I'm having a bit of an issue wrapping my head around the adjoint representation in group theory. I thought I understood the principle but I've got a practice problem which I can't even really begin to attempt. The question is this:

However, I have a couple of questions about the actual mechanics of this problem. First of all, I had understood the expression $Ad(X)$ to be something along the lines of "the image of $X$ under the adjoint representation" - that is to say, in the adjoint rep, the Lie algebra element $X$ is given by the matrix $Ad(X)$. $Ad(X)$ is just a linear operator which acts on the vector space $\mathfrak{g}$ and the matrix elements of $Ad(X)$ are just given by the structure constants $f^{abc}$. However with this in mind, I don't really appreciate the significance of the object $Ad(X)(Y)$, beyond the superficial. Clearly it's a linear transformation of a vector $Y\in \mathfrak{g}$, given by the commutator $[X,Y]$, but so what? I think the correct interpretation of $Ad(X)(Y)=[X,Y]$ is that it defines how elements of $\mathfrak{g}$ transform in the adjoint representation.

Secondly, I know that that commutator is equal to a linear combination of the remaining generators via $[T_a,T_b]=f^{abc}T_c$, but how does that help me here? A hint to the question says to consider an element $U=e^{\epsilon X_a}$ where $X_a \in \mathfrak{su}(N)$ and to subsequently consider $Ad(U)(X_b)$ for some generic $X_b$ in the Lie algebra. My initial inclination was to write something of the form $V=e^{f^{abc}X_c}$ where $V\in SU(N)$, but I don't think this is correct.

I'm a bit lost, I think because I'm looking at this in the wrong way because I haven't quite understood something. Does anybody have any pointers?

*EDIT* SORRY FOR ALL THE TERRIBLE FORMATTING ATTEMPTS.

My understanding of this question is that, given a representation $d(\mathfrak{g})$ of a Lie algebra, we say that $D(G)$ is induced by $d(\mathfrak{g})$ if $D(G)=e^{d(\mathfrak{g})}$, and so if we start with the Lie algebra in the adjoint representation, then there should be some resulting (induced) representation which emerges as a result of the exponential map when we go from the algebra to the group.Given the Lie algebra $\mathfrak{su}(N)$, with the Lie bracket given by the commutator, in the adjoint representation defined by $ad(X)(Y)=[X,Y]$, what is the induced representation on the Lie group $SU(N)$ using the exponential map?

However, I have a couple of questions about the actual mechanics of this problem. First of all, I had understood the expression $Ad(X)$ to be something along the lines of "the image of $X$ under the adjoint representation" - that is to say, in the adjoint rep, the Lie algebra element $X$ is given by the matrix $Ad(X)$. $Ad(X)$ is just a linear operator which acts on the vector space $\mathfrak{g}$ and the matrix elements of $Ad(X)$ are just given by the structure constants $f^{abc}$. However with this in mind, I don't really appreciate the significance of the object $Ad(X)(Y)$, beyond the superficial. Clearly it's a linear transformation of a vector $Y\in \mathfrak{g}$, given by the commutator $[X,Y]$, but so what? I think the correct interpretation of $Ad(X)(Y)=[X,Y]$ is that it defines how elements of $\mathfrak{g}$ transform in the adjoint representation.

Secondly, I know that that commutator is equal to a linear combination of the remaining generators via $[T_a,T_b]=f^{abc}T_c$, but how does that help me here? A hint to the question says to consider an element $U=e^{\epsilon X_a}$ where $X_a \in \mathfrak{su}(N)$ and to subsequently consider $Ad(U)(X_b)$ for some generic $X_b$ in the Lie algebra. My initial inclination was to write something of the form $V=e^{f^{abc}X_c}$ where $V\in SU(N)$, but I don't think this is correct.

I'm a bit lost, I think because I'm looking at this in the wrong way because I haven't quite understood something. Does anybody have any pointers?

*EDIT* SORRY FOR ALL THE TERRIBLE FORMATTING ATTEMPTS.

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