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.
