Undergrad Adjoint Representation Confusion

[Arcturus7]

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:

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?

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.

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.

 

[fresh_42]

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:
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.

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.

To be honest, I found it difficult to grasp a point here where an answer could start. I have written some insight articles which might answer your questions, or at least sort out the language.

Representations and language:
https://www.physicsforums.com/insights/representations-precision-important/

The example(s) $SU(2)$ and $\mathfrak{su}(2)$:
https://www.physicsforums.com/insights/journey-manifold-su2-part-ii/

Lie derivative and left invariant vector fields on the example of $GL(n)$:
https://www.physicsforums.com/insights/pantheon-derivatives-part-iv/

 

[Arcturus7]

To be honest, I found it difficult to grasp a point here where an answer could start. I have written some insight articles which might answer your questions, or at least sort out the language.

Representations and language:
https://www.physicsforums.com/insights/representations-precision-important/

The example(s) $SU(2)$ and $\mathfrak{su}(2)$:
https://www.physicsforums.com/insights/journey-manifold-su2-part-ii/

Lie derivative and left invariant vector fields on the example of $GL(n)$:
https://www.physicsforums.com/insights/pantheon-derivatives-part-iv/

Hi - yes sorry, reading that post back it's very much word salad. Firstly, thank you for the links - I'll be sure to give them a read.

Let me be more concrete with my question. I followed the hint as suggested in the question, and in so doing ended up with:

$$U\approx 1+\epsilon X$$

And so:

$$Ad(U)(Y)=Ad(X)(Y)=[X,Y]$$

But from here I'm not really sure what to do, because I'm not sure how to interpret $Ad(X)(Y)$ in a way that helps me with the question. Does that make sense?

 

[fresh_42]

Firstly, it is convenient to distinguish group and algebra elements. So let's say we have $u\in SU(2)$ and $X,Y\in \mathfrak{su}(2)$.

The adjoint representation of the Lie group is then given as $Ad(u) = (X \longmapsto uXu^{-1})$ that is $Ad\, : \,SU(2) \longrightarrow GL(\mathfrak{su}(2))$ defined by conjugation on the vector space $\mathfrak{su}(2)$.

The adjoint representation of the Lie algebra is simply its left multiplication, given as $ad(X) = (Y \longmapsto [X,Y])$ that is $ad\, : \,\mathfrak{su(2)} \longrightarrow \mathfrak{gl}(\mathfrak{su}(2))$ defined by commutation on itself.

Both are related by $Ad(e^X)=e^{\mathfrak{ad}(X)}$. Exponentiation gets you from the Lie algebra to the Lie group, one in the case of $\mathfrak{su}(2)$ to $SU(2)$ and the other one in the case of $\mathfrak{gl}(V)$ to $GL(V)$ where the vector space of the representation is again $V=\mathfrak{su}(2)$. It is a sort of integration since we recover group elements by their tangents, that is we solve a differential equation and that's where the exponential function comes into play; same as it does in ordinary differential equation systems if we make an ansatz of a solution. The details are bit more complicated, but the principle is that.

This isn't very practical, as most matrices are hard to exponentiate, and a short view on the sizes of the matrices here doesn't help. So what we can do is, define a path through the identity matrix of $SU(2)$ within $SU(2)$, say $\varepsilon \longmapsto \gamma(\varepsilon)$ and differentiate it along a direction $X$ since we know that the elements of the Lie algebra are the tangents there - more specifically, the left invariant vector fields evaluated at $I=1$. This gives us a first approximation $1+\varepsilon \cdot X$.
I think the example in the last link I gave is a bit better explained (sec. B).