Is G a Connected Group? Proving Connectivity in a Matrix Set

[Spriteling]

Homework Statement

Show that the group G where $G = \left(\begin{array}{ccc} \cos\theta & - sin\theta & u \\ \sin\theta & \cos\theta & v \\ 0 & 0 & 1 \end{array} \right) u,v \in \Re, \theta \in \Re/2\pi Z$

Homework Equations

I know that a set is connected if it is not the disjoint union of two non-empty sets, if any two elements in G can be joined by a $C^k$ path in G, and if G is generated by a neighbourhood of 1.

The Attempt at a Solution

I thought that it would be easiest to show that G is not the disjoint union of two non-empty sets, and I was trying to do it by contradiction, but got nowhere.

Any hints/nudges in the right direction would be appreciated.

 

[Office_Shredder]

Path connectivity is often the easiest direction to go in, especially when your set is parametrized by real numbers

 

[Spriteling]

Would it then work to say that you can take the function f: [0,1] -> G, where f = tg + (1-t)g' with g, g' members of G. f(0) =g and f(1) = g' so G is indeed connected?

Or am I missing something here?

 

[Office_Shredder]

That's not quite right. Imagine
$$g = \left(\begin{array}{ccc} 1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right)$$

$$g' = \left(\begin{array}{ccc} 0 & -1 & 0\\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{array} \right)$$

then the halfway point is

$$g = \left(\begin{array}{ccc} 1/2 & - 1/2 & 0\\ 1/2 & 1/2 & 0 \\ 0 & 0 & 1 \end{array} \right)$$

which isn't actually of the desired form. You need to be a little more careful about how you deal with the sin and cos terms

Homework Equations

I know that a set is connected if it is not the disjoint union of two non-empty sets, if any two elements in G can be joined by a $C^k$ path in G, and if G is generated by a neighbourhood of 1.

The Attempt at a Solution

I thought that it would be easiest to show that G is not the disjoint union of two non-empty sets, and I was trying to do it by contradiction, but got nowhere.

Any hints/nudges in the right direction would be appreciated.[/QUOTE]

 

[Spriteling]

Yeah, I realized that last night when I was showering.

However, I can't seem to find a curve that will actually suit all of the conditions, namely that:
f(t) = a(t)g + b(t)g' with b(t) + a(t) = 1 from the bottom right corner and that a(t)cos(x) + b(t)cos(y) = cos(z) and a(t)sin(x) + b(t)sin(y) = sin(z).

I think it may just be easier to show that any g in G can be written as an exponential, and so then all points are joined together.