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Isomorphism of matrix groups

  1. Sep 24, 2005 #1
    Can isomorphism of matrix groups
    [tex]\phi: G_1 \rightarrow G_2[/tex]
    always be expressed by
    [tex]\phi(M) = S M S^{-1}[/tex]?
  2. jcsd
  3. Sep 24, 2005 #2
    I don't know if I got the question right (had linear algebra a long time ago). Can we assume [tex] \phi(\lambda_1 w_1 + \lambda_2 w_2) = \lambda_1 \phi(w_1) + \lambda_2 \phi(w_2) [/tex] and that there is a basis for [tex] G_1, G_2 [/tex]?

    I think it works something like this then:

    If [tex] { (a_1, a_2, \dots , a_n) } [/tex] is a basis to [tex] G_1 [/tex], what can you say about [tex] { (\phi(a_1), \dots, \phi(a_n))} [/tex] ? What does this say about the linear combination you get by mapping [tex] g_1 = \sum \limits_{i=1}^n \alpha_n a_n [/tex] to [tex]G_2[/tex] by [tex] \phi [/tex] ? What does this mean for an arbitrary basis of G_2?
    Last edited: Sep 24, 2005
  4. Sep 24, 2005 #3
    Well, since groups are not vector spaces, I think this was just glibberish. Sorry.
  5. Sep 24, 2005 #4


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    You have to make a big assumption for φ(M) = SMS^-1 to even have a chance at working -- is it a valid assumption?
  6. Sep 25, 2005 #5
    I mean matrix goups of same dimension
    let me rephrase the question:
    Can isomorphism of matrix groups of same dimension
    [tex]\phi: G_1 \rightarrow G_2[/tex]
    always be expressed by
    [tex]\phi(M) = S M S^{-1}[/tex]?
    S is any invertible matrix
  7. Sep 26, 2005 #6
    What does "same dimension" mean here?
    Would the group of matrices [tex] (A_i) [/tex] have the same dimension as the group of Matrices [tex] \left ( \begin{array}{*{2}{c}}
    A_i & 0 \\
    0 & 0
    \end{array} \right)[/tex] ?
  8. Sep 26, 2005 #7
    if dimension of a matrix group is n, then each element is an n x n matrix.
  9. Jun 30, 2008 #8
    I think what you're remembering is the following: Let K be one of the fields R, C, or H. Then, up to equivalence, the only irreducible real representation of K(n) is the natural representation on Kn. Hence, if you have two irreducible real representations r1 and r2 of K(n), then there exists a matrix S such that r2(M) = Sr1(M)S-1 for all M in K(n).

    You can find details of this in Lang's Algebra.
  10. Jul 1, 2008 #9

    matt grime

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    Resurrecting a 3 year old thread?

    As it happens, I don't think you're supposed to assume that G_i is the group of all nxn invertible matrices.

    The answer to (my take on) the original question is no, since it is equivalent to the assertion that all (faithful) representations of a group are uniquely characterised by dimension, which is not true, eg S_n has two faithful n-1 dim reps (for n>3): the natural permutation representation breaks up as the direct sum of trivial plus an n-1 dim rep V. V\otimes sign is the second, which is different from V if n>3.
  11. Jul 24, 2008 #10
    Right, this thread is the closest to the topic I could find.

    My question is more convention choice than anything else.

    The finite Cyclic [tex]C_3[/tex] group is defined by:


    where [tex]e[/tex] is the identity, [tex]c[/tex] is rotation through [tex]\frac{2\pi}{3}[/tex] etc. I'm keeping formalities to a minimum here. Clearly, we are rotating a triangle with directed sides in one plane through three angles, yeah?

    Now, my question has to do with representations of this group.

    Let me begin with how I am learning about groups.

    The definition I'm working from (and taking the example from) is in line with H F Jones' "Groups, Representations and Physics Second Ed".

    A representation of a Group [tex]G[/tex] is the pair [tex]{\iota, V}[/tex] where [tex]V[/tex] is the vector space which is also a group and

    [tex]\iota : G \mapsto V[/tex]

    is a homomorphism. With the usual condition that [tex]\iota[/tex] preserves the group structure.

    So, some books understandably skip the generality of [tex]V[/tex] and claim that [tex]V=M_{n \times n}[/tex] the set of invertible nxn matrices. And to represent [tex]C_3[/tex] we will use the finite subspace

    [tex]R(\theta)= \begin{bmatrix}\cos \theta & -\sin \theta & {0} \\ \sin \theta & \cos \theta & {0} \\ {0} & {0} & {1}\end{bmatrix} [/tex]

    where [tex]\theta=0, \frac{2 \pi}{3},\frac{4\pi}{3}[/tex]

    A representation of [tex]C_3[/tex] is the pair [tex]{\iota, M}[/tex] where [tex]M[/tex] is the vector space with three elements

    [tex]M=R(0),R(\frac{2 \pi}{3}),R(\frac{4\pi}{3})[/tex]

    [tex]\iota : G \mapsto M[/tex]

    Now, finally, I can ask my question.

    According to the book [tex]R(\frac{2 \pi}{3})[/tex] would be the "representation" of [tex]c[/tex] in [tex]C_3[/tex]. I find this kind of confusing.

    So what would ([tex]\textbf{x'},\textbf{x}[/tex] are just Cartesian column vectors)



    Can someone shed some light on how to view this definition as intuitive. I understand that a representation shouldn't demand a co-od system. I suppose, to me, it just seems like we're identifying an 'operator' ([tex]R(\theta)\in M_{n \times n}[/tex]) with a 'state' ([tex]e,c,b \in C_3[/tex]).
  12. Jul 24, 2008 #11
    That isn't really a minimum. A cyclic group with 3 elements is just <g: g^3=e>

    That is an example of a cyclic group with 3 elements. It is not *the* example.

    There are 3 simple ones over the complex numbers....

    If you're just learning about groups, then it may be best to ignore representations to begin with.

    That is not a representation.

    See, here's a problem. A vector space V comes equipped with a group operation, sure. Addition. But representations map from G to a set of invertible matrices with the group operation matrix composition.

    Representing G, and working out *a* representation are probably not the same thing. Be careful.

    M is not a vector space.

    You should think of it as identifying a group G with a group of operators, though really we aren't identifying since representations are not isomorphisms from G to something but homomorphisms from G to matrices.

    Shall we start again? Let G be a group. A representation of the GROUP G is a group homomorphism from G to the invertible matrices M_n for some n. There are infinitely many such representations.

    If G were C_3 as above then one representation of G would send a generator of G to rotation by 2pi/3 in R^2. Another represenatation would be to send a generator of G to the identity matrix in M_n for any n.
  13. Jul 24, 2008 #12
    when you say

    "send a generator of G to rotation by 2pi/3"

    what does "to rotation" mean, like an operator on R^2?
  14. Jul 24, 2008 #13
    If f is a function from X to Y, and f(x)=y for some x in X, then it is a common abuse of language to say f sends x to y.

    A representation is a group homomorphism. A function from G to End(V).

    If G=C_3 = <g : g^3=e> , then V=R^2, and the function f(g)=R(2pi/3) - the 2x2 rotation matrix by the angle 2pi/3 - extended in the obvious manner (i.e. f(g^2) is rotation by 4pi/3 etc) is a representation of G.
  15. Jul 24, 2008 #14
    May I redefine in more familiar language(to my course).

    A representation of C_3 is the pair {i,R^2}, where

    i : C_3 -> R^2

    is the function taking the following form:

    i(g) = R(2pi/3)

    where, g \in C_3 and R is the 2X2 rotation matrix.
  16. Jul 24, 2008 #15
    No! That is not a representation. That is just a group hom (presumably) into the vector space R^2.

    That rotation matrix isn't in R^2, so that makes no sense. As you wrote it, i ought to be a map from C_3 to End(R^2) - the matrices of the linear maps from R^2 to itself- and that is a representation.

    I'll say it again: a representation of G is a group hom to the linear maps of a vector space. It is not a map to the vector space itself.

    There are several ways of referring to a representation which may be confusing you.

    To define a representation of G we need _in full_ a vector space V, and a group homomorphism f:g--> End(V). We may call either f, or V the representation, but the function f is matrix valued.
  17. Jul 25, 2008 #16
    I apologise, it was very late last night. It all makes sense now. Further, you have solved the puzzle as to why I thought I was dealing with 'operators'.

    I hope that I have it right:

    A representation of [tex]C_3[/tex] is the pair [tex]\left\{ \iota,\mathbb{R}^2\right\}[/tex] such that;

    [tex]\iota : g \mapsto End(\mathbb{R}^2)[/tex]

    [tex]\forall g \in C_3[/tex]. The set of all linear maps over [tex]\mathbb{R}^2[/tex].


    [tex]\iota(g)=R(2\pi/3) [/tex]
    [tex]\iota(g^2)=R(4\pi/3) [/tex]
    [tex]\iota(g^3)=\iota(e)=R(0) [/tex]


    [tex]R(\theta)= \begin{bmatrix}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{bmatrix}[/tex]

    is just the rotation matrix. The necessary condition of preservation of group structure are inherited from matrix algebra.

    And of course,

    [tex]R(\theta) : \mathbb{R}^2 \mapsto \mathbb{R}^2[/tex]

    which is what gives us our representation.
  18. Jul 25, 2008 #17
    Looking at the definition, I didn't realise that V in his definition was not "the space to visualise things in" [tex]\mathbb{C}^{n \times n}[/tex] but the space of all linear maps over that space. And usually it's just taken to be [tex]M_{n \times n}(\mathbb{C})[/tex]

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