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Is there a ($\mathrm{T}_0$) topological group that is connected and locally connected but is not path-connected?

This is a cross-post from MSE, since my question there was posted over three weeks ago and hasn't gotten anything useful. An earlier question of mine from MSE did not specify local connectedness.

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1 Answer 1

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A similar question was answered here: http://mathoverflow.net/a/119962/2926. The idea is to start with your favorite non-trivial abelian group $A$, say $A = \mathbb{Z}/(2)$ (viewed as a group object in $\mathbf{Set}$) and apply to it a sequence of product-preserving functors

$$\mathbf{Set} \stackrel{K}{\to} \mathbf{Cat} \stackrel{\text{nerve}}{\to} \mathbf{Set}^{\Delta^{op}} \stackrel{\text{Real}_{\text{long}}}{\to} \mathbf{CGHaus}$$

where $K$ takes a set $S$ to the category whose objects are elements of $S$ and exactly one element in each $\hom(x, y)$, where $\text{nerve}$ is the usual nerve functor to simplicial sets, and where $\text{Real}_{\text{long}}$ is akin to the geometric realization functor on simplicial sets, except that instead of gluing together ordinary affine simplices $\sigma_n = \{0 = x_0 \leq x_1 \leq \ldots \leq x_n \leq x_{n+1} = 1\}$ based on configurations of points in the ordinary unit interval $I = [0, 1]$, one glues together "long affine simplices"

$$\Sigma_n = \{\bot = x_0 \leq x_1 \leq \ldots \leq x_n \leq x_{n+1} = \top\}$$ based on configurations of points in the end compactification of the long line (with endpoints $\bot, \top$). It is important here that the long line is connected and locally connected, but not path connected.

Because the functors are product-preserving, the result is an abelian group object in the category $\mathbf{CGHaus}$ of compactly generated Hausdorff spaces. A technical argument mentioned in the cited answer shows that the output (call it $L(A)$) is in fact a topological abelian group (i.e., continuity of the multiplication $L(A) \times L(A) \to L(A)$ is with respect to the usual topological product, not just the product in $\mathbf{CGHaus}$ which in general has a topology finer than that of the topological product).

The cited answer also showed that $L(A)$ is connected but not path-connected. All one needs to see now is that $L(A)$ is locally connected. Let $U$ be an arbitrary open set; it suffices to show that any connected component $V$ of $U$ is open. Since the topology on $L(A)$ is a colimit topology (given by a coend described in the cited answer), to check openness of $V$ it is enough that its intersection with any (long) simplex $\Sigma$ is open. Let $x \in V \cap \Sigma$; a fortiori $x \in U \cap \Sigma$, and since a long simplex $\Sigma$ is locally connected, there is a connected open neighborhood $O_x$ of $x$ relative to $\Sigma$ contained in $U \cap \Sigma$; given that $O_x \subset U$ and $V$ is the maximal connected subset of $U$ containing $x$, we must have $O_x \subset V$ and so $O_x \subset V \cap \Sigma$. Thus $V \cap \Sigma$ is a union of such open $O_x$ in $\Sigma$, as was to be shown.

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