# Are there nonlinear operators that have the group property?

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.

• Are you asking "are there any non-linear operators from $V$ to $V$ that are group homomorphisms?" or are you asking "are there any collections of non-linear operators from $V$ to $V$ that form a group?" ? May 25 '13 at 20:40
• I'm asking if there are any collections of nonlinear operators from V to V that form a group.
– Mr X
May 25 '13 at 20:51
• Does "nonlinear operator" have any particular meaning here I'm not aware of, or is it just an arbitrary map (of sets)? Pick your favourite three points in V, say a, b and c, and permute them around. (This is just a non-linear action of $S_3$ on V - you can do this with any group you like on any V that has enough points.) May 25 '13 at 21:18
• Yes it does, billy. If V is our n-dimensional vector space,f is a mapping from V to V, and A & B are any 2 vectors in V,then f is linear iff f(A+B)=f(A)+f(B) which is called the superposition principle. A nonlinear map f from V to V is one where f(A+B)!=f(A)+f(B).
– Mr X
May 25 '13 at 21:52
• Jun 4 '13 at 13:44

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.

• Based on Mr X's clarification in the comments, the $\mathbb{R}$ action on itself by left translation is "linear". (Mr X, perhaps nonstandardly, doesn't mention scalar multiplication!). That said, examples still abound. Pick your favorite nonlinear automorphism $f$ (in whatever category you wish, so long as the category contains the linear maps). Then conjugating $Gl(V)$ by $f$ will givea group of nonlinear things. May 26 '13 at 4:42
• Er... no it's not? May 26 '13 at 4:45
• Wow - I really shouldn't be doing math after midnight I apologize! I can delete my whole rant if you'd prefer. (And that would be my preference, but whatever.) May 26 '13 at 5:06
• Thanks, Jason DeVito! And Quiaochu Yuan, I was asking for examples of nonlinear transformations on a vector space $V^{n}\;$ where n > 1. I'm aware that a nonlinear maps of the form $f:\mathbb{R}^{1} \to \mathbb{R}^{1} \;$ form a translation group.
– Mr X
Jun 5 '13 at 14:36
• Examples are just as easy to write down in the higher-dimensional case. Jason DeVito's comment works in all dimensions. Jun 5 '13 at 18:18

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 $\hat{I}(\hat{v})= e^{ln(\hat{v})}=\hat{v}$.

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.