User:Jack

The Join of the Boundaries of two Simplicial Complexes

Lemma If m is any natural-number or zero, then \[ \mathbf{S^0}*\mathbf{S^m} \] is homeomorphic to \[ \mathbf{S^{m+1}} \]

Proof By the definition of Join and Cone in Armstrong, the join of a space X with a space Y is precisely the union of all cones with base X and vertex in Y.

So, the join of the m-sphere and 0-sphere is, by definition, {cone with base m-sphere, apex a point x}U{cone with base m-sphere, apex some other point y} as long as we choose x and y such that the two cones do not intersect except, of course, at their common base the m-sphere. This space, in turn, can simply be written, by the definition of the cone, as

\[ \mathbf{S^0}*\mathbf{S^m} = ({\mathbf{S^m}\times\mathbf{I}})/[({\mathbf{S^m}}\times{\mathbf{0}})\cup({\mathbf{S^m}}\times{\mathbf{1}})] = ({\partial\mathbf{B^{m+1}}\times\mathbf{I}})/[({\mathbf{S^m}}\times{\mathbf{0}})\cup({\mathbf{S^m}}\times{\mathbf{1}})] \]

However, \[ {\partial\mathbf{B^{m+1}}\times\mathbf{I}} \] can be rewritten as

\[ \partial(\mathbf{B^{m+1}}\times\mathbf{I}) - [(int(\mathbf{B^{m+1}})\times{0})\cup(int(\mathbf{B^{m+1}})\times{1})] \]

Because the boundary of [(m+1-ball)XI] is the boundary of the ball times I, together with the interior of the ball times the endpoints of I, 0 and 1.

So,

\[ \mathbf{S^0}*\mathbf{S^m} = [\partial(\mathbf{B^{m+1}}\times\mathbf{I}) - [(int(\mathbf{B^{m+1}})\times{1})\cup(int(\mathbf{B^{m+1}})\times{0})]]/[({\mathbf{B^{m+1}}}\times{\mathbf{0}})\cup({\mathbf{B^{m+1}}}\times{\mathbf{1}})] \]

This can be simplified to,

\[ \mathbf{S^0}*\mathbf{S^m} = [\partial(\mathbf{B^{m+1}}\times\mathbf{I})]/[({\mathbf{B^{m+1}}}\times{\mathbf{0}})\cup({\mathbf{B^{m+1}}}\times{\mathbf{1}})] \]

since our equivalence relation collapses int(m+1-ball)X{1} down to a point along with (m+1-ball)X{1}, and same for int(m+1-ball)X{0}

Now, \[ {\mathbf{B^{m+1}}\times\mathbf{I}} \]

is homeomorphic to

\[ \mathbf{B^{m+2}} \]

so,

\[ \partial(\mathbf{B^{m+1}}\times\mathbf{I}) = \partial\mathbf{B^{m+2}} = \mathbf{S^{m+1}} \]

\[ \mathbf{S^0}*\mathbf{S^m} = [\partial(\mathbf{B^{m+1}}\times\mathbf{I})]/[({\mathbf{B^{m+1}}}\times{\mathbf{0}})\cup({\mathbf{B^{m+1}}}\times{\mathbf{1}})]\] is homeomorphic to \[ \mathbf{S^{m+1}}/\mathbf{R} \]

Where R is composed of two disjoint homeomorphic copies of the m-ball within the (m+1)-sphere.

However, because each of these disjoint copies of the m-ball are obviously simply-connected, the (m+1)-sphere modulo R is simply still equivalent to the (m+1)-sphere, if we model the (m+1)-sphere, say, using a triangulation, and simply take the equivalence relation to be the collapsing of two different k-simplex faces to points.

And so, we have proven that

\[\mathbf{S^0}*\mathbf{S^m} = \mathbf{S^{m+1}} \]

From this, it is easily proven by induction that

Corollary of Lemma The m-sphere is the join of (m+1) copies of the 0-sphere

Using induction either from this lemma or its corollary above, it is then easily shown that:

'THEOREM' \[ \mathbf{S^m}*\mathbf{S^n} = \mathbf{S^{m+n+1}} \]

How do we show this using induction? Well, by our above corollary, \[ \mathbf{S^m}*\mathbf{S^n} = \] The join of m+1 and n+1, or (m+n+2) 0-Spheres. But also by our above results, the join of (m+n+2) 0-spheres is simply \[ \mathbf{S^{m+n+1}} \]

Now, I would very much like to find a theorem concerning the expression of joins of the realizations of two complexes, or perhaps of chains in-general. Of course, in order to do this, it will first be necessary to rigorously establish a definition of "join" in a combinatorial-sense, on chains, which is completely consistent with the point-set definition.