# Number of triple points on complete intersection Calabi-Yau threefolds

Kacper Grzelakowski
Wydział Matematyki i Informatyki
Uniwerstytet Łódzki
Banacha 22
90-238 Łódz, Poland
E-mail<EMAIL_ADDRESS>
###### Abstract

We discuss bounds for the number of ordinary triple points on complete
intersection Calabi-Yau threefolds in projective spaces and for Calabi-Yau
threefolds in weighted projective spaces. In particular, we show that in
$\mathbb{P}^{5}$ the intersection of a quadric and a quartic cannot have more
than $10$ ordinary triple points. We provide examples of complete intersection
Calabi-Yau threefolds with multiple triple points. We obtain the exact bound
for a sextic hypersurface in $\mathbb{P}[1:1:1:1:2]$, which is $10$. We also
discuss Calabi-Yau threefolds that cannot admit triple points.

††2020 _Mathematics Subject Classification_ : 14J17; 14J30.†† _Key words and
phrases_ : intersection theory, Calabi-Yau threefolds.

##  1 Introduction

By an _ordinary triple point_ on a threefold $X$ we mean an isolated
singularity $O$ such that the tangent cone to $X$ at $O$ is a cone over a
smooth degree three surface. In this paper we consider bounds on the number of
ordinary triple points on Calabi-Yau threefolds. Our interest was sparked off
by the results of Kloosterman and Rams regarding the possible number of triple
points on a Calabi-Yau quintic threefold in $\mathbb{P}^{4}$ [9]. They have
proven that a quintic threefold with a reducible hyperplane section cannot
have more than $10$ ordinary triple points as the only singularities and
constructed an example where this bound is reached. The question whether or
not a quintic threefold can have $11$ ordinary triple points remains open, but
it is known that the number $12$ is impossible.

Our interest in Calabi-Yau threefolds admitting ordinary triple points as the
only singularities comes from the fact that the resolution of such
singularities is canonical - it does not modify the canonical class of a
variety and leads to new smooth Calabi-Yau threefolds. For double points we
can obtain such resolution only if there exists a so-called small resolution
of singularities, that is, replacing a point by a $\mathbb{P}^{1}$. There
already exists an extensive literature on the number of double points on
Calabi-Yau threefolds, for example ([7][8][13]), yet the exact bounds are
still difficult to obtain. For instance, in the case of a quintic threefold
the best result is given by van Straten in [13] with $130$ double points and
it is not known whether or not it is indeed the least upper bound. The case of
ordinary triple points should be simpler, yet it is much less discussed.

The threefolds we consider are a natural next step after a quintic
hypersurface in $\mathbb{P}^{4}$. We analyse Calabi-Yau threefolds that are
complete intersection manifolds in projective spaces, namely
$X_{2,4},X_{3,3}\subset\mathbb{P}^{5}$, $X_{2,2,3}\subset\mathbb{P}^{6}$ and
$X_{2,2,2,2}\subset\mathbb{P}^{7}$ where the subscripts are the degrees of the
hypersurfaces we intersect. We also provide a bound for the number of ordinary
triple points for a sextic in weighted projective space. Some of the results
are quite straightforward but we have not been able to locate them in the
existing literature. Numerical bounds are presented in Table 1. We include
there the results of Rams and Kloosterman regarding the quintic.

We always assume that the threefold in question has only ordinary triple
points (OTPs) as singularities. Let $\mu_{3}(X)$ denote the maximal number of
ordinary triple points that $X$ can have under this condition. We drop $X$ if
the context is clear. The analysis of $\mu_{3}(X_{2,4})$ reduces to the
analysis of a quintic threefold as we perform a projection from one of the
triple points of $X_{2,4}$ to obtain a quintic hypersurface in
$\mathbb{P}^{4}$. In doing so we obtain a particular quintic, namely one
containing a cubic surface and also admitting $24$ double points. Because of
those additional singularities we are not able to use all the methods of [9]
to analyse this quintic, yet we still make use of the Varchenko bound to find
an upper bound on $\mu_{3}(X_{2,4})$ which is $10$. We have only been able to
construct $X_{2,4}$ admitting $7$ ordinary triple points as singularities and
so the problem to determine the exact bound remains open.

The case of $X_{3,3}$ proves to be more complex. In our discussion we assume
that all triple points of $X_{3,3}$ are inherited from some cubic fourfold,
that is, for each $O$ triple on $X_{3,3}$ there exists a cubic fourfold
$X_{O}$ such that $X_{3,3}\subset X_{O}$ and $O$ is triple on $X_{O}$. In that
case we find the exact bound $\mu_{3}$ for a complete intersection threefold
${X_{3,3}}$, which is $9$, and we provide a construction of a $X_{3,3}$
admitting that many. The points lie in an interesting configuration as they
come in three collinear triples. We discuss the geometry of a resolution of
singularities of a threefold $X_{3,3}$ containing only inherited triple
points, in particular the fact that it can be realised as a double cover of a
cubic threefold.

We include one example of $X_{2,2,3}$ admitting multiple ordinary triple
points.

The next part of this paper is devoted to finding $\mu_{3}$ of a sextic
hypersurface $X_{6}\subset\mathbb{P}(1:1:1:1:2)$. To find it we have to look
at its hyperplane section, as it turns out that all the triple points are
contained in one. We use the fact that the weighted projective space
$\mathbb{P}(1:1:1:1:2)$ can be realized as a Veronese embedding of
$\mathbb{P}^{3}$ in $\mathbb{P}^{9}$ and $X_{6}$ as its trisection and thus a
triple cover of $\mathbb{P}^{3}$. We also employ the techniques used in
finding the polar bound for the possible number of singular points.

The paper ends with a discussion of Calabi-Yau threefolds of degree $8$ and
$10$ in weighted projective spaces and of the complete intersection of four
quadrics in $\mathbb{P}^{7}$ which do not admit any ordinary triple points.

Table 1: Bound on the number of triple points on Calabi-Yau threefolds Calabi-Yau threefold $X$ | $\mu_{3}(X)$
---|---
$X_{5}\subset\mathbb{P}^{4}$ |  $10$ or $11$
$X_{2,4}\subset\mathbb{P}^{5}$ | $7\leq\dots\leq 10$
$X_{3,3}\subset\mathbb{P}^{5}$ | $9\leq$
$X_{2,2,3}\subset\mathbb{P}^{6}$ | $4\leq$
$X_{2,2,2,2}\subset\mathbb{P}^{7}$ | 0
$X_{6}\subset\mathbb{P}(1:1:1:1:2)$ | 10
$X_{8}\subset\mathbb{P}(1:1:1:1:4)$ | 0
$X_{10}\subset\mathbb{P}(1:1:1:2:5)$ | 0

##  2 Triple points on $X_{2,4}\subset\mathbb{P}^{5}$

We consider a threefold that is an intersection of quadric and quartic
fourfolds in $\mathbb{P}^{5}$. We use $[x:y:z:t:u:w]$ as variables.

######  Theorem 2.1.

A complete intersection threefold $X_{2,4}\subset\mathbb{P}^{5}$ with only
ordinary triple points as singularities can contain at most $10$ such points.

###### Proof.

At an ordinary triple point the complete intersection threefold in
$\mathbb{P}^{5}$ locally looks like the intersection of a hyperplane and a
fourfould with an ordinary triple point. Let us assume the point in question
in $O=[1:0:0:0:0:0]$. Let $G_{i}$ be a homogeneous polynomial of degree $i$
independent of the first variable $x$. Then it is easy to see that the
complete intersection $X_{2,4}$ of a quadric $X_{2}$ and a quartic $X_{4}$
with respective equations $F_{2}:=xG_{1}+G_{2}$ and $F_{4}:=xG_{3}+G_{4}$ does
indeed have a triple point at $O$. Note that $G_{3}$ needs to be a polynomial
defining a smooth threefold in $\mathbb{P}^{4}$ for this singularity to be an
ordinary one. In the next paragraph we show that for an ordinary triple point
$O$ on $X_{2,4}$ we can always find $X_{2}$ and $X_{4}$ with equations of the
above form.

We fix $X_{2}=V(F_{2})=V(xG_{1}+G_{2})$ as a quadric that defines $X_{2,4}$.
Let $\hat{X}_{4}$ be a quartic fourfold that forms a regular sequence with
$X_{2}$. Let $\hat{F}_{4}$ be the polynomial defining $\hat{X}_{4}$ and let us
group the terms of this polynomial with respect to $x$, that is,
$\hat{F}_{4}=\alpha x^{4}+x^{3}\hat{G}_{1}+\dots+\hat{G}_{4}$ with
$\hat{G}_{i}$ a homogeneous polynomial independent of $x$ of degree $i$ and
$\alpha$ a constant. We want $O$ to be a triple point, and so in particular
$\hat{X}_{4}$ has to pass through $O$, which means $\alpha=0$. If $G_{1}$
defined a hyperplane different from $\hat{G}_{1}$ then the complete
intersection of $X_{2}$ and $\hat{X}_{4}$ would be smooth at $O$ and so
$G_{1}=\beta\hat{G}_{1}$ with $\beta$ being a constant. (It may happen in
particular that $G_{1}$ is the zero polynomial but this does not cause any
problems.) For the complete intersection threefold to have a triple point at
$O$ we need $\hat{G}_{2}=\beta G_{2}+G_{1}H_{1}$ with $H_{1}$ being a degree
one polynomial independent of $x$ and $\beta$ a constant. Otherwise, our
intersection would have a double point at $O$. Consequently, we can write
$\hat{F}_{4}=x^{3}\beta G_{1}+x^{2}(\beta
G_{2}+G_{1}H_{1})+x\hat{G}_{3}+\hat{G}_{4}$. We can always divide by a nonzero
constant and assume $\beta=1$ (and abuse notation by leaving $H_{1}$
unchanged). In that case $F_{4}=\hat{F}_{4}-x^{2}F_{2}-xH_{1}F_{2}$ is the
polynomial defining another fourfold $X_{4}$ containing $X_{2,4}$ and it is
indeed of the desired form, thus proving the assertion from the previous
paragraph.

From now on we assume that $X_{2,4}$ is defined by $F_{2}=xG_{1}+G_{2}$ and
$F_{4}=xG_{3}+G_{4}$ as above. We can perform a projection from $O$ onto a
$\mathbb{P}^{4}$ and obtain a quintic threefold. Indeed, let $E$ denote the
exceptional divisor of a blowup of $O$. The projection is given by a linear
system $|H-E|$, that is a system of strict transforms of hyperplanes passing
through $O$. Since $(H-E)^{3}=5$, as a result of this projection we obtain a
threefold of degree $5$ which we denote $X_{5}$. We want to show that $X_{5}$
is normal. Assume to the contrary that it is not. Then by [11, p. 254] there
would exist a normalization $\bar{X}_{5}$ with a two-dimensional ramification
locus $D$ on $X_{5}$. Let $P$ be a point on the threefold $X_{5}$ and let
$L_{P}$ be the line in $\mathbb{P}^{5}$ passing through $O$ that projects to
$P$. Note that $P\in D$ if $L_{p}$ cuts $X_{2,4}$ either in two points other
that $O$ or if it is tangent to $X_{2,4}$ at some point $\neq O$. We see that
lines cutting $X_{2,4}$ twice or tangent to it need to be contained both in
$X_{2}$ and $X_{4}$. Indeed, counting the number of intersection points (with
multiplicities in case of tangency), these lines cut the fourfolds in $3$
points (in case of $X_{2}$) or at a triple point and $2$ additional points (in
case of $X_{4}$). Thus they are contained in $X_{2,4}$, which means that the
projection from $O$ contracts a three-dimensional cone ruled over $D$, a
contradiction.

Every triple point $O_{i}$ other than $O$ on $X$ yields a triple point
$O^{\prime}_{i}$ on $X_{5}$, thus if $\mu_{3}(X)=n$ then $\mu_{3}(X_{5})=n-1$.
Also note, that by the choice of equations for $X_{2}$ and $X_{4}$ we can
consider $G_{1}G_{4}-G_{2}G_{3}$ to be the polynomial defining $X_{5}$ in
$\mathbb{P}^{4}$ and thus obtain $24$ double points on $X_{5}$, namely those
for which $G_{1}=G_{2}=G_{3}=G_{4}=0$. By an argument similar to [9, 2.6] we
can consider Varchenko’s spectral bound for $Q$. For $\alpha=2/5$ the spectrum
of the fivefold point has length $155$ in the interval $(2/5,7/5)$ whilst the
spectrum of the triple point has length $14$ and the spectrum of the double
point has length $1$. Thus from $155-24=131$ and $10\cdot 14=140$ a quintic
threefold containing $24$ double points can have at most $9$ ordinary triple
points as the remaining singularities. As then $n-1=9$ we conclude that
$X_{2,4}$ can contain at most $n=10$ ordinary triple points. ∎

######  Lemma 2.2.

There exitsts a threefold $X_{2,4}$ containing $7$ ordinary triple points as
the only singularities.

###### Proof.

We can easily produce a fourfold $X_{4}$ admitting $6$ ordinary triple points
in general linear position. Indeed the space of quartic fourfolds is of
dimension $125$ and a triple point imposes $20$ linear conditions. Let $X_{4}$
be a quadric fourfold having triple points $[1:1:1:1:1:1]$ and
$[1:0:0:0:0:0],\dots,[0:0:0:0:1:0]$. Let $\tilde{F}_{4}(x,y,z,t,u,w)$ be the
polynomial defining $X_{4}$. We can write
$\tilde{F}_{4}=w^{3}G_{1}+w^{2}G_{2}+wG_{3}+G_{4}$ where $G_{i}$ are
homogeneous degree $i$ polynomials independent of $w$ such that $V(G_{1})$ and
$V(G_{2})$ pass through the points $[1:1:1:1:1],[1:0:0:0:0],\dots,[0:0:0:0:1]$
in $\mathbb{P}^{4}$. Now lets take $X_{2}=V(\tilde{F}_{2})$ passing through
these points with $\tilde{F}_{2}=wG_{1}+G_{2}$. For $G_{i}$ general enough the
complete intersection $X_{2,4}$ will have an ordinary triple point at
$[0:0:0:0:0:1]$ and inherit all the ordinary triple points of $X_{4}$
producing an $X_{2,4}$ with $7$ OTPs. We are unaware of any $X_{2,4}$ having
more than $7$ OTPs as the only singularities. ∎

##  3 Triple points on $X_{3,3}\subset\mathbb{P}^{5}$

We consider a degree $9$ threefold that is an intersection of two degree $3$
hypersurfaces in $\mathbb{P}^{5}$. As before, we assume that $X_{3,3}$ has
ordinary triple points as the only singularities. Let $X_{3,3}$ be
$V(F_{3},G_{3})$ where $F_{3}$ and $G_{3}$ are the degree $3$ polynomials
defining fourfolds in $\mathbb{P}^{5}$. Any other cubic fourfold containing
$X_{3,3}$ is an element of the pencil $\alpha F_{3}+\beta G_{3}$ for
$\alpha,\beta\in\mathbb{C}$. We denote this pencil by $\mathscr{L}$ and the
fourfold defined by the above equation by $X_{\alpha,\beta}$.

In our discussion we distinguish between the triple points that we call
inherited and noninherited. For the former, we assume that in the pencil of
cubic fourfolds there exists a fourfold $X_{O}$ for which the point $O$ is
triple and thus the triple point on $X_{3,3}$ can be obtained by intersecting
this singular fourfold with a smooth one. The latter case is when there is no
fourfold in the pencil having an (ordinary) triple point at $O$ and thus the
triple point on $X_{3,3}$ is specifically obtained because of the intersection
of fourfolds, and not inherited from one of them. We only analyse the first
case.

######  Theorem 3.1.

Let $X_{3,3}$ be a complete intersection threefold in $\mathbb{P}^{5}$ having
only ordinary triple points as singularities. Furthermore, assume that for
every triple point $O_{i}$ on $X_{3,3}$ there exists a cubic fourfold $X_{i}$
containing $X_{3,3}$ such that $O_{i}$ is triple on $X_{i}$. Then $X_{3,3}$
can have at most $9$ ordinary triple points.

###### Proof.

We begin by the analysis of points in general linear position. First, notice
that a cubic fourfold having a triple point is necessarily a cone. Let us
consider two triple points on $X_{3,3}$ denoted by $O_{1}$, $O_{2}$ and assume
thery are inherited from $X_{1}$ and $X_{2}$ respectively. We observe that the
line $L_{1,2}$ passing through $O_{1}$ and $O_{2}$ is contained in $X_{3,3}$
as it has at least a degree $4$ intersection with $X_{1}$ and $X_{2}$. On the
other hand, it may happen that both points $O_{1}$ and $O_{2}$ are triple on
the same fourfold $X_{3}$, which makes the line $L_{1,2}$ triple on $X_{3}$.
Hence, there exists a point $O_{3}$ lying on $L_{1,2}$ which is triple on
$X_{3,3}$ as the other cubic fourfold yielding the complete intersection
threefold has to cut $L_{1,2}$ in three points in total. In short, we see that
any two triple points on $X_{3,3}$ span a triple line contained in $X_{3,3}$
unless there is a third triple point of $X_{3,3}$ collinear with them.
Analogously, we find that $3$ triple points in general linear position span a
plane contained in $X_{3,3}$ and $4$ triple points in general linear position
would span a $\mathbb{P}^{3}$ contained in $X_{3,3}$, which is a contradiction
as this makes $X_{3,3}$ reducible. Thus there can be at most $3$ ordinary
triple points in general linear position on $X_{3,3}$.

Let us see what happens when the points are not in general linear position.
Assume there are $4$ OTPs on a line. Then the line is triple (by assumption)
on one of the cubic fourfolds and contained in both (by degree count) and so
singular on $X_{3,3}$, a contradiction. We show that there cannot be more than
$8$ OTPs on the plane (the number $8$ is sufficient for our calculations yet
it seems absurdly high and we believe $4$ triple points on a plane is the
maximum). Remember our assumption that all the triple points are inherited
from cubic fourfolds and thus the third triple point $O_{3}$ causes a plane
$\Pi$ spanned by $O_{1},O_{2}$ and $O_{3}$ to be contained in the intersection
$X_{3,3}$. Assume that $\Pi=V(t,u,w)$. Since the plane has to be contained in
every cubic fourfold $X_{\alpha,\beta}$ in the pencil $\mathscr{L}$ we can
write the equation of the cubic in $\mathbb{P}^{5}\times\mathbb{P}^{1}$ as
$X_{\alpha,\beta}=\alpha(tT_{2}+uU_{2}+wW_{2})+\beta(tT^{\prime}+uU^{\prime}+wW^{\prime})$
with $T,\dots,W^{\prime}$ homogeneous in degree $2$. As every triple point $P$
has to be inherited from some fourfold, for each such point we need to be able
to find parameters $\alpha_{P}$, $\beta_{P}$ such that the first and second
derivatives of the equation $F$ defining $X_{\alpha_{P},\beta_{P}}$ vanish.
Moreover, we can restrict our analysis to the product of plane $\Pi$ with
$\mathbb{P}^{1}$ since we assume that all the triple points lie on this plane.
For brevity we use $\alpha_{i_{m}}$ to denote $\alpha$ or $\beta$, $x_{j_{m}}$
to denote variables from the set $\\{x,y,z\\}$, and $t_{k_{m}}$ to denote
variables from the set $\\{t,u,w\\}$ for $m\in\\{1,2\\}$. Note that
$\alpha_{i_{1}}$ may not be different from $\alpha_{i_{2}}$ etc. After
restriction to $\Pi$ we have:

$\frac{\partial^{2}{F}}{\partial\alpha_{i_{1}}\partial\alpha_{i_{2}}}=\frac{\partial^{2}{F}}{\partial
x_{j_{1}}\partial
x_{j_{2}}}=\frac{\partial^{2}{F}}{\partial\alpha_{i_{1}}\partial
x_{j_{1}}}=0,$
$\frac{\partial^{2}{F}}{\partial{t_{k_{1}}}\partial{t_{k_{2}}}}=\alpha(\frac{\partial{T_{k_{1}}}}{\partial{t_{k_{2}}}}+\frac{\partial{T_{k_{2}}}}{\partial{t_{k_{1}}}})+\beta(\frac{\partial{T^{\prime}_{k_{1}}}}{\partial{t_{k_{2}}}}+\frac{\partial{T^{\prime}_{k_{2}}}}{\partial{t_{k_{1}}}}),$
$\frac{\partial^{2}{F}}{\partial{\alpha_{i_{1}}}\partial{t_{k_{1}}}}=T_{k_{1}},$
$\frac{\partial^{2}{F}}{\partial{x_{j_{1}}}\partial{t_{k_{1}}}}=\alpha{\frac{\partial
T_{k_{1}}}{\partial x_{j_{1}}}}+\beta{\frac{\partial
T^{\prime}_{k_{1}}}{\partial x_{j_{1}}}}.$

In this way we obtain $21$ degree $2$ equations that need to vanish
simultaneously on $\Pi\times\mathbb{P}^{1}$. As three of them should cut out a
finite set of points we see that a maximum number of triple points we can
obtain that way is $8$. For reasons stated above any point outside the plane
containing $3$ or more noncollinear triple points would cause $X_{3,3}$ to be
reducible, a contradiction.

Assume now that in the pencil $\mathscr{L}$ there are two fourfolds $X_{1}$
and $X_{2}$ containing triple lines $L_{1}$ and $L_{2}$ respectively such that
these lines do not meet. Then $X_{1}$ cuts $L_{2}$ in three points and
similarly for $X_{2}$ and $L_{1}$. Each of these intersection points yields a
triple point on $X_{3,3}$ and so we obtain $6$ triple points on $X_{3,3}$. As
$L_{1}$ and $L_{2}$ are in general linear position we can assume
$L_{1}=V(z,t,u,w)$ and $L_{2}=V(x,y,z,t)$. This means that
$X_{1}=V(F(z,t,u,w))$ and $X_{2}=V(G_{3}(x,y,z,t))$, with each of the defining
equations being independent of two out of six variables. Then the elements of
the pencil $\mathscr{L}$ are as follows: $\bar{X}_{a,b}=V(aF_{3}+bG_{3})$.
Note that for $F(z,t,u,w)=a_{1}z^{3}+a_{2}t^{3}+a_{3}u^{3}+a_{4}w^{3}$,
$G_{3}(x,y,z,t)=b_{1}x^{3}+b_{2}y^{3}+b_{3}z^{3}+b_{4}t^{3}$ with $a=b=1$ and
$a_{1}=-b_{3}$ $a_{2}=-b_{4}$ we obtain a fourfold $X_{1,1}$ containing a
triple line $L_{3}=V(x,y,u,w)$ providing another three triple points on
$\bar{X}_{3,3}$ yielding a threefold $X_{3,3}$ with $9$ ordinary triple points
as the only singularities. We cannot hope to obtain a $10$th triple point this
way as we would always be able to find a plane $\Pi$ passing through one
triple point of each of the lines $L_{1},L_{2},L_{3}$ and not containing the
putative $P_{10}$ and thus giving $\mathbb{P}^{3}$ contained in $X_{3,3}$, a
contradiction. ∎

In particular, we have shown that $\mu_{3}(X_{3,3})=9$ if we assume that all
the triple points are inherited, and $\mu_{3}(X_{3,3})\geq 9$ in general.

In the case when at least one triple point of $X_{3,3}$ is inherited from the
triple point of some fourfold we have a nice characterization of $X_{3,3}$ as
a double cover of a cubic threefold.

######  Lemma 3.2.

Assume that in the pencil of cubic fourfolds containing a complete
intersection threefold $X_{3,3}\subset\mathbb{P}^{5}$ with only triple points
as singularities there is a fourfold with a triple point. Then a threefold
$\tilde{X}_{3,3}$ obtained by blowing up one of the triple points of $X_{3,3}$
is a double cover of a cubic threefold $Y_{3}$ ramified over a degree $12$
surface.

###### Proof.

Without loss of generality we may assume that $X_{3,3}$ is the complete
intersection of a fourfold $X=V(x^{2}G_{1}+xG_{2}+G_{3})$ and a cone
$\bar{Y}=V(F_{3})$ where $G_{i}$ and $F_{i}$ are homogeneous polynomials of
degree $i$ independent of $x$. Note that in this case $P$ is exactly the
vertex of the cone $\bar{Y}$ through which $X$ passes smoothly. A general line
in $\mathbb{P}^{5}$ cuts $X$ in three points. As $X$ passes through $P$, this
means that it cuts a general line of the ruling of the cone $\bar{Y}$ in $2$
other points. As we blow up $P$ and obtain $\tilde{X}_{3,3}$ we can treat it
as a double cover of the base of the cone, that is $Y\subset\mathbb{P}^{4}$.
We still have to check the ramification locus. This is exactly the vanishing
locus of the discriminant, $D=V(G_{2}^{2}-4G_{1}G_{3})$, calculated with
respect to $x$ from the equation defining $X$. The intersection of $D$ and $Y$
in $\mathbb{P}^{4}$ is a degree $12$ surface $S_{12}$. This concludes the
proof. ∎

##  4 Triple points on $X_{2,2,3}$ in $\mathbb{P}^{6}$

We provide an easily obtainable example of na $X_{2,2,3}$ threefold with $4$
ordinary triple points.

######  Theorem 4.1.

There exists a complete intersection of two quadrics and a cubic in
$\mathbb{P}^{6}$ with exactly $4$ ordinary triple points as the singular
locus.

###### Proof.

It is enough to choose a cubic fivefold having a triple plane and intersect it
with two general quadric fourfolds. The complete intersection will have $4$
triple points lying on this plane as the only singularities. ∎

There is an interesting link between the geometry of the complete intersection
threefolds $X_{3,3}$ and $X_{2,2,3}$, analogous to the one between $X_{2,4}$
and $X_{5}$ shown in the proof of the Theorem 2.1.

######  Lemma 4.2.

The projection from a triple point of a $X_{2,2,3}$ having $n$ triple points
is a complete intersection $X_{3,3}$ threefold in $\mathbb{P}^{5}$ with at
least $n-1$ triple points and $12$ double points.

###### Proof.

Let $F_{3}$ and $F_{2}=xA_{1}+A_{2}$, $G_{2}=xB_{1}+B+2$ denote homogeneous
polynomials defining $X_{2,2,3}$ in $\mathbb{P}^{5}$. Note that $F_{3}$ is
independent of $x$. Then $F_{3}=0$ and $A_{2}B_{1}-B_{2}A_{1}=0$ define the
threefold $X_{3,3}$ that is the image of the projection from the triple point
$O$ of $X_{2,2,3}$. Any triple points other than $O$ project to triple points
on $X_{3,3}$ and we also additionally obtain $12$ double points whenever
$A_{1}=A_{2}=B_{1}=B_{2}=F_{3}=0$, as the image of lines passing through $O$
contracted by this projection. ∎

##  5 Triple points on a sextic in $\mathbb{P}{[1:1:1:1:2]}$

For background on weighted projective spaces (WPS) see for example [2] or
[10].

######  Theorem 5.1.

A sextic hypersurface $X_{6}$ in $\mathbb{P}[1:1:1:1:2]$ with only ordinary
triple points as singularities can have at most $10$ ordinary triple points
and this bound is attainable.

We work in the weighted projective space $\mathbb{P}[1:1:1:1:2]$ with
variables $x,y,z,t$ of weight $1$ and $u$ of weight $2$. We can thus write the
equation of a sextic hypersurface as
$F(x,y,z,t,u)=u^{3}+u^{2}G_{2}(x,y,z,t)+uG_{4}(x,y,z,t)+G_{6}(x,y,z,t)$ where
$G_{i}$ is a homogeneous polynomial of degree $i$. Note that the term $u^{3}$
has to appear in this equation as otherwise the sextic would pass through the
singular point of $WPS$ and inherit its singularity.

######  Lemma 5.2.

All triple points on a sextic hypersurface $X_{6}$ in $\mathbb{P}[1:1:1:1:2]$
have to lie on a sextic surface that is a hyperplane section of $X$.

###### Proof.

For $O$ to be an (ordinary) triple point of $X_{6}$ we need all the second
derivatives of $F$ to vanish. In particular, we need
$\frac{\partial^{2}F}{\partial u^{2}}=6u+2G_{2}(x,y,z,t)=0$ and thus $O$ has
to lie on $V(3u+G_{2}(x,y,z,t))$ which is isomorphic to $\mathbb{P}^{3}$. This
$\mathbb{P}^{3}$ cuts $X_{6}$ in a sextic surface. ∎

We can perform a change of variables so the hyperplane section of $X_{6}$ is
$u=0$. Thus we can consider triple points of $X_{6}$ as lying on a sextic
surface in $\mathbb{P}^{3}$ which is the vanishing locus of $G_{6}(x,y,z,t)$.

######  Fact 5.3.

Triple points of a sextic threefold $X_{6}$ are also triple points of its
hyperplane section $S_{6}=V(G_{6})$.

From Proposition 3.2 in [4] we know that if $S_{6}$ is a normal surface it can
have at most $10$ (ordinary) triple points. Thus if we want to obtain more
than $10$ triple points on $X_{6}$ we need to assume that $S_{6}$ is not
normal. Then $S_{6}$ must have at least one-dimensional singular locus. We
first assume that $\dim(\text{Sing}S_{6})=2$. This is for example the case
when $S_{6}$ is a triple quadric.

######  Lemma 5.4.

A sextic threefold $X_{6}$ in $\mathbb{P}(1:1:1:1:2)$ whose triple points lie
on a hyperplane section with a two-dimensional singular locus admits a
singular curve.

###### Proof.

We have $X_{6}=V(u^{3}+u^{2}G_{2}+uG_{4}+G_{6})$. Note that on the curve
$C=V(u)\cap V(G_{4})\cap\text{Sing}S_{6}\subset X_{6}$ all partials
derivatives of the equation defining $X_{6}$ vanish. This finishes the proof.
∎

From now on we are only concerned with the case where $S_{6}$ is a non-normal
surface with one-dimensional singular locus.

###### Proof of 5.1.

Let the defining equaton of $X_{6}$ be
$F(x,y,z,t,u)=u^{3}+u^{2}G_{2}(x,y,z,t)+uG_{4}(x,y,z,t)+G_{6}(x,y,z,t)$. Let
$G_{6}$ be the equation of a non-normal sextic surface with at least $11$
triple points and denote by $\Sigma$ the set of those triple points. We modify
the argument regarding the polar bound to encompass the case where $S_{6}$ is
not normal. Let us consider $|S_{5}|$, the linear system of polar surfaces of
$S_{6}$ that is, the system of quintic surfaces generated by the partial
derivatives of $G_{6}$. The base locus of this system is the union of a
singular curve of $S_{6}$ and a set $\Sigma$ taken with multiplicity two, thus
the general member $X_{5}$ of $|S_{5}|$ will be a quintic containing the
singular curve of $S_{6}$ and having double points at $\Sigma$. Note that the
dimension of intersection of $S_{6}$ and a general polar surface is $1$ as
expected, and contains the singular curve of $S_{6}$. Now, $S_{6}$, $X_{5}$
and $X_{4}=V(G_{4})$ should intersect only in a finite number of points. As we
know that all these surfaces contain points of $\Sigma$ as triple or double
points, we can calculate the degree of this intersection to be at least
$11\cdot 3\cdot 2\cdot 2=132$ which is more than expected $120$ and thus there
has to be at least a common curve of intersection for all these surfaces. As
this calculation is valid for all the polar surfaces of $S_{6}$, this curve
has to be contained in the singular curve of $S_{6}$, but then, from
calculating the partials, it would be singular on $X_{6}$, contradicting the
assumption that the singular locus of $X_{6}$ has only isolated ordinary
triple points.

To prove the final statement of the theorem it is enough to use the equation
$X_{6}=V(u^{3}+G_{6})$ where $G_{6}$ is any of the sextics with $10$ OTPs as
in [4] or [12]. It is clear that $X_{6}$ cannot be singular for $u\neq 0$, and
for $u=0$ the only singularities are those on $V(G_{6})=S$. Also, $X$ misses
the singular point of the WPS as the monomial $u^{3}$ appears in its equation.
∎

######  Remark 5.5.

In [1, 2,2] Cynk calculates the defect $\delta$ and the Hodge numbers of the
resolution of singularities of an exemplary sextic threefold
$X\subset\mathbb{P}(1:1:1:1:4)$ with equation of the form $G=u^{3}+G_{6}$
where $V(G_{6})$ is one of the sextic surfaces with $10$ triple points as
described in [4]. In particular, he obtains $\delta=10$,
$h^{1,1}(\tilde{X})=21$ and $h^{1,2}(\tilde{X})=3$ where $\tilde{X}$ is the
resolution of singularities of $X$.

##  6 Calabi-Yau threefolds with no triple points

We briefly discuss the complete intersection threefold
$X_{2,2,2,2}\subset\mathbb{P}^{7}$. In a similar spirit we mention two
hypersurfaces in weighted projective space that cannot admit triple points.

######  Proposition 6.1.

A complete intersection threefold $X_{2,2,2,2}\subset\mathbb{P}^{7}$ cannot
have triple points.

###### Proof.

We cannot obtain an ordinary triple point $O$ because the tangent cone of
$X_{2,2,2,2}$ at $O$ would have to be a cone over a smooth cubic surface. This
means that there could exist a change of variables such that locally around
$O$ the equation of $X_{2,2,2,2}$ would be $F_{3}=x=y=z=0$. As all the
sixfolds are of degree $2$, this cannot happen. ∎

######  Proposition 6.2.

There can be no ordinary triple points on
$X_{8}\subset\mathbb{P}{(1:1:1:1:4)}$ or
$X_{10}\subset\mathbb{P}{(1:1:1:2:5)}$.

###### Proof.

Both $X_{8}$ and $X_{10}$ can be considered a double cover of $\mathbb{P}^{3}$
and $\mathbb{P}(1:1:1:2)$ via a certain Veronese embedding. The ambient
weighted projective space can be seen as a cone with the hyperplane section
being isomorphic to the $\mathbb{P}^{3}$ (respectively $\mathbb{P}(1:1:1:2)$)
and $X_{8}$ (resp. $X_{10}$) being the double section. Any line passing
through a triple point of a cover would have to be contained in $X_{8}$ (resp.
$X_{10}$). As this line passes through the vertex of the ambient cone, that
means that $X_{8}$ (resp. $X_{10}$) would have to pass through this point as
well, providing an additional, much more complicated singularity on $X_{8}$
(resp. $X_{10}$).

This can also be seen algebraically, because the equation of
$X_{8}\subset\mathbb{P}{(1:1:1:1:4)}$ (resp.
$X_{10}\subset\mathbb{P}{(1:1:1:2:5)}$) necessarily involves a term with the
last variable $u$ in the second power (otherwise the singular point of WPS
would be contained in the threefold). As we can choose coordinates so that the
putative triple point is $O=[1:0:0:0:0]$, we see that locally the equation of
$X_{8}$ (resp. $X_{10}$) begins with $u^{2}$, meaning $O$ cannot be a triple
point, a contradiction. ∎

### Acknowledgements

The author wishes to thank G. Kapustka, M. Kapustka and S. Cynk for helpful
discussions. The author is also grateful to the reviewers for insightful and
highly useful comments. The author is supported by the project Narodowe
Centrum Nauki 2018/30/E/ST1/00530.

## References

  * [1] S. Cynk, _Hodge numbers of hypersurfaces in $\mathbb{P}^{4}$ with ordinary triple points_, Adv. Geom., 21 (2) 2021, pp. 293–298
  * [2] I. Dolgachev, _Weighted projective varieties_ , Group actions and vector fields, Vancouver, B.C., 1981, pp. 34-71
  * [3] I. Dolgachev, _Corrado Segre and nodal cubic threefolds_ , In: G.Casnati, A.Conte, L.Gatto, L.Giacardi, M.Marchisio, A.Verra (eds), From Classical to Modern Algebraic Geometry. Trends in the History of Science. Birkhäuser, 2016, pp. 429-450
  * [4] S. Endrass, U. Persson, J. Stevens, _Surfaces with triple points_ , J. Algebraic Geom., 12, 2003, pp. 367-404
  * [5] H. Finkelnberg, J. Werner, _Small resolutions of nodal cubic threefolds_ , Indag. Math., 92 (2), June 19, 1989, pp. 185-198
  * [6] E. Fortuna, R. Frigerio, R. Pardini, _Projective Geometry. Solved Problems and Theory Review_ , Springer International, 2016
  * [7] M. Gross, S. Popescu, _Calabi-Yau threefolds and moduli of Abelian Spaces I_ , Compos. Math., 127, 2001, pp. 169-228
  * [8] G. Kapustka, M. Kapustka, _Primitive contractions of Calabi-Yau threefolds_ , Comm. Algebra, 37, Issue 2, 2009, pp. 482-502
  * [9] R. Kloosterman, S. Rams, _Quintic threefolds with triple points_ , Commun. Contemp. Math., 23, No. 01, 2021, art.. 1950085
  * [10] M. Reid, _Graded rings and varieties in weighted projective space_ , https://homepages.warwick.ac.uk, 2002
  * [11] J. Roberts, _Hypersurfaces with nonsingular normalization and their double loci_ , J. Algebra, 53, 1978, pp. 253-267
  * [12] J. Stevens, _Sextic surfaces with ten triple points_ , arXiv:math/0304060v1 [math.AG], 2003
  * [13] D. van Straten, _A quintic hypersurface in $\mathbb{P}^{4}$ with 130 nodes_, Topology, 32, No. 4, 1993, pp. 857-864,


# On flagged $K$-theoretic symmetric polynomials

Shinsuke Iwao  Faculty of Business and Commerce, Keio University, Hiyosi
4-1-1, Kohoku-ku, Yokohama, Kanagawa 223-8521, Japan, ORCID:
https://orcid.org/0000-0002-6847-7433<EMAIL_ADDRESS>
###### Abstract.

We provide a fermionic description of flagged skew Grothendieck polynomials,
which can be seen as a $K$-theoretic counterpart of flagged skew Schur
polynomials. Our proof relies on the Jacobi-Trudi type formula established by
Matsumura. This result generalizes the author’s previous works on a fermionic
description of skew Grothendieck polynomials and multi-Schur functions.

###### Key words and phrases:

Boson-Fermion correspondence, Grothendieck polynomial, $K$-theoretic symmetric
polynomial

###### 1991 Mathematics Subject Classification:

05E05, 05E14, 13M10, 14N15

##  1\. Introduction

###  1.1. Overview

Grothendieck polynomials [6] are a family of polynomials that represent the
structure sheaf of a Schubert variety in the $K$-theory of the flag variety.
Since each Schubert variety is naturally associated with a permutation,
Grothendieck polynomials are indexed by a permutation.

A flagged Grothendieck polynomial [5] is a Grothendieck polynomial that
associates with a vexillary permutation. Being a $K$-theoretic analog of
flagged Schur polynomials, flagged Grothendieck polynomials possess various
interesting combinatorial and algebraic properties. Knuston-Miler-Yong [5]
showed that flagged Grothendieck polynomials coincide with generating
functions of flagged set-valued tableaux. Hudson-Matsumura [2] provided a
Jacobi-Trudi type formula for these polynomials.

For a permutation $w\in S_{n}$, the inversion set (see [7, 10]) of $w$ is the
subset $I_{i}(w)=\\{j\,|\,i<j\mbox{ and }w(i)>w(j)\\}$ of $\\{1,2,\dots,n\\}$.
A permutation $w$ is called vexillary if the family
$\\{I_{i}(w)\\}_{i=1,2,\dots,n}$ forms a chain by inclusion. For a vexillary
permutation $w$, we associate a partition $\lambda(w)$ by arranging the
cardinalities of inversion sets in decreasing order. A flagging of $w$ is the
sequence obtained by arranging $\min I_{i}(w)-1$ in increasing order. The
flagged Grothendieck polynomial $G_{w}(x)$ is often written as
$G_{\lambda,f}(x)$, where $\lambda=\lambda(w)$ and $f$ is the flagging of $w$.

In the work [8], Matsumura generalized these polynomials for skew shapes
$\lambda/\mu$ with a flagging $f/g$, where $f=(f_{1},\dots,f_{r})$ and
$g=(g_{1},\dots,g_{r})$ are sequences of natural numbers. He proved that a
flagged skew Grothendieck polynomial, which was defined as a generating
function of flagged skew set-valued tableaux, admits a Jacobi-Trudi type
formula. For $n,p,q\in\mathord{\mathbb{Z}}$, define $G_{n}^{[p/q]}(x)$ by the
generating function

(1)
$\sum_{n\in\mathord{\mathbb{Z}}}G_{n}^{[p/q]}(x)z^{n}=\begin{cases}\frac{1}{1+\beta
u^{-1}}\prod_{k=q}^{p}\frac{1+\beta x_{k}}{1-x_{k}u}&(p\geq q)\\\
\frac{1}{1+\beta u^{-1}}&(p<q)\end{cases}.$

Then, Matsumura’s determinant formula [8, §4] is expressed as

(2)
$G_{\lambda/\mu,f/g}(x)=\det\left(\sum_{s=0}^{\infty}\binom{i-j}{s}\beta^{s}G^{[f_{i}/g_{j}]}_{\lambda_{i}-\mu_{j}-i+j+s}(x)\right).$

In this paper, we study such “flagged skew Grothendieck polynomials”. However,
we should adopt the slightly different definition

(3)
$\sum_{n\in\mathord{\mathbb{Z}}}G_{n}^{[[p/q]]}(x)z^{n}=\begin{cases}\frac{1}{1+\beta
u^{-1}}\prod_{k=q}^{p}\frac{1+\beta x_{k}}{1-x_{k}u}&(p\geq q)\\\
\frac{1}{1+\beta u^{-1}}&(p=q-1)\\\ \frac{1}{1+\beta
u^{-1}}\prod_{k=p+1}^{q-1}\frac{1-x_{k}u}{1+\beta x_{k}}&(p<q-1)\end{cases}$

and consider the polynomial

(4)
$G_{\lambda/\mu,[[f/g]]}(x)=\det\left(\sum_{s=0}^{\infty}\binom{i-j}{s}\beta^{s}G^{[[f_{i}/g_{j}]]}_{\lambda_{i}-\mu_{j}-i+j+s}(x)\right)$

because the definition (3) is more suitable for our algebraic calculations.
The polynomials $G_{\lambda/\mu,f/g}(x)$ and $G_{\lambda/\mu,[[f/g]]}(x)$ are
different in general but coincide with each other if $f_{i}+\lambda_{i}-i\geq
g_{j}+\mu_{j}-j$ whenever $f_{i}<g_{j}-1$. In particular, if
$g_{1}=g_{2}=\dots=g_{r}=1$, they coincide with each other for any skew shape
$\lambda/\mu$ and a flagging $f$.

Our aim is to construct a new algebraic description of
$G_{\lambda/\mu,[[f/g]]}(x)$ using concepts from vertex algebras, including
fermion Fock spaces, vertex operators, and vacuum expectation values. In the
previous work [3, §4], the author of the paper presented a fermionic
description of skew Grothendieck polynomials. Generalizing this approach, we
show the main theorem (Theorem 3.2) which provides a fermionic description for
the flagged Grothendieck polynomial. We emphasize the fact that our
descriptions (13), (19) are not found in the previous work [4] of multi-Schur
functions. This suggests that there is still room for generalization of
fermionic descriptions of $K$-theoretic polynomials.

### Acknowledgements

This work was partially supported by JSPS KAKENHI Grant Numbers 19K03065,
22K03239, and 23K03056.

##  2\. Preliminaries

###  2.1. Fermion Fock space

Let $\mathcal{A}$ be the $\mathord{\mathbb{C}}$-algebra generated by the free
fermions $\psi_{n},\psi_{n}^{\ast}$ $(n\in\mathord{\mathbb{Z}})$ with anti-
commutation relations

$[\psi_{m},\psi_{n}]_{+}=[\psi^{\ast}_{m},\psi^{\ast}_{n}]_{+}=0,\qquad[\psi_{m},\psi^{\ast}_{n}]_{+}=\delta_{m,n},$

where $[A,B]_{+}=AB+BA$ is the anti-commutator.

Let $\mathcal{F}=\mathcal{A}\cdot|0\rangle$ be the Fock space, the left
$\mathcal{A}$-module generated by the vacuum vector

$\psi_{m}|0\rangle=\psi^{\ast}_{n}|0\rangle=0,\quad m<0,\ n\geq 0.$

We also use the dual Fock space $\mathcal{F}^{\ast}:=\langle
0|\cdot\mathcal{A}$, the right $\mathcal{A}$-module generated by the dual
vacuum vector

$\langle 0|\psi_{n}=\langle 0|\psi^{\ast}_{m}=0,\quad m<0,\ n\geq 0.$

There uniquely exists an anti-algebra involution on $\mathcal{A}$

${}^{\ast}:\mathcal{A}\to\mathcal{A};\quad\psi_{n}\leftrightarrow\psi_{n}^{\ast},$

satisfying $(ab)^{\ast}=b^{\ast}a^{\ast}$ and $(a^{\ast})^{\ast}=a$ for
$a,b\in\mathcal{A}$, which induces the $\mathord{\mathbb{C}}$-linear
involution

$\omega:\mathcal{F}\to{\mathcal{F}}^{\ast},\quad X|0\rangle\mapsto\langle
0|{X}^{\ast}.$

The vacuum expectation value [9, §4.5] is the unique
$\mathord{\mathbb{C}}$-bilinear map

(5) ${\mathcal{F}}^{\ast}\otimes_{k}\mathcal{F}\to k,\quad\langle
w|\otimes|v\rangle\mapsto\langle{w}|v\rangle,$

satisfying $\langle 0|0\rangle=1$, $(\langle w|\psi_{n})|v\rangle=\langle
w|(\psi_{n}|v\rangle)$, and $(\langle w|\psi_{n}^{\ast})|v\rangle=\langle
w|(\psi_{n}^{\ast}|v\rangle)$. For any expression $X$, we write $\langle
w|X|v\rangle:=(\langle{w}|X)|v\rangle=\langle w|(|X|v\rangle)$. The
expectation value $\langle 0|X|0\rangle$ is often abbreviated as $\langle
X\rangle$.

######  Theorem 2.1 (Wick’s theorem (see [1, §2], [9, Exercise 4.2]) ).

Let $\\{m_{1},\dots,m_{r}\\}$ and $\\{n_{1},\dots,n_{r}\\}$ be sets of
integers. Then we have

$\langle\psi_{m_{1}}\cdots\psi_{m_{r}}\psi^{\ast}_{n_{r}}\cdots\psi^{\ast}_{n_{1}}\rangle=\det(\langle\psi_{m_{i}}\psi^{\ast}_{n_{j}}\rangle)_{1\leq
i,j\leq r}.$

For an integer $m$, we define the shifted vacuum vectors
$|m\rangle\in\mathcal{F}$ and $\langle m|\in\mathcal{F}^{\ast}$ by

$|m\rangle=\begin{cases}\psi_{m-1}\psi_{m-2}\cdots\psi_{0}|0\rangle,&m\geq
0,\\\
\psi^{\ast}_{m}\cdots\psi^{\ast}_{-2}\psi^{\ast}_{-1}|0\rangle,&m<0,\end{cases}\qquad\langle
m|=\begin{cases}\langle
0|\psi^{\ast}_{0}\psi^{\ast}_{1}\dots\psi^{\ast}_{m-1},&m\geq 0,\\\ \langle
0|\psi_{-1}\psi_{-2}\dots\psi_{m},&m<0.\end{cases}$

###  2.2. Vertex operators and commutation relations

For any monomial expression $M$ in $\psi_{n}$ and $\psi^{\ast}_{n}$, the
normal ordering

$:M:\quad\in\mathcal{A}$

is defined by moving the annihilation operators

$\psi_{m},\quad\psi^{\ast}_{n},\qquad m<0,\,n\geq 0$

to the right, and multiplying $-1$ for each move (See [1, §2], [9, §5.2]). For
example, we have $:\psi_{1}\psi^{\ast}_{1}:=\psi_{1}\psi^{\ast}_{1}$ and
$:\psi^{\ast}_{1}\psi_{1}:=-\psi_{1}\psi^{\ast}_{1}$. The normal ordering
extends to the $\mathord{\mathbb{C}}$-linear map

$\\{\mbox{polynomial expressions in $\psi_{n}$ and $\psi^{\ast}_{n}$ with
coefficients in $\mathord{\mathbb{C}}$}\\}\to\mathcal{A};\quad X\mapsto{:X:}$

Let $a_{m}$ ($m\in\mathord{\mathbb{Z}}$) be the current operator
$a_{m}=\sum_{k\in\mathord{\mathbb{Z}}}:\psi_{k}\psi^{\ast}_{k+m}:$, which
satisfies

(6)
$[a_{m},a_{n}]=m\delta_{m+n,0},\qquad[a_{m},\psi_{n}]=\psi_{n-m},\qquad[a_{m},\psi^{\ast}_{n}]=-\psi^{\ast}_{n+m},$

where $[A,B]=AB-BA$ (see [9, §5.3]). If $|v\rangle)=\langle v^{\ast}|$, we
have $\omega(a_{i}|v\rangle)=\langle v^{\ast}|a_{-i}$ for any
$n\in\mathord{\mathbb{Z}}$.

Let $X=(X_{1},X_{2},\dots)$ be a set of (commutative) variables. We define the
Hamiltonian operator

$H(X)=\sum_{n>0}\frac{p_{n}(X)}{n}a_{n},\qquad
p_{n}(X)=X_{1}^{n}+X_{2}^{n}+\cdots$

and its dual

$H^{\ast}(X)=\sum_{n>0}\frac{p_{n}(X)}{n}a_{-n}.$

We define the vertex operators by

$e^{H(X)}=\sum_{n=0}^{\infty}\frac{H(X)^{n}}{n!},\qquad
e^{H^{\ast}(X)}=\sum_{n=0}^{\infty}\frac{H^{\ast}(X)^{n}}{n!}$

Let $\psi(z)=\sum_{n\in\mathord{\mathbb{Z}}}\psi_{n}z^{n}$ and
$\psi^{\ast}(z)=\sum_{n\in\mathord{\mathbb{Z}}}\psi^{\ast}_{n}z^{n}$ be the
fermion fields. Here, we summarize useful commutation relations concerning the
vertex operators and the fermion fields. For proofs, see the textbook [9].

(7) $\displaystyle
e^{H(X)}\psi(z)e^{-H(X)}=\left(\prod_{i}\frac{1}{1-X_{i}z}\right)\psi(z),$ (8)
$\displaystyle
e^{H^{\ast}(X)}\psi(z)e^{-H^{\ast}(X)}=\left(\prod_{i}\frac{1}{1-X_{i}z^{-1}}\right)\psi(z),$
(9) $\displaystyle
e^{H(X)}e^{-H^{\ast}(Y)}=\left(\prod_{i,j}(1-X_{i}Y_{j})\right)e^{-H^{\ast}(Y)}e^{H(X)},$
(10)
$\displaystyle\langle-r|\psi^{\ast}(w)\psi(z)|-r\rangle=\frac{z^{-r}w^{-r}}{1-zw},$

where $\frac{z^{-r}w^{-r}}{1-zw}=\sum_{p=-r}^{\infty}z^{p}w^{p}$.

##  3\. Flagged Skew Grothendieck polynomial

###  3.1. $G^{[[f/g]]}_{n}(x)$

In this section, we give a fermionic presentation of flagged skew Grothendieck
polynomials. For brevity, we adopt the convention

$\prod_{k=q}^{p}X_{k}=\begin{cases}X_{q}X_{q+1}\cdots X_{p}&(p\geq q)\\\
1&(p=q-1)\\\ X^{-1}_{p+1}X^{-1}_{p+2}\cdots X^{-1}_{q-1}&(p<q-1)\end{cases}$

for a sequence $X_{1},X_{2},\dots$ of (commutative) functions.

Let $x^{[f]}=(x_{1},x_{2},\dots,x_{f})$ and

$H(x^{[f/g]})=H(x^{[f]})-H(x^{[g-1]}).$

If $f\geq g$, $H(x^{[f/g]})$ coincides with $H(x_{g},x_{g+1},\dots,x_{f})$.
From (7–9), we have

$\displaystyle e^{H(x^{[f/g]})}\psi(z)e^{-H^{\ast}(-\beta)}$
$\displaystyle=\left(\frac{1}{1+\beta z^{-1}}\prod_{j=g}^{f}\frac{1+\beta
x_{i}}{1-x_{i}z}\right)e^{-H^{\ast}(-\beta)}\psi(z)e^{H(x^{[f/g]})},$

where the rational function on the right hand side expands in the ring111Note
that the two rings $\mathord{\mathbb{C}}((z))[[\beta]]$ and
$\mathord{\mathbb{C}}[[\beta]]((z))$ are different. In fact,
$\mathord{\mathbb{C}}((z))[[\beta]]$ contains
$1+\frac{\beta}{z}+\frac{\beta^{2}}{z^{2}}+\cdots,$ while
$\mathord{\mathbb{C}}[[\beta]]((z))$ does not.

$\mathord{\mathbb{C}}[x_{1},x_{2},\dots]((z))[[\beta]].$

Comparing this equation with (3), we obtain

(11)
$e^{H(x^{[f/g]})}\psi(z)e^{-H^{\ast}(-\beta)}=\left(\sum_{n\in\mathord{\mathbb{Z}}}G_{n}^{[[f/g]]}(x)z^{n}\right)e^{-H^{\ast}(-\beta)}\psi(z)e^{H(x^{[f/g]})}.$

A similar calculation leads

(12)
$e^{H^{\ast}(-\beta)}\psi^{\ast}(w)e^{-H(x^{[f/g]})}=\left(\sum_{n\in\mathord{\mathbb{Z}}}G_{n}^{[[f/g]]}(x)w^{-n}\right)^{-1}e^{-H(x^{[f/g]})}\psi^{\ast}(w)e^{H^{\ast}(-\beta)}.$

######  Lemma 3.1.

$G_{n}^{[[f/g]]}(x)$ has the fermionic description

$G_{n}^{[[f/g]]}(x)=\langle
0|e^{H(x^{[f/g]})}\psi_{n-1}e^{-H^{\ast}(-\beta)}|-1\rangle.$

###### Proof.

Let $F_{n}=\langle
0|e^{H(x^{[f/g]})}\psi_{n-1}e^{-H^{\ast}(-\beta)}|-1\rangle$. Since $\langle
0|e^{H^{\ast}(-\beta)}=\langle 0|$ and
$e^{H(x^{[f/g]})}|-1\rangle=|-1\rangle$, we have

$\displaystyle\sum_{n\in\mathord{\mathbb{Z}}}F_{n}z^{n-1}$
$\displaystyle=\langle
0|e^{H(x^{[f/g]})}\psi(z)e^{-H^{\ast}(-\beta)}|-1\rangle$
$\displaystyle=\left(\sum_{m\in\mathord{\mathbb{Z}}}G_{m}^{[[f/g]]}(x)z^{m}\right)\langle
0|e^{-H^{\ast}(-\beta)}\psi(z)e^{H(x^{[[f/g]]})}|-1\rangle\qquad\mbox{(Eq.~{}\eqref{eq:comm_Groth})}$
$\displaystyle=\left(\sum_{m\in\mathord{\mathbb{Z}}}G_{m}^{[[f/g]]}(x)z^{m}\right)\langle
0|\psi(z)|-1\rangle.$

Since $\langle 0|\psi(z)|-1\rangle=\langle
0|\psi(z)\psi^{\ast}_{-1}|0\rangle=z^{-1}$, we conclude
$F_{n}=G_{n}^{[[f/g]]}(x)$. ∎

###  3.2. Fermionic description

We introduce a fermionic description of skew Flagged Grothendieck polynomials
in this section. For a sequence of noncommutative elements
$P_{1},P_{2},\dots$, denote

$\prod_{i:1\to N}P_{i}:=P_{1}P_{2}\cdots P_{N},\qquad\prod_{i:N\to
1}P_{i}:=P_{N}\cdots P_{2}P_{1}.$

For any $X,Y$, we use the notation $\mathrm{Ad}_{e^{X}}(Y)=e^{X}Ye^{-X}$.

Let $f=(f_{1},f_{2},\dots,f_{r})$ and $g=(g_{1},g_{2},\dots,g_{r})$ be
sequences of positive integers. Let $G_{\lambda/\mu,[[f/g]]}(x)$ be the
polynomial defined by the determinantal formula (4). The following is the main
theorem of the paper:

######  Theorem 3.2.

Let $\lambda/\mu$ be a skew partition. Then, the flagged skew Grothendieck
polynomial $G_{\lambda/\mu,[[f/g]]}(x)$ is expressed as

(13) $\langle-r|\left(\prod_{j:r\to
1}\psi^{\ast}_{\mu_{j}-j}e^{H^{\ast}(-\beta)}e^{-H(x^{[g_{j}-1/g_{j-1}]})}\right)\left(\prod_{i:1\to
r}e^{H(x^{[f_{i}/f_{i-1}+1]})}\psi_{\lambda_{i}-i}e^{-H^{\ast}(-\beta)}\right)|-r\rangle,$

where $x^{[f_{1}/f_{0}+1]}=x^{[f_{1}]}$ and
$x^{[g_{1}-1/g_{0}]}=x^{[g_{1}-1]}$.

###### Proof.

Since

$H(x^{[f_{i}/f_{i-1}+1]})=H(x^{[f_{i}]})-H(x^{[f_{i-1}]}),\quad
H(x^{[g_{j}-1/g_{j-1}]})=H(x^{[g_{j}-1]})-H(x^{[g_{j-1}-1]}),$

and $e^{H(x)}|-r\rangle=|-r\rangle$, the expectation value (13) equals to

(14) $\begin{split}\langle-r|e^{-H(x^{[g_{r}-1]})}\left(\prod_{j:r\to
1}e^{H(x^{[g_{j}-1]})}\psi^{\ast}_{\mu_{j}-j}e^{H^{\ast}(-\beta)}e^{-H(x^{[g_{j}-1]})}\right)\\\
\cdot\left(\prod_{i:1\to
r}e^{H(x^{[f_{i}]})}\psi_{\lambda_{i}-i}e^{-H^{\ast}(-\beta)}e^{-H(x^{[f_{i}]})}\right)|-r\rangle.\end{split}$

By using the equations

$\begin{gathered}\left(\prod_{j:r\to 1}P_{j}\right)e^{-rX}=\prod_{j:r\to
1}(e^{(r-j)X}P_{j}e^{-(r-j+1)X})=\prod_{j:r\to
1}\mathrm{Ad}_{e^{(r-j)X}}(P_{j}e^{-X}),\\\ e^{rX}\prod_{i:1\to
r}P_{i}=\prod_{i:1\to r}(e^{(r-i+1)X}P_{i}e^{-(r-i)X})=\prod_{i:1\to
r}\mathrm{Ad}_{e^{(r-i)X}}(e^{X}P_{i}),\end{gathered}$

(14) is rewritten as

(15) $\begin{split}\langle-r|e^{-H(x^{[g_{r}-1]})}\left(\prod_{j:r\to
1}\mathrm{Ad}_{e^{(r-j)H^{\ast}(-\beta)}}\left(e^{H(x^{[g_{j}-1]})}\psi^{\ast}_{\mu_{j}-j}e^{H^{\ast}(-\beta)}e^{-H(x^{[g_{j}-1]})}e^{-H^{\ast}(-\beta)}\right)\right)\\\
\cdot\left(\prod_{i:1\to
r}\mathrm{Ad}_{e^{(r-i)H^{\ast}(-\beta)}}\left(e^{H^{\ast}(-\beta)}e^{H(x^{[f_{i}]})}\psi_{\lambda_{i}-i}e^{-H^{\ast}(-\beta)}e^{-H(x^{[f_{i}]})}\right)\right)|-r\rangle.\end{split}$

Let

$\begin{gathered}A_{i}(z_{i}):=\mathrm{Ad}_{e^{(r-i)H^{\ast}(-\beta)}}\left(e^{H^{\ast}(-\beta)}e^{H(x^{[f_{i}]})}\psi(z_{i})e^{-H^{\ast}(-\beta)}e^{-H(x^{[f_{i}]})}\right),\\\
B_{j}(w_{j}):=\mathrm{Ad}_{e^{(r-j)H^{\ast}(-\beta)}}\left(e^{H(x^{[g_{j}-1]})}\psi^{\ast}(w_{j})e^{H^{\ast}(-\beta)}e^{-H(x^{[g_{j}-1]})}e^{-H^{\ast}(-\beta)}\right)\end{gathered}$

Then, by Wick’s theorem (Theorem 2.1), the expectation value (15) equals to
the coefficient of $z_{1}^{\lambda_{1}-1}\cdots z_{r}^{\lambda_{r}-r}\cdot
w_{1}^{\mu_{1}-1}\cdots w_{r}^{\mu_{r}-r}$ in the determinant

(16)
$\det\left(\langle-r|e^{-H(x^{[g_{r}-1]})}B_{j}(w_{j})A_{i}(z_{i})|-r\rangle\right)_{i,j}.$

From (11), $A_{i}(z_{i})$ satisfies

$\displaystyle A_{i}(z_{i})$
$\displaystyle=\left(\sum_{n\in\mathord{\mathbb{Z}}}G_{n}^{[f_{i}]}(x)z_{i}^{n}\right)\cdot\mathrm{Ad}_{e^{(r-i)H^{\ast}(-\beta)}}(\psi(z_{i}))$
$\displaystyle=\left(\sum_{n\in\mathord{\mathbb{Z}}}G_{n}^{[f_{i}]}(x)z_{i}^{n}\right)\cdot(1+\beta
z_{i}^{-1})^{-(r-i)}\cdot\psi(z_{i}).$

On the other hand, since

$\displaystyle
e^{H(x^{[g_{j}-1]})}\psi^{\ast}(w_{j})e^{H^{\ast}(-\beta)}e^{-H(x^{[g_{j}-1]})}e^{-H^{\ast}(-\beta)}$
$\displaystyle=(1+\beta w_{j})^{-1}\cdot
e^{H(x^{[g_{j}-1]})}e^{H^{\ast}(-\beta)}\psi^{\ast}(w_{j})e^{-H(x^{[g_{j}-1]})}e^{-H^{\ast}(-\beta)}$
$\displaystyle=\left(\sum_{n\in\mathord{\mathbb{Z}}}G_{n}^{[g_{j}-1]}(x)w_{j}^{-n}\right)^{-1}(1+\beta
w_{j})^{-1}\cdot\psi^{\ast}(w_{j})\qquad\mbox{(Eq.~{}\eqref{eq:comm_dual_Groth})},$

we obtain

$\displaystyle B_{j}(w_{j})$
$\displaystyle=\left(\sum_{n\in\mathord{\mathbb{Z}}}G_{n}^{[g_{j}-1]}(x)w_{j}^{-n}\right)^{-1}\cdot(1+\beta
w_{j})^{-1}\mathrm{Ad}_{e^{(r-j)H^{\ast}(-\beta)}}(\psi^{\ast}(w_{j}))$
$\displaystyle=\left(\sum_{n\in\mathord{\mathbb{Z}}}G_{n}^{[g_{j}-1]}(x)w_{j}^{-n}\right)^{-1}\cdot(1+\beta
w_{j})^{r-j-1}\cdot\psi^{\ast}(w_{j}).$

Therefore, we have

$\displaystyle\langle-r|e^{-H(x^{[g_{r}-1]})}B_{j}(w_{j})A_{i}(z_{i})|-r\rangle$
$\displaystyle=\frac{\sum_{n}G_{n}^{[f_{i}]}(x)z_{i}^{n}}{\sum_{n}G_{n}^{[g_{j}-1]}(x)w_{j}^{-n}}\frac{(1+\beta
w_{j})^{r-j-1}}{(1+\beta
z_{i}^{-1})^{r-i}}\langle-r|e^{-H(x^{[g_{r}-1]})}\psi^{\ast}(w_{j})\psi(z_{i})|-r\rangle$
$\displaystyle=\frac{\sum_{n}G_{n}^{[f_{i}]}(x)z_{i}^{n}}{\sum_{n}G_{n}^{[g_{j}-1]}(x)w_{j}^{-n}}\frac{(1+\beta
w_{j})^{r-j-1}}{(1+\beta
z_{i}^{-1})^{r-i}}\frac{\prod_{k=1}^{g_{r}-1}(1-x_{k}z_{i})}{\prod_{k=1}^{g_{r}-1}(1-x_{k}w_{j}^{-1})}\langle-r|\psi^{\ast}(w_{j})\psi(z_{i})|-r\rangle$
$\displaystyle=\frac{\sum_{n}G_{n}^{[f_{i}]}(x)z_{i}^{n}}{\sum_{n}G_{n}^{[g_{j}-1]}(x)w_{j}^{-n}}\frac{(1+\beta
w_{j})^{r-j-1}}{(1+\beta
z_{i}^{-1})^{r-i}}\frac{\prod_{k=1}^{g_{r}-1}(1-x_{k}z_{i})}{\prod_{k=1}^{g_{r}-1}(1-x_{k}w_{j}^{-1})}\frac{z_{i}^{-r}w_{j}^{-r}}{1-z_{i}w_{j}}\qquad\mbox{(Eq.~{}\eqref{eq:comm4})}$
$\displaystyle=\frac{\prod_{k=1}^{f_{i}}(1+\beta
x_{k})}{\prod_{k=1}^{g_{j}-1}(1+\beta x_{k})}\frac{(1+\beta
w_{j})^{r-j}}{(1+\beta
z_{i}^{-1})^{r-i+1}}\frac{\prod_{k=f_{i}+1}^{g_{r}-1}(1-x_{k}z_{i})}{\prod_{k=g_{j}}^{g_{r}-1}(1-x_{k}w_{j}^{-1})}\frac{z_{i}^{-r}w_{j}^{-r}}{1-z_{i}w_{j}}$
$\displaystyle=:F(z_{i},w_{j}).$

To take the coefficient of $z^{\lambda_{i}-i}w^{\mu_{j}-j}$ in $F(z,w)$, we
use the complex line integral. Note that the expansion of the rational
function $F(z,w)$ in the field

$\mathord{\mathbb{C}}[x_{1},x_{2},\dots]((w^{-1}))((z))[[\beta]]$

coincides with the Laurent expansion on the domain
$\\{|\beta|<|z|<|w^{-1}|<|x_{k}^{-1}|\,;\,\forall k\\}$. Then, we have

$\displaystyle[w^{\mu_{j}-j}]\langle-r|e^{-H(x^{[g_{r}-1]})}B_{j}(w)A_{i}(z)|-r\rangle$
$\displaystyle=\frac{1}{2\pi\sqrt{-1}}\oint_{|\beta|<|z|<|w^{-1}|<|x_{k}^{-1}|}F(z,w)\cdot(w^{-1})^{\mu_{j}-j}\frac{d(w^{-1})}{(w^{-1})}$
(17)
$\displaystyle=\frac{1}{2\pi\sqrt{-1}}\oint_{|\beta|<|z|<|t|<|x_{k}^{-1}|}F(z,t^{-1})\cdot
t^{\mu_{j}-j-1}dt.$

Since $F(z,t^{-1})\cdot t^{\mu_{j}-j-1}dt$ is expressed as

$\frac{\prod_{k=1}^{f_{i}}(1+\beta x_{k})}{\prod_{k=1}^{g_{j}-1}(1+\beta
x_{k})}\frac{(t+\beta)^{r-j}}{(1+\beta
z^{-1})^{r-i+1}}\frac{\prod_{k=f_{i}+1}^{g_{r}-1}(1-x_{k}z)}{\prod_{k=g_{j}}^{g_{r}-1}(1-x_{k}t)}\frac{z_{i}^{-r}}{t-z}\cdot
t^{\mu_{j}}dt,$

the contour integral (17) coincides with the residue of the differential form
at $t=z$. Finally, it equals to

(18) $\displaystyle\frac{1}{1+\beta
z^{-1}}\prod_{k=g_{j}}^{f_{i}}\frac{1+\beta
x_{k}}{1-x_{k}z}\cdot(z+\beta)^{i-j}z^{\mu_{j}-i}=\left(\sum_{n\in\mathord{\mathbb{Z}}}G_{n}^{[[f_{i}/g_{j}]]}(x)z^{n}\right)(1+\beta
z^{-1})^{i-j}z^{\mu_{j}-j}.$

Since the coefficient of $z^{\lambda_{i}-i}$ in (18) is
$\sum_{s=0}^{\infty}\binom{i-j}{s}\beta^{s}G^{[[f_{i}/g_{j}]]}_{\lambda_{i}-\mu_{j}-i+j+s}(x)$,
the coefficient of $z_{1}^{\lambda_{1}-1}\cdots z_{r}^{\lambda_{r}-r}\cdot
w_{1}^{\mu_{1}-1}\cdots w_{r}^{\mu_{r}-r}$ in the determinant (16) is

$\det\left(\sum_{s=0}^{\infty}\binom{i-j}{s}\beta^{s}G^{[[f_{i}/g_{j}]]}_{\lambda_{i}-\mu_{j}-i+j+s}(x)\right)_{i,j}.$

Comparing it with (4), we conclude the theorem. ∎

###  3.3. Remarks

If $g_{1}=g_{2}=\dots=g_{r}=1$, $G_{\lambda/\mu,[[f/g]]}(x)$ reduces to the
(usual) flagged Grothendieck polynomial $G_{\lambda/\mu,f}(x)$. In this case,
our main Theorem 3.2 reduces to

(19)
$\begin{split}G_{\lambda/\mu,f}(x)=&\langle-r|\psi^{\ast}_{\mu_{r}-r}e^{H^{\ast}(-\beta)}\dots\psi^{\ast}_{\mu_{2}-2}e^{H^{\ast}(-\beta)}\psi^{\ast}_{\mu_{1}-1}e^{H^{\ast}(-\beta)}\\\
&\cdot(e^{H(x^{[f_{1}]})}\psi_{\lambda_{1}-1}e^{-H^{\ast}(-\beta)})\cdots(e^{H(x^{[f_{r}/f_{r-1}]})}\psi_{\lambda_{r}-r}e^{-H^{\ast}(-\beta)})|-r\rangle.\end{split}$

By taking $f_{1}=f_{2}=\dots=f_{r}=n$, we recover the fermionic presentation
of the skew Grothendieck polynomial given in [3, §4.2]. This expression is not
found in the previous work [4] of fermionic presentations for multi-Schur
functions.

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