# Are spinors representations of the Lorentz group or its associated algebra?

I got confused about the true origin of spinors with regards to their connection to the proper orthochronous Lorentz group $$SO^+(3,1)$$. In the book on QFT by Maggiore that I'm following, we reached the Lie algebra corresponding to $$SO^+(3,1)$$ and then showed how it's isomorphic to $$su(2)\oplus su(2)$$*. Then it was simply stated that representations of $$su(2) \oplus su(2)$$ are representations of the Lorentz group (which I take to mean representations that can be exponentiated to get representations of the Lorentz group).

My questions are the following:

1. Are spinors representations of the Lorentz group (which would mean the Lorentz algebra) or the Lorentz algebra (since I've read that not all representations of an algebra correspond to a particular group since many groups can have the same algebra) or $$su(2) \oplus su(2)$$?

2. If they are supposed to be representations of the Lorentz group, why is Maggiore working on a Lie algebra level to deduce representations of the group?

3. In general, fields in classical or quantum field theory are a representation of the Lie group or interest, the Lie algebra of interest (that might be related to the group) or the Lie algebra and any Lie algebra isomorphic to that?

*In page 162 of Schwartz's Quantum Field Theory and the Standard model, the author writes $$so(3,1)=su(2) \oplus su(2)$$

• Spinors in 1+3 Minkowski spacetime (the one of special relativity and its Lorentz group) are finite-dimensional vector representations of $\text{SL}(2,\mathbb C)$. – DanielC Nov 28 '20 at 19:44

Spinors are projective representations of the $$SO(m,n)$$, i.e. they provide representations up to some phases: $$U(g)U(h)=e^{i\theta_{g,h}}U(gh)$$ These phases turns out to represent the group of closed loops in the $$SO(m,n)$$, also known as the fundamental group of $$SO(m,n)$$. Loops in the $$SO(m,n)$$ manifold from $$0$$ to $$2\pi$$ are not continuously connected to arbitrarily small loops, they are stretched by topological reasons. The phases $$\theta_{g,h}$$ capture those loops. It is $$n\pi$$ for loops obtained by continuous rotations from $$0$$ to $$2\pi n$$. You can see more about that here

The spinors will be true representations -not projetive- of the double cover of $$SO(m,n)$$ also known as $$\text{spin}(m,n)$$. It turns out that the double cover is also the universal cover if $$m+n>2$$.

We have some "accidental" isomorphisms where for small enough $$m$$ and $$n$$ the spin group is isomorphic to some classical groups such as

$$\text{Spin}(4)\cong SU(2)\times SU(2),\quad \text{Spin}(3,1)\cong SL(2,\mathbb{C}),$$ $$\text{Spin}(2,2)\cong SL(2,\mathbb{R})\times SL(2,\mathbb{R})$$

Particle physicists usually make the mistake of confusing $$SU(2)\times SU(2)$$ with $$SL(2,\mathbb{C})$$. If the reality conditions for the spinors are not important than this mistake does not harm you. However, if you want to do things a little more sophisticated like spinor helicity variables and twistors this mistake will harm you really bad.

Why projective representations are relevant for physics?

It turns out that in Quantum Mechanics the group symmetry is represented projectivelly by the quantum system Hilbert space. This is so because two states that differ by a phase should describe the same physics, so we must consider class of equivalences

$$|\psi\rangle\cong e^{i\theta}|\psi\rangle$$

and this will reflect in a weaker notion of group multiplication

$$U(g)|\psi\rangle = e^{i\theta_{g}}|g\psi\rangle,\quad U(h)|\psi\rangle = e^{i\theta_{h}}|h\psi\rangle \quad U(g)U(h)|\psi\rangle = e^{i\theta_{gh}}|gh\psi\rangle$$

sometimes we can absorb all those phases by redefinitions of the $$U$$'s but sometimes we cannot. There might be algebraic and/or topological obstructions to do so. The presence of central charges are algebraic obstructions while non-trivial fundamental group are topological obstructions.

Projective representations are algebraically hard to manipulate but we can use a trick. It turns out that it is always possible to modify the symmetry group $$G$$ to $$G'$$ in order to have:

All the projective representations of $$G$$ are non-projective representations of $$G'$$

For situations where we have central charges the modification is to promote the central charge to be a new generator and consider your theory in a superselection sector for this charge, i.e. all physical operators should commute with this new generator and all states should be an particular eigenstate of this new generator.

For situations like the case for spin, where there is a topological obstruction, the modification is given by promoting the original group $$G$$ to its universal cover $$G'$$, which is basically accepting all the representations of the Lie algebra and exponentiate it, which gives all the representations of $$G'$$.

You can see more about it in chapter 2 and 5 of Quantum Theory of Fields: Foundations by S. Weinberg.

• Thank you for the answer. So, when books talk about finding spinor representations of the Lorentz group, they actually mean that they're seeking representations of its universal cover? If so, since the true symmetry transformations come from the Lorentz group itself (and not its universal cover), why would representations of its universal cover which are not representations of the Lorentz group (only projective) be physically relevant? – TheQuantumMan Nov 12 '20 at 2:46
• @TheQuantumMan because Wigner theorem tell us that in a quantum system the symmetries are represented projectively. This is so because in quantum mechanics a quantum state is only defined up to an overal phase. – Nogueira Nov 12 '20 at 2:49
• This is well explained in Weinberg QFT volume 1 textbook chapter 2. – Nogueira Nov 12 '20 at 2:50
• I see. It seems that a lot of QFT textbooks are missing these details. Will check out Weinberg. Thank you! – TheQuantumMan Nov 12 '20 at 2:51
• Thank you, the second answer there helps a lot! – TheQuantumMan Nov 12 '20 at 3:09