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The symplectic volume of spatial p olygon spaces

Yasuhiko Kamiyama University of the Ryukyus

30 May 2006

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Let S 2 b e the unit sphere in R3. For n 3, we define An by

An = {P = (a1, . . . , an) (S 2)n : a1 + · · · + an = 0}. a3

a2 a1 O an an-1 · An · Set S O (3)

the diagonal action

Mn = An/S O (3)
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Basic prop erties

· Mn is compact. · When n is o dd or n = 4, Mn is a complex manifold such that dimC Mn = n - 3. · When n is even 6, Mn has singular p oints. P = (a1, . . . , an) Mn is a singular p oint a2 = ±a1, a3 = ±a1, . . . , an = ±a1. a3 O a1 a2

Such a singular p oint has a neighb orho o d

C (S n-3 â S n-3).
S1

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Example. (1) M3 = {one p oint}. (2) M4 = S 2. (3) M5

biholomorphic

=

the surface obtained from CP 2 by blowing up 4 p oints in general p osition

diffeo.

=

2 CP 2 # 4 CP .

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The symplectic form We define the canonical symplectic form n 2(Mn) as follows. (Actually n is a K¨ ahler form.) For P = (a1, . . . , an) Mn, TP Mn is given as follows:

TpMn = {u = (u1, . . . , un) (R3)n : (ui, ai) = 0 (1 i n) and u1 + · · · + un = 0}/ .
For

u = (u1, . . . , un) TpMn, v = (v1, . . . , vn)
n

set

n(u, v ) =
i=1

det(ui, vi, ai)

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We consider the following: Question. What is the symplectic volume of Mn? Main theorem (Tezuka-K). For n 3, set

vn =
Then we have

Mn

n n -3.

[ n ]-1 2 j n - 1 (n - 2 - 2j )n-3. vn = (-1) j
j =0

Example.

v3 = 1, v4 = 2, v5 = 5, v6 = 24, v7 = 154, v8 = 1280 and v9 = 13005.

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The Duistermaat-Heckman theorem Since the main theorem is proved using the Duistermaat-Heckman theorem, we recall this.

· (X, ) a symplectic manifold with dimR X = 2k. · T k = (S 1)k · µ : X Rk · Vol(µ(X )) (X, )
preserving .

a moment map.

the volume of µ(X ).

Then

X

k = k! Vol(µ(X ))

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Pro of of the main theorem Unfortunately, we do not have an action T n-3 Mn. However, there is an op en dense subspace Mn Mn such that T n-3 Mn as follows: (1) Define a map µn : Mn Rn-3 by
n-2

µn(P ) = (|a1+a2|, |a1+a2+a3|, . . . , |
i=1

ai|).

That is, the lengths of the diagonals connecting the vertices to the origin.

a2 a1 O an |a1 + a2|

a3

|a1 + a2 + a3| |
n-2 i=1

ai|
8

an-1


(2) Set

Mn = {P Mn: none of these n - 3 lengths vanishes}. · Mn is op en and dense in Mn.
(3) Define T n-3 Mn as follows: The i-th circle acts by rotating the part of a p olygon, formed by the first i + 1 edges, around the i-th diagonal.

a2 a1 O an an-1

a3

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Our situation is as follows.

· (Mn, n) · T n-3 Mn

a symplectic manifold.

preserving n.

· There is a map µn|Mn : Mn Rn-3.
Theorem (Kap ovich-Millson).

µ|Mn is a moment map for the action T n-3 (Mn, n).

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We can apply the Duistermaat-Heckman theorem and we have
Mn n n -3 = (n - 3)! Vol(µn(Mn)).

We set

n = µn(Mn).
Since Mn is op en and dense in Mn, we have

vn = (n - 3)! Vol(n)
that n Rn-3 is a convex p olywhich is defined by triangle inequalHence we can calculate the rightside by multiple integral.

Note top e ities. hand

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Example. Consider the case n = 5.

a2 a1 O a5 x y a3

a4

x = |a1 + a2|, y = |a1 + a2 + a3| and µ5(P ) = (x, y ).
By triangle inequalities, we have

0 x, y 2
and

1 x + y x1+y y 1 + x
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y 2

1

5

O

1
Vol(5) = 4 - 3 ·

2 1 2

x

=
Hence

5 2

v5 = 2! Vol(5) = 5.

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Another pro of of the main theorem Hereafter n is o dd and all cohomology groups are with R-co efficients. We can calculate vn = cohomology ring H Let
n n -3 from the

Mn (Mn).

z1 , . . . , z n
b e the generators of H 2(Mn).

· They are essentially the generators of H 2(S 2 â · · · â S 2).
n

· The cohomology classes z1, . . . , zn generate the cohomology ring H (Mn). (Similarly to toric varieties, the cohomology ring H (Mn) is generated by 2-dimensional classes.)
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Theorem (Hausmann-Knutson).

[n] = z1 + · · · + zn H 2(Mn).
Theorem (Tezuka-K). For all i1, . . . , in 0 with i1 + · · · + in = n - 3, the intersection pairings
Mn in z11 . . . zn i

are determined. In particular, we can calculate

vn =

Mn

(z1 + · · · + zn)n-3

Example. Consider the case n = 5.

M5

z i zj =

1 -3

i=j i = j.

Hence

v5 = -3 · 5 + 1 · 20 = 5.
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Cob ordism Since we can calculate all the Chern numb ers of Mn, it is p ossible to describ e Mn in the complex cob ordism group U -6. 2n For example, (i) In U , M5 = 4 (CP 1)2 - 3 CP 2. 4 (ii) In U , 8
1)4

M7 = - 9 (CP - 33 CP
(iii) In U , 12

+ 33 (CP 1)2 â CP 2 1 â CP 3 + 10 CP 4.

M9 = 3123 (CP + 0 · (CP + (omit

- 10196 2)3 + 0 · CP 6 terms) -

1)6

(CP 1)4 â CP 2 2 â CP 4 35 CP 6.

Remark. Let n = 2m + 1.
O Then in Sn-6, we have 2

Mn = (-

1)m+1

2m - 1 m

CP n-3.
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