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Consider a 2N dimensional space, $x\in \mathbb{R}^{2N}$ is a point with constraint $||x||_2=1$

Thus $x$ is actually lies on the surface of a unit sphere.

Given that we know the fact $x$ is always on the surface on the unit sphere

With aid of trigonometric functions can we claim that only $N$ parameters is needed to describe the position of $x$?

For instance $x=[x_1,...,x_{2N}]^T$, can I claim $\theta = [\theta_1,...,\theta_N]^T$ is enough to describe $x$ since we can rewrite $x$ into $x=[sin(\theta_1) ,cos(\theta_1) ,sin(\theta_2) ,cos(\theta_2) ,... ,sin(\theta_N) ,cos(\theta_N)]^T$?

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What's a tri-function? And it seems there are lots of commas missing in your vectors; they all seem to contain only a single component. – joriki Dec 4 '12 at 14:34
Thanks for the comment. It is updated. – Rein Dec 4 '12 at 14:36
That would mean $x_1^2+x_2^2=1$, which is definitely not true. You can't parametrize an $m$-dimensional manifold smoothly with fewer than $m$ variables. In this case, $m=2N-1$ and you are trying to parametrize wit $N$ variables, which can only work if $N=1$. – Thomas Andrews Dec 4 '12 at 14:37
N parameters aren't enough. To describe surface you need 2N-1 parameters – unknown Dec 4 '12 at 14:38
You describe a point $x$ on $(S^1)^N \subseteq \mathbb R^{2N}$, where $S^1 = \{x \in \mathbb R^2\mid \|x\|_2 = 1\}$. But $S^{2N-1} \ne (S^1)^N$. – martini Dec 4 '12 at 14:40
up vote 0 down vote accepted

The proposed map doesn't even map onto the sphere, since $|x|^2=\sum_i (\cos^2\theta_i + \sin^2\theta_i) = N$.

Even if you divided your formula by $\sqrt{N}$, so that the resulting point was always on the sphere, it wouldn't be complete - there would be many many many pointson the sphere that would not bein your image, such as $(1,0,0,...0)$.

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