Matrix norms and homotopies.

**Deveno** · December 1st 2011, 10:28 AM

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

).

the question is this:

is the subgroup of $\mathrm{GL}_n(\mathbb{R})$ , given by:

$H = \{A \in \mathrm{GL}_n(\mathbb{R})\ |\text{ } \exists \text{ continuous } f:[0,1] \to \mathrm{GL}_n(\mathbb{R}) \text{ with } f(0) = A, f(1) = I \}$

normal in $\mathrm{GL}_n(\mathbb{R})$ ?

the topology used on $\mathrm{GL}_n(\mathbb{R})$ is the standard (metric) topology on $\mathbb{R}^{n^2}$ , which I believe induces the frobenius norm on the matrices. this norm is sub-multiplicative, so it seems to me, that if $P \in \mathrm{GL}_n(\mathbb{R})$ , that:

$g:[0,1] \to \mathrm{GL}_n(\mathbb{R})$ given by:

$g(t) = Pf(t)P^{-1}$ is the desired homotopy of $PAP^{-1}$ with $I$ , which would then prove the normality of $H$ .

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

suppose that for $t_0 \in [0,1], f(t_0) = B$ , and denote $f(t) = B_t$ .

if $\epsilon > 0$ , then if I choose $\delta > 0$ such that:

$|t - t_0| < \delta \implies |B_t - B| < \frac{\epsilon}{|P||P^{-1}|}$ , then

$|t - t_0| < \delta \implies |g(t) - g(t_0)| = |PB_tP^{-1} - PBP^{-1}| = |P(B_t - B)P^{-1}|$

$\leq |B_t - B||P||P^{-1}| < \epsilon$ , right?

**Drexel28** · December 1st 2011, 12:40 PM

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

).

the question is this:

is the subgroup of $\mathrm{GL}_n(\mathbb{R})$ , given by:

$H = \{A \in \mathrm{GL}_n(\mathbb{R})\ |\text{ } \exists \text{ continuous } f:[0,1] \to \mathrm{GL}_n(\mathbb{R}) \text{ with } f(0) = A, f(1) = I \}$

normal in $\mathrm{GL}_n(\mathbb{R})$ ?

the topology used on $\mathrm{GL}_n(\mathbb{R})$ is the standard (metric) topology on $\mathbb{R}^{n^2}$ , which I believe induces the frobenius norm on the matrices. this norm is sub-multiplicative, so it seems to me, that if $P \in \mathrm{GL}_n(\mathbb{R})$ , that:

$g:[0,1] \to \mathrm{GL}_n(\mathbb{R})$ given by:

$g(t) = Pf(t)P^{-1}$ is the desired homotopy of $PAP^{-1}$ with $I$ , which would then prove the normality of $H$ .

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

suppose that for $t_0 \in [0,1], f(t_0) = B$ , and denote $f(t) = B_t$ .

if $\epsilon > 0$ , then if I choose $\delta > 0$ such that:

$|t - t_0| < \delta \implies |B_t - B| < \frac{\epsilon}{|P||P^{-1}|}$ , then

$|t - t_0| < \delta \implies |g(t) - g(t_0)| = |PB_tP^{-1} - PBP^{-1}| = |P(B_t - B)P^{-1}|$

$\leq |B_t - B||P||P^{-1}| < \epsilon$ , 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 $G$ is a topological group with multiplication map $m:G\times G\to G$ and $i:G\to G$ . Then, the map $c_g:G\to G$ given by $h\mapsto ghg^{-1}$ is the composition

$h\mapsto (h,g) \mapsto (h,g^{-1})\mapsto hg^{-1}\mapsto (g,hg^{-1})\mapsto ghg^{-1}$

The first map is just the continuous map $G\to G\times\{g\}$ , the second is continuous because each coordinate map is continuous, the third is just the continuous multiplication map, the fourth is just the inclusion $G\ \{g\}\times G$ , and the last just multiplication again.

So, if you know that $\text{GL}_n(\mathbb{R})$ is a topological group then you're golden because your function $g$ is just $c_P\circ f$ , which is the composition of two continuous functions. But, $\text{GL}_n(\mathbb{R})$ 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 $\text{GL}_n(\mathbb{R})$ is a Lie group).

Remark: It's not fruitful to fret over whether a given norm is the one that induces your topology. Indeed, $\text{GL}_n(\mathbb{R})$ is a fin. dim. vector space, and so all norms induce the same topology. But, yes if you fix the usual Euclidean norm for $\mathbb{R}^{n^2}$ then this norm is carried naturally by the obvious identification $\mathbb{R}^{n^2}\approx\text{GL}_n(\mathbb{R})$ to the Frobenius norm.

**xxp9** · December 1st 2011, 12:41 PM

So the question is, is the connected component containing I, as a subgroup of $G=GL_n(R)$ , normal or not?
Note that $GL_n(R)$ 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 $g \in H$ , 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, $gh \in K$ , so gH is contained in K.
And for any $a \in K, a=g*(g^{-1}a)$ , and $det(g^{-1}a)=det(a)/det(g)>0$ , so $g^{-1}a \in H$ ,
So $a \in gH$ , that is K is contained in gH. So we have gH=K. Similarly we have Hg=K. So gH=Hg. H is normal.

**Deveno** · December 1st 2011, 02:01 PM

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 $G$ is a topological group with multiplication map $m:G\times G\to G$ and $i:G\to G$ . Then, the map $c_g:G\to G$ given by $h\mapsto ghg^{-1}$ is the composition

$h\mapsto (h,g) \mapsto (h,g^{-1})\mapsto hg^{-1}\mapsto (g,hg^{-1})\mapsto ghg^{-1}$

The first map is just the continuous map $G\to G\times\{g\}$ , the second is continuous because each coordinate map is continuous, the third is just the continuous multiplication map, the fourth is just the inclusion $G\ \{g\}\times G$ , and the last just multiplication again.

So, if you know that $\text{GL}_n(\mathbb{R})$ is a topological group then you're golden because your function $g$ is just $c_P\circ f$ , which is the composition of two continuous functions. But, $\text{GL}_n(\mathbb{R})$ 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 $\text{GL}_n(\mathbb{R})$ is a Lie group).

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

Remark: It's not fruitful to fret over whether a given norm is the one that induces your topology. Indeed, $\text{GL}_n(\mathbb{R})$ is a fin. dim. vector space, and so all norms induce the same topology. But, yes if you fix the usual Euclidean norm for $\mathbb{R}^{n^2}$ then this norm is carried naturally by the obvious identification $\mathbb{R}^{n^2}\approx\text{GL}_n(\mathbb{R})$ to the Frobenius norm.

well, when i first came across the problem, the usual topology (for $\text{GL}_n(\mathbb{R}))$ 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).

Originally Posted by xxp9

So the question is, is the connected component containing I, as a subgroup of $G=GL_n(R)$ , normal or not?
Note that $GL_n(R)$ 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 $g \in H$ , 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, $gh \in K$ , so gH is contained in K.
And for any $a \in K, a=g*(g^{-1}a)$ , and $det(g^{-1}a)=det(a)/det(g)>0$ , so $g^{-1}a \in H$ ,
So $a \in gH$ , 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 $\text{GL}_n(\mathbb{R})$ , which makes it rather simple, because det is continuous.

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