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    i'm trying to find a subgroup of the afifne group that is isomorphic to the abelian group \mathbb{R}^n. i've tried the centre and got stuck-i thought this might be right because the elements of the centre commute as do elements of \mathbb{R}^n under addition. however, rather than proceeding by trial and error, which could take ages, does anyone have any suggestions?

    thanks
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    Assuming your definition of affine group is the set of all affine transformations (linear transformations + translations), can you see how translations could be isomorphic to \mathbb{R}^n?
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    ok. i have the affine group as the group of all invertible affine transformations.

    (i) the initial question:
    firstly i was asked to find a subgroup isomorphic to the general linear group GL(n,\mathbb{R})
    after some googling i found that the answer was Stab(x) for any arbitrary x \in \mathbb{R}^n. i then tried to prove why it was isomorphic:

    consider the function f:G \rightarrow GL(n,\mathbb{R}); g=(A,a) \mapsto A where G is the affine group and g \in G

    f is a homeomorphism since (for g_1=(A_1,a_1),g_2=(A_2,a_2) \in G)
    f(g_1 * g_2)=f((A_1A_2,A_1a_2+a_1))=A_1A  _2 \in GL(n,\mathbb{R})
    and
    f(g_1)*f(g_2)=f((A_1,a_1))*f((A_  2,a_2))=A_1*A_2=A_1A_2 \in GL(n,\mathbb{R})
    \Rightarrow f(g_1*g_2)=f(g_1)*f(g_2)
    and for A_1,A_2 \in GL(n,\mathbb{R}):
    f^{-1}(A_1*A_2)=f^{-1}(A_1A_2)=(A_1A_2,y) \in G for some y \in \mathbb{R}^n
    and
    f^{-1}(A_1)*f^{-1}(A_2)=(A_1,a_1)*(A_2,a_2)=(A_1  A_2,A_1a_2+a_1) \in G
    but y=A_1a_2+a_1 as both affine transformations are in the stabiliser of x and so must leave x unchanged.
    i.e. A_1A_2x+y=x and A_1A_2x+A_1a_2+a_1=x implies y=A_1a_2+a_1
    \Rightarrow f^{-1}(A_1*A_2)=f^{-1}(A_1)*f^{-1}(A_2)
    so f,f^{-1} are homeomorphisms.

    now f is surjective because \forall A \in GL(n,\mathbb{R}), \exists g \in G such that f(g)=A e.g. take g=(A,a)

    however i'm struggling to prove injectivity: since surely (A,a_1) and (A,a_2) would both map to A???

    if i can show that is injective then i will have my isomorphism.

    (ii) the second part is to find a subgroup isomorphic to \mathbb{R}^n. i still can't see what this would be. you are suggesting that the translations alone be a subgroup. how can this be? the elements g \in G are of the form g=(A,a) so how could T=\{ a \} \subset G be possible? could i perhaps fix the matrix involved in the transformation as A and then consider all the possible transformations of the form (A,-). these would form a subgroup (i think) although i'm not sure about the addition stuff.

    thanks a lot.
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    Have A as the identity... (sorry I'm in a rush)
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    was my answer to the first part ok? can you help prove that function is bijective?

    (Original post by SimonM)
    Have A as the identity... (sorry I'm in a rush)
    so for the second part i consider the subset T=\{ ( \mathbb{I},a ) | a \in \mathbb{R}^n \} \subseteq G. yes?
    i now need to prove f:T \rightarrow \mathbb{R}^n ; t \mapsto a for t \in T is an isomorphism?
    edit: i have succesfully proved T is isomorphic to \mathbb{R}^n
    and i think i've succesfully sorted out the first part of the question as well.
    note to self: don't trust wikipedia!

    thanks.
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    i do have another question on affine groups that i can't really get started on though:

    Show that the affine group is a linear group by showing it is the subgroup of GL(n+1,\mathbb{R}) which preserves the affine hyperplane consisting of vectors in \mathbb{R}^{n+1} whose last entry is equal to 1. What is the (n+1) \times (n+1) matrix corresponding to the affine transformation (A,a)? Check that under matrix multiplication one recovers the composition law (A_1,a_1)(A_2,a_2)=(A_1A_2,A_1a_  2+a_1)

    i don't really understand how to go about the bit about the hyperplane and then for the next bit do i just add a_i to the A_{ii} entry to give the (n+1)x(n+1) matrix?
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    Ok, so that subspace is isomorphic to \mathbb{R}^n by just looking at the first n co-ordinates. So we want (A, a) to work on the first n co-ordinates like that. So we want the map

    (A,a) \mapsto ( M : e_1 \mapsto Ae_1, e_2 \mapsto Ae_2, \ldots e_n \mapsto Ae_n, e_{n+1} \mapsto a_1e_1+a_2e_2 + \ldots a_ne_n + e_{n+1} )...

    Can you see this as a matrix?
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    hi.
    (Original post by SimonM)
    Ok, so that subspace is isomorphic to \mathbb{R}^n by just looking at the first n co-ordinates. So we want (A, a) to work on the first n co-ordinates like that. So we want the map

    (A,a) \mapsto ( M : e_1 \mapsto Ae_1, e_2 \mapsto Ae_2, \ldots e_n \mapsto Ae_n, e_{n+1} \mapsto a_1e_1+a_2e_2 + \ldots a_ne_n + e_{n+1} )...

    Can you see this as a matrix?
    sorry i don't really understand your last post.

    (i) when you say that subspace is isomorphic to [latx]\mathbb{R}^n[/latex], are you referring to the previous part of the question where we had to show T \simeq \mathbb{R}^n? or are you talking about some part of this new question?

    (ii) what is meant by an affine hyperplane? i haven't come across this before. searching the internet, i can't really see any distinguishing features from what i thought was just a "normal" hyperplane.

    (iii)why do we set the last coordinate to 1.

    (iv) i understand what you're saying when you're on about having (A,a) work only on the first n coordinates. this would leave the n+1th coordinate unchanged as 1. so affine transformations just map points in the affine hyperplane to other points in the affine hyperplane i.e. the affine hyperplane is left unlatered by affine transformations. is this true?

    (v) i don't understand your notation for that map in the last post. why is the M inside those brackets?

    sorry for being thick here.
    thanks.
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    (Original post by latentcorpse)
    (i) when you say that subspace is isomorphic to [latx]\mathbb{R}^n[/latex], are you referring to the previous part of the question where we had to show T \simeq \mathbb{R}^n? or are you talking about some part of this new question?
    So the hyperplane of points consisting of points in R^(n+1) with last coordinate 1 can be identified with R^n (just ignore the last coordinate)

    (ii) what is meant by an affine hyperplane? i haven't come across this before. searching the internet, i can't really see any distinguishing features from what i thought was just a "normal" hyperplane.
    The thing defined to me looks awful like a hyperplane, so I'm just as confused.

    (iii)why do we set the last coordinate to 1.
    Stupid answer: Because we're told
    Less stupid answer: Because it works in the next part of the question. To motivate it, we could consider 'how' (A,a) behaves

    [qupte](iv) i understand what you're saying when you're on about having (A,a) work only on the first n coordinates. this would leave the n+1th coordinate unchanged as 1. so affine transformations just map points in the affine hyperplane to other points in the affine hyperplane i.e. the affine hyperplane is left unlatered by affine transformations. is this true?[/quote]

    Yes

    (v) i don't understand your notation for that map in the last post. why is the M inside those brackets?
    We're trying to get an isomorphism from the affine group to a subgroup of GL(n+1) (to show that it is linear). I am mapping (A, a) to the map M, (which is defined as M: stuff to stuff)

    Sorry if any of this isn't clear, I'm exhausted
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    (Original post by SimonM)
    Sorry if any of this isn't clear, I'm exhausted
    it's ok. i think i understand all of the above except the mapping.

    so we're going to show the affine group is linear by finding an isomorphism between it and a subgroup of GL(n,R). That makes sense.

    now you have defined the isomorphism as
    M: ((A,a),e_i) \mapsto Ae_i \forall i \in [1, \dots , n] and e_i are the basis vectors of \mathbb{R}^n i.e. M: H \times \mathbb{R}^n \rightarrow \mathbb{R}^n where H is the affine group.
    so each basis vector is just rotated by the matrix A and therefore doesn't leave the hyperplane.
    is this correct?

    i don't think this is an isomorphism though. for example if we take h1 and h2 in H,

    M( h_1 * h_2 )=M( ((A_1,a_1),e_i) * ((A_2,a_2),e_j))=M( ((A_1A_2,A_1a_2+a_1),e_i+e_j) )=A_1A_2(e_i+e_j)
    but
    M(h_1)*M(h_2)=M( ((A_1,a_1),e_i))*M( ((A_2,a_2),e_j))=A_1e_i*A_2e_j=A  _1e_i+A_2e_j
    i.e. M(h_1*h_2) \neq M(h_1)*M(h_2) so M isn't even a homomorphism.

    N.B. i have used * to represent the binary operation in the respective space i.e. composition of affine transformations in H and vector addition in \mathbb{R}^n.

    thanks again.
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    It looks like you ignored what a_{n+1} maps to
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    (Original post by SimonM)
    It looks like you ignored what a_{n+1} maps to
    don't you mean e_{n=1}?

    why did you decide to map it to a_1e_1+a_2e_2+ \dots e_{n+1}?

    is the map i posted above correct? except that i should have had the index i running from 1 to n+1, yes?
    how will the inclusion of what e_{n+1} maps to solve the problem i listed above (i.e. with regards to it not being a homomorphism)?

    cheers.
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    (Original post by latentcorpse)
    don't you mean e_{n=1}?

    why did you decide to map it to a_1e_1+a_2e_2+ \dots e_{n+1}?

    is the map i posted above correct? except that i should have had the index i running from 1 to n+1, yes?
    how will the inclusion of what e_{n+1} maps to solve the problem i listed above (i.e. with regards to it not being a homomorphism)?

    cheers.
    No. I meant what I said. It means that the linear map acting on the last basis "adds" the a part of (A,a). Read my map carefully
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    (Original post by SimonM)
    Read my map carefully
    ok. so the map is

    M : H \times \mathbb{R}^n \rightarrow \mathbb{R}^n ; \begin{cases} ((A,a),e_i)  \mapsto  Ae_i  \forall i \in [1 \dots n] \\ ((A,a),e_i) \mapsto e_{n+1} + \displaystyle \sum_{j=1}^n Ae_j , i=n+1 \end{cases}

    so this is the correct map?

    know i have to show it is an isomorphism in order to establish that the affine group is isomorphic to a subgroup of GL(n,\mathbb{R}^n) as this allows me to conclude that the affine group is linear. is that correct?

    finally, the reason for mapping e_i to Ae_i when i is in [1...n] is so that points in the hyperplane are mapped to other points in the hyperplane, thus leaving it invariant. but why did you decide to map e_{n+1} to ae_1 + ... ae_n + e_{n+1}? surely you could equally well have just sent it to itself because so long as it's sent to something containing a non-zero multiple of e_{n+1} it won't be in the hyperplane?
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    bump.
 
 
 
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