Is a locally finite simplicial complex metrizable?

Question

I was trying to give a metric on an arbitrary simplicial complex which is locally finite. Assume the space is path connected.

Naturally, one can consider a metric $d(x,y) =\inf \sum_{i=1}^{k} length(\gamma_{i})$ where $\cup \gamma_{i}$ is a path joining $x$ and $y$; each $\gamma_{i}$ belongs to a cell.

Then it would be a metric on the space, but I'm not sure if the topology derived from geometrical realization and the topology derived from the metric are equal.

Any help will be appreciated.

Answer 1

There are several topologies we can put on a simplicial complex $\cal S$. For example

• The topology of the geometric realization $|\cal S|$.
• On the product space $I^V$, consider the subspace $$\textstyle X=\{\alpha \mid \{v\mid \alpha_v>0\}\in\cal S, \sum_V\alpha_v=1\}$$ Then the map $\sum\alpha_v v \mapsto (\alpha_v)_v$ a continuous bijection $|\mathcal S|\to X$. So the topology on $X$ is finer that the quotient topology on the realization.
• There is also the topology induced by the metric $$ \textstyle d(\sum_V\alpha_v v, \sum_V\beta_v v) = \sum_V|\alpha_v-\beta_v| $$

It is not to difficult to show that this metric induces the product topology on $X$. Moreover, if $\cal S$ is locally finite, then the simplices form a locally finite closed cover of $X$, so a set $A$ is closed in $X$ if it intersects every simplex in a closed set. That means the topology is actually the final topology with respect to all simplices, i.e. that of the realization $|\cal S|$.