Number of Components of a Spinor

I'm trying to develop my understanding of spinors. In quantum field theory I've learned that a spinor is a 4 component complex vector field on Minkowski space which transforms under the chiral representation of the Lorentz group.

Now I've been reading that we can derive spinor representations by looking at the universal covering group of the proper orthochronous Lorentz group, which is $SL(2,\mathbb{C})$. Now $SL(2,\mathbb{C})$ acts on $\mathbb{C}^2$ by the fundamental representation. My book (Costa and Fogli) then calls elements of $\mathbb{C}^2$ spinors.

But the second type of spinors have a different number of components to the first! What is going on here? Could someone clearly explain the link between these two concepts in a mathematically rigorous way? I come from a maths background of group theory and topology, but don't know much representation theory at present.

Many thanks in advance!

• One way to think about spinors is that they are fundamentally used to perform rotations. As such, spinors break down into two parts: a scalar part and a bivector part (bivector meaning one component for each linearly independent 2d subspace). So it helps to look at other common vector spaces to see what their spinors are. On a 2d vector space, there is 1 scalar and 1 bivector. These spinors have 2 components and are isomorphic to complex numbers. On a 3d real vector space, there is 1 unit scalar and 3 unit bivectors. These 4 components make these spinors isomorphic to quaternions. – Muphrid Dec 21 '12 at 15:54
• @Muphrid Could you point me in the direction of a text which formalises your notion above? – Edward Hughes Dec 21 '12 at 15:58
• I don't think I have the mathy answer you want, which is why this is a comment. A so-called "spinor" has two components, $\psi_{A}$, $A=1,2$. There are two types of spinors which transform differently, the other is denoted by dotted indices, $\psi_{\dot{B}}$. A so-called "bi-spinor" is a spinor with two indices, one dotted, one not: $\Psi_{A\dot{B}}$. This object must have four components since each index runs from 1 to 2. A "dirac spinor" is a bi-spinor. – kηives Dec 21 '12 at 23:57
• @knives: Eh? Isn't a Dirac spinor a pair $(\Psi_A,\Phi_{\dot{B}})$, living in the direct sum representation $2 \oplus \bar{2}$? The object with two indices lives in the product product representation $2 \otimes \bar{2}$, which is the vector representation of the Lorentz group. – user1504 Dec 22 '12 at 0:00
• @user1504 What I've done in QFT would agree with knive's comment. Could you expand on your objection? What exactly is the representation $2$ and what's the difference between the sum and product reps? – Edward Hughes Dec 22 '12 at 0:17

There are a number of mathematical imprecisions in your question and your answer. Some advice: you will be less confused if you take more care to avoid sloppy language.

First, the term spinor either refers to the fundamental representation of $SU(2)$ or one of the several spinor representations of the Lorentz group. This is an abuse of language, but not a bad one.

A particularly fussy point: What you've described in your first paragraph is a spinor field, i.e., a function on Minkowski space which takes values in the vector space of spinors.

Now to your main question, with maximal pedantry: Let $L$ denote the connected component of the identity of the Lorentz group $SO(3,1)$, aka the proper orthochronous subgroup. Projective representations of $L$ are representations of its universal cover, the spin group $Spin(3,1)$. This group has two different irreducible representations on complex vector spaces of dimension 2, conventionally known as the left- and right- handed Weyl representations. This is best understood as a consequence of some general representation theory machinery.

The finite-dimensional irreps of $Spin(3,1)$ on complex vector spaces are in one-to-one correspondence with the f.d. complex irreps of the complexification $\mathfrak{l}_{\mathbb{C}} = \mathfrak{spin}(3,1) \otimes \mathbb{C}$ of the Lie algebra $\mathfrak{spin}(3,1)$ of $Spin(3,1)$. This Lie algebra $\mathfrak{l}_{\mathbb{C}}$ is isomorphic to the complexification $\mathfrak{k} \otimes \mathbb{C}$ of the Lie algebra $\mathfrak{k} = \mathfrak{su}(2) \oplus \mathfrak{su}(2)$. Here $\mathfrak{su}(2)$ is the Lie algebra of the real group $SU(2)$; it's a real vector space with a bracket.

I'm being a bit fussy about the fact that $\mathfrak{su}(2)$ is a real vector space, because I want to make the following point: If someone gives you generators $J_i$ ($i=1,2,3$) for a representation of $\mathfrak{su}(2)$, you can construct a representation of the compact group $SU(2)$ by taking real linear combinations and exponentiating. But if they give you two sets of generators $A_i$ and $B_i$, then you by taking certain linear combinations with complex coefficients and exponentiating, you get a representation of $Spin(3,1)$, aka, a projective representation of $L$. If memory serves, the 6 generators are $A_i + B_i$ (rotations) and $-i(A_i - B_i)$ (boosts). See Weinberg Volume I, Ch 5.6 for details.

The upshot of all this is that complex projective irreps of $L$ are labelled by pairs of half-integers $(a,b) \in \frac{1}{2}\mathbb{Z} \times \frac{1}{2}\mathbb{Z}$. The compex dimension of the representation labelled by $a$,$b$ is $(2a + 1)(2b+1)$.

The left-handed Weyl-representation is $(1/2,0)$. The right-handed Weyl representation is $(0,1/2)$. The Dirac representation is $(1/2,0)\oplus(0,1/2)$. The defining vector representation of $L$ is $(1/2,1/2)$.

The Dirac representation is on a complex vector space, but it has a subrepresentation which is real, the Majorana representation. The Majorana representation is a real irrep, but in 4d it's not a subrepresentation of either of the Weyl representations.

This whole story generalizes beautifully to higher and lower dimensions. See Appendix B of Vol 2 of Polchinski.

Figuring out how to extend these representations to full Lorentz group (by adding parity and time reversal) is left as an exercise for the reader. One caution however: parity reversal will interchange the Weyl representations.

Sorry for the long rant, but it raises my hackles when people use notation that implies that some vector spaces are spheres. (If it's any consolation, I know mathematicians who get very excited about the difference between a representation $\rho : G \to Aut(V)$ and the "module" $V$ on which the group acts.)

• Thank you so much - that's exactly the sort of detail I was looking for, yet couldn't find in any of my course books! – Edward Hughes Dec 22 '12 at 13:53

Okay - I've had a think and here's an attempt at an answer. Do people agree that my reasoning is correct?

I need to consider the whole Lorentz group $L$, which has universal cover $SU(2)\times SU(2)$. Then one spinor representation of $L$ is the fundamental representation of $SU(2) \times SU(2)$. The space $\mathbb{C}^2 \times \mathbb{C}^2$ which this representation acts on is the spinor space we talk about in QFT.

• In fact if this is right - where do left and right handed spinors come in? – Edward Hughes Dec 21 '12 at 15:46
• Are the left handed spinors just the first two components of the full spinor, and right handed spinors the second two components? – Edward Hughes Dec 21 '12 at 16:18
• Yes, but since we distinguish the two copies of $SU(2)$ (acting on the top and bottom two components), the LH and RH spinors transform under different representations. – Vibert Dec 22 '12 at 10:13