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Why does the Betti number give the measure of k-dimensional holes?
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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
asked 4 hours ago
Joe MartinJoe Martin
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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
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
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.
$endgroup$
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
algebraic-topology simplicial-complex betti-numbers
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
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
asked 4 hours ago
Joe MartinJoe Martin
987
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.
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.
Joe Martin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
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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.
answered 3 hours ago
Ethan BolkerEthan Bolker
47.1k555123
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$begingroup$
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.
answered 3 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.
answered 3 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.
answered 3 hours ago
Ethan BolkerEthan Bolker
47.1k555123
$endgroup$
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.
answered 3 hours ago
Ethan BolkerEthan Bolker
47.1k555123
edited 3 hours ago
answered 3 hours ago
Ethan BolkerEthan Bolker
47.1k555123
answered 3 hours ago
Ethan BolkerEthan Bolker
47.1k555123
answered 3 hours ago
Ethan BolkerEthan Bolker
47.1k555123
47.1k555123
add a comment |
add a comment |
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