Scaling action and one parameter group action

Cosnider the following scaling action: $$\epsilon . x = e^{2\epsilon}x, \ \ \ \epsilon. y = e^{\epsilon}y.$$

I have to prove that this is a one paraemter group action. What does this mean? I can show that this is invariant under the scaling transformation i.e. scaling symmetry but I don't know how to show it's a one parameter group action. Any help appreciated.

Thanks

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A one parameter group action is an action of $\mathbb{R}$ on some space (usually linear action on a vector space). In this case, it's a representation from $\mathbb{R}$ into the invertible linear transformations of $\mathbb{R}^2$, since you have a real number acting on pairs of real numbers.

So this problem amounts to showing that the map $$\rho:\epsilon\mapsto \begin{pmatrix}e^{2\epsilon} & 0 \\ 0 & e^{\epsilon}\end{pmatrix}$$ is a homomorphism from $\mathbb{R}$ to $GL(2;\mathbb{R})$. That is, show that $\rho(s+t) = \rho(s)\rho(t)$.

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I find the notation rather odd, but I'll run with it, and further extend it so that $\epsilon.(x,y)=(e^{2\epsilon}x,e^\epsilon y)$. Then you need to prove that $(\epsilon+\eta).z=\epsilon.(\eta.z)$, and also that $0.z=z$. A more fancy way to say this is that the action is a group homomorphism from the additive group of real numbers to a group of transformations of the pairs $(x,y)$.

I said I find the notation odd: As you cansee from the equation I gave, there is a clashed between addition and the dot, as the latter looks more multiplicative than additive. So you suffer some cognitive upset when you use this notation.

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I can see that the notation is odd. When I write $\epsilon . x$ I have meant $\epsilon$ acting over $x$. – user45503 Jan 12 '13 at 16:54
Yes, I gathered as much. It is, however, unhelpful that $\epsilon.$ means different things depending on whether it acts on $x$ or on $y$. Finding good notation is often difficult, but always important, as bad notation gets in the way of clear thinking! (I edited my answer a bit, as an earlier edit had put the additional $0.z=z$ in the wrong place. Also clarified my variant of the notation, which is marginally better than the original.) – Harald Hanche-Olsen Jan 12 '13 at 17:00