# Problem concerning smooth manifolds

1. Sep 25, 2007

### ala

A={ {{cos x, -sin x},{sin x, cos x}}|x $$\in$$R}, show that set A is smooth manifold in space of 2x2 real matrix. What is tangent space in unity matrix?

1. What is topology here? (Because I need topology to show that this is manifold)
2. In solution they say that mapping x$$\rightarrow$${{cos x, -sin x},{sin x, cos x}} is diffeomorphism between circle and given set and from this follows that given set is manifold.
How to prove that this is diffeomorphosm? (proving by definition is pretty hard, although it's somehow obvious that this is true if use proper topology)
Diffeomorphism is special kind of mapping between two smooth manifolds - so I really don't know to show that this mapping is diffeomorphism (because before that I must show that A is smooth manifold).
3. They also say that in coords of this one dimensional smooth manifold, tangent vector is $$\frac{\partial}{\partial x}$$?

2. Sep 25, 2007

### HallsofIvy

Staff Emeritus
Do you recognize that matrix as the rotation, in R2 about the origin through an angle $\theta$? In other words, every matrix in A corresponds to a point on the unit circle- the point to which (1, 0) would be rotated by this matrix. This is a "many to one" correspondence since sine and cosine are periodic with period $2\pi$. you might take, as your "coordinate neighborhoods", the intervals $(n\pi -\delta, (n+1)\pi - \delta)$, with the obvious coordinate functions from R to A.

3. Sep 25, 2007

### ala

I know that, but in problem no topology is given. It is not justified to use topology that seem most natural. How can somebody ask question is some set manifold? (you can ask is some topological space manifold or is set A with topology T manifold)
Someone can say, ok matrices 2x2 can be understood as points in R^4, and given set of matrices as one dimensional subset of R^4 so I will use topology from R^4 (i.e. topology induced by metric in R^4). Using this topology given set maybe isn't manifold.
Do you see ambiguity?