I recently asked a question regarding the representations of $\mathfrak{u}(1)$ on $\mathbb{C}$ and $\bar{\mathbb{C}}$ here, and I think my main issue stems from a lack of clarity regarding what an isomorphism of a representations actually is. The book this problem is from is Mathematical Gauge Theory by Hamilton, and in it he defines a morphism (or intertwining map) of representations as follows:
Given a Lie algebra $\mathfrak{g}$, and representations of $\mathfrak{g}$, $\phi_V,\phi_W$, on vector spaces $V$ and $W$, a $\textbf{morphism}$ of the represenations is a $\mathfrak{g}$-equivariant linear map $f:V\rightarrow W$ such that:
$$f(\phi_v(X)v)=\phi_W(X)f(v)$$
An $\textbf{isomorphism}$ of representations is a $\mathfrak{g}$-equivariant isomorphism of vector spaces.
This definition makes sense to me as is, however when I try to apply it in practice I get quite confused. For example, Part $A$ of the problem is to show that there exists a matrix $A\in GL(2,\mathbb{C})$ such that for every $M\in\mathfrak{su}(2)$ we have that:
$$AMA^{-1}=\bar{M}$$
and to then conclude that the representations of $\mathfrak{su}(2)$ on ${V}\cong\mathbb{C}^2$ and $\bar{V}$ are isomorphic.
Finding $A$ was, in my opinion, easy. By examining the fundamental representation of $\mathfrak{su}(2)$ we see that for a matrix $M\in \mathfrak{su}(2)$ we have:
$$M=\begin{pmatrix}
ia&-\bar{z}\\
z&-ia
\end{pmatrix}$$
For some $a\in\mathbb{R}$ and $z\in\mathbb{C}$. It is then clear that:
$$A=\begin{pmatrix}
0&1\\
-1&0
\end{pmatrix}$$
With inverse given by:
$$A^{-1}=\begin{pmatrix}
0&-1\\
1&0
\end{pmatrix}$$
However from this point onwards I am completely lost. How can I say that this implies that these two representations are isomorphic from this? If $f$ were an isomorphism I would be able to write:
$$\phi_W(X)=f\circ\phi_V(X)\circ f^{-1}$$
But with $f=A$ I do not obtain the complex conjugate of a vector $v\in \mathbb{C}^2$, so I don't see how this works here. Are they just isomorphic representations because the representation of $G$ on $\mathfrak{g}$ via the Ad map has a linear transformation that takes $M$ to $\bar{M}$? So the existence of such a map on $\mathbb{C}^2$ is unimportant? but then how would that satisfy the original definition of isomorphic representations?
Any help or clarity would be appreciated, my exposure to representation theory is limited to this book however, so I assume I am missing something obvious.
Best Answer
There are two equivalent but subtly different ways to define the conjugate representation. It seems that your source mixed them together in an unfortunate way.
Let's say we have a real Lie algebra $\mathfrak g$, and a representation $(\pi, V)$ of it, i.e. an $n$-dimensional $\mathbb C$-vector space $V$ and an $\mathbb R$-linear map
$$\pi: \mathfrak g \rightarrow \mathrm{End}_{\mathbb C}(V) $$
such that $\pi([x,y]) = \pi(x) \circ \pi(y) - \pi(y) \circ \pi(x)$ for all $x,y \in \mathfrak g$.
Method 1 to define the conjugate representation: Define a new $\mathbb C$-vector space $V'$ which, as set and additive group, is $V$, but where complex scalars $z \in \mathbb C$ act via scalar multiplication $\ast: \mathbb C \times V' \rightarrow V'$: $z \ast v := \bar z v$.
Check that each element $y \in \mathrm{End}_{\mathbb C}(V)$ also commutes with the $\ast$-scalar multiplication, meaning it is also an element of $\mathrm{End}_{\mathbb C}(V')$. This is in particular true for all $\pi(x)$ for all $x \in \mathfrak g$. Conclude that $(\pi, V')$ is a representation of $\mathfrak g$. This is the conjugate of $(\pi, V)$, as per method 1.
Method 2 to define the conjugate representation: Fix a basis $\mathcal{B}$ of $V$. With respect to that basis, we have $\mathrm{End}_{\mathbb C}(V) \simeq M_n(\mathbb C)$ and can prolong $\pi$ to a map
$$\pi: \mathfrak g \rightarrow M_n(\mathbb C)$$
Define $\pi' (x) = \overline{\pi(x)}$, i.e. complex-conjugate each entry of the matrix $\pi(x)$ for all $x \in \mathfrak g$. Then $\pi'$ is an $\mathbb R$-linear map
$$\pi': \mathfrak g \rightarrow M_n(\mathbb C)$$
and via identifying those matrices with $\mathrm{End}_{\mathbb C}(V)$ again, through the basis $\mathcal{B}$, we view it as map $\mathfrak g \rightarrow\mathrm{End}_{\mathbb C}(V)$. Check that $(\pi', V)$ is a representation of $\mathfrak g$. This is the conjugate of $(\pi, V)$, as per method 2.
You see either one twists the $V$ or one twists the $\pi$.
How to show that a representation is equivalent to its conjugate in method 1: Show equivalence of $(\pi, V)$ and $(\pi, V')$. That is, show that there is a bijective $\mathbb C$-linear map $\alpha: V \rightarrow V'$ such that for all $x \in \mathfrak g$,
$$\alpha \circ \pi(x) = \pi(x) \circ \alpha$$
However, note that a $\mathbb C$-linear map $\alpha: V \rightarrow V'$ is the same as a conjugate-linear (i.e. $\alpha(zv) = \bar z v$) map $\alpha:V \rightarrow V$.
How to show that a representation is equivalent to its conjugate in method 2: Show equivalence of $(\pi, V)$ and $(\pi', V)$. That is, show that there is a bijective $\mathbb C$-linear map $\beta: V \rightarrow V$ such that for all $x \in \mathfrak g$, in the chosen basis $\mathcal B$,
$$\beta \circ \pi(x) = \overline{\pi(x)} \circ \beta.$$
Again notice that either one twists with a conjugate-linear map which leaves the matrices of $\mathfrak g$ as they are; or one twists with a $\mathbb C$-linear map which turns the matrices into conjugates.
(One can actually show from this that the approaches are equivalent: After fixing a basis, and knowing a little "semilinear algebra", one gets the map $\alpha$ from $\beta$, or conversely, by just using the same matrix in the same basis, just that one time one interprets this matrix as giving a linear, and one time as giving a conjugate-linear map.)
I usually do not recommend math videos, but this one discusses the same two methods, except that it talks about representations of groups instead of Lie algebras. But you see that otherwise the ideas are identical.
Now you see why in this question you are done once you show $AMA^{-1}=\bar M$. We are using method 2, $A$ is the matrix of $\beta$ and $M$ is the matrix $\pi(x) \in \mathfrak{su}_2$, in the standard basis. In this method 2, there is no need for what you call $f$ (i.e. what I called $\beta$, i.e. the $\mathbb C$-linear map represented by the matrix $A$) to give out some "conjugate of a vector" (whatever that would be).
The problem in your source, as quoted by you in comments, is that it mixes the two methods together. It first defines the underlying vector space $\bar V$ like in method 1, but then it also wants you to conjugate matrices entrywise as in method 2. That is too much! As you see above, to show equivalence of a representation to its conjugate, either you give a conjugate-linear map $\alpha$ which commutes with all $\pi(x)$ (method 1); or you give a linear map $\beta$, matrix-conjugation with which turns all $\pi(x)$ (as matrices) into their conjugates.
Here in this question, your error (very understandable from the unclarity in your source) is that you expect $\beta$ to be something like conjugate-linear, which it should not be.
In your previous question, you actually do not mix the methods, but stay very consistent in method 1 (although commenters there are confused about your proposed map $F$, which is my $\alpha$ above, being conjugate-linear; but if one stays inside method 1, that is OK, because as said, a conjugate-linear map $V \rightarrow V$ is the same as a $\mathbb C$-linear map $V \rightarrow V'$).
The mistake there is a different one, and might be related to the annoying covention among physicists to take out the imaginary unit $i$. Quite simply, your map $F$ is not $\mathfrak g$-equivariant for $k \neq 0$: Because $\mathfrak g = \mathfrak{u}(1) = i \mathbb R$ operates on $V_k$ via $(\pi_k (ix)) (v) = ikx v$ (for any $x \in \mathbb R$) so that e.g. $$(\pi_k(i) \circ F )(1)= \pi_k(i) (1) = ki$$ but $$(F \circ \pi_k(i))(1)=F(ki)=-ki.$$
As a last remark, to show that the two methods are really equivalent, I claim there always is an equivalence between $(\pi', V)$ and $(\pi, V')$, i.e. a $\mathbb C$-linear map $\gamma: V \rightarrow V'$ (which again is the same as a conjugate-linear map $V \rightarrow V$) such that $$ \gamma \circ \pi'(x) = \pi(x) \circ \gamma$$ for all $x \in \mathfrak g$. Namely, for the basis $\mathcal B = \{v_1, ..., v_n\}$ fixed in method 2 to define $\pi'$, the map $\gamma$ is defined by $\gamma(\sum z_i v_i) := \sum \bar z_i v_i$.
So again one has to be careful: To show equivalence of a representation $(\pi, V)$ to its conjugate, one should not search for a map $\gamma$ like here, which is both conjugate linear and twists the matrices to their conjugates. Because such a map always exists, and unlike the maps $\alpha$ or $\beta$ above, does not tell us anything about the relation of $(\pi, V)$ to its conjugate, but just confirms that the two ways to define the conjugate representation are equivalent.