I have asked this question in Math Stackexchange as well. However, given that it is closely related to gauge theories studied by physicists who will probably be more familiar with the language and will have most likely encountered this before, I post it here as well.

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 - 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.

This post imported from StackExchange Physics at 2014-12-19 23:07 (UTC), posted by SE-user Prahar