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From: Lie_Algebra on 16 Jul 2010 17:45
> On 16-07-2010 16:40, Lie_Algebra wrote:
>
> >>> This is an Example 4.15 of the book "An
> >> introduction to Lie groups and Lie Algebras" by
> >> Kirillov.
> >>>
> >>> link: (p52)
> >>>
> >>
> http://www.math.sunysb.edu/~kirillov/liegroups/liegrou
> >> ps.pdf
> >>>
> >>>
> ===================================================
> >>> Example 4.15.
> >>> Let B be a bilinear form on a representation V.
> >> Then B is invariant under the action of G defined
> in
> >> Example 4.12 (please see the link) iff
> >>>
> >>> B(gv, gw)= B(v, w) (1)
> >>>
> >>> for any g \in G, v,w \in V. Similarly, B is
> >> invariant under the action of \mathfrak{g} iff
> >>>
> >>> B(x.v, w) + B(v, x.w) = 0 (2)
> >>>
> >>> for any x \in \mathfrak{g}, v, w \in V.
> >>>
> >>> (*) We leave it to the reader to check that B is
> >> invariant iff the linear map V-->V^* defined by v
> |->
> >> B(v, -) is a morphism of representations.
> >>>
> >>
> ====================================================
> >>> The definition of (1) makes sense to me. Anyhow,
> I
> >> am having hard time understanding how (2) is
> derived
> >> or defined.
> >>
> >> Consider the equality (1) and see it as an
> equality between two
> >> functions from G into R (or C). The second
> function is constant, of
> >> course. Now, derive this equality. On the left you
> get
> >>
> >> B(X.v,w) + B(v,X.w)
> >>
> >> and on the right you get 0, of course.
> >
> > What are your first and second functions? I still
> can't figure out how the equality is derived.
>
> My functions are functions from G into R. The first
> function is the
> function defined by g |-> B(g.v,g.w), whereas the
> second one is defined
> by g |-> B(v,w) (_v_ and _w_ are fixed).
>
> Now, take X in _g_. Consider the map from R into R
> defined by
>
> t |-> B(exp(t X).v,exp(t X).w).
>
> It is a constant map. Therefore, its derivative at 0
> is 0. But the
> derivative at 0 is B(X.v,w) + B(v,X.W).
>
> >>> Can anyone give some hints or thoughts including
> (*)?
> >>
> >> Do you know what's the action of _g_ on V^*
> induced from the action of
> >> _g_ on V? If you do, then a very short computation
> (three equalities)
> >> will solve this problem.
> >
> > For the action of _g_ on V^* induced from the
> action of
> > _g_ on V, the proof in the Lemma 4.10 in the link
> says (quote),
> >
> ====================================================
> > "To define the action of G, _g_ on V^*, we require
> that the natural
> > pairing V (x) V^* -->C be a morphism of
> representations, considering
> > C as the trivial representation, This gives, for v
> \in V, v* \in V^*,
> > <\rho(g)v, \rho(g)v^*> =<v, v^*>, so the action of
> G in V^* is
> > given by \rho_{V^*}(g) = \rho(g^-1)^t, where for
> A:V-->V, we denote
> > A^t in the adjoint operator V^*-->V^*. Similarly,
> for the action of
> > _g_, we get<\rho(x)v, v^*> +<v, \rho(x)v^*> = 0,
> so rho_{V^*}(g) =
> > -(\rho_V(x))^t "
> > ==================================================
> >
> > 1. Why those pairings<\rho(g)v, \rho(g)v^*> =<v,
> v^*> work?
>
> I am not sure about what you meant with the word
> "work". Anyway, the
> author wants to define \rho(g)v^* in a such a way
> that the equality
>
> <\rho(g)v, \rho(g)v^*> =<v, v^*>
>
> holds.
>
> > 2. Why \rho_{V^*}(g) = \rho(g^-1)^t
>
> Do you know what "adjoint operator" means?


You mean the "hermitian adjoint" in http://en.wikipedia.org/wiki/Hermitian_adjoint ?

Do you have any example or details of matrix calculation why \rho_{V^*}(g) = \rho(g^-1)^t ?

Thanks



> Best regards,
>
> Jose Carlos Santos
From: Timothy Murphy on 17 Jul 2010 06:53
José Carlos Santos wrote:

>>>> This is an Example 4.15 of the book "An
>>> introduction to Lie groups and Lie Algebras" by
>>> Kirillov.
>>>>
>>>> link: (p52)
>>>>
>>> http://www.math.sunysb.edu/~kirillov/liegroups/liegrou
>>> ps.pdf
>>>>
>>>> ===================================================
>>>> Example 4.15.
>>>> Let B be a bilinear form on a representation V.
>>> Then B is invariant under the action of G defined in
>>> Example 4.12 (please see the link) iff
>>>>
>>>> B(gv, gw)= B(v, w) (1)
>>>>
>>>> for any g \in G, v,w \in V. Similarly, B is
>>> invariant under the action of \mathfrak{g} iff
>>>>
>>>> B(x.v, w) + B(v, x.w) = 0 (2)
>>>>
>>>> for any x \in \mathfrak{g}, v, w \in V.
>>> ====================================================
>>>> The definition of (1) makes sense to me. Anyhow, I
>>> am having hard time understanding how (2) is derived
>>> or defined.
>>>
>>> Consider the equality (1) and see it as an equality between two
>>> functions from G into R (or C). The second function is constant, of
>>> course. Now, derive this equality. On the left you get
>>>
>>> B(X.v,w) + B(v,X.w)
>>>
>>> and on the right you get 0, of course.
>>
>> What are your first and second functions? I still can't figure out how
>> the equality is derived.
>
> My functions are functions from G into R.

Are you sure?
I assume \mathfrac{g} is the Lie algebra LG of G.
So the second function is a function from LG to R.

Maybe I have misunderstood.

--
Timothy Murphy
e-mail: gayleard /at/ eircom.net
tel: +353-86-2336090, +353-1-2842366
s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland
From: José Carlos Santos on 18 Jul 2010 10:03
On 17-07-2010 11:53, Timothy Murphy wrote:

>>>>> This is an Example 4.15 of the book "An
>>>> introduction to Lie groups and Lie Algebras" by
>>>> Kirillov.
>>>>>
>>>>> link: (p52)
>>>>>
>>>> http://www.math.sunysb.edu/~kirillov/liegroups/liegrou
>>>> ps.pdf
>>>>>
>>>>> ===================================================
>>>>> Example 4.15.
>>>>> Let B be a bilinear form on a representation V.
>>>> Then B is invariant under the action of G defined in
>>>> Example 4.12 (please see the link) iff
>>>>>
>>>>> B(gv, gw)= B(v, w) (1)
>>>>>
>>>>> for any g \in G, v,w \in V. Similarly, B is
>>>> invariant under the action of \mathfrak{g} iff
>>>>>
>>>>> B(x.v, w) + B(v, x.w) = 0 (2)
>>>>>
>>>>> for any x \in \mathfrak{g}, v, w \in V.
>>>> ====================================================
>>>>> The definition of (1) makes sense to me. Anyhow, I
>>>> am having hard time understanding how (2) is derived
>>>> or defined.
>>>>
>>>> Consider the equality (1) and see it as an equality between two
>>>> functions from G into R (or C). The second function is constant, of
>>>> course. Now, derive this equality. On the left you get
>>>>
>>>> B(X.v,w) + B(v,X.w)
>>>>
>>>> and on the right you get 0, of course.
>>>
>>> What are your first and second functions? I still can't figure out how
>>> the equality is derived.
>>
>> 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.

No. My first function is f(g) = B(g.v,g.w); my second function is the
constant function _f*_ defined by f*(g) = B(v,w). They are equal and
therefore their derivatives at the identity element of G are equal. The
derivative of _f*_ is 0, of course. And the derivative of _f_ is the
linear map from mathfrak{g} into R defined by X |-> B(X.v,w) + B(v,X.w).

So, again, my first and second functions are functions from G into R.
Of course, their derivatives at the identity of G are indeed functions
from mathfrak{g} into R.

Best regards,

Jose Carlos Santos

Best regards,

Jose Carlos Santos
From: José Carlos Santos on 18 Jul 2010 10:08
On 17-07-2010 2:45, Lie_Algebra wrote:

>>>>> Can anyone give some hints or thoughts including (*)?
>>>>
>>>> Do you know what's the action of _g_ on V^* induced from the action of
>>>> _g_ on V? If you do, then a very short computation (three equalities)
>>>> will solve this problem.
>>>
>>> For the action of _g_ on V^* induced from the action of
>>> _g_ on V, the proof in the Lemma 4.10 in the link says (quote),
>>> ====================================================
>>> "To define the action of G, _g_ on V^*, we require that the natural
>>> pairing V (x) V^* -->C be a morphism of representations, considering
>>> C as the trivial representation, This gives, for v \in V, v* \in V^*,
>>> <\rho(g)v, \rho(g)v^*> =<v, v^*>, so the action of G in V^* is
>>> given by \rho_{V^*}(g) = \rho(g^-1)^t, where for A:V-->V, we denote
>>> A^t in the adjoint operator V^*-->V^*. Similarly for the action of
>>> _g_, we get<\rho(x)v, v^*> +<v, \rho(x)v^*> = 0, so rho_{V^*}(g) =
>>> -(\rho_V(x))^t "
>>> ==================================================
>>>
>>> 1. Why those pairings<\rho(g)v, \rho(g)v^*> =<v,
>> v^*> work?
>>
>> I am not sure about what you meant with the word
>> "work". Anyway, the
>> author wants to define \rho(g)v^* in a such a way
>> that the equality
>>
>> <\rho(g)v, \rho(g)v^*> =<v, v^*>
>>
>> holds.
>>
>>> 2. Why \rho_{V^*}(g) = \rho(g^-1)^t
>>
>> Do you know what "adjoint operator" means?
>
>
> You mean the "hermitian adjoint" in http://en.wikipedia.org/wiki/Hermitian_adjoint ?

Yes.

> 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).

Best regards,

Jose Carlos Santos
From: Lie_Algebra on 18 Jul 2010 08:05
> On 17-07-2010 2:45, Lie_Algebra wrote:
>
> >>>>> Can anyone give some hints or thoughts
> including (*)?
> >>>>
> >>>> Do you know what's the action of _g_ on V^*
> induced from the action of
> >>>> _g_ on V? If you do, then a very short
> computation (three equalities)
> >>>> will solve this problem.
> >>>
> >>> For the action of _g_ on V^* induced from the
> action of
> >>> _g_ on V, the proof in the Lemma 4.10 in the link
> says (quote),
> >>>
> ====================================================
> >>> "To define the action of G, _g_ on V^*, we
> require that the natural
> >>> pairing V (x) V^* -->C be a morphism of
> representations, considering
> >>> C as the trivial representation, This gives, for
> v \in V, v* \in V^*,
> >>> <\rho(g)v, \rho(g)v^*> =<v, v^*>, so the action
> of G in V^* is
> >>> given by \rho_{V^*}(g) = \rho(g^-1)^t, where for
> A:V-->V, we denote
> >>> A^t in the adjoint operator V^*-->V^*. Similarly
> for the action of
> >>> _g_, we get<\rho(x)v, v^*> +<v, \rho(x)v^*> =
> 0, so rho_{V^*}(g) =
> >>> -(\rho_V(x))^t "
> >>>
> ==================================================
> >>>
> >>> 1. Why those pairings<\rho(g)v, \rho(g)v^*>
> =<v,
> >> v^*> work?
> >>
> >> I am not sure about what you meant with the word
> >> "work". Anyway, the
> >> author wants to define \rho(g)v^* in a such a way
> >> that the equality
> >>
> >> <\rho(g)v, \rho(g)v^*> =<v, v^*>
> >>
> >> holds.
> >>
> >>> 2. Why \rho_{V^*}(g) = \rho(g^-1)^t
> >>
> >> Do you know what "adjoint operator" means?
> >
> >
> > You mean the "hermitian adjoint" in
> http://en.wikipedia.org/wiki/Hermitian_adjoint ?
>
> Yes.
>
> > 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).
>
> Best regards,
>
> Jose Carlos Santos

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 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^*.

In a similar vein, if g^-1 (or \rho(g^-1)) is a matrix for linear transformation between V-->V then (g^-1)^t is a matrix for linear transformation between their dual spaces V^*-->V^*, (its induced map is contravariant).

Since gf(v)=f(g^-1v), we have gf=(g^-1)^tf.
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