Talk:Alexander–Spanier cohomology

De Rahm vs. Alexander-Spanier
If the article admits that the description is not generally called "Alexander-Spanier" but rather "De Rahm cohomology with compact support," then why isn't the page titled the latter? This page could then give the more general case. The current situation seems decidedly inane. —Preceding unsigned comment added by 67.190.102.81 (talk) 01:03, 17 February 2010 (UTC)

Formula error?
The article defines the differential as


 * $$df(x_0,\ldots,x_p)= \sum_i(-1)^if(x_0,\ldots,x_{i-1},x_{i+1},\ldots,x_p).$$

It seems to me, just based on counting arguments, that the formula should be:


 * $$df(x_0,\ldots,x_p)= \sum_i(-1)^if(x_0,\ldots,x_{i-1},x_{i+1},\ldots,x_{p+1}).$$

The function f on the left hand side has p+1 arguments, therefore so should f in the sum on the right, and we are going to a space of one higher dimension.

However, I don't know the subject well enough to make this change without confirmation by someone more knowledgeable.--agr (talk) 00:57, 18 January 2022 (UTC)
 * I think there may be a mild mistake in the article, but not where you have it. The function $$df$$ is in $$C^{p+1}$$, so should take $$p+2$$ arguments.  It is calculated by an alternating sum of functions taking $$p+1$$ arguments.  So the formula given is actually from $$C^{p-1} \to C^p$$, although this is unlikely to cause much confusion.  (I'm familiar with cohomology in general, but not this specific topic, so there's a chance that I'm missing something.) Russ Woodroofe (talk) 13:04, 18 January 2022 (UTC)


 * agr: The left side function is not f but df.  There is no $$x_{p+1}.$$
 * Russ Woodroofe: I think you are right, so that's two of us; I will make this change.  If any true expert thinks it's wrong, they can revert it.  Zaslav (talk) 18:55, 18 January 2022 (UTC)
 * That clears things up, thanks.--agr (talk) 19:16, 18 January 2022 (UTC)