$(\Bbb R,+)$ is a topological group. Is there any other group structure on $\Bbb R$ such that it is still a topological group and this group is not isomorphic to $(\Bbb R,+)$ ?

Refer to Non-isomorphic Group Structures on a Topological Group for the general problem. In that problem I have conjectured that there is no such group however I couldn't prove it.

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    $\begingroup$ But you mean with the same topology, don't you? $\endgroup$ – Stefan Hamcke Mar 5 '14 at 17:08
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    $\begingroup$ @StefanHamcke If so, I think L.E.J. Brouwer proved this in his thesis, IIRC. $\endgroup$ – Henno Brandsma Mar 5 '14 at 18:30
  • $\begingroup$ @HennoBrandsma could you provide the link or some detail of the thesis. $\endgroup$ – viplov_jain Mar 5 '14 at 22:52
  • $\begingroup$ @StefanHamcke yes it has the same topology $\endgroup$ – viplov_jain Mar 5 '14 at 22:53
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    $\begingroup$ @viplov_jain It's a thesis from 1907, in Dutch ("Over de Grondslagen der Wiskunde"), I'm not sure whether there is a paper somewhere else that also covers it as well. I think the general problem (for all dimensions) was an open problem at the time, and Brouwer did the first dimension. $\endgroup$ – Henno Brandsma Mar 6 '14 at 6:30

No, non-isomorphic group structures on the same topology of the real line are not possible. For the sake of clarity I’ll refer to the topological space as to the real line, not ℝ, to emphasize we don’t rely on a known arithmetical structure, only on the known topological structure, and have some (unknown) group structure.

First of all, let’s define the family of left group operations (currying):

Gx(y) = x • y .

For each x: Gx must be a homeomorphism of the real line. Such homeomorphism may be, in principle, either order-preserving or order-reversing. Since for the identity element e: Ge must be the identity map, Gx depends on x continuously, and the real line is connected, for any x: Gx is order-preserving.

Let’s go further. Except for Ge, any Gx may not have fixed points. It is easy to check that when x > e it shifts the line to positive side and for x < e it shifts the line to negative side. We assume that inherited certain total order structure from ℝ, but it is generally not important which of two possible order structures on the real line are we using.

From this follows that our group is a totally ordered group that implies that is torsion-free.

Now you can think of e as if 0 and of an arbitrary distinct element as of 1, and tedious 19-century mathematical analysis-fashioned reasonings can demonstrate that we haven’t anything but addition of real numbers up to isomorphism.


Incnis Mrsi's nice argument from 2014 has not actually been completed so let's complete it, since this came up recently, twice (actually a third time on MO, too).

We have a topological group $G$ with identity $e$ homeomorphic to $\mathbb{R}$. Incnis has established that this group is totally ordered with respect to the usual order on $\mathbb{R}$ (in a way compatible with the topology; that is, $G$ is totally ordered, the total order is isomorphic to $\mathbb{R}$, and the isomorphism is a homeomorphism also), and in particular torsion-free. Thanks to the total order it follows that for any positive integer $n$, the $n^{th}$ power function $g \mapsto g^n$ is strictly monotonic and hence injective.

Proposition: $g \mapsto g^n$ is unbounded from above and below.

Proof. If the image of $g \mapsto g^n$ contains some positive $L > e$ then it also contains $L^2 > L$, which is strictly larger. Similarly for negatives. $\Box$

Corollary: The function $g \mapsto g^n$ is bijective; equivalently, every element has a unique $n^{th}$ root.

Proof. By the above, the image of $g \mapsto g^n$ contains $e$ and is unbounded from above and below. Since it is also connected, the image is all of $G$, so $g \mapsto g^n$ is surjective. Since it is also strictly monotonic, it is injective (as observed above), hence bijective. $\Box$

Pick any $g > e$ in $G$ and define $f : \mathbb{Q} \to G$ by sending $\frac{p}{q}$ to the unique $q^{th}$ root of $g^p$. This is well-defined because if $\frac{p}{q} = \frac{r}{s}$ then $g^{ps} = g^{qr}$ has a unique $qs$-th root. It is a homomorphism for the same reason. And it is strictly monotonic, hence extends continuously to a homomorphism $F : \mathbb{R} \to G$ by a limiting argument. (This should all be very familiar if you've ever constructed the exponential $a^x$ in terms of limits of rational exponentials in a real analysis class, and in fact that argument is a special case of this one, applied to $G = (\mathbb{R}_{+}, \times)$.)

$F$ is strictly monotonic, hence injective. We will write $F(r) = g^r$.

Proposition: $r \mapsto g^r$ is unbounded from above and below.

Proof. If the image of $r \mapsto g^r$ contains any positive $L > e$ then it also contains $gL > L$, which is strictly larger. Similarly for negatives. $\Box$

Corollary: $r \mapsto g^r$ is surjective, hence an isomorphism of topological groups $G \cong (\mathbb{R}, +)$.

Proof. The image of $r \mapsto g^r$ is unbounded from above and below, contains $e$, and is connected, hence is all of $G$. So $r \mapsto g^r$ is surjective, hence a bijection. Since it's strictly increasing (hence continuous), its inverse is also strictly increasing (hence continuous), so it's a homeomorphism. $\Box$

Corollary: $G$ is abelian.


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