I was wondering the following. If I take a simplicial complex $X$, I can form the cochain complex $C^{\bullet}(X, \mathbb{C})$, along with the usual differential $d(\alpha)_{i_{0}, \dots, i_{k+1}} = \sum\limits_{j = 0}^{k+1} (-1)^{j}\alpha_{i_{0}, \dots, \hat{i_{j}}, \dots i_{k+1}}$. The homology of this chain complex leads to the standard cohomology groups of $X$.
What do I get if I use the same chain groups, but use the differential defined by: $d(\alpha)_{i_{0}, \dots, i_{k+1}} = \sum\limits_{j = 1}^{k} (-1)^{j}\alpha_{i_{0}, \dots, \hat{i_{j}}, \dots i_{k+1}}$
That is, we leave out the first and last terms in the usual differential.
Thank you very much!
Elliot
As $d(\alpha)_{i_0,i_1} = 0$ we have $$H^1(X)\approx \text{ker} \, d(\alpha)_{i_0,i_1,i_2} /\text {Im} \, d(\beta)_{i_0,i_1} \approx \text{ker}\, d(\alpha)_{i_0,i_1,i_2}$$ Then we have enough problems that $H^*$ is not a homology theory. To prove this we proceed by contradiction. Assume $H^*$ defined in this way is indeed a homology theory. As the cocomplex is constructed by $$C^n(X;\mathbb C)\approx \text{Hom}(C^n,\mathbb C) \approx \text{Hom}(\mathbb Z^l,\mathbb C)$$ it will be enough to consider coefficients in $\mathbb Z$. Now consider the space $X$ with triangulation the 2-simplex. We have the chain complex $$0\rightarrow \mathbb Z \rightarrow \mathbb Z^3 \rightarrow \mathbb Z^3 \rightarrow 0$$ Then we construct the cochain complex $$0 \leftarrow \mathbb Z \xleftarrow{d_1} \mathbb Z^3 \xleftarrow 0 \mathbb Z^3 \leftarrow 0$$ So we have the differential $d_1$ given by $$d(a,b,c) = -b $$ Thus for the kernel $$\text{ker} \, d(a,b,c) \approx \langle(1,0,0),(0,0,1)\rangle \approx \mathbb Z^2 \approx H^1(X;\mathbb Z)$$ But $X$ is contractible so the identity map $i:X \rightarrow X$ is homotopic to a constant map $c:X \rightarrow X$. Then by the Homotopy Axiom of the Eilenberg-Steenrod axioms for homology theories we have that the induced homomorphisms $i_*:H^1(X;\mathbb Z) \rightarrow H^1;\mathbb Z)$ and $c_*:H^1(X;\mathbb Z) \rightarrow H^1(X;\mathbb Z)$ the zero map are the same map. This implies $H^1(X;\mathbb Z) \approx \mathbb Z^2 \approx 0$ a contradiction thus $H^*$ is not a homology theory.