Are there nonlinear operators that have the group property?

Question

To be clear: What I am actually talking about is a nonlinear operator on a finitely generated vector space V with dimension $d(V)\;\in \mathbb{N}>1$. I can think of several nonlinear operators on such a vector space but none of them have the requisite properties of a group. In particular, but not exclusively, are there any such nonlinear operator groups that meet the definition of a Lie Group.

Answer 1

As some guy once said, dividing operators into linear and nonlinear is like dividing the world into bananas and nonbananas. If you're ignoring the vector space structure on $V$ then it's just a set, and you're just asking for a group action. It is easy to write down group actions, even Lie group actions (see homogeneous space), on a vector space that just don't happen to be linear. Probably the simplest example is $\mathbb{R}$ acting on itself by translation.

Answer 2

Oh I've got one....

Let A by a non-singular, invertible square matrix whose elements are components of vectors in $V^{n}$. Now consider the operator $T(\hat{v}) = e^{A\hat{v}}, \; \;\forall \hat{v} \in V^{n}$. So if $\alpha,\beta$ are any such operators with corresponding invertible non-singular matrices (A,B) respectively, than the binary group operation is $\phi(\alpha,\beta) = e^{AB\hat{v}} = (e^{A\hat{v}})^{B}$ where AB is the product matrix A*B. The inverse operator is $T^{-1}(\hat{v}) = A^{-1} ln(\hat{v})$.The identity mapping is $\hat{I}(\hat{v})= e^{ln(\hat{v})}=\hat{v}$.

Answer 3

How about the following construction. Let $M \in R_{d}(G)$ be a $d$ dimensional representation of some Lie group (e.g. $SL(2,R)$ and $V$ is some $d$ dimensional vector space. Now define: $$ f(x,M,\epsilon)= \frac{M x}{(1+\epsilon w.x)} $$ where the vector $w$ is chosen so that $w.M= w$ for all $M\in SL(2,\mathbb{R})$. Such vectors exist, for instance, if $d=4$ and $R_4(G)=R_2(G) \otimes R_2(G)$ then $w=\begin{pmatrix} 0 \\ 1\\ -1\\ 0 \end{pmatrix}$ is one such vector. It is now not too difficult to convince oneself that: $$ \begin{eqnarray} f(x,I,0) = x \\ f(f(x,M_1,\epsilon_1),M_2,\epsilon_2) = f(x,M_1 M_2,\epsilon_1+\epsilon_2) \end{eqnarray}$$ so that the mapping $f$ is nonlinear and satisfies the group property.