Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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
Do you allow your manifolds to have boundary? – JSchlather Nov 27 '12 at 22:06

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 $\Bbb R$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.