Expanding the exponent of the adjoint map

Question

I am reading the following: https://i.stack.imgur.com/wY7Yk.jpg and am having trouble understand the definition of $$B_{b,n} = e^{-ad_{u}}(b\lambda^{n})$$

I know that $ad_{u} = [u,\cdot]$ and I know that the Lie bracket is skew-symmetric. Then, $-ad_{u} = -[u,\cdot] = [\cdot, u] \stackrel{?}{=} ad_{\mathbf{\cdot}}(u)$. Now examining $B_{b,n}$ I write $B_{b,n} = e^{-ad_{u}}(b\lambda^{n}) \stackrel{?}{=} e^{ad_{b\lambda^{n}}}(u)$? Is this right? If so, does this expand to: $$u + [b\lambda^{n}, u] + \frac{1}{2}[b\lambda^{n},[b\lambda^{n}, u]] + \cdots$$ ??

Answer 1

With help from a colleague I write the following.

My question involves the exponential map. A map from a Lie algebra $g$ into its Lie group $\mathcal{G}$. In the case of a matrix Lie group and matrix Lie algebra, exp coincides with the matrix exponential: $$e^{X} := I + X + \frac{1}{2}X^2 + \cdots + \frac{1}{n!}X^{n} + \cdots$$ where $X \in \mathcal{g}$ and $I$ the identity matrix. If $X = ad_{u}$, in this case we then have $X \in End(\mathcal{g})$. To obtain $$e^{ad_{u}}(L) = L + [u,L] + \frac{1}{2}[u,[u,L]] + \cdots $$ we simply apply $L$ to the adjoint operator which yields the above.

Now let $X = -ad_{u}$. Then we obtain: $$e^{-ad_{u}} = I - ad_{u} + \frac{1}{2}(-ad_{u})^{2} - \cdots + \cdots$$ Now applying $b\lambda^{n}$ to the above we obtain: $$e^{-ad_{u}}(b\lambda^{n}) = b\lambda^{n} - [u, b\lambda^{n}] + \frac{1}{2}[u,[u, b\lambda^{n}]] - \cdots + \cdots$$