#  $\mathrm{GL}_{n}$-structure and principal $\mathfrak{sl}_{2}$-triple on the
cohomology ring of complex Grassmannian

Nhok Tkhai Shon Ngo  V.N.Karazin Kharkiv National University
Svobody Square 4
Kharkiv
61022
Ukraine<EMAIL_ADDRESS>
###### Abstract.

In this note we describe the cohomology ring of the Grassmannian of $k$-planes
in $n$-dimensional complex vector space as an $\mathrm{GL}_{n}$-module. We
give explicit formulas for the operators of its principal
$\mathfrak{sl}_{2}$-triple. It is proved that one of these operators
corresponds to the shifted cohomology degree operator and the second operator
coincides with the Lefschetz map on cohomology (as in the hard Lefschetz
theorem). We check that the cohomology ring of the complex Grassmannian as a
$\mathrm{GL}_{n}$-representation is isomorphic to the $k$-th exterior power of
the standard $n$-dimensional representation.

##  1\. Introduction

###  1.1. Cohomology ring.

Fix positive integers $n$ and $k$ such that $1\leq k\leq n-1$. Consider the
corresponding complex Grassmannian $\operatorname{Gr}(k,n)$ of $k$-dimensional
subspaces in a $n$-dimensional complex vector space which we will denote as
$W$. We will consider the cohomology ring
$H^{*}(\operatorname{Gr}(k,n),\mathbb{C})$ of $\operatorname{Gr}(k,n)$ with
complex coefficients. Since all odd cohomology groups of the
$\operatorname{Gr}(k,n)$ are trivial its cohomology ring is commutative.

Recall that over $\operatorname{Gr}(k,n)$ there are three canonical vector
bundles: the _trivial bundle_ $T=\operatorname{Gr}(k,n)\times\mathbb{C}^{n}$
of rank $n$, the _universal subbundle_ $S$ of rank $k$, whose fiber at each
point $\Lambda$ is the $k$-plane $\Lambda$ itself and the _universal quotient
bundle_ $Q$ of rank $n-k$ defined as quotient $Q=S/T$.

Then, for the total Chern classes $c(S)$, $c(Q)$ and $c(T)$ we have the
following equalities:

(1.1) $\begin{split}&c(S)=c_{0}(S)+c_{1}(S)+\ldots+c_{k}(S),\\\
&c(Q)=c_{0}(Q)+c_{1}(Q)+\ldots+c_{n-k}(Q),\\\ &c(T)=c(S)c(Q)=1.\end{split}$

Note that here $c_{0}(S)=c_{0}(Q)=1$ and $c(T)=c_{0}(T)=1$. Moreover, it is
known that in terms of Chern classes of bundles $S$ and $Q$ the cohomology
ring $H^{*}(\operatorname{Gr}(k,n),\mathbb{C})$ can be described in the
following way (see [1, Proposition 23.2]).

######  Proposition 1.1.

As a ring, $H^{*}(\operatorname{Gr}(k,n),\mathbb{C})$ is isomorphic to the
quotient algebra

(1.2)
$\mathbb{C}[c_{1}(S),\ldots,c_{k}(S),c_{1}(Q),\ldots,c_{n-k}(Q)]/(c(S)c(Q)-1).$

Denote for brevity $p_{j}=c_{j}(S)$ for $1\leq j\leq k$ and $q_{l}=c_{l}(Q)$
for $1\leq l\leq n-k$. We also let $p_{0}=q_{0}=1$, $p_{j}=0$ if $j>k$ or
$j<0$ and $q_{l}=0$ if $l>n-k$ or $l<0$. Then, the proposition above states
that cohomology ring of the complex Grassmannian is isomorphic to the quotient
algebra $\mathcal{A}=\mathbb{C}[p_{1},\ldots,p_{k},q_{1},\ldots,q_{n-k}]/I$.
Here $I$ is the ideal in the polynomial algebra
$\mathbb{C}[p_{1},\ldots,p_{k},q_{1},\ldots,q_{n-k}]$ generated by elements
$R_{m}$, which are defined via the polynomial identity

(1.3)
$x^{n}+R_{1}x^{n-1}+\ldots+R_{n-1}x+R_{n}=(x^{k}+p_{1}x^{k-1}+\ldots+p_{k})(x^{n-k}+q_{1}x^{n-k-1}+\ldots+q_{n-k}),$

or explicitly as

$R_{m}=\sum_{r+s=m}p_{r}q_{s}\leavevmode\nobreak\
\text{for}\leavevmode\nobreak\ 1\leq m\leq n.$

######  Remark 1.1.

Cohomological degree on $H^{*}(\operatorname{Gr}(k,n),\mathbb{C})$ produces a
grading on algebra $\mathcal{A}$. Since the Chern classes $p_{j}=c_{j}(S)$ and
$q_{j}=c_{j}(Q)$ have cohomological degree $2j$ the degree of a monomial
$a=\prod_{j}p_{j}^{\alpha_{j}}\prod_{l}q_{l}^{\beta_{l}}$ in $\mathcal{A}$
equals

(1.4) $\deg(a)=2\sum_{j=1}^{k}j\alpha_{j}+2\sum_{l=1}^{n-k}l\beta_{l}.$

###  1.2. $\mathbf{GL_{n}(\mathbb{C})}$-structure.

According to the Geometric Satake isomorphism of [12] there is a canonical
isomorphism between the monoidal categories of irreducible representations of
a complex reductive group and of perverse sheaves on the affine Grassmanian
for the Langlands dual group. A refined version [12, 7.4 Corollary] identifies
the weight space decomposition of a maximal torus on the irreducible
representation with the decomposition of the intersection cohomology of the
matching affine Schubert cell in the affine Grassmannian into the classes of
certain Mirkovic-Vilonen cycles. In the case of $\mathrm{GL}_{n}$ and its
$k$-th fundamental representation on $\bigwedge^{k}\mathbb{C}^{n}$ the
corresponding affine Schubert cell is our classical Grassmanian
$\operatorname{Gr}(k,n)$. The Mirkovic-Vilonen cycles correspond to the
classical Schubert cells in $\operatorname{Gr}(k,n)$. [12, 7.4 Corollary] in
this case implies that the canonical isomorphism
$H^{*}(\operatorname{Gr}(k,n),\mathbb{C})\cong\bigwedge^{k}\mathbb{C}^{n}$
matches the cohomology classes of the classical Schubert cells with pure
$k$-forms in $\bigwedge^{k}\mathbb{C}^{n}$ as explained in Subsection 3.2.
Namely, this isomorphism sends the Schubert class $\sigma_{\lambda}\in
H^{*}(\operatorname{Gr}(k,n),\mathbb{C})$ to the basis element
$\bigwedge_{j=1}^{k}e_{\lambda_{j}+k-j}$ of $\bigwedge^{k}\mathbb{C}^{n}$,
where $\lambda$ is a partition of the length at most $k$ and whose parts are
at most $n-k$ and $\\{e_{j}\\}_{j=0}^{n-1}$ is a basis of $\mathbb{C}^{n}$. We
review this isomorphism in Section 3 of this paper.

In particular, we can regard the cohomology ring of the complex Grassmannian
as a $\mathrm{GL}_{n}$-representation.

######  Proposition 1.2.

The cohomology ring of the complex Grassmannian can be endowed with the
structure of a $\mathrm{GL}_{n}$-module and as a
$\mathrm{GL}_{n}$-representation it is isomorphic to the $k$-th exterior power
of the standard $n$-dimensional $\mathrm{GL}_{n}$-representation.

######  Remark 1.2.

Of course, in order to prove the proposition in this formulation we only need
to check that $H^{*}(\operatorname{Gr}(k,n),\mathbb{C})$ as a vector space has
the dimension $\binom{n}{k}$ (the dimension of the $k$-th exterior power of an
$n$-dimensional vector space). However, we will show that there are some
connections between the inner structure of the cohomology ring and the
$\mathrm{GL}_{n}$-structure defined in Subsection 3.2. Specifically, we will
show that the Dynkin’s grading (see Subsection 1.3 and Proposition 1.5) on the
weight spaces corresponds to the cohomological degree.

###  1.3. Dynkin’s grading and principal $\mathfrak{sl}_{2}$-triples.

Dynkin in his paper [3] considered irreducible representations of semisimple
Lie groups and proved several properties of their weight systems. In that
paper he also introduced what is now called ‘a principal
$\mathfrak{sl}_{2}$-triple’. In this paragraph, we will give a short overview
of Dynkin’s results from [3] which will be used in the sequel.

Let $\mathfrak{g}$ be a complex semisimple Lie algebra and let $\mathfrak{h}$
be its Cartan subalgebra. Denote the set of simple roots of the Lie algebra
$\mathfrak{g}$ by $S$ and the Killing form by $B(\cdot,\cdot)$.

######  Definition 1.3.

An _$\mathfrak{sl}_{2}$ -triple_ in the Lie algebra $\mathfrak{g}$ is a triple
$(e,f,h)$ of elements of $\mathfrak{g}$ which satisfy the following
commutation relations:

(1.5) $[h,e]=2e,\leavevmode\nobreak\ [h,f]=-2f,\leavevmode\nobreak\ [e,f]=h.$

In other words, elements $e$, $f$ and $h$ span a representation of
$\mathfrak{sl}_{2}$ in $\mathfrak{g}$.

Elements $e$ and $f$ are often referred to as _nilpotent_ while operator $h$
is often referred to as _semisimple_. These names are justified by the fact
that in every $\mathfrak{sl}_{2}$-triple as linear operators $e$ and $f$ are
nilpotent and operator $h$ is semisimple. The latter is one of consequences of
the representation theory of $\mathfrak{sl}_{2}$.

######  Definition 1.4.

An $\mathfrak{sl}_{2}$-triple is called _principal_ if its elements are
regular elements of the Lie algebra $\mathfrak{g}$, that is, their
centralizers has the least possible dimension (see [6, Appendix D, Def. D.2]).
The corresponding three-dimensional Lie subalgebra spanned by the elements of
the principal $\mathfrak{sl}_{2}$-triple is called _principal sublagebra_.

Let $\mathbb{V}_{\lambda}$ be the irreducible representation of $\mathfrak{g}$
with the highest weight $\lambda$ and
$\rho\colon\mathfrak{g}\to\mathfrak{gl}(\mathbb{V}_{\lambda})$ be the
corresponding homomorphism. Dynkin proved that every weight $\mu$ of the
representation $\mathbb{V}_{\lambda}$ can be represented as

(1.6) $\mu=\lambda-\alpha_{1}-\ldots-\alpha_{k},\leavevmode\nobreak\
\text{where}\leavevmode\nobreak\ \alpha_{1},\ldots,\alpha_{k}\in S.$

Moreover, for any such decomposition the number $k$ is the same. Following
Dynkin, we will say that weight $\mu$ is a weight of the level $k$ if the
equality (1.6) holds for some $\alpha_{1},\ldots,\alpha_{k}\in S$.

Dynkin constructed in $\mathfrak{g}$ three special elements $e$, $f$ and $h$,
where $h\in\mathfrak{h}$, which after multiplying by some constants satisfy
the following relations:

(1.7) $[h,e]=2e,\leavevmode\nobreak\ [h,f]=-2f,\leavevmode\nobreak\ [e,f]=h$

and also

(1.8) $\alpha(h)=2\leavevmode\nobreak\ \text{for all}\leavevmode\nobreak\
\alpha\in S.$

The explicit construction of $e$, $f$ and $h$ is discussed in more detail in
the Section 2 of this note. Note that triple $\\{e,f,h\\}$ is not defined
uniquely by these conditions (see Section 2 and Remark 2.1).

In particular, triple $\\{e,f,h\\}$ spans a three-dimensional subalgebra of
$\mathfrak{g}$ and form a principal $\mathfrak{sl}_{2}$-triple. Denote this
subalgebra by $\mathfrak{s}$ and note that $\mathfrak{s}$ is isomorphic to
$\mathfrak{sl}_{2}$. It is clear that $\mathbb{V}_{\lambda}$ can be also
considered as a representation of the Lie algebra $\mathfrak{s}$.

The connection between weight levels and triple $\\{e,f,h\\}$ is described by
the following observation: the level $k$ of a weight $\mu$ in
$\mathbb{V}_{\lambda}$ can be computed as

(1.9) $k=\frac{\lambda(h)-\mu(h)}{2}.$

In other words, the element $h$ produces a certain grading on the set of
weights of $\mathbb{V}_{\lambda}$. Moreover, it turns out that this grading
corresponds exactly to the weight decomposition of $\mathbb{V}_{\lambda}$ as
an $\mathfrak{s}$-representation. Namely, consider decomposition of
$\mathbb{V}_{\lambda}$ into eigenspaces of $\rho(h)$:

(1.10) $\mathbb{V}_{\lambda}=\bigoplus_{k=0}^{d}\mathbb{V}_{\lambda,d-2k},$

where $\mathbb{V}_{\lambda,r}$ is the $r$-eigenspace of $\rho(h)$, i.e.

$\mathbb{V}_{\lambda,r}=\ker(\rho(h)-r\cdot\operatorname{id}).$

Dynkin proved that $d=\lambda(h)$ and that $\mathbb{V}_{\lambda,d-2k}$ is a
direct sum of the $\mathfrak{g}$-weight spaces of $\mathbb{V}_{\lambda}$
corresponding to the weights of level $k$.

Besides that, it follows from the representation theory of $\mathfrak{sl}_{2}$
that eigenspaces $\mathbb{V}_{\lambda,2j-d}$ and $\mathbb{V}_{\lambda,d-2j}$
has the same dimension and moreover, the map
$\rho(e)^{2(d-2j)}\colon\mathbb{V}_{\lambda,2j-d}\to\mathbb{V}_{\lambda,d-2j}$
produces a linear isomorphism between these vector spaces. Similarly, the map
$\rho(f)^{2(d-2j)}\colon\mathbb{V}_{\lambda,d-2j}\to\mathbb{V}_{\lambda,2j-d}$
is also a linear isomorphism. We will observe similar effects on
$H^{*}(\operatorname{Gr}(k,n),\mathbb{C})$ which are the consequences of the
hard Lefschetz theorem (see Remark 1.3).

###  1.4. Main results.

Proposition 1.2 states that $H^{*}(\operatorname{Gr}(k,n),\mathbb{C})$ is
isomorphic as a representation of $\mathrm{GL}_{n}$ to its $k$-th fundamental
representation $\bigwedge^{k}\mathbb{C}^{n}$. Recall that any representation
of a Lie group produces a representation of the corresponding Lie algebra.
Namely, if $G$ is a Lie algebra, $\mathfrak{g}$ is the corresponding Lie
algebra and $\rho\colon G\to\mathrm{GL}(V)$ is a representation of $G$, then
differential $d\rho\colon\mathfrak{g}\to\mathfrak{gl}(V)$ is a representation
of $\mathfrak{g}$. In particular, Proposition 1.2 implies there is an action
of the Lie algebra $\mathfrak{gl}_{n}(\mathbb{C})$ on the cohomology ring
$H^{*}(\operatorname{Gr}(k,n),\mathbb{C})$ of the Grassmannian. Thus, we can
also regard $H^{*}(\operatorname{Gr}(k,n),\mathbb{C})$ as an
$\mathfrak{sl}_{n}(\mathbb{C})$-representation.

Let $\\{e,f,h\\}$ be the Dynkin’s principal $\mathfrak{sl}_{2}$-triple for
$\mathfrak{sl}_{n}(\mathbb{C})$ defined as in Subsection 2.3. Denote the
operators which correspond to the actions of $e$, $f$ and $h$ on cohomology by
$E$, $F$ and $H$, respectively. It turns out that the actions $H$ and $F$ on
the cohomology ring have clear geometrical meaning.

######  Proposition 1.5.

The actions of the elements $h$ and $f$ of the Dynkin’s principal
$\mathfrak{sl}_{2}$-triple of $\mathfrak{sl}_{n}(\mathbb{C})$ correspond to
the operators $H$ and $F$ on the cohomology ring
$H^{*}(\operatorname{Gr}(k,n),\mathbb{C})$ which act as follows:

(1.11) $\begin{split}H([\omega])&=(k(n-k)-2j)\cdot[\omega],\\\
F([\omega])&=-c_{1}(S)\cdot[\omega]\end{split}$

for every cohomology class $[\omega]\in
H^{2j}(\operatorname{Gr}(k,n),\mathbb{C})$, where $0\leq j\leq k(n-k)$.
Moreover, Dynkin’s grading on $H^{*}(\operatorname{Gr}(k,n),\mathbb{C})$
considered as an $\mathfrak{sl}_{n}(\mathbb{C})$-representation corresponds to
the cohomological degree.

In other words, $H=k(n-k)\cdot\operatorname{id}-2\cdot\mathrm{Deg}$, where
$\mathrm{Deg}$ is the operator of multiplication by the cohomology degree and
$F$ is the operator of multiplication by the negative of the first Chern class
(i.e. by $-p_{1}$ in terms of algebra $\mathcal{A}$).

######  Remark 1.3.

Note that $H^{2}(\operatorname{Gr}(k,n),\mathbb{C})$ is one-dimensional, so it
is generated by the first Chern class $c_{1}(S)=-c_{1}(Q)$ or by
$p_{1}=-q_{1}$ in terms of $\mathcal{A}$. Thus, it follows from the hard
Lefschetz theorem for the complex Grassmannian that for each
$j\leq\frac{k(n-k)}{2}$ the operator $F^{k(n-k)-2j}$ produces an isomorphism
between $H^{j}(\operatorname{Gr}(k,n),\mathbb{C})$ and
$H^{k(n-k)-j}(\operatorname{Gr}(k,n),\mathbb{C})$. This exactly the same
effect that we have seen on weight spaces of the representations of the Lie
algebra $\mathfrak{s}$ (see Subsection 1.3).

One can ask a natural question: does the action of the operator $E$ have a
geometrical meaning? Unfortunately, it seems that the answer for this question
is unknown. However, it is possible to describe actions of $E$, $F$ and $H$ on
$H^{*}(\operatorname{Gr}(k,n),\mathbb{C})$ completely in terms of the quotient
algebra $\mathcal{A}=\mathbb{C}[p_{1},\ldots,p_{k},q_{1},\ldots,q_{n-k}]/I$
and its generators $p_{i},q_{j}$ (recall that $\mathcal{A}$ is isomorphic to
the $H^{*}(\operatorname{Gr}(k,n),\mathbb{C})$). The following proposition is
the main result of this note.

######  Proposition 1.6.

The actions of the operators $E$, $F$ and $H$ in terms of algebra
$\mathcal{A}$ can be described via the following differential operators:

(1.12) $\displaystyle H=$ $\displaystyle\leavevmode\nobreak\
-2\sum_{j=1}^{k}jp_{j}\frac{\partial}{\partial
p_{j}}-2\sum_{j=1}^{n-k}jq_{j}\frac{\partial}{\partial q_{j}}+k(n-k),$
$\displaystyle E=$
$\displaystyle-\sum_{j=1}^{k}(k-j+1)(n-k+j-1)p_{j-1}\frac{\partial}{\partial
p_{j}}+$
$\displaystyle+\sum_{j=1}^{n-k}(k+j-1)(n-k-j+1)q_{j-1}\frac{\partial}{\partial
q_{j}}+$
$\displaystyle+\sum_{j,l=1}^{k}\left(\sum_{d=0}^{j-1}(j+l-2d-1)p_{d}p_{j+l-d-1}\right)\frac{\partial^{2}}{\partial
p_{j}\partial p_{l}}-$
$\displaystyle-\sum_{j,l=1}^{n-k}\left(\sum_{d=0}^{j-1}(j+l-2d-1)q_{d}q_{j+l-d-1}\right)\frac{\partial^{2}}{\partial
q_{j}\partial q_{l}}+$
$\displaystyle+2\sum_{j=1}^{k}\sum_{l=1}^{n-k}(j+l-1)\left(\sum_{d=0}^{j-1}p_{d}q_{j+l-d-1}\right)\frac{\partial^{2}}{\partial
p_{j}\partial q_{l}},$ $\displaystyle F=$ $\displaystyle\leavevmode\nobreak\
-p_{1}.$

Here the elements of the algebra $\mathcal{A}$ are considered as the
corresponding scalar multiplication operators, i.e. $a\in\mathcal{A}$
correspond to the operator $L_{a}\colon\mathcal{A}\to\mathcal{A}$, where
$L_{a}(b)=a\cdot b$.

We prove propositions 1.5 and 1.6 in Section 4 of this note.

######  Remark 1.4.

In particular, differential operators given by these formulas are well defined
on the quotient $\mathbb{C}[p_{1},\ldots,p_{k},q_{1},\ldots,q_{n-k}]/I$, i.e.
as operators on the whole polynomial algebra
$\mathbb{C}[p_{1},\ldots,p_{k},q_{1},\ldots,q_{n-k}]$ they preserve the ideal
$I$. Moreover, these operators form an $\mathfrak{sl}_{2}$-triple.

###  1.5. Overview of the proofs of propositions 1.5 and 1.6.

On order to prove Proposition 1.5 we will use the explicit formulas for $e$,
$f$ and $h$ given in Subsection 2.3 and the isomorphism between
$\bigwedge^{k}\mathbb{C}^{n}$ and $H^{*}(\operatorname{Gr}(k,n),\mathbb{C})$.
After that, it is easy to check that actions of $H$ and $F$ on cohomology
correspond to the differential operators in Proposition 1.6. However, the
similar check for $E$ requires much more complicated computations involving
many tools from the theory of symmetric functions such as Pieri, Giambelli and
Jacobi-Trudi identities. This is because we do not have an analogue of
formulas (1.11) for the operator $E$. Initially, only the action of $E$ on
Schubert classes is known (see Proposition 3.2).

Instead of performing these computations, we first check that differential
operators in Proposition 1.6 are well defined on $\mathcal{A}$ and satisfy the
$\mathfrak{sl}_{2}$ relations. The proof is straight-forward and rather
computational. Since we have explicit formulas it remains to check some
algebraic identities.

Next, in order to prove that the action of the differential operator
corresponding to $E$ coincides with the action of $e$ on $\mathcal{A}$, we use
the Jacobson-Morozov theorem. The latter claims that any nonzero nilpotent
endomorphism of the semisimple Lie algebra $\mathfrak{g}$ can be extended to
an $\mathfrak{sl}_{2}$-triple. Moreover, such triple is unique up to
transformation from the centralizer of that nilpotent element, see [13,
Chapter 6, Proposition 2.1]. In particular, for given $f$ and $h$ the third
element $e$ of $\mathfrak{sl}_{2}$-triple, if exists, is defined uniquely.
Applying this fact to the Lie algebra of linear operators on $\mathcal{A}$, we
obtain the desired. Indeed, since $H$ and $F$ are already the images of
elements $h$ and $f$ of the Dynkin’s principal $\mathfrak{sl}_{2}$-triple and
$\\{E,F,H\\}$ satisfy $\mathfrak{sl}_{2}$ commutation relations, the operator
$E$ must coincide with the image of the element $e$.

One drawback of this proof is that the differential expression for $E$ is
appearing from nowhere. It raises the following interesting question: given a
linear operator on the quotient of the polynomial algebra, how can we get the
differential expression for it? We discuss the general approach for solving
such problems in Section 5 of this note.

###  1.6. Contents.

Structure of the paper is as follows. In Section 2 we review the general
construction of Dynkin’s principal $\mathfrak{sl}_{2}$-triples. In Section 3
we review the explicit isomorphism between the cohomology ring and the $k$-th
exterior power of $\mathbb{C}^{n}$ which is given by means of the Schubert
classes. Besides that, we also give actions of $\\{e,f,h\\}$ on the Schubert
classes. Section 4 is devoted to the proofs of propositions 1.6 and 1.5. In
Section 5 we discuss general approach for finding differential expression for
a given linear operator acting on polynomial algebra (or its quotient).
Finally, in Section 6 we give concluding remarks to this paper.

Acknowledgements. This work was done during the IST Austria Summer Program and
with the support of OeAD (Austria’s Agency for Education and
Internationalisation). The author is grateful to his supervisor Tamás Hausel
for the constant guidance and useful advice and also to Anton Mellit for
fruitful discussions.

##  2\. Construction of Dynkin’ principal $\mathfrak{sl}_{2}$-triples

In this section we discuss the construction of Dynkin’ principal
$\mathfrak{sl}_{2}$-triples. It will be useful to consider more general
context.

###  2.1. Notation and conventions.

Let $\mathfrak{g}$ be a semisimple Lie algebra. Denote by $\operatorname{ad}$
the adjoint map
$\operatorname{ad}(\cdot)(\cdot)\colon\mathfrak{g}\times\mathfrak{g}\to\mathfrak{g}$
which is defined as $\operatorname{ad}(X)(Y)=[X,Y]$.

Let $\mathfrak{h}$ be a Cartan subalgebra of the Lie algebra $\mathfrak{g}$.
Consider the corresponding root system and the decomposition

(2.1) $\mathfrak{g}=\mathfrak{h}\oplus\bigoplus_{\alpha\in
R}\mathfrak{g}_{\alpha},$

where $R$ is the set of roots of the Lie algebra $\mathfrak{g}$. Here
$\mathfrak{g}_{\alpha}$ is a root space corresponding to the root
$\alpha\in\mathfrak{h}^{*}$, i.e. for any $H\in\mathfrak{h}$ and
$X\in\mathfrak{g}_{\alpha}$ we have

(2.2) $\operatorname{ad}(H)(X)=\alpha(H)\cdot X,$

so the $\operatorname{ad}(H)$ acts diagonally on $\mathfrak{g}_{\alpha}$. In
fact, each subspace $\mathfrak{g}_{\alpha}$ is one-dimensional.

It is known that the set $R$ of roots is symmetric with respect to the origin
and one can choose an element $l\in(\mathfrak{h}^{*})^{*}$ such that $0\notin
l(R)$ and then decompose $R$ into two subsets: $R^{+}$ and $R^{-}$ depending
on the sign of $l(\alpha)$. These roots are called positive and negative,
respectively. We define then the set of _simple roots_ as the set of those
$\alpha\in R^{+}$ which cannot be represented as sum of two elements in
$R^{+}$. We will denote the set of simple roots as $S$.

Let $B(\cdot,\cdot)\colon\mathfrak{g}\times\mathfrak{g}\to\mathbb{C}$ be the
Killing form which is defined via the formula
$B(X,Y)=\operatorname{tr}(\operatorname{ad}(X)\circ\operatorname{ad}(Y))$. It
can be shown that $B(\cdot,\cdot)$ is a nondegenerate symmetric bilinear from
on the semisimple Lie algebra $\mathfrak{g}$ (and on $\mathfrak{h}$ in
particular) and, hence, the Killing form produces a natural isomorphism
between $\mathfrak{h}^{*}$ and $\mathfrak{h}$ given by the correspondence

$\alpha\in\mathfrak{h}^{*}\leftrightarrow T_{\alpha}\in\mathfrak{h},$

where $T_{\alpha}$ is defined by

(2.3) $B(T_{\alpha},H)=\alpha(H)\leavevmode\nobreak\ \text{for
all}\leavevmode\nobreak\ H\in\mathfrak{h}.$

Using this isomorphism we can define the Killing form on $\mathfrak{h}^{*}$
via formula

(2.4) $B(T_{\alpha},T_{\beta})=B(\alpha,\beta).$

###  2.2. Dynkin’s construction.

Now we can construct a principal $\mathfrak{sl}_{2}$-triple which satisfies
conditions (1.7) and (1.8). For each $\alpha\in S$ choose an
$X_{\alpha}\in\mathfrak{g}_{\alpha}$ and $Y_{\alpha}\in\mathfrak{g}_{-\alpha}$
such that

$B(X_{\alpha},Y_{\alpha})=1.$

Next, consider elements $e$, $f$, $h$ defined by the equalities

(2.5) $h=\sum_{\alpha\in S}p_{\alpha}T_{\alpha},\leavevmode\nobreak\
e=\sum_{\alpha\in S}u_{\alpha}X_{\alpha},\leavevmode\nobreak\
f=\sum_{\alpha\in S}v_{\alpha}Y_{\alpha}$

where complex numbers $u_{\alpha}$, $v_{\alpha}$ and $p_{\alpha}$ satisfy the
following conditions

(2.6) $\displaystyle B(h,T_{\alpha})=2\leavevmode\nobreak\ \text{for
all}\leavevmode\nobreak\ \alpha\in S$ (2.7) $\displaystyle
u_{\alpha}v_{\alpha}=p_{\alpha}\leavevmode\nobreak\ \text{for
all}\leavevmode\nobreak\ \alpha\in S.$

In other words, $u_{\alpha},v_{\alpha}$ and $p_{\alpha}$ satisfy the following
system

(2.8) $\begin{cases}\sum\limits_{\beta\in S}B(T_{\alpha},T_{\beta})\cdot
p_{\alpha}=2\\\ u_{\alpha}v_{\alpha}=p_{\alpha}\end{cases}$

for all $\alpha\in S$.

######  Remark 2.1.

It is clear that this principal $\mathfrak{sl}_{2}$-triple is not unique since
the only condition on $u_{\alpha}$ and $v_{\alpha}$ is that their product
equals $p_{\alpha}$. However, the element $h$ is defined uniquely by the
system (2.8) because the matrix
$\\{B(T_{\alpha},T_{\beta})\\}_{\alpha,\beta\in S}$ is nondegenerate.

Dynkin stated in his paper [3] the following fact:

######  Proposition 2.1.

Operators $E,F,H$ defined via (2.8) satisfy conditions (1.7) and (1.8). In
particular, for any finite-dimensional representation
$\rho\colon\mathfrak{g}\to\mathfrak{gl}(V)$ the triple
$\\{\rho(E),\rho(F),\rho(H)\\}$ is an $\mathfrak{sl}_{2}$-triple acting on
$V$.

###  2.3. Explicit formulas for the case
$\mathfrak{g}=\mathfrak{sl}_{n}(\mathbb{C})$.

We will use the standard matrix representation of $\mathfrak{sl}_{n}$. It
would be convenient for us to enumerate rows and columns by numbers from $0$
to $n-1$. For any $i,j\in\\{0,1,\ldots,n-1\\}$ let $E_{ij}$ be the $(i,j)$-th
matrix unit. It can be computed that elements of Dynkin’s triple in the case
$\mathfrak{g}=\mathfrak{sl}_{n}(\mathbb{C})$ are as follows

(2.9) $\displaystyle h=\sum_{j=0}^{n-1}(n-1-2j)E_{jj},$ $\displaystyle
e=\sum_{j=1}^{n-1}u_{j}\cdot E_{j-1j},$ $\displaystyle
f=\sum_{j=1}^{n-1}u_{j}^{-1}\cdot j(n-j)E_{jj-1},$

where $\\{u_{j}\\}_{j=1}^{n-1}$ are arbitrary nonzero scalars. From now on we
set $u_{j}=j(n-j)$ for all $j=\overline{1,n-1}$ in further results. In this
case we have

(2.10) $\displaystyle h=\sum_{j=0}^{n-1}(n-1-2j)E_{jj},$ $\displaystyle
e=\sum_{j=1}^{n-1}j(n-j)\cdot E_{j-1j},$ $\displaystyle
f=\sum_{j=1}^{n-1}E_{jj-1}.$

We will often refer to this particular triple as to the Dynkin’s principal
$\mathfrak{sl}_{2}$-triple of $\mathfrak{sl}_{n}(\mathbb{C})$.

######  Remark 2.2.

One can also choose $u_{j}=1$ instead of $u_{j}=j(n-j)$. This option is used
for example, in [2, Chapter 3, 3.7].

Now, using this formulas one can find the action of $\\{e,f,h\\}$ on the basis
of the $k$-th exterior power $\bigwedge^{k}\mathbb{C}^{n}$. Let
$V=\mathbb{C}^{n}$ be the standard $n$-dimensional representation of
$\mathfrak{sl}_{n}(\mathbb{C})$ with the basis $\\{e_{0},\ldots,e_{n-1}\\}$
and $\rho\colon\mathfrak{g}\to\mathfrak{gl}(\bigwedge^{k}V)$ be the
corresponding representation of $\mathfrak{g}$. Then, it easy to check using
(2.10) that for any basis vector $e_{j_{1}}\wedge\ldots\wedge e_{j_{k}}$ in
$\bigwedge^{k}V$ we have

(2.11) $\begin{split}&\rho(h)(e_{j_{1}}\wedge\ldots\wedge
e_{j_{k}})=\left(k(n-1)-2\sum_{l=1}^{k}j_{l}\right)\cdot
e_{j_{1}}\wedge\ldots\wedge e_{j_{k}},\\\ &\rho(e)(e_{j_{1}}\wedge\ldots\wedge
e_{j_{k}})=\sum_{l=1}^{k}j_{l}(n-j_{l})\cdot e_{j_{1}}\wedge\ldots\wedge
e_{j_{l-1}}\wedge e_{j_{l}+1}\wedge e_{j_{l+1}}\wedge\ldots\wedge
e_{j_{k}},\\\ &\rho(f)(e_{j_{1}}\wedge\ldots\wedge
e_{j_{k}})=\sum_{l=1}^{k}e_{j_{1}}\wedge\ldots\wedge e_{j_{l-1}}\wedge
e_{j_{l}+1}\wedge e_{j_{l+1}}\wedge\ldots\wedge e_{j_{k}}.\end{split}$

##  3\. $\mathrm{GL}_{n}$-structure on the cohomology ring

In this section we discuss the $\mathrm{GL}_{n}$-structure on the cohomology
ring of the complex Grassmannian.

###  3.1. Schubert classes.

The cohomology ring of the Grassmannian can be also described via the Schubert
calculus. We refer the reader to [4, Chapter 14, Sections 14.6 and 14.7], [5,
Chapter 9, Section 9.4] and [8, Sections 3 and 4] for more details.

Let us first recall some notions from the theory of symmetric functions. A
partition is a nonincreasing sequence of nonnegative integers
$\lambda=(\lambda_{1},\ldots,\lambda_{m})$. The size of the partition
$\lambda$, denoted $|\lambda|$, is $\sum_{j}\lambda_{j}$, the entries of
$\lambda$ are called its parts and the number of nonzero parts, denoted
$\ell(\lambda)$, is called the length of $\lambda$. The Young diagram of a
partition $\lambda$ is the left-aligned partial grid of boxes in which the
$i$-th row from the top has $\lambda_{i}$ boxes. Let $\mathcal{P}_{k,n}$ be
the set of partitions $\lambda$ of the length at most $k$ and whose parts are
at most $n-k$.

######  Remark 3.1.

In terms of the Young diagrams the set $\mathcal{P}_{k,n}$ consists of all
partitions $\lambda$, whose Young diagram can be embedded into the rectangle
$k\times(n-k)$.

Fix an arbitrary complete flag $\mathcal{F}=(F_{j})_{j=1}^{n}$ in the initial
$n$-dimensional vector space $W$:

(3.1) $0\subset F_{1}\subset\ldots\subset F_{n-1}\subset F_{n}=W.$

For each partition $\lambda\in\mathcal{P}_{k,n}$ the Schubert subvariety
$\Omega_{\lambda}$ associated to $\mathcal{F}$ is defined as

(3.2) $\Omega_{\lambda}=\\{U\in\operatorname{Gr}(k,n)\colon\dim(U\cap
F_{r})=j\leavevmode\nobreak\ \text{for}\leavevmode\nobreak\
n-k+j-\lambda_{j}\leq n-k+j-\lambda_{j+1}\\}.$

The corresponding Schubert class $\sigma_{\lambda}$ is defined as the
cohomology class dual to $\Omega_{\lambda}$.

The following proposition describes multiplication in the cohomology ring in
terms of Schubert classes.

######  Proposition 3.1.

Schubert classes $\\{\sigma_{\lambda}\\}_{\lambda\in\mathcal{P}_{k,n}}$ form
$\mathbb{C}$-basis in the cohomology ring
$H^{*}(\operatorname{Gr}(k,n),\mathbb{C})$. The ring structure is given by the
Littlewood-Richardson coefficients $c_{\lambda\mu}^{\nu}$:

(3.3)
$\sigma_{\lambda}\cdot\sigma_{\mu}=\sum_{\nu}c_{\lambda\mu}^{\nu}\cdot\sigma_{\nu}.$

Moreover, there is an isomorphism

(3.4)
$H^{*}(\operatorname{Gr}(k,n),\mathbb{C})\simeq\Lambda[x_{1},x_{2},\ldots]/(s_{\lambda}\mid\lambda\notin\mathcal{P}_{k,n}),$

where $\Lambda[x_{1},x_{2},\ldots]$ is the ring of the symmetric functions in
infinitely many variables; under this isomorphism the Schubert class
$\sigma_{\lambda}$ maps to the Schur function $s_{\lambda}$ which corresponds
to the partition $\lambda$.

In this description of the cohomology ring of the Grassmannian the Chern
classes correspond to the elementary and complete symmetric functions. Namely,
under the isomorphism mentioned in Proposition 3.1 we have the following
correspondence:

(3.5)
$\begin{split}&p_{j}\longleftrightarrow(-1)^{j}e_{j}=(-1)^{j}s_{(1,\ldots,1)},\\\
&q_{j}\longleftrightarrow h_{j}=s_{(j)},\end{split}$

where $e_{j}$ is the $j$-th elementary symmetric polynomial and $h_{j}$ is the
$j$-th complete symmetric polynomial (see [10] for more details).

###  3.2. Isomorphism with the $k$-th exterior power of $\mathbb{C}^{n}$.

Let $V=\mathbb{C}^{n}$ be an $n$-dimensional vector space with basis
$\\{e_{0},\ldots,e_{n-1}\\}$. Now, following [7, Section 1.3], let us assign
to each partition $\lambda=(\lambda_{1},\ldots,\lambda_{k})$, where
$\lambda\in\mathcal{P}_{k,n}$, the basis element
$\bigwedge\limits_{j=1}^{k}e_{\lambda_{j}+k-j}$ of the $k$-th exterior power
$\bigwedge^{k}V$. Note that while $\lambda$ runs over the set
$\mathcal{P}_{k,n}$, the sequence $(\lambda_{j}+k-j)_{j=1}^{k}$ runs over all
strictly decreasing sequences with elements from the set
$\\{0,1,\ldots,n-1\\}$.

Thus, we defined a linear isomorphism between
$H^{*}(\operatorname{Gr}(k,n),\mathbb{C})$ and $\bigwedge^{k}\mathbb{C}^{n}$,
so now we can endow the cohomology ring of $\operatorname{Gr}(k,n)$ with a
structure of a $\mathrm{GL}_{n}$-representation via this correspondence.

As was mentioned in Subsection 2.3, the elements of the Dynkin’s
$\mathfrak{sl}_{2}$-triple of $\mathfrak{sl}_{n}(\mathbb{C})$ act on basis
vectors of $\bigwedge^{k}V$ in the following way:

(3.6) $\begin{split}&\rho(h)(e_{j_{1}}\wedge\ldots\wedge
e_{j_{k}})=(k(n-1)-2(j_{1}+\ldots+j_{k}))\cdot e_{j_{1}}\wedge\ldots\wedge
e_{j_{k}},\\\ &\rho(e)(e_{j_{1}}\wedge\ldots\wedge
e_{j_{k}})=\sum_{l=1}^{k}j_{l}(n-j_{l})\cdot e_{j_{1}}\wedge\ldots\wedge
e_{j_{l-1}}\wedge e_{j_{l}-1}\wedge e_{j_{l+1}}\wedge\ldots\wedge
e_{j_{k}},\\\ &\rho(f)(e_{j_{1}}\wedge\ldots\wedge
e_{j_{k}})=\sum_{l=1}^{k}e_{j_{1}}\wedge\ldots\wedge e_{j_{l-1}}\wedge
e_{j_{l}+1}\wedge e_{j_{l+1}}\wedge\ldots\wedge e_{j_{k}}.\end{split}$

Here we define $e_{-1}=e_{n}=0$.

Now using the isomorphism between $\bigwedge^{k}V$ and
$H^{*}(\operatorname{Gr}(k,n),\mathbb{C})$ we can describe the action of the
triple $\\{E,F,H\\}$ on Schubert classes. The following proposition is an
immediate consequence of formulas (3.6) and the aforementioned isomorphism.

######  Proposition 3.2.

Operators $E$, $F$ and $H$ act on Schubert classes $\sigma_{\lambda}$ in the
following way: for every $\lambda\in\mathcal{P}_{k,n}$

(3.7)
$\begin{split}&H(\sigma_{\lambda})=(-2|\lambda|+k(n-k))\cdot\sigma_{\lambda},\\\
&E(\sigma_{\lambda})=\sum_{l}(k+\lambda_{l}-l)(n-k-\lambda_{l}+l)\cdot\sigma_{\lambda^{(l)}},\\\
&F(\sigma_{\lambda})=\sum_{\mu}\sigma_{\mu}.\end{split}$

Here $\mu$ runs over all partitions from $\mathcal{P}_{k,n}$ which can be
obtained from $\lambda$ by adding one box and $l$ runs over all indices from
$\\{1,2,\ldots,k\\}$ for which the sequence

$\lambda^{(l)}=(\lambda_{1},\ldots,\ldots,\lambda_{l-1},\lambda_{l}-1,\lambda_{l+1},\ldots,\lambda_{k})$

is a partition from $\mathcal{P}_{k,n}$.

######  Remark 3.2.

We can use this isomorphism in different way. Observe that we can endow the
$k$-th exterior power of the standard representation $V$ of $\mathrm{GL}_{n}$
(or $\mathrm{SL}_{n}$) with a certain ring structure. It turns out that there
is a class of irreducible representations of semisimple algebraic groups
called weight multiplicity free representations that also possess a similar
ring structure. Moreover, weight multiplicity free representations often arise
as cohomology rings of certain projective varieties. For instance, for a
minuscule representation the corresponding projective variety is a generalised
flag variety. We refer the reader to the paper [14] for more details. It can
be verified that our ring strucure on $\bigwedge^{k}V$ coincides with the one
defined in [14, Section 5 and Theorem 5.5].

##  4\. Proofs of Propositions 1.5 and 1.6

In this section we prove the propositions 1.5 and 1.6. We will need the
following simple computational lemma for both propositions (here
$[\cdot,\cdot]$ denotes the commutator and $\delta_{km}$ is the Kronecker
delta):

######  Lemma 4.1.

a) For all possible indices $j$, $l$, $r$ and $s$ the following commutation
relations hold

$\displaystyle\left[\frac{\partial}{\partial
p_{j}},p_{r}q_{s}\right]=\delta_{jr}q_{s},\leavevmode\nobreak\
\left[\frac{\partial^{2}}{\partial p_{j}\partial
p_{l}},p_{r}q_{s}\right]=\delta_{lr}q_{s}\frac{\partial}{\partial
p_{j}}+\delta_{jr}q_{s}\frac{\partial}{\partial p_{l}},$
$\displaystyle\left[\frac{\partial}{\partial
q_{j}},p_{r}q_{s}\right]=\delta_{js}p_{r},\leavevmode\nobreak\
\left[\frac{\partial^{2}}{\partial q_{j}\partial
q_{l}},p_{r}q_{s}\right]=\delta_{ls}p_{r}\frac{\partial}{\partial
q_{j}}+\delta_{js}p_{r}\frac{\partial}{\partial q_{l}},$
$\displaystyle\left[\frac{\partial^{2}}{\partial p_{j}\partial
q_{l}},p_{r}q_{s}\right]=\delta_{ls}p_{r}\frac{\partial}{\partial
p_{j}}+\delta_{jr}q_{s}\frac{\partial}{\partial
q_{l}}+\delta_{jr}\delta_{ls}.$

b) For any nonnegative integers $\alpha$ and $\beta$

$\left[x\frac{\partial}{\partial x},x^{\alpha}\frac{\partial^{\beta}}{\partial
x^{\beta}}\right]=(\alpha-\beta)\cdot
x^{\alpha}\frac{\partial^{\beta}}{\partial x^{\beta}}\leavevmode\nobreak\
\text{and}\leavevmode\nobreak\ \left[\frac{\partial^{\alpha}}{\partial
x^{\alpha}},x\right]=\alpha\cdot\frac{\partial^{\alpha-1}}{\partial
x^{\alpha-1}}.$

###### Proof of Proposition 1.5..

Proposition 3.2 gives the explicit formulas for the actions of the elements
$h$ and $f$ on the Schubert classes. Now we will simply check that the action
of operators from Proposition 1.5 on Schubert classes gives the same result as
in Proposition 3.2. Denote temporarily the operators from Proposition 1.5 as
$H^{\prime}$ and $F^{\prime}$. We need to prove that $H=H^{\prime}$ and
$F=F^{\prime}$. Since $\\{\sigma_{\lambda}\\}_{\lambda\in\mathcal{P}_{k,n}}$
form a basis of $H^{*}(\operatorname{Gr}(k,n),\mathbb{C})$ (see Proposition
3.1) it is enough to check that the actions of $H,H^{\prime}$ and
$F,F^{\prime}$ on $\sigma_{\lambda}$ are the same.

For $H$ it is sufficient to notice that according to Proposition 3.2 we have

(4.1) $H(\sigma_{\lambda})=(-2|\lambda|+k(n-k))\cdot\sigma_{\lambda}.$

Note that the Schubert class $\sigma_{\lambda}$ has cohomological degree
$2|\lambda|$ and hence we have
$H^{\prime}(\sigma_{\lambda})=H(\sigma_{\lambda})$ (see 1.5).

Similarly, the action of $F^{\prime}$ is the multiplication by $-p_{1}$ which
is the Schubert class of the partition $(1,0,\ldots,0)$. Hence,

(4.2)
$F^{\prime}(\sigma_{\lambda})=\sigma_{(1,0,\ldots,0)}\cdot\sigma_{\lambda}.$

Now note that for the corresponding Schur functions we have the following
identity which is a particular case of the Pieri’s rule (see [10, Chapter I,
5.16]):

(4.3) $s_{(1,0,\ldots,0)}\cdot s_{\lambda}=\sum_{\mu}s_{\mu},$

where $\mu$ runs over all partitions which can be obtained by adding one box
to $\lambda$. In view of Proposition 3.1 it means that

(4.4) $\sigma_{(1,0,\ldots,0)}\cdot\sigma_{\lambda}=\sum_{\mu}\sigma_{\mu},$

where $\mu$ runs over all partitions from $\mathcal{P}_{k,n}$ which can be
obtained by adding one box to $\lambda$. Thus,

(4.5)
$F^{\prime}(\sigma_{\lambda})=\sum_{\mu}\sigma_{\mu}=F(\sigma_{\lambda})$

for all $\lambda\in\mathcal{P}_{k,n}$ and consequently $F=F^{\prime}$.

Finally, to check that Dynkin’s grading on
$H^{*}(\operatorname{Gr}(k,n),\mathbb{C})$ corresponds to the cohomological
degree it suffices to observe that $H^{2j}(\operatorname{Gr}(k,n),\mathbb{C})$
is the $(k(n-k)-2j)$-eigenspace for the operator $H$. Therefore, the
decomposition
$H^{*}(\operatorname{Gr}(k,n),\mathbb{C})=\bigoplus_{j=0}^{k(n-k)}H^{2j}(\operatorname{Gr}(k,n),\mathbb{C})$
is the weight decomposition of $H^{*}(\operatorname{Gr}(k,n),\mathbb{C})$ as
an $\mathfrak{s}$-representation. ∎

######  Remark 4.1.

At first glance, it might seem that coefficients of
$\frac{\partial^{2}}{\partial p_{j}\partial p_{l}}$ and
$\frac{\partial^{2}}{\partial q_{j}\partial q_{l}}$ in the expression of $E$
(see (1.12)) are not symmetric with respect to $j$ and $l$. However, these
coefficients $\sum\limits_{d=0}^{j-1}(j+l-2d-1)p_{d}p_{j+l-d-1}$ and
$\sum\limits_{d=0}^{j-1}(j+l-2d-1)q_{d}q_{j+l-d-1}$ are indeed symmetric with
respect to $j$ and $l$ since

(4.6)
$\begin{split}\sum_{d=0}^{j-1}(j+l-2d-1)p_{d}p_{j+l-d-1}&=\sum_{d=0}^{j-1}(j+l-d-1)p_{d}p_{j+l-d-1}-\sum_{d=0}^{j-1}dp_{d}p_{j+l-d-1}=\\\
&=\sum_{d=0}^{l}dp_{d}p_{j+l-d-1}-\sum_{d=0}^{j-1}dp_{d}p_{j+l-d-1}=\\\
&=\sum_{d=0}^{j}dp_{d}p_{j+l-d-1}-\sum_{d=0}^{l}dp_{d}p_{j+l-d-1}=\\\
&=\sum_{d=0}^{l-1}(j+l-2d-1)p_{d}p_{j+l-d-1}.\end{split}$

In addition, we can make the expression
$\sum\limits_{d=0}^{j-1}p_{d}q_{j+l-d-1}$, which is the coefficient of
$\frac{\partial^{2}}{\partial p_{j}\partial q_{l}}$ in $E$, symmetric with
respect to the variables $p_{j}$ and $q_{l}$ in $\mathcal{A}$ since we have

(4.7)
$\begin{split}\sum_{d=0}^{j-1}p_{d}q_{j+l-d-1}&=\sum_{d=0}^{j+l-1}p_{d}q_{j+l-d-1}-\sum_{d=j}^{j+l-1}p_{d}q_{j+l-d-1}=\\\
&=R_{j+l-1}-\sum_{d=0}^{l-1}p_{d}q_{j+l-d-1}\equiv-\sum_{d=0}^{l-1}p_{d}q_{j+l-d-1}\pmod{I}.\end{split}$

Thus, the coefficient of $\frac{\partial^{2}}{\partial p_{j}\partial q_{l}}$
in $E$ can be rewritten as

$\sum_{d=0}^{j-1}p_{d}q_{j+l-d-1}\equiv\frac{1}{2}\left(\sum_{d=0}^{j-1}p_{d}q_{j+l-d-1}-\sum_{d=0}^{l-1}p_{d}q_{j+l-d-1}\right)\pmod{I}.$

We use this observations in the proof of Proposition 1.6.

###### Proof of Proposition 1.6.

Denote differential operators defined in Proposition 1.6 which correspond to
$E$, $F$ and $H$ as $D_{e}$, $D_{f}$ and $D_{h}$, respectively. Firstly, note
that $D_{e}$, $D_{f}$ and $D_{h}$ are linear operators on the polynomial
algebra $\mathbb{C}[p_{1},\ldots,p_{k},q_{1},\ldots,q_{n-k}]$. We will say
that a differential operator $D$ on
$\mathbb{C}[p_{1},\ldots,p_{k},q_{1},\ldots,q_{n-k}]$ is zero modulo $I$ if
all its coefficients are zero modulo $I$. In other words, $D$ is zero modulo
$I$ if it can be represented as linear combination

(4.8)
$\sum_{\alpha,\beta}a_{\alpha\beta}\cdot\prod_{j=1}^{k}\frac{\partial^{\alpha_{j}}}{\partial
p_{j}^{\alpha_{j}}}\prod_{l=1}^{n-k}\frac{\partial^{\beta_{l}}}{\partial
q_{l}^{\beta_{j}}},$

where $a_{\alpha\beta}\in I$ for all $\alpha=(\alpha_{1},\ldots,\alpha_{k})$
and $\beta=(\beta_{1},\ldots,\beta_{n-k})$. In particular, if $D$ is zero
modulo $I$, then its image is contained in $I$. The proof of Proposition 1.6
is divided into three steps.

Step 1. We prove that differential operators $D_{e}$, $D_{f}$ and $D_{h}$ are
well defined operators on the quotient
$\mathcal{A}=\mathbb{C}[p_{1},\ldots,p_{k},q_{1},\ldots,q_{n-k}]/I$.

In order to prove that $D_{e},D_{f},D_{h}$ are well defined on the quotient
$\mathbb{C}[p_{1},\ldots,p_{k},q_{1},\ldots,q_{n-k}]/I$ we need to check that
they preserve the ideal $I$. Clearly, for this it is sufficient to check that
for all $m$ the commutators $[D_{e},R_{m}]$, $[D_{f},R_{m}]$ and
$[D_{h},R_{m}]$ are zero modulo $I$ (here we consider $R_{m}$ as a scalar
multiplication operator). Indeed, in this case for any elements
$f_{m}\in\mathbb{C}[p_{1},\ldots,p_{k},q_{1},\ldots,q_{n-k}]$ we have

(4.9)
$D_{h}\left(\sum_{m}f_{m}R_{m}\right)=\sum_{m}D_{h}(R_{m}f_{m})=\sum_{m}\big{(}[D_{h},R_{m}](f_{m})+R_{m}\cdot
D_{h}(f_{m})\big{)}\equiv\sum_{m}R_{m}\cdot D_{h}(f_{m})\equiv 0\pmod{I}$

which implies that $D_{h}(I)\subset I$ (similarly for $D_{e}$ and $D_{f}$).

Now we proceed to the computation of these commutators. Since $D_{f}$ is a
scalar operator, the equality $[D_{f},R_{m}]=0$ is trivial. It remains to
compute $[D_{h},R_{m}]$ and $[D_{f},R_{m}]$ (here we use Lemma 4.1).

(4.10)
$\begin{split}[D_{h},R_{m}]=&-2\sum_{j=1}^{k}\left[jp_{j}\frac{\partial}{\partial
p_{j}},R_{m}\right]-2\sum_{j=1}^{n-k}\left[jq_{j}\frac{\partial}{\partial
q_{j}},R_{m}\right]=-2\sum_{j=1}^{k}\left[jp_{j}\frac{\partial}{\partial
p_{j}},p_{j}q_{m-j}\right]-\\\
&-2\sum_{j=1}^{n-k}\left[jq_{j}\frac{\partial}{\partial
q_{j}},p_{m-j}q_{j}\right]=-2\sum_{j=1}^{k}jp_{j}q_{m-j}-2\sum_{j=1}^{n-k}jp_{m-j}q_{j}=\\\
&=-2\sum_{j=0}^{m}jp_{j}q_{m-j}-2\sum_{j=0}^{m}jp_{m-j}q_{j}=-2m\sum_{j=0}^{m}p_{j}q_{m-j}=-2m\cdot
R_{m}\equiv 0\pmod{I}.\end{split}$

Similar computation for $[D_{e},R_{m}]$ is more complicated. Since
$[D_{f},R_{m}]=\sum\limits_{r+s=m}[D_{f},p_{r}q_{s}]$ it is useful to compute
$[D_{f},p_{r}q_{s}]$ first. Lemma 4.1 implies that

(4.11)
$\begin{split}[D_{e},p_{r}q_{s}]=&-\sum_{j=1}^{k}(k-j+1)(n-k+j-1)\delta_{jr}p_{j-1}q_{s}+\sum_{j=1}^{n-k}(k+j-1)(n-k-j+1)\delta_{js}p_{r}q_{j-1}+\\\
&+\sum_{j,l=1}^{k}\left(\sum_{d=0}^{j-1}(j+l-2d-1)p_{d}p_{j+l-d-1}\right)\left(\delta_{lr}q_{s}\frac{\partial}{\partial
p_{j}}-\delta_{jr}q_{s}\frac{\partial}{\partial p_{l}}\right)-\\\
&-\sum_{j,l=1}^{n-k}\left(\sum_{d=0}^{j-1}(j+l-2d-1)q_{d}q_{j+l-d-1}\right)\left(\delta_{ls}p_{r}\frac{\partial}{\partial
q_{j}}-\delta_{js}p_{r}\frac{\partial}{\partial q_{l}}\right)+\\\
&+2\sum_{j=1}^{k}\sum_{l=1}^{n-k}(j+l-1)\left(\sum_{d=0}^{j-1}p_{d}q_{j+l-d-1}\right)\left(\delta_{ls}p_{r}\frac{\partial}{\partial
p_{j}}+\delta_{jr}q_{s}\frac{\partial}{\partial
q_{l}}+\delta_{jr}\delta_{ls}\right).\end{split}$

Thus, modulo $I$ we have (here we use the symmetry mentioned in Remark 4.1)

$\begin{split}[D_{e},p_{r}q_{s}]=&-(k-r+1)(n-k+r-1)p_{r-1}q_{s}+(k+s-1)(n-k-s+1)p_{r}q_{s-1}+\\\
&+2\sum_{j=1}^{k}\sum_{d=0}^{j-1}(j+r-2d-1)p_{d}p_{j+r-d-1}q_{s}\frac{\partial}{\partial
p_{j}}-2\sum_{l=1}^{n-k}\sum_{d=0}^{l-1}(l+s-2d-1)q_{d}q_{l+s-d-1}p_{r}\frac{\partial}{\partial
q_{l}}+\\\
&+2\sum_{j=1}^{k}(j+s-1)\left(\sum_{d=0}^{j-1}p_{d}q_{j+s-d-1}\right)p_{r}\frac{\partial}{\partial
p_{j}}-2\sum_{l=1}^{n-k}(r+l-1)\left(\sum_{d=0}^{l-1}p_{r+l-d-1}q_{d}\right)q_{s}\frac{\partial}{\partial
q_{l}}+\\\ &+2(r+s-1)\sum_{d=0}^{r-1}p_{d}q_{r+s-d-1}.\end{split}$

Here we used the fact that
$\sum\limits_{d=0}^{r-1}p_{d}q_{r+l-d-1}=-\sum\limits_{d=0}^{l-1}p_{r+l-d-1}q_{d}$
modulo $I$. Rearranging the previous sum, we see that

(4.12) $[D_{e},p_{r}q_{s}]=2\sum_{j=1}^{k}A^{(rs)}_{j}\frac{\partial}{\partial
p_{j}}-2\sum_{l=1}^{n-k}B^{(rs)}_{l}\frac{\partial}{\partial q_{l}}+C^{(rs)},$

where $A^{(rs)}_{j}$, $B^{(rs)}_{j}$ and $C^{(rs)}_{j}$ are defined as

(4.13) $\displaystyle A^{(rs)}_{j}$
$\displaystyle=\sum_{d=0}^{j-1}(j+r-2d-1)p_{d}p_{j+r-d-1}q_{s}+(j+s-1)\sum_{d=0}^{j-1}p_{d}q_{j+s-d-1}p_{r},$
$\displaystyle B^{(rs)}_{j}$
$\displaystyle=\sum_{d=0}^{l-1}(l+s-2d-1)q_{d}q_{l+s-d-1}p_{r}+(r+l-1)\sum_{d=0}^{l-1}p_{r+l-d-1}q_{d}q_{s},$
$\displaystyle C^{(rs)}$
$\displaystyle=-(k-r+1)(n-k+r-1)p_{r-1}q_{s}+(k+s-1)(n-k-s+1)p_{r}q_{s-1}+$
$\displaystyle+2(r+s-1)\sum_{d=0}^{r-1}p_{d}q_{r+s-d-1}.$

Now, in order to prove that $[D_{e},R_{m}]\equiv 0\pmod{I}$, it suffices to
check that

(4.14) $\sum_{r+s=m}A^{(rs)}_{j}\equiv 0,\leavevmode\nobreak\
\sum_{r+s=m}B^{(rs)}_{j}\equiv 0,\leavevmode\nobreak\
\sum_{r+s=m}C^{(rs)}\equiv 0\pmod{I}.$

We compute the sum of $C^{(rs)}$ first. Note that

(4.15)
$\begin{split}\sum_{r+s=m}C^{(rs)}=&-\sum_{r+s=m}(k-r+1)(n-k+r-1)p_{r-1}q_{s}+\\\
&+\sum_{r+s=m}(k+s-1)(n-k-s+1)p_{r}q_{s-1}+2(m-1)\sum_{d=0}^{r-1}p_{d}q_{m-d-1}.\end{split}$

Since

(4.16)
$\begin{split}&-\sum_{r+s=m}(k-r+1)(n-k+r-1)p_{r-1}q_{s}+\sum_{r+s=m}(k+s-1)(n-k-s+1)p_{r}q_{s-1}=\\\
&=-\sum_{u+v=m-1}(k-u)(n-k+u)p_{u}q_{v}+\sum_{u+v=m-1}(k+v)(n-k-v)p_{u}q_{v}=\\\
&=\sum_{u+v=m-1}(n-2k+u-v)(u+v)p_{u}q_{v}=(m-1)\sum_{u+v=m-1}(n-2k+u-v)p_{u}q_{v}=\\\
&=(m-1)(n-2k)R_{m-1}+(m-1)\sum_{u+v=m-1}(u-v)p_{u}q_{v}\equiv(m-1)\sum_{u+v=m-1}(u-v)p_{u}q_{v}\pmod{I}\end{split}$

and

(4.17)
$\sum_{r+s=m}\sum_{d=0}^{r-1}p_{d}q_{m-d-1}=\sum_{d=0}^{m-1}(m-d)p_{d}q_{m-d-1}=\sum_{u+v=m-1}(v+1)p_{u}q_{v},$

we have

(4.18)
$\begin{split}\sum_{r+s=m}C^{(rs)}&\equiv(m-1)\sum_{u+v=m-1}(u-v)p_{u}q_{v}+2(m-1)\sum_{u+v=m-1}(v+1)p_{u}q_{v}=\\\
&=(m-1)\sum_{u+v=m-1}(u+v+2)p_{u}q_{v}=(m^{2}-1)R_{m-1}\equiv
0\pmod{I}\end{split}$

for all $m\geq 1$.

Next we check that $\sum\limits_{r+s=m}A^{(rs)}_{j}\equiv 0\pmod{I}$ for all
$m\geq 1$ (the proof for $B^{(rs)}_{j}$ is similar). Indeed, we have

(4.19)
$\sum_{r+s=m}A^{(rs)}_{j}=\sum_{r+s=m}\sum_{d=0}^{j-1}(j+r-2d-1)p_{d}p_{j+r-d-1}q_{s}+\sum_{r+s=m}(j+s-1)\sum_{d=0}^{j-1}p_{d}q_{j+s-d-1}p_{r}.$

Now note that modulo $I$ we have (here we make substitutions in summations:
$u=s-l-1$, $v=m-s$, $w=l$)

(4.20)
$\begin{split}&\sum_{r+s=m}(j+s-1)\sum_{d=0}^{j-1}p_{d}q_{j+s-d-1}p_{r}\equiv-\sum_{r+s=m}(j+s-1)\sum_{l=0}^{s-1}p_{j+s-l-1}q_{l}p_{r}=\\\
&=[u=s-l-1,\leavevmode\nobreak\ v=m-s,\leavevmode\nobreak\
w=l,\leavevmode\nobreak\ u+v+w=m-1,\leavevmode\nobreak\ u,v,w\geq 0]=\\\
&=-\sum_{u+v+w=m-1}(j+u+w)p_{j+u}p_{v}q_{w}\pmod{I}.\end{split}$

On the other hand,

(4.21)
$\begin{split}\sum_{r+s=m}\sum_{d=0}^{j-1}(j+r-2d-1)p_{d}p_{j+r-d-1}q_{s}&=\sum_{r+s=m}q_{s}\left(\sum_{d=0}^{j-1}(j+r-d-1)p_{d}p_{j+r-d-1}-\sum_{d=0}^{j-1}dp_{d}p_{j+r-d-1}\right)=\\\
&=\sum_{r+s=m}q_{s}\left(\sum_{d=r}^{j+r-1}dp_{d}p_{j+r-d-1}-\sum_{d=0}^{j-1}dp_{d}p_{j+r-d-1}\right)=\\\
&=\sum_{r+s=m}q_{s}\left(\sum_{d=j}^{j+r-1}dp_{d}p_{j+r-d-1}-\sum_{d=0}^{r-1}dp_{d}p_{j+r-d-1}\right)=\\\
&=\sum_{r+s=m}\sum_{d=0}^{r-1}(j+d)p_{j+d}q_{r-d-1}q_{s}-\sum_{r+s=m}\sum_{d=0}^{r-1}dp_{d}p_{j+r-d-1}q_{s}.\end{split}$

Now we can rearrange the last expression in a way similar to (4.20). Note that

(4.22)
$\begin{split}\sum_{r+s=m}\sum_{d=0}^{r-1}(j+d)p_{j+d}q_{r-d-1}q_{s}&=[u=d,\leavevmode\nobreak\
v=s,\leavevmode\nobreak\ w=r-d-1,\leavevmode\nobreak\ u,v,w\geq 0]=\\\
&=[u+v+w=m-1]=\sum_{u+v+w=m-1}(j+u)p_{j+u}q_{v}p_{w}\end{split}$

and

(4.23)
$\begin{split}\sum_{r+s=m}\sum_{d=0}^{r-1}dp_{d}p_{j+r-d-1}q_{s}&=[u=r-d-1,\leavevmode\nobreak\
v=d,\leavevmode\nobreak\ w=s,\leavevmode\nobreak\ u,v,w\geq 0]=\\\
&=[u+v+w=m-1]=\sum_{u+v+w=m-1}vp_{j+u}q_{v}p_{w}.\end{split}$

Finally, combining equalities (4.19) and $\eqref{A_aux_1}-\eqref{A_aux_4}$ we
obtain

(4.24)
$\begin{split}\sum_{r+s=m}A^{(rs)}_{j}\equiv&\sum_{u+v+w=m-1}p_{j+u}q_{v}p_{w}(j+u-v-
j-u-w)=\\\ &=\sum_{u+v+w=m-1}p_{j+u}q_{v}p_{w}(-v-w)=\\\
&=\sum_{u=0}^{m-1}p_{j+u}(u-m+1)\sum_{v+w=m-1-u}p_{w}q_{v}=\\\
&=\sum_{u=0}^{m-1}(u-m+1)p_{j+u}\cdot R_{m-1-u}\equiv 0\pmod{I},\end{split}$

as desired. The congruence $\sum\limits_{r+s=m}B^{(rs)}_{j}\equiv 0\pmod{I}$
can be checked in the same way.

Thus, we proved that differential operators $[D_{h},R_{m}]$, $[D_{e},R_{m}]$
and $[D_{f},R_{m}]$ are zero modulo $I$. Hence, $D_{e}$, $D_{f}$ and $D_{h}$
are well defined on the quotient
$\mathcal{A}=\mathbb{C}[p_{1},\ldots,p_{k},q_{1},\ldots,q_{n-k}]/I$.

Step 2. We check that triple $\\{D_{e},D_{f},D_{h}\\}$ satisfy the
$\mathfrak{sl}_{2}$-commutation relations (1.5). In other words, we need to
check the following eqialities:

(4.25) $[D_{e},D_{f}]=D_{h},\leavevmode\nobreak\
[D_{h},D_{e}]=2D_{e},\leavevmode\nobreak\ [D_{h},D_{f}]=-2D_{f}.$

Commutators $[D_{e},D_{f}]$ and $[D_{h},D_{f}]$ can be easily computed
directly. This is because $D_{f}$ is the scalar multiplication by $-p_{1}$, so
we only need to take terms containing differentiation with respect to $p_{1}$
into account. Indeed, Lemma 4.1 implies that

$[D_{h},D_{f}]=\left[-2p_{1}\frac{\partial}{\partial
p_{1}},-p_{1}\right]=2p_{1}=2D_{f}\leavevmode\nobreak\ \text{and}$ (4.26)
$\begin{split}[D_{e},D_{f}]=&-[D_{e},p_{1}]=\sum_{j=1}^{k}(k-j+1)(n-k+j-1)p_{j-1}\left[\frac{\partial}{\partial
p_{j}},p_{1}\right]-\\\
&-\sum_{j,l=1}^{k}\left(\sum_{d=0}^{j-1}(j+l-2d-1)p_{d}p_{j+l-d-1}\right)\left[\frac{\partial^{2}}{\partial
p_{j}\partial p_{l}},p_{1}\right]-\\\
&-2\sum_{j=1}^{k}\sum_{l=1}^{n-k}(j+l-1)\left(\sum_{d=0}^{j-1}p_{d}q_{j+l-d-1}\right)\left[\frac{\partial^{2}}{\partial
p_{j}\partial q_{l}},p_{1}\right]=\\\
&=k(n-k)-2\sum_{l=1}^{k}lp_{l}\frac{\partial}{\partial
p_{l}}-2\sum_{l=1}^{n-k}lq_{l}\frac{\partial}{\partial
q_{l}}=D_{h}.\end{split}$

In order to prove that $[D_{h},D_{e}]=2D_{e}$ let us make the following
observation first. Lemma 4.1 implies that for any $2n$ nonnegative integers
$\alpha_{1}$, …, $\alpha_{k}$, $\beta_{1}$, …, $\beta_{k}$ and $\gamma_{1}$,
…, $\gamma_{n-k}$, $\delta_{1}$ …, $\delta_{n-k}$ for the operator

$D=\left(\prod_{j=1}^{k}p_{j}^{\alpha_{j}}\prod_{l=1}^{n-k}q_{l}^{\gamma_{l}}\right)\cdot\prod_{j=1}^{k}\left(\frac{\partial}{\partial
p_{j}}\right)^{\beta_{j}}\prod_{l=1}^{n-k}\left(\frac{\partial}{\partial
q_{l}}\right)^{\delta_{l}}$

we have

(4.27)
$[D_{h},D]=\left(\sum_{j=1}^{k}2j(\beta_{j}-\alpha_{j})+\sum_{l=1}^{n-k}2l(\delta_{l}-\gamma_{l})\right)\cdot
D.$

Thus, in order to check the equality $[D_{h},D_{e}]=2D_{e}$ it suffices to
check that for each term of $D_{e}$ the value of
$\sum\limits_{j=1}^{k}2j(\beta_{j}-\alpha_{j})+\sum\limits_{l=1}^{n-k}2l(\delta_{l}-\gamma_{l})$
equals $2$ which is clear from the definition of $D_{e}$ (see (1.12)).

Step 3. We prove that $D_{e}$, $D_{f}$ and $D_{h}$ coincide with operators
$E$, $F$ and $H$ defined in Subsection 3.2. We will check this for $D_{f}$ and
$D_{h}$ first. According to Proposition 3.2 the action of $F$ corresponds to
the multiplication by the negative of the first Chern class, i.e. by
$-c_{1}(S)=-p_{1}$. Hence, the action of $F$ coincides with the action of
$D_{f}=-p_{1}$.

In order to prove that operators $D_{h}$ and $H$ coincide it suffices to check
that the image of any element $a\in\mathcal{A}$ of the degree $\deg(a)=d$
under the differential operator $D_{h}$ equals $H(a)=(-d+k(n-k))\cdot a$.
Indeed, for any monomial
$a=\prod_{j=1}^{k}p_{j}^{\alpha_{j}}\prod_{l=1}^{n-k}q_{l}^{\beta_{l}}$ we
have

(4.28)
$D_{h}(a)=D_{h}\left(\prod_{j=1}^{k}p_{j}^{\alpha_{j}}\prod_{l=1}^{n-k}q_{l}^{\beta_{l}}\right)=\left(-2\sum_{j=1}^{k}j\alpha_{j}-2\sum_{l=1}^{n-k}l\beta_{l}+k(n-k)\right)\cdot\left(\prod_{j=1}^{k}p_{j}^{\alpha_{j}}\prod_{l=1}^{n-k}q_{l}^{\beta_{l}}\right),$

so due to (1.4) we have $D_{h}(a)=(-d+k(n-k))\cdot a=H(a)$, as desired.

Finally, in order to prove that $D_{e}$ and $E$ coincide we use the corollary
of the Jacobson-Morozov theorem. Namely, we use the following fact: for given
semisimple and nilpotent elements of the semisimple Lie algebra the third
element which complements the first two to $\mathfrak{sl}_{2}$-triple, if
exists, is unique. Now note that $\\{E,F,H\\}$ and $\\{D_{e},D_{f},D_{h}\\}$
are $\mathfrak{sl}_{2}$-triples. Therefore, equalities $D_{h}=H$ and $D_{f}=F$
imply $D_{e}=E$. ∎

######  Remark 4.2.

It is evident that the essential part of the proof consisted of computations.
Namely, the proof of Proposition 1.6 reduces to the computation of commutators
$[D_{e},R_{m}]$, $[D_{f},R_{m}]$, $[D_{h},R_{m}]$, $[D_{e},D_{f}]$,
$[D_{h},D_{e}]$ and $[D_{h},D_{f}]$.

##  5\. Construction of the operator $D_{e}$

In this section we outline the way in which we obtained the expression for the
operator $D_{e}$. From the formal point of view, there this is unnecessary for
the proof of Proposition 1.6 since we have already shown that $D_{e}$
corresponds to the action of $E$ on cohomology. However, it might be
beneficial to describe the general approach which might be useful for similar
problems.

Suppose that we have a linear operator $L$ on polynomial algebra
$\mathbb{C}[x_{1},\ldots,x_{m}]$. We will show an algorithm which represents
$T$ as a differential operator, i.e. in form

(5.1) $L=P\left(\frac{\partial}{\partial
x_{1}},\ldots,\frac{\partial}{\partial x_{m}}\right),$

where $P(y_{1},\ldots,y_{m})$ is a polynomial in $m$ variables with
coefficients in $\mathbb{C}[x_{1},\ldots,x_{m}]$:

(5.2)
$P(y_{1},\ldots,y_{m})=\sum_{k_{1},\ldots,k_{m}}p_{k_{1},\ldots,k_{m}}(x_{1},\ldots,x_{m})\prod_{j=1}^{m}y_{j}^{k_{j}}.$

In other words, our goal is to express $T$ in the form

(5.3)
$L=\sum_{k_{1},\ldots,k_{m}}p_{k_{1},\ldots,k_{m}}(x_{1},\ldots,x_{m})\frac{\partial^{k_{1}+\ldots+k_{m}}}{\partial
x_{1}^{k_{1}}\ldots\partial x_{m}^{k_{m}}}$

for some polynomials
$p_{k_{1},\ldots,k_{m}}\in\mathbb{C}[x_{1},\ldots,x_{m}]$.

######  Definition 5.1.

Let $\mathcal{S}$ be a complex commutative algebra with unit. An operator
$L\colon\mathcal{S}\to\mathcal{S}$ is called a differential operator of order
0 if $L=L_{a}$ (or simply $L=a$) for some $a\in S$ where $L_{a}$ is an
operator of multiplication by $a$: $L_{a}(b)=a\cdot b$. An operator
$L\colon\mathcal{S}\to\mathcal{S}$ is called a differential operator of order
$d\geq 1$ if for any $a\in S$ the commutator $[L,L_{a}]=[L,a]$ is a
differential operator of order $d-1$ (or less). (Here we consider elements of
$\mathcal{S}$ as corresponding scalar operators.)

This definition is motivated by the polynomial algebra: if
$\mathcal{S}=\mathbb{C}[x_{1},\ldots,x_{m}]$, then this definition is
equivalent to the standard definition of the differential operator of order
$d$ in variables $x_{1},\ldots,x_{m}$.

######  Remark 5.1.

If $\mathcal{S}$ is generated by elements $\\{a_{j}\\}_{j\in A}$, then to
check that $L$ is a differential operator of order $k$ it is sufficient to
check that commutators $[L,a_{j}]$ is a differential operator. In particular,
for $\mathcal{S}=\mathbb{C}[x_{1},\ldots,x_{m}]$ we only need to compute
commutators of the form $[L,x_{j}]$.

For the sake of simplicity we will consider only the case when
$\mathcal{S}=\mathbb{C}[x_{1},\ldots,x_{m}]$. However, everything also works
in case when $\mathcal{S}$ is a quotient of polynomial a algebra.

The ring of linear operators on $\mathbb{C}[x_{1},\ldots,x_{m}]$ possess a
natural structure of the Lie algebra in which the Lie bracket is the
commutator $[\cdot,\cdot]$ of two linear operators. We will denote the
operator $[\cdot,x]$ as $\operatorname{Adj}_{x}$. Next, note that for any
$p(x_{1},\ldots,x_{m})\in\mathbb{C}[x_{1},\ldots,x_{m}]$ and nonnegative
integers $k_{1},\ldots,k_{m}$

(5.4)
$\operatorname{Adj}_{x_{j}}\left(p(x_{1},\ldots,x_{m})\frac{\partial^{k_{1}+\ldots+k_{m}}}{\partial
x_{1}^{k_{1}}\ldots\partial
x_{m}^{k_{m}}}\right)=k_{j}p(x_{1},\ldots,x_{m})\frac{\partial^{k_{1}+\ldots+k_{m}-1}}{\partial
x_{1}^{k_{1}}\ldots\partial x_{j-1}^{k_{j-1}}\partial x_{j}^{k_{j}-1}\partial
x_{j+1}^{k_{j+1}}\ldots\partial x_{m}^{k_{m}}}.$

Thus, for any polynomial $P$ in variables $y_{1},\ldots,y_{m}$ with
coefficients in $\mathbb{C}[x_{1},\ldots,x_{m}]$ we have

(5.5) $\operatorname{Adj}_{x_{j}}\left(P\left(\frac{\partial}{\partial
x_{1}},\ldots,\frac{\partial}{\partial x_{m}}\right)\right)=\frac{\partial
P}{\partial y_{j}}\left(\frac{\partial}{\partial
x_{1}},\ldots,\frac{\partial}{\partial x_{m}}\right),$

which implies that for any nonnegative integers $k_{1},\ldots,k_{n}$ the
following equality holds:

(5.6)
$\left(\operatorname{Adj}_{x_{1}}^{k_{1}}\circ\ldots\circ\operatorname{Adj}_{x_{m}}^{k_{m}}\right)\left(P\left(\frac{\partial}{\partial
x_{1}},\ldots,\frac{\partial}{\partial
x_{m}}\right)\right)=\left(\frac{\partial^{k_{1}+\ldots+k_{m}}P}{\partial
y_{1}^{k_{1}}\ldots\partial
y_{m}^{k_{m}}}\right)\left(\frac{\partial}{\partial
x_{1}},\ldots,\frac{\partial}{\partial x_{m}}\right).$

The last formula provides a way to compute coefficients of a differential
operator if its action on $\mathbb{C}[x_{1},\ldots,x_{m}]$ is given. If $L$ is
a differential operator of order $d$, then for any nonnegative numbers
$\\{k_{j}\\}_{j=1}^{m}$ whose sum grater than $d$, we have

(5.7)
$\left(\operatorname{Adj}_{x_{1}}^{k_{1}}\circ\ldots\circ\operatorname{Adj}_{x_{m}}^{k_{m}}\right)(L)=0.$

Similarly, for any nonnegative $\\{k_{j}\\}_{j=1}^{m}$ whose sum equals $d$ we
have

(5.8) $p_{k_{1},\ldots,k_{m}}(x_{1},\ldots,x_{m})=\frac{1}{k_{1}!\ldots
k_{m}!}\left(\operatorname{Adj}_{x_{1}}^{k_{1}}\circ\ldots\circ\operatorname{Adj}_{x_{m}}^{k_{m}}\right)(L),$

where $p_{k_{1},\ldots,k_{m}}(x_{1},\ldots,x_{m})$ is the coefficient of
$\prod_{j=1}^{m}(\partial/\partial x_{j})^{k_{j}}$ in $L$. In this way we can
find the terms of order $d$ in $L$ if we know the action of $L$ on
$\mathbb{C}[x_{1},\ldots,x_{m}]$. After that we can subtract these terms from
$L$ and then repeat this process. Therefore, the algorithm for finding the
differential expression for $L$ is as follows:

  * •

Find the minimal $d\geq 0$ such that for all $j_{1},\ldots,j_{j+1}$ all
operators
$(\operatorname{Adj}_{x_{j_{1}}}\circ\ldots\circ\operatorname{Adj}_{x_{j_{d+1}}})(L)$
are zero. If $L$ is indeed a differential operator, then such $d$ exists and
equals to the order of $L$.

  * •

Using formula (5.8) find all terms of the (highest) order $d$ in $L$. After
that subtract all these terms from $L$ and denote the resulting operator as
$L^{\prime}$.

  * •

Now $L^{\prime}$ is a differential operator whose order is at most $d-1$.
Repeat the second part of the algorithm for $L^{\prime}$.

  * •

After $d$ steps we will find the differential expression for $L$ which is the
sum of all terms found in the second part of the algorithm.

######  Remark 5.2.

It is possible to find a closed formula for $L$ in terms of commutators
$(\operatorname{Adj}_{x_{j_{1}}}\circ\ldots\circ\operatorname{Adj}_{x_{j_{n}}})(L)$
but it seems that such formula would be quite complicated and not suitable for
any direct computations.

In our particular problem we knew the action of the operator $E$ on Schubert
classes which form a basis of $H^{*}(\operatorname{Gr}(k,n),\mathbb{C})$. With
the help of computer it was found out that for all $i$ and $j$ the operator
$(\operatorname{Adj}_{x_{i}}\circ\operatorname{Adj}_{x_{j}})(E)$ is zero. The
latter means that $E$ is an differential operator of the order at most 2.
After that it remained to compute coefficients via the formula (5.8). In order
to do that we also needed to express Schubert classes in terms of Chern
classes. This was done via the correspondence from the Proposition 3.1,
formula (3.5) and classical formulas from the theory of symmetric functions.

##  6\. Concluding remarks

In this section we give final remarks and discuss several questions related to
our work which remain unanswered in this paper but still seem to be important.

Our description of $\mathrm{GL}_{n}$-structure on the cohomology ring of the
complex Grassmannian is quite complicated and uses a nontrivial isomorphism.
This raises the following question.

######  Question 6.1.

Is it possible to describe the $\mathrm{GL}_{n}$-structure on
$H^{*}(\operatorname{Gr}(k,n),\mathbb{C})$ directly, i.e without reference to
the Schubert classes?

However, we have already seen that even the expression for $E$ is fairly
complex, so it is unlikely that $\mathrm{GL}_{n}$-action on cohomology has a
simple description.

Next, since $H$ and $F$ can be described in terms of cohomology it is natural
to expect that the same is true for $E$. Nevertheless, it is not clear how the
differential expression of $E$ can be translated to the language of
cohomology.

######  Question 6.2.

Does the operator $E$ have a geometrical meaning similar to $H$ and $F$ (see
(1.11))?

We also could consider the action of $\\{E,F,H\\}$ on Schubert classes.
Proposition 3.2 suggests that action of $F$ on $\sigma_{\lambda}$ corresponds
to the addition of one box to $\lambda$ in all possible ways. Similarly,
action of $E$ corresponds to the deletion of boxes from $\lambda$ but the
meaning of the coefficients $(k+\lambda_{l}-l)(n-k-\lambda_{l}+l)$ is unclear.
It would be interesting to find a combinatorial interpretation of the action
of $E$ on Schubert classes.

Finally, our algebra $\mathcal{A}$ arose as the cohomology ring of the compact
Kähler manifold (complex Grassmannian in our case). Such algebras possess
special properties because of the Poincare duality and the hard Lefschetz
theorems. There are similar objects in commutative algebra called the Artinian
Gorenstein algebras (see [11, Section 2] for the details). One particular
class of examples of Artinian Gorenstein algebras consists of the quotients
$\mathbb{C}[x_{1},\ldots,x_{n}]/I$, where the ideal $I$ is generated by some
polynomial $P\in\mathbb{C}[y_{1},\ldots,y_{n}]$ in the following way

(6.1) $I=\operatorname{Ann}P,\leavevmode\nobreak\
\text{where}\leavevmode\nobreak\
\operatorname{Ann}P=\left\\{f\in\mathbb{C}[x_{1},\ldots,x_{n}]\colon
f\left(\frac{\partial}{\partial x_{1}},\ldots,\frac{\partial}{\partial
x_{n}}\right)P=0\right\\}.$

In this case the polynomial $P$ is called the cogenerator of the initial
algebra (see also [11, Theorem 2.1]). It can be checked that for our algebra
$\mathcal{A}$ the cogenerator equals

(6.2)
$P(x_{1},\ldots,x_{k},y_{1},\ldots,y_{n-k})=\sum_{\alpha,\beta}\frac{1}{\alpha_{1}!\,\ldots\,\alpha_{k}!\,\beta_{1}!\,\ldots\,\beta_{n-k}!}\left(\prod_{j=1}^{k}c_{j}(S)^{\alpha_{j}}\prod_{l=1}^{n-k}c_{l}(Q)^{\beta_{l}}\right)\cdot
x_{1}^{\alpha_{1}}\ldots x_{k}^{\alpha_{k}}\,y_{1}^{\beta_{1}}\ldots
y_{n-k}^{\beta_{n-k}},$

where summation is over all nonnegative integers $\alpha_{j}$ and $\beta_{l}$
that satisfy

(6.3)
$\alpha_{1}+2\alpha_{2}+\ldots+k\alpha_{k}+\beta_{1}+2\beta_{2}+\ldots+(n-k)\beta_{n-k}=k(n-k).$

Thus, the ideal $I$ generated by elements
$R_{m}\in\mathbb{C}[p_{1},\ldots,p_{k},q_{1},\ldots,q_{n-k}]$ can be also
represented as

(6.4)
$I=\operatorname{Ann}P=\left\\{f\in\mathbb{C}[p_{1},\ldots,p_{k},q_{1},\ldots,q_{n-k}]\colon
f\left(\frac{\partial}{\partial p_{1}},\ldots,\frac{\partial}{\partial
p_{k}},\frac{\partial}{\partial q_{1}},\ldots\frac{\partial}{\partial
q_{n-k}}\right)P=0\right\\}.$

The proof of this fact is similar to the proof of the Theorem 2.1 in [9].

######  Remark 6.1.

Here we identify the one-dimensional top cohomology group
$H^{k(n-k)}(\operatorname{Gr}(k,n),\mathbb{C})$ with $\mathbb{C}$. Hence, $P$
is defined up to nonzero scalar.

It appears that closed form for this polynomial $P$ is unknown as well as its
connection to the complex Grassmannian, so we have the following question.

######  Question 6.3.

How does the polynomial $P$ relate to the complex Grassmannian?

The cogenerators are also related to the so called volume polynomials, see [9,
Section 4]. We suspect that $P$ might be related to the hypersymplex
$\Delta_{kn}$ which is the moment polytope of the Grassmannian with respect to
the action of the maximal torus. Specifically, it seems that the cogenerator
of the cohomology ring of $\operatorname{Gr}(k,n)$ coincides with the volume
polynomial of the hypersimplex.

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