Problems with simplicial space

Question

So I'm going through "Elementary Topology Problem Textbook" by Viro and all, and have problems with 23.3x paragraph which is devoted to simplicial schemes. More concretely I can't tackle 23.4x problem which is also on an attached image. My thoughts on that are:

So (open) simplex is just a set $ \{ c \in S : Supp(c) = \sigma\} $, where $\sigma \in \Sigma$. I know that open sets in $S(V,\Sigma)$ are just open balls $ B_r(c) := \{ c' \in S(V,\Sigma) : \rho(c,c') < r \} $ and closed balls are $ B_r(c) := \{ c' \in S(V,\Sigma) : \rho(c,c') \le r \} $. By that I can conclude that $c' \in B_r(c) $ iff $ \sigma_{c'} \subseteq \sigma_c $ (but I can't work through that correctly, I've done this step on my own feelings).

Can anyone give me a hint how to tackle this problem, I feel like I either misunderstood something or can't see an obvious step. Thanks.

Answer 1

What is called a simplicial scheme in your book is usually denoted as an abstract simplicial complex. See https://en.wikipedia.org/wiki/Abstract_simplicial_complex. The space $S(V,\Sigma)$ is the geometric realization. You can give it the metric topology as in your book, but normally it is endowed with the so-called weak topology.

Some remarks concerning the definition in your book:

1. Usually one requires that the elements of $\Sigma$ are finite non-empty subsets of $V$. It is perhaps questionable whether on wants to consider the empty set as a simplex (I do not think it makes sense), but that is a minor issue. The major distiction is that your book allows simplices which are infinite subsets of $V$. I doubt that this makes sense, but "definitions are free".

2. The "intersection condition" is redundant because each subset of an element of $\Sigma$ belongs to $\Sigma$.

3. The support of a function $c : V \to I$ such that $\sum_{v \in V} c(v) = 1$ can be at most countable, otherwise $\sum_{v \in V} c(v)$ would be divergent. This means that simplices which are uncountable sets do not have a geometric realization. This indicates that it does really make no sense to admit arbitrary subsets of $V$ as simplices.

4. Given $\sigma \in \Sigma$, the set $\lvert \sigma \rvert = \{c : V \to I \mid \text{Supp}(c) \subset \sigma, \sum_{v \in V} c(v) = 1 \}$ is called the closed simplex associated to $\sigma$ or the geometric realization of $\sigma$. The open simplex associated to $\sigma$ is defined as the set $\lvert \sigma \rvert^\circ = \{c : V \to I \mid \text{Supp}(c) = \sigma, \sum_{v \in V} c(v) = 1 \}$. Let us give a simple example. If $\sigma = \{v_0,v_1\}$, then the set $\lvert \sigma \rvert$ is nothing else than set of functions $c_t : V \to I$, $t \in I$, such that $c_t(v_0) = t, c_t(v_1) = 1-t$ and $c_t(v) = 0$ else. The function $t \mapsto c_t$ gives a bijection between $I$ and $\lvert \sigma \rvert$. It is easy to see that this is a homeomorphism if $\lvert \sigma \rvert$ is endowed with the metric topology. Under this identification $\lvert \sigma \rvert^\circ$ corresponds to the open interval $(0,1)$. For $\sigma = \{v_0,v_1,v_2\}$ the space $\lvert \sigma \rvert$ can similarly be identified with a triangle (= geometric $2$-simplex) which is the convex hull of three points $v'_0,v'_1,v'_2$ in general position in some $\mathbb R^n$. Each point in the convex hull has the form $\sum_{i=0}^2t_iv'_i$ with $t_i \in I$ and $\sum_{i=0}^2t_i = 1$. Each function $c \in \lvert \sigma \rvert$ is identified with the point $\sum_{i=0}^2c(v_i)v'_i$. Also see the Appendix.

I recommend to consult also another textbook, for example

Spanier, Edwin H. Algebraic Topology. Springer Science & Business Media, 1989.

Note that always

a) $\lvert \sigma \rvert^\circ \subset \lvert \sigma \rvert$

b) $ \lvert \sigma \rvert \subset \lvert \tau \rvert$ if $\sigma \subset \tau$.

Your question is completely answered by the following:

1. The set $\lvert \sigma \rvert$ is closed in $S(V,\Sigma)$: Consider a sequence of points $c_n$ in $\lvert \sigma \rvert$ which converges to some $c \in S(V,\Sigma)$. For $v \notin \sigma$ we have $c_n(v) = 0$, thus $c(v) = \lvert c(v) - c_n(v) \rvert \le \rho(c,c_n)$ which shows $c(v) = 0$ because $\rho(c,c_n)$ becomes arbitrarily small. Thus $\text{Supp}(c) \subset \sigma$ which means $c \in \lvert \sigma \rvert$.

2. Define $\dim(\sigma) = n$ if $\sigma$ contains $n+1$ elements (= vertices). Then $\lvert \sigma \rvert^\circ = \lvert \sigma \rvert$ if and only if $\dim(\sigma) = 0$: If $\dim(\sigma) = 0$, i.e. $\sigma = \{ v \}$, then both $\lvert \sigma \rvert^\circ, \lvert \sigma \rvert$ contain precisely the function $\delta_v$ given by $\delta_v(v) =1, \delta_v(w) = 0$ for $w \ne v$. If $\dim(\sigma) = n > 0$, then $\sigma = \{v_0,\ldots, v_n\}$. The functions $\delta_{v_i}$ belong to $\lvert \sigma \rvert$, but not to $\lvert \sigma \rvert^\circ$.

3. $\lvert \sigma \rvert$ is the closure of $\lvert \sigma \rvert^\circ$: Let $c \in \lvert \sigma \rvert$, where $\sigma = \{v_0,\ldots, v_n\}$. Define $b : V \to I, b(v) = 1/(n+1)$ for $v \in \sigma$ and $b(v) = 0$ else. Then $b \in \lvert \sigma \rvert^\circ$ and for $m \in \mathbb N$ the functions $c_m = (1-2^{-m})c +2^{-m}b$ belong to $\lvert \sigma \rvert^\circ$. We have $$\rho(c,c_m) = \sup_{v \in V} \lvert c(v) - c_m(v) \rvert = \max_{i=0,\ldots,n} \lvert 2^{-m}c(v) - 2^{-m}b(v) \rvert \le 2\cdot 2^{-m} ,$$ thus $c_m \to c$ which shows that $c$ is contained in the closure of $\lvert \sigma \rvert^\circ$.

4. Thus an open simplex is closed if and only if its dimension is $0$.

5. If $\Sigma$ contains a simplex $\tau \supsetneqq \sigma$, then $\lvert \sigma \rvert^\circ$ is not open: Let $c \in \lvert \sigma \rvert^\circ$ and $\epsilon > 0$. We have $\lvert \sigma \rvert^\circ \subset \lvert \sigma \rvert \subset \vert \tau \rvert$, thus $c$ is contained in the closure of $\subset \vert \tau \rvert^\circ$. Choose $c' \in \vert \tau \rvert^\circ$ such that $\rho(c,c') < \epsilon$. We have $\text{Supp}(c') = \tau$, thus $c' \notin \lvert \sigma \rvert^\circ$. This shows that no open ball $B_\epsilon(c)$ can be contained in $\lvert \sigma \rvert^\circ$.

6. If $\Sigma$ contains no simplex $\tau \supsetneqq \sigma$, then $\lvert \sigma \rvert^\circ$ is open: Let $c \in \lvert \sigma \rvert^\circ$, where $\sigma = \{v_0,\ldots, v_n\}$. We have $c(v_i) > 0$. Let $\epsilon = \min c(v_i)$. We claim that $B_\epsilon(c) \subset \lvert \sigma \rvert^\circ$. So let $c' \in B_\epsilon(c)$, i.e. $\rho(c,c') < \epsilon$. Then $\lvert c(v_i) - c'(v_i) \rvert \le \rho(c,c') < \epsilon$ which is impossible if $c'(v_i) = 0$. Let $\tau = \text{Supp}(c') \in \Sigma$. We have shown that $\sigma\subset \text{Supp}(c') = \tau$, hence by assumption $\sigma = \tau$. Thus $c' \in \lvert \sigma \rvert^\circ$.

Appendix:

The geometric realization of a simplicial scheme (abstract simplicial complex) is defined as the set of all functions $c : V \to I$ such that $\text{Supp}(c)$ is a simplex in $\Sigma$ and $\sum_{v \in V} c(v) = 1$. Equivalently it can be defined as the set of all functions $c : V \to \mathbb R$ such that $\text{Supp}(c)$ is a simplex in $\Sigma$, $c(v) \ge 0$ for all $v$ and $\sum_{v \in V} c(v) = 1$.

Define an $\mathbb R$-vector space $$\ell^\infty(V) = \{f : V \to \mathbb R \mid \sup_{v \in V} \lvert f(v) \rvert < \infty \}$$ and give it the norm $\lVert f \rVert_\infty = \sup_{v \in V} \lvert f(v) \rvert $. Then Viro's set $S = S (V,\Sigma)$ is a subset of $\ell^\infty(V)$ which receives its metric by $\rho(c,c') = \lVert c - c' \rVert_\infty$.

For each $v \in V$ let $\delta_v : V \to \mathbb R, \delta_v(v) = 1$ and $\delta_v(w) = 0$ else. The set of these $\delta_v$ is a linearly independent subset of $\ell^\infty(V)$. For each $f \in \ell^\infty(V)$ with finite support we have $f = \sum_{v \in V}f(v)\delta_v$. Note that $\sum_{v \in V}f(v)\delta_v$ is actually a finite sum since $f(v) = 0$ for almost all $v$.

For $\dim(\sigma) = n < \infty$, $\sigma = \{v_0,\ldots,v_n\}$, the closed simplex $\lvert \sigma \rvert$ is nothing else than the convex hull of the $n+1$ basis vectors $\delta_{v_0},\ldots,\delta_{v_n}$, i.e. an $n$-dimensional simplex in the $n+1$-dimensional real vector space spanned by $\delta_{v_0},\ldots,\delta_{v_n}$. To see this let $c \in \lvert \sigma \rvert$ and $t_i = c(v_i)$. Then $t_i \ge 0$, $\sum_{i=0}^n t_i = 1$ and $c = \sum_ {v \in V} c(v) \delta_v = \sum_{i=0}^n t_i\delta_{v_i}$ since $\text{Supp}(c) \subset \sigma$.

For $\dim(\sigma) = \infty$ (if that is admitted as a simplex), this no longer true: The convex hull of $B(\sigma) =\{ \delta_v \mid v \in \sigma \}$ is a strict subset of $\lvert \sigma \rvert$ since there are functions $c \in \lvert \sigma \rvert$ with infinite support.

The sets $\lvert \sigma \rvert$ are always closed (see the proof above) and for $\dim(\sigma) = \infty$ one can easily show that $\lvert \sigma \rvert$ is the closure of the convex hull of $B(\sigma)$.

What about open simplices $\lvert \sigma \rvert^\circ$ with infinite dimension (i.e. with an infinite set $\sigma$)?

If $\sigma$ is an uncountable set, then $\lvert \sigma \rvert^\circ$ is empty. This is due to the fact that an uncountable sum of positive real numbers (which we can interpret as the supremum over all finite subsums) cannot be $< \infty$.

If $\sigma$ is a countable infinite set, $\sigma = \{v_n \mid n \in \mathbb N\}$, then $\lvert \sigma \rvert^\circ$ is non-empty. For example, the function $c$ given by $c(v_n) = 2^{-n}$ and $c(v) = 0$ for $v \notin \sigma$ is in $\lvert \sigma \rvert^\circ$. However, it is never an open subset. In fact, it is not even open in the subspace $\lvert \sigma \rvert$. For each $C \in \lvert \sigma \rvert^\circ$ and each $\epsilon > 0$ the open ball $B_\epsilon (c)$ contains an element of $\lvert \sigma \rvert \setminus \lvert \sigma \rvert^\circ$. I leave this as an exercise.