Simplicial complexes vs. geometric realization of abstract simplicial complexes - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T15:29:15Z http://mathoverflow.net/feeds/question/23713 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/23713/simplicial-complexes-vs-geometric-realization-of-abstract-simplicial-complexes Simplicial complexes vs. geometric realization of abstract simplicial complexes ferret 2010-05-06T14:11:37Z 2010-05-07T03:06:14Z <p>A <em>finite abstract simplicial complex</em> is a pair $D=(S,D)$ where $S$ is a finite set and $D$ is a non-empty subset of the power set of $S$ closed under the subset operation, e.g. $(\{a,b,c\},\{\emptyset,\{a\},\{b\},\{c\},\{a,b\}\})$.</p> <p>For $n\geq 0$ the topological space $\Delta^n=\{(x_0,...,x_n)\in\mathbb{R}^{n+1}\mid x_i\geq 0, \sum x_i =1\}$ is called the <em>standard $n$-simplex</em>. A topological space homeomorphic to the standard $n$-simplex is called an <em>$n$-simplex</em>. For $n\geq 1$ the $n+1$ faces of any $n$-simplex are $n-1$ simplices.</p> <p>A <em>finite topological simplicial complex</em> is a pair $(X,F)$ where $X$ is a topological space and $F=(F_1,...,F_m)$ is a finite sequence of embeddings $F_k:\Delta^{i_k}\to X$ such that</p> <ul> <li>$X=\cup_k F_k(\Delta^{i_k})$</li> <li>$F_k\neq F_l$ if $k\neq l$ </li> <li>for every $1\leq k\leq m$ with $i_k\geq 1$ and for every face $A$ of the $i_k$-simplex $F_k(\Delta^{i_k})$ there is a $1\leq l\leq m$ with $F_l(\Delta^{i_l})=A$</li> <li>for every $1\leq k\neq k'\leq m$ the simplex $F_k(\Delta^{i_k})\cap F_{k'}(\Delta^{i_{k'}})$ is a face of each of them.</li> </ul> <p>I hope, the definitions are correct.</p> <p>There is the notation of a <em>geometric realization</em> of a finite abstract simplicial complex: Let $D=(S,D)$ be a finite abstract simplicial complex. Then choose a total order on $S$, w.l.o.g. $S=\{1,...,M\}$. The colimit of the functor sending an element $\{0,...,n\}$ of the poset $D$ (considered as a category) to $\Delta^n$ is the geometric realization $|D|$ of $D$.</p> <p><s>If I am not mistaken there are finite topological simplicial complexes which are not the geometric realization of a finite abstract simplicial complex. This is because the choice of the total order determines an orientation of the realization. I think the projective plane for example is not in the image of the realization functor.</s></p> <p><s>My question is: Is there a reasonable notation of a geometric realization for abstract simplicial complexes which has exactly the topological simplicial complexes as its image or do I have a wrong understanding somehow?</s></p> <p>I have realized that the original question does not make sense. Please let me ask if this is the right way to understand the situation:</p> <p>A finite triangulation of a space is the same as a "finite topological simplicial complex". Every finite triangulation is the realization of a finite abstract simplicial complex. The realization of a finite abstract simplicial complex comes with a "direction" of each 1-simplex such that the neighbouring edges are pointing in the <strong>same</strong> directions (they are glued together in this way). The triangulation is orientable if and only if one can permute these "directions" of the 1-simplices such that all the neighbouring edges are pointing in <strong>opposite</strong> directions. How can I see that this condition gives the right concept of orientability? Why "opposite"?</p> http://mathoverflow.net/questions/23713/simplicial-complexes-vs-geometric-realization-of-abstract-simplicial-complexes/23718#23718 Answer by Robin Chapman for Simplicial complexes vs. geometric realization of abstract simplicial complexes Robin Chapman 2010-05-06T14:27:40Z 2010-05-06T14:27:40Z <p>I don't follow your comments about the projective plane. Surely the geometric realization of the simplicial complex consisting of $$\{a,b,c\},\{a,c,d\},\{a,d,e\},\{a,e,f\},\{a,f,b\},$$ $$\{b,c,e\},\{c,d,f\},\{d,e,b\},\{e,f,c\},\{f,b,d\}$$ and their subsets is the real projective plane (I got these by identifying antipodal faces of a regular icosahdron).</p> http://mathoverflow.net/questions/23713/simplicial-complexes-vs-geometric-realization-of-abstract-simplicial-complexes/23785#23785 Answer by David Speyer for Simplicial complexes vs. geometric realization of abstract simplicial complexes David Speyer 2010-05-06T21:17:06Z 2010-05-06T21:17:06Z <p><img src="http://www-math.mit.edu/~speyer/Orientation.png" alt="Two compatibly oriented triangles"></p> <p>The image shows two triangle in an oriented surface. An orientation is a compatible choice of "clockwise" everywhere on the surface; observe that both triangles are oriented clockwise. The edges which are glued together point in opposite directions. A triangulated oriented surface looks locally like this picture at each pair of triangles.</p> http://mathoverflow.net/questions/23713/simplicial-complexes-vs-geometric-realization-of-abstract-simplicial-complexes/23807#23807 Answer by Allen Hatcher for Simplicial complexes vs. geometric realization of abstract simplicial complexes Allen Hatcher 2010-05-07T01:46:40Z 2010-05-07T01:46:40Z <p>At least in the realm of topology an abstract simplicial complex is equivalent to a topological (or geometric) simplicial complex, and neither of these two notions involves anything about orienting the simplices or ordering the vertices. If one has a simplicial complex of either type, one can choose a partial ordering of the vertices that restricts to a linear ordering of the vertices of each simplex, and this gives the notion of an ordered simplicial complex. For a simplicial complex that is a manifold (or more generally a pseudo-manifold) one can define an orientation to consist of a choice of orientations of the top-dimensional simplices such that the two induced orientations on each codimension one face are opposite. For a general simplicial complex the only definition of an orientation that I have seen is the trivial one where an orientation is chosen for each simplex with no compatibility assumptions on these orientations.</p> http://mathoverflow.net/questions/23713/simplicial-complexes-vs-geometric-realization-of-abstract-simplicial-complexes/23814#23814 Answer by Peter LeFanu Lumsdaine for Simplicial complexes vs. geometric realization of abstract simplicial complexes Peter LeFanu Lumsdaine 2010-05-07T03:06:14Z 2010-05-07T03:06:14Z <p>To answer just a small part of your question, three small picture-examples that can help see why "edges match up in <strong>opposite</strong> directions" is the right condition for "orientability":</p> <ol> <li><p>Analogy with how vertices match up when you glue paths together along vertices. Each path has one vertex oriented as "source", and another oriented "target". To glue paths together in a consistently oriented way, we do it like this: $\quad \cdot \longrightarrow \cdot \longrightarrow \cdot \quad$ where a "source" vertex is matched to a "target" vertex; in other words, a vertex should have <em>opposite orientations</em> in the edges that it joins. (Matching them in the same orientation would give $\quad \cdot \longrightarrow \cdot \longleftarrow \cdot \qquad$ ...ouch!)</p></li> <li><p>(Working from the picture in David Sayer's example.) Each triangle in his picture has the edges in its boundary oriented clockwise, which makes them match up consistently at their vertices, in the sense of the previous point. When we paste the triangles together along an edge, that edge is no longer part of the boundary of the resulting quadrilateral. So looking at just the resulting boundary, we want that still to be consistently oriented, with each vertex occurring as a source of one edge and a target of another. Having the glued edge matched in opposite directions is what will make that work... contemplate how the vertices of the "gluing edge" occur in the remaining boundary edges, as sources/targets, and what will make sure that they still cancel out.</p></li> <li><p>Now, imagine you're standing somewhere on the surface of either triangle; the notion of orientation you can see, locally, is the idea of "clockwise around you". What should it mean to say the edges are consistent with this? It seems reasonable to say that as you walk up to any edge of the triangle, its direction should agree with your idea of "clockwise around you". So let's you're on the right-hand triangle and you sidle up to the "gluing edge"; it's on your left, and it points forward &mdash; that's good, that's clockwise! Now you step over it; but now it's on your right, so to still agree with your orientation, it should point <em>backwards</em>, seen from this side... and it does! Awesome!</p></li> </ol> <p>As I see it, these (and a few other such examples) are the intuitively natural seeds from which the various abstractions and generalisations of orientation grow, and whenever I have trouble seeing why some choice is made the way it is, chasing it back to something like this usually resolves the question.</p> <p>Looking outward: the ideas in 1 and 2 are the first steps down the path of combinatorial approaches to orientation, where the orientaion is on entire simplices, as seen in simplicial complexes like you sketch. No. 3 points down a different path, to more geometric notions of <em>local</em> orientation at a point, which can be expressed in terms of local homology groups, among other ways; Hatcher's excellent free <a href="http://www.math.cornell.edu/~hatcher/AT/ATpage.html" rel="nofollow">book</a> has a nice treatment of this, with really good examples/exercises, in Section 3.3 under "Orientations and Homology".</p>