# Proving a set is connected

## 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.

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Office_Shredder
Staff Emeritus
Gold Member
Path connectivity is often the easiest direction to go in, especially when your set is parametrized by real numbers

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
Staff Emeritus
Gold Member
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]

Yeah, I realised 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.