whereLis the lower-triangular matrix that defines the stabilisation map, see Definition 19. Computing the determinants gives
Det(sn−1(v)) = (±v1, . . . ,±vm,±1, . . . ,±1),
with the signs dependent onnandm. Using the action of Tn to change the lastn−1 determinants as in the proof of Lemma 18, we can get an element ofYm,nwith any prescribed determinant. Altogether this defines a lift,f, of the identity map over the determinant,
Ym,n
Det
Tm+n−1
f 88
Id //Tm+n−1. By taking first cohomology, we get maps
H1(Ym,n)
f∗..
H1(Tm+n−1)
Det∗
mm ,
with
f∗◦Det∗= Id.
This shows that Det∗ is injective andf∗ is surjective. But by the previous theorem, f∗ is a map between free abelian groups of the same rank, hence it must also be injective and so an isomorphism with inverse Det∗. This shows that the determinant map is an isomorphism on the first cohomology groups and immediately extends to the quotiented spaces, with
Det∗:H1(Tm+n−1/Tn)→H1(Xm,n)
an isomorphism. Note that the action of Tn on Tm+n−1 is not the obvious one, but is given by how the determinants change when Tn acts on Ym,n. The quotient is still isomorphic to Tm−1. Furthermore the stabilisation maps on degree one cohomology are surjective, since the stabilisation maps
s:Ym,n→Ym,n+1
change the signs of some determinants and add an extra determinant that is always plus or minus one. So if the class di ∈H1(Ym,n+1) represents the image of moving once around zero in thei’th coordinate of Tn+1, then
s∗(di) =
(di 1≤i≤n, 0 otherwise.
3.3. THE SPECTRAL SEQUENCE OFY3,3 37 The way we will calculate this is by using the spectral sequence defined in Section 3.1. Some of the calculations have been done in Sage and the relevant source code can be found in Appendix A.
The groups we need to compute for the first page in the spectral sequence are:
E1p,q ∼=
Hq(Y∅) p= 0,
Hq+1(Y{1}×(D, D−0))⊕Hq+1(Y{2}×(D, D−0)) p= 1, Hq+2(Y{1,2}×(D2, D2−0)) p= 2
0 otherwise.
The four spaces that appear will be handled separately.
The cohomology ofY{1,2}
Consider the spaceY{1,2}. This consists of the elements ofY3,3that have two zeroes in the last column:
Y{1,2}=
a b 0
c d 0
e f i
a6= 0, ad−bc6= 0 adi−bci6= 0
di6= 0, i6= 0
.
The space is homeomorphic to (C∗)3×C2×
(b, c)∈C2|bc6= 1 , with the home- omorphism given by
Y{1,2}→(C∗)3×C2×
(b, c)∈C2|bc6= 1
a b 0
c d 0
e f i
7→
a, d, i,e
i,f i,b
a,c d
.
We already computed the cohomology ofY ={(b, c)|bc6= 1}in Example 11, so we can calculate the cohomology ofY{1,2}:
Hq(Y{1,2}) =
Z q= 0, Z4 q= 1, Z7 q= 2, Z7 q= 3, Z4 q= 4, Z q= 5, 0 otherwise.
Since we know the product structure from the Künneth formula, we can name the generators and get the ring
H∗(Y{1,2}) = Λ∗[a00, d00, i00]⊗H∗(Y).
The two generators ofY will be denotedy00,1for the degree one generator and y00,2
for the degree two generator. All other generators have degree one. The ring structure is the usual one on a tensor product, with the added relation y00,1·y00,2 = 0. The elementy00,1∈H1(Y) is given by the map
Y 3(b, c)7→1−bc∈C∗
in cohomology, since there is another map
C∗3λ7→(1,1−λ)∈Y
and the composition is the identity on C∗. ApplyingH1 shows that the two maps give isomorphisms in degree 1.
The cohomology ofY{1}
InY{1} we have the first entry of the last column equal to zero, which is the space
Y{1}=
a b 0
c d h
e f i
a6= 0, ad−bc6= 0 adi+beh−bci−af h6= 0
di−f h6= 0, h6= 0, i6= 0 .
This space is homeomorphic to (C∗)5×C×Y:
a b 0
c d h
e f i
7→
a, i, h,d h−f
i,d h− bc
ah,e
i, bhi adi−af h, c
h−e i
.
Since the cohomology ofY is already known, applying the Künneth formula gives
Hq(Y{1}) =
Z q= 0, Z6 q= 1, Z16 q= 2, Z25 q= 3, Z25 q= 4, Z16 q= 5, Z6 q= 6, Z q= 7, 0 otherwise.
As before, the ring structure is given by
H∗(Y{1}) = Λ∗[a01, d01, f01, h01, i01]⊗H∗(Y).
The cohomology ofY{2}
The space is
Y{2}=
a b g
c d 0
e f i
a6= 0, ad−bc6= 0 adi+cf g−bci−deg6= 0
di6= 0, g6= 0, i6= 0
,
which can also be identified with (C∗)5×C×Y using the homeomorphism
a b g
c d 0
e f i
7→
a
g, d, i, g,1 + cf g adi −eg
ai − bc ad,f
i,b a,c
d
.
3.3. THE SPECTRAL SEQUENCE OFY3,3 39 Since we already know the cohomology of this space, we know
Hq(Y{2}) = (Λ∗[a10, d10, f10, g10, i10]⊗H∗(Y))q=
Z q= 0, Z6 q= 1, Z16 q= 2, Z25 q= 3, Z25 q= 4, Z16 q= 5, Z6 q= 6, Z q= 7, 0 otherwise.
The cohomology ofY∅
This space is slightly more complicated than the others. The space consists of the elements inY3,3 where all entries of the last vector are non-zero.
Y∅=
a b g
c d h
e f i
a6= 0, ad−bc6= 0, di−f h6= 0, adi+beh+cf g−deg−bci−af h6= 0,
g6= 0, h6= 0, i6= 0
.
The space is homeomorphic to (C∗)7×
(x, y, z)∈C3|y−z−xyz= 1 , with the homeomorphism given by
a b g
c d h
e f i
7→
g, h, i,a g,f
i − d h, D= ad−bc
ah , C= c
h−a
g + cdi−deh
f h2−dhi+ beh−bci f gh−dgi, bC
aD, c hC − a
gC,D C
eh−ci f h−di
.
To get the cohomology, we refer back to Example 12, where we computed the cohomology of
Z =
(x, y, z)∈C3|y−z−xyz= 1 ,
and found it to be the same as the cohomology ofS2. The generator of H2(Z) will
be calledz11, withz211= 0. If we apply this, we get
Hq(Y∅) =
Z q= 0, Z7 q= 1, Z22 q= 2, Z42 q= 3, Z56 q= 4, Z56 q= 5, Z42 q= 6, Z22 q= 7, Z7 q= 8, Z q= 9, 0 otherwise.
and the generators are
H∗(Y∅) = Λ∗[a11, d11, g11, h11, i11, C11, D11]⊗H∗(Z).
A note on the cohomology ofZ
The spaceZ defined above is a manifold, which we will prove by applying the implicit function theorem. Consider the formula defining Z,
(x, y, z)7→y−z−xyz−1.
Taking partial derivatives gives the three functions (x, y, z)7→yz, (x, y, z)7→1−xz, (x, y, z)7→ −1−xy.
If the first function is zero, then at least one of the two others must be non-zero.
By the implicit function theorem, the coordinate that has non-zero derivative can be written as a smooth function of the other coordinates. This shows thatZ is locally the graph of a smooth function, so it is a smooth manifold.
Later on it will be necessary to have a description of H2(Z). This is done by considering the subspace
R= (
(x, y, z)∈R3
x >0, y <0, xy <−1, z= y−1 1 +xy >0
)
⊂Z.
This space is closed in Z since any sequence in R that converges in Z must also converge in R. Consider a subbundleN R∼=R×D of the normal bundle ofR in the tangent bundle T Z, such that the exponential map is a diffeomorphism onN R. This gives us a diagram
H2(N R, N R0)oo ∼= H2(Z, Z−R) j∗ //H2(Z),
3.3. THE SPECTRAL SEQUENCE OFY3,3 41 whereN R0 isN Rwithout the zero section,N R0∼=R×(D−0). The isomorphism is given by excision, since Z=N R∪(Z−R). The space Ris homeomorphic toR2, so we have
H2(Z, Z−R)∼=H2(N R, N R0)∼=H2(R×(D, D−0))∼=Z.
From the diagram,j∗(1) gives us an element ofH2(Z). To see that this is a generator, consider the torus
T =b (
x, y, z= y−1 1 +xy
∈Z
|x|=|y|= 2 )
⊂Z.
This intersectsRin exactly one point, namelyp= (2,−2,1). By taking the intersection withT we can extend the previous diagram,b
H2(N R, N R0)
i∗
H2(Z, Z−R)
i∗
∼=
oo j∗ //H2(Z)
i∗
H2(p×(D, D−0))oo ∼= H2(bT,Tb−p) ∼= //H2(bT).
The lefti∗ is an isomorphism since the cohomology in degree two is given by the pair (D, D−0), which does not change under the inclusion. This shows that the inclusion Tb →Z is an isomorphism in cohomology, soj∗(1) is a generator ofH2(Z). Taking the same diagram with cohomology replaced with homology reverses all arrows but is otherwise exactly the same. This shows that to calculate the homology class of a cycle σinH2(Z), we can count the number of points of intersection betweenσand R with orientation, since each of these will contribute either 1 or−1, depending on orientation, to the homology class in (Z, Z−R).
There is a smiliar argument with the space Y from before and the subspace RY =
(b, c)∈R2|b >0, c >0, bc >1 . In particular, the generator ofH2(Y) can be taken asi∗
[bT]
, where [T] is a generator ofb H2(T) for the torusb
T =b {(b, c)| |b|=|c|= 2} ⊂RY. The first page
With all of the above, we can draw theE1-page of the spectral sequence. Remember that this is
E1p,q ∼=
Hq(Y∅) p= 0,
Hq+1(Y{1}×(D, D−0))⊕Hq+1(Y{2}×(D, D−0)) p= 1, Hq+2(Y{1,2}×(D2, D2−0)) p= 2,
0 otherwise.
By inserting the groups that were calculated above into this, we get Figure 3.4 with the non-zero differentials moving directly to the right. To get theE2-page of the spectral sequence, we need to compute the homology of the differentials.
OO //
q p
9 Z 0 0
8 Z7 //Z2 0
7 Z22 //Z12 //Z 6 Z42 //Z32 //Z4 5 Z56 //Z50 //Z7 4 Z56 //Z50 //Z7 3 Z42 //Z32 //Z4 2 Z22 //Z12 //Z
1 Z7 //Z2 0
0 Z 0 0
0 1 2
Figure 3.4: TheE1-page with differentials.