles did that for me. I wanted to include some messaging with the notebook to make clear its purpose and intent, but I didn't want to junk up the illustration with a lot of text, so it seemed like a good packaging solution. It's one-color silver foil on a red cover stock, and I hand-applied the crystal myself.
What are your goals for 2018? Any projects on the horizon that you can share?
Sure! I have set both goals and intentions for myself in 2018. This year in particular, my intentions are important because they are really influencing and motivating my goals. There is one particular big project goal (that I can't talk about just yet but I'm SUPER excited about!) and it's taking a lot of bravery on my part to move forward with it. So I am really focusing my intention on being bold in my decision-making and remembering to dream BIG, which has been a challenge for me in the past to help achieve those goals. I want to create more products using collage and design. I want to have more collage illustration projects for clients. I want to have projects that are purpose-driven and give back in some way. I want to be a part of more gallery shows. I also reset the goal every year to meditate every day, which helps me stay focused and remember my daily intentions. Oh! One project I am working on that I can talk about is with photographer Stephanie Vovas – I love her work so much and we are developing her first product right now that is going to be awesome. So, I'm helping someone else realize their goal too, which is exciting!
How can we get our hands on a copy of your notebook?
I sent out a bunch for holiday and self-promotion, but I still have some in my shop for sale, Scout readers can enter SCOUTGOALS at checkout for 20% off! Or if you are in a position to hire me and want to work together I might just send you one free. It's easy to find me through Somsara.com. Thanks again Scout, and Happy New Year!
Big thanks to Somsara for working with on this project with us and sharing her process!


text:
Language comes alive
J.M. Dent , Toronto
English language -- Composition and exercises., English language -- Gr
Statement F.L. Carver, K. Mead Elliott
Contributions Carver, F. L., Elliott, K. Mead.
Pagination 296 p. :
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English Comes Alive. Dynamic, Brain-Building Ways to Teach ESL and EFL [Witherspoon Ph.D., Jim] on *FREE* shipping on qualifying offers. English Comes Alive. Dynamic, Brain-Building Ways to Teach ESL and EFL/5(40). Alive: The Story of the Andes Survivors Mass Market Paperback – December 3, the simplicity of the language had a bigger impact on me and as I read I felt that I was there.
The book starts with a bit of background on the boys, their school and rugby team and the decision to charter a plane. Then the journey, the crash, their time on /5(). AA Speakers Joe and Charlie - The Big Book Comes Alive AA Speakers - Joe and Charlie - "The Doctors Opinion" - The Big Book Comes Alive by AA Speakers and More.
An evil school comes alive in this exciting book for early elementary readers.
Description Language comes alive EPUB
Sam the hall monitor must defeat the school with the help of his best friends and the grizzled old custodian. This book was fun to read and I really liked the illustrations and design (fun fact, the illustrator is also an alumni of the University of Baltimore).4/5.
User Review - Flag as inappropriate Google Books is including both the and editions in the same search/database, because they have the same title. Type in 54 (the page), and you will see two different authors/edition dates.5/5(4).
English Alive for CSEC is a 2-year course designed to meet the needs of students in Caribbean schools studying for English A, the language component of the CSEC English exam. The book covers the key areas of reading comprehension, listening and speaking, writing, grammar and language usage and vocabulary development.
Particular attention is given to the often challenging area of poetry. English language comes alive. 3 min read. Strategies for English Language Learning and Reading, better known as Stellar, aims to make English more interactive and fun to learn.
The program was first started inwith Primary 1 pupils in 30 primary school, and eventually rolled out to. Language Comes Alive at the Library Written on: October 22nd, by: in Blog Posts, Informal learning Libraries offer fun and engaging programs for all ages, and language programs are an important part of these offerings.
This book has to be one of the best examples of an author able to relate an account containing culturally taboo subject matter without judgment or sensationalism. Alive: The Story of the Andes Survivors recounts the survival story of a rugby team whose plane crashes in the Andes Mountains.
Those who live find themselves confronted with awful Cited by: Teacher's manual accompanies vols. Addeddate Approval_date Approval Approval_statement.
Details Language comes alive PDF
"English Alive. is a series designed specifically to meet the needs of English students in Caribbean secondary schools. The book covers the key areas of reading comprehension, listening and speaking, writing grammar and language usage and vocabulary development.
Particular attention is given to the often challenging area of poetry. The book also covers common errors made by students.
Make Books Come Alive for Your Kids. Many great children's books stand out not just because they are fun to read, but because the illustrations are very eye-catching. Why not recreate the main character and make a craft based on the book with your kid.
Incorporate pretend play and try to recreate a scene from the book. Alive: The Story of the Andes Survivors is a book by the British writer Piers Paul Read documenting the events of Uruguayan Air Force Flight Alive tells the story of an Uruguayan Rugby team (who were alumni of Stella Maris College), and their friends and family who were involved in the airplane crash of Uruguayan Air Force Flight Author: Paul Read.
Cornish (Standard Written Form: Kernewek or Kernowek, Old English: Cornwealisċ) is a Southwestern Brittonic language of the Celtic language is a revived language that became extinct as a first language in Cornwall in the late 18th century.
A revival began in the early 20th century. Some have expressed the opinion that the language is an important part of Cornish identity, culture Language family: Indo-European, CelticInsular. The Big Book Comes Alive This series of talks came about from a friendship that started in Alcoholics Anonymous over 30 years ago.
When AA members talk about the program, they will often speak the names Bill and Bob. The point of fiction is to cast a spell, a momentary illusion that you are living in the world of the story.
But as a writer, how do you suck your readers into your stories in this way. Nalo Hopkinson shares some tips for how to use language to make your fiction really come alive. How to write descriptively - Nalo Hopkinson - YouTube. Garth Greenwell 's first novel, " What Belongs to You " (), uses mesmeric cadences.
The world observed and the interior life are slowed down, the rhythms in the prose given a strange and. Accounting Comes Alive: Introducing Color Accounting is a new kind of accounting book — color, movement, language, graphics, video and audio narration make accounting come alive in this interactive iPad book.
Balance sheets, income statements, ledgers, journals, debits. The Bible Comes ALIVE. The articles and books on this website are formatted in PDF.
If you cannot open up these documents then you need "Adobe Acrobat Reader." Click on the following link to download "Adobe Acrobat Reader" for FREE: Adobe Acrobat Reader Books to Read Books to Order Articles Doctrines Book Covers Donations.
English Alive. is a four-book series designed specifically to meet the needs of English students in Caribbean secondary schools. The series adopts a lively and exciting new approach to the study of English, helping students to become more confident in their.
Hello there. It's been awhile. We're happy to announce the release of MCA beta for Minecraft. This is a much more robust version of MCA rewritten from the ground up to play nicer with servers and other mods. Some notable features: RadixCore is no longer a requirement. Chores are smarter and more mod-friendly.
Minecraft Comes Alive is a hugely detailed mod that turns your Minecraft experience, both single and multiplayer, into an RPG in which you can interact with villagers in a whole new way. Villagers are either males or females who look and act like humans. You can interact with them in a lot of different ways, like chatting, asking them to follow /5.
Browse over educational resources created by History Comes Alive in the official Teachers Pay Teachers store. "His groundbreaking work has changed the very ways we consider our health and examine disease." —Barack Obama.
From Dr. Francis Collins, director of the National Institute of Health, recipient of the Presidential Medal of Freedom, and year head of the Human Genome Project, comes one of the most important medical books of the year: The Language of Life.
The Second Coming: Revelations Comes Alive depicts the end time in language easily relatable to readers today. Sonia McGary has written these words so that all who read them will have the opportunity to be open to the Lord and His message.
About the Author. Just the facts about Latin Alive 1 by Classical Press: The bundle pack includes the teacher guide, student book, 7 Dvd lessons and a pronunciation cd. 29 lessons classically taught, in seven units, with tests.
An average lesson is 30 minutes. $ for the bundle pack, products may be purchased separately. Latin Alive 2 available. Shel Silverstein Comes Alive in a New Book, 12 Years After His Death.
Sarah Weinman. transposing consonants and generally embracing the. Define come alive. come alive synonyms, come alive pronunciation, come alive translation, English dictionary definition of come alive.
Language: Twitter. Get our app. Flashcards. My Daniel Morgan Comes Alive. Heidi and Daniel Morgan "Alive" in one sense, for Christ is in fact dead, though the poet alludes to his resurrection.
Accounting Comes Alive is a new kind of accounting book — color, movement, language, graphics and breathtaking simplicity make true accounting literacy easy. Balance sheets, income statements, ledgers, journals, debits & credits will make perfect sense after you've delighted in this powerful book.
Download Language comes alive FB2
'The format of this book is alive with visuals and packed with persuasive language.' 'The town will be alive with an abundance of colour and plenty of entertainment for all the family.' 'Obscure slices of history and allegories abound and every spot comes alive with some parable or other.'.
Make Reading Come Alive Through Food. Almost every book you read will mention the characters eating or drinking. You can bring the story to life by making those foods as a snack or even as your family meal one night. One great example is the book The BFG. A whole chapter of the book describes a fizzy, delicious drink that the giant enjoys.Where English comes alive.
Here is some information regarding the new IELTS test on computer. These are just some of the plus and minus points.This The Gingerbread Boy Comes Alive! Lesson Plan is suitable for Kindergarten - 1st Grade. Students read "The Gingerbread Boy" and recall the details of the 4/5.
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Eden Pools and Spa is a leading Garden Route swimming pool builder. The company builds a variety of beautiful swimming pools and spas of all shapes and sizes. Eden Pools and Spa will help you improve the investment in your home.
PoolsnSpas is an all-round service provider in the pool contracting industry, from the design to construction as well as the maintenance process. Based in Pretoria, we have a reputation for quality and diversity by supplying a wide range of styles, finishes and accessories.


I don't think there was anything that I wish I would have known before coming here. At orientation and visits they informed us of almost everything. They explained the class sizes, events, living arrangements, teachers, grading, financial aid, and more. It really took my nervousness away and made me more excited and comfortable to come to this school. K-state is really a great school! It offers many different events, tutoring sessions, financial aid options, and more. If you like a medium sized school and love to be social then this is the right place for you.
Before attending my first semester at Kansas State, I wish I would have known to bring my entire collection of purple wear and gear. People take their purple seriously here! It is the only proper attire for Monday thorugh Saturday, especially for games. You can never have too much purple pride and this really reflects the attitude and love Wildcats have for our school. Every chance there is to shout "EMAW!" or "Go Cats!" is observed in endless ways, wearing t-shirts, sweaters, socks, and face paint is just the beginning.
Pay the little bit of extra money to live closer to campus. Save money and don't get a student parking pass. While parking seems doable, once everyone is on campus, it's nearly impossible to find. I was late to class several times when I lived off-campus simply because it took me half an hour to find parking and, usually, I ended up parking off-campus anyway. However, don't pay for dorms. They're not worth it, and you can actually be closer to your classes when you live in an apartment off-campus.
This doesn't apply to only my school, but college in general. I wish I had known how to study. In high school I was able to get by with just reading over my notes before a test or writing an entire paper the day before it was due. However, in college I have not been able to do that. If you can get into a rhythm of reading all of the necessary material and studying a day or two ahead of time, you should be fine with the workload you'll face in college.
I wish I would have gotten more involved right off the bat my freshman year. I kind of layed low my first semester here. I loved getting to know the people in my residence hall, but I would have loved meeting a larger variety of people. I am now very involved and I really enjoy knowing more people. Also, I wish that I would have utilized my academic advisor more. They are really there to help and are great people.
How useful and awesome studying abroad is. I studied in Guatemala in the summer and taught English. I never thought so much could be learned outside of a classroom! Studying abroad is really easy to do early, most students exchange their sophomore year. I had to squeeze it in, but still was able to do it. So I wish someone would tell me to just look into it and do it! It is a life-changing experiment.
College is a different experience, my first semester in college was all about figuring out my new life schedule. It took me awhile to figure out how to appropriate my time between all my activities. It would have been nice to have a better idea about the diferent stresses in college life ahead of time and to have better time management skills.
The level of expectations, especially from the junior college in which I transferred from. I wish I would have pushed myself harder in the years before coming to Kansas State University. Not only in junior college but also in high school. After my first semester here I already see a change in my drive and aspirations to do better academically.
I wish I had comprehended just how much school actually costs. If I had, I wouldn't have spent so much money on books and movies and hanging out with my friends during high school. I would have definitely saved more money. It is hard when your financial aide is determined by how much your parents make, but they don't help pay for college.
You will get out of college what you put into it. If you spend time studying, you will get good grades; if you spend time socializing, you will make lots of friends; if you go out and party then you will make lots of memories and mistakes. Whatever you nurture and pay attention to the most will flourish and that you neglect will wilt.


text:
package com.revinate.ship.gueststay;

import com.fasterxml.jackson.databind.JsonNode;
import com.github.fge.jsonschema.core.report.ProcessingReport;
import com.github.fge.jsonschema.main.JsonSchema;
import com.github.fge.jsonschema.main.JsonSchemaFactory;
import com.jayway.jsonpath.DocumentContext;
import com.jayway.jsonpath.JsonPath;
import com.revinate.ship.TestUtils;
import com.revinate.ship.profile.ProfileSerializationTest;
import org.junit.Test;

import java.time.LocalDate;
import java.util.Collections;

import static com.revinate.assertj.json.JsonPathAssert.assertThat;
import static org.assertj.core.api.Assertions.assertThat;

public class GuestStaySerializationTest {

    public static final GuestStay GUEST_STAY = GuestStay
            .builder()
            .action(GuestStay.Action.ADD)
            .property("Avertine")
            .interfaceType("SHIP")
            .remoteSystemName("avertine_ship")
            .confirmationCode("12345")
            .statusCode(GuestStay.StatusCode.RESERVED)
            .checkinDate(LocalDate.of(2016, 8, 29))
            .checkoutDate(LocalDate.of(2016, 9, 4))
            .profiles(Collections.singletonList(ProfileSerializationTest.PROFILE))
            .build();

    @Test
    public void serialize_shouldContainValues() throws Exception {
        String guestStayString = TestUtils.MAPPER.writeValueAsString(GUEST_STAY);
        DocumentContext context = JsonPath.parse(guestStayString);

        assertThat(context).jsonPathAsString("$.action").isEqualTo("ADD");
        assertThat(context).jsonPathAsString("$.property").isEqualTo("Avertine");
        assertThat(context).jsonPathAsString("$.interfaceType").isEqualTo("SHIP");
        assertThat(context).jsonPathAsString("$.remoteSystemName").isEqualTo("avertine_ship");
        assertThat(context).jsonPathAsString("$.confirmationCode").isEqualTo("12345");
        assertThat(context).jsonPathAsString("$.statusCode").isEqualTo("RESERVED");
    }

    @Test
    public void serialize_shouldConformToSchema() throws Exception {
        JsonNode JsonNode = TestUtils.MAPPER.valueToTree(GUEST_STAY);
        JsonSchema schema = JsonSchemaFactory.byDefault().getJsonSchema(
                TestUtils.classpathResourceAsUrlString("schemata/guest-stay-schema.json"));
        ProcessingReport report = schema.validate(JsonNode);
        assertThat(report.isSuccess()).isTrue();
    }
}


text:
The U.S. National Soccer Team Players Association (USNSTPA) is the players union for the members of the United States men's national soccer team. Though the organization has existed since 1996, they rose to prominence within American soccer in late 2004 when their collective bargaining negotiations with the United States Soccer Federation (USSF) became public.  At the time, the collective bargaining agreement had been expired for two years, and negotiations during that period hadn't resulted in a new agreement.

History
According to various sources, either the players went on strike or were locked out of a December preliminary camp and later the January 2005 camp. In response, the USSF held a camp with replacement players.  These players came from the United Soccer Leagues (USL) and Major Indoor Soccer League (MISL). Major League Soccer's players union (MLSPU) stood with the national team players, advising their membership not to accept invitations to the January camp. With World Cup Qualifying games at risk, the two sides agreed to an interim deal that would guarantee that the first choice players appeared through the qualifying campaign.

The two sides continued negotiations throughout the summer and into the fall of 2005, before finally reaching a deal in November 2005.  That deal was approved by both parties in early December, running through 2010 and assuring that labor issues would not threaten World Cup participation.

The U.S. National Soccer Team Players Association is governed by a five-player board of representatives elected annually, and run by an executive director. The women's national team has a separate organization, the U.S. Women's National Soccer Team Players Association.

Aside from negotiating for the players, the USNSTPA owns USSoccerPlayers.com a large soccer-specific website.  That site was formerly InternetSoccer.com — one of the major independent soccer sites to emerge in the late 1990s.

The USNSTPA is also heavily involved in fundraising for the Leukemia & Lymphoma Society through their own Time In program and the national Soccer Kicks for Cancer program.

References

External links

Players Association
Trade unions established in 1996
Sports trade unions of the United States

redpajama_set_name: RedPajamaWikipedia


text:
package com.zml.eclipse.powermode;

import org.eclipse.jdt.internal.ui.javaeditor.CompilationUnitEditor;
import org.eclipse.jface.text.ITextViewerExtension2;
import org.eclipse.jface.text.source.ISourceViewer;
import org.eclipse.swt.widgets.Composite;

import com.zml.eclipse.powermode.explosion.ExplosionPainter;

@SuppressWarnings("restriction")
public class PowerModeEditor extends CompilationUnitEditor {

    public void createPartControl(Composite parent) {
        super.createPartControl(parent);
        ISourceViewer sourceViewer = getSourceViewer();

        if (sourceViewer instanceof ITextViewerExtension2) {
            ITextViewerExtension2 extension = (ITextViewerExtension2)sourceViewer;
            ExplosionPainter fExplodePainter = new ExplosionPainter(sourceViewer);
            extension.addPainter(fExplodePainter);
        }
    }

}