SlipEternal
Junior Member
- Joined
- Jan 4, 2012
- Messages
- 114
I am pretty sure I understand this, but I want to make sure.
Let \(\displaystyle X\) be the \(\displaystyle \Delta\)-complex obtained from \(\displaystyle \Delta^n\) by identifying all faces of the same dimension. Thus \(\displaystyle X\) has a single \(\displaystyle k\)-simplex for all \(\displaystyle k\le n\). If I am to compute the homology groups, I begin with the following:
\(\displaystyle \displaystyle \text{Ker}(\partial_k) = \left\{\begin{array}{cc} \Delta^k & \text{ if }k\text{ is even}\\ 0 & \text{ if }k\text{ is odd}\end{array}\right.\)
\(\displaystyle \displaystyle \text{Im}(\partial_{k+1}) = \left\{\begin{array}{cc} \Delta^k & \text{ if }k\text{ is even}\\ 0 & \text{ if }k\text{ is odd}\end{array}\right.\)
Thus \(\displaystyle H_k(X) = 0\) for all \(\displaystyle k\in \mathbb{Z}^+\) (nonnegative integers).
Or am I missing something?
Let \(\displaystyle X\) be the \(\displaystyle \Delta\)-complex obtained from \(\displaystyle \Delta^n\) by identifying all faces of the same dimension. Thus \(\displaystyle X\) has a single \(\displaystyle k\)-simplex for all \(\displaystyle k\le n\). If I am to compute the homology groups, I begin with the following:
\(\displaystyle \displaystyle \text{Ker}(\partial_k) = \left\{\begin{array}{cc} \Delta^k & \text{ if }k\text{ is even}\\ 0 & \text{ if }k\text{ is odd}\end{array}\right.\)
\(\displaystyle \displaystyle \text{Im}(\partial_{k+1}) = \left\{\begin{array}{cc} \Delta^k & \text{ if }k\text{ is even}\\ 0 & \text{ if }k\text{ is odd}\end{array}\right.\)
Thus \(\displaystyle H_k(X) = 0\) for all \(\displaystyle k\in \mathbb{Z}^+\) (nonnegative integers).
Or am I missing something?