8

QUESTION: Is there a homomorphism

$$f :(\Bbb R, +) \to (\Bbb R, +)$$

that does not have the form $x \mapsto ax$, where $a\in\mathbb{R}$?

Gerry Myerson
  • 179,216
Enigma
  • 3,909
  • 1
    I take it from the answers that you meant a homomorphism of additive groups, not a linear map? – Seth Oct 11 '14 at 21:29
  • Yes, that is correct. – Enigma Oct 11 '14 at 21:32
  • This has been asked more than a handful of times before. See http://math.stackexchange.com/questions/423492/overview-of-basic-facts-about-cauchy-functional-equation for details. – Asaf Karagila Oct 11 '14 at 23:07
  • There is also the other side. If such a homomorphism is measurable, then it does have the form $ax$. – GEdgar Oct 12 '14 at 00:53

1 Answers1

13

Yes, using the Axiom of Choice. The abelian group $\mathbb R$ is an infinite-dimensional vector space over the field $\mathbb Q$. Unfortuately, one cannot give an explicit basis of this vector space, but the linear map of "swapping" two of the basis vectors, say, is a nontrivial automorphism of $\mathbb R$ as abelian group that is not of the form $x\mapsto ax$.

Hagen von Eitzen
  • 374,180
  • 1
    Neat, never thought of it like this – Alan Oct 11 '14 at 21:23
  • 1
    If we can't give a construction of the basis of this infinite-dimensional vector space, and thus can't know the actual results of the group operation, how do we know that this isn't just a more complicated way of mapping $x\mapsto ax$? – Peter Olson Oct 11 '14 at 22:01
  • 3
    @PeterOlson Because we can specify an initial part of the basis, say $1, \sqrt 2, \sqrt{3}, \pi, e,\ldots$. Then the linear map "swap the $1$ and $\sqrt 2$ components" would map $42\mapsto 42\sqrt 2$ (so $a$ must be $\sqrt 2$) and $\sqrt 3+7\pi-4+\sqrt 2\mapsto \sqrt 3+7\pi-4\sqrt 2+1$ (which doesn't match $a=\sqrt 2$). Alternatively, one could ppick a nontrivial map with nontrivial kernel. – Hagen von Eitzen Oct 11 '14 at 22:25
  • 1
    Another way to view it is that a homomorphism of the form $x\mapsto ax$ preserves order. If you swap two basis elements, as suggested by Peter Olson, the newly constructed homomorphism no longer preserves order. – Enigma Oct 12 '14 at 03:01