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is $M=\{ (\cos t,\cos2t):t\in \mathbb{R}\}$ a differential manifold

I know that for small intervals is manifold(is a parabola), but in all $\mathbb{R}$ i dont know

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I think you're right. It's just a piece of parabola bounded by the square $[-1, 1]^2$. –  Tunococ Nov 27 '12 at 20:26
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Do you allow your manifolds to have boundary? –  JSchlather Nov 27 '12 at 22:06

1 Answer 1

As $\cos(2t)=2\cos^2t-1$ and $\cos t$ can be any real number in $[-1,1]$, we have that $$M=\{(x,2x^2-1) \mid x\in [-1,1]\}$$ So, this all is a piece of parabola, with endpoints $P=(-1,1)$ and $Q=(1,1)$. In other view, $M$ is just the range of the continuous path $t\mapsto (\cos t,\cos (2t))$, which tour across $M$, from $Q$ to $P$ then to $Q$ again, and so on, touching each point infinitely many times. The crucial thing is that $M$ is homeomorphic to a line segment (i.e. $(x,2x^2-1)\mapsto x$ gives a homeomorphism).

So, the answer is yes in case we are talking about manifolds with boundary, and in this case $\partial M=\{P,Q\}$. And the answer is no if only manifolds without boundary are allowed, in this case you can argue that $P$ and $Q$ don't have neighborhood in $M$ homeomorphic to an open set of $\Bbb R$.

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