1
$\begingroup$

I am very sorry but apparently I am really weak in cohomology flavored questions. I try to reformulate my problem in a very simple and hopefully clear way. This question is related with a problem in hyperplane arrangement theory, too.

Let $d,h\geq1$ be integer numbers and let $F_{1},\ldots,F_{h}$ be elements of the polynomial ring $\mathbb{C}\left[t_{1},\ldots,t_{d}\right].$ For $1\leq j\leq h$ let $X_{j}=\lbrace P\in\mathbb{C}^{d}\mid F_{j}(P)=0\rbrace$ where $F_{j}(P)$ denotes the evaluation of the polynomial $F_{j}$ at the point $P.$ Finally, set $X=\bigcap_{1\leq j\leq d}X_{j}.$

$\textbf{Central hypothesis:}$ Here everything has the classical topology.

Let $H_{k}(X)$ be the singular homology groups of $X.$

$\textbf{Question:}$ Is there any explicit formula which relate $H_{0}(X),$ i.e., the number of connected components of $X,$ with the polynomials $F_{1},\ldots,F_{h}$ or the spaces $X_{j}=\lbrace P\in\mathbb{C}^{d}\mid F_{j}(P)=0 \rbrace$?

Improve this question
$\endgroup$
0

1 Answer 1

Reset to default
2
$\begingroup$

Rather than a closed formula, which in my opinion would be too optimistic, you have algorithms. In the following paper:

Bürgisser, Peter(D-PDRB); Scheiblechner, Peter(1-PURD) On the complexity of counting components of algebraic varieties. (English summary) J. Symbolic Comput. 44 (2009), no. 9, 1114–1136.

that you can find at

http://www.scheiblechner.ch/files/count-comp.pdf

you can find a nice one.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Thank you for the suggestion. For sure, I also think that a closed formula is too optimistic. However, in my case I am already interested in some low dimensional explicit computation, too. So that, your suggestion is really appropriate. Thank you! $\endgroup$
    – snaleimath
    Commented May 8, 2015 at 11:31

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .