Is the conformal group in $2+0$D isomorphic to the Lorentz group $SO(3,1)$ in $3+1$D? In the Wikipedia article on conformal field theory, it is said the following:

The conformal group is locally isomorphic to $SO(1,d+1)$ (Euclidean)
or $SO(2,d)$ (Minkowski).

Does that mean that the conformal group in Euclidean $d=2$ is isomorphic to $SO(3,1)$, the group of proper Lorentz transformations in $3+1$D?
 A: To see why it's true, consider the set of all lightlike rays through the origin in Minkowski spacetime. This set has the topology of a two-dimensional sphere $S^2$, with each point on the sphere corresponding to one lightlike ray through the origin in Minkowski spacetime. What do Lorentz transformations in 4d Minkowski spacetime do to the points on that sphere?
You can find the answer to that question in several places in the literature. The keywords "celestial sphere" might help in a keyword search. One of the most entertaining reviews is arXiv:1508.00920, which uses the insight to explain what Han Solo and Chewbacca should have seen in the "jump to
lightspeed" scenes in Star Wars... I mean, right before they violate the laws of physics by exceeding the speed of light.
How is the effect of these transformations of $S^2$ related to conformal transformations of the Euclidean plane? The usual answer is that the sphere $S^2$ can be regarded as the Euclidean plane with the point at infinity added. That's correct, but I'll be a little more explicit. Consider any 3d hyperplane $P$ in Minkowski spacetime that contains a lightlike ray but that does not pass through the origin, and let $E$ be the intersection of $P$ with the set of light rays through the origin. I'm using the letter $E$ for this 2d intersection, because it will correspond to the Euclidean plane. The hyperplane $P$ intersects all lightlike rays through the origin except one, which I'll call $L$, so $E$ represents all but this one point of the sphere $S^2$. (For some intuition, you can draw this setup in 3d Minkowski spacetime, but then $P$ is 2d and $E$ is only 1d.) Given this correspondence between lightlike rays through the origin and points of $E$, we can define the usual Euclidean metric on $E$ in such a way that conformal transformations of $E$ are Lorentz transformations in the ambient 4d Minkowski spacetime. Some of those transformations mix $L$ with other lightlike rays through the origin, so they mix points of $E$ with a point that is not in $E$. That missing point $L$ is the "point at infinity" that we need to add to $E$ in order for conformal transformations to be properly defined.
By the way, when $d=2$, people sometimes say that the group of conformal transformations is infinite-dimensional, but that's an abuse of the word group. The space of conformal transformations is infinite-dimensional when $d=2$, but most of those transformations are not invertible everywhere, so they don't form a group in the long-established sense of the word. Those that are invertible form a finite-dimensional group, just like they do when $d>2$.
A: One important point to add to the obvious answer "yes, that is right" is that you should be careful about your signature.
Specifically, Euclidean conformal group $SO(1,d+1)$ describes symmetries of Euclidean correlation functions and has nothing to do with unitary operators on the Hilbert space of the CFT. On the other hand, the Lorentz group $SO(1,d-1)$ is a group of actual symmetries of the theory, not just of Euclidean correlators, and is represented by unitary operators on the Hilbert space of the QFT.
The thing that has to do with unitary operators on Hilbert space of a CFT is the (universal cover of) the Lorentzian conformal group $SO(2,d)$. You can see that it is not the same as the Lorentz group $SO(1,d-1)$ that appears in the QFT Hilbert space, so there is really not much surprise here that the physics is different in the two cases.
P.S. Similarly you can compare Euclidean conformal group $SO(1,d+1)$ and the "Euclidean Lorentz group" $SO(d)$ -- again different.
