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.

# Questions tagged [betti-numbers]

56 questions

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### 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…

Angry Euler

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### 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…

Math-Nerd

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### 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

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### 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…

Vrouvrou

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### 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…

Akatsuki

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### 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…

Joe Martin

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### 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…

wonderich

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### 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

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### 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

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Michael jordan

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### 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…

David Silva

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### 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

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### 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…

Alessio Di Lorenzo

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### 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…

abenthy

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### 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…

Brian D'Souza

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### 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 $\deg f_{1}\leq\cdots\leq\deg f_{m}$.…

user118827

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