Matrix norms and homotopies.

I came across a question which I think I know the answer to, but I'd like some confirmation, or a counter-example to show me where my reason fails me (as it often does....darn unreliable brain (Surprised)).

the question is this:

is the subgroup of , given by:

normal in ?

the topology used on is the standard (metric) topology on , which I believe induces the frobenius norm on the matrices. this norm is sub-multiplicative, so it seems to me, that if , that:

given by:

is the desired homotopy of with , which would then prove the normality of .

the fly in the ointment being, that has to be continuous, which is where the frobenius norm comes in.

suppose that for , and denote .

if , then if I choose such that:

, then

, right?

Re: Matrix norms and homotopies.

Quote:

Originally Posted by

**Deveno** I came across a question which I think I know the answer to, but I'd like some confirmation, or a counter-example to show me where my reason fails me (as it often does....darn unreliable brain (Surprised)).

the question is this:

is the subgroup of

, given by:

normal in

?

the topology used on

is the standard (metric) topology on

, which I believe induces the frobenius norm on the matrices. this norm is sub-multiplicative, so it seems to me, that if

, that:

given by:

is the desired homotopy of

with

, which would then prove the normality of

.

the fly in the ointment being, that

has to be continuous, which is where the frobenius norm comes in.

suppose that for

, and denote

.

if

, then if I choose

such that:

, then

, right?

Right, this looks fine to me. I think the more general thing, that if someone would have said to you, you would have understood is the conjugation maps are continuous in topological groups. Indeed, suppose that is a topological group with multiplication map and . Then, the map given by is the composition

The first map is just the continuous map , the second is continuous because each coordinate map is continuous, the third is just the continuous multiplication map, the fourth is just the inclusion , and the last just multiplication again.

So, if you know that is a topological group then you're golden because your function is just , which is the composition of two continuous functions. But, is clearly a topological group since the multiplication maps and inversion maps are just rational functions in each coordinate (in fact, this clearly implies that is a Lie group).

*Remark:* It's not fruitful to fret over whether a given norm is the one that induces your topology. Indeed, is a fin. dim. vector space, and so all norms induce the same topology. But, yes if you fix the usual Euclidean norm for then this norm is carried naturally by the obvious identification to the Frobenius norm.

Re: Matrix norms and homotopies.

So the question is, is the connected component containing I, as a subgroup of , normal or not?

Note that has only two components, H={det>0} and K={det<0}, where det is the determinant function.

A subgroup H of G is said to be normal if Hg=gH for any g in G. If , we have gH=Hg=H certainly.

If g is not in H, det(g)<0, we have det(gh)=det(g)det(h)<0, that is, , so gH is contained in K.

And for any , and , so ,

So , that is K is contained in gH. So we have gH=K. Similarly we have Hg=K. So gH=Hg. H is normal.

Re: Matrix norms and homotopies.

Quote:

Originally Posted by

**Drexel28** Right, this looks fine to me. I think the more general thing, that if someone would have said to you, you would have understood is the conjugation maps are continuous in topological groups. Indeed, suppose that

is a topological group with multiplication map

and

. Then, the map

given by

is the composition

The first map is just the continuous map

, the second is continuous because each coordinate map is continuous, the third is just the continuous multiplication map, the fourth is just the inclusion

, and the last just multiplication again.

So, if you know that

is a topological group then you're golden because your function

is just

, which is the composition of two continuous functions. But,

is clearly a topological group since the multiplication maps and inversion maps are just rational functions in each coordinate (in fact, this clearly implies that

is a Lie group).

d'oh! conjugation is continuous, right. should i start dyeing my hair blonde?

Quote:

*Remark:* It's not fruitful to fret over whether a given norm is the one that induces your topology. Indeed,

is a fin. dim. vector space, and so all norms induce the same topology. But, yes if you fix the usual Euclidean norm for

then this norm is carried naturally by the obvious identification

to the Frobenius norm.

well, when i first came across the problem, the usual topology (for was indicated, so it seemed appropriate to use the usual norm for the vectorization of a matrix. since i was trying to prove the continuity of g directly, i needed the sub-multiplicative property, and not all norms are sub-multiplicative (although that can be fixed by re-scaling).

Quote:

Originally Posted by

**xxp9** So the question is, is the connected component containing I, as a subgroup of

, normal or not?

Note that

has only two components, H={det>0} and K={det<0}, where det is the determinant function.

A subgroup H of G is said to be normal if Hg=gH for any g in G. If

, we have gH=Hg=H certainly.

If g is not in H, det(g)<0, we have det(gh)=det(g)det(h)<0, that is,

, so gH is contained in K.

And for any

, and

, so

,

So

, that is K is contained in gH. So we have gH=K. Similarly we have Hg=K. So gH=Hg. H is normal.

great answer. i wasn't thinking in terms of homotopies being paths in , which makes it rather simple, because det is continuous.