Questions tagged [betti-numbers]

This tag is for questions about Betti numbers. In algebraic topology, the Betti numbers are used to distinguish topological spaces based on the connectivity of n-dimensional simplicial complexes.

56 questions
10
votes
2 answers

Euler characteristic: dependence on coefficients

Let $X$ be a finite CW complex and $\chi(X)$ its Euler characteristic (defined using integer coefficients). When is it true that $\chi(X)=\sum (-1)^i \dim H_i(X;F)$, where $F$ is a field? I thought it would be true for all fields, but I noticed that…
7
votes
1 answer

What is the Betti number of a group?

I'm studying the Fundamental Theorem of finitely generated Abelian group, and it says that the number of factors equal to $\mathbb Z$ (textbook says it is the Betti number of the group) is unique up to isomorphism. So what is "the number of…
7
votes
2 answers

Betti numbers of complex "sphere"

Let $X$ be the set of solutions to $x_1^2+\ldots+x_n^2=1$ in $\mathbb{C}^n$. This has real dimension $2(n-1)$, but since $X$ is an affine algebraic variety, the only possible non-zero topological Betti numbers of $X$ are $b_0,\ldots,b_{n-1}$. What…
User
  • 742
  • 3
  • 10
6
votes
1 answer

Question about the Betti numbers

Definition of Betti number at http://en.wikipedia.org/wiki/Betti_number The $n^{th}$ Betti number represents the rank of the $n^{th}$ homology group, denoted $H_n$ "which tells us the maximum amount of cuts that can be made before separating a…
6
votes
1 answer

Hodge numbers of singular varieties

I just realized that Hodge numbers can be defined for every $\mathbb C$-variety, not only the smooth proper ones. At least we can define them using the Grothendieck ring $K_0(\text{Var}/\mathbb C)$ since it is generated by smooth projective…
6
votes
1 answer

Why does the Betti number give the measure of k-dimensional holes?

I was reading Paul Renteln "MANIFOLDS, TENSORS, AND FORMS An Introduction for Mathematicians and Physicists" p.145, where he defined the Betti number as $dim H_m(K)$, where $H_m(K)$ is the quotient space of cycles modulo boundary $Z_m(K)/B_m(K)$. He…
4
votes
1 answer

Poincaré's take on Poincaré duality before the advent of cohomology?

Poincaré duality was first stated, without proof, by Henri Poincaré in 1893. It was stated in terms of Betti numbers: The $k$th and $n-k$th Betti numbers, $b_k$ and $b_{n-k}$ of a closed orientable n-manifold are equal. $$b_k = b_{n-k}.$$ From…
4
votes
1 answer

on the definition of graded Betti numbers

Let's use as reference the slides 19-31. Let $S=k[x_1,\cdots,x_n]$ and $M$ a finitely generated graded $S$-module. Then by Hilbert's Syzygy Theorem, $M$ has a minimal, graded, free resolution of length at most $n+1$, i.e. $0 \rightarrow F_m…
Manos
  • 24,205
  • 4
  • 51
  • 145
4
votes
1 answer

Betti numbers of connected sum of real projective spaces

I know that $\beta_{1}(\sharp_{h}\mathbb{RP}^{2})=h-1$. Also it is clear: $$\beta_{i}(\sharp_{h}\mathbb{RP}^{2n})=0 \mbox{ for }0
Michael jordan
  • 324
  • 2
  • 8
4
votes
1 answer

Closed oriented manifold with middle Betti is one with odd degree.

The rational cohomology ring of complex projective plane $\mathbb{CP}^{2}$ is truncated polynomial ring $\frac{\mathbb{Q}[X]}{(X)^{3}},\,\,deg(X)=2$. In this case, the degree of a generator is 2. Is there any closed oriented 2m-manifold with the…
4
votes
1 answer

calculate Betti numbers of a specific polynomial variety

My question is: I am interested in calculating the Betti numbers of a specific polynomial variety (w.r.t. singular cohomology) whose zeros I am looking at over $\mathbb{C}$ (it has integer coefficients). I know the polynomial explicitly. Please…
Jyothi
  • 87
  • 4
4
votes
1 answer

A compact flat manifold whose first Betti number is equal to the dimension is a flat torus

I know the following to be true: If $(M,g)$ is a compact flat Riemannian manifold whose first Betti number ($= \dim H_{dR}^1(M)$) is equal to the dimension, then it is (isometric to) a flat torus. In the notes I am following it is stated that this…
4
votes
0 answers

Do manifolds have the homotopy type of finite-dimensional CW-complexes?

It is well know that a topological manifold $M$ is homotopy equivalent to a CW-complex $X$. In addition, if $M$ is compact, one even has $M \simeq X$ for some finite CW-complex. In a Mathoverflow thread, it was discussed if one can take an…
4
votes
1 answer

Betti Numbers with coefficients in reals, rationals & integers.

One knows from the Universal Coefficient Theorem that Integral Homology can be used to derive homology with coefficients in any other groups like e.g. Reals, Rationals, Z/2Z etc. Suppose you have access to Homology group computed over real…
3
votes
0 answers

Graded Betti Numbers of a Graded Ideal with Linear Quotients

Exercise 8.8(a) in Monomial Ideals by Herzog and Hibi: Let $I\subset S=K[x_{1},...,x_{n}]$ be a graded ideal which has linear quotients with respect to a homogeneous system of generators $f_{1},...,f_{m}$ with $\deg f_{1}\leq\cdots\leq\deg f_{m}$.…
1
2 3 4