# Why there are so many spinor components in higher dimensions if the number of degrees of freedom is only 2?

In the book of Freedman & van Proeyen on Supergravity a table (3.2) can be found which shows for dimensions from 2-11 the number of components of Majorana spinors.

For instance in 4 dimensions we have 4 components for a spin-1/2 Majorana spinor (represented by a bispinor), whereas in 8 dimensions there are 16 components.

Actually, in 4 dimensions all components have a physical meaning, 2 correspond to the spin 1/2 and another 2 to the particle anti-particle symmetry. The latter two get kind of superfluous once the reality condition for a Majorana spinor is imposed.

However, in 8 dimensions there are 16 components whereas only 2 degrees of freedom for the spin 1/2 are needed. What is the physical meaning of the other components ? Are they also "superfluous" due to some conditions as the reality condition ? Or do they get "lost" in the higher-dimensional space ?

EDIT:

In order to make my point clearer let's assume a Majorana spinor in Weyl-respresentation

$$\Psi_M = \left( \begin{array}{c} \xi \\ \xi^\dagger \end{array}\right)$$

(Other representations can be found by multiplying $$\Psi_M$$ with an appropiate matrix, for instance one which makes all components real.)

where $$\xi$$ (if considered in the rest frame of the concerned particle and the axes (s,t,u,v,w,x,y,z) of the coordinate system are appropiately chosen) is an eigenmode of the z-component of the spin (or helicity) operator -- apparently there is no further condition:

$$S_z \left( \begin{array}{c} \xi_1 \\ \xi_2 \\ \xi_3 \\ \xi_4 \\ \xi_5 \\ \xi_6\\ \xi_7 \\ \xi_8 \end{array}\right) = \pm \frac{1}{2} \left( \begin{array}{c} \xi_1 \\ \xi_2 \\ \xi_3 \\ \xi_4 \\ \xi_5 \\ \xi_6\\ \xi_7 \\ \xi_8 \end{array}\right)$$

Why in 8 dimensions do I need an 8-component long spinor for fulfilling such a simple eigenmode equation whereas in 4D spacetime I only need a 2-component long spinor ? May be there is a misconception in my thinking, so I would appreciate if such a misconception could be cleared up.

• Doesn't a spin state have 4 extra degrees of freedom in 8 dimensions? Feb 8, 2022 at 10:44
• @Felicia The question is not about the spin states. Moreover, the number of components of a (Majorana)-spinor in $n$-dimensional space is not equal to $n$. Feb 8, 2022 at 12:26
• @FredericThomas I don't mean the number of components (which is 8), but the number of degrees of extra freedom. So in 4D space up or down (thanks to 3 space and 1 time dimension), and for every extra space dimension an up and down degree extra. Feb 8, 2022 at 12:37
• related/possible dup? physics.stackexchange.com/q/381752/84967 Feb 8, 2022 at 17:44

There is no such a thing as "the z-component of spin $$S_z$$" in higher dimensions. There is no z-axis for general $$d$$, rather there are multiple "$$z$$-axes" all of which we need to diagonalize simultaneously, so to speak.

Of course I cannot give you a crash course on representation theory of simple groups here (the book on CFT by di Francesco et all does a good job, it has a physics-oriented chapter on Lie groups, see chapter 13). But I can give you a flavor of some of the ingredients that are important for your question.

Consider the Lie algebra $$g=\mathfrak{so}(d)$$. It has some commutation relations, which are probably familiar to you. There is a nice basis of the algebra, known as the Cartan-Weyl basis, which takes the form $$[H^i,H^j]=0,\quad [H^i,E^\alpha]=\alpha^i E^\alpha,\qquad [E^\alpha,E^\beta]\propto E^{\alpha+\beta}$$ Here, the set $$H^i$$ (where $$i=1,2,\dots,\text{rank}(g)$$) is a set of commuting operators, which replace $$S^z$$ for general $$d$$. Note that the rank of $$SO(3)$$ is 1, hence a single $$S^z$$, but the rank of $$SO(d)$$ is $$\lfloor d/2\rfloor$$ for higher $$d$$, so you need multiple $$S^z$$ that you have to diagonalize.

It just so happens that the spinor representation (i.e., the representation that you get when you simultaneously diagonalize all the $$H^i$$), has dimension $$\sim 2^d$$. Roughly speaking, this representation has $$H^i=0$$ for all $$i<\text{rank}(g)$$, and $$H^i=1/2$$ for $$i=\text{rank}(g)$$. So this plays the role of a higher dimensional fermion, this is the smallest representation that has half-integral eigenvalues. It is a fact of life that this representation requires $$\sim 2^d$$ components, no smaller vector spaces is able to diagonalize all the require operators.

(Note: of course, sometimes you can half the size of the representation by imposing suitable reducibility conditions, such as Majorana. This depends on $$d\mod8$$, and the reference in the OP does a good job at explaining when and why can you impose such conditions.)

• That looks interesting. I will in any case go through it. Feb 9, 2022 at 18:01

For a spin $$j=1/2$$ spinor field, the number of DOF is not just 2 as OP writes in the title. Instead

$$\text{Off-shell DOF = #(components)},$$

while

$$\text{On-shell DOF = #(helicity states) = #(components)/2}.$$

A Dirac spinor has $$2^{[D/2]}$$ complex components in $$D$$ spacetime dimensions, while a Majorana spinor has $$2^{[D/2]}$$ real components, cf. e.g. my related Phys.SE answer here.