In another thread I promised a question on this topic. I have been struggling with it off and on ever since (very much more off than on, it has to be said). First let me admit I asked a similar question on another board but got short shrift. in return. Second, I assume a lot of prior knowledge - sorry about that. So. Let \(G\) be a Lie group, and \(\mathfrak{G} \equiv T_eG\) its algebra. Then I define the adjoint rep of \(G \) on \( \mathfrak{G}\) by \(Ad: G \to Aut(\mathfrak{G}),\,\,\,Ad(g):=Ad_g: \mathfrak{G} \to \mathfrak{G} \) and invertibly. Notice this is an operator on the vector space. So the dual to this rep will be \(Ad^*: G \to \mathfrak{G}^* \equiv T^*_eG,\,\,\, Ad^*_g:\mathfrak{G}^* \to \mathfrak{G}^*\), also an operator. Suppose that some \( X \in \mathfrak{G},\,\, \phi \in \mathfrak{G}^*\) and consider the natural pairing \( \langle\cdot\,,\,\cdot\rangle\). I will denote this naturality by, say, \(\langle \phi_X, X\rangle\). Now consider the pullback, that is, for \(Ad_gX =Y \in \mathfrak{G}\) then I will have that \((Ad_g)^*\phi_Y = \phi_X \in \mathfrak{G}^*\), so that, by my construction, the natural pairing \(\langle (Ad_g)^*\phi_Y, Ad_gX\rangle = \langle\phi_X, Y\rangle \ne \langle \phi_Y, Y\rangle \ne \langle \phi_X, X \rangle \). But it would seem that using the inverse I have that \(\langle (Ad^{-1}_g)^*\phi_Y, Ad_gX\rangle = \langle \phi_Y Ad_gX\rangle =\langle \phi_Y,Y\rangle\). Is this correct? Now it is a requirement of the coadjoint (i.e. dual) rep that \( \langle Ad^*_g\phi_X, Ad_gX\rangle =\langle\phi_X, X\rangle\), so I conclude that \(Ad^*_g \equiv (Ad^{-1}_g)^*\). Is this correct? So (correct if I am wrong) it seems to follow that \(\langle Ad^*_g \phi, X\rangle =\langle \phi, Ad^{-1}_gX\rangle\), that is..... the inverse of the operator induced by the adjoint rep is adjoint to that induced by the coadjoint rep. Wow. More adjoints than a houseful of hippies! (just as hand-wavey as well). My question, such as it now is: is my sloppy reasoning correct?
I assume you mean \(Ad^*: G \to Aut(\mathfrak{G}^*) \equiv Aut(T^*_eG)\) since you then say \(Ad^*_g:\mathfrak{G}^* \to \mathfrak{G}^*\) which means \(Ad^{\ast}_{g}\) is an automorphism on \(\mathfrak{G}^{\ast}\). Yep, with you so far. I don't think so, I can't see how you went from the first to the second expression. By definition of the adjoint operator we have that \(\langle (Ad^{-1}_g)^*\phi_Y, Ad_gX\rangle = \langle \phi_Y, Ad^{-1}_g( Ad_gX)\rangle\) and then using \((Ad_{g})^{-1} = Ad_{g^{-1}}\) and \(Ad_{g}Ad_{h} = Ad_{gh}\) we have \( \langle \phi_Y, Ad^{-1}_g( Ad_gX)\rangle = \langle \phi_Y, X\rangle\). This isn't what you have but maybe its just a typo on your part. Perhaps we're using different definitions but I've been using the definition \(\langle Ax,y\rangle = \langle x,A^{\ast}y\rangle\) as the definition of an adjoint operator. The condition \(\langle Ax,A^{\ast}y\rangle = \langle x,y\rangle\) is something else. For instance, in quantum Hilbert spaces a transformation on a state must be unitary to get inner product invariance \(UU^{\dag} = \mathbb{I}\), it is not automatic. Since we're talking about the adjoint rep, the coadjoint and adjoint operators maybe we're talking across one another. :shrug: How come you aren't defining the adjoint operator by the usual way, as I just used? If you do that you are led to your pull back action \((Ad_g)^*\phi_Y = \phi_X \in \mathfrak{G}^*\) pretty quickly. Perhaps some wires are crossed.
I think the origin of your confusion is here. Try writing \(\phi_X\) without the subscript X, i.e., \(\langle \phi, X\rangle\). There is nothing that connects \(\phi\) with X, it is just an arbitrary element of the dual space. On the rest of the matter, my view is the same as AlphaNum's.
Oops-a-daisy, I fluffed here. Thanks, you are right Well, on the LHS: this is not a definition of the adjoint operator I have ever seen before, which is usually given as the conjugate transpose. Am I missing something? On the RHS: this gives me the willies. However, I can live with it I guess, since \(Ad_g\) is an automorphism, so that \(Ad_g^{-1}(Ad_g) = Ad_e\) I shall not be so dishonest as to claim that. How is it that you find \((Ad^{-1}_g)^*\phi_Y=\phi_Y\)? I think I am being confused by the different uses of the asterisk here: on the one hand it denotes the dual, on another the pullback and on the "third hand" (Huh?) the adjoint operator. This is not unique to you, it is universal, or so it would appear Yeah, but you don't need me o tell that it;s child's play to show that an equivalent definition is \(\langle x, Ay \rangle =\langle A^{\dag}x,y\rangle\) Sure it is, it seems that the adjoint and codjaint reps preserve the natural pairing, as I claimed. At least that's what Fulton & Harris claim probably not; it seems more likely I am just being thick temur I chose that notation to emphasize the pullback. I probably fooled myself thereby into assuming that for each vector there a unique covector. I think this is true in an IPS (under certain conditions), but not in the way I offered it. Golly I find this hard going. Any more help either of you tough guys (or anyone else, for that matter) can offer? OK here's the deal: I am given that, when the asterisk on the RHS denotes the coadjoint rep, and that on the RHS the adjoint mapping, that \(Ad^*_g = (Ad^{-1}_g)^*\). I was trying to see why.
Let \(\bar{\mathrm{Ad}}:G\to\mathrm{Aut}(\mathfrak{G}^*)\) be the co-adjoint representation, which is defined by \(\langle\bar{\mathrm{Ad}}_g\phi,X\rangle=\langle\ph,\mathrm{Ad}_{g^{-1}}X\rangle\) for all \(\phi\in\mathfrak{G}^*\) and \(X\in\mathfrak{G}\). Now we use \(\mathrm{Ad}_{g^{-1}}=(\mathrm{Ad}_{g})^{-1}\) to infer \(\langle\bar{\mathrm{Ad}}_g\phi,X\rangle=\langle\ph,\mathrm{Ad}_{g^{-1}}X\rangle=\langle\ph,(\mathrm{Ad}_{g})^{-1}X\rangle=\langle((\mathrm{Ad}_g)^{-1})^*\phi,X\rangle\).
OK, so the bit I bolded crops up in all my texts. I hate the blithe expression "defined by", but I guess we all have to be non-constructivist at times More stars!! May I re-write your last identity as \(\langle((\mathrm{Ad}_g)^{-1})^{\dag}\phi,X\rangle\), where the "dagger" indicates conjugate-transpose? If so, forgiving the non-constructivist approach, it seems to settle the score. Thank you many times.
Are you distinguishing between \(A^{\ast}\) and \(A^{\dag}\)? I was working as them being equivalent in your discussion but obviously complex conjugation and hermitian conjugation are different so that might be a crossing of wires I referred to. I don't think I did, did I? Please Register or Log in to view the hidden image! I think a clarification of whats pull back, whats adjoint, whats coadjoint and what's anything else which we might be tempted to label with an asterix might be in order. Off the top of my head I can't think of any case where I've done any work with something which is the TM version of the adjoint map on T*M so I am going on some vague memory of having read a bit about it. There's a book called 'Gauge Theory and Variational Principles' by Bleecker which is a little heavy going in its notation (Nakahara is infinitely better in that regard) but covers this kind of stuff from the point of view of highly mathematical physics and hence is more a "Let's learn how to use this stuff in specific cases" rather than the pure mathematician's "Bring on the rigorous nightmare of analysis!" view.
I assume by "Hermitian conjugation" you mean conjugate-transpose? But yes, I introduced the "dagger" as, while it is normally taken to be what you call Hermitian conjugation on a complex space reduces to the transpose over a real space. I also introduced it to avoid too many bloody stars, so that..... I agree, but don't tempt me. Mary-Bernadette will be away for the weekend, so a) I am likely to drink a lot of beer and b) will be bored, a lethal cocktail on boards such as this!!
