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Suppose you know nothing about CFT, and suppose you have found in (closed bosonic) String Theory that \begin{equation} [L_n , L_m ]=(n-m) L_{n+m} \;\;\;\;(\mathrm{"right"\;Witt\;Algebra}\; \mathfrak{w}_{R}) \\ \mathrm{same\;for}\;\bar{L}_n\\ [L_{n},\bar{L}_{m}]=0 \end{equation} Thus the complete Witt Algebra is \begin{equation} \mathfrak{W}=\mathfrak{w}_{R}\oplus \mathfrak{w}_{L} \end{equation} You can readily see that $L_{\{-1,0,1\}}$ is a closed subalgebra defined by \begin{equation} [L_1 , L_{-1} ]=2 L_{0}\\ [L_0 , L_1 ]=- L_{1}\\ [L_0 , L_{-1} ]= L_{-1}\\ \mathrm{same\;for}\;\bar{L}_{\{-1,0,1\}} \end{equation} Now, we know that $SL(n,\mathbb{R}/\mathbb{C})=\{M_{n\times n}(\mathbb{R}/\mathbb{C})\;\mathrm{such\;that}\; \det M=1\}$ and that its Lie Algeabra is $\mathfrak{sl}(n,\mathbb{R}/\mathbb{C})=\{X_{n\times n}\;\mathrm{such\;that}\;$Tr(X)$=0\}$. In the case $n=2$, we can write simply a general matrix in $\mathfrak{sl}(n,\mathbb{R}/\mathbb{C})$ and (after a bit of work to write it in a smart way) a basis as \begin{equation} A=\begin{pmatrix} -a & b \\ -c & a \end{pmatrix}\;\;\;\;\mathfrak{sl}(n,\mathbb{R}/\mathbb{C})=\bigg\{ e_{-1}=\begin{pmatrix} 0 & 0 \\ -1 & 0 \end{pmatrix},\; e_{0}=\frac{1}{2}\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix},\; e_{1}=\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}\bigg\} \end{equation} You can see that, for $i,j=\{-1,0,1\}$, \begin{equation} [e_{i},e_{j}]=(i-j)e_{i+j} \end{equation}

Question 1 How can I understand if this last commutator provides an isomorphism between $\mathfrak{w}$-$\mathfrak{sl}(2,\mathbb{R})$ OR $\mathfrak{w}$-$\mathfrak{sl}(2,\mathbb{C})$? Polchinski (pag.56 of the first volume) seems to say $\mathfrak{sl}(2,\mathbb{R})$ (even if it is not true for me that $\mathfrak{sl}(2,\mathbb{R})$ differs from $\mathfrak{su}(2)$ only for signs.)

Question 2 I have read everywhere that the full $\mathfrak{W}$ is isomorphic ($\sim$) to $\mathfrak{sl}(2,\mathbb{C})$. How can this possible? If the answer to quest.1 is "$\mathfrak{sl}(2,\mathbb{R})$", then I would find $\mathfrak{W}\sim \mathfrak{sl}(2,\mathbb{R}) \oplus \mathfrak{sl}(2,\mathbb{R})$. This, however, is not $\mathfrak{sl}(2,\mathbb{C})$, because \begin{equation} \mathfrak{sl}(2,\mathbb{C})\sim\mathfrak{sl}(2,\mathbb{R})_{\mathbb{C}}=\mathfrak{sl}(2,\mathbb{R})\oplus i \mathfrak{sl}(2,\mathbb{R}) \end{equation} If the answer is instead "$\mathfrak{sl}(2,\mathbb{C})$", the full $\mathfrak{W}$ is isomorphic to two copies of $\mathfrak{sl}(2,\mathbb{C})$.

Question 3 (remember you know nothing about CFT, where I know how to answer to the following question). To be precise, the Witt "group" $W$ (restricted to $\{-1,0,1\}$) is \begin{equation} W|_{\{-1,0,1\}}\sim\frac{SL(2,\mathbb{C})}{\mathbb{Z}_{2}}\equiv PSL(2,\mathbb{C})\sim SO(1,3) \end{equation} How to see this in this context? Where does it enter $\mathbb{Z}_{2}$?

Question 4 Now you know CFT. In (2d) CFT, where $L_n \sim \partial_{z}$ I would say that with $L_{-1},\;L_0,\;L_1$ I can construct all the transformation \begin{equation} z\longrightarrow z'=\frac{az-b}{cz+d},\;\;\;\; ad-bc=1,\;\;\; (a,b,c,d)\;in\;\mathbb{C} \end{equation} This is exactly $PSL(2,\mathbb{C})$. However we have also the transformation of $\bar z$, given by $\bar L$. So it seems that the 2-dimensional conformal "group" is $PSL(2,\mathbb{C})_{L}\times PSL(2,\mathbb{C})_{R}$. However, at the very end of the day, $\bar z =z*$, so the second part of the product $\times$ is "false", and the group is only $PSL(2,\mathbb{C})$, with only one factor (as is confirmed in "Blumenhagen, Lust, Theisen: Basic Concepts of String Theory, pag.72"). How to deal with this if the answer in (2) is $\mathfrak{sl}(2,\mathbb{C})$? How to treat all this stuff?

Sorry for this tedious question; I would really appreciate a pedantic answer, or corrections to my hypothesis.

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    $\begingroup$ The right Witt algebra is the set of real linear combinations of $L_{0,\pm 1}$ so your explicit representation proves that it's isomorphic to $\mathfrak{sl}(2, \mathbb{R})$. The exponential of a general element, which acts as $z^\prime = \frac{az + b}{cz + d}$, is the same transformation when $a,b,c,d$ go to minus themselves so that's where the $\mathbb{Z}_2$ comes in. $\endgroup$ Sep 8 at 17:07
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    $\begingroup$ Related: Global conformal group in 2D Euclidean space. $\endgroup$
    – Qmechanic
    Sep 8 at 17:46
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    $\begingroup$ Might be interested in this. $\endgroup$ Sep 8 at 20:41
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As a mathematically careful and reasonably rigorous discussion of the mathematical structures of conformal field theory, I highly recommend Schottenloher's "A Mathematical Introduction to Conformal Field Theory". In particular, its chapters 2 and 5 deal almost exclusively with the correct notions of "conformal group" and "conformal algebra" in various dimensions and signatures, and the relation between them.

In addition to these points, we also need to pay attention to the distinction between a real and a complex Lie algebra. There are no complex numbers involved in the commutation relations of the abstract Witt algebra, so we are free to view it as a real or complex algebra as we choose.

  1. The subalgebra of $L_{-1},L_0,L_1$ is isomorphic to $\mathfrak{sl}(2,\mathbb{R})$ as a real Lie algebra, and $\mathfrak{sl}(2,\mathbb{C})$ as a complex Lie algebra.

  2. Hence, of course, the subalgebra of $L_{-1},L_0,L_1,\bar{L}_{-1},\bar{L}_0,\bar{L}_1$ is isomorphic to $\mathfrak{sl}(2,\mathbb{R})\oplus\mathfrak{sl}(2,\mathbb{R})$ as a real algebra and the same with $\mathbb{C}$ as a complex algebra. You are correct in pointing out that $\mathfrak{sl}(2,\mathbb{R})\oplus\mathfrak{sl}(2,\mathbb{R})$ and $\mathfrak{sl}(2,\mathbb{C})$ are not isomorphic as real Lie algebras (if they were, then by the same token they would also be isomorphic to $\mathfrak{su}(2)\oplus\mathfrak{su}(2)$, since the complexification of both $\mathfrak{su}(2)$ and $\mathfrak{sl}(2,\mathbb{R})$ yields $\mathfrak{sl}(2,\mathbb{C})$). See point 4 below why both $\mathfrak{sl}(2,\mathbb{C})$ and $\mathfrak{sl}(2,\mathbb{R})\oplus\mathfrak{sl}(2,\mathbb{R})$ might appear in the context of discussing the infinitesimal versions of global conformal transformations, depending on the signature we're looking at.

  3. You cannot "see" discrete quotients from the algebra alone. Without knowing "CFT", i.e. what symmetry group we are actually looking for, asking what the group corresponding to the subalgebra of $L_{-1},L_0,L_1$ is an unanswerable question: It is the Lie algebra of both $\mathrm{SL}(2,\mathbb{C})$ and $\mathrm{PSL}(2,\mathbb{C})$! This is just a specific manifestation of the general principle that Lie groups that are coverings of each other have the same Lie algebra.

  4. When we look at the conformal group in two dimensions, we must specify whether we are looking at the Euclidean or Minkowskian metric.

    In the Euclidean case $\mathbb{R}^{2,0}$, the global conformal group is the complex Lie group of Möbius transformations $\mathrm{PSL}(2,\mathbb{C})$, given by holomorphic functions. Their infinitesimal version (i.e. algebra), are the $L_{-1},L_0,L_1$ as a complex algebra. The $\bar{L}_i$ generate anti-holomorphic functions and do not contribute to the Euclidean conformal group.

    In the Minkowskian case $\mathbb{R}^{1,1}$, the global conformal group is an infinite-dimensional real Lie group $\mathrm{Diff}_+(S^{1,1})\times \mathrm{Diff}_+(S^{1,1})$ where $S^{1,1}$ is the conformal compactification of the Minkowski plane and $\mathrm{Diff}_+$ denotes the set of orientation-preserving diffeomorphisms on it. There is no natural complex structure on this group, and so the restriction to the "-1,0-1" degrees in this case yields the real Lie group $\mathrm{PSL}(2,\mathbb{R})\times\mathrm{PSL}(2,\mathbb{R})$, indeed generated by the $L_1,L_0,L_{-1}$ and $\bar{L}_{-1},\bar{L}_0,\bar{L}_1$ this time.

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  • $\begingroup$ I completely agree with your answers 1,3,4; that has been very helpful for me, so thank you a lot. I edit this comment before you edit the answer, so you don't have to spend too much time replying again. I have still 2 problems: N1 Why authors choose to use a real algebra for $L_{-1,0,1}$, saying that it is isomorphic to $\mathfrak{sl}(2,\mathbb{R})$? Are there links with the fact that $SL(2,\mathbb{C})$ is simply connected and $SL(2,\mathbb{R})$ is not? $\endgroup$ Sep 9 at 20:49
  • $\begingroup$ N2, part one <br/> Complexification. Indeed it is known that $\mathfrak{su}(2)\oplus\mathfrak{su}(2)\sim\mathfrak{so}(4)$ and also that $\mathfrak{su}(2)_{\mathbb{C}}\sim\mathfrak{sl}(2,\mathbb{R})_{\mathbb{C}}\sim\mathfrak{sl}(2,\mathbb{C})$. I understand that we can simbolically write the complexification of a real vector space $V_{\mathbb{C}}$ as a direct sum $V\oplus V$ (also without the $i$), where it is intended that we are using a definition for the multiplication for $i$. But in our case? $\endgroup$ Sep 9 at 20:52
  • $\begingroup$ N2, part two When we have the two copy of $\mathfrak{sl}(2,\mathbb{R})$ for $L_{n}$ and $\bar{L}_{n}$, we are directly summing them or complexifying them? and why? I would answer "summing" (since $[L,\bar L]=0$), but in that case I would not obtain $\mathfrak{sl}(2,\mathbb{C})$ (as in the case $\mathfrak{su}(2)\oplus\mathfrak{su}(2)\sim\mathfrak{so}(4)$) as you have said and as is right . So, what am I missing? Thank you again a lot for your time. $\endgroup$ Sep 9 at 20:53
  • $\begingroup$ @AlessioFontanarossa I'm not sure why you think you're missing anything - why some authors might choose to view the algebras as real or complex depends on the context. E.g. you have to complexify if you're thinking about the quantum theory, since quantum theory is about complex representations, but classically there's no need to. It would probably help if you cited one of the claims from these authors explicitly and explain what your issue with it is. $\endgroup$
    – ACuriousMind
    Sep 9 at 21:12

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