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From: Lie_Algebra on 16 Jul 2010 01:48
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/liegroups.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.

Can anyone give some hints or thoughts including (*)?

Thanks.
From: José Carlos Santos on 16 Jul 2010 10:32
On 16-07-2010 10:48, 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/liegroups.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.

> 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.

Best regards,

Jose Carlos Santos
From: Maarten Bergvelt on 16 Jul 2010 11:19
On 2010-07-16, Lie_Algebra <liealgebra11(a)gmail.com> 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/liegroups.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.
>
> Can anyone give some hints or thoughts including (*)?


Hint for thinking about (2): if x is in the lie algebra, g(t)=exp(tx)
should be in the corresponding Lie group. But this in (1) and
differentiate wrt t.

For (*) think about what it means to be a morphism of representations
of G or \mathfrak{k}.


--
Maarten Bergvelt
From: Lie_Algebra on 16 Jul 2010 07:40
> On 16-07-2010 10:48, 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.

> > 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.
>
> Best regards,
>
> Jose Carlos Santos

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?
2. Why \rho_{V^*}(g) = \rho(g^-1)^t

I saw the similar one in the first chapter of Fulton's Representation theory, but I am not able to fully understand it.

Thanks.
From: José Carlos Santos on 16 Jul 2010 17:59
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?

Best regards,

Jose Carlos Santos
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