# In a universe with four spatial dimensions would there be elementary particles with intrinsic isoclinic spin?

Elementary particles have an intrinsic property called spin which is different from classical spin as it does not involve actual rotation and the magnitude of spin cannot be changed but particles with intrinsic spin behave in some ways as if they were rotating such as behaving like subatomic magnets if electrically charged.

In four spatial an object with classical rotation can have isoclinic rotation in which it has two independent directions of rotation and in both directions the rate of rotation is the same. An object could also have a double rotation that is not isoclinic but isoclinic rotation is the most stable type of rotation and none isoclinic double rotation will tend to decay into isoclinic rotation through momentum transfer.

In a universe with four spatial dimensions would there be elementary particles that behave as if they have isoclinic rotation? If so would there still be elementary particles that behave as if they have simple rotation? How would a particle with intrinsic isoclinic spin behave? Would there be particles that have two none zero spin numbers that are both different values? If so how would a particle with one spin value being half integer and the other being integer behave?

## A sketch of how spin arises in particle physics.

There is a theorem in quantum mechanics, called the Coleman-Mandula theorem, that tells you that under very reasonable assumptions, the most general group of symmetries of a quantum theory is the direct product of the Poincaré group and a compact connected Lie group (called the group of internal symmetries).

As is usually the case, we can organise the spectrum of the theory in terms of irreducible representations of the symmetry group. Being a direct product, we can discuss Poincaré and internal symmetries separately. The latter gives rise to "charge" quantum numbers, such as isospin, colour, etc., which are the eigenvalues of a maximal torus of the internal group.

The former is the most interesting part. The Poincaré group is a semi-direct product of the Lorentz group and the group of translations (see this PSE post for some more details). A complete classification of its (projective, unitary) representations can be obtained via the Frobenius-Wigner method of induced representations. This method proceeds as follows:

1. We first diagonalise the normal sub-group; being abelian, we just introduce $d$ arbitrary real parameters, leading to what we usually call the momentum $p=(p_0,p_1,\dots,p_{d-1})\in\mathbb R^{1,d-1}$.

2. We next break up $\mathbb R^{1,d-1}$ into manifolds where $\mathrm{SO}(1,d-1)$ acts transitively. That is we, identify all inequivalent orbits of momenta under the Lorentz group: these are vacuum states $p\equiv 0$; massive states $p^2>0$; massless states $p^2\equiv0$; and tachyonic states $p^2<0$.

3. We pick one representative for each class. From now on, we focus on massive states only. A representative of these states is $p=(\sqrt{p^2},0,\dots,0)$. The (Wigner's) little group of such a representative is defined as the sub-group of the Lorentz group that leaves it invariant: $W\equiv\{R\in\mathrm{SO}(1,d-1)\,\mid\, Rp=p\}$, which is easily seen to be the group of rotations, $W\cong \mathrm{SO}(d-1)$.

4. Pick an arbitrary (unitary, projective) representation of the little group, $\lambda\in\mathrm{Rep}(W)$. Here we are lucky that the orthogonal group is simple; otherwise we must go back to step 1, and induce a representation of $W$ from its normal sub-group. (This is precisely what happens for massless orbits1).

5. The representation of Poincaré is finally given by the pair $(p,\lambda)$. Here, $p=(p_0,p_1,\dots,p_{d-1})$ is an arbitrary $d$-tuple of real numbers, and $\lambda$ is a finite-dimensional unitary projective representation of the little group of $p$, to wit, the orthogonal group $\mathrm{SO}(d-1)$.

In $d=3+1$, the little group is $\mathrm{SO}(3)$; its projective representations are the standard representations of its universal cover, $\mathrm{SU}(2)$. The representations of the latter are well-known in physics: they are labelled by a half-integer $j$, called spin. Therefore, the states of a relativistic quantum theory in $d=3+1$ dimensions are labelled by the following numbers: four-momenta, spin, internal charges. This nicely ties up with our intuition/experience.

In $d\ge4+1$, the little group is $\mathrm{SO}(d-1)$; its projective representations are the standard representations2 of its universal cover, $\mathrm{Spin}(d-1)$. The representations of the latter are not as common as those of $\mathrm{SU}(2)$ in physics. We claim without proof that the representations of this group are in one-to-one correspondence with the so-called highest weights of the algebra (cf. highest weight representations). These can be labelled by $r=\mathrm{rank}(\mathfrak{so}(d-1))=\lfloor(d-1)/2\rfloor$ integers $\lambda_1,\lambda_2,\dots,\lambda_r$, known as the Dynkin labels of the representation (which are defined as the coefficient of the highest weight in the basis of fundamental weights, these being the basis dual to that of simple roots). For $d=3+1$, we have a single label, that we identify with the spin, $\lambda_1=2j$. For $d\ge4+1$, we have several labels, so it doesn't make sense to speak of the spin of a particle (rather, we would have to speak of its spin quantum numbers; but this would not be very accurate, because the $\lambda_i$ are not eigenvalues of a Casimir, unlike the $d=3+1$ case).

For example, in $d=4+1$, we have two "little group" quantum numbers, $\lambda_1,\lambda_2$. In semi-classical terms, they describe the possible states of rotation in $d=4$ spatial dimensions, as in the OP. In quantum terms, it is not useful to regard this as a bona fide rotation, but the labels still describe how the particle behaves under the action of $\mathrm{SO}(4)$, that is, under spatial rotations. This is quantum mechanics after all, so classical concepts don't have a perfect translation, but they are there to some extent.

Finally, it bears mentioning that $\mathrm{Spin}(d-1)$ has a non-trivial centre. In particular, there is always a $\mathbb Z_2\subseteq Z(\mathrm{Spin}(d-1))$ sub-group, whose quotient brings us back to the $\mathrm{SO}(d-1)$ group: $$\mathrm{SO}(d-1):=\frac{\mathrm{Spin}(d-1)}{\mathbb Z_2}$$

The transformation of a state under this $\mathbb Z_2$ sub-group tells us whether it descends to a true representation of $\mathrm{SO}(d-1)$, or to a projective representation. In other words, it tells us whether it is a boson or a fermion. In terms of the Dynkin labels, if $d$ is even, the state is a boson if $\lambda_r$ is even, and a fermion if odd; and if $d$ is odd, the state is a boson if $\lambda_{r}+\lambda_{r-1}$ is even, and a fermion if odd. (Compare this to $\lambda_1=2j$ in the $d=3+1$ case). Therefore, to some extent, the last two Dynkin labels differentiates bosons and fermions; they play the role of $j\ \mathrm{mod}\ 2\mathbb Z$ in $d\ge4+1$.

1: The little group of a massless state is the so-called Euclidean group $\mathrm{ISO}(d-2):=\mathrm{SO}(d-2)\ltimes \mathbb R^{d-2}$, which is clearly non-simple. Therefore, its representations can be induced from a representation of its normal sub-group $\mathbb R^{d-2}$. A non-trivial representation of this group leads to an infinite-dimensional representation of $\mathrm{ISO}(d-2)$, which is called an infinite (or continuous) spin representation. These have been shown to be pathological (e.g., they violate causality, cf. Abbott). Thus, we must restrict ourselves to trivial representations of $\mathbb R^{d-2}$, whose little group is $\mathrm{SO}(d-2)$ itself, which is simple. Its (unitary, projective) representations induce a representation of the Poincaré group known as helicity representations, which describe massless particles, such as the photon.

2: As mentioned before, the orthogonal group is simple, and therefore its algebra has no non-trivial central extensions; thus, projective representations are purely of topological origin, cf. $\pi_1(\mathrm{SO}(n))=\mathbb Z_2$.

• @AccidentalFourierTransform any references for further reading? – LDR Aug 24 '18 at 23:36
• @Eulb Sure. Any concept in particular? There is a nice general overview in Weinberg's QFT. Vol. I, chapter 2. If you want something more specific, let me know. – AccidentalFourierTransform Aug 24 '18 at 23:41
• Would this also apply to the general case of $d=s+t$, with $s$ being the number of space like dimensions, and $t$ being the number of time dimensions, in which $s$ and $t$ can both be any positive integer? Would $d=0+5$ or $d=3+2$ be equivalent to $d=4+1$ for what you said about quantum spin? – Anders Gustafson Jan 23 '19 at 16:11
• @AndersGustafson There are some subtleties, but in general, yes. When doing representation theory, the first step is to complexify the algebra. And once you complexify, the orhotonal group is the same for any signature ($SO(r,s)_{\mathbb{C}}=SO(r+s)_{\mathbb{C}}$). So 0+5, 3+2, and 4+1 are all essentially equivalent as far as (complex) representations is concerned. When going to the real group, only the real representations remain (the complex ones can be recast as real by doubling the dimension). But we usually allow complex representations anyway, so no need to "realify". – AccidentalFourierTransform Jan 23 '19 at 16:30
• to fix: the wikipedia link for transitive group action. – AccidentalFourierTransform Aug 19 '19 at 12:06

In terms of group theory, isoclinic rotations in 4 spatial dimensions are described by the symmetry group, SO(4). It is a group that can be represented by 4x4 orthogonal matrices with unit determinants.

This group has two subgroups of left and right isoclinic rotations, respectively. They are each isomorphic to 3-sphere, $S^3$, which has a group isomorphic to SU(2), i.e., classical spin. Of course, the right-isoclinic rotation and left-isoclinic rotation would be just like left rotation and right rotation in 3D subspace.

However, since you are asking about the intrinsic spin, as far as I can tell, this property does not mean that the elementary particles would behave as if and only if they are in a isoclinic rotation. In fact, in the 3 dimensional subspace they would still behave as if they are in simple rotation. This could be due to the projection of an isoclinic rotation or a generic double rotation maybe because of the intrinsic spin had different symmetry than SU(2), or just good ol' simple rotation (similar to the projection of 3D rotation into a 2D plane) due to intrinsic SU(2) symmetry as we know it. Because $S^3_L \times S^3_R$ is not isomorphic to SO(4).

• The left and right rotations in 3d don't form a group. The left and right isoclinic rotations do. I don't think they're "just like" each other. – Peter Shor Aug 10 '18 at 11:26
• Obviously, "just like" does not mean "isomorphic to", rather it was a intuitive comparison of something not so important to this context. To be precise, I said "just like" because the quotient groups $S_L^3/C_2$ and $S_R^3 / C_2$ are both isomorphic to the group $SO(3)$, where $C_2$ is the center group of $SO(4)$ consisting of identity rotation and central inversion. Of course, 3D rotations, $SO(3)$, does form a group. – Oktay Doğangün Aug 14 '18 at 8:15