# Proving a relation for representations of gauge groups

Let ${\cal G}$ be a Lie group - possibly disconnected. Let ${\mathfrak g}$ denote the corresponding Lie algebra. Let $R_k$ be a general unitary representation of ${\cal G}$ and $R$ be the adjoint representation. We also denote the Hermitian generators of the Lie algebra ${\mathfrak g}$ in the corresponding representations as $T^a_k$ and $t^a$ respectively. Here, $a = 1, \cdots , \dim {\mathfrak g}$. These are chosen to satisfy $$[ T_k^a , T^b_k ] = i f^{abc} T^c_k,~~~ ~~~ [ t^a , t^b ] = i f^{abc} t^c,~~~ ~~~ \text{tr}[t^at^b] = \delta^{ab}$$ Let $g \in {\cal G}$. We exploit notation and denote $R(g)$ as $g$ and $R_k(g)$ as $g_k$. We now treat $g$ and $g_k$ as simple square matrices which have the same dimension as each $t^a$ and $T^a_k$ respectively and the dimension of $g$ (or $t^a$) is equal to $\dim {\mathfrak g}$.

Q. Is the following statement true? $$g_k T^a_k g_k^{-1} = (g^{-1})^{ab} T_k^b$$ I can prove this when the group element is connected to the identity, i.e. when we can write $$g = \exp [ i g^a t^a ],~~~~ g_k = \exp [ i g^a T^a_k]$$ Is true more generally, i.e. when $g$ is not connected to the identity?

The statement above seems to be saying that if the generators $(T^a_k)_{ij}$ are treated as objects that have one adjoint index and two indices in representation $R_k$, then they are invariant under the action of the gauge group.

PS - I'm a physicist as you can probably guess by the language of the question. It would be good to have an answer in a similar language as well. Thanks.

PPS - If the notation of the equation is unclear, here it is with all the matrix indices explicitly $$\left( g_k T^a_k g_k^{-1} \right)_{ij} = (g^{-1})^{ab} \left( T_k^b \right)_{ij}$$ Note that $k$ only labels the representation and is fixed, i.e. not summed over.

It seems OP's question (v1) is essentially asking:

Given a Lie group $G$, with a Lie group representation $R:G\to GL(V,\mathbb{F})$, with corresponding Lie algebra $L={\rm Lie}(G)$, and with a corresponding Lie algebra representation $\rho :L\to {\rm End}(V,\mathbb{F})$, would $$\tag{1} \forall g\in G, x\in L:~~\rho({\rm Ad}(g)x)~=~R(g)\rho(x)R(g)^{-1}~?$$

Here ${\rm Ad}:G\to GL(L,\mathbb{F})$ denotes the adjoint representation $$\tag{2} {\rm Ad}(g)x~:=~gxg^{-1}.$$

Equation (1) is, e.g., a consequence of

$$\tag{3} \forall g,h\in G:~~R(ghg^{-1})~=~R(g)R(h)R(g)^{-1},$$

by taking $h=e^{tx}$, $t\in\mathbb{F}$, under pertinent continuity assumptions.

• Thank you for restating the question in cleaner words. +1. Commented Dec 20, 2014 at 17:23
• Thanks! Your suggestion makes a lot of sense! I will try to make sense, but I think I can get it from here. Commented Dec 20, 2014 at 19:24