Adjoint representation of the Lorentz group

Question

Is it possible to construct an adjoint representation for the Lorentz group?

Answer 1

I suspect the impeccable WetSavannaAnimal answer is too abstract and general to satisfy the OP. Let me vulgarize it a bit. The Lorentz group has 6 generators, so your adjoint rep will be a 6-dimensional rep, a set of 6x6 matrices, which satisfy the same Lie algebra (commutators) as the conventional 4x4 matrices transforming the fundamental (x,y,z,t) 4-vector. As an aside, note the curvature form (parity inv. 2-form) of the Lorentz group is also 6-dimensional.

Recall the Lorentz group may be written in a pretty form, that is rotations $J_i\equiv −\epsilon_{imn} M_{mn} /2 $ and boosts, $K_i\equiv M_{0i}$, so that $[J_m,J_n] = i \epsilon_{mnk} J_k$ , $[J_m,K_n] = i \epsilon_{mnk} K_k$ , $[K_m,K_n] = -i \epsilon_{mnk} J_k$.

Might further note the important simplification $[J_m +i K_m, J_n −i K_n]$, which permits reduction of the Lorentz algebra to su(2)⊕su(2) and efficient treatment of its associated representations.

Now consider the six 6x6 matrices with the 3 J+iKs on the upper left 3x3 subspace of these 6x6 ones, and the 3 J-iKs on the lower right block, spanned by the 4,5,6 indices. Keep the indices of the upper left block to be the usual 1,2,3; and rename the indices of the lower right block from 1,2,3 to 4,5,6. The commutation relations are then manifest, and the structure constants $f_{mnl}$ are sparse, basically $\epsilon_{ijk}$ for indices (1,2,3) or (4,5,6), and zero otherwise.

So, then, as you probably learned from SU(3), these very structure constants f provide your 6x6 matrices in the adjoint, with one of the 3 indices (taking 6 values) specifying the particular generator of the Lorentz group represented.

Answer 2

Every Lie group has an adjoint representation. I'm not sure what definition you come at the adjoint representation from, but here's the fundamental one which I'm sure you'll see is always meaningful.

Think of a $C^1$ path $\sigma:[-1,1]\to\mathfrak{G}$ through the identity in a Lie group $\mathfrak{G}$ with $\sigma(0) =\mathrm{id}$ and with tangent $X$ there.

Now think of a general group member $\gamma \in\mathfrak{G}$ acting on that path so that $\sigma(t) \mapsto \gamma^{-1} \sigma(t) \gamma$. This too is naturally a path through the identity and has a transformed tangent $X^\prime$ there. This transformation is linear. We say the group "acts on its own Lie algebra $\mathfrak{g}$" in this way and write the transformation $X\mapsto\gamma^{-1} X \gamma$. This is just notation, but in a matrix Lie group it is also a literal matrix product. Perhaps, then, less confusingly we write $X\mapsto {\rm Ad}(\gamma) X$, where the linear transformation ${\rm Ad}(\gamma)$ wrought by $\gamma$ on the Lie algebra is now a member of $GL(\mathfrak{g})$, the group of invertible linear transformations of the Lie algebra $\mathfrak{g}$ thought of as a plain vector space. The association

$$\rho:\mathfrak{G}\to GL(\mathfrak{g});\;\rho(\gamma)= {\rm Ad}(\gamma)$$

is a homomorphism as is readily shown. This is sometimes called the Adjoint Representation of the Lie group. Through the big Ad Adjoint representation $\mathfrak{G}$ is mapped to a new Lie group, this time always a matrix Lie group, a subgroup of $GL(\mathfrak{g})$.

Now, then, we can look at

$${\rm ad}(X)\stackrel{def}{=}\left.\mathrm{d}_t {\rm Ad}(e^{t\,X})\right|_{t=0}$$

This too is a linear operator on $\mathfrak{g}$, although in general not an invertible one. Indeed, you can show without too much strife that:

$${\rm ad}(X) Y = [X,\,Y]$$.

So in fact the big Ad adjoint representation induces a homomorphism of Lie algebras $\rho^\prime:\mathfrak{g} \to {\rm Lie}(\rho(\mathfrak{G}))$. This too is a homomorphism of linear spaces and moreover a homomorphism that respects Lie brackets. It is thus a Lie algebra homomorphism and it is also called the adjoint representation (of the Lie algebra). I like to quaintly call it little ad adjoint representation.

Now here for me is one of the most beautiful equations around:

$${\rm ad}([X, \,Y]) = [{\rm ad}(X),\,{\rm ad}(Y)] = {\rm ad}(X) . {\rm ad}(Y)-{\rm ad}(Y) . {\rm ad}(X)$$

This is a restatement of the fact that little ad respects Lie brackets. But it is also a form of the Jacobi identity in disguise. Wow! That's the REAL meaning of the Jacobi identity: it's there so the adjoint representation of a Lie group, clearly a very basic and fundamental thing, induces a homomorphism in the corresponding Lie algebras that respects Lie brackets. Everything's exactly as we would expect and so, if you're ever designing a Universe, that's why you must remember to throw the Jacobi identity in! Write a note for yourself now so you don't forget! Now my LaTeX skills aren't up to drawing a commutative diagram from memory, so I hope you can see there is a pretty neat and simple one.

Just another couple of interesting facts. The kernel of Big Ad is the centre of the the Lie group. The kernel of little ad is then the centre of the Lie algebra. So if the Lie group is simple, i.e. contains no normal Lie subgroups, then there cannot be a continuous centre to annihilate by the homomorphism. Same is true if the group simply has no continous centre for reasons other than simplicity. So there are no bits of the Lie algebra that get wiped out. The original Lie group and the image of big Ad then have exactly the same Lie algebra. If further there is no discrete centre in a simple Lie group and if the group is connected, then the Lie group and the image of the adjoint representation are the same Lie group.

Okay. So now let's specialise to the Lorentz group. There is no continuous centre, by inspection of the commutation relationships. Therefore the Lie algebra of the image of big Ad is exactly the same as the Lie algebra of the Lorentz group. Nor is there a discrete centre in the Lorentz group. $\ker(\rho) = \{\mathrm{id}\}$, so, by the homomorphism theorem, the image of $SO(1,3)$ under the big Ad adjoint representation is the Lorentz group itself.

Answer 3

Every group has an adjoint representation. If you know differential geometry then defining this is straightforward. It is a map:

$Ad: G \rightarrow Aut(G')$

Where $G$ is a group and I write $G'$ for its Lie algebra.

More precisely, it is defined via the conjugation map:

$Cj: G \rightarrow Aut(G)$

So $Cj(g): G \rightarrow G$

And we set $Cj(g).h := g^h = hgh^{-1}$

So taking the tangent bundle gives
$TCj(g): TG \rightarrow TG$

And so the tangent space at the identity is: $T_eCj(g):T_eG \rightarrow T_eG$

But the tangent space at the identity of any Lie group is simply its Lie algebra. Hence we have, $T_eG = G'$ and so the preceding is:

$T_eCj(g): G' \rightarrow G'$

And we set:

$Ad(g) := T_eCj(g)$

It turns out that the adjoint map for $SU(2)$ is simply the double covering map of $SO(3)$.