Is there a $\Delta$-complex structure on the sphere with less than three $0$-simplices?
In Hatcher's Algebraic Topology, a $\Delta$-complex structure on a topological space $X$ is defined as follows. Here, $\Delta^n$ denotes the standard $n$-simplex in $\mathbb{R}^{n+1}$, and $\overset{\circ}{\Delta}{}^n$ denotes its interior.
A $\mathbf\Delta$-complex structure on a space $X$ is a collection of maps $\sigma_\alpha : \Delta^n \to X$, with $n$ depending on the index $\alpha$, such that:
- The restriction $\sigma_\alpha | \overset{\circ}{\Delta}{}^n$ is injective, and each point of $X$ is in the image of exactly one such restriction $\sigma_\alpha | \overset{\circ}{\Delta}{}^n$.
- Each restriction of $\sigma_\alpha$ to a face of $\Delta^n$ is one of the maps $\sigma_\beta : \Delta^{n-1} \to X$. Here we are identifying the face of $\Delta^n$ with $\Delta^{n-1}$ by the canonical linear homeomorphism between them that preserves the ordering of the vertices.
- A set $A \subset X$ is open iff $\sigma_\alpha^{-1}(A)$ is open in $\Delta^n$ for each $\sigma_\alpha$.
I can place a $\Delta$-complex structure on the sphere $S^2$ using three $0$-simplices, three $1$-simplices, and two $2$-simplices, essentially by gluing the two $2$-simplices along their boundary edges so that each simplex is a hemisphere and the (common) boundary is the equator.
Question: Is there a $\Delta$-complex structure on the sphere $S^2$ using fewer than three $0$-simplices?
I do not believe that this is possible, since I don't see how one can possibly glue the edges of some simplices onto fewer than three vertices to get a sphere; but, I don't know how to write down a rigorous argument.
1 answer
If it wasn't for the right to left implication of condition 3, you could do this with a single map out of the 0-simplex.
If there are 2 or fewer points in the image of a map from $\Delta^0$, then that means at least two vertices of the image of every map out of the 2-simplex must be the same. Since the image of a 1-simplex can't be degenerate due to condition 1 (or self-intersecting or visit an end point in the interior), the images in such cases are non-trivial closed loops which thus separate $S^2$ into at least two disconnected pieces each homeomorphic to a disc. The image of a 2-simplex will thus lead to a disc that remains to be covered. The same logic works just as well for the discs. We can never completely cover those discs with a finite number of maps from the 2-simplex.
Now, what we could almost do is use an infinite family with the areas of the discs approaching $0$ and being contained in smaller and smaller balls around one of the vertices that's the image of the 0-simplex, call it $z$. The motivating image is similar to the Hawaiian earring. This would work except that the right to left implication of condition 3 fails.
Specifically, we can consider a closed path from some point $x \neq z$ to $z$. This path is chosen such that for any point $y \neq z$ on the path, the subpath from $y$ to $z$ passes through the images of the interior of infinitely many maps from the 2-simplex and has an open intersection with the images of the interior of the 1-simplex. This path is a closed subset of $S^2$, so its complement is open. If we union $\{z\}$ into that complement, call this subset $A$, it will be non-open, because $z$ will be a boundary point. However, the inverse image of $A$ is open for all maps from simplices violating condition 3.
In more detail, for every $\sigma_\alpha$, its inverse image of the complement is open but lacks the vertices that get mapped to $z$, but for each $\sigma_\alpha$ we can always choose a sufficiently small open ball around $z$ such the inverse image of the complement of the path unioned with this open ball agrees with the inverse image of the complement alone except that it includes the inverse image of $z$. Specifically, the open ball just needs to be small enough that it doesn't include any of part of the path that intersects the image away from $z$. This shows that adding the inverse image of $z$ to the inverse image of the complement is still open, but this means the inverse image of $A$ is open.
1 comment thread
1 comment thread