#  TIT/HEP-689 May 2022 Analytic continuation for giant gravitons

Yosuke Imamura111E-mail<EMAIL_ADDRESS>

Department of Physics, Tokyo Institute of Technology,
Tokyo 152-8551, Japan

We investigate contributions of giant gravitons to the superconformal index.
We concentrate on coincident giant gravitons wrapped around a single cycle,
and each contribution is obtained by a certain variable change for fugacities
from the index of the worldvolume theory on the giant gravitons. Because we
treat the index as a series of fugacities and the variable change relates
different convergence regions, we need an analytic continuation before summing
up such contributions. We propose a systematic prescription for the
continuation. Although our argument is based on some unproved assumptions, it
passes non-trivial numerical checks for some examples. With the prescription
we can calculate the indices of the M5-brane theories from those of the
M2-brane theories, and vice versa.

###### Contents

  1. 1 Introduction
  2. 2 Plethystic exponential and plethystic logarithm
  3. 3 Variable changes and analytic continuation
  4. 4 Numerical tests
    1. 4.1 Double expansion of the Schur index for $T=D3$
    2. 4.2 Simple expansion of the Schur index for $T=D3$
    3. 4.3 Superconformal index for $T=D3$
    4. 4.4 M5 from M2
    5. 4.5 M2 from M5
  5. 5 Conclusions and Discussion
  6. A Full index of $6$-dim $A_{N-1}$ SCFT

##  1 Introduction

For the last few decades the AdS/CFT correspondence [1, 2, 3] has been playing
an important role in the progress of string theory and quantum field theories.
It provides novel approaches for investigation of different physical
quantities in various situations. It is very powerful for the analysis of
large $N$ gauge theories, and even in the finite $N$ region, where the Planck
length is not negligible, it is possible to calculate supersymmetry protected
quantities like $R$-charges of gauge invariant operators by using string
theory in the AdS background.

Generating functions of such quantities have been calculated on the gravity
side of the AdS/CFT corresponsence. For example, the superconformal index of
the ${\cal N}=4$ $U(N)$ SYM in the large $N$ limit was reproduced as the index
of the bulk supergravity modes [4]. Extended branes are important when $N$ is
finite. It is known that not only branes wrapped around topologically non-
trivial cycles [5, 6] but also branes wrapped around trivial cycles can be
stable and BPS. Such branes carry the same quantum numbers with point-like
gravitons, and are called giant gravitons [7, 8, 9, 10]. The BPS partition
function for scalar fields in 4d ${\cal N}=4$ SYM was reproduced by geometric
quantization of giant gravitons in [11], and the same quantity was also
obtained by using the dual giants in [12]. Similar analysis was done in [13]
for the 3d and 6d theories realized on M2 and M5-branes.

The contribution of giant gravitons with different wrapping numbers to the
superconformal index was suggested by a characteristic form of the analytic
result in [14] for the unrefined Schur index of ${\cal N}=4$ SYM. (See also
[15] for generalization to ${\cal N}=2$ quiver gauge theories.) It has been
confirmed by studying fluctuation modes of giant gravitons in [16, 17, 18] for
the ${\cal N}=4$ SYM, and for more general theories in [19, 20, 21, 22, 23].
In all cases the index $I_{N}^{(T)}$ of a theory $T$ with finite $N$ whose
holographic dual is string/M theory in $AdS\times M_{T}$ is given by a
multiple expansion of the form

$\displaystyle\frac{I_{N}^{(T)}}{I_{\infty}^{(T)}}=\sum_{m_{1}=0}^{\infty}\cdots\sum_{m_{d}=0}^{\infty}x_{1}^{m_{1}N}\cdots
x_{d}^{m_{d}N}F^{(T)}_{m_{1},\ldots,m_{d}},$ (1)

where $m_{1},\ldots,m_{d}$ are wrapping numbers of giant gravitons around
appropriately chosen $d$ supersymmetric cycles in $M_{T}$. The number of
cycles $d$ depends on the theory $T$. For each set of the wrapping numbers the
function $F^{(T)}_{m_{1},\ldots,m_{d}}$ is the index of the field theory
realized on the corresponding system of giant gravitons and
$F_{0,\ldots,0}^{(T)}=1$. A typical structure of the theory is shown in Figure
1.

Figure 1: A typical structure of the theory realized on a system of giant
gravitons. Each vertex represents the theory $T_{i}$ with rank $m_{i}$
realized on $m_{i}$ coincident giant gravitons wrapped around cycle $i$. Edges
connecting vertices are degrees of freedom arising on intersections of cycles.
$d=3$ case is shown.

It is the direct product of theories $T_{i}$ realized on cycles $i=1,\ldots,d$
coupling to the degrees of freedom arising on their intersections.
$F^{(T)}_{m_{1},\ldots,m_{k}}$ are $N$-independent, and the $N$-dependence
appears only through the prefactor $x_{1}^{m_{1}N}\cdots x_{d}^{m_{d}N}$.

Similar expansion of the index is also proposed in [24] based on a
complementary analysis on the gauge theory side, and such expansions were
named “the giant graviton expansions”. See also [25]. In their analysis the
correspondents for giant gravitons and fluctuations on them are baryon
operators and their modifications, respectively. Although there are some
technical subtleties about the relation between these expansions obtained on
the two sides of the AdS/CFT correspondence they are expected to be
essentially the same.

In this work we focus on the contribution of giant gravitons wrapped around a
single cycle $i$. Let $F^{(T)}_{i,m_{i}}=F^{(T)}_{0,\ldots,m_{i},\ldots,0}$ be
such contributions. It is essentially the index $I_{m_{i}}^{(T_{i})}$ of the
theory $T_{i}$ realized on the worldvolume of $m_{i}$ coincident giant
gravitons. More precisely, $F_{i,m}^{(T)}$ and $I_{m}^{(T_{i})}$ are related
by a simple variable change of the fugacities as we will explain shortly. A
purpose of this work is to study this relation in detail.

For concreteness let us consider the Schur index [26] of the ${\cal N}=4$
$U(N)$ SYM. Let $J_{1}$ and $J_{2}$ be the angular momenta and $R_{x}$,
$R_{y}$, and $R_{z}$ be the $R$ charges. The Schur index is defined by

$\displaystyle I_{N}^{(D3)}(x,y)=\mathop{\rm
tr}\nolimits[(-1)^{F}q^{J_{1}}x^{R_{x}}y^{R_{y}}],\quad q=xy,$ (2)

where we denote the ${\cal N}=4$ $U(N)$ SYM by $T=D3$. The constraint $q=xy$
is necessary to preserve the supersymmetry. We eliminate $q$ by the constraint
and treat the index as a function of $x$ and $y$.

The giant graviton expansion for the Schur index is given in [17] as the
double expansion

$\displaystyle\frac{I_{N}^{(D3)}}{I^{(D3)}_{\infty}}=\sum_{m_{x}=0}^{\infty}\sum_{m_{y}=0}^{\infty}x^{m_{x}N}y^{m_{y}N}F^{(D3)}_{m_{x},m_{y}}.$
(3)

The two non-negative integers $m_{x}$ and $m_{y}$ are wrapping numbers over
the cycles $i=x$ and $i=y$, respectively, which are respectively defined to be
the $R_{x}$ and $R_{y}$ fixed roci in $M_{D3}=S^{5}$.222We use $x$ and $y$ for
the labels of cycles. These should not be confused with the fugacities. A
special property of the Schur index is that the functions
$F_{m_{x},m_{y}}^{(D3)}$ are factorized [17]:

$\displaystyle
F_{m_{x},m_{y}}^{(D3)}=(xy)^{m_{x}m_{y}}F^{(D3)}_{x,m_{x}}F^{(D3)}_{y,m_{y}}.$
(4)

Therefore, we only need to determine $F_{i,m}^{(D3)}$ to write down the
expansion (3). $F_{i,m}^{(D3)}$ is the index of the ${\cal N}=4$ $U(m)$ SYM
realized on $m$ coincident giant gravitons wrapped around the cycle $i(=x,y)$.
The theories $T$, $T_{x}$, and $T_{y}$ are accidentally the same in this case,
and the corresponding indices $I_{N}^{(D3)}$, $F_{x,N}^{(D3)}$, and
$F_{y,N}^{(D3)}$ are also essentially the same functions. Careful analysis of
the action of superconformal algebra on the $AdS$ boundary and that on giant
gravitons wrapped around the cycle $x$ shows that they are related by the
involution [16]

$\displaystyle(H,J_{1},J_{2},R_{x},R_{y},R_{z})\rightarrow(H-2R_{x},R_{y},R_{z},-R_{x},J_{1},J_{2}).$
(5)

A similar relation also holds for cycle $y$. This means that
$I_{N}^{(D3)}(x,y)$ and $F_{i,N}^{(D3)}(x,y)$ are related by the variable
changes [16]

$\displaystyle F_{x,N}^{(D3)}(x,y)=I_{N}^{(D3)}(x^{-1},xy),\quad
F_{y,N}^{(D3)}(x,y)=I_{N}^{(D3)}(xy,y^{-1}).$ (6)

See also [24] for a derivation of (6) on the gauge theory side.

Despite the simplicity of (6), it is not straightforward to calculate
$F_{i,N}^{(D3)}$ from $I_{N}^{(D3)}$, and vice versa, because we usually do
not have analytic form of the indices, and only series expansions are
available. The variable changes relate different convergence regions of the
functions, and it is a non-trivial problem to generate one from another. In
this paper, we propose a simple prescription to obtain the functions
$F_{i,N}^{(T)}$ from $I_{N}^{(T_{i})}$.

To understand the relation, we need to carefully analyze the dependence of the
series expansion on the choice of expansion variables. Different choices of
expansion variables give different series expansions. Indeed, the expansion
found in [24] is the simple expansion over a single non-negative integer $m$:

$\displaystyle\frac{I_{N}^{(T)}}{I_{\infty}^{(T)}}=\sum_{m=0}^{\infty}x^{mN}F_{i,m}^{(T)}.$
(7)

This looks different from the multiple expansion (1). Once we have understood
the relation between $I_{N}^{(T_{i})}$ and $F_{i,N}^{(T)}$, we can give a
partial explanation to this difference.

In the process of confirming the relation (7) the analytic continuation has
been necessarily done in [24] for functions appearing in (7) based on a pole
structure commonly found in the functions. Our prescription is useful to
clarify the pole structure and enable us to study the relations like (6) in a
systematic way.

The paper is organized as follows.

In the next section we prepare tools used in the following sections. In
particular, we define “a domain” specifying the expansion variables. We also
define the plethystic exponential and the plethystic logarithm with emphasis
on the dependence on domains. In section 3 we explain the necessity of the
analytic continuation, and propose a simple prescription to realize it. We
numerically test the proposed method in Section 4 for the maximally
supersymmetric theories realized on D3, M5, and M2-branes. The last section is
devoted to conclusions and discussion.

##  2 Plethystic exponential and plethystic logarithm

The calculation of the index $I$ of a free field theory usually starts from
the analysis of single-particle states. The corresponding index is called the
letter index. It is obtained by summing up the contributions from all one-
particle states. If there are two fugacities $x$ and $y$ it is given by333We
mainly consider the case with two variables. Generalization is
straightforward.

$\displaystyle i=\sum_{(n_{x},n_{y})\in R}c_{n_{x},n_{y}}x^{n_{x}}y^{n_{y}},$
(8)

where $R$ is a certain region on the $2$-dimensional charge lattice specifying
a set of monomials appearing in the expansion. We call such a region “a
domain.” We usually adopt $R$ such that it covers all one-particle states. We
call such a domain “a physical domain”. For the Schur index of $T=D3$ we can
take

$\displaystyle R_{0}=\\{(n_{x},n_{y})|n_{x},n_{y}\geq 0\\}\backslash(0,0)$ (9)

as a physical domain. Later we will discuss other choices of $R$. We call the
series associated with a domain $R$ the $R$-series. For a free field theory
the series sums up into a simple rational function. For example, the letter
index of the ${\cal N}=4$ $U(1)$ vector multiplet is [4]

$\displaystyle i^{(D3)}_{1}=1-\frac{(1-x)(1-y)}{1-xy},$ (10)

and that of the type IIB supergravity multiplet in $AdS_{5}\times S^{5}$,
which is dual to the large $N$ SYM, is [4]

$\displaystyle
i^{(D3)}_{\infty}=\frac{1}{1-x}+\frac{1}{1-y}-\frac{1}{1-xy}-1.$ (11)

By expanding these functions into terms appearing in the domain $R_{0}$ we can
obtain the set of coefficients in (8). In the following we use the notation
$i|_{R}$ when we want to emphasize that $i$ is given as an $R$-series. We
graphically express a series by plotting the coefficients in the two-
dimensional lattice as shown in Figure 2.

Figure 2: (a) The letter index $i_{1}^{(D3)}$ of the ${\cal N}=4$ vector
multiplet. (b) The letter index $i_{\infty}^{(D3)}$ of the type IIB
supergravity multiplet in $AdS_{5}\times S^{5}$.

Once we have obtained the letter index (8), the index for multi-particle
states is uniquely determined by considering the combinatorics of the letters.
The solution to this problem is

$\displaystyle I=\mathop{\rm Pexp}\nolimits i,$ (12)

where $\mathop{\rm Pexp}\nolimits$ is the plethystic exponential defined by

$\displaystyle\mathop{\rm Pexp}\nolimits
f(x,y)=\exp\left(\sum_{m=1}^{\infty}\frac{1}{m}f(x^{m},y^{m})\right).$ (13)

For the definition (13) to make sense the condition $|x^{n_{x}}y^{n_{y}}|<1$
must hold for all terms appearing in the sum (8). This means all terms in $f$
must be contained in the domain $R_{(x,y)}$ defined by

$\displaystyle R_{(x,y)}:=\\{(n_{x},n_{y})|n_{x}\log|x|+n_{y}\log|y|<0\\}.$
(14)

For example, if we take $x$ and $y$ satisfying $|x|=|y|<1$, (14) gives the
upper half plane with the horizontal boundary excluded, and all points of
$i_{1}^{(D3)}$ and $i_{\infty}^{(D3)}$ shown in Figure 2 are contained in the
domain. (Namely, we can use $R_{(x,y)}$ as a physical domain.) If $i$ is
expanded with such a domain $R$, $I$ given by (12) can also be expanded in a
similar form;

$\displaystyle I=1+\sum_{(n_{x},n_{y})\in
R}c^{\prime}_{n_{x},n_{y}}x^{n_{x}}y^{n_{y}}$ (15)

with the same domain $R$. We can show that the map from
$\\{c_{n_{x},n_{y}}\\}$ to $\\{c^{\prime}_{n_{x},n_{y}}\\}$ is one to one if

  * •

$R$ is a sector with the center angle $\theta\leq\pi$.

  * •

$R\cap(-R)=\phi$.

The latter is the condition for the boundary of $R$. We do not include the
origin, and if $\theta=\pi$ we can include at most one of two boundary rays.
We always require $R$ to satisfy these conditions, and then we can define the
inverse operation to $\mathop{\rm Pexp}\nolimits$:

$\displaystyle i=\mathop{\rm Plog}\nolimits I.$ (16)

This is called the plethystic logarithm. It is important that we can introduce
the group structure for $\mathop{\rm Pexp}\nolimits i$ under the
multiplication, which corresponds to the addition for $i$.

Even for an interacting theory we can calculate the index by using
localization method if it is a Lagrangian theory. For the ${\cal N}=4$ SYM
with $U(N)$ gauge group the index is given by

$\displaystyle
I^{(D3)}_{N}|_{R_{(x,y)}}=\frac{1}{N!}\int\prod_{a=1}^{N}\frac{dz_{a}}{2\pi
iz_{a}}\prod_{a\neq b}\left(1-\frac{z_{a}}{z_{b}}\right)\mathop{\rm
Pexp}\nolimits\left(i_{1}^{(D3)}|_{R_{(x,y)}}\sum_{a,b}\frac{z_{a}}{z_{b}}\right),$
(17)

where $z_{a}$ ($a=1,\ldots,N$) are gauge fugacities. As is explicitly shown we
suppose $i_{1}^{(D3)}$ is given as an $R_{(x,y)}$-series. Usually $R_{(x,y)}$
is assumed to be a physical domain. This means that $|x^{n_{x}}y^{n_{y}}|<1$
is satisfied for all terms in $i_{1}^{(D3)}|_{R_{(x,y)}}$. If we take the unit
circle $|z_{a}|=1$ for the integration contours
$|x^{n_{x}}y^{n_{y}}\frac{z_{a}}{z_{b}}|<1$ is also satisfied, and
$\mathop{\rm Pexp}\nolimits$ in (17) makes sense. The contour integrals
extract the contribution of gauge invariant states and give $I_{N}$ as an
$R_{(x,y)}$-series.

##  3 Variable changes and analytic continuation

In the following we frequently use the variable changes in (6), and it is
convenient to introduce $\sigma_{x}$ and $\sigma_{y}$ to represent these
variable changes as follows.

$\displaystyle\sigma_{x}(x,y,q)=(x^{-1},q,y),\quad\sigma_{y}(x,y,q)=(q,y^{-1},x).$
(18)

If we use the triangular lattice to express the series expansion as in Figure
2, $\sigma_{x}$ and $\sigma_{y}$ are the reflections through the lines
perpendicular to $n_{x}$ and $n_{y}$ axes, respectively. With the maps
$\sigma_{i}$ ($i=x,y$) the relations in (6) are expressed in the simple form

$\displaystyle F_{i,N}^{(D3)}=\sigma_{i}I_{N}^{(D3)}\quad(i=x,y).$ (19)

We want to confirm that $F_{i,N}^{(D3)}$ given by (19) correctly reproduce
$I_{N}^{(D3)}$ via (3). Let $R$ be a physical domain used for the expansion of
$I_{N}^{(D3)}$. To confirm the relation (3) holds we need
$F_{i,N}^{(D3)}|_{R}$. However, the variable change $\sigma_{i}$ transforms
$I_{N}^{(D3)}|_{R}$ into $F_{i,N}^{(D3)}|_{\sigma_{i}R}$, where $\sigma_{i}R$
is the image of $R$ under the map $\sigma_{i}$. Therefore, to compare the left
hand side and the right hand side of (3), we need to resum or analytically
continue $F_{i,N}^{(D3)}|_{\sigma_{i}R}$ to $F_{i,N}^{(D3)}|_{R}$.

In general, if we have $i|_{R}$ for a domain $R$, it can also be regarded as
$i|_{R^{\prime}}$ for another domain $R^{\prime}$ which contains $R$ as a
subset. Therefore, it is convenient to define “a maximal domain” which cannot
be enlarged any more. $R$ is a maximal domain if

$\displaystyle R\cup(-R)\cup O=L,$ (20)

where $O$ is the origin and $L$ is the whole plane. A maximal domain $R$ is a
sector with center angle $\pi$, and only one of the boundary rays is included
in $R$.

Let us suppose that $f|_{R^{\prime}}$ for a maximal domain $R^{\prime}$ is
given and we want to obtain $(\mathop{\rm Pexp}\nolimits f)|_{R}$ for another
maximal domain $R\neq R^{\prime}$. We divide $R^{\prime}$ into the following
two parts.

$\displaystyle R_{1}=R^{\prime}\backslash R,\quad R_{2}=R^{\prime}\cap R.$
(21)

By definition, $R^{\prime}=R_{1}\cup R_{2}$ and $R=(-R_{1})\cup R_{2}$.
Correspondingly, we divide $f$ into two parts $f_{1}$ and $f_{2}$ so that
$f|_{R^{\prime}}=f_{1}|_{R_{1}}+f_{2}|_{R_{2}}$. The plethystic exponential of
$f|_{R^{\prime}}$ is factorized into two factors:

$\displaystyle\mathop{\rm Pexp}\nolimits f|_{R^{\prime}}=(\mathop{\rm
Pexp}\nolimits f_{1}|_{R_{1}})(\mathop{\rm Pexp}\nolimits f_{2}|_{R_{2}}).$
(22)

The factor $\mathop{\rm Pexp}\nolimits f_{2}|_{R_{2}}$ can be regarded as an
$R$-series. What we need to do to obtain $(\mathop{\rm Pexp}\nolimits f)|_{R}$
is to rewrite the other factor $\mathop{\rm Pexp}\nolimits f_{1}|_{R_{1}}$ as
a $(-R_{1})$-series. This can be done by using analytic continuation as
follows. Let us suppose that $f_{1}$ is given by

$\displaystyle f_{1}(x,y)=\sum_{(n_{x},n_{y})\in
R_{1}}c_{n_{x},n_{y}}x^{n_{x}}y^{n_{y}}.$ (23)

For values of $x$ and $y$ such that $R_{1}\subset R_{(x,y)}$ we can rewrite
the plethystic exponential in the product form

$\displaystyle\mathop{\rm Pexp}\nolimits
f_{1}|_{R_{1}}=\prod_{(n_{x},n_{y})\in
R_{1}}\frac{1}{(1-x^{n_{x}}y^{n_{y}})^{c_{n_{x},n_{y}}}}.$ (24)

Once we have obtained this expression, we can analytically continue this
function to values of $x$ and $y$ such that $-R_{1}\subset R_{(x,y)}$. Then,
we can rewrite (24) as

$\displaystyle\prod_{(n_{x},n_{y})\in
R_{1}}\frac{(-x^{n_{x}}y^{n_{y}})^{-c_{n_{x},n_{y}}}}{(1-x^{-n_{x}}y^{-n_{y}})^{c_{n_{x},n_{y}}}}=P(x,y)\mathop{\rm
Pexp}\nolimits f_{1}(x^{-1},y^{-1})|_{-R_{1}},$ (25)

where $P(x,y)$ is the monomial function

$\displaystyle P(x,y)=\prod_{(n_{x},n_{y})\in
R_{1}}(-x^{n_{x}}y^{n_{y}})^{-c_{n_{x},n_{y}}}.$ (26)

It is important that if $f_{1}$ has infinitely many terms (26) may not be
well-defined, and then our prescription does not work.

By combining (25) with the factor $\mathop{\rm Pexp}\nolimits f_{2}$ we obtain
the analytically continued plethystic exponential associated with the domain
$R$:

$\displaystyle\mathop{\rm Pexp}\nolimits_{R}f\equiv P\mathop{\rm
Pexp}\nolimits f^{\rm mod}|_{R}$ (27)

where $f^{\rm mod}$ is defined by

$\displaystyle f^{\rm mod}(x,y)=f_{1}(x^{-1},y^{-1})+f_{2}(x,y).$ (28)

Each point $(n_{x},n_{y})\in R_{1}$ in $f$ is moved to the opposite point
$(-n_{x},-n_{y})\in-R_{1}$ in the modified function $f^{\rm mod}$, and all
terms of $f^{\rm mod}$ are contained in $R$.

In fact, essentially the same prescription has been used in previous works.
Because giant gravitons are wrapped around topologically trivial cycles, there
are unwrapping modes with negative excitation energies. Such modes are handled
in [16, 17, 18] by applying the prescription explained above to the integrand
of the gauge fugacity integrals. The variable change of the index after the
gauge fugacity integrals has also been considered in [24]. Although detailed
explanation is not given, the pole structure caused by the factor (24) was
pointed out, and essentially the same method seems to be used.

The prescription explained above can be summarized in the following equations.

$\displaystyle F_{i,N}^{(T)}|_{R}=\mathop{\rm
Pexp}\nolimits_{R}\sigma_{i}i_{N}^{(T_{i})},\quad i_{N}^{(T_{i})}=\mathop{\rm
Plog}\nolimits I_{N}^{(T_{i})}.$ (29)

We first calculate $i_{N}^{(T)}$ as the plethystic logarithm of
$I_{N}^{(T_{i})}$, and then we calculate the analytically continued plethystic
exponential.

Unfortunately, it will turn out that this prescription works only in limited
cases. If $\sigma_{i}i_{N}^{(T_{i})}$ has infinitely many points outside the
domain $R$, then the factor (26) becomes an infinite product which in general
does not converge. Even so, we will find that (29) is quite effective for
simple expansion (7).

##  4 Numerical tests

In this section we numerically calculate $F_{i,N}^{(T)}$ from
$I_{N}^{(T_{i})}$ by (29) in some examples and confirm the consistency with
known results. The main results in 4.2 and 4.3 have been already given in
[24].

###  4.1 Double expansion of the Schur index for $T=D3$

We first consider the double giant graviton expansion of the Schur index of
$T=D3$ studied in [17]. Let us define ${\mathfrak{q}}$ and $u$ by

$\displaystyle x={\mathfrak{q}}u,\quad y={\mathfrak{q}}u^{-1},\quad
q={\mathfrak{q}}^{2}.$ (30)

In [17] functions $F_{i,N}^{(D3)}$ are treated as ${\mathfrak{q}}$-series and
the coefficients are given as rational functions of $u$.444The fugacity
${\mathfrak{q}}$ is denoted by $q$ in [17]. Here we do the further $u^{-1}$
expansion ($u$ expansion around $u=\infty$) after the
${\mathfrak{q}}$-expansion. This corresponds to the maximal physical domain
(Figure 3 (a))

$\displaystyle R_{qu}$ $\displaystyle=\lim_{{\mathfrak{q}}\rightarrow
0,|u|>1}R_{(x,y)}$
$\displaystyle=\\{(n_{x},n_{y})|n_{x}+n_{y}>0\\}\cup\\{(n_{x},n_{y})|n_{x}+n_{y}=0,n_{y}>0\\}.$
(31)

Figure 3: The maximal domains $R_{qu}$ and $R_{yx}$

Let us first consider $F_{x,1}^{(D3)}=\mathop{\rm
Pexp}\nolimits(\sigma_{x}i_{1}^{(D3)})$ and $F_{y,1}^{(D3)}=\mathop{\rm
Pexp}\nolimits(\sigma_{y}i_{1}^{(D3)})$. We find $\sigma_{x}i_{1}^{(D3)}$ has
one term $x^{-1}={\mathfrak{q}}^{-1}u^{-1}$ that is not contained in $R_{qu}$,
and we have to move it to the opposite point to define the modified letter
index. For $\sigma_{y}i_{1}^{(D3)}$ we need to move two points corresponding
to $xy^{-1}=u^{2}$ and $y^{-1}={\mathfrak{q}}^{-1}u$. See Figure 4.

Figure 4: The modification of letter indices $\sigma_{x}i_{1}^{(D3)}$ and
$\sigma_{y}i_{1}^{(D3)}$

Following the prescription explained in the previous section, we can calculate
$F_{x,1}^{(D3)}|_{R_{qu}}$ and $F_{y,1}^{(D3)}|_{R_{qu}}$. The results are
shown in Figure 5.

Figure 5: (a) $F_{x,1}^{(D3)}|_{R_{qu}}$ (b) $F_{y,1}^{(D3)}|_{R_{qu}}$

These agree with the results in [17]:555$F_{x,N}(x,y)$ is denoted by
$F_{N}(q,u)$ in [17] and $F_{y,N}(x,y)$ corresponds to $F_{N}(q,u^{-1})$.
$F_{N}$ are given in [17] as ${\mathfrak{q}}$-series. We need further $u^{-1}$
expansion to obtain (32), (33), (34), and (35).

$\displaystyle F_{x,1}^{(D3)}$
$\displaystyle=(-u-u^{-1}-u^{-3}-\cdots){\mathfrak{q}}$
$\displaystyle+(-u^{2}+1){\mathfrak{q}}^{2}+(-u^{3}+u^{-3}){\mathfrak{q}}^{3}+(-u^{4}+1-u^{-2}+u^{-6}){\mathfrak{q}}^{4}+\cdots,$
$\displaystyle F^{(D3)}_{y,1}$
$\displaystyle=(u^{-3}+u^{-5}+u^{-7}+\cdots){\mathfrak{q}}$
$\displaystyle+(1-u^{-2}){\mathfrak{q}}^{2}+(u^{3}-u^{-3}){\mathfrak{q}}^{3}+(u^{6}-u^{2}+1-u^{-4}){\mathfrak{q}}^{4}+\cdots.$
(32)

This is not surprising because essentially the same prescription is used in
[17] for the single giant graviton sector.

Next, let us apply (29) to $F_{i,N}^{(D3)}$ with $N\geq 2$. We first calculate
$I_{N}^{(D3)}$ by (17), and calculate their plethystic logarithms
$i_{N}^{(D3)}=\mathop{\rm Plog}\nolimits I_{N}^{(D3)}$. $i_{N}^{(D3)}$ have
information not only about letters generating the operator spectra but also
about non-trivial constraints among letters called “syzygies” [27]. The
results for $N=2,3,4$ are shown in Figure 6.

Figure 6: $i_{N}^{(D3)}$ for $N=2,3,4$ are shown. The distributions are left-
right symmetric, and we show only the left half ($n_{x}\geq n_{y}$) of each
plot. The dotted lines show a numerical cut-off $n_{x}+n_{y}\leq 20$.

The asymptotic behavior of the distribution of dots is important for the
following analysis. We find in the plots in Figure 6 that except for finite
number of dots near the origin almost all dots are contained in the sectors
bounded by $E_{1}:n_{y}=Nn_{x}$ and $E_{2}:n_{x}=Nn_{y}$.

For $N=2$ the image of $E_{2}$ under $\sigma_{x}$, $\sigma_{x}E_{2}$,
coincides with the right half of the horizontal line, which is the boundary of
$R_{qu}$ (Figure 7).

Figure 7: The modification of $\sigma_{x}i_{2}^{(D3)}$ for $R_{qu}$. The
dotted line shows a numerical cut-off.

There are three points outside $R_{qu}$ corresponding to terms
$x^{-1}={\mathfrak{q}}^{-1}u$, $x^{-2}={\mathfrak{q}}^{-2}u^{2}$, and
$x^{-2}y={\mathfrak{q}}^{-1}u^{3}$. These give the factor
$P=-{\mathfrak{q}}^{4}u^{-6}=-x^{5}y^{-1}$, and the analytically continued
plethystic exponential is shown in Figure 8.

Figure 8: $F_{x,2}^{(D3)}|_{R_{qu}}$ is shown. The dotted line shows a
numerical cut-off.

Again, we find agreement with the corresponding result in [17]:

$\displaystyle F_{x,2}^{(D3)}$
$\displaystyle=(-u^{6}+u^{4}+2+u^{-2}+3u^{-4}+2u^{-6}+\cdots){\mathfrak{q}}^{4}$
$\displaystyle+(-u^{9}+2u^{5}){\mathfrak{q}}^{5}+(-u^{12}+2u^{6}+2){\mathfrak{q}}^{6}+\cdots.$
(33)

Next, let us consider $F_{y,2}^{(D3)}$. This time, $\sigma_{y}E_{1}$ is the
semi-infinite horizontal line, and unlike the previous case the line is not
contained in $R_{qu}$. This means that there are infinitely many points
outside $R_{qu}$, and the factor (26) becomes infinite product. Therefore,
(29) does not work. In fact, this is also expected from $F_{y,2}^{(D3)}$ in
[17]:

$\displaystyle F_{y,2}^{(D3)}({\mathfrak{q}},u)$
$\displaystyle=(2u^{-10}+u^{-12}+3u^{-14}+\cdots){\mathfrak{q}}^{4}+\cdots.$
(34)

As we see, the leading term has coefficient $2$, and it is impossible to
obtain such a series by the plethystic exponential.

For $N=3$ and $4$ the situation is worse. Both $\sigma_{x}$ and $\sigma_{y}$
maps one of $E_{1}$ and $E_{2}$ below the boundary of $R_{qu}$, and there are
infinitely many points outside $R_{qu}$, which make the factor (26) ill-
defined. We can also see that the leading term coefficients of the functions
in [17] are not $\pm 1$.

$\displaystyle F_{x,3}^{(D3)}$
$\displaystyle=(-2u^{15}+u^{13}+\cdots){\mathfrak{q}}^{9}+\cdots,$
$\displaystyle F_{y,3}^{(D3)}$
$\displaystyle=(5u^{-21}+2u^{-23}+\cdots){\mathfrak{q}}^{9}+\cdots,$
$\displaystyle F_{x,4}^{(D3)}$
$\displaystyle=(-5u^{28}+2u^{26}+\cdots){\mathfrak{q}}^{16}+\cdots,$
$\displaystyle F_{y,4}^{(D3)}$
$\displaystyle=(14u^{-36}+5u^{-38}+\cdots){\mathfrak{q}}^{16}+\cdots.$ (35)

These expansions cannot be reproduced by the relation (29).

###  4.2 Simple expansion of the Schur index for $T=D3$

Let us consider the simple giant graviton expansion investigated in [24]. The
index is treated in [24] as a series obtained by the successive expansions in
which $y$ expansion is carried out first and then $x$ expansion is done for
the coefficients of the first expansion. The associated maximal physical
domain is (Figure 3 (b))

$\displaystyle R_{yx}=\lim_{y\rightarrow
0,|x|<1}R_{(x,y)}=\\{(n_{x},n_{y})|n_{y}>0\\}\cup\\{(n_{x},0)|n_{x}>0\\}.$
(36)

Let us start with the analysis of $F_{x,1}^{(D3)}|_{R_{yx}}$ and
$F_{y,1}^{(D3)}|_{R_{yx}}$.

Figure 9: (a) The modification of $\sigma_{x}i_{1}$. (b) $\sigma_{y}i_{1}$

For $\sigma_{x}i_{1}^{(D3)}$, there is one dot outside $R_{yx}$ (Figure 9
(a)). This is the same as in Figure 4 (a), and hence we obtain the same result
for $F_{x,1}^{(D3)}$ as the previous subsection (Figure 5 (a)). However, for
$\sigma_{y}i_{1}^{(D3)}$, there are infinitely many points outside $R_{yx}$
corresponding to terms $y^{-1}x^{k}$ ($k=0,1,2,\ldots$) (Figure 9 (b)), and
(26) becomes

$\displaystyle P=\prod_{k=0}^{\infty}(-yx^{-k}).$ (37)

Although this is an infinite product, we can give it significance. $P$
includes the factor $y^{+\infty}$, and the contribution decouples. We can
simply treat $F_{y,1}^{(D3)}|_{R_{yx}}$ as $0$. Namely, as far as single-
wrapping giant graviton contributions are concerned, the difference between
the multiple expansion (1) and the simple expansion (7) comes from the
different choice of the domains.

Let us proceed to $N\geq 2$ contributions. For $\sigma_{y}i_{N}^{(D3)}$, we
find infinitely many points outside $R_{qu}$, just like the $N=1$ case.
Because both positive and negative coefficients appear, it is not clear
whether they decouple like $F_{y,1}^{(D3)}$. Here, based on the analysis in
[24], we simply assume their decoupling and focus only on $F_{x,N}^{(D3)}$.

In $i_{N}^{(D3)}$ shown in Figure 6 we find $N$ dots on the positive part of
the $n_{x}$ axis corresponding to the $1/2$ BPS operators $\mathop{\rm
tr}\nolimits X^{k}$ ($k=1,2,\ldots,N$). By $\sigma_{x}$ they are mapped to the
$N$ points on the negative part of the $n_{x}$ axis. Because they are not
contained in $R_{yx}$ we need to move them back to the opposite points on the
positive part of the $n_{x}$ axis (Figure 10).

Figure 10: The modification of $i_{N}^{(D3)}$ for $R_{yx}$ for $N=2,3,4$. (a)
$\sigma_{x}i_{2}$. (b) $\sigma_{x}i_{3}$. (c) $\sigma_{x}i_{4}$. The dotted
lines show a numerical cut-off

The factorization corresponding to (22) is

$\displaystyle
I_{N}^{(D3)}(x,y)=\frac{1}{\prod_{k=1}^{N}(1-x^{k})}I^{\prime(D3)}_{N}(x,y),$
(38)

The fractional factor coming from the $N$ points on the $n_{x}$ axis and
produces poles on the $x$-plane at $k$-roots of unity with $1\leq k\leq N$. As
is pointed out in [24] these are only poles we need to take care of in the
analytic continuation. The function $I^{\prime(D3)}_{N}$ corresponds to all
other points with $n_{y}\geq 1$ in the range $0\leq n_{x}\leq(N+1)n_{y}$. The
coefficient of $y$-expansion $I^{\prime(D3)}_{N}$ at order $n_{y}$ is a
Laurant polynomial of $x$ consisting of terms in this range. Therefore, we can
safely perform the variable change $\sigma_{x}$ for the function
$I^{\prime(D3)}_{N}$. After the variable change we obtain

$\displaystyle
F_{x,N}^{(D3)}=\frac{(-1)^{N}x^{\frac{N(N+1)}{2}}}{\prod_{k=1}^{N}(1-x^{k})}\sigma_{x}I^{\prime(D3)}_{N}$
(39)

where $\sigma_{x}I^{\prime(D3)}_{N}$ has terms in the range $-Nn_{y}\leq
n_{x}\leq n_{y}$ for each order of its $y$-expansion. Including the fractional
factor with the numerator $(-1)^{N}x^{\frac{N(N+1)}{2}}$, we obtain
$F_{x,N}^{(D3)}|_{R_{yx}}$ consisting of terms in the region (Figure 11)

$\displaystyle\frac{N(N+1)}{2}-Nn_{y}\leq n_{x}.$ (40)

Figure 11:  (a) $F_{x,2}^{(D3)}|_{R_{yx}}$. (b) $F_{x,3}^{(D3)}|_{R_{yx}}$.
(c) $F_{x,4}^{(D3)}|_{R_{yx}}$. The dotted lines show a numerical cut-off.

As is confirmed in [24], the simple expansion (7) correctly reproduces the
index $I_{N}^{(D3)}$ for different values of $N$. See Figure 12.

Figure 12:  The result of numerical check for $N=3$. (a) The ratio
$I_{3}^{(D3)}/I_{\infty}^{(D3)}$. (b)
$(I_{3}^{(D3)}/I_{\infty}^{(D3)})-\sum_{m=0}^{4}x^{mN}F_{x,m}^{(D3)}$. The
dotted lines show numerical cut-offs. The red lines labeled by $m$ show
thresholds for respective wrapping numbers. Up to expected errors due to the
$m=5$ contribution indicated by the line labeled by $m=5$, all terms are
correctly canceled.

###  4.3 Superconformal index for $T=D3$

In the analysis of the Schur index we could focus on $F_{i,N}^{(D3)}|_{R}$
thanks to the factorization (4). This is not the case for the superconformal
index without the Schur limit taken. The superconformal index of the ${\cal
N}=4$ SYM is defined by [4]

$\displaystyle I^{(D3)}=\mathop{\rm
tr}\nolimits[(-1)^{F}q^{J_{1}}p^{J_{2}}x^{R_{x}}y^{R_{y}}z^{R_{z}}],\quad
qp=xyz.$ (41)

The giant graviton expansion investigated in [18] is a triple expansion with
functions $F_{m_{x},m_{y},m_{z}}^{(D3)}$ in the summand, and the involution
(5) gives the following relation between $F_{x,N}^{(D3)}=F_{N,0,0}^{(D3)}$ and
$I_{N}^{(D3)}$ [16, 24]:

$\displaystyle F_{x,N}^{(D3)}(p,q,x,y,z)=I_{N}^{(D3)}(y,z,x^{-1},p,q).$ (42)

Let $\sigma_{x}$ denote this variable change. $F_{0,N,0}^{(D3)}$ and
$F_{0,0,N}^{(D3)}$ are also related to $I_{N}^{(D3a)}$ via similar variable
changes, which we denote by $\sigma_{y}$ and $\sigma_{z}$, respectively.

Due to the absence of the factorization, we need to calculate functions
$F_{m_{x},m_{y},m_{z}}^{(D3)}$ directly by using localization formula similar
to (17). The letter index appearing in the integrand of the localization
formula contains $\sigma_{x}i_{1}^{(D3)}$, $\sigma_{y}i_{1}^{(D3)}$, and
$\sigma_{z}i_{1}^{(D3)}$ at the same time, where $i_{1}^{(D3)}$ is given by
[4]

$\displaystyle i_{1}^{(D3)}=1-\frac{(1-x)(1-y)(1-z)}{(1-q)(1-p)}.$ (43)

Even if we use a maximal domain we cannot cover three functions
$\sigma_{i}i_{1}^{(D3)}$ with a single domain, and we need to use deformed
contours for the gauge fugacity integrals [18, 25].

Despite the absence of the factorization, the analysis in [24] suggests that
only $F_{x,m}^{(D3)}$ contribute to the index if we use an appropriate domain,
and then our relation (29) is still useful to obtain $F_{x,N}^{(D3)}$ from
$I_{N}^{(D3)}$. Corresponding to four independent variables we need to use a
four-dimensional lattice to express full expansion of the index. To make two-
dimensional plot possible we project the four-dimensional lattice onto the
two-dimensional lattice corresponding to the unrefinement $y=z$ and $p=q$.
Then, the index becomes function of $x$ and $y$ just like the Schur
index.666The projection is used only when we show functions as two-dimensional
plots, and we do not take any unrefinement in the calculation. We regard the
index as a function of $x$, $\sqrt{yz}$, $y/z$, and $q/p$, and treat
$\sqrt{yz}$ as if $y$ in the previous subsection. Although we neglect $y/z$
and $q/p$ in the graphical expression, we keep all variables in the
calculation.  The letter indices $i_{1}^{(D3)}$ and $\sigma_{i}i_{1}^{(D3)}$
($i=x,y,z$) after the projection are shown in Figure 13.

Figure 13: (a) The letter index $i_{1}^{(D3)}$ (b) The modification of
$\sigma_{x}i_{1}^{(D3)}$ for $R_{yx}$. (c) $\sigma_{y}i_{2}^{(D3)}$ and
$\sigma_{z}i_{2}^{(D3)}$ are identical after the projection to the two-
dimensional lattice.

We see that if we use the domain $R_{yx}$ $F_{y,1}^{(D3)}$ and
$F_{z,1}^{(D3)}$ decouple just like the Schur index, and $F_{x,1}^{(D3)}$ is
the only contribution with $m_{x}+m_{y}+m_{z}=1$. Although we cannot prove the
decoupling of all $F_{m_{x},m_{y},m_{z}}^{(D3)}$ with $m_{y}+m_{z}\geq 1$, the
analysis in [24] shows that we can reproduce the finite $N$ index by the
simple expansion containing only $F_{x,m}^{(D3)}$. See Figure 14 for the
result of a numerical test.

Figure 14: (a) The ratio of the superconformal indices
$I_{N}^{(D3)}/I_{\infty}^{(D3)}$ for $N=3$. (b)
$(I_{N}^{(D3)}/I_{\infty}^{(D3)})-\sum_{m=0}^{4}x^{mN}F_{x,N}^{(D3)}$ for
$N=3$. The dotted lines show numerical cut-offs. The red lines labeled by $m$
show thresholds for respective wrapping numbers. Only small number of dots,
which are consistent with the higher order contributions with $m\geq 5$, are
left.

###  4.4 M5 from M2

Let us apply our method to the theory realized on $N$ coincident M5-branes,
which we denote by $T=M5$. It is the $6$-dim ${\cal N}=(2,0)$ theory of
$A_{N-1}$ type together with a free tensor multiplet.

The superconformal index is defined by [28]

$\displaystyle I_{N}^{(M5)}=\mathop{\rm
tr}\nolimits[(-1)^{F}\check{q}_{1}^{\check{J}_{12}}\check{q}_{2}^{\check{J}_{34}}\check{q}_{3}^{\check{J}_{56}}\check{x}_{1}^{\check{R}_{12}}\check{x}_{2}^{\check{R}_{34}}]\quad\check{q}_{1}\check{q}_{2}\check{q}_{3}=\check{x}_{1}\check{x}_{2},$
(44)

where $\check{J}_{12}$, $\check{J}_{34}$, and $\check{J}_{56}$ are Cartan
generators of $SO(6)_{\rm spin}$ and $\check{R}_{12}$ and $\check{R}_{34}$ are
Cartan generators of $SO(5)_{R}$.

The index is given as a double giant graviton expansion associated with two
two-cycles in [21], and the single giant graviton sector has been studied. The
cycles are defined as the fixed loci of $\check{R}_{12}$ and $\check{R}_{34}$
in $M_{M5}=S^{4}$. We call $\check{R}_{12}$ ($\check{R}_{34}$) fixed locus
“12-cycle” (“34-cycle”). M2-branes wrapped around these cycles contribute to
the index.

The superconformal index of the $3$d ${\cal N}=8$ SCFT realized on coincident
M2-branes is defined by [28]

$\displaystyle I_{N}^{(M2)}=\mathop{\rm
tr}\nolimits[(-1)^{F}\hat{q}^{\hat{J}_{12}}\hat{x}_{1}^{\hat{R}_{12}}\hat{x}_{2}^{\hat{R}_{34}}\hat{x}_{3}^{\hat{R}_{56}}\hat{x}_{4}^{\hat{R}_{78}}]\quad\hat{q}=\hat{x}_{1}\hat{x}_{2}\hat{x}_{3}\hat{x}_{4},$
(45)

where $\hat{J}_{12}$ is the spin and $\hat{R}_{12}$, $\hat{R}_{34}$,
$\hat{R}_{56}$, and $\hat{R}_{78}$ are $SO(8)_{R}$ Cartan generators. We can
calculate this index for an arbitrary $N$ by applying the localization method
[29] to the ABJM theory with Chern-Simons level $k=1$ [30].

The Cartan generators of the $3$d ${\cal N}=8$ superconformal algebra acting
on M2-branes wrapped around $12$-cycle and those of the $6$d ${\cal N}=(2,0)$
superconformal algebra acting on the $AdS_{7}$ boundary are related by [21]

$\displaystyle(\hat{H},\hat{J}_{12},\hat{R}_{12},\hat{R}_{34},\hat{R}_{56},\hat{R}_{78})=(\tfrac{1}{2}\check{H}-\tfrac{3}{2}\check{R}_{12},\check{R}_{34},\check{J}_{12},\check{J}_{34},\check{J}_{56},-\check{R}_{12}).$
(46)

Correspondingly, the fugacities in (44) and those in (45) are related by

$\displaystyle\sigma_{12}(\hat{q},\hat{x}_{1},\hat{x}_{2},\hat{x}_{3},\hat{x}_{4})=(\check{x}_{2},\check{q}_{1},\check{q}_{2},\check{q}_{3},\check{x}_{1}^{-1}).$
(47)

The variable change associated with $34$-cycle, $\sigma_{34}$, is given by
swapping $\check{x}_{1}$ and $\check{x}_{2}$ after $\sigma_{12}$. the M2-giant
contributions $F_{12,N}^{(M5)}$ are obtained from the index $I_{N}^{(M2)}$ by
the relation

$\displaystyle F_{12,N}^{(M5)}=\sigma_{12}I_{N}^{(M2)}.$ (48)

Again, there are four independent variables and the lattice is four-
dimensional. We want to define a projection to two-dimensional lattice to show
an expansion as a two-dimensional plot. We introduce variables $x$, $y$, and
$u_{i}$ ($i=1,2,3$) by

$\displaystyle\check{q}_{i}=yu_{i}\quad(u_{1}u_{2}u_{3}=1),\quad\check{x}_{1}=x,\quad\check{x}_{2}=x^{-1}y^{3}.$
(49)

We focus on $x$ and $y$ to show series in figures.

We also rewrite the fugacities for M2-branes as follows:

$\displaystyle\hat{q}=xy^{3},\quad\hat{x}_{i}=yu_{i}\quad(u_{1}u_{2}u_{3}=1),\quad\hat{x}_{4}=x.$
(50)

Then, the variable change (47) becomes

$\displaystyle\sigma_{12}(x,y,u_{i})=(x^{-1},y,u_{i}).$ (51)

Namely, (48) becomes

$\displaystyle F_{12,N}^{(M5)}(x,y,u_{i})=I_{N}^{(M2)}(x^{-1},y,u_{i}).$ (52)

As in the previous examples, let us first look at the letter index for a
single giant graviton. The theory on a single M2-brane is the free theory of
an ${\cal N}=8$ scalar multiplet with the letter index [28]

$\displaystyle
i_{1}^{(M2)}=\frac{\hat{x}_{1}+\hat{x}_{2}+\hat{x}_{3}+\hat{x}_{4}-\hat{q}(\hat{x}_{1}^{-1}+\hat{x}_{2}^{-1}+\hat{x}_{3}^{-1}+\hat{x}_{4}^{-1})}{1-\hat{q}}.$
(53)

We show $i_{1}^{(M2)}$, $\sigma_{12}i_{1}^{(M2)}$, and
$\sigma_{34}i_{1}^{(M2)}$ projected on the two-dimensional plane in Figure 15.

Figure 15: (a) The letter index $i_{1}^{(M2)}$. (b) The modification of
$\sigma_{1}i_{1}^{(M2)}$ for $R_{yx}$. (c) Infinitely many dots of
$\sigma_{2}i_{1}^{(M2)}$ are out of $R_{yx}$.

If we take the domain $R_{yx}$ defined in (36), $\sigma_{34}i_{1}^{(M2)}$ has
infinitely many points outside $R_{yx}$. They give the factor

$\displaystyle
P=\prod_{k=1}^{\infty}(-y^{3}x^{-k})\prod_{k=1}^{\infty}(-yx^{-k})^{-3}.$ (54)

If we take the product for each $k$ first, we obtain
$\prod_{k=1}^{\infty}x^{2k}=x^{\infty}$ and the contribution decouples. Just
like the previous examples, we simply assume that $F_{m_{12},m_{34}}^{(M5)}$
with $m_{34}\geq 1$ decouple, and let us consider the simple expansion
associated with the $12$-cycle

$\displaystyle\frac{I_{N}^{(M5)}}{I_{\infty}^{(M5)}}=\sum_{m=0}^{\infty}x^{mN}F_{12,m}^{(M5)}.$
(55)

Let us numerically confirm (55) holds for small $N$. $I_{N}^{(M5)}$ with
$N=-1,0,1$ and $\infty$ are given by

$\displaystyle I_{-1}^{(M5)}=0,\quad I_{0}^{(M5)}=1,\quad
I_{1}^{(M5)}=\mathop{\rm Pexp}\nolimits i_{1}^{(M5)},\quad
I_{\infty}^{(M5)}=\mathop{\rm Pexp}\nolimits i_{\infty}^{(M5)},$ (56)

where $i_{1}^{(M5)}$ and $i_{\infty}^{(M5)}$ are the letter indices of the
tensor multiplet and the supergravity multiplet in $AdS_{7}\times S^{4}$. They
are given by [28]

$\displaystyle
i_{1}^{(M5)}=\frac{\check{x}_{1}+\check{x}_{2}-\check{q}_{1}\check{q}_{2}-\check{q}_{2}\check{q}_{3}-\check{q}_{3}\check{q}_{1}+\check{x}_{1}\check{x}_{2}}{(1-\check{q}_{1})(1-\check{q}_{2})(1-\check{q}_{3})},$
(57)

and

$\displaystyle
i_{\infty}^{(M5)}=\frac{\check{x}_{1}+\check{x}_{2}-\check{q}_{1}\check{q}_{2}-\check{q}_{2}\check{q}_{3}-\check{q}_{3}\check{q}_{1}+\check{x}_{1}\check{x}_{2}(\check{q}_{1}+\check{q}_{2}+\check{q}_{3}-\check{x}_{1}-\check{x}_{2})}{(1-\check{x}_{1})(1-\check{x}_{2})(1-\check{q}_{1})(1-\check{q}_{2})(1-\check{q}_{3})}.$
(58)

By using these we can calculate the left hand side of (55). In the following
we calculate the right hand side of (55).

We first calculate the index $I_{N}^{(M2)}$ for different $N$ with the
physical domain $R_{yx}$. This is done by applying localization method [29] to
the ABJM theory [30]. The plethystic logarithms $i_{N}^{(M2)}$ for small $N$
are shown in Figure 16.

Figure 16: $i_{N}^{(M2)}=\mathop{\rm Plog}\nolimits I_{N}^{(M2)}$ for
$N=1,2,3,4$ are shown. As shown by the dotted lines the $y$ expansions are cut
off at $n_{y}=3$.

They have similar structure to $i_{N}^{(D3)}$. In particular, for each $N$
there are $N$ dots on the positive part of the vertical axis corresponding to
$N$ $1/2$ BPS operators. After the variable change (51), these dots have to be
moved to the opposite points to obtain the modified letter index. The
expansion of the analytically continued plethystic exponential
$F_{12,N}^{(M5)}|_{R_{yx}}=\mathop{\rm
Pexp}\nolimits_{R_{yx}}\sigma_{12}i_{N}^{(M2)}$ for $N=1,2,3,4$ are shown in
Figure 17.

Figure 17: M2-giant contributions $F_{12,m}^{(M5)}$ with wrapping number
$m=1,2,3,4$ are shown. As shown by the dotted lines the $y$ expansions are cut
off at $n_{y}=3$.

For $N\geq 2$ the dots are in the region above the line
$n_{x}=\frac{m(m+1)}{2}+mN-mn_{y}$. ($N=1$ case is exceptional and the line is
$n_{x}=1-\frac{2}{3}n_{y}$.)

With the functions $F_{12,m}^{(M5)}$ obtained above we can confirm that (55)
holds for $N=-1,0,1$ up to expected errors. See Figure 18.

Figure 18: The results of consistency check for $N=-1,0,1$.
$I_{N}^{(M5)}/I_{\infty}^{(M5)}$ for $N=0$ and $N=1$ are shown in (b) and (d),
respectively. Trivial one, $I_{-1}^{(M5)}/I_{\infty}^{(M5)}=0$ is not shown.
$(I_{N}^{(M5)}/I_{\infty}^{(M5)})-\sum_{m=0}^{4}x^{mN}F_{12,m}^{(M5)}$ for
$N=-1$, $0$, and $1$ are shown in (a), (c), and (e),respectively. As shown by
the dotted lines the $y$ expansions are cut off at $n_{y}=3$. In (a), (c), and
(e), all terms below the $m=5$ lines are correctly canceled, and only expected
errors due to $m\geq 5$ contributions are left.

This strongly suggests that (55) correctly gives $I_{N}^{(M5)}$ for an
arbitrary $N$. Although we only show the two-dimensional plots, we can
calculate the full superconformal index. See Appendix A.

We can also compare our results with the analytic result for the Schur-like
limit [31, 32] obtained from the analysis of five-dimensional SYM. It is known
that by setting $\check{q}_{1}=\check{x}_{1}$, the index $I_{N}^{(M5)}$
reduces to the following function of the single variable $\check{x}_{1}$:

$\displaystyle I_{N}^{(M5)}=\mathop{\rm
Pexp}\nolimits\frac{\check{x}_{1}+\check{x}_{1}^{2}+\cdots+\check{x}_{1}^{N}}{1-\check{x}_{1}}.$
(59)

If we take the same limit in our results shown in Appendix A we obtain

$\displaystyle I_{2}^{(M5)}$
$\displaystyle=1+\check{x}_{1}+3\check{x}_{1}^{2}+5\check{x}_{1}^{3}+10\check{x}_{1}^{4}+\underline{15\check{x}_{1}^{5}}+{\cal
O}(\check{x}_{1}^{6}),$ $\displaystyle I_{3}^{(M5)}$
$\displaystyle=1+\check{x}_{1}+3\check{x}_{1}^{2}+6\check{x}_{1}^{3}+12\check{x}_{1}^{4}+\underline{20\check{x}_{1}^{5}}+{\cal
O}(\check{x}_{1}^{6}),$ $\displaystyle I_{4}^{(M5)}$
$\displaystyle=1+\check{x}_{1}+3\check{x}_{1}^{2}+6\check{x}_{1}^{3}+13\check{x}_{1}^{4}+\underline{22\check{x}_{1}^{5}}+{\cal
O}(\check{x}_{1}^{6}),$ $\displaystyle I_{5}^{(M5)}$
$\displaystyle=1+\check{x}_{1}+3\check{x}_{1}^{2}+6\check{x}_{1}^{3}+13\check{x}_{1}^{4}+\underline{23\check{x}_{1}^{5}}+{\cal
O}(\check{x}_{1}^{6}).$ (60)

We only showed terms independent of $\check{q}_{2}$ and $\check{q}_{3}$. The
underlines indicate the terms are incorrect. These are consistent with (59)
because in our numerical results the $y$ expansion is cut off at $n_{y}=3$ and
the expected order of errors is ${\cal O}(\check{x}_{1}^{4})$ or higher. The
$\check{q}_{2}$ or $\check{q}_{3}$-dependent terms which we did not show in
(60) are also consistent with the expected errors.

Note that the Schur-like limit is ill-defined for each M2-giant contribution
because the limit corresponds to $\hat{x}_{1}=\hat{x}_{4}^{-1}$ via (47). The
terms $\hat{x}_{1}$ and $\hat{x}_{1}^{-1}$ appear at the same time in the
letter index $i_{1}^{(M2)}$ in the limit, and there is no domain covering both
of them. We need to take the limit after summing up the contributions of
M2-giants.

###  4.5 M2 from M5

It is possible to interchange the roles of M2 and M5 in the previous
subsection. Namely, we can calculate the finite $N$ index of M2 theory by
summing up M5-giant contributions by the relation

$\displaystyle\frac{I_{N}^{(M2)}|_{R_{yx}}}{I_{\infty}^{(M2)}|_{R_{yx}}}=\sum_{m=0}^{\infty}x^{mN}(\sigma_{12}^{-1}I_{m}^{(M5)})|_{R_{yx}},$
(61)

where the large $N$ index is given by $I_{\infty}^{(M2)}=\mathop{\rm
Pexp}\nolimits i_{\infty}^{(M2)}$ with the letter index [28]

$\displaystyle
i_{\infty}^{(M2)}=\frac{(1-\hat{q}\hat{x}_{1}^{-1})(1-\hat{q}\hat{x}_{2}^{-1})(1-\hat{q}\hat{x}_{3}^{-1})(1-\hat{q}\hat{x}_{4}^{-1})}{(1-\hat{x}_{1})(1-\hat{x}_{2})(1-\hat{x}_{3})(1-\hat{x}_{4})(1-\hat{q})^{2}}-\frac{1-\hat{q}+\hat{q}^{2}}{(1-\hat{q})^{2}}.$
(62)

The calculation to test (61) is parallel to the previous example, and we only
give a brief explanation.

In [21] four five-cycles in $M_{M5}=S^{7}$ were taken into account. As the
previous examples we can show for a single M5 giant that if we take $R_{yx}$
only one of the four cycles, $\hat{R}_{78}$ fixed locus, gives non-trivial
contribution. Hence, we focus on $F_{78,N}^{(M2)}$ associated with the cycle.

By using the results of previous subsection for $I_{N}^{(M5)}$ we can obtain
the plethystic logarithms $i_{N}^{(M5)}=\mathop{\rm Plog}\nolimits
I_{N}^{(M5)}$ for small $N$ shown in Figure 19.

Figure 19: $i_{N}^{(M5)}$ for $N=1,2,3,4,5$ are shown. As shown by the dotted
lines the $y$ expansions are cut off at $n_{y}=3$.

As the previous examples we find $N$ dots associated with the $1/2$ BPS
operators on the positive part of the vertical axis for each $N$. The
functions $F_{78,N}^{(M2)}=\mathop{\rm
Pexp}\nolimits_{R_{yx}}\sigma_{12}^{-1}i_{N}^{(M5)}$ are shown in Figure 20.

Figure 20: M5 giants contributions $F_{78,N}^{(M2)}$ for $N=1,2,3,4,5$ are
shown. As shown by the dotted lines the $y$ expansions are cut off at
$n_{y}=3$.

The results of a numerical check of (61) for $N=-1,0,1$ are shown in Figure
21.

Figure 21:  $(I_{N}^{(M2)}/I_{\infty}^{(M2)})$ for $N=0,1$ are shown in (b)
and (d), respectively. The trivial one $I_{-1}^{(M2)}/I_{\infty}^{(M2)}=0$ is
not shown.
$(I_{N}^{(M2)}/I_{\infty}^{(M2)})-\sum_{m=0}^{5}x^{mN}F_{78,m}^{(M2)}$ for
$N=-1,0,1$ are shown in (a), (c), and (e), respectively. As shown by the
dotted lines the $y$ expansions are cut off at $n_{y}=3$. In (a), (c), and (e)
all terms below $m=6$ lines are correctly canceled, and only expected errors
due to $m\geq 6$ contributions are left.

##  5 Conclusions and Discussion

We investigated the giant graviton expansions. In particular, we concentrated
on coincident giant gravitons wrapped around a single cycle. Because the
expansion domain for giant gravitons and that of the boundary theory are
different, we need to perform an analytic continuation to relate them. We
proposed a simple prescription (29) to realize it.

We explicitly showed for the Schur index of $T=D3$ that different choices of
the expansion domains give different functions for single giant graviton
contributions $F_{i,1}^{(D3)}$ ($i=x,y$). If we use $R_{qu}$ in (31) both
$F_{x,1}^{(D3)}$ and $F_{y,1}^{(D3)}$ contribute to the Schur index, while if
we use $R_{yx}$ in (36), $F_{y,1}^{(D3)}$ does not contribute. Although we
have not proved $F_{y,N\geq 2}^{(D3)}$ also vanish, this partially explains
why there are two different expansions: the simple expansion found in [24] and
the multiple expansion in [17]. It is surprising that they give the same
result even though some contributions are lost in the simple expansion. This
may be related to the fact that the set of functions $F_{m_{x},m_{y}}$ are
strongly constrained. For example, we can formally substitute negative $N$ to
expansion (1) or (7), and find that the result is vanishing. This gives very
strong constraints on the functions. This implies that the functions share
common information, and only small subset appearing in the simple expansion
may be sufficient to give the complete answer. It would be interesting to
investigate the structure of the constrained set of the functions.

An important application of our method is the calculation of the full
superconformal index of the 6d ${\cal N}=(2,0)$ $A_{N-1}$ SCFT. It is in
principle possible to calculate $I_{N}^{(M5)}$ for an arbitrary $N$ up to an
arbitrary order starting from the indices $I_{m}^{(M2)}$ of the M2-brane
theories with different $m$. Some results are shown in Appendix A.

An important merit of our method is that the theory on giant gravitons does
not have to be Lagrangian theories unlike the method adopted in [17, 18, 25],
which uses deformed contours in the gauge fugacity integrals to realize the
analytic continuation. We only need the final expression of the index of the
giant graviton theory. It enables us to apply the method to M5-giants, as was
demonstrated in 4.5. Another example with non-Lagrangian giant gravitons is
$AdS_{5}\times S^{5}$ with a $7$-brane insertion [33, 34]. As is demonstrated
in [23], the giant graviton expansion works well for this system at least for
the leading contribution. There are higher order contributions coming from
giant gravitons coincident with the $7$-brane, on which a non-Lagrangian
theory is realized. We expect our method is useful for the analysis of such
contributions.

We studied two expansion domains $R_{yx}$ and $R_{qu}$ for the Schur index of
$T=D3$. An advantage of $R_{qu}$ over $R_{yx}$ is that for each value of
$m=m_{x}+m_{y}$ the symmetry between $x$ and $y$ is manifest. Unfortunately,
it turned out that our method has limited applicability for the calculation
with the domain $R_{qu}$ due to the ill-defined factor $P$. It would be very
nice if we can somehow regularize the product in $P$ when it is infinite and
make it possible to apply the method to an arbitrary contribution
$F_{i,m}^{(T)}$.

In general, the functions $F_{m_{1},\ldots,m_{d}}^{(T)}$ do not factorize into
$F_{i,m}^{(T)}$. In the case of D3-giants it is possible to write
$F_{m_{1},\ldots,m_{d}}^{(T)}$ in the matrix integral form. Even so, it is not
clear how we should choose the integration contours. Because we cannot take a
physical domain of expansion, using unit circles for integration contours is
not justified. Although some rules for contours have been proposed [18, 25],
it is desirable to find more efficient method of calculation applicable to
non-Lagrangian giant gravitons.

Another important problem is to clarify to what extent the giant graviton
expansion works. In this work we focused only on the maximally supersymmetric
theories: 4d ${\cal N}=4$ SYM on D3-branes, 3d ${\cal N}=8$ SCFT on M2-branes,
and 6d ${\cal N}=(2,0)$ SCFT on M5-branes. The analysis on the gauge theory
side [24, 25] found the structure of the giant graviton expansions in variety
of theories. It would be interesting to study the applicability of our method
to more general class of theories which have holographic dual description.

In addition to the giant graviton expansions (1) and (7), another similar
expansion was proposed in [35]. It may be interesting to study the relation
among them.

## Acknowledgments

The author would like to thank S. Murayama and D. Yokoyama for valuable
discussions. The work of Y. I. was partially supported by Grand-in-Aid for
Scientific Research (C) (No.21K03569), Ministry of Education, Science and
Culture, Japan. This work used computational resources TSUBAME3.0
supercomputer provided by Tokyo Institute of Technology.

##  Appendix A Full index of $6$-dim $A_{N-1}$ SCFT

Let $\chi_{[a,b]}$ be the $SU(3)$ character of the representation with Dynkin
label $[a,b]$ defined such that $\chi_{[1,0]}=u_{1}+u_{2}+u_{3}$. The
superconformal indices $I_{N}^{(M5)}$ for $N=2,3,4,5$ are given as follows.

$\displaystyle I^{(M5)}_{2}$
$\displaystyle=(1+\check{x}_{1}+2\check{x}_{1}^{2}+2\check{x}_{1}^{3}+3\check{x}_{1}^{4}+3\check{x}_{1}^{5}+4\check{x}_{1}^{6}+4\check{x}_{1}^{7}+5\check{x}_{1}^{8}+5\check{x}_{1}^{9}+6\check{x}_{1}^{10}+\cdots)$
$\displaystyle+y\chi_{[1,0]}(\check{x}_{1}+2\check{x}_{1}^{2}+3\check{x}_{1}^{3}+4\check{x}_{1}^{4}+5\check{x}_{1}^{5}+6\check{x}_{1}^{6}+7\check{x}_{1}^{7}+8\check{x}_{1}^{8}+9\check{x}_{1}^{9}+10\check{x}_{1}^{10}+\cdots)$
$\displaystyle+y^{2}[\chi_{[0,1]}(-1-2\check{x}_{1}-3\check{x}_{1}^{2}-3\check{x}_{1}^{3}-4\check{x}_{1}^{4}-4\check{x}_{1}^{5}-5\check{x}_{1}^{6}-5\check{x}_{1}^{7}-6\check{x}_{1}^{8}-6\check{x}_{1}^{9}-7\check{x}_{1}^{10}+\cdots)$
$\displaystyle\quad+\chi_{[2,0]}(\check{x}_{1}+3\check{x}_{1}^{2}+5\check{x}_{1}^{3}+8\check{x}_{1}^{4}+10\check{x}_{1}^{5}+13\check{x}_{1}^{6}+15\check{x}_{1}^{7}+18\check{x}_{1}^{8}+20\check{x}_{1}^{9}+23\check{x}_{1}^{10}+\cdots)]$
$\displaystyle+y^{3}[(\check{x}_{1}^{-1}+2+\check{x}_{1}-\check{x}_{1}^{3}-2\check{x}_{1}^{4}-3\check{x}_{1}^{5}-4\check{x}_{1}^{6}-5\check{x}_{1}^{7}-6\check{x}_{1}^{8}-7\check{x}_{1}^{9}-8\check{x}_{1}^{10}+\cdots)$
$\displaystyle\quad+\chi_{[1,1]}(-1-3\check{x}_{1}-5\check{x}_{1}^{2}-6\check{x}_{1}^{3}-6\check{x}_{1}^{4}-6\check{x}_{1}^{5}-6\check{x}_{1}^{6}-6\check{x}_{1}^{7}-6\check{x}_{1}^{8}-6\check{x}_{1}^{9}-6\check{x}_{1}^{10}+\cdots)$
$\displaystyle\quad+\chi_{[3,0]}(\check{x}_{1}+3\check{x}_{1}^{2}+7\check{x}_{1}^{3}+11\check{x}_{1}^{4}+16\check{x}_{1}^{5}+21\check{x}_{1}^{6}+26\check{x}_{1}^{7}+31\check{x}_{1}^{8}+36\check{x}_{1}^{9}+41\check{x}_{1}^{10}+\cdots)]$
$\displaystyle+\cdots$ (63) $\displaystyle I^{(M5)}_{3}$
$\displaystyle=(1+\check{x}_{1}+2\check{x}_{1}^{2}+3\check{x}_{1}^{3}+4\check{x}_{1}^{4}+5\check{x}_{1}^{5}+7\check{x}_{1}^{6}+8\check{x}_{1}^{7}+10\check{x}_{1}^{8}+12\check{x}_{1}^{9}+14\check{x}_{1}^{10}+\cdots)$
$\displaystyle+y\chi_{[1,0]}(\check{x}_{1}+2\check{x}_{1}^{2}+4\check{x}_{1}^{3}+6\check{x}_{1}^{4}+9\check{x}_{1}^{5}+12\check{x}_{1}^{6}+16\check{x}_{1}^{7}+20\check{x}_{1}^{8}+25\check{x}_{1}^{9}+30\check{x}_{1}^{10}+\cdots)$
$\displaystyle+y^{2}[\chi_{[0,1]}(-1-2\check{x}_{1}-4\check{x}_{1}^{2}-5\check{x}_{1}^{3}-7\check{x}_{1}^{4}-8\check{x}_{1}^{5}-10\check{x}_{1}^{6}-11\check{x}_{1}^{7}-13\check{x}_{1}^{8}-14\check{x}_{1}^{9}-16\check{x}_{1}^{10}+\cdots)$
$\displaystyle\quad+\chi_{[2,0]}(\check{x}_{1}+3\check{x}_{1}^{2}+6\check{x}_{1}^{3}+11\check{x}_{1}^{4}+17\check{x}_{1}^{5}+25\check{x}_{1}^{6}+34\check{x}_{1}^{7}+45\check{x}_{1}^{8}+57\check{x}_{1}^{9}+71\check{x}_{1}^{10}+\cdots)]$
$\displaystyle+y^{3}[(\check{x}_{1}^{-1}+2+2\check{x}_{1}+\check{x}_{1}^{2}-\check{x}_{1}^{3}-4\check{x}_{1}^{4}-8\check{x}_{1}^{5}-12\check{x}_{1}^{6}-18\check{x}_{1}^{7}-24\check{x}_{1}^{8}-31\check{x}_{1}^{9}-39\check{x}_{1}^{10}+\cdots)$
$\displaystyle\quad+\chi_{[1,1]}(-1-3\check{x}_{1}-6\check{x}_{1}^{2}-10\check{x}_{1}^{3}-13\check{x}_{1}^{4}-16\check{x}_{1}^{5}-18\check{x}_{1}^{6}-19\check{x}_{1}^{7}-19\check{x}_{1}^{8}-19\check{x}_{1}^{9}-17\check{x}_{1}^{10}+\cdots)$
$\displaystyle\quad+\chi_{[3,0]}(\check{x}_{1}+3\check{x}_{1}^{2}+8\check{x}_{1}^{3}+15\check{x}_{1}^{4}+26\check{x}_{1}^{5}+40\check{x}_{1}^{6}+58\check{x}_{1}^{7}+79\check{x}_{1}^{8}+105\check{x}_{1}^{9}+133\check{x}_{1}^{10}+\cdots)]$
$\displaystyle+\cdots$ (64) $\displaystyle I^{(M5)}_{4}$
$\displaystyle=(1+\check{x}_{1}+2\check{x}_{1}^{2}+3\check{x}_{1}^{3}+5\check{x}_{1}^{4}+6\check{x}_{1}^{5}+9\check{x}_{1}^{6}+11\check{x}_{1}^{7}+15\check{x}_{1}^{8}+18\check{x}_{1}^{9}+23\check{x}_{1}^{10}+\cdots)$
$\displaystyle+y\chi_{[1,0]}(\check{x}_{1}+2\check{x}_{1}^{2}+4\check{x}_{1}^{3}+7\check{x}_{1}^{4}+11\check{x}_{1}^{5}+16\check{x}_{1}^{6}+23\check{x}_{1}^{7}+31\check{x}_{1}^{8}+41\check{x}_{1}^{9}+53\check{x}_{1}^{10}+\cdots)$
$\displaystyle+y^{2}[\chi_{[0,1]}(-1-2\check{x}_{1}-4\check{x}_{1}^{2}-6\check{x}_{1}^{3}-9\check{x}_{1}^{4}-11\check{x}_{1}^{5}-15\check{x}_{1}^{6}-17\check{x}_{1}^{7}-21\check{x}_{1}^{8}-23\check{x}_{1}^{9}-27\check{x}_{1}^{10}+\cdots)$
$\displaystyle\quad+\chi_{[2,0]}(\check{x}_{1}+3\check{x}_{1}^{2}+6\check{x}_{1}^{3}+12\check{x}_{1}^{4}+20\check{x}_{1}^{5}+32\check{x}_{1}^{6}+47\check{x}_{1}^{7}+68\check{x}_{1}^{8}+92\check{x}_{1}^{9}+124\check{x}_{1}^{10}+\cdots)]$
$\displaystyle+y^{3}[(\check{x}_{1}^{-1}+2+2\check{x}_{1}+2\check{x}_{1}^{2}-4\check{x}_{1}^{4}-10\check{x}_{1}^{5}-18\check{x}_{1}^{6}-30\check{x}_{1}^{7}-44\check{x}_{1}^{8}-62\check{x}_{1}^{9}-84\check{x}_{1}^{10}+\cdots)$
$\displaystyle\quad+\chi_{[1,1]}(-1-3\check{x}_{1}-6\check{x}_{1}^{2}-11\check{x}_{1}^{3}-17\check{x}_{1}^{4}-23\check{x}_{1}^{5}-30\check{x}_{1}^{6}-36\check{x}_{1}^{7}-40\check{x}_{1}^{8}-43\check{x}_{1}^{9}-43\check{x}_{1}^{10}+\cdots)$
$\displaystyle\quad+\chi_{[3,0]}(\check{x}_{1}+3\check{x}_{1}^{2}+8\check{x}_{1}^{3}+16\check{x}_{1}^{4}+30\check{x}_{1}^{5}+50\check{x}_{1}^{6}+79\check{x}_{1}^{7}+117\check{x}_{1}^{8}+168\check{x}_{1}^{9}+231\check{x}_{1}^{10}+\cdots)]$
$\displaystyle+\cdots$ (65) $\displaystyle I^{(M5)}_{5}$
$\displaystyle=(1+\check{x}_{1}+2\check{x}_{1}^{2}+3\check{x}_{1}^{3}+5\check{x}_{1}^{4}+7\check{x}_{1}^{5}+10\check{x}_{1}^{6}+13\check{x}_{1}^{7}+18\check{x}_{1}^{8}+23\check{x}_{1}^{9}+30\check{x}_{1}^{10}+\cdots)$
$\displaystyle+y\chi_{[1,0]}(\check{x}_{1}+2\check{x}_{1}^{2}+4\check{x}_{1}^{3}+7\check{x}_{1}^{4}+12\check{x}_{1}^{5}+18\check{x}_{1}^{6}+27\check{x}_{1}^{7}+38\check{x}_{1}^{8}+53\check{x}_{1}^{9}+71\check{x}_{1}^{10}+\cdots)$
$\displaystyle+y^{2}[\chi_{[0,1]}(-1-2\check{x}_{1}-4\check{x}_{1}^{2}-6\check{x}_{1}^{3}-10\check{x}_{1}^{4}-13\check{x}_{1}^{5}-18\check{x}_{1}^{6}-22\check{x}_{1}^{7}-28\check{x}_{1}^{8}-32\check{x}_{1}^{9}-38\check{x}_{1}^{10}+\cdots)$
$\displaystyle\quad+\chi_{[2,0]}(\check{x}_{1}+3\check{x}_{1}^{2}+6\check{x}_{1}^{3}+12\check{x}_{1}^{4}+21\check{x}_{1}^{5}+35\check{x}_{1}^{6}+54\check{x}_{1}^{7}+81\check{x}_{1}^{8}+116\check{x}_{1}^{9}+163\check{x}_{1}^{10}+\cdots)]$
$\displaystyle+y^{3}[(\check{x}_{1}^{-1}+2+2\check{x}_{1}+2\check{x}_{1}^{2}+\check{x}_{1}^{3}-3\check{x}_{1}^{4}-10\check{x}_{1}^{5}-20\check{x}_{1}^{6}-36\check{x}_{1}^{7}-57\check{x}_{1}^{8}-86\check{x}_{1}^{9}-123\check{x}_{1}^{10}+\cdots)$
$\displaystyle\quad+\chi_{[1,1]}(-1-3\check{x}_{1}-6\check{x}_{1}^{2}-11\check{x}_{1}^{3}-18\check{x}_{1}^{4}-27\check{x}_{1}^{5}-37\check{x}_{1}^{6}-48\check{x}_{1}^{7}-59\check{x}_{1}^{8}-69\check{x}_{1}^{9}-75\check{x}_{1}^{10}+\cdots)$
$\displaystyle\quad+\chi_{[3,0]}(\check{x}_{1}+3\check{x}_{1}^{2}+8\check{x}_{1}^{3}+16\check{x}_{1}^{4}+31\check{x}_{1}^{5}+54\check{x}_{1}^{6}+89\check{x}_{1}^{7}+138\check{x}_{1}^{8}+208\check{x}_{1}^{9}+300\check{x}_{1}^{10}+\cdots)]$
$\displaystyle+\cdots$ (66)

We can also calculate the index of $A_{N-1}$ SCFT $I_{N}^{(A)}$ by removing
the contribution of the free tensor multiplet by

$\displaystyle I_{N}^{(A)}=\frac{I_{N}^{(M5)}}{I_{1}^{(M5)}}.$ (67)

The numerical results for $N=2,3,4,5$ are as follows:

$\displaystyle I^{(A)}_{2}$
$\displaystyle=(1+\check{x}_{1}^{2}+\check{x}_{1}^{4}+\check{x}_{1}^{6}+\check{x}_{1}^{8}+\check{x}_{1}^{10}+\cdots)$
$\displaystyle+y\chi_{[1,0]}(\check{x}_{1}^{2}+\check{x}_{1}^{4}+\check{x}_{1}^{6}+\check{x}_{1}^{8}+\check{x}_{1}^{10}+\cdots)$
$\displaystyle+y^{2}[\chi_{[0,1]}(-\check{x}_{1}-\check{x}_{1}^{3}-\check{x}_{1}^{5}-\check{x}_{1}^{7}-\check{x}_{1}^{9}+\cdots)$
$\displaystyle\quad+\chi_{[2,0]}(\check{x}_{1}^{2}+2\check{x}_{1}^{4}+2\check{x}_{1}^{6}+2\check{x}_{1}^{8}+2\check{x}_{1}^{10}+\cdots)]$
$\displaystyle+y^{3}[(1-\check{x}_{1}+\check{x}_{1}^{2}-x^{3}+\check{x}_{1}^{4}-\check{x}_{1}^{5}+\check{x}_{1}^{6}-\check{x}_{1}^{7}+\check{x}_{1}^{8}-\check{x}_{1}^{9}+\check{x}_{1}^{10}+\cdots)$
$\displaystyle\quad+\chi_{[1,1]}(-\check{x}_{1}-2\check{x}_{1}^{3}+\check{x}_{1}^{4}-2\check{x}_{1}^{5}+\check{x}_{1}^{6}-2\check{x}_{1}^{7}+\check{x}_{1}^{8}-2\check{x}_{1}^{9}+\check{x}_{1}^{10}+\cdots)$
$\displaystyle\quad+\chi_{[3,0]}(\check{x}_{1}^{2}+2\check{x}_{1}^{4}+3\check{x}_{1}^{6}+3\check{x}_{1}^{8}+3\check{x}_{1}^{10}+\cdots)]$
$\displaystyle+\cdots$ (68) $\displaystyle I^{(A)}_{3}$
$\displaystyle=(1+\check{x}_{1}^{2}+\check{x}_{1}^{3}+\check{x}_{1}^{4}+\check{x}_{1}^{5}+2\check{x}_{1}^{6}+\check{x}_{1}^{7}+2\check{x}_{1}^{8}+2\check{x}_{1}^{9}+2\check{x}_{1}^{10}+\cdots)$
$\displaystyle+y\chi_{[1,0]}(\check{x}_{1}^{2}+\check{x}_{1}^{3}+\check{x}_{1}^{4}+2\check{x}_{1}^{5}+2\check{x}_{1}^{6}+2\check{x}_{1}^{7}+3\check{x}_{1}^{8}+3\check{x}_{1}^{9}+3\check{x}_{1}^{10}+\cdots)$
$\displaystyle+y^{2}[\chi_{[0,1]}(-\check{x}_{1}-\check{x}_{1}^{2}-\check{x}_{1}^{3}-2\check{x}_{1}^{4}-\check{x}_{1}^{5}-2\check{x}_{1}^{6}-2\check{x}_{1}^{7}-2\check{x}_{1}^{8}-2\check{x}_{1}^{9}-3\check{x}_{1}^{10}+\cdots)$
$\displaystyle\quad+\chi_{[2,0]}(\check{x}_{1}^{2}+\check{x}_{1}^{3}+2\check{x}_{1}^{4}+3\check{x}_{1}^{5}+4\check{x}_{1}^{6}+4\check{x}_{1}^{7}+6\check{x}_{1}^{8}+6\check{x}_{1}^{9}+7\check{x}_{1}^{10}+\cdots)]$
$\displaystyle+y^{3}[(1-\check{x}_{1}^{4}-\check{x}_{1}^{5}-\check{x}_{1}^{6}-2\check{x}_{1}^{7}-2\check{x}_{1}^{8}-2\check{x}_{1}^{9}-3\check{x}_{1}^{10}+\cdots)$
$\displaystyle\quad+\chi_{[1,1]}(-\check{x}_{1}-\check{x}_{1}^{2}-2\check{x}_{1}^{3}-3\check{x}_{1}^{4}-2\check{x}_{1}^{5}-3\check{x}_{1}^{6}-3\check{x}_{1}^{7}-2\check{x}_{1}^{8}-3\check{x}_{1}^{9}-3\check{x}_{1}^{10}+\cdots)$
$\displaystyle\quad+\chi_{[3,0]}(\check{x}_{1}^{2}+\check{x}_{1}^{3}+2\check{x}_{1}^{4}+4\check{x}_{1}^{5}+5\check{x}_{1}^{6}+6\check{x}_{1}^{7}+9\check{x}_{1}^{8}+10\check{x}_{1}^{9}+11\check{x}_{1}^{10}+\cdots)]$
$\displaystyle+\cdots$ (69) $\displaystyle I^{(A)}_{4}$
$\displaystyle=(1+\check{x}_{1}^{2}+\check{x}_{1}^{3}+2\check{x}_{1}^{4}+\check{x}_{1}^{5}+3\check{x}_{1}^{6}+2\check{x}_{1}^{7}+4\check{x}_{1}^{8}+3\check{x}_{1}^{9}+5\check{x}_{1}^{10}+\cdots)$
$\displaystyle+y\chi_{[1,0]}(\check{x}_{1}^{2}+\check{x}_{1}^{3}+2\check{x}_{1}^{4}+2\check{x}_{1}^{5}+4\check{x}_{1}^{6}+4\check{x}_{1}^{7}+6\check{x}_{1}^{8}+6\check{x}_{1}^{9}+9\check{x}_{1}^{10}+\cdots)$
$\displaystyle+y^{2}[\chi_{[0,1]}(-\check{x}_{1}-\check{x}_{1}^{2}-2\check{x}_{1}^{3}-2\check{x}_{1}^{4}-3\check{x}_{1}^{5}-3\check{x}_{1}^{6}-4\check{x}_{1}^{7}-4\check{x}_{1}^{8}-5\check{x}_{1}^{9}-5\check{x}_{1}^{10}+\cdots)$
$\displaystyle\quad+\chi_{[2,0]}(\check{x}_{1}^{2}+\check{x}_{1}^{3}+3\check{x}_{1}^{4}+3\check{x}_{1}^{5}+7\check{x}_{1}^{6}+7\check{x}_{1}^{7}+12\check{x}_{1}^{8}+12\check{x}_{1}^{9}+19\check{x}_{1}^{10}+\cdots)]$
$\displaystyle+y^{3}[(1+\check{x}_{1}^{2}-\check{x}_{1}^{3}-3\check{x}_{1}^{5}-2\check{x}_{1}^{6}-6\check{x}_{1}^{7}-5\check{x}_{1}^{8}-9\check{x}_{1}^{9}-9\check{x}_{1}^{10}+\cdots)$
$\displaystyle\quad+\chi_{[1,1]}(-\check{x}_{1}-\check{x}_{1}^{2}-3\check{x}_{1}^{3}-3\check{x}_{1}^{4}-6\check{x}_{1}^{5}-5\check{x}_{1}^{6}-8\check{x}_{1}^{7}-6\check{x}_{1}^{8}-10\check{x}_{1}^{9}-6\check{x}_{1}^{10}+\cdots)$
$\displaystyle\quad+\chi_{[3,0]}(\check{x}_{1}^{2}+\check{x}_{1}^{3}+3\check{x}_{1}^{4}+4\check{x}_{1}^{5}+9\check{x}_{1}^{6}+10\check{x}_{1}^{7}+17\check{x}_{1}^{8}+20\check{x}_{1}^{9}+30\check{x}_{1}^{10}+\cdots)]$
$\displaystyle+\cdots$ (70) $\displaystyle I^{(A)}_{5}$
$\displaystyle=(1+\check{x}_{1}^{2}+\check{x}_{1}^{3}+2\check{x}_{1}^{4}+2\check{x}_{1}^{5}+3\check{x}_{1}^{6}+3\check{x}_{1}^{7}+5\check{x}_{1}^{8}+5\check{x}_{1}^{9}+7\check{x}_{1}^{10}+\cdots)$
$\displaystyle+y\chi_{[1,0]}(\check{x}_{1}^{2}+\check{x}_{1}^{3}+2\check{x}_{1}^{4}+3\check{x}_{1}^{5}+4\check{x}_{1}^{6}+6\check{x}_{1}^{7}+8\check{x}_{1}^{8}+10\check{x}_{1}^{9}+13\check{x}_{1}^{10}+\cdots)$
$\displaystyle+y^{2}[\chi_{[0,1]}(-\check{x}_{1}-\check{x}_{1}^{2}-2\check{x}_{1}^{3}-3\check{x}_{1}^{4}-3\check{x}_{1}^{5}-5\check{x}_{1}^{6}-5\check{x}_{1}^{7}-7\check{x}_{1}^{8}-7\check{x}_{1}^{9}-9\check{x}_{1}^{10}+\cdots)$
$\displaystyle\quad+\chi_{[2,0]}(\check{x}_{1}^{2}+\check{x}_{1}^{3}+3\check{x}_{1}^{4}+4\check{x}_{1}^{5}+7\check{x}_{1}^{6}+10\check{x}_{1}^{7}+15\check{x}_{1}^{8}+19\check{x}_{1}^{9}+27\check{x}_{1}^{10}+\cdots)]$
$\displaystyle+y^{3}[(1+\check{x}_{1}^{2}-\check{x}_{1}^{4}-2\check{x}_{1}^{5}-4\check{x}_{1}^{6}-7\check{x}_{1}^{7}-10\check{x}_{1}^{8}-14\check{x}_{1}^{9}-19\check{x}_{1}^{10}+\cdots)$
$\displaystyle\quad+\chi_{[1,1]}(-\check{x}_{1}-\check{x}_{1}^{2}-3\check{x}_{1}^{3}-4\check{x}_{1}^{4}-6\check{x}_{1}^{5}-9\check{x}_{1}^{6}-10\check{x}_{1}^{7}-13\check{x}_{1}^{8}-15\check{x}_{1}^{9}-16\check{x}_{1}^{10}+\cdots)$
$\displaystyle\quad+\chi_{[3,0]}(\check{x}_{1}^{2}+\check{x}_{1}^{3}+3\check{x}_{1}^{4}+5\check{x}_{1}^{5}+9\check{x}_{1}^{6}+14\check{x}_{1}^{7}+21\check{x}_{1}^{8}+30\check{x}_{1}^{9}+42\check{x}_{1}^{10}+\cdots)]$
$\displaystyle+\cdots$ (71)

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#  A note on the value distribution of
$f^{l}(f^{(k)})^{n}$

Yan Jiang and Bin Huang Mathematics Subject Classifications (2010): Primary
30D35

###### Abstract

Let $f$ be a transcendental meromorphic function in the complex plane
$\mathbb{C}$, and $a$ be a nonzero complex number . We give quantitative
estimates for the characteristic function $T(r,f)$ in terms of
$N(r,1/(f^{l}(f^{(k)})^{n}-a))$, for integers $k$, $l$, $n$ greater than 1. We
conclude that $f^{l}(f^{(k)})^{n}$ assumes every nonzero finite value
infinitely often.

Keywords: Transcendental meromorphic function, deficiency.

##  1 Introduction

Let $f$ be a transcendental meromorphic function in the complex plane
$\mathbb{C}$. In this article, we use the standard notations in the sense of
Nevanlinna [7], such as $T(r,f)$, $N(r,f)$, $\bar{N}(r,f)$, $m(r,f)$,
$S(r,f)$, $\delta(a,f)$. In particular, $T(r,f)$ the characteristic function
and $\bar{N}(r,f)$ is a counting function with respect to poles of $f$,
ignoring multiplicities. We shall use the symbol $S(r,f)$ to denote an error
term $v(r)$ satisfying $T(r,v(r))=o\left(T(r,f)\right)$ as
$r\rightarrow\infty$, possibly outside a set of finite linear measure.
Throughout this paper a small function (with respect to $f$) means a function
$\varphi(z)$ meromorphic in $\mathbb{C}$ satisfying $T(r,\varphi)=S(r,f)$. In
addition, in this paper, we use another type of small function $S^{*}(r,f)$
which has the property $S^{*}(r,f)=o\left(T(r,f)\right)$ as
$r\rightarrow\infty$, $r\not\in E$, where $E$ is a set of logarithmic density
0.

A meromorphic function $f$ is rational if and only if $T(r,f)=O(\log r)$ (see
[6]). The quantity

$\delta(a,f)=\mathop{\lim\inf}\limits_{r\to\infty}\frac{m(r,1/(f-a))}{T(r,f)}=1-\mathop{\lim\sup}\limits_{r\to\infty}\frac{N(r,1/{(f-a)})}{T(r,f)}$

is called the deficiency of $f$ at the point $a$. Another deficiency is
defined by

$\Theta(a,f)=1-\mathop{\lim\sup}\limits_{r\to\infty}\frac{\bar{N}(r,1/{(f-a)})}{T(r,f)}.$

Note that $0\leq\delta(a,f)\leq\Theta(a,f)\leq 1$.

The First Fundamental Theorem of the value distribution theory due to
Nevanlinna is utilized frequently in this note. It is stated as the following
property:

$T\left(r,\frac{1}{f-a}\right)=T(r,f)+O(1)$

for any constant $a$. The details can be found in [6] for example. A root of
the equation $f(z)=a$ $(1/f(z)=0$ for $a=\infty)$ will be called an $a-$point
of the function $f(z)$ for $a\in\mathbb{C}\cup\\{\infty\\}$. $a$ is called a
Picard exceptional value of a function $f(z)$ if the number of its $a-$points
in $\mathbb{C}$ is finite.

The aim of this paper is to look for estimates with respect to
$f^{l}(f^{(k)})^{n}$. The following well-known estimate is due to Hayman [7,
Theorem 3.5].

######  Theorem A.

Let $f$ be a meromorphic and transcendental function in the plane, $l$ be a
positive integer, and $a$, $b$ be constants with $b\neq 0$. Then

$T(r,f)\leq\left(2+\frac{1}{l}\right)N\left(r,\frac{1}{f-a}\right)+\left(2+\frac{2}{l}\right)\bar{N}\left(r,\frac{1}{f^{(l)}-b}\right)+S(r,f).$
(1)

Hayman also concluded a corollary from the previous inequality.

######  Corollary.

Under the same assumptions as in Theorem A, either $f$ assumes every finite
value infinitely often or $f^{(l)}$ assumes every finite value except possibly
zero infinitely often.

Moreover, Hayman conjectured that if $f$ is a transcendental meromorphic
function and $l\geq 1$, then $f^{l}f^{\prime}$ takes every finite nonzero
value infinitely often. This conjecture has been confirmed by himself in [7]
for $l\geq 3$, by Mues [13] for $l=2$ and by Bergweiler and Eremenko [3] for
$l=1$. During the past decades, a sequence of related research have been made.
In 1982, Doeringer [4, Corollary 1] proved that for a transcendental
meromorphic function $f$, the only possible Picard exceptional value is zero
for a differential monomial $f^{l}(f^{(k)})^{n}$ when $l\geq 3$. In 1994, Tse
and Yang [15] gave an estimate of $T(r,f)$ for $l=1$ and $l=2$ and confirmed
the only possible Picard exceptional value is zero. In 1996, Yang and Hu [19,
Theorem 2] proved that if $\delta(0,f)>3/(3(l+n)+1)$ with positive integers
$k$, $l$, $n$, then for a nonzero finite complex number $a$,
$f^{l}(f^{(k)})^{n}-a$ has infinitely many zeros. In 2002, Li and Wu [12]
obtained that for a nonzero finite complex number $a$ and positive integers
$l$, $k$ with $l\geq 2$, there exists a constant $M>0$ such that

$T(r,f)<M\bar{N}\left(r,\frac{1}{f^{l}f^{(k)}-a}\right)+S(r,f).$

In 2003, Wang [16] studied the zeros of $f^{l}f^{(k)}-\phi$ for a small
meromorphic function $\phi(z)\not\equiv 0$, and verified that for $l\geq 2$,
$f^{l}f^{(k)}-\phi$ had infinitely many zeros if the poles of $f$ were
multiple. In 2004, Alotaibi [2] gave an estimate and showed that the function
$f(f^{(k)})^{n}-\phi$ has infinitely many zeros for a small function
$\phi(z)\not\equiv 0$, when $n\geq 2$.

We introduce a result given by Lahiri and Dewan [9, Theorem 3.2].

######  Theorem B.

Let $f$ be a transcendental meromorphic function and $a$, $\alpha$ be both
small functions of $f$ without being identically to zero and infinity. If
$\psi=\alpha f^{l}(f^{(k)})^{n}$, where $l(\geq 0)$, $n(\geq 1)$, $k(\geq 1)$
are integers, then

$\displaystyle(l+n)T(r,f)\leq\bar{N}\left(r,f\right)+\bar{N}\left(r,\frac{1}{f}\right)+nN_{(k)}\left(r,\frac{1}{f}\right)+\bar{N}\left(r,\frac{1}{\psi-a}\right)+S(r,f),$
(2)

where $N_{(k)}(r,1/f)$ is the counting function of zeros of $f$ with
multiplicity $q$ counted $\min\\{q,k\\}$ times.

Remark. Inequality (2) implies that for $l\geq 3$, $n\geq 1$, $k\geq 1$,

$T(r,f)\leq\frac{1}{l-2}N\left(r,\frac{1}{f^{l}(f^{(k)})^{n}-a}\right)+S(r,f),$
(3)

then

$\displaystyle\delta(a,\psi)\leq\Theta(a,\psi)\leq 1-\frac{l-2}{nk+n+l}.$ (4)

However, this result is still worth refining. In the current paper, we
obtained an estimate corresponding to the case $k$, $l$, $n$ all greater than
1, and in our proof, we use a very important inequality of Yamanoi.

######  Theorem C.

_[18, Yamanoi]_ Let $f$ be a meromorphic and transcendental function in the
complex plane and let $k\geq 2$ be an integer, $A\subset\mathbb{C}$ be a
finite set of complex numbers. Then we have

$(k-1)\bar{N}(r,f)+\sum_{a\in A}N_{1}\left(r,\frac{1}{f-a}\right)\leq
N\left(r,\frac{1}{f^{(k)}}\right)+S^{*}(r,f),$ (5)

where

$N_{1}\left(r,\frac{1}{f-a}\right)=N\left(r,\frac{1}{f-a}\right)-\bar{N}\left(r,\frac{1}{f-a}\right).$

Remark. Actually, this theorem is obviously a correspondent result to the
famous Gol’dberg Conjecture, which says that for a transcendental meromorphic
function $f$ and $k\geq 2$, then $\bar{N}(r,f)\leq
N\left(r,1/{f^{(k)}}\right)+S(r,f)$. There is obviously a special case of
Yamanoi’s inequality when $A$ is an empty set. The following special case is
the one we use in our proof,

$(k-1)\bar{N}(r,f)\leq N\left(r,\frac{1}{f^{(k)}}\right)+S^{*}(r,f).$ (6)

In this paper, we continue to consider the general form $f^{l}(f^{(k)})^{n}-a$
for a nonzero constant $a$. The following theorem improved Theorem B in some
sense.

######  Theorem 1.1.

Let $f$ be a transcendental meromorphic function in $\mathbb{C}$, $l$, $n$,
$k$ be integers greater than 1 and $a$ be a nonzero constant. Then

$T(r,f)\leq\frac{1}{l-1}N\left(r,\frac{1}{f^{l}(f^{(k)})^{n}-a}\right)+S^{*}(r,f).$
(7)

Remark. If the differential monomials $f^{l}(f^{(k)})^{n}$ is allowed to take
$l\geq 2$, $n\geq 2$, $k\geq 2$, then (7) is better than (3) with the except
of a finite set of logarithmic density 0. If the case $k=1$, $l\geq 3$ or
$n=1$, $l\geq 3$ occurs, (3) might be the best choice so far.

Another important remark should be made here. As we realized that for general
form $f^{l}(f^{(k)})^{n}$, except the cases stated in Theorem B and Theorem
1.1, two cases are inevitably excluded: $l=1$, $n\geq 1$, $k\geq 1$ and $l=2$,
$n\geq 1$, $k\geq 1$. We summarize the known estimates of these two cases. For
the case $l=2$, $n=k=1$, Zhang [20] obtained a quantitative result, proving
that the inequality $T(r,f)<6N(r,1/(f^{2}f^{\prime}-1))+S(r,f)$ holds. For the
case $l=2$, $n=1$, $k>1$ the the inequality is due to Huang and Gu [8]. For
the case $l=1$, $n\geq 2$, $k\geq 1$, by Li and Yang [11] and Alotaibi [2]
gave two different inequalities for the estimates independently. For the case
$l=n=1$, $k\geq 1$, again Alotaibi [1] obtained an estimate provided that
$N_{1)}\left(r,\frac{1}{f^{(k)}}\right)=S(r,f)$, where
$N_{1)}\left(r,\frac{1}{f^{(k)}}\right)$ is the counting function of simple
zeros of $f^{(k)}$, as well, Wang [16] gave an estimate but under the
additional condition that multiplicities of all poles of $f$ are at least $3$
and $N_{1)}(r,1/f)\leq\lambda T(r,f)$, where $\lambda<1/3$ is a constant.

Though these cases are excluded in Theorem 1.1, our estimate is considered to
be stronger compared to the known results so far. Furthermore, it is natural
to estimate the deficiency of $f^{l}(f^{(k)})^{n}$ by making use of Theorem
1.1. This leads us to the following.

######  Theorem 1.2.

Let $f$ be a transcendental meromorphic function in $\mathbb{C}$, $k$, $l$,
$n$ be positive integers all greater than 1 and $a$ be a nonzero constant.
Then

$\delta(a,f^{l}(f^{(k)})^{n})\leq 1-\frac{l-1}{nk+n+l}.$

Remark. Since for a nonzero constant $a$, $\delta(a,f^{l}(f^{(k)})^{n})<1$,
Theorem 1.2 also implies that the possible Picard exceptional value of
$f^{l}(f^{(k)})^{n}$ is zero for $k\geq 2$, $l\geq 2$, $n\geq 2$. We like to
state these results as a corollary here.

######  Corollary 1.3.

Under the same conditions as Theorem 1.1, $f^{l}(f^{(k)})^{n}$ assumes every
finite value except possibly zero infinitely often.

Remark. In fact, this kind of result is not brand new. There are already a
couple of known results implying that for any positive integers $k$, $l$, $n$,
the function $f^{l}(f^{(k)})^{n}$ assumes every finite value except possibly
zero infinitely often. The readers should see Lahiri and Dewan [9, 10],
Steinmetz [14], Wang [17], Alotaibi [2, 1] and Li and Wu [12] for further
details.

Lemmas used for the proof of Theorem 1.1 and Theorem 1.2 are presented in
Section 2. The proofs of Theorem 1.1 and Theorem 1.2 are placed in Section 3
and 4 respectively. In the last section, we give an application to the sum of
deficiencies.

##  2 Lemmas

Before we proceed to the proofs of the theorems, we need the following lemmas.

######  Lemma 2.1.

_[7, Theorem 3.1]_ Let $f$ be a non-constant meromorphic function in the
complex plane, $l$ be positive integer, $a_{0}(z)$, $a_{1}(z)$,$\cdots$,
$a_{l}(z)$ be meromorphic functions in the plane satisfying
$T\left(r,a_{\nu}(z)\right)=S(r,f)$ for $\nu=0$, $1$, $\ldots$, $l$ ( as
$r\rightarrow+\infty$) and

$\psi(z)=\sum_{\nu=0}^{l}a_{\nu}(z)f^{(\nu)}(z).$

Then

$m\left(r,\frac{\psi}{f}\right)=S(r,f).$

In particular, this lemma implies $m\left(r,f^{(l)}/f\right)=S(r,f)$ and
$m\left(r,f^{(l+1)}/f^{(l)}\right)=S(r,f^{(l)})$.

######  Lemma 2.2.

_[6, p. 99]_ Let $f$ be a non-constant meromorphic function in the complex
plane, $k$ be a positive integer. Then

$\displaystyle T(r,f^{(k)})\leq(k+1)T(r,f)+S(r,f).$ (8)

In particular, $S(r,f^{(k)})\leq S(r,f)$. Inequality (8) will be used often in
this note without reference.

######  Lemma 2.3.

Let $f$ be a transcendental meromorphic function in the plane. Then the
differential monomial

$\psi=f^{l}(f^{(k)})^{n}$

is transcendental, where $l$, $n$ and $k$ are positive integers.

Proof. Since we have

$\frac{1}{f^{l+n}}=\left(\frac{f^{(k)}}{f}\right)^{n}\frac{1}{\psi}.$

We obtain from Lemma 2.1 and the First Fundamental Theorem that

$\displaystyle(l+n)T(r,f)$ $\displaystyle\leq$ $\displaystyle
nT\left(r,\frac{f^{(k)}}{f}\right)+T\left(r,\frac{1}{\psi}\right)$ (9)
$\displaystyle\leq$ $\displaystyle
nN\left(r,\frac{f^{(k)}}{f}\right)+T\left(r,\frac{1}{\psi}\right)+S(r,f)$
$\displaystyle\leq$ $\displaystyle
nk\left[\bar{N}(r,f)+\bar{N}\left(r,\frac{1}{f}\right)\right]+T\left(r,\frac{1}{\psi}\right)+S(r,f).$

Since $\bar{N}(r,f)\leq\bar{N}(r,\psi)+S(r,f)$ and
$\bar{N}\left(r,\frac{1}{f}\right)\leq\bar{N}\left(r,\frac{1}{\psi}\right)+S(r,f)$,
we can simplify inequality (9) to

$(l+n)T(r,f)\leq(2nk+1)T\left(r,\frac{1}{\psi}\right)+S(r,f).$

Because $f$ is transcendental, we conclude that $\psi$ is transcendental.

######  Lemma 2.4.

Let $f$ be a transcendental meromorphic function in $\mathbb{C}$, let $k$,
$l$, $n$ be positive integers, and set

$g=f^{l}(f^{(k)})^{n}-1.$

Then,

$T(r,g)\leq O\left(T(r,f)\right),$

as $r\rightarrow\infty$, possibly outside a set of finite linear measure.

Proof. Note that $N\left(r,f^{l}(f^{(k)})^{n}\right)=O(N(r,f))$ and
$m\left(r,f^{(k)}/f\right)=S(r,f)$ by Lemma 2.1. Applying the First
Fundamental Theorem, we get

$\displaystyle T(r,g)$ $\displaystyle=$ $\displaystyle
T\left(r,f^{l}(f^{(k)})^{n}-1\right)$ $\displaystyle=$ $\displaystyle
N\left(r,f^{l}(f^{(k)})^{n}\right)+m\left(r,f^{l}(f^{(k)})^{n}\right)+O(1)$
$\displaystyle\leq$ $\displaystyle
O\left(N(r,f)\right)+lm(r,f)+nm\left(r,f^{(k)}\right)++O(1)$
$\displaystyle\leq$ $\displaystyle
O\left(N(r,f)\right)+lm(r,f)+nm(r,f)+nm\left(r,\frac{f^{(k)}}{f}\right)+O(1)$
$\displaystyle=$ $\displaystyle O\left(T(r,f)\right)+S(r,f).$

We can see that

$T(r,g^{\prime})\leq N(r,g^{\prime})+m(r,g)+S(r,g)\leq T(r,g)+S(r,g).$

Hence

$T(r,g)\leq O\left(T(r,f)\right).\@qedbox{}$

##  3 Proof of Theorem 1.1.

Without loss of generality, we assume $a=1$. $g=f^{l}(f^{(k)})^{n}-1$. By
Lemma 2.3, we know that $g$ is not constant. Since

$\frac{1}{f^{l+n}}=\left(\frac{f^{(k)}}{f}\right)^{n}-\frac{g^{\prime}}{f^{l+n}}\left(\frac{g}{g^{\prime}}\right),$

it follows that

$m\left(r,\frac{1}{f^{l+n}}\right)\leq
m\left(r,\frac{g}{g^{\prime}}\right)+m\left(r,\frac{g^{\prime}}{f^{l+n}}\right)+S(r,f).$

Note that

$\frac{{g}^{\prime}}{f^{l+n}}=l\frac{{f}^{\prime}}{f}\left(\frac{f^{(k)}}{f}\right)^{n}+n\frac{f^{(k+1)}}{f}\left(\frac{f^{(k)}}{f}\right)^{n-1},$

which implies

$m\left(r,\frac{g^{\prime}}{f^{l+n}}\right)=S(r,f).$

Therefore, we have

$m\left(r,\frac{1}{f^{l+n}}\right)\leq
m\left(r,\frac{g}{g^{\prime}}\right)+S(r,f).$

We know that the poles of $g^{\prime}/g$ come from the zeros and poles of $g$,
and all are simple. The poles of $g/g^{\prime}$ come from zeros of
$g^{\prime}$ which are not zeros of $g$, preserving multiplicity. Hence, we
get

$N\left(r,\frac{g^{\prime}}{g}\right)=\bar{N}\left(r,\frac{1}{g}\right)+\bar{N}(r,g),$
(10)

and

$N\left(r,\frac{g}{g^{\prime}}\right)=N\left(r,\frac{1}{g^{\prime}}\right)-\left(N\left(r,\frac{1}{g}\right)-\bar{N}\left(r,\frac{1}{g}\right)\right).$
(11)

By combining (10) with (11),

$N\left(r,\frac{g^{\prime}}{g}\right)-N\left(r,\frac{g}{g^{\prime}}\right)=\bar{N}(r,g)+N\left(r,\frac{1}{g}\right)-N\left(r,\frac{1}{g^{\prime}}\right).$
(12)

By Lemma 2.4, we know that

$m(r,g^{\prime}/g)=S(r,g)\leq S(r,f),\quad\quad\bar{N}(r,g)=\bar{N}(r,f).$

Applying the First Fundamental Theorem and (12),

$\displaystyle m\left(r,\frac{1}{f^{l+n}}\right)$ $\displaystyle=$
$\displaystyle(l+n)m\left(r,\frac{1}{f}\right)$ $\displaystyle\leq$
$\displaystyle
N\left(r,\frac{g^{\prime}}{g}\right)-N\left(r,\frac{g}{g^{\prime}}\right)+m\left(r,\frac{g^{\prime}}{g}\right)+S(r,f)$
$\displaystyle\leq$ $\displaystyle
N\left(r,\frac{g^{\prime}}{g}\right)-N\left(r,\frac{g}{g^{\prime}}\right)+S(r,g)+S(r,f)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ =\
\bar{N}(r,f)+N\left(r,\frac{1}{g}\right)-N\left(r,\frac{1}{g^{\prime}}\right)+S(r,f).$
(13)

Here we add $N(r,1/f^{l+n})$ to both sides of inequality (13), then

$(l+n)T\left(r,\frac{1}{f}\right)\leq\bar{N}(r,f)+N\left(r,\frac{1}{g}\right)-N\left(r,\frac{1}{g^{\prime}}\right)+N\left(r,\frac{1}{f^{l+n}}\right)+S(r,f).$
(14)

Note that
$g^{\prime}=f^{l-1}\left(f^{(k)}\right)^{n-1}\left(lf^{(k)}f^{\prime}+nff^{(k+1)}\right)$,
which implies

$(l-1)N\left(r,\frac{1}{f}\right)+(n-1)N\left(r,\frac{1}{f^{(k)}}\right)\leq
N\left(r,\frac{1}{g^{\prime}}\right).$ (15)

Substituting (15) into (14), we get

$\displaystyle T\left(r,\frac{1}{f^{l+n}}\right)$ $\displaystyle\leq$
$\displaystyle\bar{N}(r,f)+N\left(r,\frac{1}{g}\right)+N\left(r,\frac{1}{f^{l+n}}\right)-(l-1)N\left(r,\frac{1}{f}\right)$
$\displaystyle-(n-1)N\left(r,\frac{1}{f^{(k)}}\right)+S(r,f).$

Hence,

$(l+n)T(r,f)\leq
N\left(r,\frac{1}{g}\right)+\bar{N}(r,f)+(n+1)N\left(r,\frac{1}{f}\right)-(n-1)N\left(r,\frac{1}{f^{(k)}}\right)+S(r,f).$
(16)

Inequality (6) implies that for $k\geq 2$,

$\displaystyle(k-1)\bar{N}(r,f)\leq
N\left(r,\frac{1}{f^{(k)}}\right)+S^{*}(r,f).$ (17)

Now by combining inequality (16) and (17), we have

$\displaystyle(l+n)T(r,f)$ $\displaystyle\leq$ $\displaystyle
N\left(r,\frac{1}{g}\right)+\frac{1}{k-1}N\left(r,\frac{1}{f^{(k)}}\right)+(n+1)N\left(r,\frac{1}{f}\right)$
$\displaystyle-(n-1)N\left(r,\frac{1}{f^{(k)}}\right)+S(r,f)+S^{*}(r,f).$

Since $1/{(k-1)}-n+1\leq 0$ for $n\geq 2$, $k\geq 2$, then

$\displaystyle(l+n)T(r,f)\leq
N\left(r,\frac{1}{g}\right)+\left(n+1\right)N\left(r,\frac{1}{f}\right)+S^{*}(r,f).$
(18)

Since $l-1>0$ and
$\left(n+1\right)N\left(r,1/f\right)\leq\left(n+1\right)T\left(r,f\right)$,
then

$\displaystyle
T(r,f)\leq\frac{1}{l-1}N\left(r,\frac{1}{f^{l}(f^{(k)})^{n}-1}\right)+S^{*}(r,f).$
(19)

Replacing the number 1 in $f^{l}(f^{(k)})^{n}-1$ by any nonzero constant $a$,
the inequality (7) is obtained. The proof is completed.

##  4 Proof of Theorem 1.2.

Set $\psi=f^{l}(f^{(k)})^{n}$. Inequality (19) is stated that

$T(r,f)\leq\frac{1}{l-1}N\left(r,\frac{1}{\psi-a}\right)+S^{*}(r,f)$ (20)

for $l\geq 2$, $n\geq 2$, $k\geq 2$. By the definition of $\delta(a,f)$ and
the First Fundamental Theorem, we obtain

$\displaystyle T(r,\psi)$ $\displaystyle\leq$
$\displaystyle(nk+n+l)T(r,f)+S(r,f)$ (21) $\displaystyle\leq$
$\displaystyle\frac{nk+n+l}{l-1}N\left(r,\frac{1}{\psi-a}\right)+S^{*}(r,f).$

By combining two inequalities (20) and (21), we have

$N\left(r,\frac{1}{\psi-a}\right)\geq\frac{l-1}{nk+n+l}T(r,\psi)-S^{*}(r,f).$

Since

$\displaystyle T(r,f)$ $\displaystyle=$
$\displaystyle\frac{1}{l}T(r,f^{l})\leq
T\left(r,(f^{(k)})^{n}\right)+T(r,\psi)$ $\displaystyle\leq$ $\displaystyle
O\left(T\left(r,\psi\right)\right),$

then we deduce that

$\mathop{\lim\inf}\limits_{r\to\infty}\frac{S^{*}(r,f)}{T(r,\psi)}=\mathop{\lim\inf}\limits_{r\to\infty}\frac{S^{*}(r,f)}{T(r,f)}\frac{T(r,f)}{T(r,\psi)}=0.$

Therefore, by the definition of deficiency,

$\displaystyle\delta(a,\psi)$ $\displaystyle=$ $\displaystyle
1-\mathop{\lim\sup}\limits_{r\to\infty}\frac{N\left(r,\frac{1}{\psi-a}\right)}{T(r,\psi)}$
$\displaystyle\leq$ $\displaystyle
1-\mathop{\lim\sup}\limits_{r\to\infty}\frac{\frac{l-1}{nk+n+l}T(r,\psi)-S^{*}(r,f)}{T(r,\psi)}$
$\displaystyle\leq$ $\displaystyle
1-\frac{l-1}{nk+n+l}+\mathop{\lim\inf}\limits_{r\to\infty}\frac{S^{*}(r,f)}{T(r,\psi)}$
$\displaystyle=$ $\displaystyle 1-\frac{l-1}{nk+n+l}.\@qedbox{}$

##  5 An application

After Yamanoi’s result was published in 2013, there are some results about
deficieny relations came out by using his important theorem. We take a result
from Fang and Wang [5] as a good example here, and we analogue their steps to
get an estimate of the sum of deficiencies of $f^{l}(f^{(k)})^{n}$.

######  Theorem D.

_[5, Propostion 2]_ Let $f$ be a meromorphic and transcendental function in
the complex plane, $k$ be a positive integer, and $P$ be the set of all
polynomials. Then

$\sum_{b\in\mathbb{C}}\delta{(b,f^{(k)})}\leq
1-(k-1)\left(1-\Theta_{E}\left(\infty,f^{(k)}\right)\right),$

where for $r\not\in E$,

$\Theta_{E}\left(\infty,f^{(k)}\right)=1-\mathop{\lim\sup}\limits_{r\to\infty}\frac{\bar{N}(r,f^{(k)})}{T(r,f^{(k)})},$

where $E(=M(K)\cup E_{1})$, $M(K)$ is a set of finite upper logarithmic
density and $E_{1}$ is a set of finite measure.

We need the following lemma for our calculation. This lemma is as well used in
paper [5].

######  Lemma 5.5.

_[7, p. 33]_ Let $a_{1}$, $a_{2}$, $\cdots$, $a_{q}$, where $q>2$, be distinct
finite complex numbers. Then

$\sum_{i=1}^{q}m\left(r,\frac{1}{f-a_{i}}\right)\leq
m\left(r,\sum_{i=1}^{q}\frac{1}{f-a_{i}}\right)+O(1).$

######  Theorem 5.6.

Let $f$ be a transcendental meromorphic function in $\mathbb{C}$, $k$, $l$,
$n$ be positive integers all at least $2$ and $a_{i}\in\mathbb{C}$ be
constants, $i=1,2,\cdots,q$. Then

$\sum_{i=1}^{q}\delta(a_{i},f^{l}(f^{(k)})^{n})\leq 1+\frac{1}{nk+n+l}.$

Proof. By Nevanlinna theory, for constants $a_{i}\in\mathbb{C}$, the sum of
deficiencies of function $f$ are defined by

$\displaystyle\sum_{i=1}^{q}\delta{(a_{i},f)}=\mathop{\lim\inf}\limits_{r\to\infty}\sum_{i=1}^{q}\frac{m\left(r,\frac{1}{f-a_{i}}\right)}{T\left(r,f\right)}.$
(22)

Let $\psi=f^{l}(f^{(k)})^{n}$. By Lemma 5.5, we have

$\displaystyle\sum_{i=1}^{q}m\left(r,\frac{1}{\psi-a_{i}}\right)$
$\displaystyle\leq$ $\displaystyle m\left(r,\sum_{i=1}^{q}\frac{1}{\psi-
a_{i}}\right)+O(1)$ (23) $\displaystyle\leq$ $\displaystyle
m\left(r,\sum_{i=1}^{q}\frac{\psi^{\prime\prime}}{\psi-
a_{i}}\right)+m\left(r,\frac{1}{\psi^{\prime\prime}}\right)+S(r,f)$
$\displaystyle\leq$ $\displaystyle
T(r,\psi^{\prime\prime})-N\left(r,\frac{1}{\psi^{\prime\prime}}\right)+S(r,f)$
$\displaystyle\leq$ $\displaystyle
N(r,\psi^{\prime\prime})+m(r,\psi^{\prime\prime})-N\left(r,\frac{1}{\psi^{\prime\prime}}\right)+S(r,f)$
$\displaystyle\leq$ $\displaystyle
N(r,\psi)+2\bar{N}(r,\psi)+m(r,\psi)-N\left(r,\frac{1}{\psi^{\prime\prime}}\right)+S(r,f).$

By Yamanoi’s result (6), it follows from inequality (23) that

$\displaystyle\sum_{i=1}^{q}m\left(r,\frac{1}{\psi-a_{i}}\right)$
$\displaystyle\leq$ $\displaystyle
T(r,\psi)+2\bar{N}(r,\psi)-\bar{N}\left(r,\psi\right)+S(r,f)$ (24)
$\displaystyle\leq$ $\displaystyle T(r,\psi)+\bar{N}(r,f)+S(r,f)$
$\displaystyle\leq$ $\displaystyle T(r,\psi)+T(r,f)+S(r,f).$

By Theorem 1.1 and Theorem 1.2, it follows from inequality (24) that,

$\displaystyle\sum_{i=1}^{q}m\left(r,\frac{1}{\psi-a_{i}}\right)$
$\displaystyle=$
$\displaystyle\mathop{\lim\inf}\limits_{r\to\infty}\sum_{i=1}^{q}\frac{m\left(r,\frac{1}{\psi-
a_{i}}\right)}{T\left(r,\psi\right)}$ $\displaystyle\leq$ $\displaystyle
1+\mathop{\lim\inf}\limits_{r\to\infty}\frac{T(r,f)}{T(r,\psi)}+S(r,f)$
$\displaystyle\leq$ $\displaystyle
1-\frac{1}{l-1}\left(1-\mathop{\lim\sup}\limits_{r\to\infty}\frac{N\left(r,\frac{1}{\psi-a}\right)}{T(r,\psi)}-1\right)$
$\displaystyle=$ $\displaystyle
1-\frac{1}{l-1}\left(\delta\left(a,\psi\right)-1\right)$ $\displaystyle\leq$
$\displaystyle 1-\frac{1}{l-1}\left(1-\frac{l-1}{nk+n+l}-1\right)$
$\displaystyle=$ $\displaystyle 1+\frac{1}{nk+n+l}.\@qedbox{}$

##  6 Acknowledgement

The authors are very grateful to Professor Toshiyuki Sugawa and every member
in the seminars for their valuable suggestions and comments, which helped a
lot in improving the paper.

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Yan Jiang
Graduate School of Information Sciences,
Tohoku University, Sendai 980-88579, JAPAN

E-mail address<EMAIL_ADDRESS><EMAIL_ADDRESS>
Bin Huang
Department of Mathematics and Computing Science,
Changsha University of Science and Technology, Changsha 410076, P.R.China
E-mail address<EMAIL_ADDRESS>