# Examples of topological groups that are not metrizable.

I am looking for examples of topological groups but most are metrizable likes real numbers etc. Then there are Lie groups that I am not familiar with. Are there any simple enough to understand examples of non metric topological groups, or topological rings? How about ordered square or long line, can they be turned into a topological group?

• How about taking a Cartesian product of a lot of copies of a standard example? Aug 31, 2019 at 10:56
• Can look at Linear Algebraic Groups, They come with the Zariski topology which is not Hausdorff. Aug 31, 2019 at 11:02
• @Brozovic Algebraic groups are mostly not algebraic groups. Indeed, the law is not continuous (it is continuous $G\to G\to G$ when $G\times G$ is equipped with the Zariski topology, which is not the product of Zariski topologies on factors, unless $G$ is finite). Actually, a topological group in which $\{1\}$ is closed is Hausdorff, which discards infinite algebraic groups.
– YCor
Sep 5, 2019 at 22:40

$$\mathbb{R}^I$$ for any uncountable index set $$I$$ (pointwise addition), likewise $$\{0,1\}^I$$ (pointwise addition mod 2, compact as well), $$C(X)$$ (the continuous functions on an uncountable, completely regular topological space $$X$$, again with addition, with pointwise convergence topology), etc.

The ordered square and the long line are not homogeneous (the end points are distinct) so cannot be made into a topological group. Or if your versions of them are homogeneous, then they cannot be topological groups because thy're first countable and not metrisable (Birkhoff's theorem tells us that a first countable topological group is metrisable, which explains why my examples need to be "big").

• isn't the long line homogeneous? (It has different versions, going one way or going both ways, at any rate, take any version that has no endpoints, e.g. $(0,\omega_1)$ or $(-\omega_1,\omega_1)$.) So another argument might be needed to show it cannot be made into a topological group. Aug 31, 2019 at 19:24
• @Mirko, for me it starts at $0$ and has no end point: $\omega_1 \times [0,1)$ ,ordered lexicographically, in the order topology. It's first countable but not metrisable, which shows it cannot be made into a topological group. Aug 31, 2019 at 19:26
• I'm not sure what you mean by $C_p(X)$ since you don't specify the topology; if so you probably have to make assumptions on $X$.
– YCor
Sep 5, 2019 at 22:41
• @YCor the pointwise (product) topology. Hence the $p$. The space $X$ must be uncountable that’s all. Sep 6, 2019 at 4:04
• That's $C_p(X)$ as subset of $\mathbf{R}^X$. You need to assume that continuous real-valued functions on $X$ separate points (assuming $X$ Hausdorff is not enough).
– YCor
Sep 6, 2019 at 5:23

If you really just want some non-metrizable topological group without other properties, you can create some examples as follows:

Take some group $$G$$ with order $$\text{ord}(G) \geq 2$$ and equip it with the trivial topology (only the empty set $$\emptyset$$ and $$G$$ being open). As a topological space, $$G$$ cannot be metrizable then:

Suppose it is and let $$d$$ be a metric on $$G$$. As $$\text{ord}(G) \geq 2$$, we can choose $$a,b \in G$$ with $$a \neq b$$. If $$\epsilon = d(a,b)$$, then we have $$b \not\in B_{<\epsilon}(a),$$ the open ball around $$a$$ with radius $$\epsilon$$, which is a contradiction to $$G$$ being equipped with the trivial topology.

As every map into a space with trivial topology is continuous, $$G$$ is a topological group, which means we have a non-metrizable topological group.

• maybe the OP wants $T_1$ examples.., a pretty standard assumption. Aug 31, 2019 at 10:59
• I totally agree. As the OP was asking for some easy examples I felt that a rather unnatural or stupid example might still be a good starting point.
– Con
Aug 31, 2019 at 11:04