# What is adjoint representation in Lie group?

1. Jan 8, 2012

### ndung200790

What is the adjoint representation in Lie group? Where is the vector space that the ''elements of the group'' act on in this representation(adjoint representation)?
Thank you very much for your kind helping.

2. Jan 8, 2012

### ndung200790

Because Lie groups (e.g SU(3)) act on physical fields,but I do not know the ''elements of group'' act on what fields in adjoint representation.

3. Jan 8, 2012

### vanhees71

First of all a Lie group is a group with elements in a differentiable manifold for which the group multiplication and the inversion of elements are differentiable mappings of the group to itself.

For simplicity let's consider matrix groups like SU(N), SO(N), etc. These are the common cases relevant for physics. Now you can define the Lie algebra of the group as the tangent space at the group identity. This is a vector space with matrices as elements (e.g., for SU(N) the Lie algebra consists of all antihermitean traceless matrices, in physics written as $\mathrm{i} \alpha_k T^k$ with $T^k$ hermitean).

Now let $G$ be the group and $\mathcal{L}G$. Then it's easy to show that for each $g \in G$ the mappings

$$\text{Ad}_g: \; \mathcal{L}G \rightarrow \mathcal{L}G, \; \text{Ad}_g(t)=g t g^{-1}$$

are a well-defined linear mapping which respects the Lie-algebra product (given by commutators of the Lie-algebra matrices in our case), and obeys the group-representation property,

$$\text{Ad}_{g_1} \circ \text{Ad}_{g_2}=\text{Ad}_{g_1 g_2},$$

i.e., it's a linear representation of the Lie group on the Lie algebra, which is called the adjoint representation.

The nice thing is that it's constructed wholly from the Lie-group structure itself without introducing any other kind of elements.

In physics it appears in gauge symmetries, where the gauge fields are defined as affine connections of the group's Lie algebra to define covariant derivatives. Thus, under gauge transformations the gauge fields transform in the adjoint representation of the gauge group.

4. Jan 9, 2012

### ndung200790

I do not understand the gauge fields transform under the gauge group in adjoint representation,because there are exist an extra
δ$_{\mu}$ :because gauge fields transform as:
((A$^{\alpha}$)$^{a}_{\mu}$=
=A$^{\alpha}_{\mu}$+
+(1/g)$\delta$$_{\mu}$$\alpha$$^{a}$+
f$^{abc}$A$^{b}_{\mu}$$\alpha$$^{c}$)

5. Jan 9, 2012

### tom.stoer

strictly speaking the gauge field is not Lie-algebra valued b/c it does not transform as a tensor under gauge trf. (like the field strength tensor) but as a connection (like Christoffel symbols in ART do not transform as tensors)

6. Jan 9, 2012

### Dickfore

The Lie algebra is given by the commutator:
$$\left[ T^a_R, T^b_R \right] = i \, f^{a b c} \, T^c_R$$
where the structure constants $f^{a b c}$ are the same in any representation. You may form one special representation using them:
$$(T^a_A)^{b c} \equiv -i f^{a b c}$$
This is called the adjoint representation. It is g-dimensional, where g is the number of generators of the Lie group.