Group completion of topological monoids

Let $$M$$ be an abelian monoid. For sake of simplicity we shall assume that in $$M$$ the cancellation law holds true. With this last assumption we define the group completion $$G$$ of $$M$$ as $$G:=M\times M/\sim$$ where $$(a,b)\sim (a',b')$$ if and only if $$a+b'=a'+b$$. It has been quite surprizing find out, reading the paper Completions and Fibrations for Topological Monoids - Paulo Lima-Filho, that in general, if $$M$$ is a topological abelian monoid then its group completion $$G$$, endowed with the quotient topology, is not a topological group. Can anyone help me to provide an example of this fact?

Let $$(e_n)_{n\in\omega}$$ be the standard orthonormal basis of the Hilbert space $$\ell_2$$. For every $$n\in\mathbb N$$ consider the linear hull $$L_n$$ of the vectors $$e_1,\dots,e_{n}$$ in $$\ell_2$$. On the union $$L^\infty:=\bigcup_{n=1}^\infty L_n$$ consider the strongest topology that induces the Euclidean topology on each space $$L_n$$. It is well-known (and easy to see) that the space $$L^\infty$$ is a non-metrizable $$k_\omega$$-space (i.e., the direct limit of a sequence of compact Hausdorff spaces). Observe that the vector $$e_0$$ is orthogonal to the subset $$L^\infty$$ in $$\ell_2$$.
In the Hilbert space $$\ell_2$$ consider the submonoid $$S:=\bigcup_{n=0}^\infty(n e_0+L_n)$$. Now consider the product $$S\times S$$ and the map $$q_S:S\times S\to S-S=\mathbb Z+\bigcup_{n=1}^\infty L_n,\;\;q_S:(x,t)\mapsto x-y.$$ It is easy to see that the quotient topology on $$S-S$$ is not metrizable -- it coincides with the topology of the direct limit of the sequence $$(\mathbb Z+L_n)_{n=1}^\infty$$.
Finally consider the topological monoid $$M:=\ell_2\times S$$. We claim that $$M$$ is a required counterexample. Indeed, consider the subgroup $$M-M\subset \ell_2\times\ell_2$$ and the map $$q:M\times M\to M-M$$, $$q:(x,y)\mapsto x-y$$.
Since the space $$M\times M$$ is second-countable, the set $$M-M$$ endowed with the quotient topology is a sequential $$\aleph_0$$-space (see Theorem 11.3 on p.494 here) and hence $$M-M$$ has countable $$cs^*$$-character. Assuming that $$M-M$$ is a topological group, we can apply Theorem 1 from this paper of Banakh and Zdomskyy and conclude that $$M-M$$ is either metrizable or contains an open $$k_\omega$$-subgroup. But none of these two conditions applies to $$M-M$$: this space contains a topological copy of the $$k_\omega$$-space $$L^\infty$$ and hence is not metrizable and contains a copy of the Hilbert space $$\ell_2$$, so cannot contain an open (and hence closed) $$k_\omega$$-subgroup.
• Is it difficult to describe the universal continuous homomorphism of your $M$ to a topological group? – მამუკა ჯიბლაძე May 20 '18 at 8:57
• The universal homomorphism will be the natural map of $M$ to the topological group $\ell_2\times\mathbb Z\times L^\infty$, but the latter topological group is not sequential, so, not a quotient space of any metrizable space. – Taras Banakh May 20 '18 at 9:00