I want to find out the euler characteristics of the connected sum of $2n$-dimensional tori. Since we know that the euler characteristic of any odd-dimensional closed manifolds is $0$.
My Attempt:
The rank of $D_p(\mathbb{T}^{2n})$ is $\binom{n}{p}$, so we get the following chain complex with respect to $X=\mathbb{T}^{2n}$: $$0\to D_{2n}(X)\to D_{2n-1}(X)\to...\to D_1(X)\to D_0(X)\to 0$$ Then, consider the boundary operator $\partial_{p+1}:D_{p+1}(X)\to D_p(X)$ which sends every $p+1$-cell to $0$ because $\partial_{p+1}(p+1\text{-cell})=a_1+a_2+...+a_m-a_1-a_2-...-a_m=0$. which implies that $B_p(X)=\text{im}(\partial_p)=0$ and thus $Z_{p-1}(X)=D_{p-1}(X)$.
So, we conclude that $H_p(X)=D_p(X) \cong \bigoplus_{i=1}^{\binom{2n}{p}}\mathbb{Z}$.
Consider the space $\#_{j=1}^k\mathbb{T}^{2n}$ obtained by connecting finitely many $2n$-tori together. Then, we proceed by induction:
Consider the reduced homology exact sequence of a pair $(\mathbb{T}^{2n}\#\mathbb{T}^{2n},\mathbb{S}^{2n-1})$: $$...\to \tilde{H}_p(\mathbb{S}^{2n-1})\to\tilde{H}_p(\mathbb{T}^{2n}\#\mathbb{T}^{2n})\to\tilde{H}_p(\mathbb{T}^{2n}\vee\mathbb{T}^{2n})\to\tilde{H}_{p-1}(\mathbb{S}^{2n-1})\to...$$ Then, by direct computation on the reduced version of Mayer-Vietoris Sequence with respect to $\mathbb{T}^{2n}\vee\mathbb{T}^{2n}$, we can rewrite the reduced homology exact sequence as $$...\to \tilde{H}_p(\mathbb{S}^{2n-1})\to\tilde{H}_p(\mathbb{T}^{2n}\#\mathbb{T}^{2n})\to\tilde{H}_p(\mathbb{T}^{2n})\oplus\tilde{H}_p(\mathbb{T}^{2n})\to\tilde{H}_{p-1}(\mathbb{S}^{2n-1})\to...$$
If $0<p<2n-1$, then $\tilde{H}_p(\mathbb{T}^{2n}\#\mathbb{T}^{2n})\cong\tilde{H}_p(\mathbb{T}^{2n})\oplus\tilde{H}_p(\mathbb{T}^{2n})\cong\bigoplus_{i=1}^{2\binom{2n}{p}}\mathbb{Z}$.
If $p=0$, then $H_0(\mathbb{T}^{2n}\#\mathbb{T}^{2n})=\tilde{H}_0(\mathbb{T}^{2n}\#\mathbb{T}^{2n})\oplus\mathbb{Z}$.
If $p=2n-1$, then $\tilde{H}_{2n-1}(\mathbb{T}^{2n}\#\mathbb{T}^{2n})\cong\tilde{H}_{2n-1}(\mathbb{T}^{2n})\oplus\tilde{H}_{2n-1}(\mathbb{T}^{2n})\cong\bigoplus_{i=1}^{2\binom{2n}{2n-1}}\mathbb{Z}$ because both of the connected summands are compact orientable.
If $p=2n$, then $\tilde{H}_{2n-1}(\mathbb{T}^{2n}\#\mathbb{T}^{2n})\cong\mathbb{Z}$ because both of the connected summands are compact orientable.
Inductively, the homology groups for the connected sum of finitely many summands are: $$H_p(\#_j^k\mathbb{T}^{2n})= \begin{cases} \bigoplus_{i=1}^{k\binom{2n}{p}}\mathbb{Z} & 0<p\le2n-1\\ \mathbb{Z} & p=0\text{ or }p=2n\\ 0 & \text{ otherwise } \end{cases} $$ which implies that the Betti numbers are: $$\beta_p(\#_j^k\mathbb{T}^{2n})= \begin{cases} k\binom{2n}{p} & 0<p\le2n-1\\ 1 & p=0\text{ or }p=2n\\ 0 & \text{ otherwise } \end{cases} $$
Let $Y=\#_{j=1}^k\Bbb{T}^{2n}$. From the book Elements of Algebraic Topology written by Munkres, we know that $\chi(M^n)=\sum_{p=0}^n(-1)^p\beta_p$.
In our case, $\chi(Y)=\sum_{p=1}^{2n-1}(-1)^p\beta_p+2=k\sum_{p=1}^{2n-1}(-1)^p\binom{2n}{p}+2=-2k+2$.
- The euler characteristic is independent to dimensions!? so there must be some errors in my calculation, could someone help me please? (already solved by @William)
Edit:
According to William's comment, my result makes sense when we generalize the case of $2n=2$.
But still, I would like to observe this question from the following perspective (Q2) which I have no idea of how to start.
2. The wikipedia of Euler characteristics states that the case for orientable closed manifolds is a corollary of Poincare Duality. So I wonder how to use Poincare duality which relates homology with cohomology to deduce the same result. Any hint/suggestions? Thank you very much.