# The second cohomology of regular semisimple Hessenberg varieties from GKM
theory

Anton Ayzenberg  Faculty of computer science, National Research University
Higher School of Economics, Moscow, Russian Federation<EMAIL_ADDRESS>,
Mikiya Masuda  Osaka City University Advanced Mathematical Institute,
Sumiyoshi-ku, Osaka 558-8585, Japan<EMAIL_ADDRESS>and  Takashi
Sato  Research Institute for Mathematical Sciences, Kyoto University, Kyoto
606-8502, Japan, and Osaka City University Advanced Mathematical Institute,
Sumiyoshi-ku, Osaka 558-8585, Japan<EMAIL_ADDRESS>
###### Abstract.

We describe the second cohomology of a regular semisimple Hessenberg variety
by generators and relations explicitly in terms of GKM theory. The cohomology
of a regular semisimple Hessenberg variety becomes a module of a symmetric
group $\mathfrak{S}_{n}$ by the dot action introduced by Tymoczko. As an
application of our explicit description, we give a formula describing the
isomorphism class of the second cohomology as an $\mathfrak{S}_{n}$-module.
Our formula is not exactly the same as the known formula by Chow or Cho-Hong-
Lee but they are equivalent. We also discuss its higher degree generalization.

###### Key words and phrases:

Hessenberg variety, torus action, GKM theory, equivariant cohomology,
permutation module

###### 2020 Mathematics Subject Classification:

Primary: 57S12, Secondary: 14M15

##  1\. Introduction

Let $\operatorname{Fl}(n)$ denote the variety of all complete flags in
$\mathbb{C}^{n}$. A regular semisimple Hessenberg variety
$\operatorname{Hess}(S,h)$ is a smooth subvariety of $\operatorname{Fl}(n)$.
It is determined by a square matrix $S$ of size $n$ with distinct eigenvalues
and a function $h$ (called a Hessenberg function) from the set of integers
$[n]=\\{1,\dots,n\\}$ to itself satisfying the condition:

$h(1)\leq h(2)\leq\cdots\leq h(n)\qquad\text{and}\qquad h(j)\geq
j\quad(\forall j\in[n]).$

The topology of $\operatorname{Hess}(S,h)$ depends only on $h$, i.e. does not
depend on the choice of $S$. The maximal $\mathbb{C}^{*}$-torus $T$ in the
general linear group ${\rm GL}_{n}(\mathbb{C})$, which commutes with $S$,
naturally acts on $\operatorname{Hess}(S,h)$. Using this $T$-action, one can
study the cohomology $H^{*}(\operatorname{Hess}(S,h))$. Specifically, Tymoczko
([17]) constructed an $\mathfrak{S}_{n}$-action (called the dot action) on
$H^{*}(\operatorname{Hess}(S,h))$ making it an $\mathfrak{S}_{n}$-module.

A theorem of Brosnan-Chow [5] (the solution of Shareshian-Wachs conjecture
[14]) says that the graded $\mathfrak{S}_{n}$-module
$H^{*}(\operatorname{Hess}(S,h))$ is equivalent to the (graded) chromatic
symmetric function $X_{G_{h}}(\mathbf{x},t)$ of a graph $G_{h}$ associated to
$h$, where $X_{G_{h}}(\mathbf{x},t)$ is a polynomial in $t$ with symmetric
functions in infinitely many variables $\mathbf{x}=(x_{1},x_{2},\cdots)$ as
coefficients. Moreover, it is shown in [9] that the Stanley-Stembridge
conjecture on the $e$-positivity of $(3+1)$-incomparability graphs is reduced
to showing the $e$-positivity of the graph $G_{h}$ for any Hessenberg function
$h$. Thus, we are led to the study of $H^{*}(\operatorname{Hess}(S,h))$.

In this paper, we investigate the second cohomology
$H^{2}(\operatorname{Hess}(S,h))$ using GKM theory [13] when
$\operatorname{Hess}(S,h)$ is connected. We exhibit its generators explicitly
in terms of GKM theory and give a formula describing the isomorphism class of
the $\mathfrak{S}_{n}$-module in terms of $h$. Prior to our work, Chow [7]
gave a formula for the coefficient of $t$ in $X_{G_{h}}(\mathbf{x},t)$ using
$P$-tableaux. Through the theorem by Brosnan-Chow mentioned above, Chow’s
formula is equivalent to ours. After Chow’s work, Cho-Hong-Lee [6] exhibited
generators of $H^{2}(\operatorname{Hess}(S,h))$ geometrically using the
Bialynicki-Birula decomposition of $\operatorname{Hess}(S,h)$ and gave a
formula describing the isomorphism class of the $\mathfrak{S}_{n}$-module
$H^{2}(\operatorname{Hess}(S,h))$. Their formula is also equivalent to ours.
However, the methods are different and the relation between their generators
of $H^{2}(\operatorname{Hess}(S,h))$ and ours is unclear.

Since we obtain explicit generators of $H^{2}(\operatorname{Hess}(S,h))$ in
this paper, we are able to characterize Hessenberg functions $h$ for which the
whole cohomology ring $H^{*}(\operatorname{Hess}(S,h))$ is generated in degree
two as a graded ring. It turns out that the graph $G_{h}$ associated to such
$h$ is what is called a (double) lollipop ([8], [10]). We will discuss this
subject in a forthcoming paper [4].

As mentioned above, we describe $H^{2}(\operatorname{Hess}(S,h))$ in terms of
explicit generators and relations when $\operatorname{Hess}(S,h)$ is
connected. It is known that $\operatorname{Hess}(S,h)$ is connected, in other
words, the restriction map $\iota^{*}\colon H^{0}(\operatorname{Fl}(n))\to
H^{0}(\operatorname{Hess}(S,h))$ is an isomorphism if and only if $h(j)\geq
j+1$ for any $j\in[n-1]$. As a generalization of this setting, we consider the
case where $h(j)\geq j+d$ for any $j\in[n-d]$, where $d$ is an integer $\geq
2$. In this case, we show that the retriction map $\iota^{*}\colon
H^{2p}(\operatorname{Fl}(n))\to H^{2p}(\operatorname{Hess}(S,h))$ is an
isomorphism for $p<d$ and describe $H^{2d}(\operatorname{Hess}(S,h))$ in terms
of explicit generators and relations.

The organization of the paper is as follows. In Section 2, we review necessary
facts on regular semisimple Hessenberg varieties and labeled graphs associated
to them. Section 3 discusses an inductive formula to compute the Poincaré
polynomial of $\operatorname{Hess}(S,h)$ and a formula on the second Betti
number of $\operatorname{Hess}(S,h)$ in terms of $h$. In Section 4, we provide
three types of elements in the $T$-equivariant cohomology
$H_{T}^{*}(\operatorname{Hess}(S,h))$ using GKM theory, discuss relations
among them, and observe the dot action of $\mathfrak{S}_{n}$ on them. We prove
that these three types of elements together with $H^{2}(BT)$ generate
$H^{2}_{T}(\operatorname{Hess}(S,h))$. In Section 5, we explicitly describe
$H^{2}(\operatorname{Hess}(S,h))$ in terms of generators and relations, and
give a formula describing the isomorphism class of the
$\mathfrak{S}_{n}$-module $H^{2}(\operatorname{Hess}(S,h))$ in terms of $h$.
In Section 6 we discuss the generalization of the result on the second
cohomology to the higher degree cohomology mentioned above.

##  2\. Regular semisimple Hessenberg varieties

###  2.1. Hessenberg variety

The flag variety $\operatorname{Fl}(n)$ is defined as the set of nested linear
subspaces of $\mathbb{C}^{n}$:

$\operatorname{Fl}(n)=\\{V_{\bullet}=(V_{1}\subset V_{2}\subset\cdots\subset
V_{n}=\mathbb{C}^{n})\mid\dim_{\mathbb{C}}V_{i}=i\quad\forall i\in[n]\\}.$

Given a square matrix $A$ of size $n$ and a function $h\colon[n]\to[n]$
(called a Hessenberg function) satisfying

$h(1)\leq h(2)\leq\cdots\leq h(n)\qquad\text{and}\qquad h(j)\geq
j\quad(\forall j\in[n]),$

the Hessenberg variety $\operatorname{Hess}(A,h)$ is defined by

$\operatorname{Hess}(A,h):=\\{V_{\bullet}\in\operatorname{Fl}(n)\mid
A(V_{j})\subset V_{h(j)}\quad\text{for $\forall j\in[n]$}\\},$

where the matrix $A$ is regarded as a linear transformation on
$\mathbb{C}^{n}$. We often express the Hessenberg function $h$ as a vector
$(h(1),\dots,h(n))$ by listing the values of $h$. When $h=(n,\dots,n)$, it is
obvious from the definition that $\operatorname{Hess}(A,h)$ is the flag
variety $\operatorname{Fl}(n)$ regardless of the choice of $A$.

As illustrated in the following example, we can visualize a Hessenberg
function $h$ by drawing a configuration of the shaded boxes on a square grid
of size $n\times n$, which consists of boxes in the $i$-th row and the $j$-th
column satisfying $i\leq h(j)$. Since it is assumed that $j\leq h(j)$ for any
$j\in[n]$, the essential part is the shaded boxes below the diagonal.

######  Example 2.1.

Let $n=5$. The Hessenberg function $h=(3,3,4,5,5)$ corresponds to the
configuration of the shaded boxes drawn on the left grid and the essential
shaded boxes (i.e. below the diagonal) are drawn on the right grid in Figure
1.

Figure 1. The configuration corresponding to $h=(3,3,4,5,5)$.

The Hessenberg variety $\operatorname{Hess}(S,h)$ for a square matrix $S$ of
size $n$ with distinct eigenvalues is called regular semisimple.

######  Theorem 2.1 ([11]).

  1. (1)

$\operatorname{Hess}(S,h)$ is smooth.

  2. (2)

$\dim_{\mathbb{C}}\operatorname{Hess}(S,h)=\sum_{j=1}^{n}(h(j)-j)$.

  3. (3)

$\operatorname{Hess}(S,h)$ is connected if and only if $h(j)\geq j+1$ for
$\forall j\in[n-1]$.

  4. (4)

$H^{odd}(\operatorname{Hess}(S,h))=0$ and the $2k$-th Betti number of
$\operatorname{Hess}(S,h)$ is given by

$\\#\\{w\in\mathfrak{S}_{n}\mid\ell_{h}(w)=k\\}$

where

(2.1) $\ell_{h}(w)=\\#\\{1\leq j<i\leq n\mid w(j)>w(i),\ i\leq h(j)\\}.$

Since $S$ commutes with a maximal torus $T$ of ${\rm GL}_{n}(\mathbb{C})$, the
restricted action of $T$ on $\operatorname{Fl}(n)$ leaves
$\operatorname{Hess}(S,h)$ invariant. One sees that

(2.2)
$\operatorname{Hess}(S,h)^{T}=\operatorname{Fl}(n)^{T}=\mathfrak{S}_{n}.$

###  2.2. Equivariant cohomology

We shall briefly review equivariant cohomology. For a $T$-space $X$, the
equivariant cohomology $H^{*}_{T}(X)$ is defined as

$H^{*}_{T}(X):=H^{*}(ET\times_{T}X)$

where $ET\to BT$ is the universal principal $T$-bundle and $ET\times_{T}X$ is
the orbit space of $ET\times X$ by the diagonal $T$-action. Since $T$ is
isomorphic to $(\mathbb{C}^{*})^{n}$, $BT$ is homeomorphic to
$(\mathbb{C}P^{\infty})^{n}$ and hence $H^{*}(BT)$ is a polynomial ring in $n$
elements of $H^{2}(BT)$ which form a generator of $H^{2}(BT)$. Note that for a
one-point space $pt$, we have

$H^{*}_{T}(pt)=H^{*}(BT).$

Since the $T$-action on $ET$ is free, the projection $ET\times X\to ET$ on the
first factor induces a fibration

$X\xrightarrow{\iota}ET\times_{T}X\xrightarrow{\pi}ET/T=BT.$

The equivariant cohomology $H^{*}_{T}(X)$ is not only a ring but also an
algebra over $H^{*}(BT)$ through $\pi^{*}\colon H^{*}(BT)\to H^{*}_{T}(X)$. As
easily seen, the restriction map $\iota^{*}\colon H^{*}_{T}(X)\to H^{*}(X)$
sends $H^{2}(BT)$ to zero. Therefore, it induces a ring homomorphism

(2.3) $H^{*}_{T}(X)/(H^{2}(BT))\to H^{*}(X),$

where $(H^{2}(BT))$ denotes the ideal generated by $\pi^{*}(H^{2}(BT))$. If
$H^{odd}(X)=0$ (this is the case when $X=\operatorname{Hess}(S,h)$), then the
map (2.3) above is an isomorphism.

###  2.3. GKM theory and labeled graph

We choose and fix a set of generators of $H^{2}(BT)$ and denote them by
$t_{1},\dots,t_{n}$, they correspond to the choice of coordinates in the
ambient space $\mathbb{C}^{n}$ of flags. Since
$H^{odd}(\operatorname{Hess}(S,h))=0$, the torus action is cohomologically
equivariantly formal [13]. Therefore the restriction map to the $T$-fixed
point set

$\displaystyle H^{*}_{T}(\operatorname{Hess}(S,h))\to
H^{*}_{T}(\operatorname{Hess}(S,h)^{T})$
$\displaystyle=\bigoplus_{w\in\mathfrak{S}_{n}}H^{*}_{T}(w)=\bigoplus_{w\in\mathfrak{S}_{n}}\mathbb{Z}[t_{1},\dots,t_{n}]$
$\displaystyle={\rm Map}(\mathfrak{S}_{n},\mathbb{Z}[t_{1},\dots,t_{n}])$

is injective, where ${\rm Map}(P,Q)$ denotes the set of all maps from $P$ to
$Q$. Since the restriction map above is injective, we think of
$H^{*}_{T}(\operatorname{Hess}(S,h))$ as a subset of ${\rm
Map}(\mathfrak{S}_{n},\mathbb{Z}[t_{1},\dots,t_{n}])$.

######  Proposition 2.2 ([17]).

An element $f\in{\rm Map}(\mathfrak{S}_{n},\mathbb{Z}[t_{1},\dots,t_{n}])$ is
in $H^{*}_{T}(\operatorname{Hess}(S,h))$ if and only if

$f(v)\equiv f(w)\pmod{t_{w(i)}-t_{w(j)}}\quad\text{whenever $v=w\cdot(i,j)$
for $j<i\leq h(j)$}$

where $(i,j)$ denotes the transposition interchanging $i$ and $j$.

To a Hessenberg function $h$, one associates a graph $(V,E)$ with a label on
the edge set $E$

$\alpha\colon E\to H^{2}(BT)\backslash\\{0\\}$

where

  1. (1)

$V=\mathfrak{S}_{n}$,

  2. (2)

$E=\\{\\{v,w\\}\mid v,w\in V,\ v=w\cdot(i,j)\text{ for some }j<i\leq h(j)\\}$,

  3. (3)

$\alpha(\\{v,w\\})=t_{w(i)}-t_{w(j)}$ up to sign for $v=w\cdot(i,j)$.

We denote the triple $(V,E,\alpha)$ by $\Gamma(h)$ and call a labeled graph
associated to $h$. This is a slight variant of the notion of GKM graph. The
set of elements in ${\rm Map}(\mathfrak{S}_{n},\mathbb{Z}[t_{1},\dots,t_{n}])$
satisfying the congruence relation in Proposition 2.2 is sometimes called the
graph cohomology of $\Gamma(h)$. Proposition 2.2 says that
$H^{*}_{T}(\operatorname{Hess}(S,h))$ agrees with the graph cohomology of
$\Gamma(h)$. In particular, $H^{*}_{T}(\operatorname{Hess}(S,h))$ is
independent of the choice of the regular semisimple matrix $S$. Notice, that
even the equivariant diffeomorphism type of $\operatorname{Hess}(S,h)$ is
independent of $S$, see [2].

We often think of $t_{i}$ as an element of ${\rm
Map}(\mathfrak{S}_{n},\mathbb{Z}[t_{1},\dots,t_{n}])$ by regarding it as a
constant map. Obviously, $t_{i}$ satisfies the congruence relation in
Proposition 2.2, so it is in $H^{2}_{T}(\operatorname{Hess}(S,h))$. Then

$H^{*}(\operatorname{Hess}(S,h))=H^{*}_{T}(\operatorname{Hess}(S,h))/(t_{1},\dots,t_{n})$

because $H^{odd}(\operatorname{Hess}(S,h))=0$, where $(t_{1},\dots,t_{n})$ is
the ideal generated by $t_{1},\dots,t_{n}$.

######  Example 2.2.

Let $n=3$. For $h=(2,3,3)$ and $h^{\prime}=(3,3,3)$, the corresponding labeled
graphs $\Gamma(h)$ and $\Gamma(h^{\prime})$ are depicted in Figure 2 where we
use the one-line notation for each vertex.

$123$$321$$132$$312$$213$$231$$\Gamma(h)$$123$$321$$132$$312$$213$$231$$\Gamma(h^{\prime})$labels=
$t_{1}-t_{2}$= $t_{2}-t_{3}$= $t_{1}-t_{3}$ Figure 2. The labeled graphs
$\Gamma(h)$ and $\Gamma(h^{\prime})$

Both graphs $\Gamma(h)$ and $\Gamma(h^{\prime})$ in Example 2.2 are connected.
In general, it is not difficult to see that $\Gamma(h)$ is connected if and
only if $h(j)\geq j+1$ for any $j\in[n-1]$.

##  3\. The second Betti number of $\operatorname{Hess}(S,h)$

In this section, we give an inductive formula to compute the Poincaré
polynomial of $\operatorname{Hess}(S,h)$ using Theorem 2.1(4) and apply it to
obtain an explicit formula of the second Betti number of
$\operatorname{Hess}(S,h)$ in terms of the Hessenberg function $h$.

###  3.1. An inductive formula

For a space $X$ such that $H^{odd}(X)=0$ and the rank of $H^{*}(X)$ over
$\mathbb{Z}$ is finite, we define

$\operatorname{Poin}(X,\sqrt{q}):=\sum_{r=0}^{\infty}b_{2r}(X)q^{r}$

where $b_{2r}(X)$ denotes the $2r$-th Betti number over $\mathbb{Z}$. Let
$h^{j}$ be the Hessenberg function obtained by removing all the boxes in the
$j$-th row and all the boxes in the $j$-th column (See Figure 3). To be
precise,

$h^{j}(i)=\begin{cases}h(i)&(i<j,\ h(i)<j)\\\ h(i)-1&(i<j,\ h(i)\geq j)\\\
h(i+1)-1&(i\geq j)\end{cases}$

$j$-th row $\rightarrow$$\downarrow$$j$-th column$h$
$\leadsto$remove$\leftarrow$$\nwarrow$$\uparrow$ $\leadsto$$h^{j}$ Figure 3.
The configuration corresponding to $h^{j}$.

######  Proposition 3.1.

Under the above understanding, we have

$\operatorname{Poin}(\operatorname{Hess}(S,h),\sqrt{q})=\sum_{j=1}^{n}q^{h(j)-j}\operatorname{Poin}(\operatorname{Hess}(S^{\prime},h^{j}),\sqrt{q})$

where $S^{\prime}$ denotes a matrix of size $n-1$ with distinct eigenvalues.

###### Proof.

It follows from Theorem 2.1(4) that

(3.1)
$\operatorname{Poin}(\operatorname{Hess}(S,h),\sqrt{q})=\sum_{w\in\mathfrak{S}_{n}}q^{\ell_{h}(w)}$

where

$\ell_{h}(w)=\\#\\{1\leq j<i\leq n\mid w(j)>w(i),\ i\leq h(j)\\}.$

For $j\in[n]$ we set

$\mathfrak{S}_{n}^{j}:=\\{w\in\mathfrak{S}_{n}\mid w(j)=n\\}$

and consider the following decomposition of $\mathfrak{S}_{n}$:

$\mathfrak{S}_{n}=\mathfrak{S}_{n}^{1}\sqcup\mathfrak{S}_{n}^{2}\sqcup\cdots\sqcup\mathfrak{S}_{n}^{n}.$

If $w$ is in $\mathfrak{S}_{n}^{j}$, then $w(j)=n$ so that $w(j)>w(i)$ for any
$j<i\leq h(j)$. Therefore, if we denote by $w^{j}$ the permutation on $[n-1]$
obtained by removing $w(j)=n$ from $w$, then we have

$\ell_{h}(w)=h(j)-j+\ell_{h^{j}}(w^{j}).$

This together with (3.1) implies the formula in the proposition. $\Box$

For a Hessenberg function $h$, we consider the following two sets:

(3.2) $\begin{split}\bot(h):&=\\{j\in[n-1]\mid h(j-1)=h(j)=j+1\\}\\\
\textsf{L}(h):&=\\{j\in[n-1]\mid h(j-1)=j\text{ and }h(j)=j+1\\}\end{split}$

where we understand $h(0)=1$.

######  Lemma 3.2.

Suppose that $h(j)\geq j+1$ for $j\in[n-1]$. Then the second Betti number
$b_{2}(\operatorname{Hess}(S,h))$ of $\operatorname{Hess}(S,h)$ can be
expressed as

$b_{2}(\operatorname{Hess}(S,h))=\sum_{j\in\textsf{L}(h)}\binom{n}{j}+(n-1)|\bot(h)|-|\textsf{L}(h)|.$

######  Example 3.1.

For the Hessenberg function $h=(3,3,4,5,5)$ in Example 2.1, we have

$\bot(h)=\\{2\\}\quad\text{and}\quad\textsf{L}(h)=\\{3,4\\}.$

Therefore

$b_{2}(\operatorname{Hess}(S,h))=\binom{5}{3}+\binom{5}{4}+(5-1)\cdot 1-2=17.$

###### Proof of Lemma 3.2.

We prove the lemma by induction on $n$. When $n=2$, $h=(2,2)$ since $h(j)\geq
j+1$ for $j\in[n-1]$ by assumption. Therefore, $\operatorname{Hess}(S,h)$ is
$\mathbb{C}P^{1}$ while $\bot(h)=\emptyset$ and $\textsf{L}(h)=\\{1\\}$. This
shows that the lemma holds when $n=2$.

Suppose that $n\geq 3$. It follows from Proposition 3.1 that

(3.3)
$b_{2}(\operatorname{Hess}(S,h))=\sum_{h(j)=j+1}b_{0}(\operatorname{Hess}(S^{\prime},h^{j}))+b_{2}(\operatorname{Hess}(S^{\prime},h^{n})).$

Here, $\operatorname{Hess}(S^{\prime},h^{j})$ is connected if $h(j-1)=j+1$ and
has $\binom{n-1}{j-1}$ many connected components if $h(j-1)=j$ (the detailed
description of connected components can be found in [16]). This means that

(3.4)
$b_{0}(\operatorname{Hess}(S^{\prime},h^{j}))=\begin{cases}1\quad&(\text{if
}j\in\bot(h))\\\ \binom{n-1}{j-1}\quad&(\text{if }j\in\textsf{L}(h))\\\
\end{cases}$

since $h(j)=j+1$ for $j$ in the sum of (3.3).

On the other hand, by the induction assumption, we have

(3.5)
$b_{2}(\operatorname{Hess}(S^{\prime},h^{n}))=\sum_{j\in\textsf{L}(h^{n})}\binom{n-1}{j}+(n-2)|\bot(h^{n})|-|\textsf{L}(h^{n})|.$

Here, by looking at the $(n-2,n-1)$ and $(n-1,n)$ boxes in the configuration
associated to $h$, one sees that

(3.6) $\bot(h^{n})=\begin{cases}\bot(h)\ &(\text{if }h(n-2)=n-1)\\\
\bot(h)\backslash\\{n-1\\}\ &(\text{if }h(n-2)=n,\ h(n-3)=n-2)\\\
\left(\bot(h)\backslash\\{n-1\\}\right)\cup\\{n-2\\}\ &(\text{if }h(n-2)=n,\
h(n-3)\geq n-1)\end{cases}$ (3.7)
$\textsf{L}(h^{n})=\begin{cases}\textsf{L}(h)\backslash\\{n-1\\}\quad&(\text{if
}h(n-2)=n-1)\\\ \textsf{L}(h)\cup\\{n-2\\}\quad&(\text{if }h(n-2)=n,\
h(n-3)=n-2)\\\ \textsf{L}(h)\quad&(\text{if }h(n-2)=n,\ h(n-3)\geq
n-1).\end{cases}$

We consider three cases according to the cases above and plug (3.6) and (3.7)
in the right hand side of (3.5) in each case. Then, this together with (3.4)
will show that the right hand side of (3.3) agrees with the right hand side of
the identity in the lemma. For instance, when $h(n-2)=n-1$, it follows from
(3.4), (3.5), (3.6), and (3.7) that the right hand side of (3.3) turns into

$\displaystyle\sum_{j\in\textsf{L}(h)}\binom{n-1}{j-1}+|\bot(h)|$
$\displaystyle+\sum_{j\in\textsf{L}(h)\backslash\\{n-1\\}}\binom{n-1}{j}+(n-2)|\bot(h)|-|\textsf{L}(h)\backslash\\{n-1\\}|$
$\displaystyle=$
$\displaystyle\sum_{j\in\textsf{L}(h)}\left(\binom{n-1}{j-1}+\binom{n-1}{j}\right)-\binom{n-1}{n-1}+(n-1)|\bot(h)|-|\textsf{L}(h)|+1$
$\displaystyle=$
$\displaystyle\sum_{j\in\textsf{L}(h)}\binom{n}{j}+(n-1)|\bot(h)|-|\textsf{L}(h)|,$

which coincides with the right hand side of the identity in the lemma. The
other two cases can be proved similarly, so we omit the proof. $\Box$

##  4\. Generators of $H^{2}(\operatorname{Hess}(S,h))$

###  4.1. Three types of elements in $H^{2}_{T}(\operatorname{Hess}(S,h))$

The following three types of elements play a role in our argument.

######  Lemma 4.1.

The elements $x_{i},y_{j,k},\tau_{A}$ in ${\rm
Map}(\mathfrak{S}_{n},\mathbb{Z}[t_{1},\dots,t_{n}])$ defined below are in
$H^{2}_{T}(\operatorname{Hess}(S,h))$.

  1. (1)

For $i\in[n]$, $x_{i}(w):=t_{w(i)}$.

  2. (2)

For $j\in\bot(h)$ and $k\in[n]$,

$y_{j,k}(w):=\begin{cases}t_{k}-t_{w(j+1)}\quad&(\text{if
}k\in\\{w(1),\dots,w(j)\\})\\\ 0\quad&(\text{otherwise}).\end{cases}$

  3. (3)

For $A\subset[n]$ with $|A|\in\textsf{L}(h)$,

$\tau_{A}(w):=\begin{cases}t_{w(j)}-t_{w(j+1)}\quad&(\text{if
}\\{w(1),\dots,w(j)\\}=A)\\\ 0\quad&(\text{otherwise}).\end{cases}$

###### Proof.

What we have to do is to check that the elements in the lemma satisfy the
congruence relation in Proposition 2.2. The check for $x_{i}$ is
straightforward and that for $y_{j,k}$ is done in [1, Lemma 10.2] in a more
general setting. Therefore, we shall check the congruence relation for
$\tau_{A}$. We set $j=|A|$.

Suppose that $v=w\cdot(r,s)$ for $r<s\leq h(r)$. We consider three cases.

Case 1. The case where $r=j$. In this case, $s=j+1$ because $h(r)=h(j)=j+1$,
so that $t_{w(r)}-t_{w(s)}=t_{w(j)}-t_{w(j+1)}$. Therefore, looking at the
definition of $\tau_{A}$, we see that the congruence relation is obviously
satisfied in this case.

Case 2. The case where $r\leq j-1$. In this case, $s\leq j$ since $s\leq
h(r)\leq h(j-1)=j$ where $h(j-1)=j$ is because $j\in\textsf{L}(h)$. Therefore,
$\\{w(1),\dots,w(j)\\}=\\{v(1),\dots,v(j)\\}$ and hence
$\tau_{A}(w)=\tau_{A}(v)$, thereby the congruence relation is trivially
satisfied.

Case 3. The case where $r\geq j+1$. In this case, $s\geq j+2$ as $s>r$.
Therefore, $\\{w(1),\dots,w(j)\\}=\\{v(1),\dots,v(j)\\}$ in this case, too.
Hence the congruence relation is satisfied. $\Box$

######  Remark 4.1.

(1) The elements $x_{i}$ lie in $H^{2}_{T}(\operatorname{Hess}(S,h))$ for any
Hessenberg function. Indeed, when $h=(n,\dots,n)$, $\operatorname{Hess}(S,h)$
is the flag variety $\operatorname{Fl}(n)$ and $x_{i}$ is the equivariant
first Chern class of the line bundle $E_{i}/E_{i-1}$ where $E_{i}$ is the
$i$-th tautological vector bundle over the flag variety
$\operatorname{Fl}(n)$:

$E_{i}:=\\{(V_{\bullet},v)\in\operatorname{Fl}(n)\times\mathbb{C}^{n}\mid v\in
V_{i}\\}.$

(2) The element $y_{j,k}$ exists in $H^{2}_{T}(\operatorname{Hess}(S,h))$ for
any $j$ with $h(j)=j+1$ by the same definition as above but for our purpose it
suffices to consider $y_{j,k}$ for $j\in\bot(h)$.

######  Lemma 4.2.

For $x_{i},y_{j,k},\tau_{A}$ in Lemma 4.1, the following holds.

  1. (1)

$\sum_{i=1}^{n}x_{i}=\sum_{i=1}^{n}t_{i}$.

  2. (2)

$\sum_{k=1}^{n}y_{j,k}=\sum_{i=1}^{j}i(x_{i}-x_{i+1})=x_{1}+\cdots+x_{j}-jx_{j+1}$.

  3. (3)

$\sum_{|A|=j}\tau_{A}=x_{j}-x_{j+1}$.

  4. (4)

Let $m$ be the maximum element in $\bot(h)$ if $\bot(h)\not=\emptyset$ and $0$
otherwise. Then

$y_{m,k}+\sum_{k\in A,\,m<|A|\leq n-1}\tau_{A}=t_{k}-x_{n}\qquad\text{for any
$k\in[n]$}$

where we understand $y_{0,k}=0$ and the sum is $0$ when $m=n-1$.

###### Proof.

For each identity, we check that the left hand side and the right hand side
take the same value at every $w\in\mathfrak{S}_{n}$. The check for the
identities in (1)–(3) is straightforward, so we leave it for the reader and
shall check the identity in (4).

It follows from the definition of $\tau_{A}$ in Lemma 4.1 that

$\left(\sum_{k\in
A,\,|A|=j}\tau_{A}\right)(w)=\begin{cases}t_{w(j)}-t_{w(j+1)}\quad&(\text{if
}k\in\\{w(1),\dots,w(j)\\})\\\ 0\quad&(\text{otherwise}).\end{cases}$

Therefore,

$\left(\sum_{k\in A,\,m<|A|\leq
n-1}\tau_{A}\right)(w)=\begin{cases}t_{w(m+1)}-t_{w(n)}\quad&(\text{if
}k\in\\{w(1),\dots,w(m)\\})\\\ t_{k}-t_{w(n)}\quad&(\text{if
}k\in\\{w(m+1),\dots,w(n-1)\\})\\\ 0\quad&(\text{otherwise}).\end{cases}$

This together with the definition of $y_{m,k}$, that is

$y_{m,k}(w)=\begin{cases}t_{k}-t_{w(m+1)}\quad&(\text{if
}k\in\\{w(1),\dots,w(m)\\})\\\ 0\quad&(\text{otherwise}),\end{cases}$

shows that the left hand side at the identity in (4) evaluated at $w$ agrees
with $t_{k}-t_{w(n)}$, proving the desired identity. $\Box$

###  4.2. Dot action

We consider an action of $\sigma\in\mathfrak{S}_{n}$ on
$\mathbb{Z}[t_{1},\dots,t_{n}]$ sending $t_{i}$ to $t_{\sigma(i)}$ for
$i\in[n]$ and define an action of $\sigma\in\mathfrak{S}_{n}$ on $f\in{\rm
Map}(\mathfrak{S}_{n},\mathbb{Z}[t_{1},\dots,t_{n}])$ by

(4.1) $(\sigma\cdot f)(w):=\sigma(f(\sigma^{-1}w)).$

One can check that if $f$ is in $H^{*}_{T}(\operatorname{Hess}(S,h))$, then so
is $\sigma\cdot f$. The action of $\mathfrak{S}_{n}$ on
$H^{*}_{T}(\operatorname{Hess}(S,h))$ preserves the ideal
$(t_{1},\dots,t_{n})$ generated by $t_{1},\dots,t_{n}$, so the action descends
to an action of $\mathfrak{S}_{n}$ on

$H^{*}(\operatorname{Hess}(S,h))=H^{*}_{T}(\operatorname{Hess}(S,h))/(t_{1},\dots,t_{n}).$

This action, called the _dot action_ , was introduced by Tymoczko [17].

######  Lemma 4.3.

Let $x_{i},y_{j,k},\tau_{A}$ be as in Lemma 4.1. Then, for
$\sigma\in\mathfrak{S}_{n}$, we have

$\sigma\cdot x_{i}=x_{i},\quad\sigma\cdot
y_{j,k}=y_{j,\sigma(k)},\quad\sigma\cdot\tau_{A}=\tau_{\sigma(A)}.$

###### Proof.

The proof is straightforward. Indeed, we have

$(\sigma\cdot
x_{i})(w)=\sigma(x_{i}(\sigma^{-1}w))=\sigma(t_{\sigma^{-1}w(i)})=t_{\sigma\sigma^{-1}w(i)}=t_{w(i)},$

proving $\sigma\cdot x_{i}=x_{i}$. As for $y_{j,k}$, we have

$\displaystyle(\sigma\cdot y_{j,k})(w)$
$\displaystyle=\sigma(y_{j,k}(\sigma^{-1}w))$
$\displaystyle=\begin{cases}\sigma(t_{k}-t_{\sigma^{-1}w(j+1)})\quad&(\text{if
}k\in\\{\sigma^{-1}w(1),\dots,\sigma^{-1}w(j)\\})\\\
0\quad&(\text{otherwise})\end{cases}$
$\displaystyle=\begin{cases}t_{\sigma(k)}-t_{w(j+1)}\quad&(\text{if
}\sigma(k)\in\\{w(1),\dots,w(j)\\})\\\ 0\quad&(\text{otherwise}),\end{cases}$

proving $\sigma\cdot y_{j,k}=y_{j,\sigma(k)}$. Similarly, as for $\tau_{A}$
with $|A|=j$, we have

$\displaystyle(\sigma\cdot\tau_{A})(w)$
$\displaystyle=\sigma(\tau_{A}(\sigma^{-1}w))$
$\displaystyle=\begin{cases}\sigma(t_{\sigma^{-1}w(j)}-t_{\sigma^{-1}w(j+1)})\quad&(\text{if
}\\{\sigma^{-1}w(1),\dots,\sigma^{-1}w(j)\\}=A)\\\
0\quad&(\text{otherwise})\end{cases}$
$\displaystyle=\begin{cases}t_{w(j)}-t_{w(j+1)}\quad&(\text{if
}\\{w(1),\dots,w(j)\\}=\sigma(A))\\\ 0\quad&(\text{otherwise}),\end{cases}$

proving $\sigma\cdot\tau_{A}=\tau_{\sigma(A)}$. $\Box$

###  4.3. Generators of $H^{2}_{T}(\operatorname{Hess}(S,h))$

Remember that

$\displaystyle\bot(h)$ $\displaystyle=\\{j\in[n-1]\mid h(j-1)=h(j)=j+1\\}$
$\displaystyle\textsf{L}(h)$ $\displaystyle=\\{j\in[n-1]\mid h(j-1)=j,\
h(j)=j+1\\}$

where $h(0)=1$, and $y_{j,k}$ is defined for $j\in\bot(h)$ and $\tau_{A}$ is
defined for $A\subset[n]$ with $|A|\in\textsf{L}(h)$.

######  Proposition 4.4.

Suppose that $h(j)\geq j+1$ for any $j\in[n-1]$. Then
$H^{2}_{T}(\operatorname{Hess}(S,h))$ is generated by $t_{i},x_{i}$
$(i\in[n])$, $y_{j,k}$ $(j\in\bot(h),k\in[n])$, and $\tau_{A}$
$(|A|\in\textsf{L}(h))$.

###### Proof.

Recall that $\Gamma(h)$ is the labeled graph introduced at the end of Section
3. As before, we consider the decomposition of $\mathfrak{S}_{n}$:

$\mathfrak{S}_{n}=\mathfrak{S}_{n}^{1}\sqcup\mathfrak{S}_{n}^{2}\sqcup\dots\sqcup\mathfrak{S}_{n}^{n}\qquad(\mathfrak{S}_{n}^{j}:=\\{w\in\mathfrak{S}_{n}\mid
w(j)=n\\})$

and the Hessenberg function $h^{j}$ obtained by removing all the boxes in the
$j$-th row and all the boxes in the $j$-th column from the configuration
corresponding to $h$. Note that the full subgraph of $\Gamma(h)$ with vertices
$\mathfrak{S}_{n}^{j}$ is $\Gamma(h^{j})$.

We prove the proposition by induction on $n$ following the idea developed in
[12]. The idea is also used in [3]. Let $z$ be an arbitrary element of
$H^{2}_{T}(\operatorname{Hess}(S,h))$.

Step 1. Since $\mathfrak{S}_{n}^{n}$ is isomorphic to $\mathfrak{S}_{n-1}$, it
follows from the inductive assumption that
$H^{2}_{T}(\operatorname{Hess}(S^{\prime},h^{n}))$ is generated by $t_{i}$’s
and elements corresponding to $x_{\bullet},y_{\bullet},\tau_{\bullet}$, where
$S^{\prime}$ is a square matrix of size $n-1$ with distinct eigenvalues.
Indeed, those elements in $H^{2}_{T}(\operatorname{Hess}(S^{\prime},h^{n})$,
denoted with $(n)$ as superscript, are defined as follows:

$\displaystyle x_{i}^{(n)}(w)$ $\displaystyle:=x_{i}(w)\quad(i\in[n-1])$
$\displaystyle y_{j,k}^{(n)}(w)$
$\displaystyle:=y_{j,k}(w)\quad(j\in\bot(h^{n}))$
$\displaystyle\tau_{B}^{(n)}(w)$
$\displaystyle:=\tau_{B}(w)\quad(B\subset[n-1],\ |B|\in\textsf{L}(h^{n}))$

where $w\in\mathfrak{S}_{n}^{n}$. This shows that any element of
$H^{2}_{T}(\operatorname{Hess}(S^{\prime},h^{n}))$ is the restriction image of
a linear combination of $t_{\bullet},x_{\bullet},y_{\bullet},\tau_{\bullet}$
to $H^{2}_{T}(\operatorname{Hess}(S^{\prime},h^{n})$. Therefore, we may assume
that $z=0$ on $\mathfrak{S}_{n}^{n}$ by subtracting an appropriate linear
combination of $t_{\bullet},x_{\bullet},y_{\bullet},\tau_{\bullet}$ from $z$.

Step 2. Suppose that $z=0$ on
$\mathfrak{S}_{n}^{r+1}\sqcup\cdots\sqcup\mathfrak{S}_{n}^{n}$ for some $1\leq
r\leq n-1$. Then we shall show that $z$ minus an appropriate linear
combination of $t_{\bullet},x_{\bullet},y_{\bullet},\tau_{\bullet}$ vanishes
on $\mathfrak{S}_{n}^{r+1}\sqcup\cdots\sqcup\mathfrak{S}_{n}^{n}$. We consider
two cases.

Case 1. The case where $h(r)\geq r+2$ (so $r<n-1$). In this case, the
$(r+1,r)$ and $(r+2,r)$ boxes are shaded in the configuration associated to
$h$. This means that at each vertex $w\in\mathfrak{S}_{n}^{r}$, there are
edges in $\Gamma(h)$ emanating from $w$ to
$w\cdot(r,r+1)\in\mathfrak{S}_{n}^{r+1}$ and
$w\cdot(r,r+2)\in\mathfrak{S}_{n}^{r+2}$, where the labels on those edges are
respectively $t_{w(r)}-t_{w(r+1)}$ and $t_{w(r)}-t_{w(r+2)}$ up to sign. Since
$z=0$ on $\mathfrak{S}_{n}^{r+1}\sqcup\cdots\sqcup\mathfrak{S}_{n}^{n}$,
$z(w)$ must be divisible by these linear polynomials. However, since the
cohomological degree of $z$ is two, this implies that $z$ must be $0$ on
$\mathfrak{S}_{n}^{r}$.

Case 2. The case where $h(r)=r+1$. In this case, each vertex $w$ in
$\mathfrak{S}_{n}^{r}$ is joined by an edge to a vertex $w\cdot(r,r+1)$ of
$\mathfrak{S}_{n}^{r+1}$. Since the label on the edge is
$t_{w(r)}-t_{w(r+1)}=t_{n}-t_{w(r+1)}$ up to sign and $z=0$ on
$\mathfrak{S}_{n}^{r+1}$, $z(w)$ for $w\in\mathfrak{S}_{n}^{r}$ is a constant
multiple of $t_{n}-t_{w(r+1)}$. Note that $h(r-1)\leq h(r)=r+1$ and

$t_{n}-t_{w(r+1)}=\begin{cases}y_{r,n}(w)\quad&\text{(when $h(r-1)=r+1$)}\\\
\tau_{\\{w(1),\dots,w(r)\\}}(w)\quad&\text{(when $h(r-1)=r$)}.\end{cases}$

We distinguish two cases according to the above.

(i) The case where $h(r-1)=r+1$. In this case, $h^{r}(j)\geq j+1$ for any
$j\in[n-2]$, so the graph $\Gamma(h^{r})$ is connected. By the observation
above, we have $z(w)=c_{w}y_{r,n}(w)$ $(w\in\mathfrak{S}_{n}^{r})$ with some
integer $c_{w}$.

Claim. $c_{w}=c_{v}$ for any $w,v\in\mathfrak{S}_{n}^{r}$.

Proof of the claim. Since $\Gamma(h^{r})$ is connected, it suffices to prove
the identity for a pair of $w$ and $v$ joined by an edge of $\Gamma(h^{r})$.
Then, $v=w\cdot(p,q)$ for some $p,q\in[n]\backslash\\{r\\}$ and

(4.2) $\begin{split}z(w)-z(v)&=c_{w}y_{r,n}(w)-c_{v}y_{r,n}(v)\\\
&=c_{w}(t_{n}-t_{w(r+1)})-c_{v}(t_{n}-t_{v(r+1)})\\\
&=(c_{w}-c_{v})t_{n}-c_{w}t_{w(r+1)}+c_{v}t_{v(r+1)},\end{split}$

which must be divisible by $t_{w(p)}-t_{w(q)}=t_{v(q)}-t_{v(p)}$. Here,
$n=w(r)=v(r)$, $p\not=r$ and $q\not=r$, so the coefficient $c_{w}-c_{v}$ of
$t_{n}$ in (4.2) must be zero. This proves the claim.

By the claim, we may write $c_{w}$ as $c$ and $z-cy_{r,n}=0$ on
$\mathfrak{S}_{n}^{r}$. Furthermore, $y_{r,n}=0$ on
$\mathfrak{S}_{n}^{r+1}\sqcup\cdots\sqcup\mathfrak{S}_{n}^{n}$ because
$y_{r,n}(v)=0$ for $v\in\mathfrak{S}_{n}$ with $n\in\\{v(r+1),\dots,v(n)\\}$
by the definition of $y_{r,n}$ in Lemma 4.1 (2). Therefore, $z-cy_{r,n}=0$ on
$\mathfrak{S}_{n}^{r}\sqcup\cdots\sqcup\mathfrak{S}_{n}^{n}$.

(ii) The case where $h(r-1)=r$. In this case, the graph $\Gamma(h^{r})$ is
disconnected since $h^{r}(r-1)=r-1$. Indeed, there are $\binom{n-1}{r-1}$ many
connected components because $h^{r}(j)\geq j+1$ for any $1\leq j\leq n-2$ with
$j\not=r-1$. The vertex set of a connected component of $\Gamma(h^{r})$ is

$\mathfrak{S}_{n}^{r}(A):=\\{w\in\mathfrak{S}_{n}^{r}\mid\\{w(1),\dots,w(r)\\}=A\\}$

for some $n\in A\subset[n]$ with $|A|=r$. On this connected component,
$z(w)=c_{A}\tau_{A}(w)$ $(w\in\mathfrak{S}_{n}^{r}(A))$ with some integer
$c_{A}$, where $c_{A}$ is independent of $w\in\mathfrak{S}_{n}^{r}(A)$ by a
similar argument to the claim above, and $\tau_{A}$ vanishes on
$\bigsqcup_{B\not=A}\mathfrak{S}_{n}^{r}(B)=\mathfrak{S}_{n}^{r}\backslash\mathfrak{S}_{n}^{r}(A)$.
Therefore,

(4.3) $z-\sum_{n\in A\subset[n],\,|A|=r}c_{A}\tau_{A}=0\quad\text{on
$\mathfrak{S}_{n}^{r}(A)$}.$

Furthermore, since $n\in A$ and $|A|=r$, $\tau_{A}$ in (4.3) vanishes on
$\mathfrak{S}_{n}^{r+1}\sqcup\cdots\sqcup\mathfrak{S}_{n}^{n}$ by the
definition of $\tau_{A}$ in Lemma 4.1(3). Therefore, the left hand side in
(4.3) vanishes on
$\mathfrak{S}_{n}^{r}\sqcup\cdots\sqcup\mathfrak{S}_{n}^{n}$.

The inductive argument developed above shows that one can change $z$ into $0$
on the whole set $\mathfrak{S}_{n}$ by subtracting an appropriate linear
combination of $t_{\bullet},x_{\bullet},y_{\bullet},\tau_{\bullet}$. Since $z$
is an arbitrary element of $H^{2}_{T}(\operatorname{Hess}(S,h))$, this proves
the proposition. $\Box$

##  5\. Structure of $H^{2}(\operatorname{Hess}(S,h))$

###  5.1. Explicit presentation of $H^{2}(\operatorname{Hess}(S,h))$

The generators of $H^{2}_{T}(\operatorname{Hess}(S,h))$ in Proposition 4.4 can
be decreased. Indeed, it follows from Lemma 4.2(4) that we can drop
$y_{n-1,k}$ when $n-1\in\bot(h)$ and $\tau_{A}$ with $|A|=n-1$ when
$n-1\in\textsf{L}(h)$, i.e. it suffices to consider
$\bot(h)\backslash\\{n-1\\}$ and $\textsf{L}(h)\backslash\\{n-1\\}$ as index
sets of $y_{j,k}$ and $\tau_{A}$. This is also true for the ordinary
cohomology $H^{2}(\operatorname{Hess}(S,h))$ because

$H^{2}(\operatorname{Hess}(S,h))=H^{2}_{T}(\operatorname{Hess}(S,h))/\mathbb{Z}\langle
t_{1},\dots,t_{n}\rangle$

where $\mathbb{Z}\langle t_{1},\dots,t_{n}\rangle$ denotes the module
generated by $t_{1},\dots,t_{n}$ over $\mathbb{Z}$. Thus, we have a surjective
homomorphism

$\Phi\colon\mathbb{Z}\langle X_{i},Y_{j,k},T_{A}\rangle\to
H^{2}(\operatorname{Hess}(S,h))$

sending $X_{i}$ to $x_{i}$, $Y_{j,k}$ to $y_{j,k}$, and $T_{A}$ to $\tau_{A}$,
where

  1. (1)

$i\in[n]$,

  2. (2)

$j\in\bot(h)\backslash\\{n-1\\},\ k\in[n]$,

  3. (3)

$A\subset[n],\ |A|\in\textsf{L}(h)\backslash\\{n-1\\}$.

By Lemma 4.2, the submodule $U$ of the free module $\mathbb{Z}\langle
X_{i},Y_{j,k},T_{A}\rangle$ generated by the following elements maps to zero
by $\Phi$:

  1. (R1)

$X_{1}+\cdots+X_{n}$,

  2. (R2)

$\displaystyle{\sum_{k=1}^{n}Y_{j,k}-(X_{1}+\cdots+X_{j}-jX_{j+1})}$
$(j\in\bot(h)\backslash\\{n-1\\})$,

  3. (R3)

$\displaystyle{\sum_{|A|=j}T_{A}-(X_{j}-X_{j+1})}$
$(j\in\textsf{L}(h)\backslash\\{n-1\\})$.

Therefore, the map $\Phi$ induces a surjective homomorphism

(5.1) $\bar{\Phi}\colon\mathbb{Z}\langle X_{i},Y_{j,k},T_{A}\rangle/U\to
H^{2}(\operatorname{Hess}(S,h)).$

With this understanding, we have the following.

######  Theorem 5.1.

Suppose that $h(j)\geq j+1$ for any $j\in[n-1]$. Then, the map $\bar{\Phi}$ in
(5.1) is an isomorphism.

###### Proof.

First we note that the quotient module $\mathbb{Z}\langle
X_{i},Y_{j,k},T_{A}\rangle/U$ is free. Indeed, if $V$ is the submodule of
$\mathbb{Z}\langle X_{i},Y_{j,k},T_{A}\rangle$ generated by

  1. (1)

$X_{i}$ $(i\not=n)$,

  2. (2)

$Y_{j,k}$ $(k\not=n)$,

  3. (3)

$T_{A}$ $(A\not=[|A|])$,

then $\mathbb{Z}\langle X_{i},Y_{j,k},T_{A}\rangle=U\oplus V$. Therefore,
$\mathbb{Z}\langle X_{i},Y_{j,k},T_{A}\rangle/U\cong V$, which is free.

The rank of the free module $\mathbb{Z}\langle X_{i},Y_{j,k},T_{A}\rangle$ is

$n+n(|\bot(h)\backslash\\{n-1\\}|)+\sum_{j\in\textsf{L}(h)\backslash\\{n-1\\}}\binom{n}{j}=n|\bot(h)|+\sum_{j\in\textsf{L}(h)}\binom{n}{j}$

while the elements in (R1), (R2), (R3) are linearly independent so that the
rank of $U$ is

$1+|\bot(h)\backslash\\{n-1\\}|+|\textsf{L}(h)\backslash\\{n-1\\}|=|\bot(h)|+|\textsf{L}(h)|.$

Therefore, the rank of the source module $\mathbb{Z}\langle
X_{i},Y_{j,k},T_{A}\rangle/U$ of $\bar{\Phi}$ is

$(n-1)|\bot(h)|+\sum_{j\in\textsf{L}(h)}\binom{n}{j}-|\textsf{L}(h)|$

which agrees with the second Betti number of $\operatorname{Hess}(S,h)$ by
Lemma 3.2. This implies that $\bar{\Phi}$ is an isomorphism because
$\bar{\Phi}$ is surjective and both the source and target modules of
$\bar{\Phi}$ are free. $\Box$

Following Lemma 4.3, we define an action of $\mathfrak{S}_{n}$ on the
variables $X_{i},Y_{j,k},T_{A}$ by

(5.2) $\sigma\cdot X_{i}:=X_{i},\quad\sigma\cdot
Y_{j,k}:=Y_{j,\sigma(k)},\quad\sigma\cdot T_{A}:=T_{\sigma(A)}$

and extend the action to the free module $\mathbb{Z}\langle
X_{i},Y_{j,k},T_{A}\rangle$ linearly. Then, $\mathfrak{S}_{n}$ acts on the
submodule $U$ trivially, so that the $\mathfrak{S}_{n}$-action on
$\mathbb{Z}\langle X_{i},Y_{j,k},T_{A}\rangle$ descends to an
$\mathfrak{S}_{n}$-action on the quotient $\mathbb{Z}\langle
X_{i},Y_{j,k},T_{A}\rangle/U$ and the isomorphism $\bar{\Phi}$ becomes
$\mathfrak{S}_{n}$-equivariant.

###  5.2. $\mathfrak{S}_{n}$-module structure on
$H^{2}(\operatorname{Hess}(S,h))$

Let $\lambda=(\lambda_{1},\dots,\lambda_{\ell})$ be a partition of $n$,
denoted by $\lambda\vdash n$, and let
$\mathfrak{S}_{\lambda}=\mathfrak{S}_{\lambda_{1}}\times\cdots\times\mathfrak{S}_{\lambda_{\ell}}$
the Young subgroup of $\mathfrak{S}_{n}$ associated to $\lambda$. We set

$M^{\lambda}=\mathbb{C}[\mathfrak{S}_{n}]\otimes_{\mathbb{C}[\mathfrak{S}_{\lambda}]}\mathbb{C}$

where $\mathbb{C}[G]$ denotes the group ring of a finite group $G$ over
$\mathbb{C}$. As is well-known, $\\{M^{\lambda}\mid\lambda\vdash n\\}$ forms
an additive basis of the complex representation ring $R(\mathfrak{S}_{n})$ of
$\mathfrak{S}_{n}$. Therefore, any $\mathfrak{S}_{n}$-module over $\mathbb{C}$
can be expressed uniquely as a linear combination of $M^{\lambda}$’s over
$\mathbb{Z}$.

######  Theorem 5.2.

Suppose that $h(j)\geq j+1$ and for $1\leq j\leq n-2$, let

$\beta_{j}:=\begin{cases}(n-j,j)\quad&\text{if }h(j-1)=j,\ h(j)=j+1,\\\
(n-1,1)\quad&\text{if }h(j-1)=h(j)=j+1,\\\
(n)\quad&\text{otherwise},\end{cases}$

where $h(0)=1$. Then

$H^{2}(\operatorname{Hess}(S,h))\otimes\mathbb{C}=\sum_{j=1}^{n-2}M^{\beta_{j}}+M^{(n)}\quad\text{in
}R(\mathfrak{S}_{n}).$

###### Proof.

Since the action of $\mathfrak{S}_{n}$ on $X_{i},Y_{j,k},T_{A}$ is given by
(5.2), $\mathbb{C}\langle X_{i},Y_{j,k},T_{A}\rangle$ decomposes into a direct
sum

$\displaystyle\mathbb{C}\langle X_{i}\mid i\in[n]\rangle$
$\displaystyle\oplus\ $ $\displaystyle\mathbb{C}\langle Y_{j,k}\mid
j\in\bot(h)\backslash\\{n-1\\},\ k\in[n]\rangle$ $\displaystyle\oplus\ $
$\displaystyle\mathbb{C}\langle T_{A}\mid A\subset[n],\
|A|\in\textsf{L}(h)\backslash\\{n-1\\}\rangle.$

as an $\mathfrak{S}_{n}$-module and the submodule $U\otimes\mathbb{C}$ of
$\mathbb{C}\langle X_{i},Y_{j,k},T_{A}\rangle$ is trivial as an
$\mathfrak{S}_{n}$-module. The space $\mathbb{C}\langle Y_{j,k}\mid
k\in[n]\rangle$ is isomorphic to $M^{(n-1,1)}$ for $j\in\bot(h)$ while the
space $\mathbb{C}\langle T_{A}\mid|A|=j\rangle$ is isomorphic to $M^{(n-j,j)}$
for $j\in\textsf{L}(h)$. There is no more non-trivial
$\mathfrak{S}_{n}$-module in $\mathbb{C}\langle
X_{i},Y_{j,k},T_{A}\rangle/U\otimes\mathbb{C}$ and one can see that the
dimension of the complementary module is the number of
$j\in[n-2]\backslash(\bot(h)\sqcup\textsf{L}(h))$ plus $1$ (this $1$ comes
from $j=n-1$ and corresponds to $M^{(n)}$ in the last part of the right hand
side in the theorem). This proves the theorem. $\Box$

######  Example 5.1 (cf. Example 6.2 in [6]).

Let $n=8$ and $h=(2,3,6,6,6,7,8,8)$. Then

$\beta_{1}=(7,1),\ \beta_{2}=(6,2),\ \beta_{3}=(8),\ \beta_{4}=(8),\
\beta_{5}=(7,1),\ \beta_{6}=(2,6).$

Therefore,

$H^{2}(\operatorname{Hess}(S,h))\otimes\mathbb{C}=3M^{(8)}+2M^{(7,1)}+2M^{(6,2)}\quad\text{in
}R(\mathfrak{S}_{8}).$

##  6\. A generalization

In the previous sections, we studied $H^{2}(\operatorname{Hess}(S,h))$ under
the condition that $h(j)\geq j+1$ for any $j\in[n-1]$. In this section, we
will study $H^{2d}(\operatorname{Hess}(S,h))$ under the condition that
$h(j)\geq j+d$ for any $j\in[n-d]$, where $d\geq 2$.

For $j\in[n]$ and $k\in[n]$, there is an element $y_{j,k}\in
H_{T}^{2(h(j)-j)}(\operatorname{Hess}(S,h))$ defined by

(6.1)
$y_{j,k}(w):=\begin{cases}\prod_{\ell=j+1}^{h(j)}(t_{k}-t_{w(\ell)})\quad&(\text{if
}k\in\\{w(1),\dots,w(j)\\})\\\ 0\quad&(\text{otherwise}).\end{cases}$

The element $y_{j,k}$ is introduced in [1] and proved to be in
$H^{*}_{T}(\operatorname{Hess}(S,h))$ ([1, Lemma 10.2]). Note that when $d=1$
and $h(j)=j+1$, the element $y_{j,k}$ above agrees with the $y_{j,k}$ in Lemma
4.1(2) (see also Remark 4.1(2)). We use the same notation $y_{j,k}$ for its
image in the ordinary cohomology $H^{2d}(\operatorname{Hess}(S,h))$. The dot
action on $y_{j,k}$ is the same as before, i.e.

(6.2) $\sigma\cdot y_{j,k}=y_{j,\sigma(k)}\quad\text{for
$\sigma\in\mathfrak{S}_{n}$}.$

Therefore, $\sum_{k=1}^{n}y_{j,k}$ is $\mathfrak{S}_{n}$-invariant. In fact,
the sum has the following expression:

(6.3)
$\sum_{k=1}^{n}y_{j,k}=\sum_{i=1}^{j}\prod_{\ell=j+1}^{h(j)}(x_{i}-x_{\ell}),$

which can be proved by checking that both sides take the same value at each
$w\in\mathfrak{S}_{n}$.

We set

$\Lambda_{d}(h):=\\{j\in[n-d]\mid h(j)=j+d\\}.$

Our main result in this section is the following.

######  Theorem 6.1.

Suppose that $d\geq 2$ and $h(j)\geq j+d$ for any $j\in[n-d]$. Then, the
restriction map

$\iota^{*}\colon H^{2p}(\operatorname{Fl}(n))\to
H^{2p}(\operatorname{Hess}(S,h))$

is an isomorphism for $p<d$. For $p=d$, the restriction map is an injective
and we have an isomorphism

$\displaystyle H^{2d}(\operatorname{Hess}(S,h))$ $\displaystyle\cong$
$\displaystyle\left(\iota^{*}(H^{2d}(\operatorname{Fl}(n)))\oplus\mathbb{Z}\langle
Y_{j,k}\mid
j\in\Lambda_{d}(h),k\in[n]\rangle\right)/\mathbb{Z}\langle\sum_{k=1}^{n}Y_{j,k}-y_{j}\mid
j\in\Lambda_{d}(h)\rangle$

where $Y_{j,k}$ corresponds to $y_{j,k}$ in $H^{*}(\operatorname{Hess}(S,h))$
and
$\displaystyle{y_{j}=\sum_{i=1}^{j}\prod_{\ell=j+1}^{h(j)}(x_{i}-x_{\ell})}$.

######  Remark 6.1.

(1) Theorem 2.1(3) says that $\operatorname{Hess}(S,h)$ is connected, in other
words, the restriction map

$\iota\colon H^{0}(\operatorname{Fl}(n))\to H^{0}(\operatorname{Hess}(S,h))$

is an isomorphism if and only if $h(j)\geq j+1$ for any $j\in[n-1]$.
Therefore, Theorem 6.1 can be regarded as a generalization of Theorem 2.1(3)
and Theorem 5.1.

(2) When $d=1$, the elements $\tau_{A}$ appear in
$H^{2}(\operatorname{Hess}(S,h))$ as shown in Theorem 5.1 but such type of
elements does not appear when $d\geq 2$.

Since the action of $\mathfrak{S}_{n}$ on $x_{i}$’s is trivial, so is that on
$\iota^{*}(H^{2d}(\operatorname{Fl}(n)))$ while that on $y_{j,k}$ is given by
(6.2). Therefore, we obtain the following corollary from Theorem 6.1.

######  Corollary 6.2.

Let the situation be as in Theorem 6.1. Then, the action of $\mathfrak{S}_{n}$
on $H^{2p}(\operatorname{Hess}(S,h))$ is trivial while

$H^{2d}(\operatorname{Hess}(S,h))\otimes\mathbb{C}=m_{d}M^{(n)}+|\Lambda_{d}(h)|M^{(n-1,1)}\quad\text{in
$R(\mathfrak{S}_{n})$}$

where $m_{d}=b_{2d}(\operatorname{Fl}(n))-|\Lambda_{d}(h)|$.

######  Remark 6.2.

Since

$\operatorname{Poin}(\operatorname{Fl}(n),\sqrt{q})=\prod_{i=1}^{n}\frac{1-q^{i}}{1-q},$

we have $b_{2d}(\operatorname{Fl}(n))\geq n-1$ while $|\Lambda_{d}(h)|\leq
n-2$. Therefore, $m_{d}$ in the corollary is positive.

The idea of the proof of Theorem 6.1 is the same as before. We compute the
Betti numbers of $\operatorname{Hess}(S,h)$ up to degree $2d$ using
Proposition 3.1 and observe that $H^{*}(\operatorname{Hess}(S,h))$ is
generated by $x_{i}$’s and $y_{j,k}$’s as a graded ring up to degree $2d$.

For the first part of Theorem 6.1, we have a homotopical version which is
proved easier and may be of independent interest.

######  Proposition 6.3.

Suppose that $d\geq 1$ and $h(j)\geq j+d$ for any $j\in[n-d]$. Then, the
natural inclusion $\iota\colon\operatorname{Hess}(S,h)\to\operatorname{Fl}(n)$
induces isomorphisms of homotopy groups $\pi_{q}$ in degrees $q\leq 2d-1$.

To prove this, it is sufficient to construct cellular structures on
$\operatorname{Hess}(S,h)$ and $\operatorname{Fl}(n)$ which are consistent
under $\iota$ and coincide in small dimensions. Instead of cellular
structures, one can use affine pavings. There is an affine paving of
$\operatorname{Fl}(n)$ by even-dimensional Bruhat cells indexed by
permutations $w\in\mathfrak{S}_{n}$. Intersecting a Bruhat cell $C_{w}$ of
$\operatorname{Fl}(n)$ with the subvariety $\operatorname{Hess}(S,h)$ gives an
affine cell $C^{\prime}_{w}$ of $\operatorname{Hess}(S,h)$. The dimensions of
cells $C_{w}$ and $C_{w}^{\prime}$ can be computed from Bialynicki-Birula
theory, see [11]. The dimension $\dim_{\mathbb{C}}C_{w}$ equals the number
$\ell(w)$ of inversions of $w$, while $\dim_{\mathbb{C}}C_{w}^{\prime}$ equals
$\ell_{h}(w)$, the number of inversions $j<i$, $w(j)>w(i)$, satisfying $i\leq
h(j)$, see (2.1).

######  Lemma 6.4.

Let $h(j)\geq j+d$ for any $j\in[n-d]$. Then for any permutation $w$, each one
of the conditions $\ell_{h}(w)<d$ or $\ell(w)<d$ implies
$\ell_{h}(w)=\ell(w)$.

###### Proof.

Since $\ell_{h}(w)\leq\ell(w)$, it suffices to prove $\ell_{h}(w)=\ell(w)$
when $\ell_{h}(w)<d$. Suppose that $\ell_{h}(w)\not=\ell(w)$. Then there is an
inversion $\\{j,i\\}$ $(j<i)$ in $w$ which contributes to $\ell(w)$ but does
not contribute to $\ell_{h}(w)$. This means $j+d\leq h(j)<i$.

We assume that the difference $i-j(>d)$ is minimum among those inversions. Any
number $j^{\prime}$ such that $j<j^{\prime}\leq j+d$ either produces an
inversion $\\{j,j^{\prime}\\}$ or $\\{j^{\prime},i\\}$. Obviously
$j^{\prime}-j\leq d$ and if $\\{j^{\prime},i\\}$ is an inversion, then
$i-j^{\prime}\leq d$ which follows from the minimality of $i-j$. In any case,
each $j^{\prime}$ produces an inversion which contributes to $\ell_{h}(w)$.
Since there are $d$ many such $j^{\prime}$, we have $\ell_{h}(w)\geq d$.
However, this contradicts the condition $\ell_{h}(w)<d$. Therefore
$\ell_{h}(w)=\ell(w)$. $\Box$

Let us prove Proposition 6.3.

###### Proof.

Since the cell $C_{w}^{\prime}$ of $\operatorname{Hess}(S,h)$ is the
intersection of the Bruhat cell $C_{w}$ of $\operatorname{Fl}(n)$ with
$\operatorname{Hess}(S,h)$, Lemma 6.4 implies that the spaces
$\operatorname{Fl}(n)$ and $\operatorname{Hess}(S,h)$ have the same
$(2d-1)$-skeleta and that any $2d$-dimensional cell $C_{w}$ of
$\operatorname{Fl}(n)$ agrees with the cell $C_{w}^{\prime}$ of
$\operatorname{Hess}(S,h)$. Therefore

$\iota_{*}\colon H_{q}(\operatorname{Hess}(S,h))\to
H_{q}(\operatorname{Fl}(n))$

an isomorphism for $q\leq 2d-1$ and an epimorphism for $q=2d$. Moreover, both
$\operatorname{Hess}(S,h)$ and $\operatorname{Fl}(n)$ are simply connected.
Therefore the proposition follows from the Whitehead theorem ([15, Theorem in
p.399]). $\Box$

We now proceed with the more detailed analysis of homology needed to prove
Theorem 6.1.

###  6.1. Betti numbers

The Betti numbers $b_{2i}(\operatorname{Hess}(S,h))$ of
$\operatorname{Hess}(S,h)$ for $i\leq d$ are given as follows.

######  Lemma 6.5.

Suppose that $d\geq 2$ and $h(j)\geq j+d$ for any $j\in[n-d]$. Then

$b_{2i}(\operatorname{Hess}(S,h))=\begin{cases}b_{2i}(\operatorname{Fl}(n))\quad&(\text{if
}i<d)\\\ b_{2i}(\operatorname{Fl}(n))+(n-1)|\Lambda_{d}(h)|\quad&(\text{if
}i=d).\end{cases}$

###### Proof.

For a polynomial $f(q)$ in $q$, we denote by $f(q)^{\leq d}$ the polynomial
obtained from $f(q)$ by truncating terms of degree $>d$. Then, the lemma is
equivalent to

(6.4) $\operatorname{Poin}(\operatorname{Hess}(S,h),\sqrt{q})^{\leq
d}=\operatorname{Poin}(\operatorname{Fl}(n),\sqrt{q})^{\leq
d}+(n-1)|\Lambda_{d}(h)|q^{d}.$

We prove the identity (6.4) by induction on $n+d$ where $d\geq 2$. Since
$d\geq 2$ and $h(j)\geq j+d$ for any $j\in[n-d]$ by assumption, $n$ is greater
than or equal to $3$ and when $(n,d)=(3,2)$, $h$ must be $(3,3,3)$. In this
case, $\operatorname{Hess}(S,h)=\operatorname{Fl}(3)$,
$\Lambda_{d}(h)=\emptyset$ and hence lemma holds.

Suppose that $n+d\geq 6$ and the lemma holds for any pair
$(n^{\prime},d^{\prime})$ such that $n^{\prime}+d^{\prime}<n+d$. It follows
from Proposition 3.1 that we have

(6.5)
$\begin{split}\operatorname{Poin}(\operatorname{Hess}(S,h),\sqrt{q})^{\leq
d}=&\sum_{j=1}^{n-d-1}\left(q^{h(j)-j}\operatorname{Poin}(\operatorname{Hess}(S^{\prime},h^{j}),\sqrt{q})\right)^{\leq
d}\\\
&+\sum_{j=n-d}^{n}\left(q^{h(j)-j}\operatorname{Poin}(\operatorname{Hess}(S^{\prime},h^{j}),\sqrt{q})\right)^{\leq
d}\\\
=|\Lambda_{d}(h)|q^{d}+\sum_{j=n-d}^{n}&\left(q^{n-j}\operatorname{Poin}(\operatorname{Hess}(S^{\prime},h^{j}),\sqrt{q})\right)^{\leq
d}\end{split}$

because $h(j)\geq j+d$ for any $j\in[n-d]$ and $h(j)=j+d<n$ if and only if
$j\in\Lambda_{d}(h)$. Applying the induction assumption (and Lemma 3.2 when
$d=2$) to the last sum in (6.5), we obtain

(6.6)
$\begin{split}&\sum_{j=n-d}^{n}\left(q^{n-j}\operatorname{Poin}(\operatorname{Hess}(S^{\prime},h^{j}),\sqrt{q})\right)^{\leq
d}\\\
=&\sum_{j=n-d}^{n-2}\left(q^{n-j}\operatorname{Poin}(\operatorname{Fl}(n-1),\sqrt{q})\right)^{\leq
d}\\\
&+\left(q\Big{(}\operatorname{Poin}(\operatorname{Fl}({n-1}),\sqrt{q})+(n-2)|\Lambda_{d-1}(h^{n-1})|q^{d-1}\Big{)}\right)^{\leq
d}\\\
&+\left(\operatorname{Poin}(\operatorname{Fl}(n-1),\sqrt{q})+(n-2)|\Lambda_{d}(h^{n})|q^{d}\right)^{\leq
d}\\\
=&\sum_{j=n-d}^{n}\left(q^{n-j}\operatorname{Poin}(\operatorname{Fl}(n-1),\sqrt{q})\right)^{\leq
d}\\\
&+(n-2)\left(|\Lambda_{d-1}(h^{n-1})|+|\Lambda_{d}(h^{n})|\right)q^{d}\\\
=&\operatorname{Poin}(\operatorname{Fl}(n),\sqrt{q})^{\leq
d}+(n-2)\left(|\Lambda_{d-1}(h^{n-1})|+|\Lambda_{d}(h^{n})|\right)q^{d},\end{split}$

where we can see the last identity above by applying Proposition 3.1 to the
flag variety $\operatorname{Fl}(n)$. Thus, if we prove

(6.7) $|\Lambda_{d-1}(h^{n-1})|+|\Lambda_{d}(h^{n})|=|\Lambda_{d}(h)|,$

then the identity (6.4) follows from (6.5) and (6.6). However, one can easily
see that

$(|\Lambda_{d-1}(h^{n-1})|,|\Lambda_{d}(h^{n})|)=\begin{cases}(0,|\Lambda_{d}(h)|)\quad&(\text{if
}h(n-d-1)=n)\\\ (1,|\Lambda_{d}(h)|-1)\quad&(\text{if
}h(n-d-1)=n-1)\end{cases}$

and this implies (6.7). $\Box$

###  6.2. Complementary elements

We introduce complementary elements $y^{*}_{i,k}$ which will make our argument
perspective.

The element $y_{j,k}$ in (6.1) is defined by looking at the $j$-th column of
the configuration associated to the Hessenberg function $h$. Similarly, one
can define an element $y^{*}_{i,k}$ of $H^{*}_{T}(\operatorname{Hess}(S,h))$
by looking at the $i$-th row of the configuration as follows. For $1<i\leq n$,
we define

$h^{*}(i):=\min\\{j\in[n]\mid h(j)\geq i\\}.$

This definition tells us that the shaded boxes in the $i$-th row and under the
diagonal in the configuration associated to $h$ are at positions $(i,\ell)$
$(h^{*}(i)\leq\ell<i)$. Looking at those shaded boxes, we define

(6.8)
$y^{*}_{i,k}(w):=\begin{cases}\prod_{\ell=h^{*}(i)}^{i-1}(t_{k}-t_{w(\ell)})\quad&(k\in\\{w(i),\dots,w(n)\\})\\\
0\quad&(\text{otherwise}).\end{cases}$

One can see that $y^{*}_{i.k}$ is in $H^{*}_{T}(\operatorname{Hess}(S,h))$
similarly to $y_{j,k}$.

Note that

  1. (1)

$h(j)\geq j+d$ for $\forall j\in[n-d]$ $\Longleftrightarrow$ $h^{*}(i)\leq
i-d$ for $d+1\leq\forall i\leq n$.

  2. (2)

Under the assumption that $h(j)\geq j+d$ for any $j\in[n-d]$, we have

$h(j)=j+d<n\ (\text{i.e.}\ j\in\Lambda_{d}(h))\Longleftrightarrow
h^{*}(j+1+d)=j+1.$

Based on this observation, we define

$\Lambda^{*}(h)=\\{i\mid d+2\leq i\leq n,\ h^{*}(i)=i-d\\}.$

Then, (2) above can be restated that $j\in\Lambda_{d}(h)\Longleftrightarrow
j+1+d\in\Lambda^{*}_{d}(h)$.

The elements $y^{*}_{i,k}$ $(i\in\Lambda^{*}_{d}(h))$ do not provide new
elements as is seen from the following lemma.

######  Lemma 6.6.

Suppose that $h(j)\geq j+d$ for any $j\in[n-d]$. Then, for
$j\in\Lambda_{d}(h)$ we have

$y_{j,k}+y^{*}_{j+1+d,k}=\prod_{\ell=j+1}^{j+d}(t_{k}-x_{\ell}).$

###### Proof.

It immediately follows from the definitions of $y_{j,k}$, $y^{*}_{i,k}$, and
$x_{i}$ that the both sides in the lemma take the same value at every
$w\in\mathfrak{S}_{n}$. $\Box$

###  6.3. Generators of $H^{*}_{T}(\operatorname{Hess}(S,h))$

We show that the elements $t_{i},x_{i},y_{j,k}$ generate
$H^{*}_{T}(\operatorname{Hess}(S,h))$ up to $*\leq 2d$ as a graded ring under
our assumption.

######  Lemma 6.7.

Suppose that $h(j)\geq j+d$ for any $j\in[n-d]$ and $d\geq 2$. Then
$H_{T}^{*}(\operatorname{Hess}(S,h))$ is generated by $t_{i}$, $x_{i}$
$(i\in[n])$ and $y_{j,k}$ $(j\in\Lambda_{d}(h),k\in[n])$ up to $*\leq 2d$ as a
graded ring, where the degree of $x_{i}$ is $2$ while that of $y_{j,k}$ is
$2d$.

###### Proof.

We prove the lemma by induction on $n+d$ similarly to the proof of Lemma 6.5.
When $(n,d)=(3,2)$, $\operatorname{Hess}(S,h)=\operatorname{Fl}(3)$ and the
lemma holds since $H_{T}^{*}(\operatorname{Fl}(n))$ is generated by
$t_{i},x_{i}$ $(i\in[n])$.

Suppose that $n+d\geq 6$ and the lemma holds for any pair
$(n^{\prime},d^{\prime})$ such that $n^{\prime}+d^{\prime}<n+d$. Similarly to
the proof of Proposition 4.4, we consider the decomposition

$\mathfrak{S}_{n}=\mathfrak{S}_{n}^{1}\sqcup\mathfrak{S}_{n}^{2}\sqcup\cdots\sqcup\mathfrak{S}_{n}^{n}\qquad(\mathfrak{S}_{n}^{j}=\\{w\in\mathfrak{S}_{n}\mid
w(j)=n\\}).$

Let $z$ be an arbitrary element of $H^{2p}_{T}(\operatorname{Hess}(S,h))$ for
$p\leq d$.

Step 1. By the same reasoning as Step 1 in the proof of Proposition 4.4, we
may assume $z=0$ on $\mathfrak{S}_{n}^{n}$ by subtracting an appropriate
polynomial in $t_{\bullet},x_{\bullet},y_{\bullet}$.

Step 2. Suppose that $z=0$ on $\mathfrak{S}_{n}^{n}$. We will show that $z$
minus an appropriate polynomial of $t_{\bullet},x_{\bullet},y_{\bullet}$
vanishes on $\mathfrak{S}_{n}^{n-1}\sqcup\mathfrak{S}_{n}^{n}$.

Since $h(n-1)=n$, each vertex $w\in\mathfrak{S}_{n}^{n-1}$ is connected to the
vertex $w\cdot(n,n-1)$ of $\mathfrak{S}_{n}^{n}$ by an edge of the labeled
graph $\Gamma(h)$. Since $z=0$ on $\mathfrak{S}_{n}^{n}$, $z(w)$ must be
divisible by the label $t_{w(n)}-t_{w(n-1)}=t_{w(n)}-t_{n}$ on the edge.
Therefore, there is a homogeneous element
$g(w)\in\mathbb{Z}[t_{1},\dots,t_{n}]$ of degree $2(p-1)$ such that

(6.9) $z(w)=(t_{w(n)}-t_{n})g(w)\quad\text{for }w\in\mathfrak{S}_{n}^{n-1}.$

We express

(6.10) $g(w)=\sum_{\ell=0}^{p-1}g_{\ell}(w)t_{n}^{\ell}$

with homogeneous polynomial $g_{\ell}(w)$ in $\mathbb{Z}[t_{1},\dots,t_{n-1}]$
of degree $2(p-1-\ell)$.

Claim. If $v,w\in\mathfrak{S}_{n}^{n-1}$ are joined by an edge of the labeled
graph $\Gamma(h^{n-1})$, i.e. $v=w\cdot(i,j)$ for some transposition $(i,j)$
with $j<i\leq h(j)$, $j\not=n-1$ and $i\not=n-1$, then

$g_{\ell}(v)\equiv g_{\ell}(w)\mod{t_{w(i)}-t_{w(j)}}.$

Proof of Claim. Since $z$ is an element of
$H^{*}_{T}(\operatorname{Hess}(S,h))$, it satisfies the congruence relation

(6.11) $z(v)\equiv z(w)\mod{t_{w(i)}-t_{w(j)}}.$

Since $v=w\cdot(i,j)$, we have $v(i)=w(j)$, $v(j)=w(i)$ and $v(s)=w(s)$ for
$s\not=i,j$. Moreover, $w(i)$ and $w(j)$ are not equal to $n$ because $i$ and
$j$ are not equal to $n-1$ and $w\in\mathfrak{S}_{n}^{n-1}$. Therefore,

$t_{v(n)}-t_{n}\equiv t_{w(n)}-t_{n}\not\equiv 0\mod{t_{w(i)}-t_{w(j)}}.$

This together with (6.9), (6.10), and (6.11) implies the congruence relation
in the claim.

By the claim above, each $g_{\ell}$ is an element of
$H^{*}_{T}(\operatorname{Hess}(S^{\prime},h^{n-1}))$ by Proposition 2.2. Since
$\ell\geq 0$ and $p\leq d$, we have

(6.12) $\text{the degree of $g_{\ell}$}=2(p-1-\ell)\leq 2(d-1).$

We note that $h(n-d-1)=n$ or $n-1$ by the assumption $h(j)\geq j+d$ for any
$j\in[n-d]$. Now we take two cases according to the value of $h(n-d-1)$.

Case 1. The case where $h(n-d-1)=n$. In this case, $h^{n-1}(j)\geq j+d$ for
any $j\in[n-d-1]$. Therefore, by the induction assumption and (6.12), any
$g_{\ell}$ can be written as a polynomial in $t_{i}$, $x_{i}^{(n-1)}$
$(i\in[n-1])$, where for $w\in\mathfrak{S}_{n}^{n-1}$,

$x_{i}^{(n-1)}(w)=\begin{cases}x_{i}(w)\quad&(i\leq n-2)\\\
x_{n}(w)\quad&(i=n-1).\end{cases}$

This shows that there is a polynomial $G_{\ell}$ in $t_{i}$, $x_{i}$’s whose
restriction to $\mathfrak{S}_{n}^{n-1}$ agrees with $g_{\ell}$. Therefore, it
follows from (6.9) that

$z=(x_{n}-t_{n})\sum_{\ell=0}^{p-1}G_{\ell}t_{n}^{\ell}\quad\text{on
}\mathfrak{S}_{n}^{n-1}.$

Both sides above vanish on $\mathfrak{S}_{n}^{n}$, so they agree on
$\mathfrak{S}_{n}^{n-1}\sqcup\mathfrak{S}_{n}^{n}$. Therefore, subtracting the
right hand side above from $z$, we may assume $z=0$ on
$\mathfrak{S}_{n}^{n-1}\sqcup\mathfrak{S}_{n}^{n}$.

Case 2. The case where $h(n-d-1)=n-1$. In this case, $h^{n-1}(j)\geq j+(d-1)$
for any $j\in[n-1-(d-1)]$ and $\Lambda_{d-1}(h^{n-1})=\\{n-d-1\\}$. Therefore,
by the induction assumption and (6.12), any $g_{\ell}$ can be written as a
polynomial in $t_{i},x_{i}^{(n-1)}$ and $y^{(n-1)}_{n-d-1,k}$, where
$k\in[n-1]$ and

$y_{n-d-1,k}^{(n-1)}(w)=y_{n-d-1,k}(w)/(t_{k}-t_{w(n-1)})\quad\text{ for
$w\in\mathfrak{S}_{n}^{n-1}$}.$

By Lemma 6.6, we may use $y_{n-1,k}^{*(n-1)}$ instead of
$y_{n-d-1,k}^{(n-1)}$, where

$y_{n-1,k}^{*(n-1)}=y^{*}_{n,k}/(t_{k}-t_{w(n-1)}).$

We note that since

$\deg g_{\ell}=2(p-1-\ell),\quad\deg y_{n-1,k}^{*(n-1)}=2(d-1),\quad p\leq d,$

$y_{n-1,k}^{*(n-1)}$ does not appear in the polynomial expression of
$g_{\ell}$ unless $p=d$ and $\ell=0$. Therefore, it follows from (6.9) and
(6.10) that we can write

(6.13)
$\begin{split}z(w)&=(t_{w(n)}-t_{n})\left(\sum_{k=1}^{n-1}c_{k}y_{n-1,k}^{*(n-1)}(w)+\sum_{\ell=0}^{p-1}f_{\ell}(w)t_{n}^{\ell}\right)\\\
&=\sum_{k=1}^{n-1}c_{k}(t_{w(n)}-t_{n})y_{n-1,k}^{*(n-1)}(w)+(t_{w(n)}-t_{n})\sum_{\ell=0}^{p-1}f_{\ell}(w)t_{n}^{\ell}\end{split}$

where $c_{k}\in\mathbb{Z}$ and $f_{\ell}$ is a polynomial in
$t_{i},x_{i}^{(n-1)}$’s $(i\in[n-1])$.

Similarly to Case 1, $f_{\ell}\in
H^{*}(\operatorname{Hess}(S^{\prime},h^{n-1}))$ in (6.13) is the image of some
$F_{\ell}\in H^{*}(\operatorname{Hess}(S,h))$. Although $y_{n-1,k}^{*(n-1)}$
may not be in the image of the restriction map, we have

$\displaystyle(t_{w(n)}-t_{n})y_{n-1,k}^{*(n-1)}(w)$
$\displaystyle=\begin{cases}\prod_{\ell=h^{*}(n)}^{n}(t_{k}-t_{w(\ell)})\quad&(k=w(n))\\\
0\quad&(\text{otherwise})\end{cases}$ $\displaystyle=y^{*}_{n,k}(w)\quad$

for $w\in\mathfrak{S}_{n}^{n-1}$ (so $t_{n}=t_{w(n-1)}$), where $h^{*}(n)=n-d$
because $(n-d-1)=n-1$ and $h(j)=n$ for $j\geq n-d$. This observation and
(6.13) show that

(6.14)
$z=\sum_{k=1}^{n-1}c_{k}y^{*}_{n,k}+(x_{n}-t_{n})\sum_{\ell=0}^{p-1}F_{\ell}t_{n}^{\ell}\quad\text{on
$\mathfrak{S}_{n}^{n-1}$}.$

Here, $z=0$ on $\mathfrak{S}_{n}^{n}$ by assumption and the right hand side
above also vanishes on $\mathfrak{S}_{n}^{n}$. Indeed, since $w(n)=n$ for
$w\in\mathfrak{S}_{n}^{n}$, we have $y^{*}_{n,k}(w)=0$ for $k\not=n$ and
$(x_{n}-t_{n})(w)=t_{w(n)}-t_{n}=0$. Thus, the identity (6.14) holds on
$\mathfrak{S}_{n}^{n-1}\sqcup\mathfrak{S}_{n}^{n}$. This together with Lemm
6.6 shows that $z$ minus an appropriate polynomial in
$t_{\bullet},x_{\bullet},y_{\bullet}$ vanishes on
$\mathfrak{S}_{n}^{n-1}\sqcup\mathfrak{S}_{n}^{n}$.

Step 3. Suppose that $z=0$ on
$\mathfrak{S}_{n}^{r+1}\sqcup\cdots\sqcup\mathfrak{S}_{n}^{n}$ for some $r$
with $n-d\leq r\leq n-2$. Then, since $h(r)=n$, there are shaded boxes at
positions $(r+1,r),(r+2,r),\dots,(n,r)$ in the configuration associated to
$h$. This means that each vertex $w\in\mathfrak{S}_{n}^{r}$ is connected to
$\mathfrak{S}_{n}^{\ell}$ for $r+1\leq\ell\leq n$ by an edge with label
$t_{w(\ell)}-t_{w(r)}=t_{w(\ell)}-t_{n}$. Since $z=0$ on
$\mathfrak{S}_{n}^{r+1}\sqcup\cdots\sqcup\mathfrak{S}_{n}^{n}$, $z(w)$ for
$w\in\mathfrak{S}_{n}^{r}$ must be divisible by
$\prod_{\ell=r+1}^{n}(t_{w(\ell)}-t_{n})$. Therefore, there is a homogeneous
element $g(w)\in\mathbb{Z}[t_{1},\dots,t_{n}]$ such that

(6.15)
$z(w)=\left(\prod_{\ell=r+1}^{n}(t_{w(\ell)}-t_{n})\right)g(w)\quad\text{for
$w\in\mathfrak{S}_{n}^{r}$}.$

The same argument as in the claim in Step 2 shows that the $g$ in (6.15)
satisfies the congruence relation for the labeled graph $\Gamma(h^{r})$. Since
$p\leq d$ and $n-r\geq 2$, we have

$\text{the degree of $g$}=2(p-n+r)\leq 2(d-2).$

Moreover, $h^{r}(j)\geq j+(d-1)$ for any $j\in[n-1-(d-1)]$. Therefore, by the
induction assumption, $g$ can be expressed as a polynomial in $t_{i}$ and
$x_{i}^{(r)}$ where

$x_{i}^{(r)}(w):=\begin{cases}x_{i}(w)\quad&(i<r)\\\ x_{i+1}(w)\quad&(r\leq
i)\end{cases}$

for $w\in\mathfrak{S}_{n}^{r}$. Since $x_{i}^{(r)}$ is in the image of the
restriction map from $H^{*}(\operatorname{Hess}(S,h))$ to
$H^{*}(\operatorname{Hess}(S^{\prime},h))$, there is an element $G\in
H^{*}(\operatorname{Hess}(S,h))$ whose restriction to
$H^{*}(\operatorname{Hess}(S^{\prime},h^{r}))$ agrees with $g$. It follows
from (6.15) that

(6.16) $z=\left(\prod_{\ell=r+1}^{n}(x_{\ell}-t_{n})\right)G\quad\text{on
}\mathfrak{S}_{n}^{r}.$

Here, $z=0$ on $\mathfrak{S}_{n}^{r+1}\sqcup\cdots\sqcup\mathfrak{S}_{n}^{n}$
by assumption and the right hand side above also vanishes on
$\mathfrak{S}_{n}^{r+1}\sqcup\cdots\sqcup\mathfrak{S}_{n}^{n}$. Indeed, since
$x_{\ell}(w)=t_{w(\ell)}=t_{n}$ for $w\in\mathfrak{S}_{n}^{\ell}$,
$\prod_{\ell=r+1}^{n}(x_{\ell}-t_{n})(w)=0$ for
$w\in\mathfrak{S}_{n}^{r+1}\sqcup\cdots\sqcup\mathfrak{S}_{n}^{n}$. Thus, the
identity (6.16) holds on
$\mathfrak{S}_{n}^{r}\sqcup\cdots\sqcup\mathfrak{S}_{n}^{n}$, so that $z$
minus an appropriate polynomial in $t_{\bullet},x_{\bullet},y_{\bullet}$
vanishes on $\mathfrak{S}_{n}^{r}\sqcup\cdots\sqcup\mathfrak{S}_{n}^{n}$.

Step 4. Suppose that $z=0$ on
$\mathfrak{S}_{n}^{r+1}\sqcup\cdots\sqcup\mathfrak{S}_{n}^{n}$ for some $r$
with $1\leq r\leq n-d-1$. Then, similarly to Step 3, $z(w)$ for
$w\in\mathfrak{S}_{n}^{r}$ must be divisible by
$\prod_{\ell=r+1}^{h(r)}(t_{n}-t_{w(\ell)})$. Here, the degree of $z$ is
$2p\leq 2d$ and $h(r)-r\geq d$, so $z(w)=0$ unless $p=d$ and $h(r)=r+d$. When
$p=d$ and $h(r)=r+d$, we have

(6.17)
$z(w)=c\left(\prod_{\ell=r+1}^{r+d}(t_{n}-t_{w(\ell)})\right)\quad\text{for
}w\in\mathfrak{S}_{n}^{r},$

where $c\in\mathbb{Z}$.

Since $r\leq n-d-1$ by assumption, $h(r)=r+d<n$. Therefore,
$r\in\Lambda_{d}(h)$ so that we have an element $y_{r,k}\in
H^{2d}_{T}(\operatorname{Hess}(S,h))$ for any $k\in[n]$. We take $k=n$. Since
$h(r)=r+d$, we have

$y_{r,n}(w)=\begin{cases}\prod_{\ell=r+1}^{r+d}(t_{n}-t_{w(\ell)})\quad&(n\in\\{w(1),\dots,w(r)\\})\\\
0\quad&(\text{otherwise})\end{cases}$

by definition. This together with (6.17) shows that

$z=cy_{r,n}\quad\text{on }\mathfrak{S}_{n}^{r}.$

Here, $z=0$ on $\mathfrak{S}_{n}^{r+1}\sqcup\cdots\sqcup\mathfrak{S}_{n}^{n}$
by assumption and the right hand side above also vanishes because
$y_{r,n}(w)=0$ if $n\in\\{w(r+1),\dots,w(n)\\}$ by definition. Therefore, $z$
minus $cy_{r,n}$ vanishes on
$\mathfrak{S}_{n}^{r}\sqcup\cdots\sqcup\mathfrak{S}_{n}^{n}$.

This completes the induction step and the lemma has been proven. $\Box$

###  6.4. Proof of Theorem 6.1

Under these preparations, we prove Theorem 6.1. When $p<d$, the restriction
map

$\iota^{*}\colon H^{2p}(\operatorname{Fl}(n))\to
H^{2p}(\operatorname{Hess}(S,h))$

is surjective by Lemma 6.7 and indeed an isomorphism by Lemma 6.4 (or Lemma
6.5).

When $p=d$, we consider the homomorphism

$\Phi\colon\iota^{*}(H^{2d}(\operatorname{Fl}(n)))\oplus\mathbb{Z}\langle
Y_{j,k}\mid j\in\Lambda_{d}(h),k\in[n]\rangle\to
H^{2d}(\operatorname{Hess}(S,h))$

sending $Y_{j,k}$ to $y_{j,k}$. The map $\Phi$ is surjective by Lemma 6.7 and
$\sum_{k=1}^{n}Y_{j,k}-y_{j}$ for $j\in\Lambda_{d}(h)$ are in the kernel of
$\Phi$ by (6.3). Therefore, the map $\Phi$ induces a surjective homomorphism

$\displaystyle\bar{\Phi}\colon\Big{(}\iota^{*}$
$\displaystyle(H^{2d}(\operatorname{Fl}(n)))\oplus\mathbb{Z}\langle
Y_{j,k}\mid
j\in\Lambda_{d}(h),k\in[n]\rangle\Big{)}/\mathbb{Z}\langle\sum_{k=1}^{n}Y_{j,k}-y_{j}\rangle$
$\displaystyle\to H^{2d}(\operatorname{Hess}(S,h)).$

Here, the rank of the source module is at most the rank of the target module
by Lemma 6.5. Moreover, one can easily see that the source module is torsion
free and we know that $H^{*}(\operatorname{Hess}(S,h))$ is also torsion free.
Thus, the surjective homomorphism $\bar{\Phi}$ must be an isomorphism, proving
the theorem.

Acknowledgement. The work of first and second authors was done within the
framework of the HSE University Basic Research Program.

## References

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# On a boundary property of Blaschke products1112010 Mathematics Subject
Classification: Primary 30J10, 30H05, 30H10. Key words: Blaschke products,
bounded analytic function, Fatou’s theorem, radial limit.

Arthur A. Danielyan and Spyros Pasias

###### Abstract

A Blaschke product has no radial limits on a subset $E$ of the unit circle $T$
but has unrestricted limit at each point of $T\setminus E$ if and only if $E$
is a closed set of measure zero.

##  1 Introduction.

Below $\Delta$ and $T$ denote the open unit disk and the unit circle in the
complex plane, respectively. Let $m$ be the Lebesgue measure on $T$. The
Blaschke products are forming an important subclass of the well-known space
$H^{\infty}$ of all bounded analytic functions in $\Delta$. In this paper we
consider a question on the boundary behavior of Blashke products.

Let $f$ be a function defined in $\Delta$ and let $z_{0}\in T$. Except the
radial and non-tangential (angular) limits of $f$ at $z_{0}$, we also consider
the unrestricted limit of $f$ at $z_{0}$. Recall that the unrestricted limit
of $f$ at $z_{0}$ is the limit of $f$ when $z\in\Delta$ approaches to $z_{0}$
arbitrarily, if the limit exists.

By a known theorem of Lindelöf, if $f\in H^{\infty}$ then the existence of the
radial limit at $z_{0}$ implies the existence of the non-tangential limit at
the same point. But, of course, the unrestricted limit of $f$ at $z_{0}$ does
not have to exist if just the radial limit exists at $z_{0}$.

The following theorem of Fatou is fundamental in the theory of boundary
behavior of analytic functions.

Theorem A. Let $f\in H^{\infty}$. Then the radial limits of $f$ exist on $T$
except perhaps for a subset $E$ of measure zero.

For Blaschke products, F. Riesz has proved the following more precise (than
Theorem A) result.

Theorem B. The moduli of the radial limits of a Blashke product $B$ are equal
to 1 a.e. on $T$.

It is well known that the exceptional set $E$ in Theorem A is a
$G_{\delta\sigma}$ set. But if we assume that the function $f$ has
unrestricted limits at the points of the set $T\setminus E$, then $E$
necessarily becomes an $F_{\sigma}$ set (in fact this statement is obvious; cf
[4]). The converse statement was proved in [4] on the base of the method of
[3], and the result can be formulated as follows.

Theorem C. Let $E$ be a set on $T$. Then there exists a function $f\in
H^{\infty}$ which has no radial limits on $E$ but has unrestricted limit at
each point of $T\setminus E$ if and only if $E$ is an $F_{\sigma}$ set of
measure zero.

Note that the proof of Theorem C is elementary and its sufficiency part
implies a well-known theorem of Lohwater and Piranian [6] as an obvious
corollary.

The purpose of the present paper is proving the analogues of Theorem C for
Blaschke products. The result is the following:

Theorem 1. Let $E$ be a set on $T$. Then there exists a Blaschke product which
has no radial limits on $E$ but has unrestricted limit at each point of
$T\setminus E$ if and only if $E$ is a closed set of measure zero.

The proof of Theorem 1 uses in particular Theorem C and some results on the
boundary behavior of Blaschke products due to R. D. Berman [1] and A. Nicolau
[7].

##  2 Some auxiliary results.

We follow the presentation of Nicolau [7] to formulate the results of Berman
[1] and Nicolau [7], respectively. The next theorem is due to Berman (cf. [7],
p. 250).

Theorem D. Let E be a subset of the unit circle of zero Lebesgue measure and
of type $F_{\sigma}$ and $G_{\delta}$. Then there exist Blaschke products
$B_{0}$ and $B_{1}$ such that:

(i) $B_{0}$ extends analytically to $T\setminus\overline{E}$ and
$\lim_{r\rightarrow 1}B_{0}(re^{it})=0$ if and only if $e^{it}\in E$;

(ii) $\lim_{r\rightarrow 1}B_{1}(re^{it})=1$ if and only if $e^{it}\in E$.

Now we formulate a theorem of Nicolau (see [7], Proposition on p. 251).

Theorem E. Let E be a subset of the unit circle. Assume that there exist a
Blaschke products $B_{0}$ that extends analytically to
$T\setminus\overline{E}$ with $\lim_{r\rightarrow 1}B_{0}(re^{it})=0$ for
$e^{it}\in E$, and an analytic function $f_{1}$ in the unit ball of
$H^{\infty}$, $f_{1}\not\equiv 1,$ such that $\lim_{r\rightarrow
1}f_{1}(re^{it})=1$ for $e^{it}\in E$. Then for each analytic function $g$ in
the unit ball of $H^{\infty}$, there exists a Blaschke product $I$ that
extends analytically to $T\setminus\overline{E}$, such that

$\lim_{r\rightarrow 1}[I(re^{it})-g(re^{it})]=0\ \ {\textit{f}or}\ \ e^{it}\in
E.$

In this result, of course, the set $E$ in fact is of Lebesgue measure zero
since if $m(E)>0$, then already the condition $\lim_{r\rightarrow
1}B_{0}(re^{it})=0$ for $e^{it}\in E$ will imply that $B_{0}\equiv 0$, which
is impossible as $B_{0}$ is a Blaschke product.

As noted in [7], p. 251, Theorem D shows that the hypothesis of Theorem E are
satisfied if the set $E$ is a subset of $T$ of Lebesgue measure zero and of
type $F_{\sigma}$ and $G_{\delta}$.

##  3 Proof of Theorem 1.

Necessity. Assume that there exists a Blaschke product $B$ such that it has
unrestricted limits at each point of $T\setminus E$ and does not have radial
limits at the points of $E$. First of all, Theorem A implies that $m(E)=0$. To
prove that $E$ is closed, let $\\{a_{n}\\}\subset\Delta$ be the set (sequence)
of zeros of $B$. We complete the proof in two steps (propositions).

Step 1. If the unrestricted limit of $B$ at $z_{0}\in T$ exists and is equal
to $d$, then $|d|=1$.

Indeed, by Theorem B in each neighborhood of $z_{0}$ on $T$ there exists a
point $w$ at which the radial limit $B(w)$ of $B$ exists and has modulus $1$.
This implies that $|d|=1$.

Step 2. The set $T\setminus E$ is open.

Indeed, if $z_{0}\in T\setminus E$, then $z_{0}\notin\overline{\\{a_{n}\\}}$
because otherwise the unrestricted limit of $B$ at $z_{0}$ would be equal to
$0$, which contradicts to Step 1. Thus, there exists an open interval
$I\subset T$ such that $z_{0}\in I$ and $I$ is disjoint of the (closed) set
$\overline{\\{a_{n}\\}}$. Then by a known theorem (see [5], p. 68) the
Blaschke product $B$ is analytic at the points of $I$. Thus $B$ is continuous
on $I$ and so it possesses unrestricted limits at the points of $I$. This
means that $I\subset T\setminus E$ and thus $T\setminus E$ is open. Hence $E$
is closed. The proof of necessity is completed.

Sufficiency. Let $E$ be a closed set of measure zero on $T$. Since $E$ is
closed, it is of type $F_{\sigma}$ and $G_{\delta}$. Thus for $E$ the
hypothesis of Theorem D are satisfied and (by Theorem D) there exist Blaschke
products $B_{0}$ and $B_{1}$ with properties (i) and (ii) listed in Theorem D.

This means that the hypothesis of Theorem E indicated in the second sentence
of its formulation are satisfied. Note that the hypothesis that there exists
an analytic function $f_{1}$ in the unit ball of $H^{\infty}$,
$f_{1}\not\equiv 1,$ such that $\lim_{r\rightarrow 1}f_{1}(re^{it})=1$ for
$e^{it}\in E$, is satisfied even regardless of Theorem D as such $f_{1}$
exists already in the disc algebra by a theorem of Fatou (cf. [5], p. 81).

By the sufficiency part of Theorem C, there exists a function $g\in
H^{\infty}$ which has no radial limit at each point of $E$ but has
unrestricted limit at each point of $T\setminus E$ (obviously, $g$ can be
extended continuously on the open set $T\setminus E$). By dividing to a
constant if needed, we may assume that the function $g$ is in the unit ball of
$H^{\infty}$ and thus one can apply for it Theorem E. Then, by Theorem E,
there exists a Blaschke product $I$ that extends analytically to $T\setminus
E$, such that for $e^{it}\in E$ we have

$\lim_{r\rightarrow 1}[I(re^{it})-g(re^{it})]=0.$

Since $\lim_{r\rightarrow 1}g(re^{it})$ does not exist for $e^{it}\in E,$
$\lim_{r\rightarrow 1}I(re^{it})$ does not exist for $e^{it}\in E.$ Since $I$
extends analytically to $T\setminus E$, it has unrestricted limits on
$T\setminus E$, and the Blaschke product $I$ has the needed properties. This
completes the proof of the sufficiency.

The theorem is proved.

## References

  * [1] R. D. Berman, The sets of fixed radial limit value for inner functions, Illinois J. Math. (2) 29 (1985) 191-219
  * [2] E.F. Collingwood and A.J. Lohwater, The Theory of Cluster Sets, Cambridge University Press, Cambridge, 1966.
  * [3] A.A. Danielyan, A theorem of Lohwater and Piranian, Proc. of AMS, 144 (2016), 3919-3920.
  * [4] A.A. Danielyan, On Fatou’s theorem, Anal. Math. Phys. 10, (2020), Article number: 28.
  * [5] K. Hoffman, Banach Spaces of Analytic Functions, Prentice Hall, Englewood Cliffs, New Jersey, 1962.
  * [6] A.J. Lohwater and G. Piranian, The boundary behavior of functions analytic in a disk, Ann. Acad. Sci. Fenn., Series A, Mathematica, 239, 1957.
  * [7] A. Nicolau, Blaschke product with prescribed radial limits, Bull. London Math. Soc. 23 (1991) 249 - 255.

Arthur A. Danielyan

Department of Mathematics

and Statistics

University of South Florida

Tampa, Florida 33620

USA

e-mail<EMAIL_ADDRESS>

Spyros Pasias

Department of Mathematics

and Statistics

University of South Florida

Tampa, Florida 33620

USA

e-mail<EMAIL_ADDRESS>