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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.

I solved to get that the integral curve is given by $$\gamma(t)=(\cos(t-a),\sin(t-a),\cos(b-\lambda t),\sin(b-\lambda t))$$ for arbitrary constants $a,b$. Given an arbitrary point $x=(\cos a,\sin a,\cos b,\sin b)\in S^1\times S^1$, I want to say that the one-parameter group is $$f_t(x)=(\cos(t-a),\sin(t-a),\cos(b-\lambda t),\sin(b-\lambda t))$$

But there is a problem: $f_0(x)=(\cos(-a),\sin(-a),\cos b,\sin b)=(\cos a,-\sin a,\cos b,\sin b)\neq x.$ What can I do here?

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PJ Miller
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  • There's some kind of symmetry in this question. So I think that the first two components of $\gamma(t)$ shall be $\cos(a-t),\sin(a-t)$. Then the solution makes sense. – Shuchang Nov 14 '13 at 02:28
  • @ShuchangZhang The first two components $(\gamma_1(t),\gamma_2(t))$ need to satisfy $d\gamma_1/dt=-\gamma_2$ and $d\gamma_2/dt=\gamma_1$, so I think it must be $\cos(t-a),\sin(t-a)$ rather than $\cos(a-t),\sin(a-t)$. I'm quite sure I solved the system correctly. – PJ Miller Nov 14 '13 at 02:52
  • So, why do you have to set intial value $\cos a,\sin a$. Why not $\cos a,-\sin a$? – Shuchang Nov 14 '13 at 03:10
  • @ShuchangZhang Yeah, I think I should be able to fix it by changing the initial values. Thank you. – PJ Miller Nov 14 '13 at 03:42
  • Also, I'd better set $\gamma(t)$ to $(\cos(t+a), \sin(t+a), \cos(b-\lambda t), \sin (b - \lambda t))$. Of course it's a solution and satisfies initial condition $\gamma(0) = (\cos a, \sin a, \cos b, \sin b)$. – Evgeny Nov 14 '13 at 11:09

1 Answers1

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According to your own solution, just let $a$ and $b$ equal to zero.

You may also want to consider the torus model (as in$\mathbb C^2$), and the one-parameter group is geometrically more tangible: $$ \{(e^{i2\pi\theta},e^{i2\pi\lambda\theta})|\theta\in\mathbb R\} $$ Coming to this point, you may want to investigate whether $\lambda$ is a rational number or not. If $\lambda$ is a rational number, say $\frac{m}{n}$, then your one-parameter group looks like (after reparametrization): $$ \{(e^{i2\pi\theta/m},e^{i2\pi\theta/n})|\theta\in\mathbb R\} $$ which is easily seen to be periodical. And therefore, it is diffeomorphic to the $S^1\simeq\mathbb R/[m,n]\mathbb Z$ (where $[m,n]$ is the least common multiple of $m$ and $n$). If $\lambda$ is not a rational number, your one-parameter group will go on and on without even coming to back to the start. And therefore it is diffeomorphic to $\mathbb R$.

Troy Woo
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