# General Linear Group

1. Nov 30, 2009

### latentcorpse

If you create a topological space by taking the set of 2x2 matrics of the form $A=\left( \begin{array}{cc} a & b \\ c & d \end{array} \right)$ and considering them as vectors $(a,b,c,d) \in \mathbb{R}^4$. Let $GL_2(\mathbb{R}) \subset \mathbb{R}^4$ be the subset of 2x2 matrices that are invertible i.e. $ad-bc \neq 0$

is:

(i) $GL_2(\mathbb{R}) \subset \mathbb{R}^4$ an open subspace?
(ii) $GL_2(\mathbb{R})$ compact?
(iii) $GL_2(\mathbb{R})$ connected?

i'm inclined to say no for (i) but i can't really explain it. if you take an invertible matrix and give it a small perturbation (i.e. change one of the entries by a small amount) it will not necessarily be invertible any more....i.e. it could now be outside the subspace, meaning it's a closed subspace. but i don't know how to put that into maths

for compactness, it needs to be closed and bounded, surely if my answer to (i) is correct then it cannot be compact

finally, i am not sure how to go about (iii) - any advice?

2. Nov 30, 2009

### Dick

You don't really think GL2 is a subspace, do you? As for open, if A is invertible then if a=det(A) is a finite distance from 0. Changing A by a little bit can't change that, can it? And isn't det(A) a continuous function on R^4?

3. Nov 30, 2009

### latentcorpse

hi. you've lost me a bit.

so we're saying it cannot be a subspace. this can be proved by finding a suitable counterexample

e.g.

$\left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right) + \left( \begin{array}{cc} -1 & 0 \\ 0 & -1 \end{array} \right) = \left( \begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array} \right)$

the two matrices on the left are in the general linear group and in the topological space $GL_2 ( \mathbb{R})$ they are the vectors $\left( 1,0,1,0 \right) , \left(-1,0,-1,0 \right)$. The matrix on the right has $ad-bc=0$ and therefore $\left( 0,0,0,0 \right) \notin GL_2 ( \mathbb{R} )$. Thus $GL_2 ( \mathbb{R} )$ isn't a subspace of $\mathbb{R}^4$. i guess i'd also have to put in a bit about how in $\mathbb{R}^4, \left( 1,0,1,0 \right) + \left( -1,0,-1,0 \right) = \left( 0,0,0,0 \right) \in \mathbb{R}^4$.

so we've established it's not a subspace. isn't that sufficient to prove (i) is false?
anyway, suppose we also want to show it's either open or closed (whichever it turn out to be) in order to be rigorous about (i)...

why would perturbing A not change a? surely if you changed an entry in A by say 1/2 this could have a drastic change on a?
also, how do we know det is a continuous funciton on $\mathbb{R}^4$? i assume we use this in an argument to do with "continuous functions take open sets to open sets"?

thanks.

4. Nov 30, 2009

### Dick

Sure, a change of 1/2 can be a big change, but to show it's open you can perturb by an arbitrarily small amount. You know det(A) is continuous just by looking at ad-bc. Isn't that a continuous function of four variables without even bothering with a formal proof? And "continuous functions take open sets to open sets" isn't even true. The inverse image of open sets are open. det:R^4->R. What open set are we talking about?

5. Nov 30, 2009

### latentcorpse

so we can infact prove $GL_2 ( \mathbb{R} )$ is open by saying that if det(A)=a and we change A by an arbitrarily small perturbation to A' then a' will be arbitrarily close to a and so in the set $P=\mathbb{R} \backslash \{ 0 \}=(-\infty,0) \cup (0,\infty)$. therefore as det is continuous, $det:\mathbb{R}^4 \rightarrow \mathbb{R}$, we will have $GL_2 ( \mathbb{R} )$ open if $P$ is open, which it clearly is.

Therefore we can say $GL_2 ( \mathbb{R} )$ is an open set in $\mathbb{R}^4$ but not a subspace. Therefore (i) will be false?

6. Nov 30, 2009

### Dick

You can also prove det(A) is continuous by saying fa(A)=a, fb(A)=b, fc(A)=c and fd(A)=d are clearly continous. det(A)=fa(A)*fd(A)-fb(A)*fc(A) is continuous since products and differences of continuous functions are continuous. But sure GL2 is the inverse image of R\{0}, therefore it's open. Is it compact? And what about iii)?

7. Nov 30, 2009

### latentcorpse

firstly, in post 5, is that a suitbale way to write an answer in an exam do you reckon? is there not a more rigorous way of writing something like "we change A by an arbitrarily small perturbation to A' then a' will be arbitrarily close to a"?

moving on:

well a subspace $X \subseteq \mathbb{R}^n$ is compact if and only if it is closed and bounded.

so $GL_2 ( \mathbb{R} ) \subseteq \mathbb{R}^4$ is compact if and only if it's closed and bounded. However, we just showed $GL_2 ( \mathbb{R} )$ is an open subset of $\mathbb{R}^4$ and therefore it cannot be compact as the if and only if statement breaks down. correct?

for connectedness, i'm inclined to say that it's not connected.
A topological space is connected if it cannot be writeen as a union $A \cup B$ of open, disjoint, non-empty sets $A,B \subseteq X$.

But we can write $GL_2 ( \mathbb{R} ) = SL_2 ( \mathbb{R} ) \cup \left( GL_2 ( \mathbb{R} \backslash SL_2 ( \mathbb{R} )$

now let the two sets in the union be A and B respectively (for ease of typing), A and B are certainly disjoint and non-empty. Are they open?

I imagine they are. I can't remember if the subset of an open set is automatically open (probably not - could you confirm please?), but surely A will be open by a similar argument to before :

all these matrices have det A =1. an arbitrarily small perturbation to A will mean the det is no longer exactly equal to 1 though. so i'm a bit confused....

Last edited: Nov 30, 2009
8. Nov 30, 2009

### Dick

Just because a set is open doesn't mean it also isn't closed. Try a find a sequence of invertible matrices that converge to a noninvertible matrix. As for connected, SL2 isn't open. Think more simply. The determinant of a matrix in GL2 is either positive or it's negative.

9. Nov 30, 2009

### Hurkyl

Staff Emeritus
Question (i) is a little ambiguous. Are they asking about topological spaces, or are they asking about topological vector spaces? When I first read it, I assumed the former.

10. Nov 30, 2009

### Dick

Good point. Since they were embedding in R^4, I took it to mean vector subspace. Probably wrongly.

11. Dec 1, 2009

### latentcorpse

hi again.

(i) does having taken it to be a vector subspace mean my answer is going to be wrong?

(ii) i assume that i'm trying to show no such sequence of matrices exists because of it did then the non invertible matrix would be included in $GL_2{ \mathbb{R} )$ which isn't allowed by defn. however if no such sequence exists, then we know GL_2 ( \mathbb{R} ) is going to be closed as sequences of matrices with no zero determinants converge to sequences of matrices with non zero determinants. i'm not really sure how to go about this, does it start something like:
assume $A_1, A_2, \dots$ is a sequence of matrices with $A_i \in GL_2 ( \mathbb{R} )$ that converges to $A_k$ ???

(iii) i think i can show that if $P : = \{ A \in GL_2 ( \mathbb{R} ) | det (A) >0 \}$ and $Q : = \{ A \in GL_2 ( \mathbb{R} ) | det (A) <0 \}$ then $GL_2 ( \mathbb{R} ) = P \cup Q$ where P and Q are certainly non empty and disjoint. if i can show that they are open then i can show $GL_2 ( \mathbb{R} )$ is disconnected.
do i show these are open using another sequence of matrices as above?

thanks.

12. Dec 1, 2009

### Dick

As Hurkyl pointed out, they almost certainly mean 'open subspace' in the sense of 'open subset'. What does that make your answer to i)? Use that det is continuous. Sequences of matrices with nonzero determinant CAN converge to a matrix with zero determinant! Give me an example. Why don't you try and finish this up and give your answers and reasons for each part.

13. Dec 1, 2009

### Hurkyl

Staff Emeritus
You can always fill in the missing detail to clarify what question you're answering. Even better, you can give both answers.

14. Dec 1, 2009

### latentcorpse

(i) do i show it's open by showing $\mathbb{R}^4 \backslash GL_2 ( \mathbb{R} )$ is closed.
All matrices ( or vectors really ) in $\mathbb{R}^4 \backslash GL_2 ( \mathbb{R} )$ have $ad-bc=0$ so any sequence of such matrices will converge to a matrix $A_k$ which also has det = 0 i.e. $\mathbb{R}^4 \backslash GL_2 ( \mathbb{R} )$. That is, $\mathbb{R}^4 \backslash GL_2 ( \mathbb{R} )$ contains it's limit points and is therefore closed. This means $GL_2 (\mathbb{R} )$ is an open subset.

(ii) ok such an example would be something like

$\left( \begin{array}{cc} 2 & 2 \\ 1 & -1 \end{array} \right) , \left( \begin{array}{cc} 1 & 1 \\ \frac{1}{2} & -\frac{1}{2} \end{array} \right) , \left( \begin{array}{cc} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{4} & -\frac{1}{4} \end{array} \right), \dots , \left( \begin{array}{cc} 0 & 0 \\ 0 & -0 \end{array} \right) , \end{array} \right)$

so $GL_2 ( \mathbb{R} )$ doesn't contain it's limit points and is therefore open (just realised this is exactly what i just did above!) anyway. to me i would then say it can't be compact because it's not closed. but you're telling me it could also be closed. how do i test this?

(iii) i could show P is open by showing GL2(R)\Q is closed using a sequnce of matrices and similarly Q is open by showing GL2(R)\P is closed using a sequence of matrices. then this would show GL2(R) is connected.

15. Dec 1, 2009

### Dick

i) Ok, if you want to go that way. But how do you know if Ak converges to A, that det(Ak) converges to det(A)?

ii) NOT containing a limit point DOES NOT mean a set is open. Being open DOES NOT mean a set is not closed. You have a convergent sequence in GL2 with a limit point that is not in GL2. What does that tell you about compactness (without digressing into whether the set is open or closed).

iii) How can you show a set is open using a sequence???? I thought you were trying to show P and Q are open? Now you are going to show that GL2(R)\P is closed???? GL2(R)\P=Q! If you want to use a complement is closed argument you want to show R^4\P is closed.

16. Dec 1, 2009

### latentcorpse

for (i) the det of any matrix $A_i$ in the sequence $\{ A_k \}$ has determinant 0 and so the sequnce of dets is just 0,0,0,....,0 i.e. if $A_k$ converges to $A$, $A$ will also have det=0 as the $A_i \in \mathbb{R}^4 \backslash GL_2 ( \mathbb{R} )$. As you may have noticed, my understanding of openness is not great. Here we show the complement is closed as it contains a limit point. this implies the set iteslef, GL2(R) is open and by open we mean topologically open i.e. is contained within the topology, yes?

(ii) In my lecture notes, the only reference to limit points is that an infinite subset $A \subseteq X$ of a compact space has limit points. But i don't see how i can use this because it only talks about the compactness of X.
Perhaps, if we take X=GL2(R) and then A= {matrices in that convergent sequence above} then A is infinite. This means GL2(R) will be compact if the limit point of the sequence is in A. The limit point is not in A hence X is not compact. How's that?

(iii) i take R^4\P and a sequence of matrices in that such as

$\left( \begin{array}{cc} 2 & 2 \\ -1 & 1 \end{array} \right) , \left( \begin{array}{cc} 1 & 1 \\ -\frac{1}{2} & \frac{1}{2} \end{array} \right) , \left( \begin{array}{cc} \frac{1}{2} & \frac{1}{2} \\ -\frac{1}{4} & \frac{1}{4} \end{array} \right) , \dots \left( \begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array} \right)$

not sure this is going to work. i need to use a sequence that converges to a positive number surely. what about if i used $\displaystyle \sum_{n=1}^{\infty} \frac{2}{n}+1$ on the entries rather than $\displaystyle \sum_{n=1}^{\infty} \frac{2}{n}$??

17. Dec 1, 2009

### Dick

You are vaguely there in ii). Take your sequence from iii) and move it into ii). Now you have a convergent sequence in GL2 whose limit is NOT is GL2. That's IT. GL2 is not compact. You might want to amplify that be saying why the sequence converges in the topology of R^4, but I'll pass on that. The rest of it is really a mess. You really need to use that det is a continuous function and use the properties of continuous functions. Like for a continuous function, the inverse image of open sets is open. Will you try and find the statement of that theorem in your book and understand it? And understand why det:R^4->R is continuous? Until you do that you are just going to keep thrashing around and getting nowhere. If it helps, try and figure out what GL1(R) would be. 1x1 invertible matrices. All of these statements are true for GL1(R) as well. Can you explain why? That would be good practice.

18. Dec 2, 2009

### latentcorpse

ok. so would (i) just be that $det: \mathbb{R}^4 \rightarrow \mathbb{R}$.
Now we have $det ( GL_2 ( \mathbb{R} ) ) = ( -\infty,0) \cup (0, \infty)$
which is clearly open and since det is a continuous function, $GL_2 ( \mathbb{R} )$ is open. we can show det is continuous using what you said in post #6 (we only have to concern ourselves with the 2 dimensional case).
Hopefully that's looking a bit better now...
can i just ask though: we are trying to prove $GL_2 ( \mathbb{R} )$ is topologically open in $\mathbb{R}^4$ i.e. here open means it's a set that's contained in the topology. but when i said $( -\infty,0) \cup (0, \infty)$ was open i was taking about open sets i.e. it has a ( bracket not a [ bracket. Surely when we say the inverse image of open sets is open (when we have a continuous functions) we are talking about topologically open in both cases?

19. Dec 2, 2009

### Dick

A=(-infinity,0) union (0,infinity) is open in R. B=GL2(R) is open in R^4. Yes, that's the way it's supposed to work. The inverse image under det:R^4->R of A (which is open in R) is B (which is open in R^4). In think you are ready for the 'connected' question.

20. Dec 2, 2009

### latentcorpse

ok but before i have a stab at (iii), why is $(-\infty,0) \cup (0, \infty)$ open in $\mathbb{R}$. surely we need to prove it's in the topology? not just say it's open because it has ) brackets instead of ] brackets.

for (iii), should i be looking for subsets A,B, or lack thereof, of GL2(R) s.t. GL2(R)=AuB where A and B are open, disjoint and connected?