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

Euler number in terms of Betti numbersHow does one show that definition of Betti number and its “informal definition” are equal?What is the Betti number of a group?Question about the Betti numbersHomology class and Betti number for a compact manifold with boundariesTwo ways to split the second Betti numberAn example of infinite Betti number refers to G-covering, and the motivation of G-coveringThe Betti number of complex projective spacesWhat are the Betti numbers of a double pinched torus?The second Betti number of a group

5

2

$begingroup$

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 said it is true that the Betti number counts the number of m-dimensional holes on K, but I don't see why.

This is similar to what Wikipedia said at https://en.wikipedia.org/wiki/Simplicial_homology, the last sentence of the definition section. Can anyone provide me with an intuitive explanation of what is going on?

algebraic-topology simplicial-complex betti-numbers

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asked 4 hours ago

Joe MartinJoe Martin

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5

2

$begingroup$

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 said it is true that the Betti number counts the number of m-dimensional holes on K, but I don't see why.

This is similar to what Wikipedia said at https://en.wikipedia.org/wiki/Simplicial_homology, the last sentence of the definition section. Can anyone provide me with an intuitive explanation of what is going on?

algebraic-topology simplicial-complex betti-numbers

share|cite|improve this question

asked 4 hours ago

Joe MartinJoe Martin

987

New contributor

Joe Martin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

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 said it is true that the Betti number counts the number of m-dimensional holes on K, but I don't see why.

This is similar to what Wikipedia said at https://en.wikipedia.org/wiki/Simplicial_homology, the last sentence of the definition section. Can anyone provide me with an intuitive explanation of what is going on?

algebraic-topology simplicial-complex betti-numbers

share|cite|improve this question

asked 4 hours ago

Joe MartinJoe Martin

987

New contributor

Joe Martin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

asked 4 hours ago

Joe MartinJoe Martin

987

New contributor

Joe Martin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

asked 4 hours ago

Joe MartinJoe Martin

987

New contributor

Joe Martin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 4 hours ago

Joe MartinJoe Martin

987

New contributor

Joe Martin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 4 hours ago

Joe MartinJoe Martin

987

1 Answer
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4

$begingroup$

This sentence in the paragraph just before the line from wikipedia you quote

It follows that the homology group Hk(S) is nonzero exactly when there
are k-cycles on S which are not boundaries. In a sense, this means
that there are k-dimensional holes in the complex.

provides an answer to your question. A $k$-cycle is (intuitively speaking) a geometric structure which potentially surrounds a $k+1$-dimensional region. $1$-cycles are easiest to visualize - think loops. On the surface of a sphere every loop can be thought of as having an inside (two, in fact, since there's no topological distinction between inside and outside). Every $1$-cycle is the boundary of a part of the surface and the first Betti number is $0$. There are no holes. On the surface of a torus there are two kinds of loops that don't have interiors - those are $1$-cycles that aren't boundaries, so a torus has $2$ "holes" where there might be disks but aren't. Those are conceptual holes - there is no hole that looks as if was cut out by a cookie cutter.

In higher dimensions you can ask whether something that looks like the surface of an ordinary sphere - a $2$-cycle - actually surrounds something like the interior of a sphere. If not, that $2$-cycle contributes to the second Betti number.

I hope this serves as the intuition you need. The technical details are, of course, technical.

share|cite|improve this answer

edited 3 hours ago

answered 4 hours ago

Ethan BolkerEthan Bolker

47.1k555123

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add a comment | 

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4

$begingroup$

This sentence in the paragraph just before the line from wikipedia you quote

It follows that the homology group Hk(S) is nonzero exactly when there
are k-cycles on S which are not boundaries. In a sense, this means
that there are k-dimensional holes in the complex.

provides an answer to your question. A $k$-cycle is (intuitively speaking) a geometric structure which potentially surrounds a $k+1$-dimensional region. $1$-cycles are easiest to visualize - think loops. On the surface of a sphere every loop can be thought of as having an inside (two, in fact, since there's no topological distinction between inside and outside). Every $1$-cycle is the boundary of a part of the surface and the first Betti number is $0$. There are no holes. On the surface of a torus there are two kinds of loops that don't have interiors - those are $1$-cycles that aren't boundaries, so a torus has $2$ "holes" where there might be disks but aren't. Those are conceptual holes - there is no hole that looks as if was cut out by a cookie cutter.

In higher dimensions you can ask whether something that looks like the surface of an ordinary sphere - a $2$-cycle - actually surrounds something like the interior of a sphere. If not, that $2$-cycle contributes to the second Betti number.

I hope this serves as the intuition you need. The technical details are, of course, technical.

share|cite|improve this answer

edited 3 hours ago

answered 4 hours ago

Ethan BolkerEthan Bolker

47.1k555123

$endgroup$

add a comment | 

4

$begingroup$

This sentence in the paragraph just before the line from wikipedia you quote

It follows that the homology group Hk(S) is nonzero exactly when there
are k-cycles on S which are not boundaries. In a sense, this means
that there are k-dimensional holes in the complex.

provides an answer to your question. A $k$-cycle is (intuitively speaking) a geometric structure which potentially surrounds a $k+1$-dimensional region. $1$-cycles are easiest to visualize - think loops. On the surface of a sphere every loop can be thought of as having an inside (two, in fact, since there's no topological distinction between inside and outside). Every $1$-cycle is the boundary of a part of the surface and the first Betti number is $0$. There are no holes. On the surface of a torus there are two kinds of loops that don't have interiors - those are $1$-cycles that aren't boundaries, so a torus has $2$ "holes" where there might be disks but aren't. Those are conceptual holes - there is no hole that looks as if was cut out by a cookie cutter.

In higher dimensions you can ask whether something that looks like the surface of an ordinary sphere - a $2$-cycle - actually surrounds something like the interior of a sphere. If not, that $2$-cycle contributes to the second Betti number.

I hope this serves as the intuition you need. The technical details are, of course, technical.

share|cite|improve this answer

edited 3 hours ago

answered 4 hours ago

Ethan BolkerEthan Bolker

47.1k555123

$endgroup$

add a comment | 

$begingroup$

This sentence in the paragraph just before the line from wikipedia you quote

It follows that the homology group Hk(S) is nonzero exactly when there
are k-cycles on S which are not boundaries. In a sense, this means
that there are k-dimensional holes in the complex.

provides an answer to your question. A $k$-cycle is (intuitively speaking) a geometric structure which potentially surrounds a $k+1$-dimensional region. $1$-cycles are easiest to visualize - think loops. On the surface of a sphere every loop can be thought of as having an inside (two, in fact, since there's no topological distinction between inside and outside). Every $1$-cycle is the boundary of a part of the surface and the first Betti number is $0$. There are no holes. On the surface of a torus there are two kinds of loops that don't have interiors - those are $1$-cycles that aren't boundaries, so a torus has $2$ "holes" where there might be disks but aren't. Those are conceptual holes - there is no hole that looks as if was cut out by a cookie cutter.

In higher dimensions you can ask whether something that looks like the surface of an ordinary sphere - a $2$-cycle - actually surrounds something like the interior of a sphere. If not, that $2$-cycle contributes to the second Betti number.

I hope this serves as the intuition you need. The technical details are, of course, technical.

share|cite|improve this answer

edited 3 hours ago

answered 4 hours ago

Ethan BolkerEthan Bolker

47.1k555123

$endgroup$

share|cite|improve this answer

edited 3 hours ago

answered 4 hours ago

Ethan BolkerEthan Bolker

47.1k555123

answered 4 hours ago

Ethan BolkerEthan Bolker

47.1k555123

answered 4 hours ago

Ethan BolkerEthan Bolker

47.1k555123