Talk:Supplement I (Ex)
(Created page with "<wikitex>; Before stating the exercise, we recall the definition of $\Sigma^m$, and explain why it is indeed a simplicial complex. {{beginrem|Definition}} The simplicial comp...")
Latest revision as of 01:15, 1 June 2012
Before stating the exercise, we recall the definition of , and explain why it is indeed a simplicial complex.
Definition 0.1. The simplicial complex is defined to have -simplices equal to the set of -simplices of , denoted by for . The face relations are given by iff .
Lemma 0.2. This is a simplicial complex.
Proof. The following is an equivalent definition of :
- Its vertices are the -dimensional faces in .
- A collection of vertices of -- i.e. -dimensional faces in -- is a simplex of iff the intersecion of all the faces is non-trivial.
This is an equivalent definition because of the following special ('miraculous'!) property of :
- The -simplices are precisely the -fold intersections of -simplices.
This equivalent description makes it clear that is a simplicial complex, since a subcollection of a non-trivially intersecting collection is again non-trivially intersecting.
Remark 0.3. The first description of gives us a 'map' of simplicial complexes
This is of course not a real map of simplicial complexes, since it sends -simplices to -simplices. However, it does induce a map of barycentric subdivisions
which is an isomorphism of simplicial complexes. So we may view subcomplexes of as living in via this isomorphism.
Exercise 0.4. Let be a subcomplex of , and consider the barycentric subdivision as a subcomplex of . Show that
Solution 0.5. First consider what happens if we view everything as living inside :
- The barycentric subdivision is the subcomplex with -simplices with a simplex of for each .
- The dual complex is by definition the subcomplex of whose -simplices are with a simplex of and for each .
- Finally, is the subcomplex with -simplices with for all .
So we clearly have , since we have different conditions and on the LHS and RHS respectively. This is solved by passing from to and replacing by . Considered as subcomplexes of , we have:
- has -simplices with a simplex of for each .
- has -simplices with a simplex of and , equivalently , for all .
- has -simplices with for all .
Hence,
Remark 0.6. Once we know that the dual cell decomposition of is actually a simplicial complex, in other words that is a simplicial complex, then the argument is very simple and formal. The main geometric input to the argument is the special property of mentioned above, that -simplices are precisely the -fold intersections of -simplices.