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

• Spriteling
In summary, the conversation discusses how to show that the group G, defined as a matrix with certain parameters, is connected. The conversation explores different approaches, such as using path connectivity or showing that G is not the disjoint union of two non-empty sets. Ultimately, it is suggested that the best approach may be to show that any element in G can be written as an exponential, thus connecting all points in G.
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.

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?

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 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.

## 1. How can you prove that a set is connected?

To prove that a set is connected, you must show that it cannot be divided into two non-empty disjoint subsets.

## 2. What is the definition of a connected set?

A connected set is a set where any two points in the set can be connected by a continuous path without leaving the set.

## 3. What is the difference between connected and disconnected sets?

A connected set cannot be divided into two non-empty disjoint subsets, while a disconnected set can be divided into two or more non-empty disjoint subsets.

## 4. Can a set be both connected and disconnected?

No, a set can only be either connected or disconnected. A set cannot have both properties simultaneously.

## 5. What are some common techniques for proving a set is connected?

Some common techniques for proving a set is connected include using the definition of connectedness, the Intermediate Value Theorem, and the Bolzano-Weierstrass Theorem.

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