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From: Lie_Algebra on 18 Jul 2010 15:39
> On 18-07-2010 17:05, Lie_Algebra wrote:
>
> >>> Do you have any example or details of matrix
> calculation why \rho_{V^*}(g) = \rho(g^-1)^t ?
> >>
> >> No need to do any calculations. Just use the
> definition of "adjoint"
> >> and the fact that you want to have
> >>
> >> <g.f,g.v> =<f,v>
> >>
> >> whenever _v_ belongs to V and _f_ belongs to its
> dual. (Of course,
> >> <f,v> = f(v) by definition).
> >
> > I think I figure that out. It is basically from the
> book Fulton's "Representation theory: A first course"
> (p 5), and Dummit&Foote's "Algebra" 3rd edition
> (p434), Please let me know if you find any error
> below. I used the same notations with the book except
> replacing \phi with f, etc.
> >
> > If V is a representation of G, then its dual
> representation is defined as
> >
> > gf(v)=f(g^-1v) where f:V-->C, v \in V (To satisfy
> the condition for an action on a dual space,
> gf(v)=f(g^-1v) rather than gf(v)=f(gv)).
>
> If you are assuming this, then you are assuming the
> fact that you wish
> to prove.
>
> > If M is a matrix for a linear transformation
> T:V-->W, then M^t is a matrix for a linear
> transformation between their dual spaces from W^* to
> V^*.
>
> Yes, but it is a specific linear transformation. It
> is the linear
> transformation such that, whenever _v_ is in V and
> _f_ is in its dual,
>
> (M^t(f))(v) = f(M(v)).
>
> Therefore, if you want to have
>
> <g.f,g.v> = <f,v> = f(v)
>
> whenever _g_ belong to G, _v_ belongs to V, and _f_
> belongs to V^*,
> then you must have
>
> (g.f)(v) = <g.f,v> = <g.f,g.(g^{-1}.v)> =
> )> = <f,g^{-1}.v> = f(g^{-1}.v).
>
> and therefore,
>
> (g.f)(v) = f(g^{-1}.v).
>
> By _definition_ of adjoint, this means that g.f is
> the adjoint of the
> linear map v |-> g^{-1}.v.
>
> Best regards,
>
> Jose Carlos Santos

Thank you so much for your help.
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