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Intertwining Maps

  1. Jun 27, 2014 #1
    1. The problem statement, all variables and given/known data
    My online class notes:
    "Along the same vein as linear maps between vector spaces and group homomorphisms between groups we have maps between group representations that respect the algebraic structure.

    Definition 3.1: Let (p,V) and (q,W) be two representations of a group G. a lineaer transformation
    ø: V → W is an intertwining map if ø(p(g)v) = q(g)ø(v) for all v in V and g in G."

    okay so my first question is what exactly does the arguments in the notation for the representations mean? For example, the (p,V) representation; I know the V is a vector space, but what is the p? Is it the permutation or mapping that results from the group action on the vector space? If so, why do two representations of the same group (p,V) and (q,W) have different mappings if the group acting on them only has one binary operation?

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Jun 27, 2014 #2


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    If I'm reading correctly, ##p## and ##q## are group homomorphisms that map the group onto the vector spaces ##V## and ##W##.

    ##\phi: V → W \space | \space \phi(p(g)v) = q(g)\phi(v), \forall v \in V, g \in G##

    It looks like the linear map is mapping ##(p(g)v) \in V → (q(g) \phi(v)) \in W##. So ##\phi## is intertwining when it takes the mapped representation of an element ##g \in G##, multiplied by some vector ##v \in V## to its representation in ##W##, where ##\phi (v) \in W## as well.
  4. Jul 1, 2014 #3
    So what exactly are p and q? What are they defined by? Is the binary operation in which G is closed under represented anywhere in this notation? Are p and q the unique mapping for each vector space yet they are "guided" by the same binary operation in G? I'm new to this stuff if it's not obvious >.<.
  5. Jul 1, 2014 #4
    So this is the condition for it to be considered an intertwining map --> ø(p(g)v) = q(g)ø(v)
    ø is the mapping between representations
    g is an element of group G
    v is an element of vector space V
    p and q are doing something to each element of G.....

    and then it's confusing why it's ø(p(g)v) = q(g)ø(v) rather than say ø(p(g)v) = ø(q(g)v)
  6. Jul 1, 2014 #5


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    A representation of a group [itex]G[/itex] on a vector space [itex]V[/itex] is a group homomorphism [itex]p : G \to \mathrm{Aut}(V)[/itex], where [itex]\mathrm{Aut}(V)[/itex] is the set of invertible linear maps from [itex]V[/itex] to itself ("automorphisms") and its group operation is composition of functions. Thus each [itex]g \in G[/itex] is associated with an invertible linear map [itex]p(g) : V \to V[/itex]. Since [itex]p(g)[/itex] is itself a function it is common to denote it by [itex]p_g[/itex] so that the image of [itex]v \in V[/itex] is [itex]p_g(v)[/itex]. Since [itex]p[/itex] is a homomorphism one has [tex]
    p_{g_1g_2} = p_{g_1} \circ p_{g_2}
    [/tex] for every [itex]g_1 \in G[/itex] and every [itex]g_2 \in G[/itex].

    Here you have not only a representation of [itex]G[/itex] on [itex]V[/itex], but also a representation of [itex]G[/itex] on the vector space [itex]W[/itex], this second representation being denoted [itex]q : G \to \mathrm{Aut}(W)[/itex] so that [itex]q_g : W \to W[/itex] is an invertible linear map.

    A linear map [itex]\phi : V \to W[/itex] is then intertwining if and only if [tex]
    \phi \circ p_g = q_g \circ \phi\qquad\mbox{(*)}
    [/tex] for every [itex]g \in G[/itex]. Hence on right composition by [itex]p_h[/itex], [itex]h \in G,[/itex] the left hand side of (*) becomes [tex]\phi \circ p_g \circ p_h = \phi \circ p_{gh}[/tex] and the right hand side of (*) becomes [tex]q_g \circ \phi \circ p_h
    = q_g \circ q_h \circ \phi = q_{gh} \circ \phi[/tex] as required.
  7. Jul 1, 2014 #6


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    ##p: G → GL(V)## and ##q: G → GL(W)##

    Where ##GL(n)## is the general linear group.
    Last edited: Jul 1, 2014
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