The Chow ring of the moduli space and its related homogeneous space of
  bundles on P^2 with charge 1
THE CHOW RING OF THE MODULI SPACE
AND ITS RELATED HOMOGENEOUS SPACE
OF BUNDLES ON P2 WITH CHARGE 1
YASUHIKO KAMIYAMA AND MICHISHIGE TEZUKA
Abstract. For an algebraically closed field K with ch(K) 6= 2, let
OM(1, SO(n,K)) denote the moduli space of holomorphic bundles
on P2 with the structure group SO(n,K) and half the first Pon-
tryagin index being equal to 1, each of which is trivial on a fixed
line l
and has a fixed holomorphic trivialization there. In this
paper we determine the Chow ring of OM(1, SO(n,K)).
1. Introduction
Let G be one of the classical groups SU(n), SO(n) or Sp(n), and
let k ≥ 0 be half the first Pontryagin index of a G-bundle P over
S4 = R4 ∪ {∞}. Denote by M(k,G) the framed moduli space whose
points represent isomorphism classes of pairs:
(self-dual G-connections on P , isomorphism P∞ ≃ G).
Let OM(k,GC) denote the moduli space of holomorphic bundles on
2 for the associated complex group, trivial on a fixed line l∞ and
with a fixed holomorphic trivialization there. Then Donaldson ([7])
showed a diffeomorphism M(k,G) ≃ OM(k,GC).
In [12] the topology ofM(1, SO(n)) ≃ OM(1, SO(n,C)) was studied
in detail. The result was used in [11] to prove the fact that the natural
homomorphism J : H·(M(1, SO(n)),Z/2) → H·(Ω30Spin(n),Z/2) is
injective. Moreover, the image of J was determined. To prove this, the
following description of OM(1, SO(n,C)) by a homogeneous space was
used: We set
Wn = SO(n)/(SO(n− 4)× SU(2)).
Then there is a diffeomorphism
(1.1) OM(1, SO(n,C)) ≃ R5 ×Wn.
2000 Mathematics Subject Classification. 14M17 (14N10).
Key words and phrases. Moduli space, homogeneous space, Chow ring, cycle
http://arxiv.org/abs/0704.1938v1
The purpose of this paper is to generalize the definition ofOM(1, SO(n,C))
for any algebraically closed field K with ch(K) 6= 2 and to determine
the Chow ring of this. The Chow ring of a classifying space was studied
by Totaro [19]. A loop space is considered to be a dual situation of a
classifying space in a certain sense. Our result and the result of [11]
are the first step for a loop space.
Definition 1.1. Let K be an algebraically closed field with ch(K) 6= 2.
Let OM(1, SO(n,K)) denote the moduli space of holomorphic bundles
on P2 with the structure group SO(n,K) and half the first Pontryagin
index being equal to 1, each of which is trivial on a fixed line l∞ and
has a fixed holomorphic trivialization there.
The moduli space OM(1, SO(n,K)) is a quasi-projective variety and
defines the Chow ring. More explicitly, the diffeomorphism (1.1) is
generalized (in the sense of a biregular map) as follows: We set
Xn = SO(n,K)/(SO(n− 4, K)× SL(2, K)) · Pu,
where Pu denotes the unipotent radical. (Recall that for a parabolic
subgroup P of an algebraic group G, P is a semidirect product of a
reductive group and its unipotent radical Pu.) Then there is a biregular
(1.2) OM(1, SO(n,K)) ≃ A2 ×Xn.
(For the proof of (1.2), see Proposition 2.1.) A formula of Grothendieck
[3] shows that
CH ·(OM(1, SO(n,K))) ≃ CH ·(Xn).
The purpose of this paper is to determine the Chow rings of Xn and
its related algebraic variety Yn explicitly.
The Schubert cell approach of the Chow ring of Yn by using a Young
diagram is done in [15], [16]. However it needs further work to deter-
mine the Chow ring of Xn from this. Hence we first calculate the Chow
ring of Yn more explicitly by a different method. Then we calculate the
Chow ring from the results. Our results for the Chow ring of Xn are
This paper is organized as follows. In Sect. 2 we first prove (1.2).
Then we recall basic facts on the Chow ring. In Sect. 3 we determine
an integral basis and the ring structure of CH ·(Yn), where Yn is an
algebraic variety which is related to Xn. (See Theorems 3.7 and 3.9.)
The ring structure of CH ·(Yn) proved in Theorem 3.9 is one of our main
results. Since the results are long, we give them in tables in Sect. 5.
(See 5.2-5.5.) Using the results of Sect. 3, we determine CH ·(Xn) in
Sect. 4. (See Theorem 4.1.)
We thank N. Yagita for turning our interest to the Chow ring and
explaining the paper [17].
2. Preliminaries
We first prove (1.2):
Proposition 2.1. For an algebraically closed field K with ch(K) 6= 2,
there is a biregular map
OM(1, SO(n,K)) ≃ A2 ×Xn.
Proof . Recall that a monad description of OM(k, SO(n,C)) was in-
dicated in [7] and given explicitly in [14] and [18]. It is easy to see that
the description remains valid for any algebraically closed field K. In
particular, OM(1, SO(n,K)) is given as follows:
Lemma 2.2. Let Cn be the space of n× 2 matrices
z1 w1
z2 w2
zn wn
with coefficients in K satisfying:
a) cT c = O, that is:
z2i = 0,
w2i = 0 and
ziwi = 0,
b) The rank of c over K is 2.
The group SL(2, K) acts on Cn from the right by the multiplication of
matrices. Then there is a biregular map
OM(1, SO(n,K)) ≃ A2 × (Cn/SL(2, K)).
From the lemma, it suffices to prove Xn ≃ Cn/SL(2, K). We prove
this for the case n = 2m. (The case n = 2m + 1 can be proved
similarly.) Recall that in [2], SO(n,K) was defined as follows: Let
q(x) be a quadratic form on An defined by q(x) =
i=1 xixm+i, and
let B(x, y) be the associated bilinear form. Then SO(n,K) is defined
SO(n,K) = {σ ∈ Aut(An) : B(σ(x), σ(y)) = B(x, y) for x, y ∈ An}.
We set
xj = zj+
−1zj , xm+j = zj−
−1zj , yj = wj+
−1wj and ym+j = wj−
−1wj,
where 1 ≤ j ≤ m. Then the defining equations of Cn are given by
q(x) = q(y) = 0 and B(x, y) = 0.
Clearly SO(n,K) acts on Cn. It is easy to prove the following lemma.
(See [2, V 23.4].)
Lemma 2.3.
SO(n,K)/SO(n− 4, K) · Pu ≃ Cn,
where Pu is the unipotent radical of a parabolic subgroup with a Levi
factor SO(n− 4, K)×GL(2, K).
Now Proposition 2.1 follows from Lemma 2.3. This completes the
proof of Proposition 2.1. �
Next we recall basic facts on the Chow ring. We suppose that an
algebraic variety V is defined over K. Let CH ·(V ) denote the Chow
ring and CH i(V ) the subgroup of CH ·(V ) generated by the cycles of
codimension i.
Theorem 2.4 ([3]). (i) Let V be a nonsingular variety, X a nonsin-
gular closed subvariety of V , and U = X − V . Then there exists an
exact sequence
CH ·(X)
i∗→ CH ·(V ) j
→ CH ·(U) → 0,
where i : X → V (resp. j : U → V ) is a closed immersion (resp. an
open immersion).
For the definitions of i∗ and j
∗, see also [10].
(ii) Let π : E → V be a fiber bundle with an affine space An as a fiber.
Then the induced map π∗ : CH ·(V ) → CH ·(E) is an isomorphism.
The Chow ring of the following projective variety is well-known.
Theorem 2.5 ([1], [6]). Let G be a reductive algebraic group and P a
maximal parabolic subgroup. Then
(i) a quotient G/P is a nonsingular projective variety.
(ii) CH ·(G/P ) is generated by the Schubert varieties.
(iii) CH ·(G/P ) is independent of ch(K). Moreover, CH ·(G/P ) ≃
H ·(G/P,Z) for K = C.
3. The ring structure of CH ·(Yn)
Before describing the results, we need some notations and results.
We set
Yn = SO(n,K)/(SO(n− 4, K)×GL(2, K)) · Pu.
Then we have a principal bundle
(3.1) Gm → Xn
π→ Yn.
In this section we determine an integral basis and the ring structure of
CH ·(Yn). By Theorem 2.5 (ii), (iii), we obtain the following theorem:
Theorem 3.1 ([12]). We have an isomorphism as modules:
(1) For n = 2m,
CH ·(Yn)⊗ Z/2 ≃ Z/2[c1, c2]/(bm−1, c2bm−2)⊗∆(v2m−4, v2m−2).
(2) For n = 2m+ 1,
CH ·(Yn)⊗ Z/2 ≃ Z/2[c1, c2]/(bm−1, c2bm−2)⊗∆(v2m−2, v2m),
where |c1| = 2, |c2| = 4, |bi| = 2i and |vi| = i.
Theorem 3.2 ([12]). Let p be an odd prime. Then we have a ring
isomorphism:
(1) For n = 2m,
CH ·(Yn)⊗ Z/p ≃ Z/p[c1, c2, χ2m−4]/(c2χ2m−4, χ22m−4 − dm−2, dm−1),
where χ2m−4 ∈ H2m−4(BSO2m−4,Z/p) is the Euler class.
(2) For n = 2m+ 1,
CH ·(Yn)⊗ Z/p ≃ Z/p[c1, c2]/(dm−1, c22dm−2).
We recall the definitions of bi, di and vi. In a polynomial ring Z[α, β],
we set c1 = α+ β and c2 = αβ. Then bk and dk are defined by
bk = (−1)k
αiβk−i
dk = (−1)k
α2iβ2k−2i.
The element v2r ∈ CH2r(Yn) is defined by
(1) For n = 2m,
(3.2)
2v2m−4 = χ2m−4 − bm−2
2v2m−2 = bm−1.
(2) For n = 2m+ 1,
(3.3)
2v2m−2 = bm−1
2v2m = c2bm−2.
The following formulas are easily proved.
Lemma 3.3. We have
bk = (−1)k
(−1)µ
k − µ
dk = (−1)k
(−1)µ
2k − µ+ 1
2k−2µ
The following lemmas are also easily shown.
Lemma 3.4.
(−1)µch−µ2 b2µ = dh.
Lemma 3.5. We set fn(x) = (1 + x)
n − (1 + xn) and write fn(x) as
fn(x) =
[n2 ]
µ(1 + x)n−2µ.
Then we have
aµ = (−1)µ+1
n− 1− µ
Especially, the last term is given by
(−1)s+12xs for n = 2s
(−1)s+1(2s+ 1)xs(1 + x) for n = 2s+ 1.
For n = 2m or 2m+ 1, we define a subgroup An of CH
·(Yn) by
(3.4) An =
Z[c1]/(c
m−1−i
1 ){ci2}
where we set
∆Z(v2m−4, v2m−2) n = 2m
∆Z(v2m−2, v2m) n = 2m+ 1.
The generators v2i is specified in (3.2) and (3.3).
The following lemma is proved in the same way as in [12, Lemma
3.8].
Lemma 3.6. For a prime p, we abbreviate CH ·(Yn)⊗Z(p) as CH ·(Yn)(p).
If p is odd, we have the following isomorphism of modules:
(i) For n = 2m,
CH ·(Yn)(p) ≃ Z(p)[c1]/(c2(m−1)1 ){1, χ2m−4}⊕
Z(p)[c1]/(c
2(m−1−i)
1 ){c2i−12 , c2i2 }.
(ii) For n = 2m+ 1,
CH ·(Yn)(p) ≃
Z(p)[c1]/(c
2(m−1−i)
1 ){c2i2 , c2i+12 }.
Theorem 3.7. An integral basis of CH ·(Yn) is constructed from the
monomial basis of An in (3.4). The results are summed up in Sect. 5,
The proof is based on a rather complicated calculation. Its outline is
as follows: We construct a set of suitable generators starting from the
basis of An. It is easily verified that it is a basis of CH
·(Yn)⊗ Z/p by
using the presentation of Lemma 3.6. Then it is a Z-basis of CH ·(Yn).
We only prove the case n = 2m + 1 and m is even. To simplify the
proof, we set:
C = CH ·(Yn)
(3.5)
bm−1 = (−1)m−1bm−1 =
i+j=m−1
αiβj, em = c2bm−2
dm−1 = (−1)m−1dm−1
Z[c1]/(c
1 )⊕ · · · ⊕ Z[c1]/(cm−1−i1 ){ci2} ⊕ · · · ⊕ {cm−22 }
⊗ Z{1, bm−1, em, bm−1em}.
Let An the set of a basis
{ci1c
m : i+ j ≤ m− 2, ǫk = 0 or 1 (k = 1, 2)}.
Then we have
C ⊂ Q[c1, c2]/(dm−1, c22dm−2), A ⊂ Q[c1, c2]/(dm−1, c22dm−2)
and c2i2 dm−1−i ∈ (dm−1, c22dm−2), i ≥ 0.
Our concern is a homogeneous polynomial algebra SZ(V ) of V =
{c1, c2}. It is identified with an inhomogeneous ring Z[x] by putting
x = β
. Then we have
c1 = 1 + x, c2 = x, bm−1 =
xm − 1
, and dm−1 =
x2m − 1
x2 − 1
Using the identification, the next formulas are directly checked: In
Z[c1, c2],
− c2i+1+j2 bm−2−2i = c
2 b2i−1bm−1 − c
2b2iem
− c2i+2+j2 bm−3−2i = c
2 b2ibm−1 − c
2b2i+1em.
We set
c2i−11 c
2 bm−1
2 b2i−1bm−1 − c
2b2iem = −c
2i+j+1
2 bm−2−2i(3.6)
c2i1 c
2 bm−1
2 b2ibm−1 − c
2b2i+1em = −c
2i+j+2
2 bm−3−2i.
Noting c22dm−2 = 0, we have
c22bm−2bm−1−c22c1dm−2 =
(x− 1)2
(xm−2−1)(xm−1−1) = c32bm−3bm−2.
Using c2i2 dm−1−i = 0, 1 ≤ i ≤ m, we repeat the argument and get
(3.7) ci2bm−1em = c
2 bm−2−ibm−1−i.
In Yn, we see that dm−1 = 0. Recall that we set f2i+3(x) = (1 +
x)2i+3 − (1 + x2i+3) (see Lemma 3.5). Then we have
c2i+11 c
2bm−1em − c
2 didm−1
xj+1(xm − 1)
(x2 − 1)2
(1 + x)2i+3(xm−1 − 1)− (x2i+2 − 1)(xm + 1)
xj+1(xm − 1)
(x2 − 1)2
(1 + x2i+3)(xm−1 − 1)− (x2i+2 − 1)(xm + 1) + f2i+3(x)(xm−1 − 1)
x2i+j+3
(x− 1)(x2 − 1)
(xm − 1)(xm−2i−3 − 1) + x
(x2 − 1)2
(xm−1 − 1)(xm − 1)f2i+3(x).
We set
x2i+j+3
(x− 1)(x2 − 1)
(xm − 1)(xm−2i−3 − 1)
(x2 − 1)2
(xm−1 − 1)(xm − 1)f2i+3(x).
Since x2λdm−1−λ = 0 for 0 ≤ λ ≤ m−1, we see that x2i+j+3+λdm−1−(i+2)−λ =
0 for 0 ≤ λ ≤ j − 1. Hence
x2i+j+3+λdm−1−(i+2)−λ =
x2i+2j+3
(x− 1)(x2 − 1)
(xm−j−1)(xm−2i−j−3−1).
From Lemma 3.5,
2i+1−2µ
1 bm−1em + (−1)ikxi+j+2
(xm−1 − 1)(xm − 1)
(x− 1)(x2 − 1)
where k = 2i+ 3.
We set J = xi+j+2
(xm−1−1)(xm−1)
(x−1)(x2−1)
. Using xi+j+2+λdm−2−λ = 0 for
1 ≤ λ ≤ i+ j, we see that
xi+j+2+λdm−2−λ =
x2i+2j+3
(x− 1)(x2 − 1)
(xm−2−i−j−1)(xm−1−i−j−1).
Afterwards, we introduce the notation
(3.8)
c2i+11 c
2bm−1em
:= c2i+11 c
2bm−1em −
2i+1−2µ
1 bm−1em,
where f2i+3(x) =
µ=1 aµx
µ(1 + x)2i+3−2µ (see Lemma 3.5).
We sum up these arguments:
(3.9)
c2i+11 c
2bm−1em
= I1 + (−1)ikJ,
where
x2i+2j+3
(x− 1)(x2 − 1)
(xm−j − 1)(xm−2i−j−3 − 1),
x2i+2j+3
(x− 1)(x2 − 1)
(xm−2−i−j − 1)(xm−1−i−j − 1), and k = 2i+ 3.
A direct calculation shows J − I1 = x
(x−1)(x2−1)
(xi+1 − 1)(xi+2 − 1).
Hence we get a key formula
(3.10)
c2i+11 c
2bm−1em
(−1)i(2i+ 3) + 1
I1+(−1)i(2i+3)
(x− 1)(x2 − 1)
(xi+1−1)(xi+2−1).
We note that (3.9) holds for m even or odd. From now on, we as-
sume that m is even. When we put j = 1 in (3.9), we see that
I1 = c
−i−3em. We set
〈c2i+11 c2bm−1em〉 :=
c2i+11 c2bm−1em
(−1)i(2i+ 3) + 1
c2i+42 dm2 −i−3em.
Then we have
(3.11)
〈c2i+11 c2bm−1em〉
2i+ 3
= (−1)icm+12
bibi+1
We call a generator cm−3−2i1 c
2 bm−1em to be the head of the pre-
sentation An in (3.4). For the head generator, it is directly shown that
I1 = 0 by (3.9). Hence the formula (3.9) implies
(3.12)
cm−3−2i1 c
2 bm−1em
m− 1− 2i
= (−1)i
−i−2bm
Then the formulas (3.10) and (3.11) imply the following relations: We
〈c2i+11 bm−1em〉 :=
c2i+11 bm−1em
(−1)i(2i+ 3) + 1
c2i+22 dm2 −i−2em
〈c2α+1−2β1 c
2 bm−1em〉 :=
2α+1−2β
2 bm−1em
(−1)α−β(2α− 2β + 3) + 1
c2α+42 dm2 −α−3em.
Then we have
〈c2i+11 bm−1em〉 −
(−1)i(2i+ 3) + 1
(−1)i−1(2i+ 1)
〈c2i−11 c2bm−1em〉 = −
bibi+1
(3.13)
〈c2α+1−2β1 c
2 bm−1em〉 −
(−1)β(2α− 2β + 3) + (−1)α
2α + 3
〈c2α+11 c2bm−1em〉
(3.14)
m+β+2
bα−β−1bα−β
m−3−2α−2β
2α+β+1
(−1)β(m− 1− 2α+ 2β) + (−1)m2 −α
m− 1− 2α
cm−3−2α1 c
2 bm−1em
(3.15)
m+2α+β+1
bγbγ+1, where γ =
− 2− α− β.
Last we consider a generator c2i+21 c
2bm−1em. We set
〈c2i+21 c
2bm−1em〉 :=c2i+21 c
2bm−1em −
2i+2−2µ
1 bm−1em
(−1)i(2i+ 3) + 1
i+j+1
2 bm−1em,
where f2i+3(x) =
µ=1 aµx
µ(1 + x)2i+3−2µ (see Lemma 3.5). Then the
formula (3.9) implies that
(3.16) 〈c2i+11 c
2bm−1em〉 = −c
2 bibi+1.
We consider a set
{(3.6), (3.10), (3.11), (3.12), (3.13), (3.14)}
∪ {x ∈ An : x = ci1c
m, x = c
1bm−1, i+ j ≤ m− 2, ǫk = 0 or 1}.
When we reduce it to the mod p reduction for an odd prime p, the
identities from (3.6) to (3.14) show that they are linearly independent
from Lemma 3.6. Hence, replacing bm−1 and em by v2m−2 and v2m
respectively, we obtain an integral basis of CH ·(Y2m+1) for even m. An
integral basis of CH ·(Yn) for other cases is written down in a table of
Sect. 5.
For a group G and a subgroup H , let [G : H ] denote the index, i.e.
the cardinality of G/H . As a corollary of the above theorem, we have
Corollary 3.8. (i) For n = 2m,
[CH ·(Yn) : An] =
12 · 32 · . . . · (m− 3)2 · (m− 1) m: even
12 · 32 · . . . · (m− 2)2 m: odd.
(ii) For n = 2m+ 1,
[CH ·(Yn) : An] =
12 · 32 · . . . · (m− 3)2 · (m− 1) m: even
12 · 32 · . . . · (m− 2)2 ·m m: odd.
The following theorem is proved by using the integral basis of CH ·(Yn)
given in Theorem 3.7.
Theorem 3.9. The ring structure of CH ·(Yn) is determined. The
results are summed up in tables in 5.3 and 5.4 in Sect. 5. (See also
5.5.)
Proof . We only show the formula (iii) in Sect. 5, 5.4. We use the
notations of the proof of Theorem 3.7. Using x2idm−1−i = 0, we have
cm−2i−11 c
2 bm−1−x2idm−1−i =
x2 − 1
(x2i−1)+ x
x2 − 1
(xm−1)fm−2i(x).
We set I1 =
(x2i − 1) and I2 = x
(xm − 1)fm−2i(x). Then we
m−2i−1−2µ
2 bm−1 + (−1)
−i+12x
m − 1
x2 − 1
We set J = x
+i xm−1
. From (3.6), we have
2 em−
(−1)µ
+i−1−µ
2 bm−1
 = (−1)
x2i − 1
x2 − 1
Hence
cm−2i−11 c
2 bm−1−
m−2i−1−2µ
2 bm−1
(3.17)
+ (−1)
2 em +
(−1)µ
+i−1−µ
2 bm−1
 = −xmx
2i − 1
x2 − 1
On the other hand, we see from (3.13) that
(3.18) − xmx
2i − 1
x2 − 1
= 〈c2i−11 bm−1em〉+
2i+ 1
2i− 1
〈c2i−31 c2bm−1em〉.
We calculate
S := (−1)
2 em +
(−1)µ
+i−1−µ
2 bm−1
By (3.6),
S = (−1)
−i2(S1 + S2),
where
(−1)µc
+i+1−µ
2 b2µ+1bm−1 and S2 = c
2 em−
(−1)µc
+i−2−µ
2 b2µ+2em.
We have from Lemma 3.3 that
−1−µ−i
m− 1− 2i− µ
m−2i−1−2µ
2 bm−1.
Using
µ=0(−1)µc
2 b2µ = (−1)hdh (see Lemma 3.4),
S2 = c
(−1)µc
−i−1−µ
2 b2µ
 bm−1 = (−1)
−1−ic2i2 dm2 −1−ibm−1.
The formula (3.17) is
cm−2i−11 c
2 bm−1 +
(−1)µ
m− 2i− 1− µ
cm−2i−11 c
2 bm−1 − 2c2i2 dm2 −1−iem
= −xm x
x2 − 1
Comparing with (3.18), we obtain
cm−2i−11 c
2 bm−1 =
(−1)µ+1
m− 2i− 1− µ
cm−2i−11 c
2 bm−1
2i− 1
c2i2 dm2 −1−iem +
c2i−11 bm−1em
2i+ 1
2i− 1
c2i−31 c2bm−1em
This completes the proof of the formula. �
4. The Chow ring of Xn
Theorem 4.1. Let T (Xn) and F (Xn) be the torsion part and the free
part of CH ·(Xn), respectively. Then we have
(i) For n = 4t,
F (Xn) ≃ Z[c2]/(ct2){1, v4t−4}
T (Xn) ≃ Z/2[c2]/(ct2){v4t−2, v4t−4v4t−2}.
(ii) For n = 4t+ 1,
F (Xn) ≃ Z[c2]/(ct2)⊕ Z[c2]/(ct−12 ){v4t}
T (Xn) ≃ Z/2[c2]/(ct2){v4t−2, v4t−2v4t} ⊕ Z/(2t){ct−12 v4t}.
(iii) For n = 4t+ 2,
F (Xn) ≃ Z[c2]/(ct2){1, v4t} ⊕ Z{v4t−2}
T (Xn) ≃ Z/2[c2]/(ct−12 ){c2v4t−2, c2v4t−2v4t} ⊕ Z/4{v4t−2v4t}.
(iv) For n = 4t+ 3,
F (Xn) ≃ Z[c2]/(ct2){1, v4t}
T (Xn) ≃ Z/2[c2]/(ct2){v4t+2, v4tv4t+2} ⊕ Z/(2t+ 1){ct2v4t}.
Proof . Let X̃n = Xn ×Gm A1 be the associated bundle of (3.1) and s :
Yn → X̃n the 0-section. Since s∗ : CH ·(X̃n)
∼→ CH ·(Yn) by Theorem
2.4 (ii), the first assertion of the same theorem for V = X̃n and X =
s(Yn) gives an exact sequence
CH ·(Yn)
·c1→ CH ·(Yn)
π∗→ CH ·(Xn) → 0.
Theorem 4.1 follows from this and the ring structure of CH ·(Yn) in
Theorems 3.7 and 3.9. �
Next we consider the cycle map. The cohomology groups mean an
etale cohomology [9], [13]. All varieties are defined over K ′, which is
a subfield of an algebraically closed field K. Let l be a prime with
(l, ch(K)) = 1. We denote a locally constant sheaf µ⊗i
by Z/l(i).
Corollary 4.2. The homomorphism cl : CH i(Xn) → H2i(Xn,Z/l(i))
is injective.
Proof . Since (X̃n, Yn) is a smooth pair, we have the Gysin sequence as
in [5, Appendice 1.3.3] and [13, VI Remark 5.4]. Since the cycle map
and the Gysin map are commutative, we have the following commuta-
tive diagram, where each row is exact:
CH i(Yn)
·c1−−−→ CH i+1(Yn)
π∗−−−→ CH i+1(Xn) −−−→ 0
H2i(Yn,Z/l
·c1−−−→ H2(i+1)(Yn,Z/l(i+1))
π∗−−−→ H2(i+1)(Xn,Z/l(i+1))
Corollary 4.2 follows from Theorem 4.1 and this diagram. �
Remark 4.3. Assume that we have aK ′-isomorphism Yn ≃ SO(n,K)/(SO(n−
4, K) × GL(2, K)), where SO(n,K) and SO(n − 4, K) are split over
K ′, and that π : Yn → Xn is a K ′-map. Then the Galois actions
G = Gal(K
/K ′) on H ·(Yn,Z/l
(i)) and H ·(Xn,Z/l
(i)) are described by
the character group X(T ) of a K ′-split maximal torus T of SO(n,K).
It follows from a result of [5, 8-2].
5. Tables of the ring structure of CH ·(Yn)
5.1. Notations. (i) For k ∈ N ∪ {0}, we define bk and dk ∈ Z[c1, c2]
as follows:
bk = (−1)k
[ k2 ]
(−1)µ
k − µ
dk = (−1)k
(−1)µ
2k − µ+ 1
2k−2µ
(ii) For g ∈ N and µ ∈ N ∪ {0,−1}, we define ag,µ ∈ Z by
ag,µ =
(−1)1+µ g
g−1−µ
µ ≥ 1
−1 µ = 0
0 µ = −1.
Then the integers ag,µ are characterized by
(1 + x)g = 1 + xg +
[ g2 ]
ag,µx
µ(1 + x)g−2µ.
(iii) The generators v2i are given by (3.2) and (3.3).
5.2. An integral basis of CH ·(Yn). In the following (I) and (II),
we give an integral basis of CH ·(Yn). The notations are explained as
follows: Let Sn be the set of the monomial basis of An in (3.4). Let T
be a subset of Sn. Then for an element ξ ∈ T , 〈ξ〉 (resp. 〈ξ〉′) is defined
to be the right-hand side of an equation (1)-(8) below. We consider a
: ξ ∈ T
∪ {η : η ∈ Sn − T} ,
where lξ ∈ N. Following this procedure, we obtain an integral basis of
CH ·(Yn). We abbreviate this basis as
: ξ ∈ T
(I) The case n = 2m.
(i) For even m,
〈c2i+11 c2v2m−4v2m−2〉
2i+ 3
〈cm−2j−31 c
2 v2m−4v2m−2〉
m− 2j − 1
: 0 ≤ i ≤
− 2, 1 ≤ j ≤
(ii) For odd m,
〈c2i+11 c2v2m−4v2m−2〉′
2i+ 3
〈cm−2j−21 c
2 v2m−4v2m−2〉′
m− 2j
: 0 ≤ i ≤ m− 5
, 1 ≤ j ≤ m− 3
(II) The case n = 2m+ 1.
(iii) For even m,
〈c2i+11 c2v2m−2v2m〉
2i+ 3
〈cm−2j−31 c
2 v2m−2v2m〉
m− 2j − 1
: 0 ≤ i ≤ m
− 2, 1 ≤ j ≤ m
(iv) For odd m,
〈c2i+11 v2m−2v2m〉
2i+ 3
〈cm−2j−21 c
2 v2m−2v2m〉
m− 2j
: 0 ≤ i ≤
, 1 ≤ j ≤
Here 〈 〉 and 〈 〉′ are defined as follows:
〈c2i+11 c2v2m−4v2m−2〉 = c2i+11 c2v2m−4v2m−2 + (−1)
m+2i+2
(−1)i(2i+ 3) + 1
c2i+42 dm−2i−6
v2m−4
a2i+3,µc
2i+1−2µ
2 v2m−4v2m−2.
〈cm−2j−31 c
2 v2m−4v2m−2〉 = c
m−2j−3
2 v2m−4v2m−2
m−2j−4
am−2j−1,µc
m−2j−3−2µ
2j+1+µ
2 v2m−4v2m−2.
〈c2i+11 c2v2m−4v2m−2〉′ = c2i+11 c2v2m−4v2m−2 + (−1)
m+2i+1
(−1)i(2i+ 3) + 1
c2i+32 dm−2i−5
v2m−2
a2i+3,µc
2i+1−2µ
2 v2m−4v2m−2.
〈cm−2j−21 c
2 v2m−4v2m−2〉′ = c
m−2j−2
2 v2m−4v2m−2
m−2j−3
am−2j,µc
m−2j−2−2µ
2 v2m−4v2m−2.
〈c2i+11 c2v2m−2v2m〉 = c2i+11 c2v2m−2v2m + (−1)
m+2i+2
(−1)i(2i+ 3) + 1
c2i+42 dm−2i−6
a2i+3,µc
2i+1−2µ
2 v2m−2v2m.
〈cm−2j−31 c
2 v2m−2v2m〉 = c
m−2j−3
2 v2m−2v2m
m−2j−4
am−2j−1,µc
m−2j−3−2µ
2j+1+µ
2 v2m−2v2m.
〈c2i+11 v2m−2v2m〉 = c2i+11 v2m−2v2m + (−1)
m+2i+3
(−1)i(2i+ 3) + 1
c2i+32 dm−2i−5
v2m−2
a2i+3,µc
2i+1−2µ
2v2m−2v2m.
〈cm−2j−21 c
2 v2m−2v2m〉 = c
m−2j−2
2 v2m−2v2m
m−2j−3
am−2j,µc
m−2j−2−2µ
2 v2m−2v2m.
5.3. The ring structure of CH ·(Yn)(2) for n = 2m.
even m odd m
cm−11 (1)
cm−k−11 c
2 (k ≥ 1) (2)
cm−11 v2m−4 (3)
cm−2i−11 c
2 v2m−4 (i ≥ 1) (4)
cm−2i−21 c
2 v2m−4 (i ≥ 0) (5) (6)
cm−2i−11 c
2 v2m−2 (i ≥ 0) (7) (8)
cm−2i−21 c
2 v2m−2 (i ≥ 0) (9)
cm−2i−11 c
2 v2m−4v2m−2 (i ≥ 0) (10) (11)
cm−2i−21 c
2 v2m−4v2m−2 (i ≥ 0) (12) (13)
v22m−4 (14) (15)
v22m−2 (16) (17)
(1) =
[m−12 ]
(−1)1+µ
m− 1− µ
m−1−2µ
+ (−1)m+12v2m−2.
(2) =
[m−k−12 ]
(−1)1+µ
m− k − 1− µ
m−k−1−2µ
(−1)m+k2c2bk−1
v2m−4 +
(−1)m+k2c2bk−2
v2m−2.
(3) =
[m−12 ]
(−1)1+µ
m− 1− µ
m−1−2µ
v2m−4 + (−1)m+12v2m−4v2m−2.
(4) =
[m−2i−12 ]
(−1)1+µ
m− 2i− 1− µ
m−2i−1−2µ
v2m−4
(−1)m2
a2i−1,µc
2i−2−2µ
v2m−4v2m−2.
(5) =
m+2i+2
2i+ 1
c2i+22 dm−2i−4
m−2i−2
am−2i−1,µc
m−2i−2−2µ
2i+1+µ
v2m−4
a2i−1,µ +
2i− 1
2i+ 1
a2i+1,1+µ
2i−3−2µ
v2m−4v2m−2.
(6) =
m−2i−3
(−1)1+µ
m− 2i− 2− µ
m−2i−2−2µ
2i+1+µ
v2m−4
m+2i+1
2i+ 1
c2i+12 dm−2i−3
v2m−2
a2i−1,µ +
2i− 1
2i+ 1
a2i+1,1+µ
2i−3−2µ
v2m−4v2m−2.
(7) =
2i+ 1
c2i+22 dm−2i−4
v2m−4
m−2i−2
(−1)1+µ
m− 2i− 1− µ
m−2i−1−2µ
v2m−2
a2i−1,µ +
2i− 1
2i+ 1
a2i+1,1+µ
2i−3−2µ
v2m−4v2m−2.
(8) =
m+2i+3
2i+ 1
c2i+12 dm−2i−3
m−2i−1
am−2i,µc
m−2i−1−2µ
v2m−2
a2i−1,µ +
2i− 1
2i+ 1
a2i+1,1+µ
2i−3−2µ
v2m−4v2m−2.
(9) =
[m−2i−22 ]
(−1)1+µ
m− 2i− 2− µ
m−2i−2−2µ
2i+1+µ
v2m−2
(−1)m2
a2i+1,µc
2i−2µ
v2m−4v2m−2.
(10) =
m−2i−4
m− 2i+ 1
m− 2i− 1
am−2i−1,µ + am−2i+1,1+µ
m−2i−3−2µ
2i+1+µ
v2m−4v2m−2.
(11) =
m−2i−1
am−2i,µc
m−2i−1−2µ
v2m−4v2m−2.
(12) =
m−2i−2
am−2i−1,µc
m−2i−2−2µ
2i+1+µ
v2m−4v2m−2.
(13) =
m−2i−5
m− 2i
m− 2i− 2
am−2i−2,µ + am−2i,1+µ
m−2i−4−2µ
2i+2+µ
v2m−4v2m−2.
(14) = (−1)
2 dm−2
v2m−4.
(15) = − bm−2v2m−4 + (−1)
2 dm−3
v2m−2.
(16) = (−1)
2 c22dm−4
v2m−4.
(17) = (−1)
2 c2dm−3
v2m−2.
5.4. The ring structure of CH ·(Yn)(2) for n = 2m+ 1.
even m odd m
cm−11 (i)
cm−k−11 c
2 (k ≥ 1) (ii)
cm−2i−11 c
2 v2m−2 (i ≥ 0) (iii) (iv)
cm−2i−21 c
2 v2m−2 (i ≥ 0) (v)
cm−11 v2m (vi)
cm−2i−11 c
2 v2m (i ≥ 1) (vii)
cm−2i−21 c
2 v2m (i ≥ 0) (viii) (ix)
cm−2i−11 c
2 v2m−2v2m (i ≥ 0) (x) (xi)
cm−2i−21 c
2 v2m−2v2m (i ≥ 0) (xii) (xiii)
v22m−2 (xiv) (xv)
v22m (xvi) (xvii)
(i) =
[m−12 ]
(−1)1+µ
m− 1− µ
m−1−2µ
+ (−1)m+12v2m−2.
(ii) =
[m−k−12 ]
(−1)1+µ
m− k − 1− µ
m−k−1−2µ
(−1)m+k2c2bk−2
v2m−2 +
(−1)m+k+12bk−1
(iii) =
m−2i−2
(−1)1+µ
m− 2i− 1− µ
m−2i−1−2µ
v2m−2
m+2i+2
2i− 1
c2i2 dm−2i−2
2i+ 1
2i− 1
a2i−1,−1+µ + a2i+1,µ
2i−1−2µ
v2m−2v2m.
(iv) =
m+2i+3
2i+ 1
c2i+12 dm−2i−3
m−2i−1
am−2i,µc
m−2i−1−2µ
v2m−2
a2i−1,µ +
2i− 1
2i+ 1
a2i+1,1+µ
2i−3−2µ
v2m−2v2m.
(v) =
[m−2i−22 ]
(−1)1+µ
m− 2i− 2− µ
m−2i−2−2µ
2i+1+µ
v2m−2
(−1)m+12
a2i+1,µc
2i−2µ
v2m−2v2m.
(vi) =
[m−12 ]
(−1)1+µ
m− 1− µ
m−1−2µ
v2m + (−1)m+12v2m−2v2m.
(vii) =
[m−2i−12 ]
(−1)1+µ
m− 2i− 1− µ
m−2i−1−2µ
(−1)m2
a2i−1,µc
2i−2−2µ
v2m−2v2m.
(viii) =
m+2i+2
2i+ 1
c2i+22 dm−2i−4
m−2i−2
am−2i−1,µc
m−2i−2−2µ
2i+1+µ
a2i−1,µ +
2i− 1
2i+ 1
a2i+1,1+µ
2i−3−2µ
v2m−2v2m.
(ix) =
m+2i+3
2i+ 3
c2i+32 dm−2i−5
v2m−2
m−2i−3
(−1)1+µ
m− 2i− 2− µ
m−2i−2−2µ
2i+1+µ
a2i+1,µ − a2i+1,1+µ +
2i+ 1
2i+ 3
a2i+3,1+µ
2i−1−2µ
v2m−2v2m.
(x) =
m−2i−4
m− 2i+ 1
m− 2i− 1
am−2i−1,µ + am−2i+1,1+µ
m−2i−3−2µ
2i+1+µ
v2m−2v2m.
(xi) =
m−2i−1
am−2i,µc
m−2i−1−2µ
v2m−2v2m.
(xii) =
m−2i−2
am−2i−1,µc
m−2i−2−2µ
2i+1+µ
v2m−2v2m.
(xiii) =
m−2i−5
m− 2i
m− 2i− 2
am−2i−2,µ + am−2i,1+µ
m−2i−4−2µ
2i+2+µ
v2m−2v2m.
(xiv) = (−1)
2 dm−2
(xv) = (−1)
2 c2dm−3
v2m−2.
(xvi) = (−1)
2 c22dm−4
(xvii) = (−1)
c32dm−5
v2m−2 −
c1v2m−2v2m.
5.5. A remark on the ring structure of CH ·(Yn). We have given
the ring structure of CH ·(Yn)(2) in 5.3 and 5.4. But actually, it is easy
to determine the ring structure of CH ·(Yn) from 5.2, 5.3 and 5.4. For
example, by the basis in 5.2, the formula 5.3 (5) is rewritten as follows:
(5)′ cm−2i−21 c
2 v2m−4 =
m+2i+2
(−1)i(2i− 1) + 1
c2i+22 dm−2i−4
m−2i−2
am−2i−1,µc
m−2i−2−2µ
2i+1+µ
v2m−4
a2i−1,µc
2i−3−2µ
v2m−4v2m−2 −
4i− 2
2i+ 1
〈c2i−11 c2v2m−4v2m−2〉.
The other cases can be calculated similarly.
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Department of Mathematics, University of the Ryukyus, Nishihara-
Cho, Okinawa 903-0213, Japan
Department of Mathematics, University of the Ryukyus, Nishihara-
Cho, Okinawa 903-0213, Japan
E-mail address : tez@sci.u-ryukyu.ac.jp
	1. Introduction
	2. Preliminaries
	3. The ring structure of CH(Yn)
	4. The Chow ring of Xn
	5. Tables of the ring structure of CH(Yn)
	References
