# Hamiltonian $S^{1}$-manifolds which are like a coadjoint orbit of $G_{2}$

Hui Li School of mathematical Sciences
Soochow University
Suzhou, 215006, China<EMAIL_ADDRESS>
###### Abstract.

Consider a compact symplectic manifold of dimension $2n$ with a Hamiltionan
circle action. Then there are at least $n+1$ fixed points. Recent works study
the case when the fixed point set consists of precisely $n+1$ isolated points.
Motivated by these works, this paper studies a Hamiltonian $S^{1}$ action on a
$10$-dimensional compact symplectic manifold with exactly 6 isolated fixed
points. Under a condition, we study the relations of the following data: the
first Chern class of the manifold, the largest weight of the action, all the
weights of the action, the total Chern class of the manifold, and the integral
cohomology ring of the manifold. We show that if one of these data is
identical with that of a coadjoint orbit of the exceptional Lie group $G_{2}$,
equipped with a Hamiltonian action of a subcircle of the maximal torus, then
the other data are identical with this coadjoint orbit of $G_{2}$.

###### Key words and phrases:

Symplectic manifold, Hamiltonian circle action, equivariant cohomology, Chern
classes.

2010 classification. 53D05, 53D20, 55N25, 57R20

##  1\. introduction

Let the circle $S^{1}$ act on a compact $2n$-dimensional symplectic manifold
$(M,\omega)$ with moment map. The moment map is a perfect Morse-Bott function
whose critical set is precisely the fixed point set $M^{S^{1}}$ of the
$S^{1}$-action. Since the even Betti numbers of $M$, $b_{2i}(M)\geq 1$ for all
$0\leq 2i\leq 2n$, $M^{S^{1}}$ contains at least $n+1$ points. When
$M^{S^{1}}$ consists of exactly $n+1$ isolated points, we say that the action
has minimal isolated fixed points.

For a compact symplectic manifold $(M,\omega)$ of dimension $2n$ with a
symplectic $S^{1}$-action, if the fixed points are isolated, a neighborhood of
each fixed point $P$ is $S^{1}$-equivariantly diffeomorphic to a neighborhood
of the tangent space $T_{P}M$ with an $S^{1}$-linear action. The space
$T_{P}M$ splits into $n$ copies of ${\mathbb{C}}$, on each of which $S^{1}$
acts by multiplication by $\lambda^{w_{i}}$, where $\lambda\in S^{1}$ and
$w_{i}\in{\mathbb{Z}}$ for $i=1,\cdots,n$. The integers $w_{i}$s’ are well
defined, and are called the weights of the $S^{1}$-action at $P$. In this
paper, when we say the largest weight or the largest weight on $M$ and there
is no other specification, we mean the largest of the weights of the
$S^{1}$-action at all the fixed points on $M$. If $W^{+}$ and $W^{-}$ are
respectively the set of all the positive weights and all the negative weights
at all the fixed points on $M$, then it is known that (see e.g. [2, 19, 15])

$W^{-}=-W^{+}.$

The examples the author knows of compact symplectic manifolds admitting
Hamiltonian circle actions with minimal isolated fixed points are the complex
projective spaces ${\mathbb{CP}}^{n}$ with $n\geq 0$, the Grassmannian of
oriented $2$-planes in ${\mathbb{R}}^{n+2}$, denoted
$\widetilde{G}_{2}({\mathbb{R}}^{n+2})$ with $n\geq 3$ odd, $V_{5}$ and
$V_{22}$ in dimension $6$ ([20]), and a 10-dimensional coadjoint orbit of the
rank 2 exceptional Lie group $G_{2}$ ([17]). In dimension 4, the Hamiltonian
$S^{1}$-manifold is equivariantly symplectomorphic to ${\mathbb{CP}}^{2}$ ([5,
3]). In dimension $6$, we know $4$ kinds of examples, ${\mathbb{CP}}^{3}$,
$\widetilde{G}_{2}({\mathbb{R}}^{5})$, $V_{5}$ and $V_{22}$. In fact, in
dimension 6, in terms of integral cohomology rings and total Chern classes,
these are the only kinds ([20, 16]). In dimension 8, there is a unique
integral cohomology ring and total Chern class, those of ${\mathbb{CP}}^{4}$
([1, 4]). For arbitrary $2n$-dimensional manifold $(M,\omega)$, we know that
$c_{1}(M)=kx$ with $1\leq k\leq n+1$, where $x\in H^{2}(M;{\mathbb{Z}})$ is a
generator (see Lemmas 4.4 and 4.2). In [9], the author studies the cases when
$c_{1}(M)=(n+1)x$ and $c_{1}(M)=nx$. For these cases, the author uses the
largest weight to show that the integral cohomology ring and total Chern class
of the manifold respectively correspond uniquely to those of
${\mathbb{CP}}^{n}$, and of $\widetilde{G}_{2}({\mathbb{R}}^{n+2})$ with
$n\geq 3$ odd. In all the cases mentioned above, the weights of the
$S^{1}$-action at the fixed points are completely determined and are standard
as in the known examples.

In this paper, we consider 10-dimensional compact Hamiltonian
$S^{1}$-manifolds with 6 isolated fixed points. As mentioned,
${\mathbb{CP}}^{5}$ and $\widetilde{G}_{2}({\mathbb{R}}^{7})$ are such kinds
of manifolds. In this work, we consider the $S^{1}$-manifolds which have some
common feature with a coadjoint orbit of $G_{2}$ through a point in a wall of
the closed Weyl chamber. We will use $\bf{\mathcal{O}}$ to denote this
coadjoint orbit throughout the paper. The 10-dimensional coadjoint orbit of
$G_{2}$ through a point in the other wall of the closed Weyl chamber is
actually diffeomorphic to $\widetilde{G}_{2}({\mathbb{R}}^{7})$ (see [7])111I
thank R. Sjamaar for sending me this reference from where I know this fact..
We will study the interplay between the following data: the first Chern class,
the total Chern class and the integral cohomology ring of the manifold, the
largest weight and all the weights of the $S^{1}$-action at all the fixed
points.

The 10-dimensional coadjoint orbit $\mathcal{O}$ of $G_{2}$ is given in [17]
as a GKM manifold with a 2-dimensional maximal torus $T^{2}$-action. Up to
scale and $SL(2,{\mathbb{Z}})$ transformation, its 2-dimensional moment
polytope is given there, see Section 2 for Example 2.1. The fixed point set of
the $T^{2}$-action consists of $6$ isolated points. In this paper, we choose a
circle $S^{1}\subset T^{2}$ such that the 6 isolated fixed points of the
$T^{2}$-action correspond exactly to $6$ isolated fixed points of the
$S^{1}$-action. So we have an example other than ${\mathbb{CP}}^{5}$ and
$\widetilde{G}_{2}({\mathbb{R}}^{7})$ of a compact Hamiltonian
$S^{1}$-manifold $\mathcal{O}$ of dimension 10 with 6 isolated fixed points;
we can write explicitly the sets of weights of the $S^{1}$-action at all the
fixed points. For this example of Hamiltonian $S^{1}$-manifold $\mathcal{O}$,
the integer $5$ is the largest weight. Due to this reason, for our compact
Hamiltonian $S^{1}$-manifold $(M,\omega)$, we assume that each weight on $M$
is no bigger than $5$ (we are not assuming $5$ really occurs). This is the
same as assuming that the absolute value of each weight on $M$ is no bigger
than 5. The invariants the first Chern class $c_{1}(M)$, the total Chern class
$c(M)$ and the integral cohomology ring $H^{*}(M;{\mathbb{Z}})$ are intrinsic
to the manifold. We will consider suitable circle actions to make these
intrinsic invariants to have good relations with the largest weight and with
all the weights of the circle action. Without the condition that each weight
on $M$ is no bigger than $5$, we cannot get all the equivalent conditions as
in our main theorems, see Remark 5.9 (first, Theorem 1 is not true). Without
the condition that each weight on $M$ is no bigger than $5$, the relation of
the weights with the moment map values of the fixed points as in Example 2.1
is crucial for computing $H^{*}(M;{\mathbb{Z}})$ and $c(M)$, see Remarks 6.4
and 6.6. Not as $\mathcal{O}$ which admits a Hamiltonian $T^{2}$-action, for
the compact Hamiltonian $S^{1}$-manifolds we consider, we do not assume the
manifolds have more symmetry than an $S^{1}$-action. Less symmetry makes
certain data relative to the action more rigid. The integral cohomology ring
and the total Chern class of the example $\mathcal{O}$ are known, for example
see [7] and the references therein. In this paper, for the Hamiltonian
$S^{1}$-manifold we study, we give a direct computation of its integral
cohomology ring and total Chern class, the same methods yield these data for
the coadjoint orbit $\mathcal{O}$.

Let $(M,\omega)$ be a compact $10$-dimensional Hamiltonian $S^{1}$-manifold
with moment map $\phi\colon M\to{\mathbb{R}}$. Assume the fixed point set
$M^{S^{1}}$ consists of minimal isolated points, i.e.,
$M^{S^{1}}=\\{P_{0},P_{1},\cdots,P_{5}\\}$, and $[\omega]$ is an integral
class. Then $\phi(P_{i})-\phi(P_{j})\in{\mathbb{Z}}$ for all $i,j$ by Lemma
3.2, and by Lemma 3.7,

$\phi(P_{0})<\phi(P_{1})<\cdots<\phi(P_{5}).$

Note that under our assumption of minimal isolated fixed points,
$H^{2}(M;{\mathbb{Z}})={\mathbb{Z}}$, so the symplectic class $[\omega]$ is
primitive integral is not a serious assumption. Note also that the minimal
isolated fixed points assumption forces the compact manifold $M$ to be
connected. Our main results are stated below. In Theorem 1, we assume a
Hamiltonian circle action with moment map image symmetric, just as in Example
2.1 for the Hamiltonian $S^{1}$-manifold $\mathcal{O}$, to give an equivalence
condition on the invariant $c_{1}(M)$ and on the places where the largest
weight occurs.

######  Theorem 1.

Let $(M,\omega)$ be a compact $10$-dimensional Hamiltonian $S^{1}$-manifold
with moment map $\phi\colon M\to{\mathbb{R}}$. Assume
$M^{S^{1}}=\\{P_{0},P_{1},\cdots,P_{5}\\}$, $[\omega]$ is a primitive integral
class, and $\phi(P_{i})-\phi(P_{0})=\phi(P_{5})-\phi(P_{5-i})$ for $i=1,2$.
Assume each weight of the $S^{1}$-action is less than or equal to $5$. Then
the following 3 conditions are equivalent.

  1. (1)

$c_{1}(M)=3[\omega]$.

  2. (2)

The largest weight is between222See Section 3 for definition. $P_{0}$ and
$P_{5}$ and is equal to $\frac{1}{2}\big{(}\phi(P_{5})-\phi(P_{0})\big{)}$.

  3. (3)

The largest weight is between $P_{1}$ and $P_{3}$, symmetrically between
$P_{2}$ and $P_{4}$, and is equal to
$\phi(P_{3})-\phi(P_{1})=\phi(P_{4})-\phi(P_{2})$.

Under any one condition, the largest weight is exactly 5, and $(2)$ and $(3)$
are the only ways it occurs and $|\pm 5|$ only occurs once at the
corresponding fixed point.

######  Theorem 2.

Let $(M,\omega)$ be a compact $10$-dimensional Hamiltonian $S^{1}$-manifold
with moment map $\phi\colon M\to{\mathbb{R}}$. Assume
$M^{S^{1}}=\\{P_{0},P_{1},\cdots,P_{5}\\}$ and $[\omega]$ is a primitive
integral class. Assume each weight of the $S^{1}$-action is less than or equal
to $5$. If the largest weight is between $P_{0}$ and $P_{5}$ and is equal to
$\frac{1}{2}\big{(}\phi(P_{5})-\phi(P_{0})\big{)}$, then the following 3
things hold.

  1. (1)

The integral cohomology ring $H^{*}(M;{\mathbb{Z}})$ is generated by the
elements

$1,\,\,\,[\omega],\,\,\,\frac{1}{3}[\omega]^{2},\,\,\,\frac{1}{6}[\omega]^{3},\,\,\,\frac{1}{18}[\omega]^{4},\,\,\,\frac{1}{18}[\omega]^{5},$

and is isomorphic to the integral cohomology ring
$H^{*}(\mathcal{O};{\mathbb{Z}})$.

  2. (2)

The total Chern class

$c(M)=1+3[\omega]+\frac{1}{3}\big{(}13[\omega]^{2}+11[\omega]^{3}+5[\omega]^{4}+[\omega]^{5}\big{)}$
$=1+3\alpha_{1}+13\alpha_{2}+22\alpha_{3}+30\alpha_{4}+6\alpha_{5},$

where $\alpha_{i}\in H^{2i}(M;{\mathbb{Z}})$ is a generator for $1\leq i\leq
5$. The total Chern classes $c(M)$ and $c(\mathcal{O})$ are isomorphic.

  3. (3)

The sets of weights at the fixed points are the same as those of a subcircle
action of a maximal torus of $G_{2}$ on $\mathcal{O}$ (as in Example 2.1); in
particular, there is exactly one weight between any pair of fixed points. The
precise sets of weights at all the fixed points on $M$ are as follows:

$P_{0}\colon\\{1,4,2,3,5\\},\,\,\,P_{5}\colon-\\{1,4,2,3,5\\},$
$P_{1}\colon\\{-1,1,5,4,3\\},\,\,\,P_{4}\colon-\\{-1,1,5,4,3\\},$
$P_{2}\colon\\{-4,-1,1,5,2\\},\,\,\,P_{3}\colon-\\{-4,-1,1,5,2\\}.$

For a compact Hamiltonian $S^{1}$-manifold $(M,\omega)$ of dimension $2n$ with
minimal isolated fixed points, if knowing all the weights of the
$S^{1}$-action at all the fixed points, one can determine the ring
$H^{*}(M;{\mathbb{Z}})$ and $c(M)$, see [20] for a general method. For
dimensions $2n=4,6,8$, the works mentioned earlier determined all the weights
in all the possible cases. In [9], for the cases $c_{1}(M)=(n+1)x$ and
$c_{1}(M)=nx$, where $x\in H^{2}(M;{\mathbb{Z}})$ is a generator, the author
uses the largest weight to determine the ring $H^{*}(M;{\mathbb{Z}})$ without
knowing all the weights, and determine $c(M)$ using an observation. In this
paper, for Theorem 2, we use the largest weight to first determine all the
weights of the $S^{1}$ action; and then we determine the ring
$H^{*}(M;{\mathbb{Z}})$ and $c(M)$ in somewhat different ways as those in [20]
and [9].

On the other hand, for a compact Hamiltonian $S^{1}$-manifold $(M,\omega)$ of
dimension $2n$ with minimal isolated fixed points, if we only know the
integral cohomology ring $H^{*}(M;{\mathbb{Z}})$, it is not tractible whether
we can determine all the weights of the action, or the total Chern class
$c(M)$. This is already hard when the integral cohomology ring of the manifold
is relatively simple, for example that of ${\mathbb{CP}}^{n}$. In the 1970s,
Petrie posed a conjecture ([18]), known as Petrie’s conjecture, which states
that if a homotopy complex projective space $X$ admits a nontrivial circle
action, then the total Pontryagin class of $X$ is isomorphic to that of
${\mathbb{CP}}^{n}$, where $2n=\dim X$. The conjecture is solved in some
special cases, but is generally open. In the case of a compact Hamiltonian
$S^{1}$-manifold $(M,\omega)$ with minimal isolated fixed points, by [8],
$\pi_{1}(M)=\pi_{1}(\mbox{min})$, where min is the minimum of the moment map,
so $M$ is simply connected, hence is homotopy equivalent to
${\mathbb{CP}}^{n}$ if it has the integral cohomology ring of
${\mathbb{CP}}^{n}$. As mentioned, the condition $c_{1}(M)=(n+1)x$ is
stronger, namely it implies the ring isomorphism $H^{*}(M;{\mathbb{Z}})\cong
H^{*}({\mathbb{CP}}^{n};{\mathbb{Z}})$ and the total Chern class isomorphism
$c(M)\cong c({\mathbb{CP}}^{n})$, hence the conjecture holds in this case
([9]). As mentioned, the ring $H^{*}(M;{\mathbb{Z}})$ and $c(M)$ are intrinsic
to the manifold. We might have different Hamiltonian $S^{1}$-actions on the
manifold. In Theorem 3 below, we consider a Hamiltonian circle action whose
moment map image is symmetric, just as in Example 2.1 for the Hamiltonian
$S^{1}$-manifold $\mathcal{O}$. In this case, knowing the topological
condition $H^{*}(M;{\mathbb{Z}})$ alone, we can determine the stronger
condition, namely all the weights of the action at all the fixed points, hence
we can determine the total Chern class of the manifold.

######  Theorem 3.

Let $(M,\omega)$ be a compact $10$-dimensional Hamiltonian $S^{1}$-manifold
with moment map $\phi$. Assume $M^{S^{1}}=\\{P_{0},P_{1},\cdots,P_{5}\\}$,
$[\omega]$ is a primitive integral class, and
$\phi(P_{i})-\phi(P_{0})=\phi(P_{5})-\phi(P_{5-i})$ for $i=1,2$. Assume each
weight of the $S^{1}$-action is less than or equal to $5$. If the ring
$H^{*}(M;{\mathbb{Z}})$ is isomorphic to the ring
$H^{*}(\mathcal{O};{\mathbb{Z}})$, then $c_{1}(M)=3[\omega]$, and the sets of
weights at all the fixed points are exactly as those of the subcircle action
of a maximal torus of $G_{2}$ on $\mathcal{O}$ as in Example 2.1, in
particular there is exactly one weight between any pair of fixed points. Hence
$c(M)$ is as in Theorem 2 $(2)$, and is isomorphic to $c(\mathcal{O})$.

The above 3 theorems imply the following corollary.

######  Corollary 1.1.

Let $(M,\omega)$ be a compact $10$-dimensional Hamiltonian $S^{1}$-manifold
with moment map $\phi\colon M\to{\mathbb{R}}$. Assume
$M^{S^{1}}=\\{P_{0},P_{1},\cdots,P_{5}\\}$, $[\omega]$ is a primitive integral
class, and $\phi(P_{i})-\phi(P_{0})=\phi(P_{5})-\phi(P_{5-i})$ for $i=1,2$.
Assume each weight of the $S^{1}$-action is less than or equal to $5$. Then
the following conditions are equivalent.

  1. (1)

$c_{1}(M)=3[\omega]$.

  2. (2)

The largest weight is between $P_{0}$ and $P_{5}$ and is equal to
$\frac{1}{2}\big{(}\phi(P_{5})-\phi(P_{0})\big{)}$.

  3. (3)

The largest weight is between $P_{1}$ and $P_{3}$, symmetrically between
$P_{2}$ and $P_{4}$, and is equal to
$\phi(P_{3})-\phi(P_{1})=\phi(P_{4})-\phi(P_{2})$.

  4. (4)

The integral cohomology ring $H^{*}(M;{\mathbb{Z}})$ is isomorphic to the
integral cohomology ring $H^{*}(\mathcal{O};{\mathbb{Z}})$.

  5. (5)

The total Chern class $c(M)$ is isomorphic to the total Chern class
$c(\mathcal{O})$.

  6. (6)

The sets of weights of the $S^{1}$-action at the fixed points are the same as
those of a subcircle action of a maximal torus of $G_{2}$ on $\mathcal{O}$ as
in Example 2.1, in particular there is exactly one weight between any pair of
fixed points.

In Theorem 2, the assumption that the largest weight is between $P_{0}$ and
$P_{5}$ and is equal to $\frac{1}{2}\big{(}\phi(P_{5})-\phi(P_{0})\big{)}$
turns out to be strong, as the conclusions of the theorem show. In particular
it implies symmetry of the image of $\phi$. In Theorem 3, if we know the ring
$H^{*}(M;{\mathbb{Z}})$, and we want to determine $c(M)$ instead of the
weights of the circle action, we might be able to remove the symmetry
condition $\phi(P_{i})-\phi(P_{0})=\phi(P_{5})-\phi(P_{5-i})$ for $i=1,2$. The
interesting question is if we know one Hamiltonian $S^{1}$-action, then can we
have another Hamiltonian circle action whose moment map image has this
symmetry, then we can use the latter circle action to determine $c(M)$.

If there is a bigger dimensional Hamiltonian torus action than solely an
$S^{1}$-action on the manifold $(M,\omega)$, if we are interested in knowing
$H^{*}(M;{\mathbb{Z}})$ and $c(M)$, we can consider a suitable subcircle
action, the current work provides a perspective by showing an easier way of
getting $H^{*}(M;{\mathbb{Z}})$ and $c(M)$ using a subcircle action instead of
the torus action. When a bigger dimensional torus acts on the manifold, there
is more flexibility to choose a subcircle action to fit our purpose, for
example it may help with the question posed above.

Closely related to compact Hamiltonian $S^{1}$-manifold $(M,\omega)$ of
dimension $2n$ with $n+1$ isolated fixed points, is the case of $n+2$ isolated
fixed points, also called almost minimal isolated fixed points. In the latter
case, $n$ must be even, $c_{1}(M)=(n+1)x$ cannot occur, where $x\in
H^{2}(M,{\mathbb{Z}})$ is a generator (see [10, Prop. 3.6 and Lemma 3.1])
(Recall that $c_{1}(M)=kx$ with $1\leq k\leq n+1$). In [10], for
$c_{1}(M)=nx$, with the symmetry
$\phi(P_{n-1})-\phi(P_{0})=\phi(P_{n})-\phi(P_{1})$, where
$\\{P_{0},P_{1},\cdots,P_{\frac{n}{2}},P_{(\frac{n}{2})^{\prime}},\cdots,P_{n-1},P_{n}\\}$
is the set of fixed points with two index $n$ fixed points, the author shows
that the compact Hamiltonian $S^{1}$-manifold $M$ has the integral cohomology
ring and total Chern class of $\widetilde{G}_{2}({\mathbb{R}}^{n+2})$ with
$n\geq 2$ even. In dimension 10, a compact Hamiltonian $S^{1}$-manifold cannot
have 7 isolated fixed points.

The current work falls into the framework of compact Hamiltonian
$S^{1}$-manifolds $(M,\omega)$ of dimension 2n with minimal even Betti numbers
or almost minimal even Betti numbers, i.e.,
$b_{2i}(M)=b_{2i}({\mathbb{CP}}^{n})=1$ for each $0\leq 2i\leq 2n$, or when
$n$ is even, $b_{2i}(M)=1$ for each $0\leq 2i\leq 2n$ with $2i\neq n$ and
$b_{n}(M)=2$. The interest of the problem was originally motivated by a
generalization of Petrie’s conjecture in the symplectic category, posed by
Tolman ([20]). For the works which do not restrict to isolated fixed points,
see [20, 14, 13, 11, 12]. For the cases of isolated fixed points and not
necessarily isolated fixed points, the difficulties of the problem lie in
different places.

The current work suggests possible research problems, and it builds a
foundation for further classification of the manifold. One question is to
consider a generalized version of Petrie’s conjecture. For this question, let
us still focus on compact Hamiltonian $S^{1}$-manifold $M$ with minimal
isolated fixed points, then $H^{2i}(M;{\mathbb{Z}})={\mathbb{Z}}$ and
$H^{2i+1}(M;{\mathbb{Z}})=0$ for all $0\leq 2i\leq\dim(M)$ (see Lemma 3.7).
Hence if we know the generator of each $H^{2i}(M;{\mathbb{Z}})$, we know the
ring $H^{*}(M;{\mathbb{Z}})$. If we solely know the integral cohomology ring
$H^{*}(M;{\mathbb{Z}})$, can we determine the total Chern class $c(M)$?
Another question is, as we know our manifold has some important topological
and geometric invariants the same as some known manifold, then how far in
different categories is our Hamiltonian $S^{1}$-manifold to the known example?
The categories could be the smooth, the symplectic, or the complex category
when the manifold is Kähler (our known examples are all Kähler manifolds). For
instance, in [16], McDuff shows that the 6-dimensional Hamiltonian
$S^{1}$-manifold with 4 fixed points which has the integral cohomology ring
and total Chern class of $V_{5}$ or $V_{22}$ is $S^{1}$-equivariantly
symplectomorphic to $V_{5}$ or $V_{22}$; in [13], using the invariants
obtained in [14], the author, Olbermann and Stanley show that up to
$S^{1}$-equivariant diffeomorphism, in each dimension, there are only finitely
many compact Hamiltonian $S^{1}$-manifolds with minimal even Betti numbers and
with two fixed point set components; in [11], for the Kähler case, the author
shows uniqueness up to $S^{1}$-equivariant biholomorphism and
$S^{1}$-equivariant symplectomorphism for certain cases of Hamiltonian
$S^{1}$-actions with 3 fixed point set components. For the Kähler case, see
similar conclusions in [12, 9] etc. when the first Chern class of the manifold
is large, namely $c_{1}(M)=(n+1)x$ and $c_{1}(M)=nx$, where $\dim(M)=2n$. In
the Kähler category, the problem is also related to the classification of fake
complex projective spaces, i.e., compact Kähler manifolds which have the same
Betti numbers as ${\mathbb{CP}}^{n}$ but are different from
${\mathbb{CP}}^{n}$.

Now we summarize the structure of the paper. In Section 2, we give the
10-dimensional example $\mathcal{O}$. In Section 3, we state some basic
results for Hamiltonian $S^{1}$-actions, building a foundation for the next
sections. In Section 4, we give two ways of expressing the first Chern class
of the manifold. In Section 5, we study the relation between the first Chern
class of the manifold and the largest weight of the $S^{1}$-action, and prove
Theorem 1. Section 6 has 3 parts. In part 6.1, we use the weight condition to
determine all the weights of the action, proving Theorem 2 (3). In part 6.2,
we use the weight condition to determine the integral cohomology ring of the
manifold, proving Theorem 2 (1). In part 6.3, we use the weight condition to
determine the total Chern class of the manifold, proving Theorem 2 (2). In
Section 7, we use the integral cohomology ring of the manifold to determine
the first Chern class of the manifold, and the sets of weights of the
$S^{1}$-action at all the fixed points, proving Theorem 3.

##  2\. the example — a coadjoint orbit of $G_{2}$

In this section, we give an explicit example other than ${\mathbb{CP}}^{5}$
and $\widetilde{G}_{2}({\mathbb{R}}^{7})$ of a compact 10-dimensional
Hamiltonian $S^{1}$-manifold with exactly 6 isolated fixed points.

######  Example 2.1.

In [17](p42-44), Morton gives a GKM-manifold, it is a $10$-dimensional
coadjoint orbit $\mathcal{O}$ of the 14-dimensional rank 2 Lie group $G_{2}$,
passing through a point on a wall of the closed Weyl chamber. The maximal
torus $T^{2}$ of $G_{2}$ acts on $\mathcal{O}$. The $T^{2}$-action is
Hamiltonian, and up to scale and $SL(2;{\mathbb{Z}})$ transformation, the
moment polytope UVWXYZ has the following vertices:

$U=(-1,-2),\,\,\,V=(-2,-1),\,\,\,W=(-1,1),\,\,\,X=(1,2),\,\,\,Y=(2,1),\,\,\,Z=(1,-1).$

They are exactly the $T^{2}$ moment map images of the 6 isolated fixed points
of the $T^{2}$-action. The weights of the $T^{2}$-action of the edges are:

$w(UV)=(-1,1),w(UZ)=(2,1),w(UW)=(0,1),w(UY)=(1,1),w(UX)=(1,2),$
$w(VZ)=(1,0),w(VW)=(1,2),w(VY)=(2,1),w(VX)=(1,1),$
$w(ZW)=(-1,1),w(ZY)=(1,2),w(ZX)=(0,1),$ $w(WY)=(1,0),w(WX)=(2,1),$
$w(YX)=(-1,1).$

(The picture is a GKM graph in [17]. The graph has the following relation with
the vertices of the moment polytope. Let $\psi$ be the $T^{2}$ moment map, and
$\widetilde{u}$ be the equivariant extension of the symplectic class
$[\omega]$ similarly as in Lemma 3.1 for $\psi$. Since
$H^{2}(\mathcal{O};{\mathbb{R}})={\mathbb{R}}$ (generated by $[\omega]$), we
have $c_{1}^{S^{1}}(M)=k\widetilde{u}$ with $k\in{\mathbb{R}}$, then
$c_{1}^{S^{1}}(M)|_{P}=\Gamma_{P}=-k\psi(P)$ for any fixed point $P$ of
$T^{2}$, where $\Gamma_{P}$ is the sum of weights at $P$.)

333I thank Chen He for telling me how to draw these pictures.

$U$$V$$W$$X$$Y$$Z$$\phi(P_{0})$$\phi(P_{1})$$\phi(P_{2})$$\phi(P_{3})$$\phi(P_{4})$$\phi(P_{5})$

We take the subcircle $S^{1}\subset T^{2}$ generated by the direction
$UX=(1,2)$. Let $\phi$ be the moment map of the circle action on
$\mathcal{O}$. The image of $\phi$ is the projection of the polytope to the
line segment $UX$. The 6 vertices of the polytope respectively project to 6
points on the line, which correspond respectively to the $\phi$ images of the
6 fixed points of the circle action on $\mathcal{O}$. Let $P_{0}$, $P_{1}$,
$\cdots$, $P_{5}$ be the $6$ fixed points on $\mathcal{O}$ of the circle
action, in the order that

$\phi(P_{0})<\phi(P_{1})<\cdots<\phi(P_{5}).$

We project the weights of the $T^{2}$-action to the direction $(1,2)$, we get
the weights of the $S^{1}$-action at the 6 fixed points of the $S^{1}$-action
and their relations with the moment map values as follows.

$P_{0}\colon\\{1,4,2,3,5\\}$
$=\Big{\\{}\phi(P_{1})-\phi(P_{0}),\phi(P_{2})-\phi(P_{0}),\frac{1}{3}\big{(}\phi(P_{3})-\phi(P_{0})\big{)},\frac{1}{3}\big{(}\phi(P_{4})-\phi(P_{0})\big{)},\frac{1}{2}\big{(}\phi(P_{5})-\phi(P_{0})\big{)}\Big{\\}}.$
$P_{1}\colon\\{-1,1,5,4,3\\}$
$=\Big{\\{}\phi(P_{0})-\phi(P_{1}),\frac{1}{3}\big{(}\phi(P_{2})-\phi(P_{1})\big{)},\phi(P_{3})-\phi(P_{1}),\frac{1}{2}\big{(}\phi(P_{4})-\phi(P_{1})\big{)},\frac{1}{3}\big{(}\phi(P_{5})-\phi(P_{1})\big{)}\Big{\\}}.$
$P_{2}\colon\\{-4,-1,1,5,2\\}$
$=\Big{\\{}\phi(P_{0})-\phi(P_{2}),\frac{1}{3}\big{(}\phi(P_{1})-\phi(P_{2})\big{)},\frac{1}{2}\big{(}\phi(P_{3})-\phi(P_{2})\big{)},\phi(P_{4})-\phi(P_{2}),\frac{1}{3}\big{(}\phi(P_{5})-\phi(P_{2})\big{)}\Big{\\}}.$
$P_{3}\colon\\{-2,-5,-1,1,4\\}$
$=\Big{\\{}\frac{1}{3}\big{(}\phi(P_{0})-\phi(P_{3})\big{)},\phi(P_{1})-\phi(P_{3}),\frac{1}{2}\big{(}\phi(P_{2})-\phi(P_{3})\big{)},\frac{1}{3}\big{(}\phi(P_{4})-\phi(P_{3})\big{)},\phi(P_{5})-\phi(P_{3})\Big{\\}}.$
$P_{4}\colon\\{-3,-4,-5,-1,1\\}$
$=\Big{\\{}\frac{1}{3}\big{(}\phi(P_{0})-\phi(P_{4})\big{)},\frac{1}{2}\big{(}\phi(P_{1})-\phi(P_{4})\big{)},\phi(P_{2})-\phi(P_{4}),\frac{1}{3}\big{(}\phi(P_{3})-\phi(P_{4})\big{)},\phi(P_{5})-\phi(P_{4})\Big{\\}}.$
$P_{5}\colon\\{-5,-3,-2,-4,-1\\}$
$=\Big{\\{}\frac{1}{2}\big{(}\phi(P_{0})-\phi(P_{5})\big{)},\frac{1}{3}\big{(}\phi(P_{1})-\phi(P_{5})\big{)},\frac{1}{3}\big{(}\phi(P_{2})-\phi(P_{5})\big{)},\phi(P_{3})-\phi(P_{5}),\phi(P_{4})-\phi(P_{5})\Big{\\}}.$

##  3\. some basic results

In this section, we state (with minor proof) some basic results which are
important for the study in the next sections.

Let $M$ be a smooth $S^{1}$-manifold. The equivariant cohomology of $M$ in a
coefficient ring $R$ is
$H^{*}_{S^{1}}(M;R)=H^{*}(S^{\infty}\times_{S^{1}}M;R)$, where $S^{1}$ acts on
$S^{\infty}$ freely. If $p$ is a point, then
$H^{*}_{S^{1}}(p;R)=H^{*}({\mathbb{CP}}^{\infty};R)=R[t]$, where $t\in
H^{2}({\mathbb{CP}}^{\infty};R)$ is a generator. If $S^{1}$ acts on $M$
trivially, then $H^{*}_{S^{1}}(M;R)=H^{*}(M;R)\otimes R[t]=H^{*}(M;R)[t]$. The
projection map $\pi\colon S^{\infty}\times_{S^{1}}M\to{\mathbb{CP}}^{\infty}$
induces a pull back map $\pi^{*}\colon H^{*}({\mathbb{CP}}^{\infty})\to
H^{*}_{S^{1}}(M)$, so that $H^{*}_{S^{1}}(M)$ is an
$H^{*}({\mathbb{CP}}^{\infty})$-module.

Let $(M,\omega)$ be a compact $2n$-dimensional symplectic $S^{1}$-manifold
with isolated fixed points. Let $P$ be a fixed point, and let
$\\{w_{1},\cdots,w_{n}\\}$ be the set of weights at $P$. The equivariant total
Chern class of $M$ can be expressed as

$c^{S^{1}}(M)=1+c_{1}^{S^{1}}(M)+\cdots+c_{n}^{S^{1}}(M)\in
H^{*}_{S^{1}}(M;{\mathbb{Z}}),$

where $c_{i}^{S^{1}}(M)\in H^{2i}_{S^{1}}(M;{\mathbb{Z}})$ is the $i$-th
equivariant Chern class of $M$. The restriction of $c^{S^{1}}(M)$ to $P$ is

$c^{S^{1}}(M)|_{P}=1+\sum_{i=1}^{n}c_{i}^{S^{1}}(M)|_{P}=1+\sum_{i=1}^{n}\sigma_{i}(w_{1},\cdots,w_{n})t^{i},$

where $\sigma_{i}(w_{1},\cdots,w_{n})$ is the $i$-th symmetric polynomial in
the weights at $P$. The restriction of $c^{S^{1}}(M)$ to ordinary cohomology
is the total Chern class of $M$:

$c(M)=1+c_{1}(M)+\cdots+c_{n}(M)\in H^{*}(M;{\mathbb{Z}}),$

where $c_{i}(M)\in H^{2i}(M;{\mathbb{Z}})$ is the $i$-th Chern class of $M$.

Let $(M,\omega)$ be a compact Hamiltonian $S^{1}$-manifold with isolated fixed
points. The moment map is a Morse function, its critical set is the set of
fixed points. At a fixed point $P$, if $\lambda_{P}$ is the number of negative
weights and $\lambda_{P}^{+}$ is the number of positive weights, then the
Morse index of $P$ is $\bf 2\lambda_{P}$, and the Morse coindex of $P$ is
$2\lambda_{P}^{+}$.

Now we start to state the results. First, the symplectic class $[\omega]$ of a
Hamiltonian $S^{1}$-manifold can be extended to an equivariant cohomology
class $\widetilde{u}$ as in Lemma 3.1, and based on this, we can obtain Lemma
3.2.

######  Lemma 3.1.

[14] Let the circle act on a compact symplectic manifold $(M,\omega)$ with
moment map $\phi\colon M\to{\mathbb{R}}$. Then there exists
$\widetilde{u}=[\omega-\phi t]\in H_{S^{1}}^{2}(M;{\mathbb{R}})$ such that for
any fixed point set component $F$,

$\widetilde{u}|_{F}=[\omega|_{F}]-\phi(F)t.$

If $[\omega]$ is an integral class, then $\widetilde{u}$ is an integral class.

For an $S^{1}$-manifold $M$, when there exists a finite stabilizer group
${\mathbb{Z}}_{k}\subset S^{1}$ with $k>1$, the set of points,
$M^{{\mathbb{Z}}_{k}}$, which is pointwise fixed by ${\mathbb{Z}}_{k}$ but not
pointwise fixed by $S^{1}$, is called a ${\mathbb{Z}}_{k}$-isotropy
submanifold.

######  Lemma 3.2.

[9] Let the circle act on a compact symplectic manifold $(M,\omega)$ with
moment map $\phi\colon M\to{\mathbb{R}}$. Assume $[\omega]$ is an integral
class. Then for any two fixed point set components $F$ and $F^{\prime}$,
$\phi(F)-\phi(F^{\prime})\in{\mathbb{Z}}$. If ${\mathbb{Z}}_{k}$ is the
stabilizer group of some point on $M$, then for any two fixed point set
components $F$ and $F^{\prime}$ on the same connected component of
$M^{{\mathbb{Z}}_{k}}$, we have $k\,|\left(\phi(F^{\prime})-\phi(F)\right)$.

When $[\omega]$ is an integral class, we will often use the fact that
$\phi(F)-\phi(F^{\prime})\in{\mathbb{Z}}$ in Lemma 3.2 without always
referring to it.

As an application of the above lemmas, we have the following lemma.

######  Lemma 3.3.

Let the circle act on a connected compact symplectic manifold $(M,\omega)$
with moment map $\phi\colon M\to{\mathbb{R}}$. Assume $M^{S^{1}}$ consists of
isolated points and $[\omega]$ is an integral class. Let $P$ be a fixed point.
If $-a$ is a negative weight at $P$, then $a|\big{(}\phi(Q)-\phi(P)\big{)}$
for some $\phi(Q)<\phi(P)$. Similarly, if $b$ is a positive weight at $P$,
then $b|\big{(}\phi(Q)-\phi(P)\big{)}$ for some $\phi(Q)>\phi(P)$.

###### Proof.

Since $a$ is a negative weight at $P$, $P$ is not the minimal fixed point in
$M^{{\mathbb{Z}}_{a}}$ containing $P$, so there is a $Q$ in
$M^{{\mathbb{Z}}_{a}}$ containing $P$ with $\phi(Q)<\phi(P)$. Then
$a|\big{(}\phi(Q)-\phi(P)\big{)}$ by Lemma 3.2. The other claim follows
similarly. ∎

Based on the symplectic nature of the manifold, Tolman gives a claim as
follows on the index of a fixed point for a Hamiltonian $S^{1}$-manifold.

######  Lemma 3.4.

[20] Let the circle act on a connected compact symplectic manifold
$(M,\omega)$ with isolated fixed points and with moment map $\phi\colon
M\to{\mathbb{R}}$. Then for any fixed point $P$, $2\lambda_{P}\leq 2l$, where
$l$ is the number of fixed points $Q$s’ such that $\phi(Q)<\phi(P)$.

Next, for any two fixed points on a connected component of an isotropy
submanifold $M^{{\mathbb{Z}}_{k}}$, the sets of $S^{1}$-weights at them are
related as follows.

######  Lemma 3.5.

[20] Let the circle act on a compact symplectic manifold $(M,\omega)$. Let $P$
and $Q$ be fixed points which lie on the same connected component of
$M^{{\mathbb{Z}}_{k}}$ for some $k>1$. Then the weights of the $S^{1}$-action
at $P$ and $Q$ are equal modulo $k$.

A variant of Lemma 3.5 is: the weights of the $S^{1}$-action at the normal
spaces of the component of $M^{{\mathbb{Z}}_{k}}$ at $P$ and at $Q$ are equal
modulo $k$.

The next result by Jang and Tolman is a claim on the multiplicity of the
smallest positive weight on the manifold. It is given for the more general
category of almost complex manifolds than symplectic manifolds.

######  Lemma 3.6.

[4] Let the circle act on a closed $2n$-dimensional almost complex manifold
with isolated fixed points. Let $w$ be the smallest positive weight that
occurs at the fixed points on $M$. Then given any $k\in\\{0,1,...,n-1\\}$, the
number of times the weight $-w$ occurs at fixed points of index $2k+2$ is
equal to the number of times the weight $+w$ occurs at fixed points of index
$2k$.

Now, we start to consider compact Hamiltonian $S^{1}$-manifolds with minimal
isolated fixed points. We will use the following facts on the Morse indices of
the fixed points and (3.8) so often that we do not refer to the lemma.

######  Lemma 3.7.

[9] Let the circle act on a compact $2n$-dimensional symplectic manifold
$(M,\omega)$ with moment map $\phi\colon M\to{\mathbb{R}}$. Assume $M^{S^{1}}$
consists of $n+1$ isolated points, denoted
$M^{S^{1}}=\\{P_{0},P_{1},\cdots,P_{n}\\}$. Then the points in $M^{S^{1}}$ can
be labelled so that $P_{i}$ has Morse index $2i$, and we have
$H^{2i}(M;{\mathbb{Z}})={\mathbb{Z}}$ and $H^{2i-1}(M;{\mathbb{Z}})=0$ for all
$0\leq 2i\leq 2n$. Moreover,

(3.8) $\phi(P_{0})<\phi(P_{1})<\cdots<\phi(P_{n}).$

In a symplectic $S^{1}$-manifold $(M,\omega)$ with isolated fixed points, if
$w>0$ is a weight at a fixed point $P$, $-w$ is a weight at a fixed point $Q$,
and $P$ and $Q$ are on the same connected component of $M^{{\mathbb{Z}}_{w}}$,
we say that $w$ is a weight from $P$ to $Q$ or there is a weight $w$ from $P$
to $Q$. When the signs of $w$ at $P$ and at $Q$ are clear, we will also say
that there is a weight $w$ or $-w$ between $P$ and $Q$, or $w$ or $-w$ is a
weight between $P$ and $Q$.

######  Lemma 3.9.

[9] Let the circle act on a compact $2n$-dimensional symplectic manifold
$(M,\omega)$ with moment map $\phi\colon M\to{\mathbb{R}}$. Assume
$M^{S^{1}}=\\{P_{0},P_{1},\cdots,P_{n}\\}$ and $[\omega]$ is a primitive
integral class. Then there is a weight $w=\phi(P_{1})-\phi(P_{0})$ from
$P_{0}$ to $P_{1}$, and there is a weight
$w^{\prime}=\phi(P_{n})-\phi(P_{n-1})$ from $P_{n-1}$ to $P_{n}$.

######  Lemma 3.10.

[9] Let the circle act on a compact $2n$-dimensional symplectic manifold
$(M,\omega)$ with moment map $\phi\colon M\to{\mathbb{R}}$. Assume $[\omega]$
is an integral class and $M^{S^{1}}=\\{P_{0},P_{1},\cdots,P_{n}\\}$. Assume
there is a weight $a=\phi(P_{i})-\phi(P_{0})$ from $P_{0}$ to $P_{i}$ for some
$i\neq 0$. Then for any other weight $a^{\prime}$ at $P_{0}$, which could be
equal to $a$, there exists $j\neq 0,i$, such that $a^{\prime}$ is a weight
from $P_{0}$ to $P_{j}$. A similar claim holds if we replace $P_{0}$ by
$P_{n}$.

For a compact Hamiltonian $S^{1}$-manifold $M$ with minimal isolated fixed
points, by standard method in equivariant cohomology by Kirwan, and Tolman-
Weitsman (see [6] and [21]), we know that the extended classes of the
equivariant Euler classes of the negative normal bundles of the fixed points
form a basis of $H^{*}_{S^{1}}(M;{\mathbb{Z}})$ as an
$H^{*}({\mathbb{CP}}^{\infty};{\mathbb{Z}})$-module. Moreover, since
$H^{*}(M^{S^{1}};{\mathbb{Z}})$ has no torsion, we have
$H^{*}(M;{\mathbb{Z}})=H^{*}_{S^{1}}(M;{\mathbb{Z}})/(t)$ (see Sec.2 of [14]),
hence the restriction of a basis of $H^{*}_{S^{1}}(M;{\mathbb{Z}})$ to
ordinary cohomology is a basis of $H^{*}(M;{\mathbb{Z}})$. We summarize these
as follows.

######  Proposition 3.11.

Let the circle act on a compact $2n$-dimensional symplectic manifold
$(M,\omega)$ with moment map $\phi\colon M\to{\mathbb{R}}$. Assume
$M^{S^{1}}=\\{P_{0},P_{1},\cdots,P_{n}\\}$. Then as an
$H^{*}({\mathbb{CP}}^{\infty};{\mathbb{Z}})$-module,
$H^{*}_{S^{1}}(M;{\mathbb{Z}})$ has a basis
$\big{\\{}\widetilde{\alpha}_{i}\in H^{2i}_{S^{1}}(M;{\mathbb{Z}})\,|\,0\leq
i\leq n\big{\\}}$ such that

$\widetilde{\alpha}_{i}|_{P_{i}}=\Lambda_{i}^{-}t^{i},\,\,\mbox{and}\,\,\,\,\widetilde{\alpha}_{i}|_{P_{j}}=0,\,\,\forall\,\,j<i,$

where $\Lambda_{i}^{-}$ is the product of the negative weights at $P_{i}$.
Moreover, $\big{\\{}\alpha_{i}=r(\widetilde{\alpha}_{i})\in
H^{2i}(M;{\mathbb{Z}})\,|\,0\leq i\leq n\big{\\}}$ is a basis for
$H^{*}(M;{\mathbb{Z}})$, where $r\colon H^{*}_{S^{1}}(M;{\mathbb{Z}})\to
H^{*}(M;{\mathbb{Z}})$ is the natural restriction map.

A direct corollary of Proposition 3.11 is:

######  Corollary 3.12.

Let the circle act on a compact $2n$-dimensional symplectic manifold
$(M,\omega)$ with moment map $\phi\colon M\to{\mathbb{R}}$. Assume
$M^{S^{1}}=\\{P_{0},P_{1},\cdots,P_{n}\\}$. Let $\widetilde{\alpha}\in
H^{2i}_{S^{1}}(M;{\mathbb{Z}})$ be a class such that
$\widetilde{\alpha}|_{P_{j}}=0$ for all $j<i$. Then

$\widetilde{\alpha}=a_{i}\widetilde{\alpha}_{i}\,\,\,\mbox{for some
$a_{i}\in{\mathbb{Z}}$}.$

##  4\. on the first Chern class $c_{1}(M)$

In this section, we give two ways of expressing the first Chern class of the
manifold. The first way is using the sums of the weights of two fixed points
as in Lemma 4.1, the second way is using the largest weight between two fixed
points as in Lemma 4.2.

######  Lemma 4.1.

[9] Let the circle act on a connected compact symplectic manifold $(M,\omega)$
with moment map $\phi\colon M\to{\mathbb{R}}$. If $c_{1}(M)=k[\omega]$, then
for any two fixed point set components $F$ and $F^{\prime}$, we have
$\Gamma_{F}-\Gamma_{F^{\prime}}=k\big{(}\phi(F^{\prime})-\phi(F)\big{)}$,
where $\Gamma_{F}$ and $\Gamma_{F^{\prime}}$ are respectively the sums of the
weights at $F$ and $F^{\prime}$.

######  Lemma 4.2.

Let the circle act on a connected compact symplectic manifold $(M,\omega)$
with moment map $\phi\colon M\to{\mathbb{R}}$. Assume $M^{S^{1}}$ consists of
isolated points. Let $P,Q\in M^{S^{1}}$ with $P\neq Q$, where
${\operatorname{index}}(P)=2i$ and ${\operatorname{index}}(Q)=2j$ with $i\leq
j$. Assume there is a weight $w>0$ from $P$ to $Q$, $-w$ is not a weight at
$P$, $-w$ has multiplicity $s$ at $Q$, and $w$ is the largest among the
absolute values of all the weights at $P$ and $Q$. Then $w$ has multiplicity
at least $s$ at $P$. If $c_{1}(M)=k[\omega]$, then
$j-i+s=k\frac{\phi(Q)-\phi(P)}{w}$.

###### Proof.

Let

$W_{P}^{-}=\big{\\{}a_{1},\cdots,a_{i}\big{\\}}\,\,\,\mbox{and}\,\,\,W_{P}^{+}=\big{\\{}a_{i+1},\cdots,a_{n}\big{\\}}$

be respectively the set of negative weights and positive weights at $P$, and

$W_{Q}^{-}=\big{\\{}b_{1},\cdots,b_{j}\big{\\}}\,\,\,\mbox{and}\,\,\,W_{Q}^{+}=\big{\\{}b_{j+1},\cdots,b_{n}\big{\\}}$

be respectively the set of negative weights and positive weights at $Q$. By
Lemma 3.5,

(4.3) $W_{P}^{-}\cup W_{P}^{+}=W_{Q}^{-}\cup W_{Q}^{+}\mod w.$

Since there is no weight $-w$ at $P$, for (4.3) to hold, for each $-w$ at $Q$,
there should be a $w$ at $P$ such that $w=-w\mod w$. So $w$ has at least
multiplicity $s$ at $P$. Assume

$a_{n-s+1}=w,\cdots,a_{n}=w\in
W_{P}^{+},\,\,\,\mbox{and}\,\,\,b_{j-s+1}=-w,\cdots,b_{j}=-w\in W_{Q}^{-}.$

The rest of the proof is similar to that of [9, Lemma 4.2]. Since $w$ is the
largest among the absolute values of all the weights at $P$ and $Q$, up to a
reordering of indices, (4.3) can only yield the following relations:

$a_{1}=b_{1},\,\,\,\cdots,\,\,\,a_{m}=b_{m}\,\,\,\mbox{for some $m$ with
$0\leq m\leq i$},$
$a_{i+1}=b_{j+1},\,\,\,\cdots,\,\,\,a_{i+l}=b_{j+l}\,\,\,\mbox{for some $l$
with $0\leq l\leq n-j$},$
$a_{m+1}=b_{j+l+1}-w,\,\,\,\cdots,\,\,\,a_{i}=b_{n}-w,$
$a_{i+l+1}=b_{m+1}+w,\,\,\,\cdots,\,\,\,a_{n-s}=b_{j-s}+w,\,\,\,\mbox{and}$
$a_{n-s+1}=b_{j-s+1}+2w,\,\,\,\cdots,\,\,\,a_{n}=b_{j}+2w.$

Let $\Gamma_{P}$ and $\Gamma_{Q}$ be respectively the sums of weights at $P$
and at $Q$. Then

$\Gamma_{P}-\Gamma_{Q}=\sum_{r=1}^{n}a_{r}-\sum_{r=1}^{n}b_{r}=(j-i+s)w.$

Combining Lemma 4.1, we obtain the claim of the lemma. ∎

The next lemma gives a criterion of using Lemma 4.2 for the largest weight on
$M$. The lemma is stated here according to Lemma 4.4 and its proof in [9].

######  Lemma 4.4.

[9] Let the circle act on a compact symplectic manifold $(M,\omega)$ with
moment map $\phi\colon M\to{\mathbb{R}}$. Assume $M^{S^{1}}$ consists of
isolated points. Let $w>0$ be the largest weight on $M$. Take a lower fixed
point $P$ (relative to the value of $\phi$) so that $-w$ is not a weight at
$P$, and take a closest fixed point $Q$ to $P$ with $\phi(P)<\phi(Q)$ such
that there is a weight $w$ from $P$ to $Q$, then $-w$ has multiplicity $1$ at
$Q$.

######  Remark 4.5.

For the largest weight $w$ on $M$, and the choice of $P$ and $Q$ in Lemma 4.4,
the $s$ in Lemma 4.2 is $1$. In Lemma 4.2, if $s>1$, then between $P$ and $Q$,
there must be other fixed points; otherwise, the connected component of
$M^{{\mathbb{Z}}_{w}}$ containing $P$ has $P$ as the minimum, and $Q$ as the
next fixed point with index bigger than 2, contradicting to Lemma 3.4 applied
on this component.

######  Example 4.6.

Consider the coadjoint orbit $\mathcal{O}$ of $G_{2}$ in Example 2.1. By Lemma
4.1, using the fixed points $P_{0}$ and $P_{1}$, we get

(4.7)
$c_{1}(\mathcal{O})=\frac{\Gamma_{0}-\Gamma_{1}}{\phi(P_{1})-\phi(P_{0})}[\omega]=3[\omega].$

By Lemma 4.2, using the fixed points $P_{0}$ and $P_{5}$, or $P_{1}$ and
$P_{3}$, we also get

$c_{1}(\mathcal{O})=3[\omega].$

##  5\. the equivalence of the conditions on $c_{1}(M)$ and on the largest
weight — proof of Theorem 1

In this section, we prove Theorem 1. If $c_{1}(M)=3[\omega]$, then by Lemmas
4.4 and 4.2, the largest weight on $M$ may be between $P_{0}$ and $P_{2}$,
symmetrically between $P_{3}$ and $P_{5}$, between $P_{1}$ and $P_{3}$,
symmetrically between $P_{2}$ and $P_{4}$, and between $P_{0}$ and $P_{5}$. We
will study the relation between these possibilities.

We start with Lemmas 5.1 and 5.2 on the multiplicities of the weights at
$P_{0}$.

######  Lemma 5.1.

Let $(M,\omega)$ be a compact $10$-dimensional Hamiltonian $S^{1}$-manifold
with $M^{S^{1}}=\\{P_{0},P_{1},\cdots,P_{5}\\}$. Then we have the following
claims.

  1. (1)

Any weight at $P_{0}$ has maximal multiplicity $2$, and $1$ has maximal
multiplicity $1$ at $P_{0}$.

  2. (2)

If $a$ and $b$ are both weights at $P_{0}$, $a>1$, $a|b$, and if $a$ or $b$
has multiplicity $2$, then a connected component of $M^{{\mathbb{Z}}_{a}}$
containing $P_{0}$ contains all the 6 fixed points, and $a$ is a weight
between $P_{0}$ and $P_{1}$, in particular, $1$ cannot be a weight at $P_{0}$.
If $c,d,e\in{\mathbb{N}}$, $c>1$, and $c|d|e$, then the total multiplicity of
$c,d,e$ as weights at $P_{0}$ is at most 3.444We will not use this last claim
in this paper.

  3. (3)

If $a$ is a weight from $P_{0}$ to $P_{i}$ for $i\geq 3$, and it is not a
weight from $P_{0}$ to $P_{i}$ for $i\leq 2$, and the weights at $P_{0}$ and
$P_{5}$ have symmetry, then $a$ has multiplicity $1$ at $P_{0}$.

###### Proof.

(1) If $a=1$ is a weight at $P_{0}$, by Lemma 3.6, it has multiplicity $1$
since there is only one index $2$ fixed point $P_{1}$ in $M$. Assume $a>1$ is
a weight at $P_{0}$, and it has multiplicity at least $3$. Then the connected
component of the isotropy submanifold $M^{{\mathbb{Z}}_{a}}$ containing
$P_{0}$ has dimension at least $6$, and has at least 3 index $2$ fixed points
by Lemma 3.6 applied on the component; by symmetry, this component of
$M^{{\mathbb{Z}}_{a}}$ has at least 3 coindex $2$ fixed points, and $1$ top
index fixed point. This contradicts to the total number of 6 fixed points in
$M$, so $a$ has maximal multiplicity $2$ at $P_{0}$.

(2) First assume $a$ has multiplicity $2$ at $P_{0}$. Then the connected
component $C$ of $M^{{\mathbb{Z}}_{a}}$ containing $P_{0}$ has dimension at
least $6$, it has 2 index $2$ fixed points by Lemma 3.6 used on $C$; by
symmetry, it also has 2 coindex $2$ fixed points, and $1$ top index fixed
point. So $C$ contains all the 6 fixed points, $P_{1}$ being an index 2 fixed
point by Lemma 3.4, and there is a weight $a$ between $P_{0}$ and $P_{1}$ by
Lemma 3.6 used on $C$. Now assume $b$ has multiplicity $2$ at $P_{0}$. Then
the connected component $C^{\prime}$ of $M^{{\mathbb{Z}}_{a}}$ containing
$P_{0}$ has dimension at least $6$, and it has at least 1 index $2$ fixed
point with negative weight $-a$ by Lemma 3.6. By symmetry, $C^{\prime}$
contains at least 1 coindex 2 fixed point with positive weight $a$. In
$C^{\prime}$, there is a $M^{{\mathbb{Z}}_{b}}$ which is of dimension at least
$4$ and which contains 2 index $2$ fixed points each having a negative weight
$-b$ by Lemma 3.6; by symmetry, $M^{{\mathbb{Z}}_{b}}$ contains 2 coindex 2
fixed points (which could be the same as the index 2 fixed points in dimension
4) each having a positive weight $b$. So the index 2 and coindex 2 fixed
points in $M^{{\mathbb{Z}}_{b}}$ are different from the index 2 and coindex 2
fixed point in $C^{\prime}$. So $C^{\prime}$ contains all the 6 fixed points,
and $a$ is a weight between $P_{0}$ and $P_{1}$ by Lemma 3.6. If $1$ is a
weight at $P_{0}$, then $-1$ is a weight at $P_{1}$ by Lemma 3.6; in the two
cases above, this is not possible. The last claim can be proved similarly as
above.

(3) By the assumption and Lemma 3.6, $a\neq 1$. Assume $a$ has multiplicity
$2$ at $P_{0}$. By the assumption and Lemma 3.6 (used on the component), the
connected component $C$ of $M^{{\mathbb{Z}}_{a}}$ containing $P_{0}$ must
contain $P_{3}$, $P_{4}$ and $P_{5}$. Since it contains $P_{5}$, by symmetry
of the weights at $P_{0}$ and $P_{5}$, the component $C$ also contains $P_{2}$
and $P_{1}$, contradicting to that there is no weight $a$ from $P_{0}$ to
$P_{1}$ and Lemma 3.6 used on $C$. ∎

######  Lemma 5.2.

Let $(M,\omega)$ be a compact $10$-dimensional Hamiltonian $S^{1}$-manifold
with moment map $\phi\colon M\to{\mathbb{R}}$. Assume
$M^{S^{1}}=\\{P_{0},P_{1},\cdots,P_{5}\\}$, $[\omega]$ is an integral class,
and $\phi(P_{i})-\phi(P_{0})=\phi(P_{5})-\phi(P_{5-i})$ for $i=1,2$. If the
integer $a>\frac{1}{3}\big{(}\phi(P_{5})-\phi(P_{0})\big{)}$ or
$a\nmid\big{(}\phi(P_{5})-\phi(P_{0})\big{)}$, then as a weight, $a$ has
maximal multiplicity 1 at $P_{0}$.

###### Proof.

Since there is symmetry between $P_{0}$ and $P_{5}$, if $a$ is a weight at
$P_{0}$, then $-a$ is a weight at $P_{5}$. Assume $a$ is a weight at $P_{0}$,
and $a$ has multiplicity $2$ (maximal multiplicity by Lemma 5.1). Then the
connected component of the isotropy submanifold $M^{{\mathbb{Z}}_{a}}$
containing $P_{0}$ is at least 4-dimensional. By Lemma 3.6 used on this
component, it contains $P_{0}$, 2 index 2 fixed points $P_{i}$ and $P_{j}$
with $i,j\in\\{1,2,3,4\\}$ and $i<j$, and 1 top index fixed point, which must
be $P_{5}$ by symmetry (if $M^{{\mathbb{Z}}_{a}}$ has dimension at least 6,
then the component contains all the fixed points since it also contains 2
coindex 2 fixed points.). Since $[\omega]$ is an integral class, by Lemma 3.2,
we have

$\big{(}\phi(P_{j})-\phi(P_{0})\big{)}+\big{(}\phi(P_{5})-\phi(P_{j})\big{)}=l_{1}a+l_{2}a=la=\big{(}\phi(P_{5})-\phi(P_{0})\big{)},$

where $3\leq l\in{\mathbb{N}}$, since
$a|\big{(}\phi(P_{5})-\phi(P_{j})\big{)}$,
$a|\big{(}\phi(P_{j})-\phi(P_{0})\big{)}$, and
$\phi(P_{j})-\phi(P_{0})>\phi(P_{5})-\phi(P_{j})$ by the assumption. So if
$a>\frac{1}{3}\big{(}\phi(P_{5})-\phi(P_{0})\big{)}$ or
$a\nmid\big{(}\phi(P_{5})-\phi(P_{0})\big{)}$, this cannot be true. ∎

Assume $c_{1}(M)=3[\omega]$, and each weight is less than or equal to $5$, the
next two lemmas will exclude the possibility that the largest weight is
between $P_{0}$ and $P_{2}$, symmetrically between $P_{3}$ and $P_{5}$.

######  Lemma 5.3.

Let $(M,\omega)$ be a compact $10$-dimensional Hamiltonian $S^{1}$-manifold
with moment map $\phi\colon M\to{\mathbb{R}}$. Assume
$M^{S^{1}}=\\{P_{0},P_{1},\cdots,P_{5}\\}$, $[\omega]$ is a primitive integral
class, and $\phi(P_{i})-\phi(P_{0})=\phi(P_{5})-\phi(P_{5-i})$ for $i=1,2$.
Assume $c_{1}(M)=k[\omega]$ with $1<k\in{\mathbb{N}}$. Then the largest weight
on $M$ has maximal multiplicity 1 at $P_{0}$.

###### Proof.

Note that by Lemmas 4.4 and 4.2,

$c_{1}(M)=k[\omega]\,\,\,\mbox{with}\,\,\,1\leq k\leq 6.$

Let $w\in{\mathbb{N}}$ be the largest weight on $M$. Assume $w$ occurs at
$P_{0}$. By symmetry, $-w$ occurs at $P_{5}$. By Lemma 5.1, $w$ has maximal
multiplicity $2$ at $P_{0}$; assume this is the case. Then the connected
component $C$ of $M^{{\mathbb{Z}}_{w}}$ containing $P_{0}$ is $4$ dimensional,
it contains two index $2$ fixed points, $P_{i}$ and $P_{j}$ with
$i,j\in\\{1,2,3,4\\}$ and $i<j$ by Lemma 3.6, and 1 index $4$ fixed point. By
symmetry, the index $4$ fixed point is $P_{5}$. By Lemma 4.2 for $P_{0}$ and
$P_{5}$, with $s=2$, we get $7=k\frac{\phi(P_{5})-\phi(P_{0})}{w}$, which is
not possible since $k\neq 1,7$ and $w|\big{(}\phi(P_{5})-\phi(P_{0})\big{)}$.
∎

######  Lemma 5.4.

Let $(M,\omega)$ be a compact $10$-dimensional Hamiltonian $S^{1}$-manifold
with moment map $\phi\colon M\to{\mathbb{R}}$. Assume
$M^{S^{1}}=\\{P_{0},P_{1},\cdots,P_{5}\\}$, $[\omega]$ is a primitive integral
class, and $\phi(P_{i})-\phi(P_{0})=\phi(P_{5})-\phi(P_{5-i})$ for $i=1,2$. If
$c_{1}(M)=3[\omega]$, and each weight is less than or equal to $5$, then the
largest weight cannot be between $P_{0}$ and $P_{2}$, symmetrically between
$P_{3}$ and $P_{5}$, and is equal to
$\phi(P_{2})-\phi(P_{0})=\phi(P_{5})-\phi(P_{3})$.

###### Proof.

If $c_{1}(M)=3[\omega]$, then by Lemmas 4.4 and 4.2 and symmetry, the largest
weight $w$ on $M$ could be between $P_{0}$ and $P_{2}$, and between $P_{3}$
and $P_{5}$, with $\phi(P_{2})-\phi(P_{0})=\phi(P_{5})-\phi(P_{3})=w$. Assume
this is the case. By Lemma 5.3, $w$ has multiplicity $1$ at $P_{0}$, so
$P_{2}$ is the only fixed point which has a weight $w$ with $P_{0}$. By
symmetry, $\Gamma_{5}=-\Gamma_{0}$. Then by Lemma 4.1, we have that
$\phi(P_{3})-\phi(P_{2})=2k$ is even,

$2\Gamma_{0}=3(2w+2k),\,\,\,\mbox{and}\,\,\,\Gamma_{0}\geq 18\,\,\,\mbox{if
$w=5$}.$

Let

$a=\phi(P_{1})-\phi(P_{0})=\phi(P_{5})-\phi(P_{4}).$

By Lemma 3.9, $a$ is a weight between $P_{0}$ and $P_{1}$. By Lemma 3.6, if
$a=1$, then $1$ has multiplicity $1$ at $P_{0}$; if $1<a<w$, then $1$ is not a
weight at $P_{0}$. First assume the largest weight $w=5$, then besides $a$ and
$5$, the other possible distinct weights at $P_{0}$ are $2$, $3$ and $4$.
Since $P_{1}$ is of index $2$, and none of $2$, $3$, and $4$ divides
$5=\phi(P_{2})-\phi(P_{0})$ (or by Lemma 3.10), these other weights are
between $P_{0}$ and $P_{i}$ with $i\geq 3$. By Lemma 5.1 (3), these other
weights each has maximal multiplicity 1. So with $a=1,2,3,4$, the set of
weights at $P_{0}$ is possibly $\\{1,5,2,3,4\\}$, or $\\{2,5,2,3,4\\}$, or
$\\{3,5,2,3,4\\}$, or $\\{4,5,2,3,4\\}$. The case $\\{4,5,2,3,4\\}$ is not
possible by Lemma 5.1 (2). The other possibilities contradict to the above
$\Gamma_{0}\geq 18$. So $w=5$ is not possible. The above arguments on the set
of weights at $P_{0}$ show that if $w<5$, then there are not enough number of
weights at $P_{0}$, so this is also not possible. ∎

In the next two lemmas, we show that when the largest weight $w$ on $M$ is
between $P_{0}$ and $P_{5}$ with
$w=\frac{1}{2}\big{(}\phi(P_{5})-\phi(P_{0})\big{)}$, then $w\geq 5$; and when
$w=5$, then the largest weight is also between $P_{1}$ and $P_{3}$,
symmetrically, between $P_{2}$ and $P_{4}$, and the sets of weights at
$P_{0}$, $P_{5}$, $P_{1}$ and $P_{4}$ can be completely determined.

######  Lemma 5.5.

Let $(M,\omega)$ be a compact $10$-dimensional Hamiltonian $S^{1}$-manifold
with moment map $\phi\colon M\to{\mathbb{R}}$. Assume
$M^{S^{1}}=\\{P_{0},P_{1},\cdots,P_{5}\\}$ and $[\omega]$ is a primitive
integral class. Assume $w=\frac{1}{2}\big{(}\phi(P_{5})-\phi(P_{0})\big{)}$ is
a weight between $P_{0}$ and $P_{5}$, and this integer is the largest weight
on $M$. Then $w\geq 5$. When $w=5$, then the following are true.

  1. (1)

The set of weights at $P_{0}$ is $\\{5,4,3,2,1\\}$ and the set of weights at
$P_{5}$ is $-\\{5,4,3,2,1\\}$.

  2. (2)

$\phi(P_{1})-\phi(P_{0})=\phi(P_{5})-\phi(P_{4})=1$,
$\phi(P_{2})-\phi(P_{1})=\phi(P_{4})-\phi(P_{3})=3$, and
$\phi(P_{3})-\phi(P_{2})=2$.

###### Proof.

By Lemma 5.1, $w$ has maximal multiplicity 2 at $P_{0}$; similarly $-w$ has
maximal multiplicity 2 at $P_{5}$. By Lemma 4.2 used for $P_{0}$ and $P_{5}$,
we know that $-w$ has multiplicity $1$ at $P_{5}$, and

(5.6) $c_{1}(M)=3[\omega].$

By symmetry, $w$ has multiplicity $1$ at $P_{0}$. Note that $w\geq 3$ by (3.8)
and the fact that $\phi(P_{i})-\phi(P_{j})\in{\mathbb{Z}}$ for all $i$ and
$j$.

First assume $w=3$. Then the possible distinct weights at $P_{0}$ are $1$, $2$
and $3$, where $3$ occurs once by the above claim. By Lemma 3.6, $1$ occurs at
most once at $P_{0}$. By Lemma 5.1, $2$ occurs at most twice at $P_{0}$. So
$P_{0}$ does not have enough number of weights, a contradiction.

Now assume $w=4$. Then the possible distinct weights at $P_{0}$ are $1$, $2$,
$3$ and $4$, where $1$ and $4$ each occur at most once by the arguments above.
By Lemma 3.5 for $P_{0}$ and $P_{5}$, $1=-3\mod 4$, and by symmetry, $3$
occurs at $P_{0}$ at most once. By Lemma 5.1 (2), if $2$ occurs twice at
$P_{0}$, then $1$ (hence $3$) cannot occur at $P_{0}$. So there are not enough
number of weights at $P_{0}$, a contradiction.

Next assume $w=5$. Then the possible distinct weights at $P_{0}$ are $1$, $2$,
$3$, $4$ and $5$. By Lemma 3.5 for $P_{0}$ and $P_{5}$, we could have the
possiblities $1=-4\mod 5$, $2=-3\mod 5$, $3=-2\mod 5$, and $4=-1\mod 5$. By
this and symmetry, we know that $3$ and $2$ have the same multiplicity at
$P_{0}$, and $1$ and $4$ have the same multiplicity at $P_{0}$. By the
argument above, $1$ has maximal multiplicity $1$ at $P_{0}$. By symmetry of
the weights at $P_{0}$ and $P_{5}$, we may assume that

(5.7)
$\phi(P_{5})-\phi(P_{4})=\phi(P_{1})-\phi(P_{0})\quad\mbox{and}\quad\phi(P_{4})-\phi(P_{3})=\phi(P_{2})-\phi(P_{1}).$

By Lemma 5.2, $3$ has maximal multiplicity 1 at $P_{0}$. The information above
implies that the set of weights at $P_{0}$ is as claimed in (1). By symmetry,
the set of weights at $P_{5}$ is as claimed.

In the rest part, assume $w=5$. By symmetry, we have

(5.8) $\phi(P_{3})-\phi(P_{2})\geq 2\,\,\,\mbox{is even}.$

By the claim in $(1)$, Lemma 3.9, Equations (5.7) and (5.8), and
$\phi(P_{5})-\phi(P_{0})=10$, we have the following possibilities:

  * (a)

$\phi(P_{1})-\phi(P_{0})=\phi(P_{5})-\phi(P_{4})=1$,
$\phi(P_{2})-\phi(P_{1})=\phi(P_{4})-\phi(P_{3})=3$,
$\phi(P_{3})-\phi(P_{2})=2$.

  * (b)

$\phi(P_{1})-\phi(P_{0})=\phi(P_{5})-\phi(P_{4})=1$,
$\phi(P_{2})-\phi(P_{1})=\phi(P_{4})-\phi(P_{3})=2$,
$\phi(P_{3})-\phi(P_{2})=4$.

  * (c)

$\phi(P_{1})-\phi(P_{0})=\phi(P_{5})-\phi(P_{4})=1$,
$\phi(P_{2})-\phi(P_{1})=\phi(P_{4})-\phi(P_{3})=1$,
$\phi(P_{3})-\phi(P_{2})=6$.

Case (a) exists by Example 2.1. Case (b) does not exist since the weight $4$
at $P_{0}$ divides none of $\phi(P_{0})-\phi(P_{i})$ for $i\neq 0$. Now
consider (c). The values $\phi(P_{i})-\phi(P_{0})$ for $i>0$ are $1$, $2$,
$8$, $9$ and $10$. Note that the weight $4$ at $P_{0}$ can only be between
$P_{0}$ and $P_{3}$. The set of possible positive weights at $P_{3}$ are
$\\{1,1\\}$ or $\\{1,2\\}$ by Lemma 3.3. By Lemma 3.3 and the fact that $5$ is
the largest weight on $M$, the possible distinct negative weights at $P_{3}$
are $-1$, $-2$, $-3$, and $-4$. By Lemma 3.5, the set of weights at $P_{3}$
except $-4$ are equal mod $4$ to the set of weights $\\{1,2,3,5\\}$ (except
$4$) at $P_{0}$. So the possibilities are as follows, where the weights at
$P_{3}$ are on the left, and the weights at $P_{0}$ which we know are on the
right.

  * •

$1=1$, $1=5$, $-1=3$, $-2=2$, $\mod 4$.

  * •

$1=1$, $2=2$, $-1=3$, $-3=5$, $\mod 4$.

  * •

$1=5$, $2=2$, $-1=3$, $-3=1$, $\mod 4$.

So the set of weights at $P_{3}$ are possibly as follows.

  * (c1)

$\\{1,1,-1,-2,-4\\}$.

  * (c2)

$\\{1,2,-1,-3,-4\\}$.

By Lemma 4.1 for $P_{0}$ and $P_{3}$, we get that the cases (c1) and (c2)
contradict to (5.6), so (c) is not possible. ∎

######  Remark 5.9.

In Lemma 5.5, if we only assume that
$w=\frac{1}{2}\big{(}\phi(P_{5})-\phi(P_{0})\big{)}$ is a weight between
$P_{0}$ and $P_{5}$, and it is the largest among all the weights at $P_{0}$
and $P_{5}$ (instead of that it is the largest weight on $M$), then all the
arguments in the proof go through except that we cannot exclude Case (c) above
for $w=5$. We still have that the set of weights at $P_{0}$ and $P_{5}$:

$P_{0}\colon\\{1,2,3,4,5\\},\,\,\,P_{5}\colon-\\{1,2,3,4,5\\}.$

But (in (c)) we may have that the set of weights at $P_{3}$ is

$P_{3}\colon\\{1,2,-1,-7,-4\\}.$

In this case, $7$ is a weight between $P_{1}$ and $P_{3}$, it is the largest
weight on $M$. Using this fact, we can get the set of weights at $P_{1}$:

$P_{1}\colon\\{-1,1,2,3,7\\}.$

By symmetry, the set of weights at $P_{4}$ is the negative of the set of
weights at $P_{1}$, and the set of weights at $P_{2}$ is the negative of the
set of weights at $P_{3}$. So we may get the weights at all the fixed points.
By Lemma 4.2 or 4.1, we have

$c_{1}(M)=3[\omega].$

Using the method in the next section, we can get that the ring
$H^{*}(M;{\mathbb{Z}})$ is generated by the elements:

$1,\,\,\,[\omega],\,\,\,[\omega]^{2},\,\,\,\frac{1}{12}[\omega]^{3},\,\,\,\frac{1}{12}[\omega]^{4},\,\,\,\frac{1}{12}[\omega]^{5}.$

But we do not know and are not studying if this case exists.

######  Remark 5.10.

By Lemma 5.5, the condition that
$\frac{1}{2}\big{(}\phi(P_{5})-\phi(P_{0})\big{)}$ is the largest weight and
is a weight between $P_{0}$ and $P_{5}$ implies symmetry of the image of
$\phi$ and symmetry of the set of weights at $P_{0}$ and $P_{5}$.

######  Lemma 5.11.

Let $(M,\omega)$ be a compact $10$-dimensional Hamiltonian $S^{1}$-manifold
with moment map $\phi\colon M\to{\mathbb{R}}$. Assume
$M^{S^{1}}=\\{P_{0},P_{1},\cdots,P_{5}\\}$ and $[\omega]$ is a primitive
integral class. Assume $w=5=\frac{1}{2}\big{(}\phi(P_{5})-\phi(P_{0})\big{)}$
is a weight between $P_{0}$ and $P_{5}$, and this integer is the largest
weight on $M$. Then the set of weights at $P_{1}$ is

$P_{1}\colon\\{-1,1,5,4,3\\},$

where $w=5$ is a weight between $P_{1}$ and $P_{3}$. Symmetrically, the set of
weights at $P_{4}$ is the negative of the above set, and $w=5$ is a weight
between $P_{2}$ and $P_{4}$. Moreover,

$w=5=\phi(P_{3})-\phi(P_{1})=\phi(P_{4})-\phi(P_{2}).$

###### Proof.

Note that $-1$ is the negative weight at $P_{1}$ by Lemmas 3.6 and 5.5. By
Lemma 5.5 (2), the values of $\phi(P_{i})-\phi(P_{1})$ for $i>1$ are $3$, $5$,
$8$ and $9$. By Lemma 3.3 and the fact that $5$ is the largest weight on $M$,
we get that the possible distinct positive weights at $P_{1}$ are $5$, $4$,
$3$, $2$ and $1$. Note that $5$ can only divide $\phi(P_{3})-\phi(P_{1})$, so
it has maximal multiplicity $1$. Note that $4$ and $2$ can only divide
$\phi(P_{4})-\phi(P_{1})$, if any of them has multiplicity more than $1$ or
they both occur at $P_{1}$, then the connected component of
$M^{{\mathbb{Z}}_{2}}$ or $M^{{\mathbb{Z}}_{4}}$ containing $P_{1}$ only
contains $P_{1}$ and $P_{4}$ and has dimension bigger than $2$, a
contradiction. So $4$ and $2$ each has maximal multiplicity $1$ and cannot
occur at the same time at $P_{1}$. Note that $3$ can only divide
$\phi(P_{2})-\phi(P_{1})$ and $\phi(P_{5})-\phi(P_{1})$, by Lemma 3.6 used on
the component of $M^{{\mathbb{Z}}_{3}}$ containing $P_{1}$, we see that $3$
has maximal multiplicity $1$ at $P_{1}$. If $1$ occurs at $P_{1}$ with
multiplicity $2$, then $-1$ occurs at $P_{2}$ with multiplicity $2$ by Lemma
3.6. Note that the weight $4$ at $P_{0}$ only divides
$\phi(P_{2})-\phi(P_{0})$, so $4$ is a weight from $P_{0}$ to $P_{2}$. So $-1$
cannot occur at $P_{2}$ twice. So $1$ has maximal multiplicity $1$ at $P_{1}$.
These facts together yield that the set of weights at $P_{1}$ is
$\\{-1,5,4,3,1\\}$ or $\\{-1,5,3,2,1\\}$. Symmetrically, the set of weights at
$P_{4}$ is $-\\{-1,5,4,3,1\\}$ or $-\\{-1,5,3,2,1\\}$. Using Lemma 4.1 for
$P_{1}$ and $P_{4}$, we get that the case of $\pm\\{-1,5,3,2,1\\}$ contradicts
to (5.6). By Example 2.1, the case of $\pm\\{-1,5,4,3,1\\}$ exists.

From the above, the weight $5$ at $P_{1}$ can only be a weight between $P_{1}$
and $P_{3}$ (with multiplicity $1$) and $5=\phi(P_{3})-\phi(P_{1})$. The rest
is by symmetry. ∎

In Lemma 5.13, assuming each weight is less than or equal to $5$, we show that
if the largest weight on $M$ is between $P_{1}$ and $P_{3}$, symmetrically
between $P_{2}$ and $P_{4}$, then the largest weight is also between $P_{0}$
and $P_{5}$. Lemma 5.12 is a preparation for proving Lemma 5.13. Note that
Lemma 5.12 does not restrict the bound for the largest weight.

######  Lemma 5.12.

Let $(M,\omega)$ be a compact $10$-dimensional Hamiltonian $S^{1}$-manifold
with moment map $\phi\colon M\to{\mathbb{R}}$. Assume
$M^{S^{1}}=\\{P_{0},P_{1},\cdots,P_{5}\\}$ and $[\omega]$ is a primitive
integral class. Assume

$w=\phi(P_{3})-\phi(P_{1})=\phi(P_{4})-\phi(P_{2})$

is a weight from $P_{1}$ to $P_{3}$, symmetrically is a weight from $P_{2}$ to
$P_{4}$, and this integer is the largest weight on $M$. Then the $w$ or $-w$
is the only weight at $P_{1}$, $P_{3}$, $P_{2}$, and $P_{4}$ with absolute
value $w$, and

$c_{1}(M)=3[\omega].$

###### Proof.

First, we prove that $-w$ is not a weight at $P_{1}$. Suppose $-w$ is a weight
at $P_{1}$. Then $w$ is a weight between $P_{0}$ and $P_{1}$ by Lemma 3.9. So
the connected component $C$ of the isotropy submanifold $M^{{\mathbb{Z}}_{w}}$
containing $P_{1}$ is of dimension at least $4$, it contains $P_{0}$ and
$P_{3}$. So $w$ has multiplicity 2 at $P_{0}$, $C$ can only contain $P_{0}$,
$P_{1}$, $P_{3}$ and $P_{5}$ (it cannot contain $P_{2}$ and $P_{4}$). By Lemma
4.2 for $P_{0}$ and $P_{1}$ with $s=1$, we get $c_{1}(M)=2[\omega]$. By Lemma
4.2 for $P_{0}$ and $P_{5}$ with $s=2$, we get a contradiction. Hence $-w$ is
not a weight at $P_{1}$. By assumption, $P_{3}$ is the closest fixed point
which has a weight $w$ with $P_{1}$. By Lemmas 4.4 and 4.2 for $P_{1}$ and
$P_{3}$ with $s=1$, we get

$c_{1}(M)=3[\omega].$

We can similarly prove that $w$ is not a positive weight at $P_{3}$.

Since $-w$ is not a weight at $P_{1}$, and $w$ is not a weight at $P_{3}$, if
$w$ has multiplicity bigger than 1 at $P_{1}$, then the connected component of
$M^{{\mathbb{Z}}_{w}}$ containing $P_{1}$ is at least of dimension 4 and can
only contain $P_{1}$ and $P_{3}$, a contradiction. So $w$ only occurs once at
$P_{1}$. Similarly, $-w$ occurs once at $P_{3}$.

The rest follows similarly. ∎

######  Lemma 5.13.

Let $(M,\omega)$ be a compact $10$-dimensional Hamiltonian $S^{1}$-manifold
with moment map $\phi\colon M\to{\mathbb{R}}$. Assume
$M^{S^{1}}=\\{P_{0},P_{1},\cdots,P_{5}\\}$, $[\omega]$ is a primitive integral
class, and $\phi(P_{i})-\phi(P_{0})=\phi(P_{5})-\phi(P_{5-i})$ for $i=1,2$.
Assume each weight on $M$ is less than or equal to $5$. If

$w=\phi(P_{3})-\phi(P_{1})=\phi(P_{4})-\phi(P_{2})$

is a weight from $P_{1}$ to $P_{3}$, symmetrically is a weight from $P_{2}$ to
$P_{4}$, and it is the largest weight on $M$, then the following hold:

  1. (1)

$w$ is a weight between $P_{0}$ and $P_{5}$ and
$w=\frac{1}{2}\big{(}\phi(P_{5})-\phi(P_{0})\big{)}=5$.

  2. (2)

$\phi(P_{1})-\phi(P_{0})=\phi(P_{5})-\phi(P_{4})=1$,
$\phi(P_{3})-\phi(P_{2})=2$,
$\phi(P_{2})-\phi(P_{1})=\phi(P_{4})-\phi(P_{3})=3$.

###### Proof.

First assume the largest weight $w$ on $M$ occurs at $P_{0}$. Then by Lemma
5.12, this $w$ can only be a weight between $P_{0}$ and $P_{5}$; so it can
only have multiplicity 1 at $P_{0}$ ($-w$ has multiplicity 1 at $P_{5}$). By
Lemma 4.2, and the fact $c_{1}(M)=3[\omega]$ in Lemma 5.12, we get
$w=\frac{1}{2}\big{(}\phi(P_{5})-\phi(P_{0})\big{)}$. By Lemma 5.5, $w=5$, and
(2) holds. By Example 2.1, this case exists.

In the following, assume the largest weight $w$ does not occur at $P_{0}$. We
will show that this is impossible. Let

$\phi(P_{1})-\phi(P_{0})=\phi(P_{5})-\phi(P_{4})=a>0.$

By Lemma 3.9, $a$ is a weight between $P_{0}$ and $P_{1}$, and between $P_{4}$
and $P_{5}$. By the assumption above, $a<5$. By symmetry at $P_{1}$ and
$P_{4}$, Lemma 4.1 and the fact $c_{1}(M)=3[\omega]$, we have that
$\phi(P_{3})-\phi(P_{2})$ is even; since $w$ is no bigger than $5$, we have 2
possibilities:

  * (i)

$\phi(P_{3})-\phi(P_{2})=4$,
$\phi(P_{2})-\phi(P_{1})=\phi(P_{4})-\phi(P_{3})=1$.

  * (ii)

$\phi(P_{3})-\phi(P_{2})=2$,
$\phi(P_{2})-\phi(P_{1})=\phi(P_{4})-\phi(P_{3})\leq 3$.

For (i), $w=5$ and $\phi(P_{5})-\phi(P_{0})=6+2a$. If $a=1$, then besides $1$
which is a weight at $P_{0}$ with multiplicity 1 by Lemma 3.6, the other
possible distinct weights at $P_{0}$ are $2$, $3$ and $4$. If $1<a<5$, then
$1$ is not a weight at $P_{0}$ by Lemma 3.6, and the possible distinct weights
at $P_{0}$ are $2$, $3$ and $4$. For each $a$, we use Lemmas 5.1 and 5.2 to
see that there are not enough number of weights at $P_{0}$.

For (ii), $w\leq 5$, and $\phi(P_{5})-\phi(P_{0})\leq 8+2a$. First assume
$w=5$. Then $\phi(P_{5})-\phi(P_{0})=8+2a$. Similar to the above, for $a=1$,
$1$ is a weight at $P_{0}$ with multiplicity 1; for $1<a<5$, $1$ is not a
weight at $P_{0}$. The cases $a=1$, $a=3$ and $a=4$ are similar to the last
paragraph, using Lemmas 5.1 and 5.2, we see that there are not enough number
of weights at $P_{0}$. If $a=2$, we note that the values of
$\phi(P_{i})-\phi(P_{0})$ for $i>0$ are $2$, $5$, $7$, $10$ and $12$, so $3$
has maximal multiplicity 1 at $P_{0}$, $2$ has maximal multiplicity 2 at
$P_{0}$, and $4$ has maximal multiplicity 1 at $P_{0}$, we also do not have
enough number of weights at $P_{0}$. If $w<5$, similar arguments show that
there are not enough number of weights at $P_{0}$. ∎

Now, we can prove our main theorem Theorem 1 in the Introduction.

######  Theorem 1.

Let $(M,\omega)$ be a compact $10$-dimensional Hamiltonian $S^{1}$-manifold
with moment map $\phi\colon M\to{\mathbb{R}}$. Assume
$M^{S^{1}}=\\{P_{0},P_{1},\cdots,P_{5}\\}$, $[\omega]$ is a primitive integral
class, and $\phi(P_{i})-\phi(P_{0})=\phi(P_{5})-\phi(P_{5-i})$ for $i=1,2$.
Assume each weight is less than or equal to $5$. Then the following 3
conditions are equivalent.

  1. (1)

$c_{1}(M)=3[\omega]$.

  2. (2)

The largest weight is between $P_{0}$ and $P_{5}$ and is equal to
$\frac{1}{2}\big{(}\phi(P_{5})-\phi(P_{0})\big{)}$.

  3. (3)

The largest weight is between $P_{1}$ and $P_{3}$, and between $P_{2}$ and
$P_{4}$, and is equal to $\phi(P_{3})-\phi(P_{1})=\phi(P_{4})-\phi(P_{2})$.

Under any one condition, the largest weight is exactly 5, and $(2)$ and $(3)$
are the only ways it occurs and $|\pm 5|$ only occurs once at the
corresponding fixed point.

###### Proof.

First assume $c_{1}(M)=3[\omega]$. By Lemmas 4.4 and 4.2, the largest weight
could be possibly between $P_{0}$ and $P_{2}$, or between $P_{3}$ and $P_{5}$,
or between $P_{1}$ and $P_{3}$, or between $P_{2}$ and $P_{4}$, or between
$P_{0}$ and $P_{5}$. With the symmetry
$\phi(P_{i})-\phi(P_{0})=\phi(P_{5})-\phi(P_{5-i})$ for $i=1,2$, by Lemma 5.4,
it cannot be between $P_{0}$ and $P_{2}$, symmetrically between $P_{3}$ and
$P_{5}$. So (1) implies (2) or (3). By Lemma 5.11, (2) implies (3). With the
symmetry $\phi(P_{i})-\phi(P_{0})=\phi(P_{5})-\phi(P_{5-i})$ for $i=1,2$, by
Lemma 5.13, (3) implies (2). By Lemmas 5.5 and 5.13, the largest weight is
exactly 5, and by Lemmas 5.5 and 5.12, the value $|\pm 5|$ only occurs once at
the corresponding fixed point.

By Lemma 5.5 (5.6) or Lemma 5.12, (2) or (3) implies (1) $c_{1}(M)=3[\omega]$.
∎

##  6\. determining all the weights, the ring $H^{*}(M;{\mathbb{Z}})$ and the
total Chern class $c(M)$ — proof of Theorem 2

In this section, for a compact 10-dimensional Hamiltonian $S^{1}$-manifold
with 6 isolated fixed points, assuming each weight is less than or equal to 5,
under the condition that the largest weight is between $P_{0}$ and $P_{5}$ and
is equal to $\frac{1}{2}\big{(}\phi(P_{5})-\phi(P_{0})\big{)}$, we determine
the sets of weights at all the fixed points, the integral cohomology ring
$H^{*}(M;{\mathbb{Z}})$, and the total Chern class $c(M)$, proving Theorem 2.

###  6.1. Determining all the weights — proof of Theorem 2(3)

In this part, we determine the sets of weights at all the fixed points,
proving Theorem 2(3).

######  Theorem 2(3).

Let $(M,\omega)$ be a compact $10$-dimensional Hamiltonian $S^{1}$-manifold
with moment map $\phi\colon M\to{\mathbb{R}}$. Assume
$M^{S^{1}}=\\{P_{0},P_{1},\cdots,P_{5}\\}$ and $[\omega]$ is a primitive
integral class. Assume each weight is less than or equal to $5$. If the
largest weight is between $P_{0}$ and $P_{5}$ and is equal to
$\frac{1}{2}\big{(}\phi(P_{5})-\phi(P_{0})\big{)}$, then the sets of weights
at all the fixed points are:

$P_{0}\colon\\{1,4,2,3,5\\},\,\,\,P_{5}\colon-\\{1,4,2,3,5\\},$
$P_{1}\colon\\{-1,1,5,4,3\\},\,\,\,P_{4}\colon-\\{-1,1,5,4,3\\},$
$P_{2}\colon\\{-4,-1,1,5,2\\},\,\,\,P_{3}\colon-\\{-4,-1,1,5,2\\}.$

So the sets of weights at the fixed points are the same as those on
$\mathcal{O}$ in Example 2.1. In particular, there is exactly one weight
between any pair of fixed points.

###### Proof.

By Lemma 5.5, the sets of weights at $P_{0}$ and $P_{5}$ are as claimed. By
Lemma 5.11, the sets of weights at $P_{1}$ and $P_{4}$ are as claimed.

We now find the set of weights at $P_{3}$. By Lemma 5.5 (2), the weight $-4$
at $P_{5}$ can only divide $\phi(P_{5})-\phi(P_{3})$, so there is a weight $4$
between $P_{3}$ and $P_{5}$, so $4$ is a positive weight at $P_{3}$. Since
$-1$ is a weight at $P_{4}$ with multiplicity 1, $1$ is a weight at $P_{3}$
with multiplicity 1 by Lemma 3.6. Hence the set of positive weights at $P_{3}$
is $\\{1,4\\}$. By Lemmas 3.3 and 5.5 (2), except $-5$, the other two possible
negative weights at $P_{3}$ are among $\\{-1,-2,-3\\}$ (with possible
multiplicities). By Lemma 5.11, $5$ is a weight between $P_{1}$ and $P_{3}$.
By Lemma 3.5, the weights at the normal space of the $M^{{\mathbb{Z}}_{5}}$
isotropy manifold at $P_{1}$, $\\{-1,1,4,3\\}$, is equal modulo $5$ to the
weights at the normal space at $P_{3}$. So we have the following
possibilities, where the weights at $P_{3}$ are on the left which we need to
find, and the weights at $P_{1}$ are on the right.

  * •

$1=1$, $4=-1$, $-1=4$, $-2=3$, $\mod 5$.

  * •

$1=1$, $4=4$, $-1=-1$, $-2=3$, $\mod 5$.

Both cases yield the set of weights at $P_{3}\colon\\{-2,-5,-1,1,4\\}$. By
symmetry, the set of weights at $P_{2}$ is $-\\{-2,-5,-1,1,4\\}$.

Using the sets of weights at $P_{1}$, $P_{2}$, and $P_{3}$, Lemmas 5.5 (2) and
3.3, and symmetry, we can see that there is exactly one weight between any
pair of fixed points. ∎

###  6.2. Determining the ring $H^{*}(M;{\mathbb{Z}})$ — proof of Theorem 2(1)

In this part, we determine the ring $H^{*}(M;{\mathbb{Z}})$, proving Theorem
2(1).

######  Lemma 6.1.

Let $(M,\omega)$ be a compact $10$-dimensional Hamiltonian $S^{1}$-manifold
with moment map $\phi\colon M\to{\mathbb{R}}$. Assume
$M^{S^{1}}=\\{P_{0},P_{1},\cdots,P_{5}\\}$ and $[\omega]$ is a primitive
integral class. Then the ring $H^{*}(M;{\mathbb{Z}})$ is generated by the
following $\alpha_{i}\in H^{2i}(M;{\mathbb{Z}})$s’ for $0\leq i\leq 5$:

(6.2)
$\alpha_{i}=\frac{\Lambda_{i}^{-}}{\prod_{j=0}^{i-1}\big{(}\phi(P_{j})-\phi(P_{i})\big{)}}[\omega]^{i},$

where $\Lambda_{i}^{-}$ is the product of the negative weights at $P_{i}$.

###### Proof.

Let $\widetilde{u}$ be the integral class in Lemma 3.1, and
$\big{\\{}\widetilde{\alpha}_{i}\in H^{2i}_{S^{1}}(M;{\mathbb{Z}})\,|\,0\leq
i\leq 5\big{\\}}$ be the basis as in Proposition 3.11. By Corollary 3.12,

(6.3)
$\prod_{j=0}^{i-1}\big{(}\widetilde{u}+\phi(P_{j})t\big{)}=a_{i}\widetilde{\alpha}_{i},\,\,\,\mbox{where
$a_{i}\in{\mathbb{Z}}$}.$

Restricting (6.3) to the fixed point $P_{i}$, we get

$\prod_{j=0}^{i-1}\big{(}\phi(P_{j})-\phi(P_{i})\big{)}=a_{i}\Lambda_{i}^{-}.$

Restricting (6.3) to ordinary cohomology, we get

$[\omega]^{i}=a_{i}\alpha_{i}.$

These two equations yield (6.2). By Proposition 3.11,
$\big{\\{}\alpha_{i}\,|\,0\leq i\leq 5\big{\\}}$ is a basis of the ring
$H^{*}(M;{\mathbb{Z}})$. ∎

######  Remark 6.4.

Note that the $\alpha_{i}$ is determined by the product $\Lambda_{i}^{-}$ of
the negative weights at $P_{i}$ and its relation with
$\prod_{j=0}^{i-1}\big{(}\phi(P_{j})-\phi(P_{i})\big{)}$, we do not need to
know the individual weights at the fixed point $P_{i}$. For example, in
Example 2.1, knowing the weights represented by the moment map values of the
fixed points (not the numbers) is enough to determine the ring
$H^{*}(M;{\mathbb{Z}})$, as shown in the following theorem.

######  Theorem 2(1).

Let $(M,\omega)$ be a compact $10$-dimensional Hamiltonian $S^{1}$-manifold
with moment map $\phi\colon M\to{\mathbb{R}}$. Assume
$M^{S^{1}}=\\{P_{0},P_{1},\cdots,P_{5}\\}$ and $[\omega]$ is a primitive
integral class. If

$\Lambda_{1}^{-}=\phi(P_{0})-\phi(P_{1})\big{)},\,\,\,\Lambda_{2}^{-}=\frac{1}{3}\prod_{j=0}^{1}\big{(}\phi(P_{j})-\phi(P_{2})\big{)},\,\,\,\Lambda_{3}^{-}=\frac{1}{6}\prod_{j=0}^{2}\big{(}\phi(P_{j})-\phi(P_{3})\big{)},$
$\Lambda_{4}^{-}=\frac{1}{18}\prod_{j=0}^{3}\big{(}\phi(P_{j})-\phi(P_{4})\big{)},\,\,\,\Lambda_{5}^{-}=\frac{1}{18}\prod_{j=0}^{4}\big{(}\phi(P_{j})-\phi(P_{5})\big{)},$

then the ring $H^{*}(M;{\mathbb{Z}})$ is generated by the following elements:

$1,\,\,\,[\omega],\,\,\,\frac{1}{3}[\omega]^{2},\,\,\,\frac{1}{6}[\omega]^{3},\,\,\,\frac{1}{18}[\omega]^{4},\,\,\,\frac{1}{18}[\omega]^{5}.$

In particular, if each weight on $M$ is less than or equal to $5$, and if the
largest weight is between $P_{0}$ and $P_{5}$ and is equal to
$\frac{1}{2}\big{(}\phi(P_{5})-\phi(P_{0})\big{)}$, then the ring
$H^{*}(M;{\mathbb{Z}})$ is as above, and is isomorphic to the ring
$H^{*}(\mathcal{O};{\mathbb{Z}})$.

###### Proof.

By Lemma 6.1, we get the first part of the claim on the ring
$H^{*}(M;{\mathbb{Z}})$. In particular, by Theorem 2(3) and Lemma 5.5 (2),
under the condition, the $\Lambda_{i}^{-}$s’ and the
$\prod_{j=0}^{i-1}\big{(}\phi(P_{j})-\phi(P_{i})\big{)}$s’ are related in the
way stated, and so are the corresponding data on $\mathcal{O}$, so the rings
$H^{*}(M;{\mathbb{Z}})$ and $H^{*}(\mathcal{O};{\mathbb{Z}})$ are isomorphic.
∎

###  6.3. Determining the total Chern class $c(M)$ — proof of Theorem 2(2)

In this part, we determine the total Chern class $c(M)$, proving Theorem 2(2).

######  Lemma 6.5.

Let $(M,\omega)$ be a compact $10$-dimensional Hamiltonian $S^{1}$-manifold
with moment map $\phi\colon M\to{\mathbb{R}}$. Assume
$M^{S^{1}}=\\{P_{0},P_{1},\cdots,P_{5}\\}$ and $[\omega]$ is a primitive
integral class. Assume the ring $H^{*}(M;{\mathbb{Z}})$ is generated by the
following elements:

$1,\,\,\,[\omega],\,\,\,\frac{1}{3}[\omega]^{2},\,\,\,\frac{1}{6}[\omega]^{3},\,\,\,\frac{1}{18}[\omega]^{4}\,\,\,\frac{1}{18}[\omega]^{5}.$

Then $H^{*}_{S^{1}}(M;{\mathbb{Z}})$ has a basis
$\big{\\{}\widetilde{\alpha}_{i}\,|\,0\leq i\leq 5\big{\\}}$ as follows, where
$\widetilde{u}$ is as in Lemma 3.1.

$1,\,\,\,\widetilde{\alpha}_{1}=\widetilde{u}+\phi(P_{0})t,\,\,\,\widetilde{\alpha}_{2}=\frac{1}{3}\prod_{j=0}^{1}\big{(}\widetilde{u}+\phi(P_{j})t\big{)},$
$\widetilde{\alpha}_{3}=\frac{1}{6}\prod_{j=0}^{2}\big{(}\widetilde{u}+\phi(P_{j})t\big{)},\,\,\,\widetilde{\alpha}_{4}=\frac{1}{18}\prod_{j=0}^{3}\big{(}\widetilde{u}+\phi(P_{j})t\big{)},\,\,\,\widetilde{\alpha}_{5}=\frac{1}{18}\prod_{j=0}^{4}\big{(}\widetilde{u}+\phi(P_{j})t\big{)}.$

###### Proof.

By Proposition 3.11, $H^{*}_{S^{1}}(M;{\mathbb{Z}})$ has a basis
$\big{\\{}\widetilde{\alpha}_{i}\,|\,0\leq i\leq 5\big{\\}}$ satisfying the
stated properties there. By Corollary 3.12,

$\prod_{j=0}^{i-1}\big{(}\widetilde{u}+\phi(P_{j})t\big{)}=a_{i}\widetilde{\alpha}_{i}.$

Restricting it to ordinary cohomology, we get the basis $\alpha_{i}$s’ in
ordinary cohomology:

$[\omega]^{i}=a_{i}\alpha_{i}.$

By the assumption on the basis $\big{\\{}\alpha_{i}\,|\,0\leq i\leq
5\big{\\}}$ in ordinary cohomology, we get the integers $a_{i}$s’, then we get
the $\widetilde{\alpha}_{i}$s’. ∎

######  Theorem 2(2).

Let $(M,\omega)$ be a compact $10$-dimensional Hamiltonian $S^{1}$-manifold
with moment map $\phi\colon M\to{\mathbb{R}}$. Assume
$M^{S^{1}}=\\{P_{0},P_{1},\cdots,P_{5}\\}$ and $[\omega]$ is a primitive
integral class. Assume each weight is less than or equal to $5$. If the
largest weight is between $P_{0}$ and $P_{5}$ and is equal to
$\frac{1}{2}\big{(}\phi(P_{5})-\phi(P_{0})\big{)}$, then

$c(M)=1+3\alpha_{1}+13\alpha_{2}+22\alpha_{3}+30\alpha_{4}+6\alpha_{5}$
$=1+3[\omega]+\frac{1}{3}(13[\omega]^{2}+11[\omega]^{3}+5[\omega]^{4}+[\omega]^{5}).$

Here $\alpha_{i}\in H^{2i}(M;{\mathbb{Z}})$ is a generator for each $1\leq
i\leq 5$. The same conclusion holds for $c(\mathcal{O})$. So $c(M)$ and
$c(\mathcal{O})$ are isomorphic.

###### Proof.

By Theorem 2(3), under the condition, the sets of weights at the fixed points
are exactly as those on $\mathcal{O}$. The following computation works for
both $M$ and $\mathcal{O}$.

We can obtain the following data from Theorem 2(3). Here, $\Lambda_{i}^{-}$ is
the product of the negative weights at $P_{i}$.

$\Lambda_{0}^{-}=1,\,\,\,\Lambda_{1}^{-}=-1,\,\,\,\Lambda_{2}^{-}=4,\,\,\,\Lambda_{3}^{-}=-10,\,\,\,\Lambda_{4}^{-}=60.$
$c_{1}^{S^{1}}(M)|_{P_{0}}=15t,\,\,\,c_{1}^{S^{1}}(M)|_{P_{1}}=12t.$
$c_{2}^{S^{1}}(M)|_{P_{0}}=85t^{2},\,\,\,c_{2}^{S^{1}}(M)|_{P_{1}}=46t^{2},\,\,\,c_{2}^{S^{1}}(M)|_{P_{2}}=-19t^{2}.$
$c_{3}^{S^{1}}(M)|_{P_{0}}=225t^{3},\,\,\,c_{3}^{S^{1}}(M)|_{P_{1}}=48t^{3},\,\,\,c_{3}^{S^{1}}(M)|_{P_{2}}=-43t^{3},\,\,\,c_{3}^{S^{1}}(M)|_{P_{3}}=43t^{3}.$
$c_{4}^{S^{1}}(M)|_{P_{0}}=274t^{4},\,\,\,c_{4}^{S^{1}}(M)|_{P_{1}}=-47t^{4},\,\,\,c_{4}^{S^{1}}(M)|_{P_{2}}=18t^{4},\,\,\,\mbox{and}$
$c_{4}^{S^{1}}(M)|_{P_{3}}=18t^{4},\,\,\,c_{4}^{S^{1}}(M)|_{P_{4}}=-47t^{4}.$

By Theorem 2(1), we have the basis $\\{\widetilde{\alpha}_{i}\,|\,0\leq i\leq
5\\}$ of $H^{*}_{S^{1}}(M;{\mathbb{Z}})$ as in Lemma 6.5. Using this basis and
the degree of $c_{1}^{S^{1}}(M)$, we can write
$c_{1}^{S^{1}}(M)=a_{0}t+a_{1}\widetilde{\alpha}_{1}$, where
$a_{0},a_{1}\in{\mathbb{Z}}$. Restricting it to $P_{0}$ and $P_{1}$
respectively, we get $c_{1}^{S^{1}}(M)=15t+3\widetilde{\alpha}_{1}$.
Restricting it to ordinary cohomology, we get

$c_{1}(M)=3\alpha_{1}.$

Similarly, we can write
$c_{2}^{S^{1}}(M)=b_{0}t^{2}+b_{1}t\widetilde{\alpha}_{1}+b_{2}\widetilde{\alpha}_{2}$.
Respectively restricting it to $P_{0}$, $P_{1}$ and $P_{2}$, we get
$c_{2}^{S^{1}}(M)=85t^{2}+39t\widetilde{\alpha}_{1}+13\widetilde{\alpha}_{2}$.
Restricting it to ordinary cohomology, we get

$c_{2}(M)=13\alpha_{2}.$

We can write
$c_{3}^{S^{1}}(M)=c_{0}t^{3}+c_{1}t^{2}\widetilde{\alpha}_{1}+c_{2}t\widetilde{\alpha}_{2}+c_{3}\widetilde{\alpha}_{3}$.
Respectively restricting it to $P_{0}$, $P_{1}$, $P_{2}$ and $P_{3}$, we get
$c_{3}^{S^{1}}(M)=225t^{3}+177t^{2}\widetilde{\alpha}_{1}+110t\alpha_{2}+22\widetilde{\alpha}_{3}$.
Restricting it to ordinary cohomology, we get

$c_{3}(M)=22\alpha_{3}.$

We have
$c_{4}^{S^{1}}(M)=d_{0}t^{4}+d_{1}t^{3}\widetilde{\alpha}_{1}+d_{2}t^{2}\widetilde{\alpha}_{2}+d_{3}t\widetilde{\alpha}_{3}+d_{4}\widetilde{\alpha}_{4}$.
Respectively restricting it to $P_{0}$, $P_{1}$, $P_{2}$, $P_{3}$ and $P_{4}$,
we get
$c_{4}^{S^{1}}(M)=274t^{4}+321t^{3}\widetilde{\alpha}_{1}+257t^{2}\widetilde{\alpha}_{2}+90t\widetilde{\alpha}_{3}+30\widetilde{\alpha}_{4}$.
Restricting it to ordinary cohomology, we get

$c_{4}(M)=30\alpha_{4}.$

Since the top Chern number is equal to the number of fixed points, we have

$c_{5}(M)=6\alpha_{5}.$

The last equality in the claim is by Theorem 2(1). ∎

######  Remark 6.6.

In the theorem above, if we remove the condition that each weight is less than
or equal to 5, instead, as in Example 2.1, we assume that the largest weight
is equal to
$\phi(P_{3})-\phi(P_{1})=\phi(P_{4})-\phi(P_{2})=\frac{1}{2}\big{(}\phi(P_{5})-\phi(P_{0})\big{)}$,
and all the weights can be represented by the moment map values of the fixed
points as in the example, then we may let

$\phi(P_{5})-\phi(P_{4})=\phi(P_{1})-\phi(P_{0})=a,\,\,\,\phi(P_{3})-\phi(P_{2})=2a,\,\,\,\mbox{and}$
$\phi(P_{4})-\phi(P_{3})=\phi(P_{2})-\phi(P_{1})=3b,$

since $\phi(P_{4})-\phi(P_{3})=\phi(P_{2})-\phi(P_{1})$ is divisible by $3$,
where $a,b\in{\mathbb{N}}$. Using $a$ and $b$, we can write down the sets of
weights at all the fixed points as in the example. Using the method above, we
can compute the total Chern class of the manifold $c(M)$. I expect that we can
get the same result as above — the result is independant of $a$ and $b$. I
tried $c_{1}(M)$ and $c_{2}(M)$ (of course $c_{5}(M)=6\alpha_{5}$).

##  7\. knowing the ring $H^{*}(M;{\mathbb{Z}})$ — proof of Theorem 3

In this section, for a compact 10-dimensional Hamiltonian $S^{1}$-manifold
with 6 isolated fixed points, assuming each weight is less than or equal to 5,
and the moment map image has symmetry, we show that if we know certain ring
$H^{*}(M;{\mathbb{Z}})$, we can determine the sets of weights at all the fixed
points, hence the total Chern class of the manifold. This proves Theorem 3.

In the following 4 lemmas, we determine the $\phi(P_{i})-\phi(P_{j})$ for each
$0\leq j<i\leq 3$.

######  Lemma 7.1.

Let $(M,\omega)$ be a compact $10$-dimensional Hamiltonian $S^{1}$-manifold
with moment map $\phi\colon M\to{\mathbb{R}}$. Assume
$M^{S^{1}}=\\{P_{0},P_{1},\cdots,P_{5}\\}$, $[\omega]$ is a primitive integral
class, and $\phi(P_{i})-\phi(P_{0})=\phi(P_{5})-\phi(P_{5-i})$ for $i=1,2$.
Assume each weight is less than or equal to $5$. If the ring
$H^{*}(M;{\mathbb{Z}})$ is generated by

$1,\,\,\,[\omega],\,\,\,\frac{1}{3}[\omega]^{2},\,\,\,\frac{1}{6}[\omega]^{3},\,\,\,\frac{1}{18}[\omega]^{4},\,\,\,\frac{1}{18}[\omega]^{5},$

then it is not possible that $3|\big{(}\phi(P_{2})-\phi(P_{0})\big{)}$.

###### Proof.

By the assumption and (6.2) of Lemma 6.1, we have

(7.2)
$\Lambda_{2}^{-}=\frac{1}{3}\big{(}\phi(P_{1})-\phi(P_{2})\big{)}\big{(}\phi(P_{0})-\phi(P_{2})\big{)}.$

Assume $3|\big{(}\phi(P_{2})-\phi(P_{0})\big{)}$.

$\bf(a)$ Assume $\phi(P_{2})-\phi(P_{0})=3$. Then $\phi(P_{1})-\phi(P_{0})=1$
or $2$ is a weight between $P_{0}$ and $P_{1}$ by Lemma 3.9.

$\bf(a1)$ Assume $\phi(P_{1})-\phi(P_{0})=1$. Then
$\phi(P_{2})-\phi(P_{1})=2$. By Lemma 3.6, $1$ has multiplicity $1$ at
$P_{0}$. By (7.2),

$\Lambda_{2}^{-}=\frac{1}{3}(-2)(-3)=(-2)(-1).$

So the set of negative weights at $P_{2}$ is $\\{-2,-1\\}$. Since each
negative weight at $P_{2}$ divides $\phi(P_{2})-\phi(P_{1})$ or
$\phi(P_{2})-\phi(P_{0})$ by Lemma 3.3, both $2$ and $1$ are weights between
$P_{1}$ and $P_{2}$, and there is no weight between $P_{0}$ and $P_{2}$ by
Lemma 3.6. Except the above $1$, the other possible distinct weights at
$P_{0}$ are $2$, $3$, $4$ and $5$, each of which can only be possibly a weight
between $P_{0}$ and $P_{i}$ with $i\geq 3$ (by the facts on the negative
weights we know about $P_{1}$ and $P_{2}$). By Lemma 5.1 (3), any of $2$, $3$,
$4$ and $5$ has maximal multiplicity $1$ at $P_{0}$. By symmetry, there is a
weight $2$ between $P_{3}$ and $P_{4}$ (just as there is a weight 2 between
$P_{1}$ and $P_{2}$). First, if $2$ is a weight from $P_{0}$ to $P_{4}$, then
the $M^{{\mathbb{Z}}_{2}}$ containing $P_{4}$ would also contain $P_{3}$ and
$P_{0}$, where $P_{3}$ is the closest fixed point to $P_{0}$, so would have a
weight $2$ with $P_{0}$ by Lemma 3.6, then $2$ has multiplicity $2$ at
$P_{0}$, a contradiction. If $2$ is a weight from $P_{0}$ to $P_{3}$, then the
$M^{{\mathbb{Z}}_{2}}$ containing $P_{3}$ contains $P_{4}$ and $P_{0}$ and has
dimension $4$, so there is a weight $4$ from $P_{0}$ to $P_{4}$. Then
$2|\big{(}\phi(P_{3})-\phi(P_{0})\big{)}$ and
$4|\big{(}\phi(P_{4})-\phi(P_{0})\big{)}$, from which we get that
$\phi(P_{3})-\phi(P_{2})=4k-1$ with $k\geq 1$. By (6.2) in Lemma 6.1,

$\Lambda_{3}^{-}=\frac{1}{6}\prod_{j=0}^{2}\big{(}\phi(P_{j})-\phi(P_{3})\big{)}=-\frac{1}{6}(4k-1)(4k+1)(4k+2)=-\frac{1}{3}(4k-1)(4k+1)(2k+1).$

Since 2 is a weight from $P_{0}$ to $P_{3}$, we have $2|(2k+1)$, a
contradiction. So $2$ cannot be a weight from $P_{0}$ to $P_{3}$, or from
$P_{0}$ to $P_{4}$. Similarly, $4$ cannot be a weight from $P_{0}$ to $P_{3}$,
or from $P_{0}$ to $P_{4}$. If $2$ and $4$ are both weights from $P_{0}$ to
$P_{5}$, then there would be a 4-dimensional symplectic manifold containing 2
fixed points, which is not possible. So $2$ and $4$ cannot be weights at
$P_{0}$ at the same time. Then there are not enough number of weights at
$P_{0}$, a contradiction.

$\bf(a2)$ Assume $\phi(P_{1})-\phi(P_{0})=2$. Then
$\phi(P_{2})-\phi(P_{1})=1$. By Lemma 3.6, $1$ is not a weight at $P_{0}$. By
(7.2),

$\Lambda_{2}^{-}=\frac{1}{3}(-1)(-3)=(-1)(-1).$

So the set of negative weights at $P_{2}$ is $\\{-1,-1\\}$. There must be a
double weight $1$ between $P_{1}$ and $P_{2}$, and there is no weight between
$P_{0}$ and $P_{2}$. Except the above $2$, the other possible weights at
$P_{0}$ are $2$, $3$, $4$, and $5$, each of which can only possibly be a
weight between $P_{0}$ and $P_{i}$ with $i\geq 3$, and has maximal
multiplicity $1$ by Lemma 5.1 (3) and (1). It cannot be that $2,2,4$ occur at
$P_{0}$, otherwise by Lemma 5.1 (2), a connected component of
$M^{{\mathbb{Z}}_{2}}$ would contain all the 6 fixed points, then there should
be a negative weight at $P_{2}$ which is a multiple of $2$, contradicting to
the above claim on the set of negative weights at $P_{2}$. So there are not
enough number of weights at $P_{0}$.

$\bf(b)$ Assume $\phi(P_{2})-\phi(P_{0})=6$. By Lemma 3.9,
$\phi(P_{1})-\phi(P_{0})\leq 5$ is a weight between $P_{0}$ and $P_{1}$.

$\bf(b1)$ Assume $\phi(P_{1})-\phi(P_{0})=1$. By (7.2),

$\Lambda_{2}^{-}=\frac{1}{3}(-5)(-6)=(-5)(-2).$

So the set of negative weights at $P_{2}$ is $\\{-5,-2\\}$. There is a weight
$5$ between $P_{1}$ and $P_{2}$, and a weight $2$ between $P_{0}$ and $P_{2}$.
By symmetry, there is a weight $5$ between $P_{3}$ and $P_{4}$. Except the
above $1$ and $2$, the other possible weights at $P_{0}$ are $2$, $3$, $4$ and
$5$, where each of the latter is a weight between $P_{0}$ and $P_{i}$ with
$i\geq 3$, so is of maximal multiplicity $1$ by Lemma 5.1 (3) and (1). By
Lemma 5.1 (2), $\\{2,2,4\\}$ cannot occur. By Lemma 4.2 used for $P_{1}$ and
$P_{2}$, we get $c_{1}(M)=2[\omega]$. If $5$ occurs at $P_{0}$, then it has
multiplicity 1 by Lemma 5.3. So $5$ cannot be a weight between $P_{0}$ and
$P_{3}$, or between $P_{0}$ and $P_{4}$ (otherwise the $M^{{\mathbb{Z}}_{5}}$
containing $P_{3}$ or $P_{4}$ contains $P_{0}$ and is 4-dimensional). So $5$
is a weight between $P_{0}$ and $P_{5}$ if it occurs at $P_{0}$. Then by Lemma
4.2 used for $P_{0}$ and $P_{5}$, we get that $\phi(P_{5})-\phi(P_{0})=15$, so
$\phi(P_{3})-\phi(P_{2})=3$. Using (6.2) in Lemma 6.1 for $\Lambda_{4}^{-}$,
we get $\Lambda_{4}^{-}=\frac{1}{18}(-5)(-8)(-13)(-14)$, which is not
possible. So $5$ does not occur at $P_{0}$. Then there are not enough number
of weights at $P_{0}$, a contradiction.

$\bf(b2)$ Assume $\phi(P_{1})-\phi(P_{0})=2$. Then
$\phi(P_{2})-\phi(P_{1})=4$. By (7.2),

$\Lambda_{2}^{-}=\frac{1}{3}(-4)(-6)=(-4)(-2).$

Since each negative weight at $P_{2}$ divides $\phi(P_{2})-\phi(P_{1})$ or
$\phi(P_{2})-\phi(P_{0})$ by Lemma 3.3, the set of negative weights at $P_{2}$
is $\\{-4,-2\\}$, where $-2$ is a weight between $P_{2}$ and $P_{0}$, and $-4$
is between $P_{2}$ and $P_{1}$. So $2$ has multiplicity 2 at $P_{0}$ (maximal
multiplicity by Lemma 5.1 (1)). Then a connected component $C$ of
$M^{{\mathbb{Z}}_{2}}$ containing $P_{0}$ contains $P_{1}$ and $P_{2}$, where
$P_{2}$ has index 4 in $C$, so there is another index 2 fixed point $P_{i}$ in
$C$ with $i=3$ or $4$ by Lemma 3.6; by symmetry, $P_{3}$, $P_{4}$ and $P_{5}$
have the same weights connection as $P_{2}$, $P_{1}$ and $P_{0}$, so $C$
contains all the fixed points and is $6$-dimensional by Lemma 5.1 (2) (there
must be a weight $4$ at $P_{0}$). So $\phi(P_{3})-\phi(P_{2})=2c$ is even.
Note that $P_{4}$ cannot be the index 2 fixed point in $C$ since it has $2$
negative weights, so the second index 2 fixed point in $C$ is $P_{3}$. But by
(6.2) in Lemma 6.1,

$\Lambda_{3}^{-}=\frac{1}{6}\prod_{j=0}^{2}\big{(}\phi(P_{j})-\phi(P_{3})\big{)}=-2^{2}\cdot\frac{1}{3}(c)(c+2)(c+3)=-2^{2}l,\,\,\,\mbox{where
$l\in{\mathbb{N}}$}.$

So $P_{3}$ cannot have index 2 with only a negative weight $-2$ in $C$, a
contradiction.

$\bf(b3)$ Assume $\phi(P_{1})-\phi(P_{0})=3$. Then
$\phi(P_{2})-\phi(P_{1})=3$. By (7.2),

$\Lambda_{2}^{-}=\frac{1}{3}(-3)(-6).$

First note that there cannot be a weight $-6$ at $P_{2}$. So either there is a
weight $3$ between $P_{1}$ and $P_{2}$, a weight $2$ between $P_{0}$ and
$P_{2}$, or there are weights $2$ and $3$ between $P_{0}$ and $P_{2}$ and
there is no weight between $P_{1}$ and $P_{2}$. In the former case, the
$M^{{\mathbb{Z}}_{3}}$ containing $P_{1}$ contains $P_{0}$ and $P_{2}$ and has
dimension $4$ by Lemma 5.1 (1) (note that there is no weight at $P_{0}$ which
is a multiple of $3$), it has $P_{1}$ and $P_{2}$ as index $2$ fixed points,
so it contains a top index fixed point $P_{i}$ with $i\geq 3$. By symmetry,
$P_{3}$, $P_{4}$ and $P_{5}$ are related by weights the same way as $P_{0}$,
$P_{1}$ and $P_{2}$, so the above $M^{{\mathbb{Z}}_{3}}$ contains all the
fixed points, but by Lemma 3.6 it only contains 2 index 2 fixed points, this
is impossible. In the latter case, the $M^{{\mathbb{Z}}_{3}}$ containing
$P_{0}$ is 4 dimensional and it contains $P_{1}$ and $P_{2}$ as two index $2$
fixed points. Similar arguments to the last case yield that this is not
possible.

$\bf(b4)$ Assume $\phi(P_{1})-\phi(P_{0})=4$. Then
$\phi(P_{2})-\phi(P_{1})=2$. Similar argument as above shows that there is a
weight $2$ between $P_{1}$ and $P_{2}$ and a weight $2$ between $P_{0}$ and
$P_{2}$. The $M^{{\mathbb{Z}}_{2}}$ containing $P_{0}$ contains both $P_{1}$
and $P_{2}$, where $P_{1}$ is of index $2$ in it. By Lemma 3.6 used on this
$M^{{\mathbb{Z}}_{2}}$, there should be a weight $2$ between $P_{0}$ and
$P_{1}$, this contradicts to the above claim that there is only a weight $4$
between $P_{0}$ and $P_{1}$.

$\bf(b5)$ Assume $\phi(P_{1})-\phi(P_{0})=5$. Then
$\phi(P_{2})-\phi(P_{1})=1$. Similar to the above, there is a weight $2$
between $P_{0}$ and $P_{2}$ (and a weight $1$ between $P_{1}$ and $P_{2}$),
and $1$ cannot be a weight at $P_{0}$ by Lemma 3.6. By Lemma 4.2 for $P_{0}$
and $P_{1}$, $c_{1}(M)=2[\omega]$, so $5$ has multiplicity $1$ at $P_{0}$ by
Lemma 5.3. Except $5$ and $2$, the other possible weights at $P_{0}$ are $2$,
$3$ and $4$, where each of the latter can only be a weight from $P_{0}$ to
$P_{i}$ with $i\geq 3$, so each of them has maximal multiplicity $1$ by Lemma
5.1 (3) and (1). But the triple $2,2,4$ cannot occur by Lemma 5.1 (2). So
there are not enough number of weights at $P_{0}$, a contradiction.

$\bf(c)$ Assume $\phi(P_{2})-\phi(P_{0})=9$. By Lemma 3.9,
$\phi(P_{1})-\phi(P_{0})\leq 5$ is a weight between $P_{0}$ and $P_{1}$.

$\bf(c1)$ If $\phi(P_{1})-\phi(P_{0})=1,2$ or $3$, then
$\phi(P_{2})-\phi(P_{1})=8,7$ or $6$. By (7.2), we see that there is a
negative weight at $P_{2}$ with absolute value bigger than 5, a contradiction.

$\bf(c2)$ If $\phi(P_{1})-\phi(P_{0})=4$ and $\phi(P_{2})-\phi(P_{1})=5$, then
similar arguments as above yield that there is a weight $5$ between $P_{1}$
and $P_{2}$, and a weight $3$ between $P_{0}$ and $P_{2}$. Except $4$ and $3$,
the other possible weights at $P_{0}$ are $2$, $3$, $4$ and $5$. Similarly as
in $\bf(b1)$, we argue that the triple $2,4,4$ cannot occur at $P_{0}$, and
$5$ cannot occur at $P_{0}$. Then there are not enough number of weights at
$P_{0}$, a contradiction.

$\bf(c3)$ If $\phi(P_{1})-\phi(P_{0})=5$ and $\phi(P_{2})-\phi(P_{1})=4$, then
similar to the above, we get that there is a weight $3$ between $P_{0}$ and
$P_{2}$ (and a weight $4$ between $P_{1}$ and $P_{2}$), and
$c_{1}(M)=2[\omega]$. Argue similarly as above, we see that the possible set
of weights at $P_{0}$ is $\\{5,3,2,3,4\\}$. By symmetry, the set of weights at
$P_{5}$ is the negative of this set. By Lemma 4.1 used for $P_{0}$ and
$P_{5}$, this is not possible.

$\bf(d)$ Assume $\phi(P_{2})-\phi(P_{0})\geq 12$. Since
$\phi(P_{1})-\phi(P_{0})\leq 5$ is a weight at $P_{0}$ by Lemma 3.9, we must
have $\phi(P_{2})-\phi(P_{1})\geq 7$. By (7.2), $\Lambda_{2}^{-}\geq 28$, so
there is a negative weight at $P_{2}$ with absolute value bigger than $5$, a
contradiction. ∎

######  Lemma 7.3.

Let $(M,\omega)$ be a compact $10$-dimensional Hamiltonian $S^{1}$-manifold
with moment map $\phi\colon M\to{\mathbb{R}}$. Assume
$M^{S^{1}}=\\{P_{0},P_{1},\cdots,P_{5}\\}$ and $[\omega]$ is a primitive
integral class. Assume each weight is less than or equal to $5$. If the ring
$H^{*}(M;{\mathbb{Z}})$ with $*\leq 4$ is generated by

$1,\,\,\,[\omega],\,\,\,\frac{1}{3}[\omega]^{2},$

then it is not possible that $\phi(P_{2})-\phi(P_{1})=3a>3$ with
$a\in{\mathbb{N}}$.

###### Proof.

Since each weight on $M$ is no bigger than 5, $\phi(P_{1})-\phi(P_{0})\leq 5$
by Lemma 3.9.

First assume $\phi(P_{2})-\phi(P_{1})=6$. By the assumption and (6.2) in Lemma
6.1,

$\Lambda_{2}^{-}=\frac{1}{3}\prod_{j=0}^{1}\big{(}\phi(P_{j})-\phi(P_{2})\big{)}.$

So $\Lambda_{2}^{-}=\frac{1}{3}(-6)(-7)$, or $\frac{1}{3}(-6)(-8)$, or
$\frac{1}{3}(-6)(-9)$, or $\frac{1}{3}(-6)(-10)$, or $\frac{1}{3}(-6)(-11)$.
Each negative weight at $P_{2}$ divides $\phi(P_{2})-\phi(P_{1})$ or
$\phi(P_{2})-\phi(P_{0})$ by Lemma 3.3. In the above cases, there would be a
weight bigger than 5, a contradiction. (In the case
$\Lambda_{2}^{-}=\frac{1}{3}(-6)(-8)$, there cannot be a double weight $4$
between $P_{0}$ and $P_{2}$, otherwise it contradicts to Lemma 3.6 used on
$M^{{\mathbb{Z}}_{4}}$ containing $P_{0}$).

For the cases $\phi(P_{2})-\phi(P_{1})=3a\geq 9$, similar to the above
argument, $\Lambda_{2}^{-}\geq\frac{1}{3}(-9)(-10)=30$, while
$\Lambda_{2}^{-}$ is the product of 2 negative weights at $P_{2}$, which is
less than 30 by the assumption that each weight on $M$ is no bigger than $5$.
∎

######  Lemma 7.4.

Let $(M,\omega)$ be a compact $10$-dimensional Hamiltonian $S^{1}$-manifold
with moment map $\phi\colon M\to{\mathbb{R}}$. Assume
$M^{S^{1}}=\\{P_{0},P_{1},\cdots,P_{5}\\}$, $[\omega]$ is a primitive integral
class, and $\phi(P_{i})-\phi(P_{0})=\phi(P_{5})-\phi(P_{5-i})$ for $i=1,2$.
Assume each weight is less than or equal to $5$. If the ring
$H^{*}(M;{\mathbb{Z}})$ is generated by

$1,\,\,\,[\omega],\,\,\,\frac{1}{3}[\omega]^{2},\,\,\,\frac{1}{6}[\omega]^{3},\,\,\,\frac{1}{18}[\omega]^{4},\,\,\,\frac{1}{18}[\omega]^{5},$

and if $\phi(P_{2})-\phi(P_{1})=3$, then it is only possible that
$\phi(P_{1})-\phi(P_{0})=1$.

###### Proof.

By Lemma 3.9, $\phi(P_{1})-\phi(P_{0})$ is a weight between $P_{0}$ and
$P_{1}$. Since each weight is less than or equal to $5$,
$\phi(P_{1})-\phi(P_{0})\leq 5$.

If $\phi(P_{1})-\phi(P_{0})=2$, then $\phi(P_{2})-\phi(P_{0})=5$. By the
assumption and (6.2) in Lemma 6.1,
$\Lambda_{2}^{-}=\frac{1}{3}(-3)(-5)=(-1)(-5)$. So there is a weight $5$
between $P_{0}$ and $P_{2}$. By Lemmas 4.4 and 4.2, $c_{1}(M)=3[\omega]$. Then
by Lemma 5.4, the largest weight is not between $P_{0}$ and $P_{2}$, a
contradiction.

If $\phi(P_{1})-\phi(P_{0})=3$, then $\phi(P_{2})-\phi(P_{0})=6$, which is not
possible by Lemma 7.1 (see (b3) of the proof of Lemma 7.1).

If $\phi(P_{1})-\phi(P_{0})=4$, then
$\Lambda_{2}^{-}=\frac{1}{3}(-3)(-7)=(-1)(-7)$, so $7>5$ is a weight between
$P_{0}$ and $P_{2}$, a contradiction.

If $\phi(P_{1})-\phi(P_{0})=5$, then
$\Lambda_{2}^{-}=\frac{1}{3}(-3)(-8)=(-1)(-8)$. First, it is not possible that
$8$ is a weight. If $-2$ and $-4$ are weights at $P_{2}$, then they are
between $P_{0}$ and $P_{2}$, then the $M^{{\mathbb{Z}}_{2}}$ containing
$P_{0}$ contains $P_{2}$ as the closest fixed point with index $4$, this is
not possible by Lemma 3.4.

Hence only one case is left possible: $\phi(P_{1})-\phi(P_{0})=1$. ∎

######  Lemma 7.5.

Let $(M,\omega)$ be a compact $10$-dimensional Hamiltonian $S^{1}$-manifold
with moment map $\phi\colon M\to{\mathbb{R}}$. Assume
$M^{S^{1}}=\\{P_{0},P_{1},\cdots,P_{5}\\}$ and $[\omega]$ is a primitive
integral class. Assume the ring $H^{*}(M;{\mathbb{Z}})$ is generated by the
elements

$1,\,\,\,[\omega],\,\,\,\frac{1}{3}[\omega]^{2},\,\,\,\frac{1}{6}[\omega]^{3},\,\,\,\frac{1}{18}[\omega]^{4},\,\,\,\frac{1}{18}[\omega]^{5}.$

If $\phi(P_{2})-\phi(P_{1})=3$ and $\phi(P_{1})-\phi(P_{0})=1$, then there is
a weight $1$ between $P_{0}$ and $P_{1}$, a weight $1$ between $P_{1}$ and
$P_{2}$, and a weight $4$ between $P_{0}$ and $P_{2}$. If moreover
$\phi(P_{i})-\phi(P_{0})=\phi(P_{5})-\phi(P_{5-i})$ for $i=1,2$, and each
weight on $M$ is less than or equal to $5$, then it is only possible that
$\phi(P_{3})-\phi(P_{2})=2$.

###### Proof.

Assume $\phi(P_{2})-\phi(P_{1})=3$ and $\phi(P_{1})-\phi(P_{0})=1$. By Lemma
3.9, there is a weight $1$ between $P_{0}$ and $P_{1}$. By (6.2) in Lemma 6.1,

$\Lambda_{2}^{-}=\frac{1}{3}\prod_{j=0}^{1}\big{(}\phi(P_{j})-\phi(P_{2})\big{)}=\frac{1}{3}(-3)(-4)=(-1)(-4).$

There cannot be a double weight $2$ between $P_{0}$ and $P_{2}$, otherwise the
$M^{{\mathbb{Z}}_{2}}$ containing $P_{0}$ has $P_{2}$ as the closest fixed
point which has index $4$, which is not possible by Lemma 3.4. So the set of
negative weights at $P_{2}$ is $\\{-1,-4\\}$, where $-1$ cannot be a weight
between $P_{2}$ and $P_{0}$ by Lemma 3.6, so it is between $P_{2}$ and
$P_{1}$, and $-4$ is between $P_{0}$ and $P_{2}$.

Assume $\phi(P_{3})-\phi(P_{2})=l$. By the assumption and (6.2) in Lemma 6.1,

$\Lambda_{3}^{-}=\frac{1}{6}\prod_{j=0}^{2}\big{(}\phi(P_{j})-\phi(P_{3})\big{)}.$

If $l=1$, then $\Lambda_{3}^{-}=-\frac{1}{6}\cdot 1\cdot 4\cdot 5$, which is
not possible. If $l=2$, then $\Lambda_{3}^{-}=-\frac{1}{6}\cdot 2\cdot 5\cdot
6$ which is possible. If $l=3$, then $\Lambda_{3}^{-}=-\frac{1}{6}\cdot 3\cdot
6\cdot 7$, which is not possible since the prime $7>5$ would be a weight. If
$l=4$, then $\Lambda_{3}^{-}=-\frac{1}{6}\cdot 4\cdot 7\cdot 8$, which is not
possible. If $l=5$, then $\Lambda_{3}^{-}=-\frac{1}{6}\cdot 5\cdot 8\cdot
9=-5\cdot 4\cdot 3$. In this case, $3$ is a weight from $P_{0}$ to $P_{3}$.
Note that $5$ is not a weight at $P_{0}$ since 5 divides none of
$\phi(P_{i})-\phi(P_{0})$ for $i>0$. Except the above weights $1,4,3$ at
$P_{0}$, the other possible weights at $P_{0}$ are $2$ and $4$, each of which
and the above $3$ has maximal multiplicity one by Lemma 5.1 (3) and (1). But
the triple $2,4,4$ cannot occur at $P_{0}$ by Lemma 5.1 (2), so there are not
enough number of weights at $P_{0}$ and $l=5$ is not possible. If $l\geq 6$,
then $|\Lambda_{3}^{-}|\geq 90$. By assumption, the 3 negative weights at
$P_{3}$ all have absolute values no bigger than $5$. We claim that $-5$ has
maximal multiplicity $1$ at $P_{3}$. Otherwise, the $M^{{\mathbb{Z}}_{5}}$
containing $P_{3}$ contains at least 2 fixed points below $P_{3}$, and the
moment map image distance between any two of them must be divisible by $5$ by
Lemma 3.2, which is not true by the assumption. So $|\Lambda_{3}^{-}|\leq
5\cdot 4\cdot 4=80$, a contradiction, hence $l\geq 6$ is not possible. ∎

Now we can prove Theorem 3 in the Introduction.

######  Theorem 3.

Let $(M,\omega)$ be a compact $10$-dimensional Hamiltonian $S^{1}$-manifold
with moment map $\phi\colon M\to{\mathbb{R}}$. Assume
$M^{S^{1}}=\\{P_{0},P_{1},\cdots,P_{5}\\}$, $[\omega]$ is a primitive integral
class, and $\phi(P_{i})-\phi(P_{0})=\phi(P_{5})-\phi(P_{5-i})$ for $i=1,2$.
Assume each weight is less than or equal to $5$. Assume the ring
$H^{*}(M;{\mathbb{Z}})$ is generated by the elements

$1,\,\,\,[\omega],\,\,\,\frac{1}{3}[\omega]^{2},\,\,\,\frac{1}{6}[\omega]^{3},\,\,\,\frac{1}{18}[\omega]^{4},\,\,\,\frac{1}{18}[\omega]^{5}.$

Then $c_{1}(M)=3[\omega]$, and the sets of weights at all the fixed points are
exactly as the $S^{1}$-action on the coadjoint orbit $\mathcal{O}$ of $G_{2}$
in Example 2.1, in particular, there is a weight between any pair of fixed
points. Hence $c(M)$ is as in Theorem 2 $(2)$, and is isomorphic to
$c(\mathcal{O})$.

###### Proof.

By Lemmas 7.1, 7.3, 7.4 and 7.5, we may only have that

$\phi(P_{2})-\phi(P_{1})=3,\,\,\,\phi(P_{1})-\phi(P_{0})=1\,\,\,\mbox{and}\,\,\,\phi(P_{3})-\phi(P_{2})=2.$

By the assumption and (6.2) in Lemma 6.1,

$\Lambda_{3}^{-}=\frac{1}{6}\prod_{j=0}^{2}\big{(}\phi(P_{j})-\phi(P_{3})\big{)}=\frac{1}{6}(-2)(-5)(-6).$

This implies that there is a weight $5$ between $P_{1}$ and $P_{3}$ and
$\phi(P_{3})-\phi(P_{1})=5$. By symmetry, there is a weight $5$ between
$P_{2}$ and $P_{4}$ and $\phi(P_{4})-\phi(P_{2})=5$. By Theorem 1,
$c_{1}(M)=3[\omega]$. By Theorems 1 and 2, we have the claim on the sets of
weights at all the fixed points. Alternatively, we give a direct proof on the
claim for the sets of weights from the integral cohomology ring
$H^{*}(M;{\mathbb{Z}})$ as follows.

By Lemma 7.5, the set of negative weights at $P_{1}$ is $\\{-1\\}$, and it is
a weight between $P_{0}$ and $P_{1}$; the set of negative weights at $P_{2}$
is $\\{-1,-4\\}$, where $-1$ is a weight between $P_{2}$ and $P_{1}$, and $-4$
is between $P_{2}$ and $P_{0}$. By the expression of $\Lambda_{3}^{-}$ above,
the set of negative weights at $P_{3}$ is $\\{-1,-5,-2\\}$. By Lemma 3.3, each
negative weight at $P_{3}$ divides some $\phi(P_{i})-\phi(P_{3})$ with
$i=0,1,2$. By Lemma 3.6 and the weights we know about $P_{0}$, $P_{1}$ and
$P_{2}$, we get that the $-1$ is a weight between $P_{2}$ and $P_{3}$. Right
now we do not know if the $-2$ is a weight between $P_{2}$ and $P_{3}$, or
between $P_{0}$ and $P_{3}$. By the assumption, we have

$\phi(P_{4})-\phi(P_{3})=3\,\,\,\mbox{and}\,\,\,\phi(P_{5})-\phi(P_{4})=1.$

By the assumption and (6.2) in Lemma 6.1,

$\Lambda_{4}^{-}=\frac{1}{18}\prod_{j=0}^{3}\big{(}\phi(P_{j})-\phi(P_{4})\big{)}=\frac{1}{18}(-3)(-5)(-8)(-9).$

By Lemma 3.3, each negative weight at $P_{4}$ divides some
$\phi(P_{i})-\phi(P_{4})$ with $i=0,1,2,3$. Since each weight is no bigger
than $5$, and there is only one weight which is $1$ between $P_{3}$ and
$P_{4}$ by symmetry (the same as $P_{1}$ and $P_{2}$), we get that the set of
negative weights at $P_{4}$ is $\\{-1,-5,-4,-3\\}$, where each is a weight
between $P_{4}$ and a different fixed point below $P_{4}$. Similarly,

$\Lambda_{5}^{-}=\frac{1}{18}\prod_{j=0}^{4}\big{(}\phi(P_{j})-\phi(P_{5})\big{)}=\frac{1}{18}(-1)(-4)(-6)(-9)(-10).$

For similar reasons as above and symmetry on what we know about the lower
fixed points, we get that the set of (negative) weights at $P_{5}$ is
$\\{-1,-4,-2,-3,-5\\}$, where each is a weight between $P_{5}$ and a different
fixed point below $P_{5}$. Note that since there is a weight $2$ between
$P_{2}$ and $P_{5}$, by symmetry, there is a weight $2$ between $P_{0}$ and
$P_{3}$, this answers the question we do not know above. By symmetry or by
using $-\phi$, we get the sets of positive weights at all the fixed points,
hence we get the sets of weights at all the fixed points. ∎

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