LetC•,• be a double chain complex with horizontal differentialsdh and vertical differentialsdv. Then the total chain complex Tot(C) is defined to be
Tot(C)n=
M
p+q=n
Cp,q
with differentials d=dh+ (−1)pdv. Then we define the horizontal filtrationFh
•Tot(C) on Tot(C) to be the filtration Fph(Tot(C))n:= M n1+n2=n n1≤p Cn1,n2.
Similarly, the vertical filtration is defined to be
Fqv(Tot(C))n := M n1+n2=n n2≤q Cn1,n2. Proposition 3.8. Let{Er
p,q}r,p,q be a spectral sequence of the double complexC•,• with respect to the horizontal filtration (this is exactly the same for vertical filtration). Then
CHAPTER 3. SPECTRAL SEQUENCES 3.3. EXAMPLES
• Ep,q0 ∼=Cp,q,
• E1
p,q ∼=Hq(Cp,•),i.e. the homology of the columns, • E2
p,q ∼=Hp(Hqv(C•,•)), i.e. homology of the rows of the previous page.
Moreover, ifC•,• is a first quadrant double complex (0≤p, q), then the spectral sequence converges to the chain homology of the total complex:
Ep,q∞ ∼=GpHp+q(Tot(C)•).
Thus the chain homology of the total chain complex can be computed by either using a horizontal or vertical filtration, and we will get the same answer on the limit page.
Proof. Ep,q0 : =GpTot(C)p+q = FpTot(C)p+q Fp−1Tot(C)p+q = L n1+n2=p+q n1≤p Cn1,n2 L n1+n2=p+q n1≤p−1 Cn1,n2 ∼ =Cp,q. Ep,q1 ∼=Hp+q(GpTot(C)•) ∼ =Hp+q(Cp,•) ∼ =Hq(Cp,•)
For the next page, [x] ∈E1
p,q = Hq(Cp,•), so pick a representative x∈ker(Cp,q →Cp,q−1). So x∈ {z∈Cp,q|dvx= 0}, i.e. the vertical maps are zero onx. So the differentiald1which is induced
by the original differentiald=dh+dv is just the horizontal differentialdh, sod1:E1
p,q →Ep−1,q. Thus, Ep,q2 = ker(d1:E1 p,q →Ep1−1,q) im(d1:E1 p+1,q→E1p,q) ∼ =Hp(Hqv(C•,•)).
By the convergence properties stated earlier, this bounded filtration means that this sequence will converge, and it will converge to the homology of the original filtered complex.
Example 3.7. Special case when rows or columns are exact
Using the previous properties, it is clear that if a double complex has exact rows or columns, the limit pageE∞will be zero. This can be very useful as we will see for an example easy proof of the Five-lemma:
Lemma 3.1. Take the following diagram that has exact rows andα, β, δ and are isomorphisms:
A B C D E A0 B0 C0 D0 E0 α β γ δ α β γ δ α β γ δ α β γ δ Thenγ is an isomorphism.
3.3. EXAMPLES CHAPTER 3. SPECTRAL SEQUENCES
Proof. Since the rows are exact by assumption, the spectral sequence converges on the first step in the horizontal direction immediately, so thatE∞
p,q= 0. Now we compute the spectral sequence in
the vertical direction, so the first page is
0 0 kerγ 0 0
0 0 cokerγ 0 0
Since the limit page is zero, and in the spectral sequence, there will be no point where we can take homology to get rid of the final remaining terms, we must have ker(γ) = 0 and coker(γ) = 0, and thusγ is an isomorphism. Looking at this page we can see that if we let ker(α) and coker() be nonzero, the result still holds. In other words, we only needβ andδ to be isomorphisms and coker(α) = 0 and ker() = 0. This is the same as saying that we only needαto be surjective and
to be injective.
Example 3.8. Homology of tensor product of chain complexes
We can compute the homology of the tensor product of chain complexes using spectral sequences. More specifically, two chain complexes can be tensored together to give the tensor product double complex, from which we get the total chain complex for the tensor product, which is an honest chain complex (not a double complex). We will restrict the discussion to the case of chain complexes of vector spaces. More generally forR-modules the universal coefficient theorem for homology tells us we gets something a little more complicated than for vector spaces (see Remark 3.9). Consider two chain complexes of vector spacesV• andW• whereVi andWi denote the ith component of
theV andW complexes, respectively. Then the total chain complex Tot(V•⊗W•), which we will write as (V ⊗W)•, has as thenth component
(V ⊗W)n= M n=i+j Vi⊗Wj = M i≤k Vi⊗Wk−i.
We define the differential to be
dV⊗W(v⊗w) =dVv⊗w+ (−1)deg(v)v⊗dWw.
This can be quickly verified to be a differential:
(dV⊗W)2=dV⊗WdVv⊗w+ (−1)deg(v)v⊗dWw
= (dV)2v⊗w+(−1)deg(dVv)+ (−1)deg(v)(dVv⊗dWw) +v⊗(dW)2w
= 0.
In order to use spectral sequences, we define a filtration as we did before on the total complex of
V•⊗W• by
Fp(V ⊗W)k=
M
i≤p
Vi⊗Wk−i.
So, for example,
F0(V ⊗W)k =V0⊗Wk
F1(V ⊗W)k = (V0⊗Wk)⊕(V1⊗Wk−1)
and so on. The associated graded object is given by
CHAPTER 3. SPECTRAL SEQUENCES 3.3. EXAMPLES
In terms of our pages of spectral sequences,
Ep,q0 =Gp(V ⊗W)p+q =Vp⊗Wq.
The differential for this (the differential on the zeroth page) is given by d0= (−1)pid
V ⊗dW, so d0(E0 p,q)⊆Ep,q0 +1. Thus for · · · →E0p,q−1→Ep,q0 →Ep,q0 +1→ · · · we have E1p,q= ker(d:E 0 p,q−1→Ep,q0 ) im(d:E0 p,q→E0p,q+1) =Vp⊗ker(d W :W q−1→Wq) Vp⊗im(dW :Wq →Wq+1) ∼ =Vp⊗Hq(W).
The differential d1on the following pageE1
p,q is given byd⊗idW, soEp,q2 =Hp(V•⊗Hq(W•)) =
Hp(V)⊗Hq(W). So each element of the the E2 page is the tensor product of V-cycles and
W-cycles and is thus aV ⊗W-cycle. Since the differentials of the spectral sequence come from the differential onV ⊗W, all the higher differentials must vanish, soE2=E∞. Thus,
Hk(V•⊗W•)∼= M
i+j=k
Hi(V)⊗Hj(W).
Remark 3.9. This differs from the more general case. The K¨unneth formula for complexes states that ifV andW are right and leftR-module complexes, andVnandd(Vn) are flat for eachn, then
there is an exact sequence
0→ M p+q=n Hp(V)⊗Hq(W)→Hn(V ⊗RW)→ M p+q=n−1 TorR1(Hp(V), Hq(W))→0
which is noncanonically split if R=ZandV is a complex of free abelian groups. But in the case
of vector spaces, we always have TorZ
1 = 0 because there is always a length zero free resolution of V, giving us the desired isomorphism by exactness. Further discussion can be found in [Wei94].
Chapter 4
The Lee Spectral Sequence
4.1
Comparison of Khovanov’s and Lee’s differentials
The Khovanov complexCKh(L) of a linkLis an object in the categoryGKom(GVeck) of finite length graded complexes of graded vector spaces, and thus the differentials inCKh(L) are graded. On the other hand, the Lee complexCKhLee(L) of a linkLis an object in the category FilKom(GVec
k) of
finite length filtered complexes of graded vector spaces — each spaceCKhLee(L)
rin the complex
CKhLee(L) is a graded vector space, but the differentials do not preserve the q-grading:
dKh:CKh(L)q,n→CKh(L)q,n+1
dLee:CKh(L)q,n→CKh(L)q,n+1⊕CKh(L)q+4,n+1
The differentials for CKhLee(L) are not even homogeneous, for example ∆Lee(v
−) =v−⊗v−+
v+⊗v+. However, given such filtered chain complex, its spectral sequence is closely related to
Khovanov homology:
Theorem 4.1. There is a spectral sequence withE1 page Kh(L)andE∞ page KhLee(L). Proof. Defineδto be the differential so that dLee=dKh+δ. So dLee is a differential for a filtered complex, where thedKhis the degree preserving part of the differential andδ shifts the degree up by 4. Explicitly,dKh is defined as in Definition 2.22,dLee is defined as in Example 2.26, andδis defined via
δm(v+⊗v+) =δm(v+⊗v−) =δm(v−⊗v+) = 0,
δm(v−⊗v−) =v+,
δ∆(v+) = 0, δ∆(v−) =v+⊗v+.
From this, it is clear thatδ2= 0. Because we have a chain complex of filtered vector spaces, there is an associated spectral sequence withE∞ its homology, i.e. E∞∼=KhLee(L). Moreover, since the E1 page is always the homology with respect to the degree preserving part of the original filtered complex, we conclude that theE1page is the homology with respect todKh, or in other
words, it isKh(L).
In fact, each page of the spectral sequence is also a link invariant. Following Rasmussen [Ras10], we can prove this via the following Lemma that can be found in [McC01]: