Representations of $SU(1,1)$ and its isomorphism with $SL(2,\mathbb{R})$

Question

I've been reading up on the Lorentz group in $(2+1)-$dimensions, i.e. $SO(2,1)$. As I understand it, $SO(2,1)$ is isomorphic to $SL(2,\mathbb{R})$, $SU(1,1)$ and $Sp(2,\mathbb{R})$ (modulo $\mathbb{Z}_2$). I am most interested in the $SL(2,\mathbb{R})$, $SU(1,1)$ isomorphisms. In particular, there are a couple of confusions I have about the matrix representations of these groups, which I shall now elaborate on. $$$$ 1. $~$ For $SL(2,\mathbb{R})$, the anti-fundamental representation is trivially equivalent (in fact, identical), to the fundamental representation $N\in SL(2,\mathbb{R})$, since by definition, the group elements are real. I assume this means that, unlike in the case of $SL(2,\mathbb{C})$, we don't need dotted indices, and $SL(2,\mathbb{R})$ naturally acts on real two-dimensional spinors $\psi_a$, which transform under $SL(2,\mathbb{R})$ as $$\psi_a \quad\rightarrow\quad \psi'_a = N_a^{~~b}\psi_b\;,$$ i.e. we do not need to consider "barred" spinors $\bar{\psi}_{\dot{a}} = (\psi_a)^\ast$ at all?

If this is correct, then my main confusion comes when considering $SU(1,1)$. In this case, there exists an invariant matrix $$K=\begin{pmatrix}1&0\\ 0&-1\end{pmatrix} \;,$$ such that for $\zeta,\zeta^\dagger\in SU(1,1)$ we have $$\zeta^\dagger K\zeta = K \;.$$ This implies an isomorphism between the fundamental and anti-fundamental representations $\zeta$ and $\zeta^\ast$, respectively. However, does that mean that we can drop the dotted indices? In principle, we still have that $$ \chi_{\alpha} \quad\rightarrow\quad \chi'_{\alpha} = \zeta_{\alpha}^{~~\beta}\chi_{\beta} \;,\qquad \bar{\chi}_{\dot{\alpha}} \quad\rightarrow\quad \bar{\chi}'_{\dot{\alpha}} = \zeta_{\dot{\alpha}}^{~~\dot{\beta}}\bar{\chi}_{\dot{\beta}} \;.$$ What I find particularly confusing, is if we then consider a bi-spinor representation of a Lorentz vector $p^\mu$, i.e. $p_\mu\sigma^\mu$ (where $\sigma^\mu$ are the Pauli matrices, and $\mu=0,1,2$), should it have one dotted and one un-dotted index, i.e. $p_{\alpha\dot{\alpha}}$, such that it transforms under $SU(1,1)$ as $$p_{\alpha\dot{\alpha}}\quad\rightarrow\quad p'_{\alpha\dot{\alpha}} = \zeta_{\alpha}^{~~\beta}(\zeta^\ast)_{\dot{\alpha}}^{~~\dot{\beta}}p_{\beta\dot{\beta}}\;?$$

$$$$ 2.$~$ Following on from this, $SL(2,\mathbb{R})$ and $SU(1,1)$ are mapped to one another via a Cayley transformation, of the form: $$C = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & i\\ i & 1\end{pmatrix} \;,$$ such that $CNC^{-1}=CNC^{\dagger}\in SU(1,1)$ for $N\in SL(2,\mathbb{R})$. What is the consistent way to map $SL(2,\mathbb{R})$ spinors $\psi_a$ to $SU(1,1)$ spinors $\chi_{\alpha}$, particularly if the above is correct about the anti-fundamental representation? Naively, it seems like it should be $$\chi_{\alpha} = C_{\alpha}^{~~a}\psi_a\;,\qquad \bar{\chi}_{\dot{\alpha}} = (C^\ast)_{\dot{\alpha}}^{~~a}\psi_a \;,$$ but I'm concerned by the mixing of dotted and un-dotted indices, and also barred and un-barred spinors. $$$$ 3. $~$ Finally, both constructions are quite confusing in indicial form. In particular, given that the Levi-Civita tensor $\epsilon^{ab}$ ($\epsilon^{12}=1=-\epsilon_{12}$) can be used as a metric in both $SL(2,\mathbb{R})$ and $SU(1,1)$, to raise/lower spinor indices, e.g. $\psi^a = \epsilon^{ab}\psi_b$ in the $SL(2,\mathbb{R})$ case. Can $\epsilon^{ab}$ be used to raise/lower indices of the matrix representations of $SL(2,\mathbb{R})$ and $SU(1,1)$ also? For example, is $$ N^{a}_{~~b} = \epsilon^{ac}\epsilon_{db} \,N_c^{~~d}$$ correct? If so, it seems to imply odd consequences, e.g. for $N\in SL(2,\mathbb{R})$, $$\epsilon^{ac}\epsilon_{db} \,N_c^{~~d} = (N^{-1})_b^{~~a} = N^{a}_{~~b} = (N^T)_{b}^{~~a} \;,$$ but this can't be right, because $SL(2,\mathbb{R})$ isn't an orthogonal group. $$$$ Apologies for the length of this post. Any help on clearing up this matter for me would be very much appreciated. It's been bugging me for a while, and I haven't been able to find any enlightening literature on the subject (Bargmann's seminal paper doesn't really go into detail on this it seems).

$$$$

Disclaimer: Please bear with me here, as I am a physicist, with a physicist's understanding of group theory, and representation theory. Apologies for the lack of rigour, and index notation.

Answer 1

It is important to be clear on the distinction between conjugate representations, and dual representations, as they are only necessarily the same when the relevant group is compact, which is not the case here. Self-conjugacy implies the representation is either quaternionic or real, in this instance the rep is manifestly real. Self duality on the other hand gives rise to invariant forms, either orthogonal or symplectic, and this is what we use to raise and lower indices, as the invariant form induces an isomorphism between the vector space being acted on, and its dual space. Furthermore in the case that a representation has both properties, or is isomorphic to its conjugate-dual a.k.a. anti-dual (but not necessarily either on their own), this gives rise to a sesqui-linear form.

It is known that $\operatorname{SL}(2, \mathbb{F}) \cong \operatorname{Sp}(2, \mathbb{F})$ where $\mathbb{F}$ is the underlying field (This is only true for $n=2$, otherwise $Sp \subset SL$). This is important with regard to the self duality of $\operatorname{SL}(2,\mathbb{R})$ because by the aformentioned isomorphism $2 \times 2$ real matrices $N$ which satisfy the fundamental symplectic relation: \begin{equation} N^\intercal \varepsilon N = \varepsilon, \qquad \varepsilon = \begin{pmatrix}0 & -1 \\ 1 & 0 \end{pmatrix} \end{equation} are precisely the matrices of $\operatorname{SL}(2,\mathbb{R})$. Thus because $\operatorname{SL}(2,\mathbb{R})$ is a symplectic group, it respects a symplectic form and we may use that form to raise and lower indices.

Now, if we are acting upon a real vector space, which we can do by the reality of the representation, then it is true there is nothing gained by making a distinction between dotted and undotted spinors, since conjugate spinors transform identically to the fundamental representation. Dual spinors transform as expected so the above symplectic form is respected.

However, we need not necessarily act upon a real vector space. We can just as well extend the group action complex linearly to complex combinations of spinors. This is desirable if we wish to impose gauge symmetry on our theory (often gauge groups will have strictly complex representations). This brandishes us with a new operation: complex conjugation, denoted with a $\sigma(\cdot)$, which acts simply by $$\psi =\varphi+i\phi \mapsto \sigma(\psi) = \psi^c = \varphi-i\phi$$ where of course the spinors $\varphi, \phi$ are real. This new operation respects our underlying group operations because they are manifestly real. So we gain access to a new invariant form, a sesqui-linear form. I.e. both of the following inner products are $\operatorname{SL}(2,\mathbb{R})$ invariant: $$ B(\psi_1, \psi_2) = \psi^\intercal_1 \varepsilon \>\psi_2 \qquad h(\psi_1, \psi_2) = \psi^\dagger_1 \varepsilon \>\psi_2 $$ for any complex spinors $\psi_i$. This is exactly analogous to the case of Majorana spinors and $\operatorname{Spin(1,3)}$. It should be clear that we may consider $B(\sigma(\cdot), \cdot) = h(\cdot, \cdot)$ and vice versa. (It is useful to consider this relationship when dealing with transformation properties.)

With this in place, there is nothing to stop us from applying a change of basis on our vector space, say, one where the sesqui-linear product is diagonal. In general this will cause us to lose our manifest realness of the group elements, but we can still infer the real structure of the group from the structure of our complex conjugation operation (see here).

It may come as no surprise to you that the transformation which diagonalizes our sesqui-linear form $h$ is the Cayley transformation (which I will denote $U$). Without fussing about indices on the invariant structures for now, we can see if we want our invariant forms to remain invariant they must transform under the change of basis via: $$ \psi^\prime = U\psi, \quad N^\prime = U N U^{-1},\quad h^\prime = U h U^{-1}, \quad B^\prime = (U^{-1})^\intercal B U^{-1}, \quad \sigma^\prime = U(U^{-1})^*\sigma $$ or with our case noticing $U^\dagger = U^{-1}$: $$ \quad N^\prime = U N U^\dagger,\quad h^\prime = U h U^\dagger, \quad B^\prime = U^* B U^\dagger, \quad \sigma^\prime = UU^\intercal \sigma $$ Here $^*$ implies complex conjugation of the field elements in the obvious way, same to how $\sigma$ initially acts. Of note though is that $\sigma$ transforms rather non-trivially, due to its complex conjugating (formally: its anti-linearity). To be explicit, after the transformation $\sigma^\prime$ acts first by conjugation, then by the linear matrix part: $$({\psi^\prime})^c = \sigma^\prime(\psi^\prime) = UU^\intercal \psi{^\prime}^* = UU^\intercal U^*\psi{^*}= U \psi^* = (\psi^c)^\prime$$ and thus it is also clear that spinors which are $\sigma$ conjugate pairs before the transformation remain $\sigma$ conjugate pairs after the transformation, as opposed to the case of merely complex conjugating in the new basis. The transformation of $B$ falls out as a result of the compatibility with $\sigma$ and $h$.

Actually performing the transformation now we have: $$ (B^\prime) = U^* \varepsilon U^\dagger = \varepsilon, \quad \mathrm{and} \quad (h^\prime) = U \varepsilon U^\dagger = \begin{pmatrix}i & 0 \\ 0 & -i \end{pmatrix} $$ Thus, it is clear for any element $N \in \operatorname{SL}(2,\mathbb{R})$, $M = U N U^\dagger$ preserves the form: $$ M^\dagger h^\prime M = h^\prime $$ clearly this is true for any scalar multiple of $h^\prime$, namely if we multiply by $-i$ on both sides we find the defining relation for elements of $\operatorname{SU}(1,1)$: $$ M^\dagger \mathbb{I}_{1,1} M = \mathbb{I}_{1,1} $$ Therefore we have shown that the elements of $\operatorname{SL}(2,\mathbb{R})$ form a subgroup of $\operatorname{SU}(1,1)$, and either going backwards with the same argument or dimensional counting, it can be seen the subgroup must be the entire group (not a proper subgroup).

So regarding your questions: 1. There is no need for dotted spinors, as there are not two distinct spinor representations at play. The $\sigma$ conjugate of a spinor tells you how to map would be dotted spinors to undotted spinors. Or, you can always transform into the $\operatorname{SL}(2,\mathbb{R})$ basis where all spinor Lorentz transforms are manifestly real.
2. The map at hand sends undotted spinors to undotted spinors, as that's all you have to start with. 3. I think there are some confusions with indicies and conclusions thereof, but what is true for any group which respects a bi-linear form, can be seen by solving the first equation I have for $N^{-1}$: $$ \varepsilon^{-1} (N^\intercal) \varepsilon = N^{-1} $$ and while this might look the same as say, vector Lorentz transforms with $\eta \leftrightarrow \varepsilon$, it is a more general statement which is true for elements of either orthogonal or symplectic groups.