Dear Daniel,

this group is the group of all affine transformations (linear transformations + translations) of the real line R (the complex version is mentioned in Viera's answer, you may substitute any field K for R or C). Indeed, if F(x)=ax+b, F'(x)=a'x+b', then

F*F'(x)=a(a'x+b')+b=aa'x+ab'+b.

In general, let K be a field. Consider the group G=Aff(n,K) of all affine transformations F: K^n->K^n of the affine space K^n over K, F(X)=A.X+B, where A\in GL(n,K), B\in K^n. Then F*F'(X)=AA'.X+A.B'+B. Note that G is the semi-direct product of the multiplicative group GL(n,K) (linear transformations) and additive group K^n (translations).

Best regards, Yuri

PS. Note also connections between some discrete subgroups of Aff(n,R) and crystallographic groups :

https://www.encyclopediaofmath.org/index.php/Crystallographic_group

Dear Daniel, no, I don´t remember to have seen this group yet :)

However, I'd like to mention a simple but beautiful geometric interpretation in a plane.

For any fixed (a,b), a homothety h composed with a translation t is defined:

• z'=az+b, where z,z' are complex numbers.

It is possible to generalize and take a,b complex, a non-zero. 

The transformation corresponding to (a,b) will be composed of a homothety with ratio |a|, a rotation with angle arg(a) and the common centre (0,0), and a translation of the vector u=(Re b, Im b).

So you may consider this group on C*xC. The neutral element (1,0) of the group corresponds to the identity in the plane, where a=1 and u=(0,0). The inverse of (a,b) in the group is (1/a,-b/a) and corresponds to the inverse transformation (h o t)-1.

Various interesting properties of the group could be discovered due to this geometric point of view.

Dear Daniel,

this group is the group of all affine transformations (linear transformations + translations) of the real line R (the complex version is mentioned in Viera's answer, you may substitute any field K for R or C). Indeed, if F(x)=ax+b, F'(x)=a'x+b', then

F*F'(x)=a(a'x+b')+b=aa'x+ab'+b.

In general, let K be a field. Consider the group G=Aff(n,K) of all affine transformations F: K^n->K^n of the affine space K^n over K, F(X)=A.X+B, where A\in GL(n,K), B\in K^n. Then F*F'(X)=AA'.X+A.B'+B. Note that G is the semi-direct product of the multiplicative group GL(n,K) (linear transformations) and additive group K^n (translations).

Best regards, Yuri

PS. Note also connections between some discrete subgroups of Aff(n,R) and crystallographic groups :

https://www.encyclopediaofmath.org/index.php/Crystallographic_group

Thanks so much, Viera, Yuri and Peter.

[ For readability this answer has been converted to pdf format and uploaded below. ]

Perhaps my query was too general or vague, but your answers were very sharp and forces me to make the question more precise.

As justified below, on the set $G=\R^*\times \R$, with elements $g=(a,b)$, take the group operation

\[

gg'=(a,b)(a',b')=(aa',b'+a'b)

\]

which is the opposite group of ---but is isomorphic to--- the affine group I originally mentioned.

Let $E$ be a real Hilbert space with elements $v\in E$; $PE$ its associated projective space, with elements $[v]\in PE$ where $[v]=\{\mu v \rr \mu\in \R\}=$line of $v$; $TPE$ the tangent space to the manifold $PE$, equal to the (disjoint) union of orthogonals

\[

TPE = \bigcup_{[v]\in PE} [v]^\bot

\]

The orthogonal can be replaced by a quotient to obtain an alternative definition of $TPE$

\[

TPE = \bigcup_{[v]\in PE} E/[v]

\]

If Hilbert spaces are too general, take simply $E=\R^n$.

Consider the product $X$ of the punctured Hilbert space and itself

\[

X=(E-\{0\})\times E

\]

with elements $x=(v,w)\in X$. An action $G\times X\to X$ is defined as $g x=(a,b)(v,w)=(av, w+bv)$.

The relevant fact is that {\emc the quotient $X/G$ is the tangent space $TPE$}. For this one uses the second definition of $TPE$ and verifies that $w+[v]=w'+[v']$ (equality of cosets in $TPE$) if and only if there exists $(a,b)\in G$ such that $(v',w')=(a,b)(v,w)$.

When the Hilbert space is one-dimensional, say $E=\R$, then both $PE$ and $TPE$ are connected zero-dimensional manifolds (they reduce to single points). On the other hand, still with $E=\R$, we have $X=\R^*\times \R$, the map $(v,w)\to (av, w+bv)$ is affine, and I believe it has bordered matrix as explained by Peter. In the next dimension, $E=\R^2$, we have $PE=$real projective line = circle, and $TPE$ is a cylinder.

There are similar results for spheres:\\

Let $\R^+$=multiplicative group of positive real numbers. This subgroup of $\R^*$ is the connected component of the identity and $\R^*/\R^+=\Z_2$. Then $G^+=\R^+ \times \R$ (with group operation given by restriction) is the connected component of the identity in $G$.

Let $SE$ be the sphere of $E$ defined as the quotient of $E-\{0\}$ by the multiplicative action of $\R^+$ and denote $TSE$ the tangent space to the manifold $SE$. This sphere consists of rays emanating from a center, but has no radius. Then $G^+$ acts on $X$ and the quotient is the tangent space $TSE$. The two-fold antipodal covering map $SE\to PE$ is $G$-equivariant. In particular when $E=\R^3$ one obtains the Celestial Sphere $S\R^3$ which, as astronomers tell, has no radius. Note that with our procedure the sphere $SE$ is well defined for any real vector space $E$. Inner products or norms are not required. It seems that if $E$ is a t.v.s. such that each $v\neq 0$ has a closed hyperplane complement, then $SE$ is a topological manifold modeled on closed hyperplanes.

In the complex Hilbert case these remarks are valid if one takes together the following: the complex affine group, $G_\C=(\C-\{0\})\times \C$ endowed with adequate group operation; complex Hilbert spaces $E^\C$; complex projective spaces $PE^\C$. With the {\em real sphere} of the complex Hilbert space, $SE^\C=(E^\C-\{0\})/G^+$, the induced equivariant map is Hopf fibration $h:SE^\C\to PE^\C$, seen here as an example of map between homogeneous spaces.

My original question should have been, within the Hilbert space framework: {\emc Is the action of $G$ on $X$, with quotient $TPE$, already known and worked out by someone, so that an independent checking of details is available}.

Meanwhile I rechecked calculations and found the need to use the opposite group. Things seem to work fine. So far and up to undetected mistakes.

Why should anyone desire such result as $TPE=X/G$? Well, to attempt solving certain differential equations in Hilbert manifolds, but this another subject better discussed on its own.

Again, thanks very much!

Cordially,

Daniel Crespin

P.S.: Usually a homothety (or homothecy) is a linear map on a vector space defined as multiplication by a fixed non-zero scalar. That is, $v\to av$ is a homothety. Often the condition $a>0$ is imposed.