Adjoint representation of a group vs of a Lie algebra

Question

I just wanted some clarification on the adjoint representation. The definition of the adjoint representation of a Lie algebra seems absolute: taking the structure constants as generators.

The situation for the adjoint representation of a Lie group seems more relative. If I understood correctly, it is defined as a representation based on associating to each group element a mapping from the lie algebra vector space to itself. As a consequence, the definition seems relative, in the sense that it depends on the choice of representation for the Lie algebra.

In this sense, could there exist different adjoint representations of a group (associated to different representations of a Lie algebra)?

Answer 1

OK, since you expressly asked for an illustration for su(2), as free of the specialized math notation of WP, consider $$ [T_j,T_k]= i \epsilon_{jkm} T_m. $$ The three $T$s are abstract vectors in the Lie algebra, which you may represent by square matrices of arbitrary size, such as the 2×2 Pauli matrices, $\pi_2(T_j)= \sigma_j/2$.

The crucial point is that the structure constants don't really care about the specific rep. All reps satisfy the above Lie algebra—but not arbitrary relations among functions of them (like their squares!).

Thus, the linear map $$ \operatorname{ad}_{T_j} ~ T_k \equiv [T_j,T_k]=i\epsilon_{jkm} T_m $$ scrambles the three $T_k$s by multiplication through the 3×3 matrix $i\epsilon^j_{km}$.

We see below that the three linear maps $\operatorname{ad}_{T_j}$ provide a 3-dim representation of the group, regardless of what rep one would use for illustrating the Ts—which is needless! Specifically, $$ [\operatorname{ad}_{T_j}, \operatorname{ad}_{T_n}] ~ T_k= (\operatorname{ad}_{T_j}\operatorname{ad}_{T_n}-\operatorname{ad}_{T_n}\operatorname{ad}_{T_j}) ~T_k\\ = [T_j,[T_n,T_k]]-[T_n,[T_j,T_k]]=[[T_j,T_n],T_k]\\ \large = \operatorname{ad}_{(i\epsilon_{jnm}T_m)} ~ T_k, $$ the middle line by virtue of the Jacobi identity. That is, by virtue of linearity, $$ [\operatorname{ad}_{T_j}, \operatorname{ad}_{T_n}]= i\epsilon_{jnk} \operatorname{ad}_{T_k}, $$ so $\operatorname{ad}_{T_j}\mapsto \pi_3(T_j)$. Its dimension is 3, the number of generators of su(2), the dimension of the vector space that ad acts on.

Again, if you used $\pi_2(T_j)$, or $\pi_{137}(T_j)$, for the $T_j$s, you wouldn't have specified this $\pi_3(T_j)$ any differently. It's absolute, in your terminology.

To look at the group action, you exponentiate, $$ e^{it\operatorname{ad}_{T_j}} ~ T_k= e^ { itT_j }~T_k ~ e^{ -itT_j }\equiv \operatorname{Ad}_{\exp(itT_j )} ~ T_k ~~ \leadsto \\ \operatorname{Ad}_{\exp(itT_j )} ~ e^{iT_k} = e^{itT_j }~e^{iT_k} ~ e^{-itT_j }, $$ as a similarity transformation, where no summation over indices, repeated or exponentiated, is implied. This is the group automorphism of WP.

The first equality is often dubbed the Hadamard lemma; the end result may be seen to only involve nested commutators: $Y+\left[X,Y\right]+\frac{1}{2!}[X,[X,Y]]+\frac{1}{3!}[X,[X,[X,Y]]]+\cdots= e^{X}Y e^{-X}$ .

Calling the group elements, unitary operators, $U= e^{iT_k} $ and $V= e^ { itT_j }$, you see that the group "multiplication table" (infinite!) of any and all different Us is identical, by similarity, to that of the $V U V^{-1}$s; as before, nothing depends on the representation: it will be the same table.

For the doublet representation, 2×2 unitary matrices, the adjoint group automorphism $$ V\mapsto V U V^ \dagger $$ is the cornerstone of chiral dynamics. But again, you may see that, since only commutators are involved in your operations, the same combinatoric answers would obtain for all reps.

(This is the celebrated Lie's third theorem: the combined exponent in the CBH composition formula combining group elements is strictly in the Lie algebra, so the combinatorics of the nested commutators involved is identical for all reps. You might as well stick to abstract Lie algebra elements).

Answer 2

To complement Cosmas' insightful answer, it might be worthwhile to review some basic ideas. Let $G$ be a matrix Lie group for simplicity, and $\mathfrak g$ its Lie algebra.

A Lie group representation $\rho:G\rightarrow \mathrm{GL}(V)$ is a map from the group to the invertible linear maps on some vector space $V$ (called the representation space) which has the property that $$\rho(g_1 g_2) = \rho(g_1)\rho(g_2) \qquad g_1,g_2 \in G $$

Similarly, a Lie algebra representation $R:G \rightarrow \mathfrak{gl}(V)$, is a map from the algebra to the (not generally invertible) linear maps on $V$ which has the property that $$R([X_1,X_2]) = [R(X_1),R(X_2)] \qquad X_1,X_2\in \mathfrak g$$

Every Lie group representation $\rho:G\rightarrow \mathrm{GL}(V)$ induces a Lie algebra representation $\mathrm d\rho:\mathfrak g \rightarrow \mathfrak{gl}(V)$ which is defined as

$$\mathrm{d\rho}(X) := \left.\frac{d}{dt} \rho(e^{tX})\right|_{t=0}$$

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It's not a trivial task to identify and classify the different representations you might have for a given $G$, and in general you need to know what $G$ actually is before finding them. However, there are some representations which you can write down automatically - for example, the trivial (Lie group) representation $\rho_0:g\mapsto \mathrm{id}_V$, which maps every element of $g$ to the identity map on $V$, and which induces the trivial (Lie algebra) representation $\mathrm d \rho_0: X \mapsto 0$.

On the other hand, the adjoint representation $\mathrm{Ad}:G \rightarrow \mathrm{GL}(\mathfrak g)$ is a non-trivial representation that we get for free. Because we have assumed $G$ to be a matrix Lie group, then we can express this representation as $$\mathrm{Ad}(g): X \mapsto g X g^{-1} $$ where the implied multiplication is just matrix multiplication. Furthermore, $\mathrm{Ad}$ induces the adjoint representation of $\mathfrak g$ - which we call $\mathrm{ad}$ - which turns out to be $$\mathrm{ad}(X) : Y \mapsto [X,Y]$$

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If I understood correctly, [the adjoint representation of a Lie group] is defined as a representation based on associating to each group element a mapping from the lie algebra vector space to itself.

The adjoint representation $\mathrm{Ad}:G\rightarrow \mathrm{GL}(\mathfrak g)$ has a specific definition given above. Remember that each Lie group $G$ can be associated to a Lie algebra $\mathfrak g$, which is in particular a vector space; $\mathrm{Ad}$ takes advantage of this intrinsic structure by using $\mathfrak g$ as the representation space to define a special representation which arises organically from the structure already present in $G$.

As a consequence, the definition seems relative, in the sense that it depends on the choice of representation for the Lie algebra.

Both the Lie algebra $G$ and the Lie group $\mathfrak g$ are defined without reference to any specific representation. For example, the Lie group $\mathrm{SO}(3)$ consists of the $3\times 3$ real orthogonal matrices. This is not a representation; it is the definition of $\mathrm{SO}(3)$. Similarly, the Lie algebra $\mathfrak{so}(3)$ is the space of $3\times 3$ skew-symmetric matrices.

It's true that $\rho:\mathrm{SO}(3)\rightarrow \mathrm{GL}(\mathbb R^3)$ which maps each element of $\mathrm{SO}(3)$ to itself is a fairly obvious choice of representation; nevertheless, the existence of a fairly obvious representation doesn't mean that $\mathrm{SO}(3)$ requires a representation to define, and the same is true of $\mathfrak{so}(3)$.

All that is to say that $\mathrm{Ad}$ is not defined via a representation of $\mathfrak g$; it is defined on $\mathfrak g$ itself.

Answer 3

The adjoint representation of a Lie group and a Lie algebra are defined independently. Thus there is only one of each. But they are related if the group and algebra are related.

The adjoint representation of a Lie algebra $\mathfrak{g}$ is:

$ad: \mathfrak{g} \rightarrow End(\mathfrak{g})$

The adjoint representation of a Lie group $G$ is

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

Here, I'm writing $G'$ as the Lie algebra associated with $G$. The precise definitions are in the Wikipedia article. Now, when we have $\mathfrak{g} = G'$ then it turns out that:

$ad = T_{Id} Ad$

Thus there are only one adjoint representation of a Lie group and also of a Lie algebra. And the adjoint representation of the Lie algebra is the tangent space at the identity of the adjoint representation of the Lie group.