Theorem 1

A topological group $G$ in which there exists a neighbourhood of the identity element homeomorphic to an open interval of $\mathbf{R}$ is locally isomorphic to $\mathbf{R}$.

The significance of this theorem is that it allows us to deduce, from a purely topological property of a group $G$, a property of the group structure of $G$.

We are concerned here with a phenomenon which is peculiar to the group $\mathbf{R}$ and has no analogue for the groups $\mathbf{R}^n$ when $n > 1$ (cf. Chapter VIII, § 1, no. 4). Groups locally isomorphic to $\mathbf{R}$ are sometimes called one-parameter groups.

To prove Theorem 1 we shall reduce it to Proposition 2 of § 2. By hypothesis, there is a homeomorphism $\varphi$ of an open neighbourhood $U$ of the identity element $e$ of $G$ onto an open interval in $\mathbf{R}$. By means of the inverse of the mapping $\varphi$ we can transport to $U$ the linear order structure of the interval $\varphi(U)$; the topology of $U$ (induced by that of $G$) then has a base consisting of all the open intervals of $U$ (Chapter

IV, , no. 4, Proposition 5). We can find a symmetric neighbourhood V of e such that V.V ⊂ U and such that V is an open interval; for there exists an open interval V' containing e such that V'.V' ⊂ U ∩ U^{-1}, V'.V'^{-1} ⊂ U and V'^{-1}.V' ⊂ U; taking V = V' ∪ V'^{-1}, V is open and symmetric, satisfies V.V ⊂ U and is connected, hence is an interval (Chapter IV, § 2, no. 5, Theorem 4)

We show that, if x, y, z belong to V, the relation x < y implies xz < yz and zx < zy. Indeed, the functions f_1(z) = φ(yz) − φ(xz) and f_2(z) = φ(zy) − φ(zx) are continuous on V; they are > 0 for z = e and do not vanish in V [e.g., if we had φ(yz) = φ(xz), we should have yz = xz and therefore y = x]. Since f_1(V) and f_2(V) are connected (Chapter I, § 11, no. 2, Proposition 4) and are therefore intervals in R (Chapter IV, § 2, no. 5, Theorem 4), and since these intervals each contain a number > 0 and do not contain 0, they are contained in R^*_+ : that is, we have f_1(z) > 0 and f_2(z) > 0 for all z ∈ V.

If x and y are two elements of V such that x ≥ e and y ≥ e, then in particular we have xy ≥ e. Let E denote the (linearly ordered) set of elements of U which are ≥ e, and let I denote the set of elements of V which are ≥ e; then the axioms (GR_I), (GR_{II}), (GR_{IIIa}) and (GR_{IVa}) of § 2 are satisfied (taking ω to be the element e, and the law of composition to be that of the group G). This is clear for (GR_I), (GR_{II}) and (GR_{IVa}), from what precedes. As to (GR_{IIIa}), it is enough to remark that, if e < x < y (x ∈ V, y ∈ V), we have x^{-1} ∈ V, hence x^{-1} < e < x^{-1}y and x^{-1}y < y; consequently z = x^{-1}y belongs to I and we have xz = y. By Proposition 2 of § 2 there exists therefore a strictly increasing mapping f of I onto an interval of R, with left-hand end-point 0, such that f(e) = 0 and f(xy) = f(x) + f(y) whenever x, y and xy belong to I (which will be the case whenever x and y belong to W ∩ I, W being a neighbourhood of e such that W.W ⊂ V).

For every element x ∈ V which does not belong to I we have x < e, hence x^{-1} > e; consequently we can extend f to a strictly increasing mapping $\overline{f}$ of V onto an interval of R by putting $\overline{f}(x) = -f(x^{-1})$ for all x < e in V. The inverse image under $\overline{f}$ of an open interval contained in $\overline{f}(V)$ is an open interval of V, so that $\overline{f}$ is continuous on V; conversely, the image under $\overline{f}$ of an open interval of V is an open interval of $\overline{f}(V)$, and therefore $\overline{f}$ is a homeomorphism of V onto a neighbourhood of 0 in the group R. On the other hand, it is easily checked (as in Proposition 6 of § 1, no. 4, by considering the various possible cases) that we have $\overline{f}(xy) = \overline{f}(x) + \overline{f}(y)$ whenever x, y and xy all belong to V; and we therefore conclude that $\overline{f}$, restricted to a suitable neighbourhood of e in G, is a local isomorphism of G with R (Chapter III, § 1, no. 3, Proposition 3).

Q.E.D.