One-parameter group of diffeomorphisms generated by vector field 
Let $S^1$ be the unit sphere $x_1^2+x_2^2=1$ in $\mathbb{R}^2$ and let $X=S^1\times S^1\in\mathbb{R}^4$ with defining equations $f_1=x_1^2+x_2^2-1=0, f_2=x_3^2+x_4^2-1=0$. The vector field $$w=x_1\frac\partial{\partial x_2}-x_2\frac\partial{\partial x_1}+\lambda\left(x_4\frac\partial{\partial x_3}-x_3\frac\partial{\partial x_4}\right)$$ ($\lambda\in\mathbb{R}$) is tangent to $X$ and hence defines by restriction a vector field $v$ on $X$. What is the one-parameter group of diffeomorphisms that $v$ generates?

The definition of a one-parameter group of diffeomorphisms that I'm using is the following:

Let $U$ be an open subset of $\mathbb{R}^n$ and $F : U \times \mathbb{R} \rightarrow U$ a $C^{\infty}$ mapping. The family of mappings $f_t: U \rightarrow U$ , $f_t(x) = F(x, t)$
  is said to be a one-parameter group of diffeomorphisms of $U$ if $f_0$ is the identity
  map and $f_s \cdot f_t = f_{s+t}$
  for all s and t.

First of all, I'm confused how this definition can be applied to our situation. The vector field $v$ is not present anywhere in the definition of a one-parameter group of diffeomorphisms. But it has to be relevant somewhere, right?
 A: This vector field corresponds to following ODE in $\mathbb{R}^4$:
$$ \left ( \begin{array}{right}
\\ \dot{x_1}\\  
\dot{x_2} \\
\dot{x_3} \\
\dot{x_4} \end{array} \right) = 
\left (
\begin{array}{right}
\\  -x_2
\\  x_1
\\  \lambda x_4
\\  -\lambda x_3
\end{array}
\right ). $$
Actually, dynamics of $x_1$, $x_2$ is decoupled from dynamics of $x_3$, $x_4$ and equations can be easily solved. If we represent pairs $(x_1, x_2)$ and $(x_3, x_4)$ as complex numbers $z_1 =  x_1 + ix_2$ and $z_2 = x_3 + i x_4$ from unit circle in $\mathbb{C}$, then solution can be written in form $(e^{i(t + \phi_1)}, e^{i(-\lambda t + \phi_2)})$, where $\phi_1$ and $\phi_2$ can be derived from initial value problem for ODE. Well, this is it: the one-parameter group of diffeomorphisms generated by this vector field. You take an arbitratry point $x^{(0)} = (x^{(0)}_1, x^{(0)}_2, x^{(0)}_3, x^{(0)}_4)$; it corresponds to 
$(z^{(0)}_1, z^{(0)}_2)$, where $z^{(0)}_1 = e^{i\phi^{(0)}_1}$, $z^{(0)}_2 = e^{i\phi^{(0)}_2}$. The action $F(x^{(0)}, t)$ is following:
$$ F(x^{(0)}, t) = \left (
\mathrm{Re} \left ( e^{i\left(t + \phi^{(0)}_1\right)} \right),
\mathrm{Im} \left ( e^{i\left(t + \phi^{(0)}_1\right)} \right),
 \mathrm{Re} \left ( e^{i\left(-\lambda t + \phi^{(0)}_2\right)}\right),
 \mathrm{Im}\left ( e^{i\left(-\lambda t + \phi^{(0)}_2\right)} \right) \right). $$ 
Properties of complex exponenti is a key fact to show that $F(x, t)$ is a one-parameter group of diffeomorphisms.
