3
$\begingroup$

Is my following understanding correct:

A chain complex $(C,\partial)$ is a family $\{C_i\}_{i\geq 0}$ of $R$-modules ($R$ is a given ring) together with a family of $R$-module homomorphisms $\partial_i:C_i\rightarrow C_{i-1}$ such that $\partial_{i-1}\circ \partial_{i}=0$. The quotient $\ker(\partial_i)/Im(\partial_{i+1})$ is an $R$-module that we denote $H_i(C,\partial)$ and we call the $i$th homology $R$-module of the chain complex $(C,\partial)$ ( In the litterature, this last sentence is usually stated as follows: $H_i$ is the $i$th homology group of the chain complex $(C,\partial)$ with coefficients in the ring $R$).

A cochain complex is exactly the same as a chain complex with the same data except that the family of $R$-module homomorphisms are of degree $+1$ meaning that $\partial_i:C_i\rightarrow C_{i+1}$ instead of $C_i\rightarrow C_{i-1}$ in the previous case. Hence the quotient $\ker(\partial_i)/Im(\partial_{i-1})$ is also an $R$-module that we now denote $H^i(C,\partial)$ and we call the $i$th cohomology $R$ module of the cochain complex.

Now for specific chain and cochain complexes, the difference becomes clearer. Indeed, for a topological space $X$ we can construct a chain and a cochain complexes but the advantage of the cochain complex is that we can always define a product on the direct sum of the cohomology $R$-modules $H^i$ making the cohomology $R$-module $\oplus_{i \geq 0}H^i$ into an $R$-algebra.

$\endgroup$
8
$\begingroup$

There is indeed no difference from a purely algebraic point of view, once you allow degrees to be negative. Then a chain complex $(A_{\bullet}, \partial_\bullet)$ defines a cochain complex $(B^{\bullet}, d^\bullet)$ by taking $B^n = A_{-n}$ and $d^n = \partial_{-n}$, and vice versa; moreover, $H_{-i} (A) = H^i (B)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.