Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I've been wondering why the boundary of a simplex $\sigma : C_q (X) \rightarrow C_{q-1}(X)$ is defined to be $\partial \sigma = \sum (-1)^i \sigma \circ f_{i,q}$ with alternating sign.

Why can it not be the sum over all faces without alternating sign? Thanks for your help!

share|cite|improve this question
You could of course get rid of the $(-1)^i$, but then you'd have to change the parametrizations $f_{i,q}$ to get the right boundary orientation. If you know some basic manifold theory, consider the simplex to be a manifold (ignore the corners for now) and compute the orientation-form for the boundary simplices. Then, using the parametrization of the boundary simplicies given by $f_{i,q}$, determine how the manifold-theoretic boundary orientation compares to the $f_{i,q}$ orientation. You'll get your formula. – Ryan Budney Jul 31 '11 at 17:12
What do you mean by the parametrizations of $f_{i,q}$? I know nothing about manifolds yet unfortunately. – Rudy the Reindeer Aug 4 '11 at 7:59
All the definitions in simplicial homology were inspired by corresponding notions among manifolds. In particular, if you have a manifold equipped with an orientation, there is a canonical orientation associated to its boundary. Orientations can also be assigned to objects via explicit parametrization by other oriented objects. The formula you're interested in is a combination of these two things. I suggest reading a book on manifolds that covers the general Stokes theorem (using differential forms). Once you understand the proof of the theorem, you'll be ready to reinterpret that formula. – Ryan Budney Aug 4 '11 at 19:02
up vote 2 down vote accepted

Because we want to have $\partial\circ\partial=0$ to built homology.

share|cite|improve this answer
OK, and what about having coefficients in $\mathbb{Z}$? That work, too, right? – Rudy the Reindeer Jul 31 '11 at 11:49
Cources start with $\mathbb{Z}$ but later they extend to any abelian group $G$ or generalize more. You say then that it is "homology with coeffitients" if it is not necessarily $\mathbb{Z}$. – old Jul 31 '11 at 11:58
But I meant in the definition of the boundary operator. Homology with coefficients refers to the coefficients of a chain not the coefficients in the definition of the boundary... – Rudy the Reindeer Jul 31 '11 at 13:18

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.