4
$\begingroup$

Using the (trivial) CW-complex structure of $\Delta_n$, I would like to compute the homology groups of $\Delta_n$. It's obvious that (for $k \leq n$) $C_k \simeq \mathbb{Z}^{{n+1 \choose k+1}}$, and that $im \delta_{k+1} \subset ker \delta_k$. It's the reverse inclusion for $k \geq 1$ that I'm not able to show.

I know how to prove this using homotopy invariance of homology, as $\Delta_n$ is contractible, and a proof in that vein is given in a comment here Explicit calculation of simplicial homology - but I'm wondering if there is something more elementary? The CW-complex structure makes computing the homology groups of the sphere almost trivial, and I was hoping that it would be easy for the disk as well.

$\endgroup$
1
$\begingroup$

You should remember that it is difficult to construct cellular homology without singular homology (you must figure out how to define degrees of maps of spheres) and then it is even harder to show it is even a homeomorphism invariant.

However, if you wish to sweep these things under the rug: the n-simplex is homeomorphic to the n-disk which has a cell structure given by a n-cell attached along the identity to the (n-1)-sphere. The image of the n-cell under the boundary map is the (n-1)-cell, so the n-1 homology is trivial (unless n-1=0 when degrees are tricky to define). The n homology is also trivial because the boundary map is an isomorphism here. The 0 homology is $\mathbb{Z}$ since the boundary map into it is $0$ (if $n>1$ because there are no 1-cells and if $n=1$ because the orientations of the boundary points of the circle cancel each other out).

Thus it has the homology of a point.

$\endgroup$
1
$\begingroup$

It's quite easy in simplicial homology to write down a contracting homotopy for the identity map of $C(\Delta_n)$. Let the vertices of $\Delta_n$ be labelled $0,1,\ldots,n$. Then the $k$-chains are labelled by tuples $(j_0,\ldots,j_k)$ where $0\le j_0<j_1<\cdots< j_k\le n$. Define $T:C_k(\Delta_n)\to C_{k+1}(\Delta_n)$ by $T:(j_0,\ldots,j_k)\mapsto(0,j_0,\ldots,j_k)$ if $j_0\ge1$ and $T:(j_0,\ldots,j_k)\mapsto0$ if $j_0=0$. If then $\delta_k(\gamma)=0$ for $\gamma\in C_k$ (with $k\ge1$) then $\gamma=\pm\delta_{k+1}(T(\gamma))$ (I can't be bothered to check the sign), so $\ker\delta_k\subseteq\mathrm{im}\,\delta_{k+1}$.

$\endgroup$

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.