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In Direct limit, Martin rightly pointed out that my naive construction (now deleted) of the colimit (direct limit) of topological abelian groups was wrong. He shows how to do it properly (at least the coproduct) here.

Since then, I've been lurking some of the literature about the subject and this problem of the colimit of topological groups (about which I had previously no idea) seems at the same time classical and topical. For instance, this 1998's paper points out that the Encyclopedic Dictionary of Mathematics, second edition, MIT (1987), article 210, has made the same mistake that I did, stating that the direct limit of topological groups, with the inductive limit of topologies (my naive attempt) has always a continuous multiplication.

The authors of this paper show a counter-example (example 1.2, page 553) and here is my question: I must be absolutely dumb, but don't understand it. Could anyone help me?

For those who don't have access to the paper, here is the example.

Let $G_n = \mathbb{Q} \times \mathbb{R}^n$ with the usual topology. Imbed $G_n \hookrightarrow G_{n+1}$ as $x \mapsto (x,0)$. Then, as a plain abelian group, $G = \varinjlim_{n} G_n= \mathbb{Q} \times \prod'\mathbb{R}$, where $\prod'\mathbb{R}$ denotes the weak or restricted product, which is the way guys in this area call the direct sum; that is, elements of $\prod'\mathbb{R}$ are infinite tuples $( x_1, \dots , x_n, \dots )$, in which all of its components $x_n \in \mathbb{R}$ are zero, except a finite number of them.

The inductive limit topology is the finest one that makes all the inclusions $G_n \hookrightarrow G$ continuous. That is, $U \subset G$ is open if and only if $U\cap G_n$ is open for all $n$.

Let's see how they show that, with this inductive topology, the "product" (in fact, addition) $\mu : G \times G \longrightarrow G$ is not continuous ($\mu$ is the operation induced as an honest colimit of groups -no topologies- by the operations $\mu_n : G_n \times G_n \longrightarrow G_n$, which I assume are the usual additions of those linear spaces). In plain English:

$$ (x_0, x_1, \dots , x_n , \dots ) + (y_0, y_1, \dots , y_n , \dots ) = (x_0+y_0, x_1 + y_1 , \dots , x_n + y_n , \dots ) \ . $$

So, in this situation it's enough to produce an open neighbourhood $U \subset G$ of the neutral element $e \in G$ such that $V^2$ is not in $U$, for any open neighbourhood $V$ of $e$ -where, I assume, $V^2$ means $V + V$.

Ok, here is the guy that is supposed to ruin (and sure it does) my naive attempt:

$$ U = \left\{ x = (x_0, x_1, \dots , x_n , \dots ) \quad \vert \quad \vert x_j\vert < \vert \cos (jx_0) \vert \ , 1 \leq j \right\} \ . $$

This guy is an open set of $G$ because $x_0$ being a rational number guarantees $\cos (jx_0) \neq 0$ for all $j$. Assume there is an open neighbourhood $V$ of $e$ such that $V^2 \subset U$. Then, $V \cap G_j$ contains an open interval $(-\varepsilon_j , \varepsilon_j)$ in $\mathbb{R}$ with $\epsilon_j > 0$ such that

$$ (-\varepsilon_0 , \varepsilon_0) \times (-\varepsilon_j , \varepsilon_j) \subset \left\{ (x_0 , x_j) \in \mathbb{Q} \times \mathbb{R} \quad \vert \quad \vert x_j\vert < \vert \cos (jx_0) \vert \right\} \ . $$

And here come the two final sentences of the example I don't understand: This is impossible if $2j\varepsilon_0 > \pi$. A contradiction.

Any hints or remarks (even humiliating ones) will be welcome.

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I'm confused: what is $\epsilon_0$? The statement in the next to last sentence must be missing a quantifier (both here and in the original paper): there exists $\epsilon_j>0$ such that (statement involving undefined variable $\epsilon_0$) is true. And then we get a contradiction when the undefined variable is big? – Dan Ramras Sep 23 '10 at 19:25
Well, I think they should have said something like: "if $V\cap G_n$ is an open neighborhood of the neutral element in $G_n$, then there is an open "cube" $(-\varepsilon_0, \varepsilon_0) \times(-\varepsilon_1, \varepsilon_1) \times \dots \times (-\varepsilon_n, \varepsilon_n) \subset V$". The problem is (see my reply to William's answer) that we are not using anywhere the condition $V^2 \subset U$. So this same "cube" should exist for $U$ too. – a.r. Sep 24 '10 at 3:14
Has anyone looked in the Encyclopedic Dictionary of Mathematics, second edition, MIT (1987), article 210 (as referenced above) to see if they give a reference for the general result? If we're perplexed by the supposed counterexample, it might be worth figuring out why it was claimed that no such counterexample could exist... Google books doesn't seem to have the necessary pages. Actually, it's in our library, so I'll look today. – Dan Ramras Sep 24 '10 at 16:25
I have to say that the result proven by Tatsuuma-Shimomura-Hirai in the paper Agusti linked to sounds like it could be the best possible: they prove that the inductive limit topology for a (countable) system of locally compact (Hausdorff) groups gives rise to a topological group (Thm 2.7 of their paper). The reason this sounds right to me is Lemma 5.5 in Milnor and Stasheff's Characteristic Classes, which says that the direct limit topology (for sequences of spaces) commutes with Cartesian products, roughly speaking, when the spaces involved are locally compact Hausdorff. – Dan Ramras Sep 24 '10 at 16:35
I finally got around to looking in the Encyclopedic Dictionary of Mathematics. The result (colimits of topological groups are topological groups) is stated there without any reference, and without any explanation. Not very helpful... – Dan Ramras Sep 27 '10 at 0:48

This means that $j\epsilon_0 > \frac{\pi}{2}$. Hence $-j\epsilon_0 < -\frac{\pi}{2}$. By the density of rationals, this tells you that there exists a sequence $q_1, q_2,\dots,q_n$ of rational numbers in $(-\epsilon_0,\epsilon_0)$ such that $jq_n\rightarrow \frac{\pi}{2}$. Hence $|\cos(jq_n)|\rightarrow 0$. Combining this with $|x_j| < |\cos(jq_n)|$ for all $n$ gives the contradiction.

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I thought about this possibility, but I discarded it because, if this reasoning was correct, then it would apply also to $U$, not just $V$: notice that, in the case of $U$, you already have the possibility of constructing such a sequence $q_n$. That is, if $V$ is not open, $U$ is not open too, so the whole example is wrong. – a.r. Sep 24 '10 at 3:00
This does not prove that $U$ is not open, since William uses all $n$ above. – Martin Brandenburg Sep 26 '10 at 10:35

I would have added a comment, but I fear the answer would overlap the allowed character limit.

Anyway, maybe I'm missing something, but I think I see a clear contradiction. Here we go...

Let $S_j = \{(x_0, x_j)\in\mathbb{Q}\times\mathbb{R}: |x_j| < |\cos(jx_0)|\}$ Let $V_j = (-\epsilon_0,\epsilon_0)\times(-\epsilon_j,\epsilon_j)$. The claim is that for all $j$, we have $V_j\subset S_j$. Fix $j$ and assume this inclusion holds. On the other hand, if $2j\epsilon_0 > \pi$, there is a sequence $q_1,q_2,\dots$ of rationals in $(-\epsilon_0,\epsilon_0)$, such that $jq_n\rightarrow \frac{\pi}{2}$. Let $x_j\in (-\epsilon_j, \epsilon_j)$ not equal to zero. Since the inclusion $V_j\subset S_j$ holds, we have $|x_j| < |\cos(jq_n)|$ for all $n$. Clearly impossible, since $x_j > 0$.

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I believe you, William (except for this last "since $x_j > 0$" -I presume you meant $\varepsilon_j > 0$-, but this doesn't matter). The point, as Pierre-Yves Gaillard says, is: this same proof shows that also $U$ is not open. So the conclusion, again, is that the whole example is wrong. – a.r. Sep 24 '10 at 5:09

This is a new version of the answer. The comments refer to previous versions. Martin Brandenburg's comment was especially helpful. A comment of Dan Ramras, somewhere else in this thread, drew my attention to Lemma 5.5 page 64 in Milnor and Stasheff's Characteristic Classes. Thank you also to Agusti Roig for his wonderful question.

Let me state the Lemma of Milnor and Stasheff mentioned above:

(1) Let $A_1\subset A_2\subset\cdots$ and $B_1\subset B_2\subset\cdots$ be sequences of locally compact spaces with inductive limits $A$ and $B$ respectively. Then the product topology on $A\times B$ coincides with the inductive limit topology which is associated with the sequence $A_1\times B_1\subset A_2\times B_1\subset\cdots$.

The proof (which can be found here) shows in fact that the following more technical statement also holds:

(2) Let $A_1\subset A_2\subset\cdots$ and $B_1\subset B_2\subset\cdots$ be as above. For each $i$ let $C_i\subset A_i$ and $D_i\subset B_i$ be subspaces. Assume $C_i\subset C_{i+1}$ and $D_i\subset D_{i+1}$ for all $i$, and call $C$ and $D$ the respective inductive limits. Then the product topology on $C\times D$ coincides with the inductive limit topology which is associated with the sequence $C_1\times D_1\subset C_2\times D_2\subset\cdots$.

In particular, if the $C_i$ are topological groups, then so is $C$ (assuming, of course, that the topological group structure on $C_i$ is induced by that of $C_{i+1}$). This seems to indicate that the alleged counterexample of Tatsuuma et al. is not really a counterexample.

I know no example of an inductive limit of topological groups which is not a topological group. (I think that such examples do exist. It would be interesting to know if $(\mathbb R^\infty)^\infty$ is a topological group.)

Let's prove (2).

Let $W$ be a subspace of $C\times D$ which is open in the inductive limit topology. For each $i$ let $W_i$ be an open subset of $A_i\times B_i$ which has the same intersection with $C_i\times D_i$ as $W$. Let $(c,d)$ be in $W\cap(C_i\times D_i)$. There is a compact neighborhood $K_i$ of $c$ in $A_i$, and a compact neighborhood $L_i$ of $d$ in $B_i$, such that $K_i\times L_i\subset W_i$. There is also a compact neighborhood $K_{i+1}$ of $K_i$ in $A_{i+1}$, and a compact neighborhood $L_{i+1}$ of $L_i$ in $B_{i+1}$, such that $K_{i+1}\times L_{i+1}\subset W_{i+1}$. Then $K_{i+1}\cap C_{i+1}$ is a neighborhood of $K_i\cap C_i$ in $C_{i+1}$, and $L_{i+1}\cap D_{i+1}$ is a neighborhood of $L_i\cap D_i$ in $D_{i+1}$. The union $U$ of the $K_i\cap C_i$ is open in $C$, and the union $V$ of the $L_i\cap D_i$ is open in $D$, the spaces $C$ and $D$ being equipped with the inductive limit topology. Moreover we have $(c,d)\in U\times V\subset W$.

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If $x_0$ is allowed to be real, then I think I need a bit of convincing for why $U'$ is open. – Dylan Wilson Sep 26 '10 at 20:23
@Edit3: I want to see a proof that $A_{\epsilon}$ is open. Even if you replace your intervalls $I(\epsilon)$ with open ones, this is not clear to me. – Martin Brandenburg Sep 29 '10 at 8:30

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