The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

102.70 Different colimit topologies

This example is [Example 1.2, page 553, TSH]. Let $G_ n = \mathbf{Q} \times \mathbf{R}^ n$, $n \geq 1$ seen as a topological group for addition endowed with the usual (Euclidean) topology. Consider the closed embeddings $G_ n \to G_{n + 1}$ mapping $(x_0, \ldots , x_ n)$ to $(x_0, \ldots , x_ n, 0)$. We claim that $G = \mathop{\mathrm{colim}}\nolimits G_ n$ endowed with the topology

\[ U \subset G\text{ open} \Leftrightarrow G_ n \cap U\text{ open }\forall n \]

is not a topological group.

To see this we consider the set

\[ U = \{ (x_0, x_1, x_2, \ldots )\text{ such that } |x_ j| < |\text{cos}(jx_0)| \text{ for } j > 0\} \]

Using that $jx_0$ is never an integral multiple of $\pi /2$ as $\pi $ is not rational it is easy to show that $U \cap G_ n$ is open. Since $0 \in U$, if the topology above made $G$ into a topological group, then there would be an open neighbourhood $V \subset G$ of $0$ such that $V + V \subset U$. Then, for every $j \geq 0$ there would exist $\epsilon _ j > 0$ such that $(0, \ldots , 0, x_ j, 0, \ldots ) \in V$ for $|x_ j| < \epsilon _ j$. Since $V + V \subset U$ we would have

\[ (x_0, 0, \ldots , 0, x_ j, 0, \ldots ) \in U \]

for $|x_0| < \epsilon _0$ and $|x_ j| < \epsilon _ j$. However, if we take $j$ large enough such that $j \epsilon _0 > \pi /2$, then we can choose $x_0 \in \mathbf{Q}$ such that $|\text{cos}(jx_0)|$ is smaller than $\epsilon _ j$, hence there exists an $x_ j$ with $|\text{cos}(jx_0)| < |x_ j| < \epsilon _ j$. This contradiction proves the claim.

Lemma 102.70.1. There exists a system $G_1 \to G_2 \to G_3 \to \ldots $ of (abelian) topological groups such that $\mathop{\mathrm{colim}}\nolimits G_ n$ taken in the category of topological spaces is different from $\mathop{\mathrm{colim}}\nolimits G_ n$ taken in the category of topological groups.

Proof. See discussion above. $\square$


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