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

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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.
  • #1
Spriteling
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Homework Statement



Show that the group G where [itex]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 [/itex]


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 [itex]C^k[/itex] 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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  • #2
Path connectivity is often the easiest direction to go in, especially when your set is parametrized by real numbers
 
  • #3
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?
 
  • #4
That's not quite right. Imagine
[tex]g = \left(\begin{array}{ccc} 1 &
0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right) [/tex]

[tex]g' = \left(\begin{array}{ccc} 0 &
-1 & 0\\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{array} \right) [/tex]

then the halfway point is

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

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 [itex]C^k[/itex] 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]
 
  • #5
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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