From: José Carlos Santos on
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
From: Timothy Murphy on
José Carlos Santos wrote:

>>> My functions are functions from G into R.
>>
>> Are you sure?
>
> Yes.
>
>> I assume \mathfrac{g} is the Lie algebra LG of G.
>
> Indeed.
>
>> So the second function is a function from LG to R.

Apologies.
I misunderstood the function you were referring to.


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Timothy Murphy
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