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Math Help - Matrix norms and homotopies.

  1. #1
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    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 ).

    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?
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    MHF Contributor Drexel28's Avatar
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    Re: Matrix norms and homotopies.

    Quote Originally Posted by Deveno View Post
    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.
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    Re: Matrix norms and homotopies.

    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.
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    Re: Matrix norms and homotopies.

    Quote Originally Posted by Drexel28 View Post
    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).

    Quote Originally Posted by xxp9 View Post
    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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