Equivalence of two adjoint representation definitions

Question

I'm following notes that define the adjoint representation as a map $\rho$ from $\mathfrak{g}$ to $End(\mathfrak{g})$ such that, for $t_1, t_2$ $\in \mathfrak{g}$ ,

\begin{equation} \rho_{\tiny{adj}}(t_1) (t_2) = [t_1,t_2]. \end{equation}

Taking a basis $t_\alpha$ for $\mathfrak{g}$ we know that

\begin{equation} [t_\alpha,t_\beta]=f_{\alpha\beta}^{\,\,\,\,\gamma} t_\gamma. \end{equation} How do we see that this matches the alternative definition

\begin{equation} \rho_{\tiny{adj}}(t_\alpha)_{\beta}^{\,\,\,\gamma} = f_{\alpha\beta}^{\,\,\,\,\gamma}. \end{equation}

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Attempt: \begin{equation} \rho(t_\alpha) (t_\beta) = [t_\alpha,t_\beta] = f_{\alpha\beta}^{\,\,\,\,\gamma} t_\gamma. \end{equation} However I seem to get confused with the indices because it seems impossible to me to get $\rho(t_\alpha)$ to be $f_{\alpha\beta}^{\,\,\,\,\gamma}$. Writing the indices explicitly I would think we get

\begin{equation} \rho(t_\alpha)_{\mu}^{\,\,\,\nu} (t_\beta)_{\nu} = [t_\alpha,t_\beta]_{\mu} = f_{\alpha\beta}^{\,\,\,\,\gamma} (t_\gamma)_{\mu}. \end{equation}

Answer 1

The adjoint map at the level of the lie algebra is such that $$\text{ad}: \mathfrak{g} \rightarrow \text{End}(\mathfrak g),$$ taking $\lambda_a \mapsto \text{ad}_{\lambda_a}$ where $$\text{ad}_{\lambda_a}: \mathfrak{g} \rightarrow \mathfrak{g}\,\,\,\,\,\text{with}\,\,\, \lambda_b \mapsto [\lambda_a, \lambda_b].$$

So, this means that $$\lambda_b \rightarrow \text{ad}_{\lambda_a} \lambda_b = [\lambda_a, \lambda_b] = if_{abc}\lambda_c.$$

The map $\text{ad}$ is equivalent to $\rho$ for matrix lie groups so $\rho: \mathfrak{g} \rightarrow \mathfrak{gl}(V=\mathfrak{g}) \equiv \text{End}(V = \mathfrak{g}) $ is the map such that $\rho(\lambda_a)_{cb}\lambda_c = [\lambda_a, \lambda_b] = if_{abc}\lambda_c$, which is to say $\rho(\lambda_a) (\lambda_b) = [\lambda_a, \lambda_b].$ Peeling off the $\lambda_c$ we obtain $$\rho(\lambda_a)_{cb} = i f_{abc}.$$

So the indices $\left\{b,c\right\} \in \left\{1, \dots , \text{dim}(\mathfrak{g})\right\}$ attached to the $\text{dim}(\mathfrak{g}) \times \text{dim}(\mathfrak{g})$ matrix representations $\rho(\lambda_a)_{bc}$ of the lie algebra adjoint action mix around the real $\text{dim}(\mathfrak{g}) \times 1$ basis vectors $\left\{\lambda \right\}$ of the lie algebra.

Alternatively, since $\rho$ is a linear operator acting on the vector space $V = \mathfrak{g}$, an equivalent operation is one in which $\rho$ mixes around the components $(\lambda_b)_d$ of some $\lambda_b$, so we can equally rewrite the transformation law $$\rho(\lambda_a)_{cd} (\lambda_b)_d = \rho(\lambda_a)_{cd} \delta_{bd} = \rho(\lambda_a)_{cb} \overset{!}{=} [\lambda_a, \lambda_b]_c = if_{abd} (\lambda_d)_c = if_{abd} \delta_{dc} = if_{abc}$$ and thus arrive at the same conclusion $$\rho(\lambda_a)_{cb} = if_{abc},$$ where $(\lambda_i)_j = \delta_{ij}$ WLOG (if there is an overall normalisation, this cancels between the two sides of the equation) and $|\lambda_b \rangle = (\lambda_b)_d |\lambda_d \rangle$.

$\mathbf{Aside}$

One may ask why $\rho(\lambda_a)_{cb} \lambda_c$ and not, say, $\rho(\lambda_a)_{bc} \lambda_c$? We define the action of a generator $\lambda_a$ on a basis vector $|\lambda_b \rangle$ through the Lie bracket as noted above. This is to say $\lambda_a |\lambda_b \rangle = \rho(\lambda_a)_{cb} |\lambda_c \rangle$ so that $\langle \lambda_c | \lambda_a |\lambda_b \rangle = \rho(\lambda_a)_{cb}$ are the matrix elements of the matrix representation of the operator $\rho$ in the basis $\left\{\lambda\right\}$. Consider some arbitrary vector built up from the basis vectors of $\mathfrak{g}$, that is $|\Psi \rangle = \sum c_b |\lambda_b \rangle$. Then $$|\Psi \rangle' = c_d' |\lambda_d \rangle \equiv \lambda_a |\Psi \rangle = c_b \lambda_a |\lambda_b \rangle = c_b \rho(\lambda_a)_{cb} |\lambda_c \rangle $$ Relabelling indices at the first equality allows us to write the transformation of the components of the vector $|\Psi \rangle$ in the basis spanned by $\left\{\lambda \right\}$ as $c_c' = \rho_{cb} c_b$.

All of this makes contact with the familiar matrix/vector product $\mathbf v' = M \mathbf v$ or in component form $v_i' = M_{ij}v_j$ and $\mathbf v = v_i e_i$ where $v_i$ are the components of $\mathbf v$ in the basis $\left\{e_j \right\}$.

In the notation of the aside, the $\rho(\lambda_a)$, $|\Psi \rangle, |\lambda_b \rangle$ and $c_b$ would be equivalent to the $M$, $\mathbf v$, $e_j$ and $v_j$ above respectively. So, we see that with the used indicial writing of the $\rho$ we obtain the familiar transformation law for components of vectors.