In Schwinger's source theory book, he suggests if $G_a$ are the hermitian generators of the Unitary group, then we have an infinitesimal transformation is given by : $$ G = \sum_{a=1}^n \delta\lambda_a G_a $$ Now if we subject to the infinitesimal transformation operator to arbitrary unitary transformation of the group ($\{\lambda\}$ are the group parameters) , $$ U^{-1}G_aU = u(\lambda)G_b $$ Alternatively, $$ UG_aU^{-1} = G_bv(\lambda) $$ And then, these two quantities are related by $ v = (u^T)^{-1} $

Firstly, I am not able to see why the author is trying to do this kind of unitary transform on the group element, Is it possibly to get a representation? How further do we proceed with this.


2 Answers 2


As I mentioned in my comment, I believe you are talking about the adjoint representation of a Lie Group $G$ with a Lie algebra $\mathfrak{g}$: \begin{equation} \forall x \in \mathfrak{g}, \;\; \mathrm{Ad} \: D(g) : x \mapsto D(g) x D^{-1}(g) \in \mathfrak{g} \end{equation} where $D(g)$ denotes a represention of $g \in G$. One method you can see why $D(g) x D^{-1}(g) \in \mathfrak{g}$ is by considering the above transformation: \begin{equation} \begin{aligned} x' & = D(g) x D^{-1}(g) \\ &= e^{i \lambda^i G_i} x e^{-i \lambda^i G_i} \\ &= (1 + i \lambda^i G_i) x (1 - i \lambda^i G_i) + O(\lambda^2) \\ &= x - x i \lambda^i G_i + i \lambda^i G_i x + O(\lambda^2)\\ &= x + i \lambda^i [G_i, x] +O(\lambda^2) \\ &= x + i \lambda^i x^j [G_i,G_j] +O(\lambda^2)\\ &= x - \lambda^i x^j f_{ij}{}^{k} G_k +O(\lambda^2)\\ &= (x^k- \lambda^i x^j f_{ij}{}^{k})G_k +O(\lambda^2)\\ &= x'^i G_i +O(\lambda^2) \end{aligned} \end{equation} where: \begin{equation} x'^i \equiv x^i- \lambda^j x^k f_{jk}{}^{i} \end{equation} This shows that $x'$ can also be expressed in terms of the generators. In other words, the above transformation ensures that if $x$ lives in the Lie algebra formed by the generators, then $x'$ also lives in that vector space.

Furthermore, note that from the fifth line of the above equation, we can also write: \begin{equation} x' = x + i \lambda^i [G_i, x] = x+ i \mathrm{ad} \; \lambda(x) = e^{i \mathrm{ad} \; \lambda} x \end{equation} Thus: \begin{equation} \mathrm{Ad} \: D(g) (x) = e^{i \mathrm{ad} \; \lambda} x \end{equation} which shows the relation between the adjoint representation of the Lie group and the adjoint representation of the Lie algebra.

  • $\begingroup$ +1 Thanks for the clear view. I presume that from eqn.1 to the next line, you are writing $x = x^j G_j $. $\endgroup$
    – user35952
    Feb 25, 2014 at 13:21
  • $\begingroup$ @user35952 yes, that's right. $\endgroup$
    – Hunter
    Feb 25, 2014 at 13:23
  • $\begingroup$ Can you please kindly elucidate more on this Adjoint representation. I am having to see this a lot of times without actually understanding it. Also is the map $ g \rightarrow hgh^{-1} $ even true for the group elements. i.e. to say does it preserve closure $\endgroup$
    – user35952
    Feb 25, 2014 at 13:40
  • $\begingroup$ @user35952 You ask if $h g h^{-1}$ preserves closure: it does if $h,g \in G$. However, the adjoint representation is defined as $x \mapsto g x g^{-1}$, where $g \in G$, but $x$ is not a group element. $x$ is a "vector" that lives in the vector space spanned by the generators! You must be careful to distinguish between representations of group elements of the Lie group and the Lie algebra of the Lie group. $\endgroup$
    – Hunter
    Feb 25, 2014 at 13:46
  • $\begingroup$ Got it, thanks for that piece giving more clarity. Also is not the second map that you define for $x$ just a differential of the original group map about the identity ? $\endgroup$
    – user35952
    Feb 25, 2014 at 13:48

Good Lord! Is Schwinger still worth reading? A top physicist of course, but unfortunately right over my head! You're talking about the big A Adjoint representation as in Hunter's Answer, and there are a great many more modern, rigorous and way clearer texts on this topic. The Wikipedia page is a great start. Also see Rossmann, "Lie Groups, An Introduction ...", chapter 2 or Hall, section 4.3.

Here is just a short alternative (not completion or correction, mind) to Hunter's Answer, which is altogether complete in itself.

You may find it helpful (if your brain works anything like mine, which, I admit is a slim likelihood!) to think of the Lie algebra as the linear space of tangents, at the identity, to $C^1$ paths in a Lie group through the identity. I'm saying just forget about the "manifold" for a bit; try thinking instead of the thing as a group first with $C^1$ paths linking its points. A Lie group is a highly specialised manifold (e.g. its fundamental group is Abelian, just one of a host of special things) and does not need anything like the full machinery of differential geometry to see clearly.

If $\sigma_X:[-1,1]\to\mathfrak{G}$ is a $C^1$ path through the identity in the Lie group $\mathfrak{G}$ with $\sigma_X(0) = \mathrm{id}$ and with tangent $X\in\mathfrak{g}={\rm Lie}(\mathfrak{G})$ there, then clearly so is:

$$\sigma_Z:[-1,1]\to\mathfrak{G};\,\sigma_Z(\tau)=U^{-1} \sigma_X(\tau) U$$

and it will therefore have some tangent $Z\in\mathfrak{G}$. So the conjugation $\sigma_X\mapsto U^{-1}\sigma_X U$ clearly induces some mapping $Ad(U):\mathfrak{g}\to\mathfrak{g}$ on the Lie algebra. ($C^1$ follows from the preservation of differentiability class by the group product property of any Lie group).

Now, if you let $\sigma_Y:[-1,1]\to\mathfrak{G}$ be a second $C^1$ path through the identity with tangent $Y\in\mathfrak{g}$ there, then work out from first principles what the tangent of the following $C^1$ path through the identity:

$$\sigma_Z:[-1,1]\to\mathfrak{G};\,\sigma_Z(\tau) = U^{-1}\sigma_X(\alpha\,\tau)\,\sigma_Y(\beta\tau)U = U^{-1}\sigma_X(\alpha\,\tau)\,U\,U^{-1}\sigma_Y(\beta\tau)U$$

is, where $\alpha, \beta\in \mathbb{K}$ (the field $\mathbb{K}$ is the field underlying the Lie algebra). You will find that the tangent is $\alpha {\rm Ad}(U) X + \beta{\rm Ad}(U) Y$, so ${\rm Ad}(U)$ is a linear, homogeneous map from $\mathfrak{g}$ into itself: in loose words: it is a matrix.

Again, from first principles considering the $C^1$ path $V^{-1}\, U^{-1} \sigma_X U\, V$ for $U,\,V\in\mathfrak{G}$, you can show ${\rm Ad}(V\, U) = {\rm Ad}(V)\,{\rm Ad}(U)$. It also follows that ${\rm Ad}(U^{-1}) = {\rm Ad}(U)^{-1}$ so that ${\rm Ad}(U)$ is invertible. Thus inverses and products of entities of the form ${\rm Ad}(U)$ for $U\in\mathfrak{G}$ is another entity of the same kind. Therefore the set:

$${\rm Ad}(\mathfrak{G})\stackrel{def}{=}\{{\rm Ad}(U):\,U\in\mathfrak{G}\}$$

is a group of invertible matrices that act on the Lie algebra $\mathfrak{g}$, and, since ${\rm Ad}(V U) = {\rm Ad}(V){\rm Ad}(U)$, ${\rm Ad}$ is a homomorphism and the pair $({\rm Ad}, \mathfrak{g})$ is therefore a representation of the original group $\mathfrak{G}$. ${\rm Ad}(\mathfrak{G}) \cong {\rm Inn}(\mathfrak{G})$, the group of inner automorphisms of $\mathfrak{G}$ and is a matrix Lie group.

As Hunter showed you, for every $X\in\mathfrak{G}$ there is another linear map definable on the Lie algebra, the so called Lie Map of ${\rm Ad}$ (i.e. its differential at the identity):

$${\rm ad}(X):\mathfrak{g}\to\mathfrak{g};\,{\rm ad}(X) Y = \left.{\rm d}_\tau {\rm Ad}(e^{ \tau\,X}) Y\right|_{\tau=0} \stackrel{def}{=}[X,\,Y]$$

This one is never invertible (${\rm ad}(X) X = 0$). With a bit of work you can prove the following interesting facts:

  1. $\ker({\rm Ad}) = \mathscr{Z}(\mathfrak{G})$, the centre of $\mathfrak{G}$,i.e. the set of all elements $\gamma\in\mathfrak{G}$ that commute with all elements of $\mathfrak{G}$;
  2. The Jacobi indentity: ${\rm ad}({\rm ad}(X)\, Y)={\rm ad}([X,\,Y]) = {\rm ad}(X).{\rm ad}(Y) - {\rm ad}(Y).{\rm ad} (X)$, so that little ${\rm ad}$ is a homomorphism of Lie algebras.
  3. The kernel of the homomorphism ${\rm ad}$ is the centre $\mathscr{z}(\mathfrak{g})$ of $\mathfrak{g}$, the set of all $X\in \mathfrak{g}$ such that $[X,\,Y] = 0;\,\forall Y\in \mathfrak{g}$.

Point 2. is what I call a stunner: this is the REAL meaning of the Jacobi identity (you need to do a little work to show that this is indeed the Jacobi ndentity in disguise): the Jacobi identity is precisely the assertion that

the big A ${\rm Ad}$ representation induces a homomorphism ${\rm ad}$ between the Lie algebra $\mathfrak{g}$ of a Lie group $\mathfrak{G}$ and the Lie algebra ${\rm Lie}({\rm Ad}(\mathfrak{G}))$ of ${\rm Ad}(\mathfrak{G})$.

The Lie algebra ${\rm Lie}({\rm Ad}(\mathfrak{G}))$ is the algebra ${\rm Der}(\mathfrak{g})$ of all derivations on $\mathfrak{g}$, i.e. the set of all linear maps on $\mathfrak{g}$ that fulfill the Leibnitz product rule.

  • $\begingroup$ Apparently, this is from Schwinger's text book on Source Theory . There are no other or even notes available for Source Theory. $\endgroup$
    – user35952
    Feb 26, 2014 at 4:11
  • $\begingroup$ All I'm saying is that Schwinger's not the best source for the adjoint reprensentation, and that hopefully it will become clearer if you look at Lie theorist's treatments. The few things of Schwinger's I have tried to read have utterly beaten me! $\endgroup$ Feb 26, 2014 at 4:23
  • $\begingroup$ +1 Truly fascinating; one day I hope to understand group theory and differential geometry at the level that you understand it (I'm aware that my answer above is a very "physics" answer). $\endgroup$
    – Hunter
    Feb 26, 2014 at 6:52
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    $\begingroup$ @Hunter There's nothing wrong with a physicist's answer. Lie theory is a particular interest of mine. One of Roger Penrose's comments that I like that one has to learn to judge well when rigor is helpful, and when, especially in the early phases of hatching physical theories, it can lead to rigor mortis! $\endgroup$ Feb 26, 2014 at 8:16
  • 1
    $\begingroup$ @Hunter I happen to believe, deep down, that ultimately mathematics and physics are not as different as people make out, but then I also lean towards Tegmark's Mathematical Universe Hypothesis, which many feel is a little bit loopy. $\endgroup$ Feb 26, 2014 at 8:17

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