# Show that $\bigoplus_{i \text{ even}}C_i=\bigoplus_{i\text{ odd}}C_i$

Let $$C_*$$ be a chain complex such that each $$C_i$$ is a torsion-free, finite-range abelian group with $$C_i=0$$ for all $$i<0$$. Suppose that $$C_i=0$$ for all $$i$$ is sufficiently large and that for all $$i$$ there exists an integer $$N_i\geq1$$ such that $$N_i(H_i(C_*))=0$$, that is, each $$H_i(C_*)$$ is a torsion group. Show that $$\bigoplus_{i \text{ even}}C_i=\bigoplus_{i\text{ odd}}C_i$$

I do not know how to solve this problem, I would appreciate any suggestions. We have to $$...\overset{\partial}{\rightarrow}C_{n+1}\overset{\partial}{\rightarrow}C_{n}\overset{\partial}{\rightarrow}C_{n-1}\overset{\partial}{\rightarrow}...$$

where each $$C_i$$ has a base with a finite amount of elements, say $$B_i$$ and we also have $$\partial(\partial(c_i))=0$$ for all $$c_i\in B_i$$. How can I use all this to define an isomorphism between $$\bigoplus_{i \text{ even}}C_i$$ and $$\bigoplus_{i\text{ odd}}C_i$$? Thank you.

• It's a standard fact in homological algebra that the alternating sum of the ranks of the chain group in a finite chain complex of f.g. abelian groups is equal to the alternating sum of the ranks of the homology groups. This is algebraic meaning of homology-invariance of Euler characteristic, if you like. You can try to prove/use this. May 18, 2019 at 23:51
• @BalarkaSen Yes, I know this fact and here I found a proof of this math.stackexchange.com/questions/2807224/…, but I do not know how to use this to solve the problem May 19, 2019 at 1:00

Your complex is of the form

$$0 \to C_n \to \cdots \to C_1 \to C_0 \to 0$$

and each $$C_i$$ is free as a $$\mathbb{Z}$$-module, i.e. for each $$i$$ we have a non-negative integer $$r_i$$ such that $$C_i \simeq \mathbb{Z}^{r_i}$$ via some isomorphism $$f_i : C^i \to \mathbb{Z}^{r_i}$$. If we define $$d_i := f_{i-1}\partial_if_i^{-1}$$, it is easy to see that $$f_\bullet : (C_\bullet,\partial) \to (\mathbb{Z}^{r_\bullet},d_\bullet)$$ is a chain complex isomorphism. Hence we can reduce this to the case

$$0 \to \mathbb{Z}^{r_n} \to \cdots \to \mathbb{Z}^{r_1} \to \mathbb{Z}^{r_0} \to 0.$$

Note that by a rank argument, we will have $$\bigoplus_{\text{k odd}}\mathbb{Z}^{r_k} \simeq \bigoplus_{\text{k even}}\mathbb{Z}^{r_k}$$ if and only if $$\sum_{\text{k odd}}{r_k}$$ = $$\sum_{\text{k even}}{r_k}$$. This is equivalent to having that

$$\chi(C_\bullet) = \chi(\mathbb{Z}^{r_\bullet}) = \sum_{k \geq 0}(-1)^k \operatorname{rk}\mathbb{Z}^{r_k} = \sum_{k \geq 0}(-1)^k r_k = 0.$$

As said in the comments, it is a well known fact that the alternating sum of ranges of each $$C_i$$ coincides with the alternating sum of the ranges of each $$H_i(C_\bullet)$$. Thus, what we are being asked to prove is equivalent to proving $$\sum_{k \geq 0}(-1)^k \operatorname{rk}{H}_k(C_\bullet) = 0$$, and the latter follows from the fact that each homology group is of torsion.

• Hola Guido, yo creo que $d_i=f_{i-1}\partial_i f_i^{-1}$ en vez de lo que pusiste, estoy en lo correcto? La verdad no entiendo por que $\bigoplus_{\text{$k$odd}}\mathbb{Z}^{r_k} \simeq \bigoplus_{\text{$k$even}}\mathbb{Z}^{r_k}$ if and only if $\sum_{\text{$k$odd}}{r_k}$ = $\sum_{\text{$k$even}}{r_k}$, tienes algun argumento para probar esto? por que esto es verdad? May 19, 2019 at 1:21
• You're right about the first equality, my bad. I have fixed it. As for the second fact, this is the same argument one makes when proving that two $\mathbb{R}$-vector spaces are isomorphic iff they have the same dimension (cont.) May 19, 2019 at 1:31
• Note that $\bigoplus_{j}\mathbb{Z}^{r_j} = \mathbb{Z}^{(\sum_j r_j)}$, we simply map each $e_i$ in the direct sum to $e_i$ in the latter. Now, in general $\mathbb{Z}^r \simeq \mathbb{Z}^s$ iff $d = s$. Sufficiency is evident as in that case, we have equality. As for necessity, if these modules are isomorphic, then their ranks coincide, which is precisely to say that $r = s$. (I've tried to keep this in English as the site rules mandate, but feel free to ask if anything is not clear, conceptually or otherwise). May 19, 2019 at 1:34
• To expand on the equality of the last comment: an explicit map is to send the $i$-th canonical vector of $\mathbb{Z}^{r_j}$ to the $(\sum_{k=1}^{j-1}r_k + i)$-canonical vector of $\mathbb{Z}^{\sum_j r_j}$. This induces a map into the latter from the direct sum, which is bijective. May 19, 2019 at 1:38
• Thank you very much for being so explicit, I value that much, I only have one last question and that is because if the homology group is torsion, then why $\sum_{k \geq 0}(-1)^k \operatorname{rk}{H}_k(C_\bullet) = 0$? (Do you have social networks so we can talk better and solve doubts?) May 19, 2019 at 3:44