7

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?

Joe Martin
  • 348

1 Answers1

6

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.

Ethan Bolker
  • 103,433