
#1
Nov112, 08:22 PM

P: 101

Hi guys,
This is a general question that I'm thinking about now. Imagine that I've been given a set which is a group and we have defined a topology on it. how can I show that the group operation is continuous? Actually to begin with, how can I know if the group operation is really continuous? maybe it's not continuous? As an example, suppose that the group is GL(n,ℝ). How can I show that the matrix multiplication is continuous? The first thing that confuses me is that the function is defined from GL(n,ℝ)×GL(n,ℝ) → GL(n,ℝ). so for each open set in GL(n,ℝ) I should show that its preimage is also open. But the topology on GL(n,ℝ)×GL(n,ℝ) is different, no? It's the product topology I guess. so I'm really confused about how I should show that matrix multiplication is a continuous operation on GL(n,ℝ), it seems a little tricky. Any helps would be appreciated. But please by descriptive, since I'm an untalented undergraduate school who can be pretty absent minded and slow sometimes. I've also asked this question on here http://www.physicsforums.com/showthread.php?t=648847 where I've proved, with the help of haruspex, that matrix inversion is continuous. Thanks in advance 



#2
Nov112, 09:54 PM

P: 1,023

I don't know how you do it in general, beyond showing that the definition is true.
But for GL(n,ℝ), GL(n,ℝ) is topologized as a subspace of ℝ^(n^2) (the vector space of all n by n matrices). I wouldn't go directly from the definition. Just look at the definition of matrix multiplication. It's built out of a bunch of continuous things. You just add stuff and multiply stuff. Continuous. Like showing x^2 is continuous. You can show that f(x) is continuous using epsilons and deltas because that is trivial. Then products of continuous functions are continuous, so you're done. And usually, if you write down a typical function given by some formula from ℝ^n to ℝ^m, you just know it's continuous for these reasons. Just like that. You can multiply any two matrices at all and that gives you a continuous map from ℝ^(n^2) χ ℝ^(n^2) to ℝ^(n^2). The group operation on GL(n, R) is just a restriction of that map. Restricting a continuous map to a subspace gives a continuous map. So, this will show you that any real matrix group is continuous. 



#3
Nov312, 02:00 PM

Sci Advisor
P: 1,716

matrix multiplication is a polynomial in each coordinate. Polynomials are continuous on subsets of Euclidean space.




#4
Nov712, 09:54 AM

P: 101

How to show that the operation of a group is continuous?
well, yea. That's what I initially thought too. But actually let's make it a bit more general. Imagine that instead of ℝ we're working in a general field F. F could be any field even with a nonzero characteristic like Z_{p}.
Please check this proof: http://www.physicsforums.com/showthread.php?t=648847 This is what is in my head now: Nowhere in my proof I've assumed the fact that the entries of the matrix A comes from ℝ. I mean for almost every bit of my proof I haven't assumed anything just more than the fact that ℝ is a field and ℝ^{n} has been given a topological structure induced by a norm defined on it. I guess I haven't used the fact that ℝ is an ordered field with convergent Cauchy sequences in it or stuff like that (I'm talking about the entries of A, not about the norm of A which is indeed a real number). Let's assume we're working in GL(n,Z_{p}) or GL(n,ℂ) (Even though that the argument for polynomial multiplication would still work in ℂ). Can we still say that matrix multiplication and inversion in GL(n,F) are continuous operations? For example one of the bad things that could happen in a vector field of dimension n over Z_{p} is that the Euclidean norm would fail on it. For example if V is a vector space of dim=2 over Z_{p} (p=4k+1 for some k) then some Fermat's theorem tells u that [itex] (((p1)/2)!)^2 \equiv 1 [/itex]. For example in Z_{5} the 2D vector (1,3) has a Euclidean length of 0, even though it's not the zero vector. so the Euclidean norm can't be used on Z_{p}^{n}. But I think much of the things I'd proved earlier about matrix inversion would remain valid if we have a norm on Z_{p}. Am I wrong? I think I've made my point clear. My point is, in a group where we've defined a topology on it we certainly can make the product topology for G × G and study continuity of the group operation. No? Now maybe this is too general, but at least in metric spaces that there is a richer topological structure there might be tips and tricks that help me in future when I encounter some harder math stuff. That's my point. 



#5
Nov812, 10:47 AM

P: 350

1. If multiplication is continuous then inversion is continuous. This can be proved in the most general of circumstances (I think).
2. Multiplication is not automatically continuous without some kind of hypothesis on the topology. E.g. G = (Z/3Z, +). Open sets are G, the empty set, and {2}. That is a topology, but the inverse image of {2} under the map 1+x is not open. 3. What about metric topologies on vector spaces? Take R with the metric d'= d/(1+d) where d is the usual distance. Then modify d'(0,x)=2 for all x other than 0. The open sets are the same except the point 0 is removed from all of them (except from the entire real line). The inverse image of the open set 0<x<2 under the mapping x > x+1 is (1,1) which is not open. 4. Continuity is equivalent to gU being open for all g in G and all open U. 



#6
Nov912, 10:32 AM

Sci Advisor
P: 1,716





#7
Nov1612, 02:19 AM

P: 101

Sorry guys if I'm being annoying or I'm asking too much. I know that when a question is too general it becomes hard to answer it. For now even if you can answer whether polynomials with their coefficients in a given ring are always continuous or not I would be very appreciative. 



#8
Nov1612, 09:11 AM

P: 350

So I double checked my statements, which were a bit hasty and reckless :)
1. Given continuous multiplication, I remembered doing an exercise in a textbook proving that inversion is automatically continuous. I found it in Armstrong. What I forgot were the additional hypotheses on the group. G must be Hausdorff and compact. If it is not, then the statement is false in general. http://math.stackexchange.com/questi.../151889#151889 2. In the example of the metric topology on R, my explanation was weak, but it is still a counter example. The singleton {0} is open because it is the ball of radius 1 around 0. However, there are no other singletons in the topology, so left translation is not a homeomorphism. 3. Considering the first point, if gU is open for all g and all U, this is clearly not enough to guarantee that the group is topological. This just means that G acts continuously on the left on itself. Now, if G also acts continuously on the right, that might be enough to prove that multiplication is continuous, but I'm not sure... 


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