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This article is cited in 4 scientific papers (total in 4 papers) Symplectic reduction and Lagrangian submanifolds of N. A. Tyurinab a Bogoliubov Laboratory of Theoretical Physics, Joint Institute for Nuclear Research, Dubna, Moscow Region, Russia b Steklov Mathematical Institute of Russian Academy of Sciences, Moscow, Russia
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    IntroductionIn [1] Mironov proposed a new construction of submanifolds of $\mathbb{C}^n$ and $\mathbb{C} \mathbb{P}^n$ that are Lagrangian with respect to the Kähler forms induced by the constant metric and Fubini–Studi metric, respectively. He was based on the problem of the construction of minimal and Hamiltonian minimal Lagrangian submanifolds and Lagrangian immersions (that is, Lagrangian submanifolds admitting self-intersections). It was shown in [2] that Mironov’s construction can be generalized to arbitrary Kähler manifolds with an action of a torus but, by contrast with our work, the minimality of the Lagrangian cycles obtained was not a topic for discussion there. In this paper we use a simple extension of these constructions to an even more general case, when the symplectic manifold is endowed with a Hamiltonian action of some Lie group. Let $(M, \omega)$ be a compact symplectic manifold with a Hamiltonian action of a Lie group $G$, and let $\mu\colon M \to \mathfrak g^*$ be the moment map to the dual Lie algebra. Let $\mu^{-1} (t) \subset M$ be the level set corresponding to the central element $t \in \mathfrak g^*$, on which the action of $G$ is free, so that the quotient manifold $N_t=\mu^{-1}(t)/G$ with the corresponding symplectic form $\omega_t$ exists, which is the base of the principal bundle $\pi_t\colon \mu^{-1}(t) \to N_t$. Then the following result holds. Theorem 1. Let $S_0 \subset N_t$ be a Lagrangian embedding (or an immersion). Then its inverse image $S=\pi^{-1}_t (S_0) \subset \mu^{-1}(t) \subset M$ is a Lagrangian embedding (an immersion) in $(M, \omega)$. In fact, by the definition of the symplectic reduction $M /\!/ G$ (see [3]) the tangent space $T_p S$ is the span of two transversal components: $\operatorname{ker} (\mathrm d \pi_t|_p) \oplus T_{\pi_t (p)}S_0$, the first of which is naturally isomorphic to $\mathfrak g$ and lies in $\operatorname{ker} \omega_p|_{T_p \mu^{-1}(t)}$. Now the assertion of the theorem follows from the definition of $\omega_t$ because $S_0$ is Lagrangian. Remark 1. It is easy to see that if the original manifold $M$ contains a Lagrangian submanifold $K \subset M$ which is transversal to the action of $G$, then this defines a Lagrangian immersion $N_t \supset S_0=\pi_t(\mu^{-1}(t) \cap K)$. Applying Theorem 1 to this immersion we obtain a new Lagrangian submanifold (we call the transition from $K$ to this submanifold a mutation of Lagrangian cycles). Hence the man results of [2] follow from Theorem 1, provided that we take the torus $\mathrm T^k$ as $G$ and the real part of $M$ with respect to the action of the antiholomorphic involution as $K$. Thus it is natural to interpret Theorem 1 as a generalization of the constructions in [1] and [2]. It was shown in [2] that the complex Grassmannian $\operatorname{Gr}(1, n)$ of straight lines in the projective space $\mathbb{C} \mathbb{P}^n$, endowed with the standard Kähler form induced by the Plücker embedding, admits a Hamiltonian action of the torus $\mathrm T^n$, which also preserves the complex structure. It was conjectured in the same paper that $\operatorname{Gr}(1, n)$ admits submanifolds of at least $n$ distinct topological types that can be realized by smooth Lagrangian embeddings. This conjecture was based on the observation that for each $k$ between 0 and $n$ we can use the generalized construction from [2] and obtain a priori topologically distinct topological submanifolds. Theorem 1 suggests a possible description of such generalized Mironov cycles, which have different degrees of homogeneity in $\operatorname{Gr}(1, n)$, so that the bound in the conjecture holds with a surplus. Namely, in this paper we prove the following result. Theorem 2. For each $k \in (2, \dots, n-1)$ there exist functions $\widetilde \mu_0, \dots, \widetilde \mu_{k-1}$ on $\operatorname{Gr}(1, n)$ that induce a Hamiltonian action of $\mathrm T^k$, and there exists a set of values $c_0, \dots, c_{k-1} \in \mathbb{R}_{+}$ such that the result of the corresponding symplectic reduction is $\operatorname{Gr}(1, n)/\!/ \mathrm T^k \cong M_{n-k}=\operatorname{tot}(\mathbb{P}(\tau) \times \dots \times\mathbb{P}(\tau) \to \operatorname{Gr}(1, n-k))$, where the total space of the Cartesian product of $k$ copies of the projectivization of the tautological bundle $\tau \to \operatorname{Gr}(1, n-k)$ is on the right. Moreover, the other generators of the action of $\mathrm T^n$ on $\operatorname{Gr}(1, n)$ do not disappear after the reduction, but generate instead a Hamiltonian action of the torus $\mathrm T^{n-k}$ on the manifold $M_{n-k}$. This action can explicitly be described as follows. We can realize $M_{n-k}$ as a submanifold of the $(k+1)$-fold Cartesian product $\mathbb{C} \mathbb{P}^{n-k} \times \dots \times \mathbb{C} \mathbb{P}^{n-k} \times \operatorname{Gr}(1, n-k)$ given by the intersection of the $k$ ‘partial incidence cycles’ $\mathcal U_i $ in the $k$-fold Cartesian product $ \mathbb{C} \mathbb{P}^{n-k} \times \dots \times \mathbb{C} \mathbb{P}^{n-k} \times \operatorname{Gr}(1, n-k)$ that consists of the pairs (a point, a line) such that for $\mathcal U_i$ the point is taken in the $i$th direct factor. In addition, on each of the $k+1$ direct factors, once we have identified the first $k$ of them, we have the diagonal action of $\mathrm T^{n-k}$, which obviously preserves $M_{n-k}$. Hence the symplectic reduction $\operatorname{Gr}(1, n)/\!/ \mathrm T^k$ can be represented as the symplectic reduction $M_{n-i} /\!/ \mathrm T^{k-i}$ for $i=0, \dots, k$. On the other hand, for each $k$ the action of $T^k$ described above preserves the whole Kähler structure, so that the manifold $M_{n-k}$ is endowed with a Kähler structure in a natural way, and therefore the real part $M_{n-k}^{\mathbb{R}} \subset M_{n-k}$ is isotropic with respect to the Kähler form. As $M_{n-k}^{\mathbb{R}} \cong\operatorname{tot}(\mathbb{P}(\tau_{\mathbb{R}}) \times \dots\times \mathbb{P}(\tau_{\mathbb{R}}) \to \operatorname{Gr}_{\mathbb{R}} (1, n-k))$, where we have the total space of a Cartesian product of $k$ copies of the projectivization of the tautological bundle $\tau \to \operatorname{Gr}(1, n-k)$ on the right, its real dimension is precisely the complex dimension of the whole of $M_{n-k}$, so it is a Lagrangian submanifold of $M_{n-k}$. Note that topologically $M^{\mathbb{R}}_{n-k}$ is a fibre bundle of $k$-tori over $\operatorname{Gr}_{\mathbb{R}}(1, n-k)$, so Theorem 1 provides a family of Lagrangian submanifolds $\{ S_k,\, k=2, \dots, n-1 \}$ such that each $S_k$ is isomorphic to a fibre bundle with fibre $T^{2k}$ over $\operatorname{Gr}_{\mathbb{R}}(1, n-k)$. It is easy to see that for distinct $k$ the corresponding $S_k$ are topologically inequivalent, which obviously implies an affirmative answer to the conjecture stated at the end of [2]. We show below that an additional analysis of our constructions provides us with at least $[{n}/{2}] [(n-1)/{2}]$ extra topological types, which increases significantly the bound for the number of distinct topological types of smooth Lagrangian submanifolds of $\operatorname{Gr} (1, n)$. Since we have managed to discuss and prove Theorem 1 in the introduction, in the next sections we turn immediately to the proof of Theorem 2: first we examine the geometric and analytic aspects for $\operatorname{Gr}(1, 3)$, and then, on this basis, we explore the general case of $\operatorname{Gr}(1, n)$. § 1. The Grassmannian $\operatorname{Gr}(1, 3)$Consider the projective space $\mathbb{C} \mathbb{P}^3$ with fixed coordinate system $[z_0: \dotsb :z_3]$ and the compatible Kähler form $\omega_{FS}$ of the corresponding Fubini–Studi metric. Then we have the system of moment maps
$$
\begin{equation*}
\mu_i=\frac{| z_i |^2}{\sum_{i=0}^3 | z_i |^2}, \qquad i=0, \dots, 3,
\end{equation*}
\notag
$$
any three of which induce a Gamiltonian action of the torus $\mathrm T^3$. Each $\mu_i$ induces an action on the set of projective lines making up the manifold $\operatorname{Gr}(1, 3)$, hence we obtain an induced action of $\mathrm T^3$ on the latter, which is Hamiltonian with respect to the standard Kähler form (we present all explicit formulae below, in the analytic part of our construction). Consider two moment maps $\mu_0$ and $ \mu_1$ and look at the corresponding moment maps $\widetilde \mu_i$ from the Grassmannian $\operatorname{Gr}(1, 3)$. Let $p_i$, $i=0, \dots, 3$, denote the points in $\mathbb{C} \mathbb{P}^3$ corresponding to the orthonormal basis vectors $e_i$ in the space $\mathbb{C}^4$ whose projectivization is the projective space in question. Then (see [4]) the functions $\widetilde \mu_i$ can be represented as
$$
\begin{equation}
\widetilde \mu_i (l)=\max_{v \in V} \frac{|\langle e_i; v\rangle|^2}{| v |^2},
\end{equation}
\tag{1}
$$
where $V \subset \mathbb{C}^4$ is the 2-space corresponding to the projective line $l \subset \mathbb{C} \mathbb{P}^3$. Hence we see that $\widetilde \mu_i$ takes values on the interval $[0, 1]$, its critical values are $0$ and $1$, and it attains these values in the case when either $l \subset \langle p_{j_1}, p_{j_2}, p_{j_3} \rangle$, $j_k \neq i$, or $p_i \in l$, respectively (throughout, $\langle \cdots \rangle$ denotes the projective hull of the objects in brackets). Thus, the critical set of $\widetilde \mu_0$ is the union of the set of lines through $p_0$ and the set of lines lying in the plane $\langle p_1, p_2, p_3 \rangle$, that is, the corresponding $\alpha$- and $\beta$-planes in $\operatorname{Gr}(1, 3)$. Furthermore, it is easy to see that if a line $l \subset \mathbb{C} \mathbb{P}^3$ has a nontrivial intersection with $ \langle p_0, p_1 \rangle$, then $\widetilde \mu_0(l) + \widetilde \mu_1(l) \geqslant 1$. In fact, if the intersection is nontrivial, then the corresponding $V$ contains a vector $v \in V$ representable in the form $v=x_0 e_0 + x_1 e_1$, so that we see from (1) that the sum of two maxima must be at least $1$. Consider the open subset $\operatorname{Gr}^0(1, 3)$ of $ \operatorname{Gr}(1, 3)$ defined by
$$
\begin{equation*}
\operatorname{Gr}^0(1, 3)=\{l \in \operatorname{Gr}(1, 3) \mid\widetilde \mu_0(l)>0, \widetilde \mu_1(l)>0 \text{ and } \widetilde \mu_0(l) + \widetilde \mu_1(l) <1 \}.
\end{equation*}
\notag
$$
Then we have the following result. Proposition 1. The open manifold $\operatorname{Gr}^0(1, 3)$ is fibred over the Cartesian product $\langle p_2, p_3 \rangle \times \langle p_2, p_3 \rangle$ so that the Hamiltonian action of $\mathrm T^2$ induced by $\widetilde \mu_0$ and $\widetilde \mu_1$ preserves the base and the fibres. Proof. Consider an arbitrary line $l \subset \mathbb{C} \mathbb{P}^3$ disjoint from $\langle p_0, p_1 \rangle$ and not lying in the subspaces $\langle p_i, p_2, p_3 \rangle$, $i=0, 1$. Then the points $s_0, s_1 \in \langle p_2, p_3 \rangle$ obtained in intersection of the planes spanned by $p_1$ and $l$ or by $p_0$ and $l$, respectively, with the line $\langle p_2, p_3 \rangle$ are well defined. In fact, the plane $\langle p_0, l \rangle \subset \mathbb{C} \mathbb{P}^3$ cannot contain the whole of $\langle p_2, p_3 \rangle$ because otherwise $l$ would lie in the plane $\langle p_0, p_2, p_3 \rangle$, which is ruled out by assumption. The same arguments can be used for $\langle p_1, l \rangle$. Thus, the line $l$ corresponds to a pair of (possibly coinciding) points $s_0, s_1 \in \langle p_2, p_3 \rangle$. This correspondence is not symmetric (but it is clear that the symmetry can be restored by interchanging two moment maps, which in its turn corresponds to the automorphism of the original projective space $\mathbb{C} \mathbb{P}^3$ implemented by a transposition of two homogeneous coordinates).
On the other hand how can we recover the line $l$ from the pair of points $(s_0, s_1)$? Consider two pencils of planes through the lines $\langle p_0, s_0 \rangle$ and $\langle p_1, s_1 \rangle$, respectively. From each pencil we delete the pair of planes containing the lines $\langle p_0, p_1 \rangle$ and $\langle p_2, p_3 \rangle$. Then it is easy to see that $\operatorname{Gr}^0(1, 3)$ embeds in a bundle with fibre $\mathbb{C}^* \times \mathbb{C}^*$ over the Cartesian product $\langle p_2, p_3 \rangle \times \langle p_2, p_3 \rangle$ in a natural way. In fact, the pencils of planes must by construction contain the planes $\langle p_0, l \rangle$ and $ \langle p_1, l \rangle $, which lie in the respective pencils and have the intersection precisely $l$. A rough estimate shows that the image of the embedding $\operatorname{Gr}^0(1, 3)$ is uniformly distributed over the bundle: in the fibre over each point in $\langle p_2, p_3 \rangle \times \langle p_2, p_3 \rangle$ we have an open four-dimensional subset. This means that after fixing some positive quantities $c_0, c_1 \in \mathbb{R}_+$, $c_0 + c_1 < 1$, the joint level set $N_{c_0, c_1}=\{ \widetilde \mu_0=c_0,\, \widetilde \mu_1=c_1 \} \subset \operatorname{Gr}(1, 3)$ is fibred with fibre $\mathrm T^2$ over the Cartesian product $\langle p_2, p_3 \rangle \times \langle p_2, p_3 \rangle$ (this can also be seen from the formulae below). Remark 2. It is also easy to see that the fibre bundle with fibre $\mathbb{C}^* \times \mathbb{C}^*$ over $\langle p_2, p_3 \rangle \times \langle p_2, p_3 \rangle$ is not topologically trivial: it is naturally embedded in a bundle with fibre $\mathbb{C} \mathbb{P}^1 \times \mathbb{C} \mathbb{P}^1$, and these projective fibrations over the projective line $\langle p_2, p_3 \rangle$ have the following realizations, allowing us to determine their topological types. Since in both cases the answer is the same, we consider the pencil of planes through $\langle p_0, s_0 \rangle$ and look at the intersection of each plane with the projective hull $\langle p_1, p_2, p_3 \rangle$, so that we associate some line through $s_0$ with each plane. Then we obtain a fibre bundle over $\langle p_2, p_3 \rangle$ with fibre equal to the pencil of lines through $s_0$; after globalization this yields a total space isomorphic to the projectivization of the bundle $\mathcal O \oplus \mathcal O(-1)$ over the projective line $\langle p_2, p_3 \rangle$. This bundle admits a pair of global sections corresponding to the planes $\langle p_0, s_0, p_1 \rangle$ and $\langle p_0, p_2, p_3 \rangle$, and the choice of values $\widetilde \mu_0=c_0$ and $ \widetilde \mu_1=c_1$ such that $c_i>0$ and $c_0 +c_1 < 1$ distinguishes a pair of circles in each fibre, so that the total space of each fibration is isomorphic to $S^3$ (the total space of the Hopf fibration), and thus Here we stress that in the case of the Hopf fibration $\pi\colon S^3 \to S^2$ the restriction to an arbitrary smooth loop $\gamma \subset S^2$ gives rise to a 2-torus $\pi^{-1}(\gamma) \subset S^3$. Now notice the following fact: the Hamiltonian $U(1)$-action on $\operatorname{Gr}(1, 3)$ induced by $\widetilde \mu_0$ preserves the base of the fibration and acts on its fibres, preserving the structure of a Cartesian product. In fact, let the line $l$ be the intersection of two planes $\pi_0$ and $\pi_1$ from the pencils of planes containing the lines $\langle p_0, s_0 \rangle$ and $ \langle p_1, s_1 \rangle$, respectively. The Hamiltonian action of each $\widetilde \mu_i$ fixes all points $p_i$ and $s_j$, because they lie in the critical sets of $\widetilde \mu_i$. Hence the lines $\langle p_i, s_i \rangle$ are taken to themselves by this action, so that the pencils of planes are too. This shows that the torus action preserves the base $\langle p_2, p_3 \rangle \times \langle p_2, p_3 \rangle$ and the fibre bundle structure, acting fibrewise. This finishes the conceptual (coordinateless) proof of Proposition 1. Note that to deduce the assertion of Theorem 2 concerning $\operatorname{Gr}(1, 3)$ from Proposition 1 it remains to show that a joint level set of the functions $\widetilde \mu_i$ is fibred with fibre $\mathrm T^2$. We perform analytic computations and deduce explicit formulae. In the Plücker variables $w_{ij}$ on the projective space $\mathbb{P}(\Lambda^2 \mathbb{C}^4)$, where ${0 \,{\leqslant}\, i \,{<}\, j \,{\leqslant}\, 3}$, the Grassmannian $\operatorname{Gr}(1, 3)$ is realized by the quadric $Q=\{ w_{01}w_{23}-w_{02} w_{13} + w_{03} w_{12}=0 \} \subset \mathbb{C} \mathbb{P}^5$; our distinguished moment maps have the form
$$
\begin{equation}
\widetilde \mu_0=\frac{| w_{01} |^2 + | w_{02} |^2 + | w_{03} |^2}{\sum_{i<j} | w_{ij} |^2}\quad\text{and} \quad \widetilde \mu_1=\frac{| w_{01} |^2 + | w_{12} |^2 + | w_{13} |^2}{\sum_{i<j} | w_{ij} |^2}
\end{equation}
\tag{2}
$$
(see [2]). The corresponding Hamiltonian action is easily seen to preserve $Q$; in addition, there are two pencils of divisors on $Q$ each of which is invariant under the action. Consider indeed the pencils
$$
\begin{equation*}
Q_0 (\alpha)=\{ \alpha_0 w_{02} + \alpha_1 w_{03}=0 \} \cap Q\quad\text{and} \quad Q_1(\beta)=\{\beta_0 w_{12} + \beta_1 w_{13} \} \cap Q.
\end{equation*}
\notag
$$
Since each $\widetilde \mu_i$ acts trivially on $Q_j$ for $i \neq j$ and acts on $Q_i$ by multiplying all coordinates in its definition simultaneously by $e^{it}$, the corresponding pencils are invariant under the action of the whole of $\mathrm T^2$. By the definition of the Plücker embedding the intersection $N(\alpha, \beta)=Q_0(\alpha) \cap Q_1 (\beta) \subset \operatorname{Gr}(1, 3)$ consists of the lines intersecting the corresponding two lines in the pencils in $\mathbb{C} \mathbb{P}^3$ spanned by the pairs $( \langle p_0, p_2 \rangle, \langle p_0, p_3 \rangle)$ and $(\langle p_1, p_2 \rangle, \langle p_1, p_3 \rangle)$. Each line in these pencils is uniquely assigned to its point of intersection with $\langle p_2, p_3 \rangle$, so that the parametrization $N(\alpha, \beta)$ is replaced in a natural way by the parametrization $N(s_0, s_1)$ by ordered pair of points on the line $\langle p_2, p_3 \rangle$. Finally, if a line $l \subset \mathbb{C} \mathbb{P}^3$ intersects both $\langle p_0, s_0 \rangle$ and $\langle p_1, s_1 \rangle$, then it is the intersection of the planes $\langle l, p_0 \rangle$ and $\langle l, p_1 \rangle$. This shows that the pencils $Q_0$ and $Q_1$ can be replaced by the appropriate pencils of planes containing the lines $\langle p_i, s_i \rangle$, so that the analytic description associating with a point $[l] \in \operatorname{Gr}(1, 3)$ an ordered pair $(s_0, s_1)$ of points on $\langle p_2, p_3 \rangle$ in the way indicated above corresponds exactly to the above geometric description of $\operatorname{Gr}^0(1, 3)$. Now note that to identify $\operatorname{Gr}^0(1, 3)$ consistently with a fibre bundle in the analytic setup we must exclude the basis sets of the pencils $Q_0$ and $Q_1$, which have the representations $B_0=\{ w_{02}=w_{03}=0 \}$ and $B_1=\{w_{12}=w_{13}=0 \}$. It is easy to see that $B_0$ is equivalent to the condition $l \subset \langle p_0, p_2, p_3 \rangle$, while $B_1$ is equivalent to $l \subset \langle p_1, p_2, p_3 \rangle$, and both these cases were excluded from $Gr^0(1, 3)$ by the choice of the noncritical values of the moment maps. Let $N^0(s_0, s_1)=N(s_0, s_1) \cap \operatorname{Gr}^0(1,3)$; then we obviously have a fibre bundle $\tau\colon \operatorname{Gr}^0(1,3) \to \langle p_2, p_3 \rangle \times \langle p_2, p_3 \rangle$ with fibre $N^0(s_0, s_1)$. These fibres can analytically be presented as the intersection $Q_0 \cap Q_1 \cap \operatorname{Gr}^0(1, 3)$, that is, they can be defined by the three equations
$$
\begin{equation*}
\alpha_0 w_{02} + \alpha_1 w_{03}=0, \qquad \beta_0 w_{12} + \beta_1 w_{13}=0\quad\text{and} \quad w_{01}w_{23} - w_{02} w_{13} + w_{03} w_{12}=0,
\end{equation*}
\notag
$$
so that if two points $s_0, s_1 \in \langle p_2, p_3 \rangle$ are distinct, then this intersection is a nondegenerate 2-quadric, while if $s_0=s_1$, which is equivalent to $\alpha_0 \beta_1=\alpha_1 \beta_0$, then this intersection is the pair of planes $\{ w_{01}=0 \} \cup \{w_{23}=0 \}$. Note furthermore that this intersection is always toric because there exist two moment maps, $\widetilde \mu_0$ and $ \widetilde \mu_1$, such that the induced Hamiltonian action preserves each fibre. From (2) we obtain
$$
\begin{equation*}
\widetilde \mu_0 + \widetilde \mu_1=1 + \frac{| w_{01} |^2 - | w_{23} |^2}{\sum_{i<j} | w_{ij} |^2},
\end{equation*}
\notag
$$
which shows that $\widetilde \mu_0 + \widetilde \mu_1 < 1$ corresponds to the choice of the plane $ w_{01}=0$ in the degenerate fibre over the diagonal in $\langle p_2, p_3 \rangle \times \langle p_2, p_3 \rangle$. Thus the global picture looks as follows: over the general point of the product $\langle p_2, p_3 \rangle \times \langle p_2, p_3 \rangle$ there is a toric fibre, namely, a nondegenerate quadric with Delzant polygon equal to a square; over a diagonal point the toric fibre degenerates to a union of a pair of planes, and its Delzant polygon consists of the triangles arising in the decomposition of the square; the intersection of these two planes corresponds to the diagonal $\widetilde \mu_0 + \widetilde \mu_1=1$ of the square. Hence $N^0(s_0, s_1)$ is the preimage of an open triangle, so the preimage of each point in this triangle is a smooth 2-torus invariant under the Hamiltonian action of $\mathrm T^2$ induced by the commuting functions $\widetilde \mu_0$ and $ \widetilde \mu_1$. This completes the analytic proof of Proposition 1. It is easy to see that the assertion of Theorem 2 for $\operatorname{Gr}(1, 3)$ is a direct consequence of Proposition 1. § 2. General caseNow consider the general case of the Grassmannian $\operatorname{Gr}(1, n)$ of straight lines in $\mathbb{C} \mathbb{P}^n$. Fix homogeneous coordinates $[z_0: \dotsb : z_n]$ compatible with the Kähler structure and consider the $k$ standard moment maps
$$
\begin{equation*}
\mu_i=\frac{| z_i |^2}{\sum_{i=0}^n | z_i |^2}, \qquad i=0, \dots, k-1,
\end{equation*}
\notag
$$
and the corresponding moment maps $\widetilde \mu_i$ from $\operatorname{Gr}(1, n)$. Again, we distinguish in $\operatorname{Gr}(1, n)$ the open component $\operatorname{Gr}^0(1, n)$ by the conditions $\widetilde \mu_i>0$ and $\widetilde \mu_0 + \dots + \widetilde \mu_{k-1} < 1$ and show that the following result holds. Proposition 2. The open manifold $\operatorname{Gr}^0(1, n)$ is fibred over the manifold $M_{n-k}$, $k=0, \dots, n-1$, the total space of the Cartesian product of $k$ copies of the projectivization of the tautological bundle $\mathbb{P}(\tau)\mkern-1mu \to \operatorname{Gr}(1, \langle p_k,\dots, p_{n+1} \rangle )$, so that the Hamiltonian action of the torus $T^k$ induced by $\widetilde \mu_0, \dots, \widetilde \mu_{k-1}$ preserves the base and the fibres. Here and in what follows we let $\operatorname{Gr} (1, H)$ denote the manifold of projective lines in the projective space $H$, including the case when $H$ is the projective hull of some points $\langle p_{i_1}, \dots, p_{i_m} \rangle$. Proof. Since here, as in the case of $\operatorname{Gr}(1, 3)$ considered above, we have excluded the lines $l \subset \mathbb{C}\mathbb{P}^n$ intersecting the subspace $\langle p_0, \dots, p_{k-1} \rangle $ or lying in the hyperplanes from consideration, $\operatorname{Gr}^0(1, n)$ can be fibred over $M_{n-k}$: each line $l$ in $\operatorname{Gr}^0(1, n)$ is skewed with $\langle p_0, \dots, p_{k-1} \rangle$, so the projective hull $H_l=\langle l, p_0, \dots, p_{k-1} \rangle$ must be projective $(k+1)$-dimensional; on the other hand $l$ determines uniquely the subspace $H_l$ and the line $l_0 \subset H_l \cap \langle p_k, \dots, p_n \rangle$ in the last $(n-k)$-dimensional projective space. It is easy to see that $H_l$ itself can uniquely be recovered from $l_0$ because it is the projective hull $\langle l_0, p_0, \dots, p_{k-1} \rangle$.
The following observation can be made: the Hamiltonian action $\mathrm T^k$ induced by the moment maps $\widetilde \mu_0, \dots, \widetilde \mu_{k-1}$ we have selected preserves the subsets $\operatorname{Gr}^0(1, H_l) \subset \operatorname{Gr}^0(1, n)$, where $\operatorname{Gr}^0(1, H_l)$ is the set of lines in the $(n-k)$-dimensional projective space $H_l \subset \mathbb{C} \mathbb{P}^n$ that satisfy the same conditions on the functions $\widetilde \mu_i$ that distinguish the component $\operatorname{Gr}^0(1, n)$. In fact, the $(k+1)$-dimensional subspace $H_l$ is uniquely determined by the pair of projective subspaces $l_0$ and $\langle p_0, \dots, p_{k-1} \rangle$ in it. The Hamiltonian action induced by $\widetilde \mu_i$ fixes the line $l_0 \subset \langle p_k, \dots, p_n \rangle$ because $l_0$ lies fully in the critical set corresponding to the zero critical values of all moment maps $\widetilde \mu_0, \dots, \widetilde \mu_{k-1}$ under consideration; on the other hand the projective subspace $\langle p_0, \dots , p_{k-1} \rangle$ is also invariant (although not fixed), which justifies our observation. Thus, our problem is split into two: the first problem consists in the symplectic reduction by a Hamiltonian action on each Grassmannian $\operatorname{Gr}(1, H_l)$ for a fixed subspace $H_l$ invariant under the action of $T^k$ induced by $\widetilde \mu_0, \dots, \widetilde \mu_{k-1}$, and the second is the globalization of the construction and the description of a suitable variety of parameters. The solution of the first problem follows the pattern in § 1: the whole set $\operatorname{Gr}(1, H_l)$ with the system of maps $\widetilde \mu_i|_{\operatorname{Gr}(1, H_l)}$, $i=0, \dots, k-1$, is fibred with torus fibres in a natural way. In fact, let us assign to a line $l \subset H_l$ a set of points $(s_0, \dots, s_{k-1})$, $s_i \in l_0$, by the following rule: first we look at the point $s_j^0= \langle l_0, p_{m_1}, \dots, p_{m_{k-1}} \rangle \cap l$ on $l$, where $(j, m_1, \dots, m_{k-1})$ is a permutation of the integers $(0, \dots, k-1)$, and then project it from $\langle p_{m_1}, \dots, p_{m_{k-1}} \rangle$ onto $l_0$ in the projective space $\langle l_0, p_{m_1}, \dots, p_{m_{k-1}} \rangle$ and denote the result by $s_j \in l_0$. Then in $\operatorname{Gr}(1, H_l)$ we have the system of divisors $\{ D(s_j) \}$ defined by
$$
\begin{equation*}
D(s_j)=\{ [l] \in \operatorname{Gr}(1, H_l) \mid l \cap \langle s_j, p_{m_1}, \dots, p_{m_{k-1}} \rangle \neq \varnothing \}
\end{equation*}
\notag
$$
(for the standard approach to the definition of a divisor in a Grassmannian, see [3]).
Note that for each point $s_j \in l_0$ the corresponding divisor $D(s_j)$ is invariant under the Hamiltonian action induced by each of our moment maps $\widetilde \mu_i$. In fact, the projective subspace $\langle s_j, p_{m_1}, \dots, p_{m_{k-1}} \rangle \subset H_l$ is spanned by points fixed by the Hamiltonian action induced by $\mu_i$, so each line $[l] \in D(s_j)$ is taken by the Hamiltonian action of $\widetilde \mu_i$ to a line in the same submanifold. Hence the submanifold
$$
\begin{equation*}
Y(s_0, \dots, s_{k-1})=\bigcap_{j=0}^{k-1} D(s_j) \subset \operatorname{Gr}(1, H_l)
\end{equation*}
\notag
$$
is invariant under the action of the whole torus $T^k$ induced by the moment maps $\widetilde \mu_0, \dots , \widetilde \mu_{k-1}$; on the other hand the dimension of this submanifold if $k$, thus for each set $(s_0, \dots, s_{k-1})$ we have a (not necessarily smooth) toric fibre $Y(s_0, \dots, s_{k-1})$.
By the general theory of toric varieties (see [5]) the variety $Y(s_0, \dots, s_{k-1})$ endowed with the set of moment maps $\widetilde \mu_0, \dots, \widetilde \mu_{k-1}$ is uniquely defined by its convex Delzant polytop $P_Y \subset \mathbb{R}^k$, the image of the action map
$$
\begin{equation*}
F_{\mathrm{act}}=(\widetilde \mu_0, \dots \widetilde \mu_{k-1})\colon Y(s_0, \dots, s_{k-1}) \to \mathbb{R}^k.
\end{equation*}
\notag
$$
It is easy to see that, given a general set $(s_0, \dots, s_{k-1})$, the Delzant polytop $P_Y$ is as follows: consider the unit cube in $\mathbb{R}^k$ and remove from it the part cut out by the hyperplane $\sum_{i=1}^k x_i=2$ and containing the origin. In fact, the toric variety $Y(s_0, \dots, s_{k-1})$ contains by definition the lines that are the preimages of vertices of the form (all the $x_i$ are zero), (just one $x_i$ is nonzero), (precisely two $x_i$ are nonzero), but just for the set of such vertices of the unit cube their convex hull is the above polytop $P_Y$. Here are such lines: the line $l=l_0 \subset H_l$ is taken to the point of the first type; the line $\langle p_i, s_i \rangle \subset H_l$, $i=0, \dots, k-1$, is taken to the point with $i$th coordinate 1 and all the other zero, finally, the line $\langle p_i, p_j \rangle \subset H_l$, $0 \leqslant i < j \leqslant k-1$, is taken to the vertex with $i$th and $j$th coordinates 1, and the other zero. As in the above case of $\operatorname{Gr}(1, 3)$, when two points $s_i=s_j$ among the ones specifying the manifold $Y(s_0, \dots, s_{k-1})$ coincide, this toric variety is reducible and consists of toric components. In fact, in this case the intersection of the divisors $D(s_i) \cap D(s_j)$ falls into the union of two components $(D(s_i) \cap D(s_j))_a \cup (D(s_i) \cap D(s_j))_b$, where the $a$-component consists of the lines passing through the point $s_i=s_j$, and the $b$-component consists of the lines lying in the projective subspace $ \langle p_0, \dots, p_{k-1},\, s_i=s_j \rangle \subset H_l$. As in the case of $\operatorname{Gr}(1, 3)$, these components are distinguished by the values of the moment maps: the $a$-component corresponds to $\sum_{i=0}^{k-1} \widetilde \mu_i \leqslant 1$, and the $b$-component to the case when this sum is at least 1. When a larger number of the $s_i$ coincide, we obtain a more special picture of the decomposition of $Y(s_0, \dots, s_{k-1})$ into components, but a full analysis of the arising combinatorial structure is not one of our aims here, because to prove Theorem 2 it is sufficient to recover the conditions $c_i>0$ and $\sum_{i=0}^{k-1} c_i < 1$, imposed on the values of the moment maps $\widetilde \mu_0, \dots, \widetilde \mu_{k-1}$, and to verify that each such set $(c_0, \dots, c_{k-1})$ defines a $k$-dimensional Liouville torus in each toric manifold $Y(s_0, \dots, s_{k-1})$, on which the torus $T^k$ generated by the Hamiltonian action of $\widetilde \mu_0, \dots, \widetilde \mu_{k-1}$ acts freely. Thus, as a preliminary conclusion, we see that for the values of the moment maps satisfying $c_i>0$ and $\sum_{i=0}^{k-1} c_i < 1$ the symplectic reduction of $\operatorname{Gr}(1, H_l)=\operatorname{Gr}(1, k+1)$ by the action of the $k$-torus results in the Cartesian product of $k$ copies of the projective line $l_0$. Now we make the second step: clearly, in the correspondence $l \mapsto H_l \mapsto l_0 \subset \langle p_k, \dots, p_n \rangle$ the second arrow is one-to-one, so we obtain a map $\operatorname{Gr}^0(1, n) \to \operatorname{Gr}(1, \langle p_k, \dots, p_n \rangle)$. As shown above, over each point $[l_0] \in \operatorname{Gr} (1, \langle p_k, \dots, p_n \rangle)$ we have a fibre bundle of toric open components $Y^0(s_0, \dots, s_{k-1}) \subset Y(s_0, \dots, s_{k-1})$, parametrized by sets of points $(s_0, \dots, s_{n-1})$ lying on $l_0$. Hence, globalizing this picture over the whole of $ \operatorname{Gr}(1, \langle p_k, \dots, p_n \rangle)$, taking into account that the $s_i$ turn out to become points on the projectivization of the tautological bundle $\tau \to \operatorname{Gr}(1, n-k)$, we obtain the assertion of Proposition 2. We denote the corresponding manifold $\operatorname{tot}(\mathbb{P}(\tau) \times \dots \times \mathbb{P}(\tau) \to \operatorname{Gr}(1, n-k))$ by $M_{n-k}$. Now if we choose a set of values $c_0, \dots, c_{k-1} \in \mathbb{R}_+$ so that $\sum_{i=0}^{k-1} c_i < 1$, then in each toric fibre of $\operatorname{Gr}^0(1, n) \to M_{n-k}$ we distinguish a smooth torus, which yields a point after taking the quotient, providing the result of Theorem 2:
$$
\begin{equation}
\operatorname{Gr}(1, n)/\!/ \mathrm T^k \cong \operatorname{tot} (\mathbb{P}(\tau) \times \dots \times \mathbb{P}(\tau) \to \operatorname{Gr}(1, n-k)),
\end{equation}
\tag{3}
$$
where each fibre is a $k$-fold Cartesian product.
The proposition is proved. Remark 3. The above representation does not allow us to say what form arises on the reduced symplectic manifold $M_{n-k}$ by the reduction from the original Kähler form on $\operatorname{Gr}(1, n)$. Although for close values of the moment maps the symplectic reduction produces isomorphic manifolds, the symplectic forms on these manifolds are different. For instance, we cannot say in general that when the symplectic form $\omega_t$, $t=(c_0, \dots, c_{k-1})$, obtained after the symplectic reduction to $M_{n-k}$ is restricted to the projective lines $\mathbb{P}(\tau)$ over the same point in the base $\operatorname{Gr}(1, n-k)$, the result is the same symplectic form. For some submanifolds constructed we do not need this. However, as we use nonstandard antiholomorphic involutions below, we will need the following simple observation. Under an isomorphism of the original projective space $\mathbb{C} \mathbb{P}^n$ induced by a permutation $z_i \leftrightarrow z_j$ of the homogeneous coordinates we have: (1) the symmetry $\widetilde \mu_i \leftrightarrow \widetilde \mu_j$ of moment maps; (2) the symmetry $s_i \leftrightarrow s_j$ of points on the projective line $l_0$; (3) hence the Kähler symmetry of the $i$th and $j$th direct factors in $M_{n-k}$ over the base $\operatorname{Gr}(1, n-k)$. This means that if in what follows we supplement the conditions on the values of the moment maps by the additional symmetry condition $c_0=\dots=c_{k-1}$, then this defines an action of the group of permutations by Kähler isometries on $M_{n-k}$. Note however that to implement Mironov’s construction, in which the real parts of manifolds with holomorphic involution play an important role, we need not consider the fully symmetric case $c_0=\dots=c_{k-1}$. In fact, our moment maps $\widetilde \mu_i$ act on $\operatorname{Gr} (1, n)$ by Kähler isometries, and the group $\mathrm T^k$ is Abelian, so each value of the moment map (in the sense of $\mathfrak g^*$) lies in the centre of the algebra, as required for a consistent implementation of Kähler reduction (see [3]). Thus the construction itself of the manifold $M_{n-k}$ endows it with a Kähler structure; furthermore, this construction agrees with the antiholomorphic involutions $\tau\colon M_{n-k} \to M_{n-k}$ which is also induced by the standard antiholomorphic involution on $\mathbb{C} \mathbb{P}^n$. In fact, it is easy to see that the standard conjugation of a line $l \subset \mathbb{C} \mathbb{P}^n$ agrees with the conjugation of the data associated with this line: since all points $p_i \in \mathbb{C} \mathbb{P}^n$ are real, $\overline l$ corresponds to $\overline l_0 \subset \langle p_k, \dots, p_n \rangle$, and the set of points $s_i \in l_0$ is transformed into the conjugate set $\overline s_i \in \overline l_0$. Hence, for any Kähler form on $M_{n-k}$ obtained by Kähler reduction the real part of this manifold is a Lagrangian submanifold; furthermore, this real part is distinguished by the reality conditions which are imposed both on the base $\operatorname{Gr}(1, n-k)$ (to ensure that $l_0$ is real) and on points in fibres of all direct factors $\mathbb{P} (\tau)$. In other words, the quotient manifold $M_{n-k}$ is naturally embedded in the Cartesian product
$$
\begin{equation}
\operatorname{Gr}(1, \langle p_k, \dots, p_n \rangle) \times \langle p_k, \dots, p_n \rangle \times \dots \times \langle p_k, \dots, p_n \rangle,
\end{equation}
\tag{4}
$$
where we see on the right the Cartesian product of $k$ copies of a projective space with the same complex structure induced by the original complex structure on $\langle p_k, \dots, p_n \rangle$ and with the same antiholomorphic involution, but with their own Kähler (symplectic) structures. The reduced manifold $M_{n-k}$ is realised as an intersection of $k$ cycles: where $\widetilde{\mathcal U}_i=\langle p_k, \dots, p_n \rangle \times \dots \times \mathcal U_i \times \dots \times \langle p_k, \dots, p_n \rangle$ is obtained by the substitution of $\mathcal U_i \subset \operatorname{Gr}(1, \langle p_2, \dots, p_n \rangle) \times \langle p_k, \dots, p_n \rangle$, the incidence cycle of the $i$th projective space, into the Cartesian product at the $i$th place; while the reduced Kähler structure is the restriction from the Cartesian product (4) to the submanifold $M_{n-k}$ defined by (5). Remark 4. It is easy to see that, as in the case of $\operatorname{Gr}(1,3)$, the fibration $\operatorname{Gr}^0(1, n) \to M_{n-k}$ is topologically nontrivial. Let us indeed return to the divisors $D(s_i) \subset \operatorname{Gr}(1, H_l)$ for the set $s_0, \dots, s_{k-1} \in l_0$. In place of $D(s_i)$ we can consider the pencil of $k$-dimensional projective subspaces containing the $(k-1)$-dimensional subspace $ \langle s_j, p_{m_1}, \dots, p_{m_{k-1}} \rangle$: it is obvious that if a line $l$ intersects $\langle s_j, p_{m_1}, \dots, p_{m_{k-1}} \rangle$, then it must lie in one of these $k$-dimensional subspaces. Hence we obtain a fibre bundle over the Cartesian product of $k$-copes of $l_0$ with fibre equal to the Cartesian product of $k$ copies of the projective lines parametrizing the above pencils. However, such a pencil has the same form as in the case of $\operatorname{Gr}(1, 3)$; see Remark 2. In fact, on the projective plane $\langle p_j, l_0 \rangle$ the corresponding pencil of $k$-dimensional projective subspaces cuts the pencil of lines through $s_j$, and by varying $s_j$ along $l_0$ we obtain several copies of the Hopf fibration again, provided that we fix an appropriate set of values of the moment maps $\widetilde \mu_i$ satisfying $c_i>0$ and $ \sum_{i=0}^{k-1} c_i < 1$. It is obvious that this construction also globalizes over $\operatorname{Gr}(1, n-k)$, which provides the topological type of the fibration $\operatorname{Gr}^0(1, n) \to M_{n-k}$. Remark 5. In some cases (for example, in the central problem in [1] concerning the construction of minimal and Hamiltonian minimal Lagrangian submanifolds) it is necessary to consider some particular distinguished values $t \in \mathfrak g^*$ of the moment map for which the group does not act freely on the corresponding level manifold $N_t$, so that the quotient space $N_t/ G$ has singularities. For instance, in our considerations of Proposition 1 we could allow values $c_0$ and $ c_1$ of the moment maps $\widetilde \mu_0$ and $ \widetilde \mu_1$ such that $c_0 + c_1=1$. In this case, as follows from the formulae in § 1, over the diagonal in the product $\langle p_2, p_3 \rangle \times \langle p_2, p_3 \rangle$ we have 1-tori in place of 2-tori, that is, we have circles which lie on lines in the intersection $\{w_{01}=0 \} \cap \{ w_{23}=0 \}$. Hence taking the quotient for these values of the moment maps we obtain a reduced manifold with singularities, but the construction from Theorem 1 can also be used in this case if we consider Lagrangian immersions, including Lagrangian immersions in symplectic manifolds with singularities. This separate topic growing out of Theorem 1 requires a thorough separate investigation. Remark 6. Our construction cannot be transferred directly to $k=1$ or $k=n$: in the first case this is not necessary because in [4] we constructed already a relevant Lagrangian manifold, which has the same form, namely, is a fibre bundle of 2-tori over $\operatorname{Gr}(1, n-1)$; in the case $k=n$ the projective space complementary to $\langle p_0, \dots, p_{n-1} \rangle$ has dimension zero, so that we have no place to project lines onto. In this paper we use the method presented above, so we defer the case $k=n$ to future work. Now we are ready to use Theorems 1 and 2 to construct a large stock of Lagrangian submanifolds in the Grassmannian $\operatorname{Gr}(1, n)$, which is our main task. § 3. ExampleProposition 3. The manifold $\operatorname{Gr} (1, n)$ endowed with a symplectic form by means of the Plücker embedding contains a system of smooth Lagrangian submanifolds $\{ S_k \}$, $k=0, \dots, n-1$, representable as smooth fibre bundles of $T^{2k}$-tori over the real Grassmannians $\operatorname{Gr}_{\mathbb{R}} (1, n-k)$. Proof. We begin with $k=0$ and $k=1$; the first case is obvious since the corresponding assertion is tautological; the second case was treated in [4]. Now, it follows from Theorem 2 that for each $k=2, \dots, n-1$, for the choice of moment maps and their values as indicated above the symplectic reduction $\operatorname{Gr}(1, n)/\!/ T^k$ is a Kähler manifold $M_{n-k}$. Its real part $M^{\mathbb{R}}_{n-k}$ with respect to the antiholomorphic conjugation induced by the standard antiholomorphic conjugations on $\mathbb{C} \mathbb{P}^k$ and $\operatorname{Gr}(1, n-k)$ in the product (4) can be represented as a real version of the intersection (5), so we have and since a fibre of $\mathbb{P}(\tau_{\mathbb{R}}) \to \operatorname{Gr}_{\mathbb{R}}(1, n-k)$ is a circle, $M^{\mathbb{R}}_{n-k}$ is globally a topologically nontrivial smooth fibre bundle with fibre $S^1 \times \dots \times S^1=T^k$ (with the exception of the case $k=n-1$, when $\operatorname{Gr}(1, n-k)$ is just a point and $M^{\mathbb{R}}_{n-1}$ is merely isomorphic to $T^{n-1}$).
We know from Theorem 1 and Remark 4 that to each Lagrangian submanifold $L \subset M_{n-k}$ there corresponds a Lagrangian submanifold of $\operatorname{Gr}(1, n)$ whose topological type is a fibration with fibre $T^k$ over $L$; since $M^{\mathbb{R}}_{n-k} \subset M_{n-k}$ is Lagrangian, to prove Proposition 3 it is sufficient to show that, although $T^k$ is topologically nontrivial over $M_{n-k}$, the corresponding Lagrangian submanifold has the type of a bundle of $2k$-tori over $\operatorname{Gr}_{\mathbb{R}}(1, n-k)$. However, as shown in Remark 4, the fibre over a point in $\operatorname{Gr}_{\mathbb{R}}(1, n-k)$ can be represented as $k$ copies of the Hopf fibration, which, after the restriction to the real torus $\mathbb{P} (\tau_{\mathbb{R}}) \times \dots \times\mathbb{P}(\tau_{\mathbb{R}})$, yield, as noted in Remark 2, the Cartesian product of $k$ copies of $T^2$, that is, the torus $T^{2k}$. We stress that the corresponding fibration of $2k$-tori $S_k \to \operatorname{Gr}_{\mathbb{R}}(1, n-k)$ is topologically nontrivial by construction, similarly to the result for $k=1$ obtained in [4]. This completes the proof of Proposition 3. Remark 7. It is easy to see that the Lagrangian submanifolds $S_k \subset \operatorname{Gr}(1, n)$ presented above are just the ones provided by the generalized Mironov construction in the case when the moment maps $\widetilde \mu_i$ and their values are chosen as above. Thus, in this paper we have made a decent progress in the problem of the description of the types of Lagrangian submanifolds of $\operatorname{Gr} (1, n)$ with the help of the generalized Mironov construction described in [2]. Furthermore, all Lagrangian submanifolds $\{S_k \}$ are interrelated by transformations called Lagrangian mutations in Remark 2, namely, we have a chain where the $i$th arrow corresponds to the Lagrangian mutation induced by the moment map $\widetilde \mu_{i-1}$; at the same time, for each pair $0 \leqslant i < j \leqslant n-1$ the corresponding system of moment maps $\widetilde \mu_i, \dots, \widetilde \mu_{j-1}$ induces a Lagrangian mutation $S_i \mapsto S_j$. Proposition 3 turns the conjecture in [2] into an assertion; however, formulae (4) and (5) suggest us a further class of manifolds extending significantly the set of examples of Lagrangian submanifolds. As usual, we begin with the simplest situation and consider $k=2$. Then, as in (4) and (5), we realize $M_{n-2}$ as a submanifold of the threefold Cartesian product $\mathbb{C} \mathbb{P}^{n-2} \times \operatorname{Gr}(1, n-2) \times \mathbb{C} \mathbb{P}^{n-2}$, where the first an third factors are identified as complex manifolds; if, bearing Remark 3 in mind, we supplement our conditions on the values of the moment maps by the extra condition $c_0=c_1$, then the first and third factors are also identified as Kähler manifolds. Note that in this case $M_{n-2}$ is birationally isomorphic to the Cartesian product $\mathbb{C} \mathbb{P}^{n-2} \times \mathbb{C} \mathbb{P}^{n-2}$, and is the blowup of the diagonal in this product. An example of a Lagrangian submanifold representing a nontrivial homology class is known in this product: by introducing the homogeneous coordinates $[x_i]$ and $ [y_i]$ in the first and second factors consider the Lagrangian submanifold equal to the antidiagonal
$$
\begin{equation*}
\overline \Delta=\{ y_i=\overline x_i \} \subset \mathbb{C} \mathbb{P}^{n-2} \times \mathbb{C} \mathbb{P}^{n-2}.
\end{equation*}
\notag
$$
Since $\overline \Delta$ intersects $\Delta$ at the real points, it is obvious that the full preimage of $\overline \Delta$ in $M_{n-2}$ with respect to the projection
$$
\begin{equation*}
\pi\colon M_{n-2} \to \mathbb{C} \mathbb{P}^{n-2} \times \mathbb{C} \mathbb{P}^{n-2}
\end{equation*}
\notag
$$
cannot be Lagrangian in $M_{n-2}$: fibres of $\pi$ over real points are complex submanifolds. However, using the representations (4) and (5) we can construct a smooth Lagrangian submanifold of $M_{n-2}$ of a similar type. Namely, we consider the antiholomorphic involution on the Cartesian product (4) that acts in the standard way on $\operatorname{Gr} (1, n-2)$ and in the antidiagonal way, that is, by the formula $([x_i], [y_i]) \mapsto ([\overline y_i], [\overline x_i])$ on $\mathbb{C} \mathbb{P}^{n-2} \times \mathbb{C} \mathbb{P}^{n-2}$. Clearly, the fixed points of this antiholomorphic involution form a Lagrangian submanifold of (4) (it is important here that the Kähler forms coincide, which holds for $c_0=c_1$) but, remarkably, it can also be restricted to $M_{n-2}$ in the natural way and define an antiholomorphic involution there. Let us show that $M_{n-2}$ is invariant under this involution, which yields the required result. In fact, by definition a point in $M_{n-2}$ is a set $(p_1, l, p_2)$ consisting of a line $l$ and two points on it; after the above nonstandard conjugation this set goes to $(\overline p_2, \overline l, \overline p_1)$, which obviously has the same incidence properties. Hence the nonstandard involution preserves $M_{n-2}$. Therefore, if the fixed point set $M^r_{n-2} \subset M_{n-2}$ has the necessary dimension, then $M^r_{n-2}$ is a Lagrangian submanifold. However, recalling the representation
$$
\begin{equation*}
M_{n-2} \equiv \operatorname{tot}(\mathbb{P} (\tau) \times \mathbb{P}(\tau) \to \operatorname{Gr}(1, n-2))
\end{equation*}
\notag
$$
as a fibre bundle over $\operatorname{Gr}(1, n-2)$, it is easy to see that $M^r_{n-2}$ is fibred with fibre $S^2$ over $\operatorname{Gr}_{\mathbb{R}}(1, n-2)$. The real dimension of this total space is precisely equal to the complex dimension of $M_{n-2}$, which gives us a new example of a Lagrangian submanifold of $M_{n-2}$ and thus of $\operatorname{Gr} (1, n)$. Let us determine the topological type of this Lagrangian submanifold. Since for appropriate fixed values of $\widetilde \mu_0$ and $ \widetilde \mu_1$ the fibre over a point in $\operatorname{Gr}(1, n-2)$ consists of two copies of the Hopf fibration, we must find the topological type of the restriction of the Hopf fibration to the antidiagonal in the Cartesian product $\mathbb{P}(\tau) \times \mathbb{P} (\tau)$. However, this fibration has Chern class zero, while $S^2$ is simply connected, so we have a fibre $T^2 \times S^2$ over each point in $\operatorname{Gr}_{\mathbb{R}}(1, n-2)$. Thus, globally we obtain a fibration with fibre $T^2 \times S^2$ over $\operatorname{Gr}_{\mathbb{R}}(1, n-2)$. Moreover it is obvious that this topological type is distinct from the topological type of any $S_k$ in Proposition 3. Hence we have supplemented the list of topological types of smooth Lagrangian submanifolds by a new type that arises after reduction by $T^2$. But why not go the same way in the case when $k=3$? After all, then $M_{n-3}$ is isomorphic to $\operatorname{tot} (\mathbb{P}(\tau) \times \mathbb{P}(\tau) \times \mathbb{P}(\tau) \to \operatorname{Gr}(1, n-3))$, so it admits a submanifold of mixed type, namely, over the real part of the Grassmannian $\operatorname{Gr}_{\mathbb{R}}(1, n-3)$ we take the real part $\mathbb{P}(\tau_{\mathbb{R}})$ of one direct factor and treat the other two components as in the case $k=2$. As a result, we obviously obtain yet another Lagrangian submanifold of $M_{n-3}$, whose type is a topologically nontrivial bundle with fibre $S^2 \times S^1$ over $\operatorname{Gr}_{\mathbb{R}}(1, n-3)$. Next, generalizing to $k=4$, we obtain even more opportunities: over $\operatorname{Gr}_{\mathbb{R}}(1, n-4)$ we have fibrations with fibre $S^1 \times S^1 \times S^1 \times S^1$ (the generalized Mironov cycle $S_4$ from Proposition 3), with fibre $S^2 \times S^1 \times S^1$ and with fibre $S^2 \times S^2$, which are three different topological types. Quite naturally, after a pullback along the relevant $T^k$-fibration, each type defines some type of a Lagrangian submanifold of $\operatorname{Gr}(1, n)$. The only requirement for such a construction is that the corresponding values of the moment maps must match. However, setting merely $1>c_0=\dots=c_{k-1}>0$ and $c_0 < 1/k$ makes it possible to consider arbitrary pairs. From the above constructions and arguments we obtain the following statement. Corollary 1. The number of distinct possible types of Lagrangian submanifolds of the Grassmannian $\operatorname{Gr}(1, n)$ is at least $n + [{n}/{2}][(n-1)/{2}]$. Proof. According to our constructions and observations, for each fixed $k=0, \dots, {n-1}$, by Theorem 1 we have at least $1 + [{k}/{2}]$ different types of Lagrangian submanifolds of $\operatorname{Gr}(1, n)$. In fact, for $M_{n-k}$ we have the standard real part $M^{\mathbb{R}}_{n-k}$ (yielding 1 in the above formula) plus nonstandard real parts of the form $M^r_{n-k}$, where we can choose a pair, two pairs, and so on from the $k$ direct factors $\mathbb{P}(\tau)$, provided that all the $c_i$ coincide; the maximum possible number of pairs is $[{k}/{2}]$. Hence the total number of such topological types is as required. Here it must be noted that our lower bound is far from being effective: for instance, even in the simple case of $k=2$ and a particular choice of the pair $\mu_0, \mu_1$ of moment maps we have not examined in full the Lagrangian geometry of the blowup $M_{n-2}$. This problem is also interesting in itself because birational transformations of algebraic varieties are the main source of the constrictions of new manifolds and the classification of known ones; on the other hand the relationships between the Lagrangian geometry of a variety and its blowup along some subvariety are poorly understood (apart from some very simple examples, such as a Lagrangian Klein bottle in the Hirzebruch surface). We are looking forward to returning to this subject in a forthcoming paper. 
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\Bibitem{Tyu24} |
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