© 1997 Copyright by Dmitri Shakhmatov. All rights reserved.

Answer

The completion of QN provides an answer to a question of Zarichnyi Dmitri Shakhmatov Asked in Imbedding of an n-Dimensional Topological Group by Michael Zarichnyi

Can every n-dimensional topological group be imbedded as a (not necessarily closed) subgroup of an n-dimensional group whose underlying space is an absolute G_\delta? (All groups are assumed to be separable metrizable.)

If I understand this question correctly, the answer seems to be NO. Indeed, one has the following theorem:

Theorem If G can be embedded as a subgroup in an n-dimensional (countable-dimensional) metric group H whose underlying space is an absolute G_\delta, then the two-sided uniformity completion G' of G is <= n-dimensional (countable-dimensional).
Proof We use a folklore fact that an absolute G-\delta metric group is complete in its two-sided uniformity. (This follows from the fact that a G_\delta subgroup of a metric group is closed in it.) Therefore H is complete in its two-sided uniformity, and so the closure cl(G) of G in H is also complete in its two-sided uniformity. Because the two-sided uniformity completion is unique, G' coincides with cl(G). Since cl(G) is a closed subspace of n-dimensional space H, cl(G) is <= n-dimensional. In the countable-dimensional case we use the sum theorem for dimension.

Now this theorem implies that the rational points in the circle group serve as a counterexample to Zarichnyi's question.

Result of Guran and Zarichnyi cited in Zarichnyi's contribution could be restated as follows: There exists a Polish countable-dimensional group which contains every finite-dimensional Polish group. The author does not say explicitly whether it contains it a subgroup or simply as a subspace. The above theorem allows one to conclude that it MUST be barely a subspace.

Corollary Let G be a (zero-dimensional) group Q^N which lives inside S^N. [Q is the group of "rational points" in the circle S, and Q^N and S^N are countable Tychonoff products of Q and S respectively.] Then (even though that G is zero-dimensional) G cannot be embedded as a subgroup into a countable-dimensional Polish group.
Proof Completion of G is S^N, and the latter space is known for its failure to be countable-dimensional.