Cohomology-exact sequences-1

A sequence of homomorphisms of vector spaces

$A \xrightarrow{f(x)} B \xrightarrow{g(x)} C$

is called "exact sequence" if $im f = ker g$.

One way of thinking about is as follows. Assume that $f(x)=g(x)=d$. Assume that $d$ applied to any element "dirties" it and second application $d$ to "dirtied" element sends it to $0$. Thus image of $d$ or $f$ is "dirtied" and because $g$  or second $d$ sends all such "dirtied" elements to zero, clearly $im\; d (or f)  = ker\; d( or g)$.

Short exact sequence has the form $0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0$.

Note for a sequence to be exact all the terms except for the first and last need to be exact.

The sequence $0 \xrightarrow{f} A \xrightarrow{g} B$ exact means $im\;f=ker;g=0$. Since only element in $ker\;g$ is $0$, $g$ is injective.

Similarly, in case $A \xrightarrow{f} B \xrightarrow{g} 0$, then all of $B$ is $ker\;g$. Hence, $im\;f=ker\;g=B$ and $f$ is surjective.

Problem 24.1 (Tu - An Introduction to Manifolds)

Given an exact sequence,
$A \xrightarrow{f(x)} B \xrightarrow{g(x)} C$

Show that $f$ is surjective iff $g$ is zero map.

Prf:

Assume $g$ is zero map. Then $ker\;g=B$. Using this in the definition of exact sequence results in $im\;f=ker\;g=B$. Hence $f$ is surjective.

Assume $f$ is surjective. From exact sequence definition, this means $im\;f=ker\;g=B$. $ker\;g=B$ implies $g$ is a zero map.

Show that $f$ is zero map iff $g$ is injective.

Prf:

Assume $g$ is injective. This means $ker\;g=0$. Using definition of exact sequence, results in $im\;f=ker\;g=0$. This $f$ is zero map.

Assume $f$ is zero map which means $im\;f=0$. Exact sequence implies $ker\;g=0$.Hence, $g$ is injective.

Problem 24.2. Four term exact sequence.

Four term sequence of vector spaces $0 \rightarrow A \xrightarrow{f} B \rightarrow 0$ is exact iff $f: A \rightarrow B$ is an isomorphism.

Prf:

Assume $f$ is an isomorphism which means $f$ is an injective and surjective function. $f$ is injective means $ker\;f=0$. And this is clearly image of previous zero map. $f$ is surjective and subsequent map is zero map. Then $im\;f$ is same as kernel of subsequent map.Both the conclusions yield an exact sequence.

Assume sequence is exact. Then image of zero map which is zero is clearly in kernel of $f$. Image of $f$ is same as kernel of subsequent (zero map). Hence $f$ is surjective. Thus $f$ is an isomorphism.

If $A \xrightarrow{f} B \rightarrow C \rightarrow 0$ is exact, then $C=coker\;f = \frac{B}{im\;f}$.

Rewrite sequence as

$A \xrightarrow{f} B \xrightarrow{g} C \rightarrow 0$

Given the sequence is exact, $im\;f=ker\;g$. Cokernel of $f$ is defined as $B/im\;f$. This can be written as $B/ker\;g$. First isomorphism theorem yields $B/ker\;g \cong im \;g$. However $im\;g = ker(C\rightarrow 0)$ which is all of $C$. Hence,
$B/ker\;g \cong C$.

Cohomology product structures and ring structure.

The product structure of wedge forms induces a product structure on Cohomoloy classes.

If $[\omega] \in H^{k}(M)$ and $[\tau] \in H^{k}(M)$ on a manifold $M$, then natural way to define the product structure is

$[\omega] \wedge [\tau] = [\omega \wedge \tau] \in H^{k+l}(M)$.

Know that $\omega,\tau$ are closed forms. So first we need to establish that the class $[\omega \wedge \tau]$ is a closed form. Note,

$d[\omega \wedge \tau] = d\omega \wedge \tau + (-)^k \omega \wedge d \tau = 0$.

Hence, $[\omega \wedge \tau]$ is a closed form.

Since, we are dealing with classes here, we need to show that if representative $\tau$ is replaced by exact form $\tilde{\tau}= \tau + d\eta$, then we need to show that

$d[\omega \wedge \tilde{\tau}] = d\omega \wedge \tau + (-1)^{k} \omega \wedge d\eta$

Thus, $\omega \wedge \tilde{\tau}$ is equal to $d[\omega \wedge \eta]$. Hence, closed.

• For a manifold $M$ of dimension $n$, the direct sum is $H^{*}(M) = \oplus_{k=1}^{n} H^{k}(M)$
• Thus $\omega \in H^{*}(M$, can be written as $\omega=\omega_0+\omega_1+\cdots+\omega_n$ where $\omega_{i} \in H^{i}(M)$.
• Product of differential forms defined on $H^{*}(M)$ gives $H^{*}(M)$ a ring structure - called "Cohomology ring".
• Since product of differential forms is anticommutative, the ring is anticommutative.
• Direct sum gives Cohomology ring a graded algebra structure.
• Thus, $H^{*}(M)$ is anticommutative graded ring.
  

Induced Cohomology maps

For a smooth map $F:N\rightarrow M$ between manifolds, $M,N$ there exists a pullback map of differential forms $F^*:\Omega(M) \rightarrow \Omega(N)$.

Pull back operator $F^*$ has a pleasant property. It commutes with d operator. For closed forms,

$d(F^* \omega)  = F^*(d\omega) = 0$

Thus it maps closed forms from $M$ to closed forms in $N$.

Similarly,

$F^*\omega=F^*d(\eta) = dF^*\eta$ for any exact form $\omega = d\eta$.

Thus it maps exact forms to exact forms.

$F^*$ induces a cohomology map

$F^{\#} : H^{k}(M) \rightarrow H^{k}(N)$ given by
$F^{\#}(\omega)=[F^*\omega]$.

What is nice about this is that diffeomorphism between manifolds $N \rightarrow M$ results in isomorphic vector spaces between $N$ and $M$.

Cohomology of Real line.

First example of applications of Cohomology is real line $R$.

To start off a fact about differential forms:

Differential forms belong to spaces of Alternating forms - $A^k(M)$. Whenever $k>n$ where $n$ dimension of tangent space at a give point, differential $k$ forms become $0$.

Since $R$ is connected, can conclude, $H^0(R)=R$. Clearly, all two forms are zero as $n=1$. Note, two forms are generated by one forms. Since all two forms are zero, all one forms are closed.

Note a function such $h(x)$ is a zero form. A one form $f(x)dx$ on $R$ is exact if and only if there exists a $C^\infty$ function $g(x)$ on $R$ such that the following is satisfied.

$f(x)dx = dg = g^'(x) dx$

which means,

$g(x) = \int_0^x f(t) dt$

Thus,

$H^k(R) = R$ when $k=0$, and $H^k(R)=0$ when $k>0$.

Cohomology as measure of connectedness.

Connected spaces:

A Topological space consisting of distinct globs is disconnected. Space that is not disconnected is connected.

There are a few ways to characterize such spaces. Topological space consisting of disjoint union of maximal open sets is disconnected. If only open and closed sets are entire space and null set then such Topological spaces are connected. To see why, if $U,V$ are disjoint open sets whose union is whole Topological space, then clearly $U,V$ are complements of each other, hence they are closed. In such disconnected spaces, more than whole space and null set are both open and closed.

Connectedness is a Topological property as it defined using open sets.

Example: Let $Y=[-1,0)\cup(0,1]$ be topological space. Clearly, $[-1,0),(0,1]$ are open sets in this subspace topology. $Y$  is a separated space as it is an union of disjoint maximal subsets of $Y$.

Computing $H^0(M)$ yields a count of connected components.

First note that there are no exact $0$ forms, since these are $k(=0)-1$ which don't exist.

Hence, $H^0(M)=Z^0(M)$.

Suppose $f$ is a closed zero form on $M$. That is $f$ is a $C*\infty$ function such that $df=0$.

$df = \sum_{i=1}^r  \frac {\partial f}{\partial x}  dx $

$df=0$ implies

$\frac{\partial f}{\partial x_i} = 0$ for each $i$.

which means $f$ is constant in each of the components.

To see this in more detail, let $f$ be a constant function a small region on Manifold $Q=\{q \in Q|f(p)=f(q)\}$. Let $U$ be corresponding chart of $Q$. Since $f$ is constant in $Q$, $df=0$ as each of the partials vanish in this region. Thus $Q$ is open. $f$ continuous means $Q$ is closed. Since $M$ is connected only open and closed sets are whole of $M$ and empty set. Then, $f$ is constant on whole of $M$.

$r$ connected components lead to $H^0(M) \equiv R^r$.

Cohomology - futher motivation and definition.

In general, in order to classify somethings, we use an invariant. For example if one is to classify a bunch of books, one can use broad classification such as fiction vs non-fiction. And classification possibilities are more depending on the diversity of the book collection. In case of fiction vs non-fiction classification, the invariant is if a book belongs to fiction or non-fiction. This is nothing new. Similar concepts of classification is used in all fields of sciences including social sciences.

While classifying books - fiction vs non-fiction, we abstract out type of book while removing all other information such as big book, small book, physics book or bible.

In Topology, in a similar fashion we ignore actual geometric shapes, areas etc and focus on whether a space can be deformed continuously into a smaller subset of spaces. When this deformation is taking place, certain things remain invariant.

To see this, assume you have a yet to be inflated balloon where you draw letter "A". Clearly, the letter "A" has one closed loop. As you inflate the balloon, the closed loop becomes larger and gets distorted - but still very much visible. This means such closed loops are invariant under continuous deformation.

Whereas in Topology, the spaces are abstract, when you add extra structure to these spaces as is done in smooth manifolds, the machinery of Linear Algebra can be deployed to perform actual, tangible computations.

For any $k$ form $\omega$, "closed" form means $d\omega=0$. "Exact" form means $\omega=d\tau$ for a form $\tau$ which is a $k-1$ form.

One way to remember these definition is, $d\omega=0$ and since $0$ looks like a closed loop, one can remeber $d\omega=0$ as closed forms.

Let $Z^k(M)$ on a smooth manifold $M$ be vector space of closed forms (think Z for zero) and similarly, let $B^k(M)$ be vector space of exact forms.

Since $d^2\omega = 0$ for any form $\omega$, the exact forms are all closed as $d\omega=d(d(\tau)=d^2\tau=0$.

But not all closed forms are exact forms.

Since, both $B^k(M),Z^k(M)$ are vector spaces, we can form a quotient space $Z^k(M)/B^k(M)$ using the vector space equivalences.

The quotient $H^k(M)=Z^k(M)/B_k(M)$ is called de Rahm Cohomology. This is an invariant of Manifolds under certain conditions which will be noted later.

Since $H^k(M)$ is vector space quotient, for any differential forms $\omega,\omega'$, this means

$\omega' - \omega \in B^k(M)$  means $\omega' \tilde{} \omega$ in $Z^k(M)$.

This means the following relation is satisfied.

$\omega' = \omega + d\nu$.

Cohomology - motivating example

Concept of vector fields as gradients of functions is well established. However, sometimes taking curl of vectors also results in vector fields. These concepts lead to certain simplifications when computing line integrals.

The following illustrative example is from Tu's "Introduction to Manifolds".

Let $F(x,y)=(P(x,y),Q(x,y))$ be a vector field defined on an open set $U$ in $R^2$. Let $C$ represent a parameterized curve defined by $c(t)=(x(t),y(t))$ where $t \in [a,b]$ as it moves from point $A$ to $B$ on $U$. The total work done by a particle moving along this path is given by line integral $\int_C P(x,y)dx+Q(x,y)dy$.

If vector field is a gradient of a scalar function the line integral is easy to compute using Stoke's theorem.

$F=grad(f) = (f_x,f_y)$

where $f_x =\frac{\partial f}{\partial x}$ and
$f_y =\frac{\partial f}{\partial y}$

$\int_C f_x dx+f_y dy = \int_C df = f(B)-f(A)$.

A necessary condition for $F$ to be grad of a scalar function $f$ is

$P_y = f_{xy}=f_{yx}=Q_x$.

The question is now the following:

If $Q_x-P_y=0$, is the vector field $F=(P,Q)$ gradient of some scalar function $f$ on $U$?

By correspondence between vector fields and 1-forms in $R^2$, we have,

$F=(P,Q) <-> \omega = Pdx + Qdy$

$grad f = (f_x,f_y) <->d\omega = f_x dx + f_y dy$

$Q_x-P_y=0 <-> d\omega = (Q_x-P_y)dx \wedge dy =0$.

So the question is, if $\omega =f _x dx+f_ydy$ is closed ie. $d\omega=0$ is it exact?

Answer is sometimes Yes and sometimes No and depends on $U$.