Remarks on Spinors in Low Dimension

0. Introduction. The purpose of these notes is to study the orbit structure of the groups Spin(p, q) acting on their spinor spaces for certain values of n = p+q, in particular, the values

(p, q) = (8, 0), (9, 0), (10, 0), and (10, 1).

though it will turn out in the end that there are a few interesting things to say about the cases (p, q) = (10, 2) and (9, 1), as well.

1. The Octonions. Let 𝕆 denote the ring of octonions. Elements of 𝕆 will be denoted by bold letters, such as x, y, etc. Thus, 𝕆 is the unique ℝ-algebra of dimension 8 with unit 1 ∈ 𝕆 endowed with a positive definite inner product ⟨, ⟩ satisfying ⟨xy, xy⟩ = ⟨x, x⟩ ⟨y, y⟩ for all x, y ∈ 𝕆. As usual, the norm of an element x ∈ 𝕆 is denoted |x| and defined as the square root of ⟨x, x⟩. Left and right multiplication by x ∈ 𝕆 define maps L_{x} , R_{x} : 𝕆 → 𝕆 that are isometries when |x| = 1.

The conjugate of x ∈ 𝕆, denoted \overline{x}, is defined to be \overline{x} = 2⟨x, 1⟩ 1−x. When a symbol is needed, the map of conjugation will be denoted C : 𝕆 → 𝕆. The identity x \overline{x} = |x|^{2} holds, as well as the conjugation identity \overline{xy} = \overline{y} \overline{x}. In particular, this implies the useful identities C L_{x} C = R\overline{x} and C R_{x} C = L\overline{x}.

The algebra 𝕆 is not commutative or associative. However, any subalgebra of 𝕆 that is generated by two elements is associative. It follows that x (\overline{x}y) = |x|^{2} y and that (xy)x = x(yx) for all x, y ∈ 𝕆. Thus, R_{x} L_{x} = L_{x} R_{x} (though, of course, R_{x} L_{y} ≠ L_{y} R_{x} in general). In particular, the expression xyx is unambiguously defined. In addition, there are the Moufang Identities

(xyx)z = x(y(xz)),
z(xyx) = ((zx)y)x,
x(yz)x = (xy)(zx),

which will be useful below. (See, for example, Spinors and Calibrations, by F. Reese Harvey, for proofs.)

2. Spin(8). For x ∈ 𝕆, define the linear map m_{x} : 𝕆 ⊕ 𝕆 → 𝕆 ⊕ 𝕆 by the formula

m_{x} = [ \begin{matrix}0 & C R_{x} \\ −C L_{x} & 0 \end{matrix} ] .

By the above identities, it follows that (m_{x})^{2} = −|x|^{2} and hence this map induces a representation on the vector space 𝕆 ⊕ 𝕆 of the Clifford algebra generated by 𝕆 with its standard quadratic form. This Clifford algebra is known to be isomorphic to M_{16}(ℝ), the algebra of 16-by-16 matrices with real entries, so this representation must be faithful. By dimension count, this establishes the isomorphism Cℓ(𝕆, ⟨, ⟩) = End_{ℝ}(𝕆 ⊕ 𝕆).

The group Spin(8) ⊂ GL_{ℝ}(𝕆 ⊕ 𝕆) is defined as the subgroup generated by products of the form m_{x} m_{y} where x, y ∈ 𝕆 satisfy |x| = |y| = 1. Such endomorphisms preserve the splitting of 𝕆 ⊕ 𝕆 into the two given summands since

m_{x} m_{y} = [ \begin{matrix}−L\overline{x} L_{y} & 0 \\ 0 & −R\overline{x} R_{y} \end{matrix} ] .

In fact, setting x = −1 in this formula shows that endomorphisms of the form

[ \begin{matrix}L_{u} & 0 \\ 0 & R_{u} \end{matrix} ] , with |u| = 1

lie in Spin(8). In fact, they generate Spin(8), since m_{x} m_{y} is clearly a product of two of these when |x| = |y| = 1.

Fixing an identification 𝕆 ≃ ℝ^{8}, this defines an embedding Spin(8) ⊂ SO(8) × SO(8), and the projections onto either of the factors is a group homomorphism. Since neither of these projections is trivial, since the Lie algebra 𝔰𝔬(8) is simple, and since SO(8) is connected, it follows that each of these projections is a surjective homomorphism. Since Spin(8) is simply connected and since the fundamental group of SO(8) is ℤ_{2}, it follows that that each of these homomorphisms is a non-trivial double cover of SO(8). Moreover, it follows that the subsets { L_{u} |u| = 1 } and { R_{u} |u| = 1 } of SO(8) each suffice to generate SO(8).

Let H ⊂ (SO(8))^{3} be the set of triples (g_{1}, g_{2}, g_{3}) ∈ (SO(8))^{3} for which
g_{2}(xy) = g_{1}(x) g_{3}(y)

for all x, y ∈ 𝕆. The set H is closed and is evidently closed under multiplication and inverse. Hence it is a compact Lie group.

By the third Moufang identity, H contains the subset

Σ = { (L_{u}, L_{u} R_{u}, R_{u}) |u| = 1} .

Let K ⊂ H be the subgroup generated by Σ, and for i = 1, 2, 3, let ρ_{i} : H → SO(8) be the homomorphism that is projection onto the i-th factor. Since ρ_{1}(K) contains { L_{u} |u| = 1 }, it follows that ρ_{1}(K) = SO(8), so a fortiori ρ_{1}(H) = SO(8). Similarly, ρ_{3}(H) = SO(8).

The kernel of ρ_{1} consists of elements (I_{8}, g_{2}, g_{3}) that satisfy g_{2}(xy) = x g_{3}(y) for all x, y ∈ 𝕆. Setting x = 1 in this equation yields g_{2} = g_{3}, so that g_{2}(xy) = x g_{2}(y). Setting y = 1 in this equation yields g_{2}(x) = x g_{2}(1), i.e., g_{2} = R_{u} for u = g_{2}(1). Thus, the elements in the kernel of ρ_{1} are of the form (1, R_{u}, R_{u}) for some u with |u| = 1. However, any such u would, by definition, satisfy (xy)u = x(yu) for all x, y ∈ 𝕆, which is impossible unless u = ±1. Thus, the kernel of ρ_{1} is {(I_{8}, ±I_{8}, ±I_{8})} ≃ ℤ_{2}, so that ρ_{1} is a 2-to-1 homomorphism of H onto SO(8). Similarly, ρ_{3} is a 2-to-1 homomorphism of H onto SO(8), with kernel {(±I_{8}, ±I_{8}, I_{8})}. Thus, H is either connected and isomorphic to Spin(8) or else disconnected, with two components.

Now K is a connected subgroup of H and the kernel of ρ_{1} intersected with K is either trivial or ℤ_{2}. Moreover, the product homomorphism ρ_{1}×ρ_{3} : K → SO(8)×SO(8) maps the generator Σ ⊂ K into generators of Spin(8) ⊂ SO(8)×SO(8). It follows that ρ_{1}×ρ_{3}(K) = Spin(8) and hence that ρ_{1} and ρ_{3} must be non-trivial double covers of Spin(8) when restricted to K. In particular, it follows that K must be all of H and, moreover, that the homomorphism ρ_{1}×ρ_{3} : H → Spin(8) must be an isomorphism. It also follows that the homomorphism ρ_{2} : H → SO(8) must be a double cover of SO(8) as well.

Henceforth, H will be identified with Spin(8) via the isomorphism ρ_{1}×ρ_{3}. Note that the center of H consists of the elements (ε_{1} I_{8}, ε_{2} I_{8}, ε_{3} I_{8}) where ε_{i}^{2} = ε_{1}ε_{2}ε_{3} = 1 and is isomorphic to ℤ_{2} × ℤ_{2}.

Triality. For (g_{1}, g_{2}, g_{3}) ∈ H, the identity g_{2}(xy) = g_{1}(x) g_{3}(y) can be conjugated, giving

Cg_{2}C(xy) = \overline{g_{2}(\overline{y} \overline{x})} = \overline{g_{1}(\overline{y}) g_{3}(\overline{x})} = \overline{g_{3}(\overline{x})} \overline{g_{1}(\overline{y})}.

This implies that (Cg_{3}C, Cg_{2}C, Cg_{1}C) also lies in H. Also, replacing x by z\overline{y} in the original formula and multiplying on the right by \overline{g_{3}(y)} shows that

g_{2}(z)\overline{g_{3}(y)} = g_{1}(z\overline{y}),

implying that (g_{2}, g_{1}, Cg_{3}C) lies in H as well. In fact, the two maps α, β : H → H defined by

α(g_{1}, g_{2}, g_{3}) = (Cg_{3}C, Cg_{2}C, Cg_{1}C), and β(g_{1}, g_{2}, g_{3}) = (g_{2}, g_{1}, Cg_{3}C)

are outer automorphisms (since they act nontrivially on the center of H) and generate a group of automorphisms isomorphic to S_{3}, the symmetric group on three letters. The automorphism τ = αβ is known as the triality automorphism.

Notation. To emphasize the group action, denote 𝕆 ≃ ℝ^{8} by V_{i} when regarding it as a representation space of Spin(8) via the representation ρ_{i}. Thus, octonion multiplication induces a Spin(8)-equivariant projection

V_{1} ⊗ V_{3} ⟶ V_{2} .

In the standard notation, it is traditional to identify V_{1} with 𝕊_{−} and V_{3} with 𝕊_{+} and to refer to V_{2} as the ‘vector representation’. Let ρ′_{i} : 𝔰𝔭𝔦𝔫(8) → 𝔰𝔬(8) denote the corresponding Lie algebra homomorphisms, which are, in fact, isomorphisms. For simplicity of notation, for any a ∈ 𝔰𝔭𝔦𝔫(8), the element ρ′_{i}(a) ∈ 𝔰𝔬(8) will be denoted by a_{i} when no confusion can arise.

Orbit structure. Let SO(Im𝕆) ≃ SO(7) denote the subgroup of SO(𝕆) ≃ SO(8) that leaves 1 ∈ 𝕆 fixed, and let K_{i} ⊂ H be the preimage of SO(Im𝕆) under the homomorphism ρ_{i} : H → SO(𝕆). Then K_{i} is a non-trivial double cover of SO(Im𝕆) and hence is isomorphic to Spin(7). Note, in particular that K_{1} contains (I_{8}, −I_{8}, −I_{8}) and hence ρ_{3}(K_{1}) ⊂ SO(8) contains −I_{8}. This implies that ρ_{3} : K_{1} → SO(V_{3}) is a faithful representation of Spin(7) and hence K_{1} acts transitively on the unit sphere in V_{3}.

In particular, it follows that Spin(8) ⊂ SO(V_{1}) × SO(V_{3}) acts transitively on the product of the unit spheres in V_{1} and V_{3}. Consequently, it follows that the quadratic polynomials

q_{1}(x, y) = |x|^{2}  and  q_{2}(x, y) = |y|^{2}

generate the ring of Spin(8)-invariant polynomials on 𝕆 ⊕ 𝕆 and that every point of this space lies on the Spin(8)-orbit of a unique element of the form (a 1, b 1) for some pair of real numbers a, b ≥ 0. For ab ≠ 0, the stabilizer of such an element is the 14-dimensional simple group G_{2}, and this group acts transitively on the unit sphere in Im𝕆.

2. Spin(9). For (r, x) ∈ ℝ ⊕ 𝕆, define a ℂ-linear map m_{(r,x)} : ℂ ⊗ 𝕆^{2} → ℂ ⊗ 𝕆^{2} by the formula

m_{(r,x)} = i [ \begin{matrix}r I_{8} & C R_{x} \\ C L_{x} & −r I_{8} \end{matrix} ] .

Since (m_{(r,x)})^{2} is − (r^{2}+|x|^{2}) times the identity map, this defines a ℂ-linear representation on ℂ ⊗ 𝕆^{2} of the Clifford algebra generated by ℝ ⊕ 𝕆 endowed with its direct sum inner product. Since this Clifford algebra is known to be isomorphic to M_{16}(ℂ), it follows, for dimension reasons, that this representation is one-to-one and onto, establishing the isomorphism Cℓ(ℝ ⊕ 𝕆, ⟨, ⟩) = End_{ℂ}(ℂ ⊗ 𝕆^{2}).

As usual, Spin(9) is the subgroup generated by the products of the form m_{(r,x)}m_{(s,y)} where r^{2} + |x|^{2} = s^{2} + |y|^{2} = 1. Note that these products have real coefficients, and so actually lie in GL_{ℝ}(𝕆^{2}) ≃ GL(16, ℝ). In fact, these products are themselves seen to be products of the products of the special form

p_{(r,x)} = m_{(−1,0)}m_{(r,x)} = [ \begin{matrix}r I_{8} & C R_{x} \\ −C L_{x} & r I_{8} \end{matrix} ] , where r^{2} + |x|^{2} = 1,

so these latter matrices suffice to generate Spin(9). By the results of the previous section, products of an even number of the p_{(0,u)} with |u| = 1 generate Spin(8) ⊂ Spin(9).

Since the linear transformations of the form p_{(r,x)} preserve the quadratic form

q(x, y) = |x|^{2} + |y|^{2},

it follows that Spin(9) is a subgroup of SO(𝕆^{2}) = SO(16).

The Lie algebra. Since Spin(9) contains Spin(8), the containment 𝔰𝔭𝔦𝔫(8) ⊂ 𝔰𝔭𝔦𝔫(9) yields the containment

\{ ( \begin{matrix}a_{1} & 0 \\ 0 & a_{3} \end{matrix} ) a ∈ 𝔰𝔭𝔦𝔫(8) \} ⊂ 𝔰𝔭𝔦𝔫(9) .

Moreover, since Spin(9) contains the 8-sphere consisting of the p_{(r,x)} with r^{2} + |x|^{2} = 1, its Lie algebra must contain the tangent space to this 8-sphere at (r, x) = (1, 0), i.e.,

\{ ( \begin{matrix}0 & C R_{x} \\ −C L_{x} & 0 \end{matrix} ) x ∈ 𝕆 \} ⊂ 𝔰𝔭𝔦𝔫(9) .

By dimension count, this implies the equality

𝔰𝔭𝔦𝔫(9) = \{( \begin{matrix}a_{1} & C R_{x} \\ −C L_{x} & a_{3} \end{matrix} ) x ∈ 𝕆, a ∈ 𝔰𝔭𝔦𝔫(8) \} .

Let ρ : Spin(9) → SO(ℝ ⊕ 𝕆) ≃ SO(9) be the homomorphism for which the induced map on Lie algebras is

ρ′ (( \begin{matrix}a_{1} & C R_{z} \\ −C L_{z} & a_{3} \end{matrix} )) = ( \begin{matrix}0 & 2 \overline{z}^{∗} \\ −2 \overline{z} & a_{2} \end{matrix} ) .

where x^{∗} : 𝕆 → ℝ is just x^{∗}(y) = ⟨x, y⟩. (The triality constructions imply that ρ′ is, indeed, a Lie algebra homormophism. Note that, when restricted to Spin(8), this becomes the homomorphism ρ_{2} : Spin(8) → SO(𝕆) = SO(8).) Then ρ is a double cover of SO(9).

Define the squaring map σ : 𝕆^{2} → ℝ ⊕ 𝕆 by

σ (( \begin{matrix}x \\ y \end{matrix} )) = ( \begin{matrix}|x|^{2} − |y|^{2} \\ 2 x y \end{matrix} ) .

A short calculation using the Moufang Identities shows that σ is equivariant with respect to ρ, i.e., that σ(gv) = ρ(g)(σ(v)) for all v ∈ 𝕆^{2} and all g ∈ Spin(9). This will be useful below.

Orbit structure and stabilizer. Every point of 𝕆^{2} lies on an Spin(8)-orbit of an element of the form (a 1, b 1), for some pair of real numbers a, b ≥ 0. Thus, the orbits of Spin(9) on the unit sphere in 𝕆^{2} are unions of the Spin(8)-orbits of the elements (cos θ 1, sin θ 1). Now, calculation yields

p_{(cos ϕ,sin ϕ 1)} ( \begin{matrix}cos θ 1 \\ sin θ 1 \end{matrix} ) = ( \begin{matrix}cos(θ−ϕ) 1 \\ sin(θ−ϕ) 1 \end{matrix} ) .

Since all of the elements (cos θ 1, sin θ 1) lie on a single Spin(9)-orbit, it follows that Spin(9) acts transitively on the unit sphere in 𝕆^{2} and, consequently, that the quadratic form q generates the ring of Spin(9)-invariant polynomials on 𝕆^{2}.

Since the orbit of (1, 0) ∈ 𝕆^{2} is the 15-sphere and since Spin(9) is connected and simply connected, it follows that the Spin(9)-stabilizer of this element must be connected, simply connected, and of dimension 21. Since K_{1} ⊂ Spin(8) ⊂ Spin(9) lies in this stabilizer and has dimension 21, it follows that K_{1} must be equal to this stabilizer.

For use in the next two sections, it will be useful to understand the orbits of Spin(9) acting on 𝕆^{2} ⊕ 𝕆^{2} and to understand the ring of Spin(9)-invariant polynomials on this vector space of real dimension 32. The first observation is that the generic orbit has codimension 4. This can be seen as follows: Since Spin(9) acts transitively on the unit sphere in 𝕆^{2}, every element lies on the Spin(9) orbit of an element of the form

( ( \begin{matrix}a 1 \\ 0 \end{matrix} ) , ( \begin{matrix}x \\ y \end{matrix} ) ) ,

where a ≥ 0. Assuming a > 0, the stabilizer in Spin(9) of this first component is K_{1} ≃ Spin(7) and this acts transitively on the unit sphere in the second 𝕆-summand of 𝕆^{2}, so that an element of the above form lies on the orbit of an element of the form

( ( \begin{matrix}a 1 \\ 0 \end{matrix} ) , ( \begin{matrix}x \\ b 1 \end{matrix} ) ) ,

where b ≥ 0. Assuming b > 0, the stabilizer in K_{1} of 1 in this second 𝕆-summand is G_{2}, which acts transitively on the unit sphere in Im𝕆 in the first 𝕆-summand. This implies that an element of the above form lies on the orbit of an element of the form

z = ( ( \begin{matrix}a 1 \\ 0 \end{matrix} ) , ( \begin{matrix}c 1 + d u \\ b 1 \end{matrix} ) ) ,

for some c, d ≥ 0 and u ∈ Im𝕆 some fixed unit imaginary octonion. Thus, the generic Spin(9)-orbit has codimension at most 4. It is still possible that two elements of the above form with distinct values of a, b, c, d > 0 might lie on the same Spin(9)-orbit, but this will be ruled out directly.

To see that these latter elements lie on distinct Spin(9)-orbits, it will be sufficient to construct Spin(9)-invariant polynomials on 𝕆^{2} ⊕ 𝕆^{2} that separate these elements. To do so, write the typical element of 𝕆^{2} ⊕ 𝕆^{2} in the form

(v_{1}, v_{2}) = ( ( \begin{matrix}x_{1} \\ y_{1} \end{matrix} ) , ( \begin{matrix}x_{2} \\ y_{2} \end{matrix} ) ) ,

and first consider the quadratic polynomials

q_{2,0} = |x_{1}|^{2} + |y_{1}|^{2}
q_{1,1} = x_{1} · x_{2} + y_{1} · y_{2}
q_{0,2} = |x_{2}|^{2} + |y_{2}|^{2}

These polynomials are manifestly Spin(9)-invariant and satisfy

q_{2,0}(z) = a^{2},  q_{1,1}(z) = ac,  q_{0,2}(z) = b^{2} + c^{2} + d^{2}.

Evidently, these three polynomials span the vector space of Spin(9)-invariant quadratic polynomials on 𝕆^{2} ⊕ 𝕆^{2}.

Since Spin(9) contains −1 times the identity, there are no Spin(9)-invariant cubic polynomials. A representation-theoretic argument shows that the Spin(9)-invariant quartic polynomials on 𝕆^{2} ⊕ 𝕆^{2} form a vector space of dimension 7. Six of these are accounted for by quadratic polynomials in q_{2,0}, q_{1,1}, and q_{0,2}, while a seventh can be constructed as follows. Define

q_{2,2} = σ(v_{1}) · σ(v_{2}) = (|x_{1}|^{2} − |y_{1}|^{2}) (|x_{2}|^{2} − |y_{2}|^{2}) + 4 (x_{1}y_{1}) · (x_{2}y_{2}).

Using the Spin(9)-equivariance of the squaring map σ, it follows that q_{2,2} is indeed invariant under Spin(9). Note that

q_{2,2}(z) = a^{2}(c^{2} + d^{2} − b^{2}),

so that knowledge of (q_{2,0}(z), q_{1,1}(z), q_{0,2}(z), q_{2,2}(z)) suffices to recover a, b, c, d > 0 when these numbers are all non-zero. It now follows that the simultaneous level sets of these four polynomials are exactly the Spin(9)-orbits on 𝕆^{2} ⊕ 𝕆^{2}. (It seems likely that these polynomials generate the ring of Spin(9)-invariant polynomials on 𝕆^{2} ⊕ 𝕆^{2}, but such a result will not be needed, so this problem will not be discussed further.)

3. Spin(10). Rather than construct the Clifford representation for an inner product on a vector space of dimension 10, it is convenient to use the fact that Spin(10) already appears as a subgroup of Cℓ(ℝ ⊕ 𝕆, ⟨, ⟩) = End_{ℂ}(ℂ ⊗ 𝕆^{2}). In fact, by the discussion in the last section, Spin(10) is the connected subgroup of this latter algebra whose Lie algebra is

𝔰𝔭𝔦𝔫(10) = \{( \begin{matrix}a_{1} + ir I_{8} & C R_{x} + i C R_{y} \\ −C L_{x} + i C L_{y} & a_{3} − ir I_{8} \end{matrix} ) r ∈ ℝ, x, y ∈ 𝕆, a ∈ 𝔰𝔭𝔦𝔫(8) \} .

Note that 𝔰𝔭𝔦𝔫(10) appears as a subspace of 𝔰𝔲(16), so that Spin(10) acts on ℂ^{16} = ℂ ⊗ 𝕆^{2} preserving the complex structure and the quadratic form

q = q_{2,0} + q_{0,2} = |x_{1}|^{2} + |y_{1}|^{2} + |x_{2}|^{2} + |y_{2}|^{2} ,

where, now, the typical element of ℂ ⊗ 𝕆^{2} will be written as

z = ( \begin{matrix}x_{1} + i x_{2} \\ y_{1} + i y_{2} \end{matrix} ) .

Note that, because there are no connected Lie groups that lie properly between Spin(9) and Spin(10), it follows that Spin(10) is generated by Spin(9) and the circle subgroup

T = \{( \begin{matrix}e^{ir} I_{8} & 0 \\ 0 & e^{−ir} I_{8} \end{matrix} ) r ∈ ℝ/2πℤ \} ,

which lies in Spin(10), but does not lie in Spin(9). In particular, a polynomial on ℂ ⊗ 𝕆^{2} is Spin(10)-invariant if and only if it is both Spin(9)-invariant and T-invariant.

Invariant polynomials. Among the quadratic polynomials that are Spin(9)-invariant, only the multiples of q = q_{2,0} + q_{0,2} are also T-invariant. Thus, q spans the space of Spin(10)-invariant quadratic forms on ℂ ⊗ 𝕆^{2}. In particular, this implies that the action of Spin(10) on ℂ ⊗ 𝕆^{2} is irreducible (even as a real vector space).

Among the quartic polynomials that are Spin(9)-invariant, a short calculation shows that only linear combinations of q^{2} and

p = \frac{1}{2} (q_{2,2} + q_{2,0} q_{0,2} − 2 q_{1,1}^{2})
= |x_{1}|^{2}|x_{2}|^{2} + |y_{1}|^{2}|y_{2}|^{2} − (x_{1} · x_{2} + y_{1} · y_{2})^{2} + 2 (x_{1}y_{1}) · (x_{2}y_{2})
= |x_{1} ∧ x_{2}|^{2} + |y_{1} ∧ y_{2}|^{2} − 2 (x_{1} · x_{2}) (y_{1} · y_{2}) + 2 (x_{1}y_{1}) · (x_{2}y_{2}).

are invariant under the action of T. Thus, it follows that q^{2} and p span the space of Spin(10)-invariant quartics. (Note the interesting feature that, in the latter expression for p, only the final term makes use of octonion multiplication operations.)

Orbits and stabilizers. Let M ⊂ ℂ ⊗ 𝕆^{2} be the Spin(10)-orbit of z_{0} = (1 + i 0, 0 + i 0). The tangent space to M at z_{0} is the set of vectors of the form

( \begin{matrix}a_{1} + ir I_{8} & C R_{x} + i C R_{y} \\ −C L_{x} + i C L_{y} & a_{3} − ir I_{8} \end{matrix} ) ( \begin{matrix}1 + i 0 \\ 0 + i 0 \end{matrix} ) = ( \begin{matrix}a_{1}1 + i r1 \\ −\overline{x} + i \overline{y} \end{matrix} ) .

and the Lie algebra of the Spin(10)-stabilizer of z_{0} is defined by the equations a_{1}1 = r = x = y = 0. Thus, the identity component of the stabilizer of z_{0} is K_{1} ≃ Spin(7) and the full stabilizer must lie in the normalizer of K_{1} in Spin(10). Evidently, the normalizer of K_{1} in Spin(10) is K_{1} ·T. Since only the identity in the subgroup T stabilizes z_{0}, the full stabilizer of z_{0} is K_{1}. Thus, M is diffeomorphic to Spin(10)/Spin(7), which is a smooth manifold of dimension 45−21 = 24 that is 2-connected, i.e., π_{0}(M) = π_{1}(M) = π_{2}(M) = 0.

The normal space to M at z_{0} is the orthogonal direct sum of the line ℝz_{0} (which is normal to the unit sphere in ℂ ⊗ 𝕆^{2}) and the subspace of dimension 7

N_{z_{0}} = \{( \begin{matrix}0 + i x \\ 0 + i 0 \end{matrix} ) x ∈ Im𝕆 \} .

The stabilizer K_{1} acts as SO(7) on this subspace. In particular, it acts transitively on the unit sphere in N_{z_{0}}, and hence it acts transitively on the space of geodesics in the unit 31-sphere that meet M orthogonally at z_{0}. Since M is itself a Spin(10)-orbit, it follows that Spin(10) must act transitively on the the normal tube of any radius about M in the unit 31-sphere. Since, for generic radii, these normal tubes are hypersurfaces, it follows that the generic Spin(10)-orbit in the 31-sphere must be a hypersurface of dimension 30. Using the fact that such a hypersurface is an S^{6}-bundle over M, the long exact sequence in homotopy implies that these hypersurface orbits are also 2-connected, which implies that the Spin(10)-stabilizer of any point on such a hypersurface must be both connected and simply connected.

Now, the full group Spin(10) must act transitively on the space of geodesics in the unit 31-sphere that meet M orthogonally at any point while every point of the unit 31-sphere lies on some geodesic that meets M orthogonally. Thus, fixing some u ∈ Im𝕆 with |u| = 1, it follows that every element of the 31-sphere lies on the Spin(10)-orbit of an element of the form

z_{θ} = ( \begin{matrix}cos θ + i sin θ u \\ 0 + i 0 \end{matrix} ) .

Note that p(z_{θ}) = cos^{2} θ sin^{2} θ = \frac{1}{4} sin^{2}(2θ), so it follows that for 0 ≤ θ ≤ π/4, the elements z_{θ} lie on distinct orbits, and that 0 ≤ p ≤ \frac{1}{4}, with the endpoints of this interval being the only critical values of p. While M = p^{−1}(0) is one critical orbit, the other critical orbit is M^{∗} = p^{−1}(\frac{1}{4}), and consists of the points of the 31-sphere that are at geodesic distance \sqrt{2}/2 from M. It follows from this that M^{∗} is also connected and is a single orbit of Spin(10). In particular, the simultaneous level sets of q and p are exactly the Spin(10)-orbits in ℂ ⊗ 𝕆^{2}.

For 0 < θ < π/4, the nearest point on M to z_{θ} is z_{0}, so the Spin(10)-stabilizer of z_{θ} is a subgroup of K_{1} that has already been seen to be both connected and simply connected.

Also, the orbit of z_{θ} is a 6-sphere bundle over M. By dimension count, this stabilizer must be of dimension 15 and must contain the stabilizer in K_{1} of 1 and u, which is Spin(6). Thus, the stabilizer of such a z_{θ} is exactly Spin(6) ≃ SU(4). In particular, the stabilizer of any point of the 31-sphere not on M or M^{∗} must be a conjugate of SU(4).

Now, the tangent space to M^{∗} at z_{π/4} is the set of vectors of the form

( \begin{matrix}a_{1} + ir I_{8} & C R_{x} + i C R_{y} \\ −C L_{x} + i C L_{y} & a_{3} − ir I_{8} \end{matrix} ) ( \begin{matrix}1 + i u \\ 0 + i 0 \end{matrix} ) = ( \begin{matrix}(a_{1}1 − r u) + i (a_{1}u + r1) \\ −\overline{(x + yu)} + i \overline{(y − xu)} \end{matrix} ) .

Thus, the Lie algebra of the stabilizer G of z_{π/4} is defined by the relations a_{1}1 − r u = a_{1}u+r1 = y−xu = 0. (Remember that u^{2} = −1.) It follows that a_{1} ∈ 𝔰𝔬(8) must belong to the stabilizer of the 2-plane spanned by {1, u}, so that a_{1} lies in 𝔰𝔬(2)⊕𝔰𝔬(6). Conversely, if a_{1} lies in this subspace, then there exists a unique r ∈ ℝ so that a_{1}1−r u = a_{1}u+r1 = 0. From the matrix representation, it is clear that the maximal torus in 𝔰𝔬(2) ⊕ 𝔰𝔬(6) (which has rank 4) is a maximal torus in the full stabilizer algebra, which has dimension 24. The root pattern is evident from the matrix representation, implying that the stabilizer algebra is isomorphic to 𝔰𝔲(5).

Now M^{∗} has dimension 21 and is the base of a fibration whose total space is one of the hypersurface orbits and whose fiber is a 9-sphere. The 2-connectivity of the hypersurface orbits implies, by the long exact sequence in homotopy, that M^{∗} is also 2-connected, which implies that M^{∗} = Spin(10)/G where G is both connected and simply connected. Since its Lie algebra is 𝔰𝔲(5), it follows that G is isomorphic to SU(5).

4. Spin(10, 1). To construct the spinor representation of Spin(10, 1), it will be easiest to construct the Lie algebra representation by extending the Lie algebra representation of Spin(10) that was constructed in §3. It is convenient to identify ℂ ⊗ 𝕆^{2} with 𝕆^{4} explicitly via the identification

z = ( \begin{matrix}x_{1} + i x_{2} \\ y_{1} + i y_{2} \end{matrix} ) = ( \begin{matrix}x_{1} \\ y_{1} \\ x_{2} \\ y_{2} \end{matrix} ) .

Via this identification, 𝔰𝔭𝔦𝔫(10) becomes the subspace

𝔰𝔭𝔦𝔫(10) = \{( \begin{matrix}a_{1} & C R_{x} & −r I_{8} & −C R_{y} \\ −C L_{x} & a_{3} & −C L_{y} & r I_{8} \\ r I_{8} & C R_{y} & a_{1} & C R_{x} \\ C L_{y} & −r I_{8} & −C L_{x} & a_{3} \end{matrix} ) r ∈ ℝ, x, y ∈ 𝕆, a ∈ 𝔰𝔭𝔦𝔫(8) \} .

Consider the one-parameter subgroup R ⊂ SL_{ℝ}(𝕆^{4}) defined by

R = \{( \begin{matrix}t I_{16} & 0 \\ 0 & t^{−1} I_{16} \end{matrix} ) t ∈ ℝ^{+} \} .

It has a Lie algebra 𝔯 ⊂ 𝔰𝔩(𝕆^{4}). Evidently, the the subspace [𝔰𝔭𝔦𝔫(10), 𝔯] consists of matrices of the form

( \begin{matrix}0_{8} & 0_{8} & r I_{8} & C R_{y} \\ 0_{8} & 0_{8} & C L_{y} & −r I_{8} \\ r I_{8} & C R_{y} & 0_{8} & 0_{8} \\ C L_{y} & −r I_{8} & 0_{8} & 0_{8} \end{matrix} ) , r ∈ ℝ, y ∈ 𝕆 .

Let 𝔤 = 𝔰𝔭𝔦𝔫(10) ⊕ 𝔯 ⊕ [𝔰𝔭𝔦𝔫(10), 𝔯]. Explicitly,

𝔤 = \{( \begin{matrix}a_{1} + x I_{8} & C R_{x} & y I_{8} & C R_{y} \\ −C L_{x} & a_{3} + x I_{8} & C L_{y} & −y I_{8} \\ z I_{8} & C R_{z} & a_{1} − x I_{8} & C R_{x} \\ C L_{z} & −z I_{8} & −C L_{x} & a_{3} − x I_{8} \end{matrix} ) \begin{matrix}x, y, z ∈ ℝ, \\ x, y, z ∈ 𝕆, \\ a ∈ 𝔰𝔭𝔦𝔫(8) \end{matrix} \} .

Compuation using the Moufang Identities shows that 𝔤 is closed under Lie bracket and hence is a Lie algebra of (real) dimension 55 that contains 𝔰𝔭𝔦𝔫(10). The induced representation of Spin(10) on 𝔤/𝔰𝔭𝔦𝔫(10) evidently restricts to Spin(9) to preserve the splitting corresponding to the sum 𝔯 ⊕ [𝔰𝔭𝔦𝔫(10), 𝔯] ≃ ℝ ⊕ ℝ^{9} and acts as the standard (irreducible) representation on the ℝ^{9} summand. It follows that Spin(10) must act via its standard (irreducible, ten dimensional) representation on 𝔤/𝔰𝔭𝔦𝔫(10). Since the trace of the square of a non-zero element in the subspace 𝔯 ⊕ [𝔰𝔭𝔦𝔫(10), 𝔯] is positive, 𝔤 is semisimple of noncompact type. It follows that 𝔤 is isomorphic to 𝔰𝔬(10, 1) and hence is the Lie algebra of a representation of Spin(10, 1). This representation must be faithful since it is faithful on the maximal compact subgroup Spin(10).

Thus, define Spin(10, 1) to be the (connected) subgroup of SL_{ℝ}(𝕆^{4}) that is generated by Spin(10) and the subgroup R. Its Lie algebra 𝔤 will henceforth be written as 𝔰𝔭𝔦𝔫(10, 1).

Invariant Polynomials and Orbits. Consider the Spin(10)-invariant polynomial

p = |x_{1}|^{2}|x_{2}|^{2} + |y_{1}|^{2}|y_{2}|^{2} − (x_{1} · x_{2} + y_{1} · y_{2})^{2} + 2 (x_{1}y_{1}) · (x_{2}y_{2}) ,

Evidently, p is invariant under R and is therefore invariant under Spin(10, 1). In particular, it follows that the orbits of Spin(10, 1) on 𝕆^{4} ≃ ℝ^{32} must lie in the level sets of p.

Also from the previous section, it is known that every element of 𝕆^{4} lies on the Spin(10)-orbit of exactly one of the elements

z_{a,b} = ( \begin{matrix}a 1 \\ 0 \\ b u \\ 0 \end{matrix} ) where 0 ≤ b ≤ a.

and where u ∈ Im𝕆 is a fixed unit imaginary octonion. However, all of the elements of the form

( \begin{matrix}at 1 \\ 0 \\ (b/t) u \\ 0 \end{matrix} )  (where 0 < t)

lie on the same R-orbit and, hence, on the same Spin(10, 1)-orbit. Since p(z_{a,b}) = a^{2}b^{2}, it now follows that each of the nonzero level sets of p is a single Spin(10, 1)-orbit while the zero level set is the union of the origin and a single Spin(10, 1)-orbit, say, the orbit of z_{1,0}. Moreover, it follows that p generates the ring of Spin(10, 1)-invariant polynomials on 𝕆^{4}.

Stabilizers. Multiplication by positive scalars acts transitively on the non-zero level sets of p, so they are all diffeomorphic. In fact, each such level set is contractible to the Spin(10)-invariant locus where q reaches its minimum on this level set and this is a manifold of dimension 21 that is diffeomorphic to M^{∗}. In particular, it follows that each of the non-zero level sets of p is 2-connected, so that the stabilizer in Spin(10, 1) of a point on such a level set must be connected and simply connected.

If z ∈ 𝕆^{4} has p(z) ≠ 0, then the Spin(10, 1)-orbit of z has dimension 31 and so its stabilizer in Spin(10, 1) must be of dimension 55−31 = 24. Moreover all of these stabilizers must be conjugate in Spin(10, 1). Since the Spin(10)-stabilizer of the point z_{1,1} is already known to be SU(5), which has dimension 24, it follows that this must be the Spin(10, 1)- stabilizer as well.

The Spin(10, 1)-orbit consisting of nonzero vectors in the zero locus of p is just the deleted cone on M, and so has dimension 25. Since it is contractible to M, it is 2-connected, so that the stabilizer in Spin(10, 1) of a point in this orbit must be connected and simply connected and of dimension 55−25 = 30. In fact, the Lie algebra of this stabilizer is just

( \begin{matrix}a_{1} & 0 & y I_{8} & C R_{y} \\ 0 & a_{3} & C L_{y} & −y I_{8} \\ 0 & 0 & a_{1} & 0 \\ 0 & 0 & 0 & a_{3} \end{matrix} )

y ∈ ℝ, y ∈ 𝕆, a ∈ 𝔨_{1}

where 𝔨_{1} is the Lie algebra of K_{1} ⊂ Spin(8). Thus, the stabilizer is a semi-direct product of Spin(7) with a copy of ℝ^{9}.

5. Spin(10, 2). It might be tempting to conjecture that Spin(10, 1) could be defined directly as the stabilizer of p. However, this is not the case, as the stabilizer of p is larger. One can see this directly by looking at the alternative expression

p = |x_{1} ∧ x_{2}|^{2} + |y_{1} ∧ y_{2}|^{2} − 2 (x_{1} · x_{2}) (y_{1} · y_{2}) + 2 (x_{1}y_{1}) · (x_{2}y_{2}) .

which makes it evident that p is invariant under the 6-dimensional Lie group

G = \{( \begin{matrix}a I_{8} & 0 & b I_{8} & 0 \\ 0 & a′ I_{8} & 0 & b′ I_{8} \\ c I_{8} & 0 & d I_{8} & 0 \\ 0 & c′ I_{8} & 0 & d′ I_{8} \end{matrix} ) ad−bc = a′d′−b′c′ = ±1 \} .

Since G does not lie in Spin(10, 1), the invariance group of p must be properly larger than Spin(10, 1).

In particular, consider the G-subgroup R′ ≃ ℝ^{+} consisting of matrices of the form

( \begin{matrix}t I_{8} & 0 & 0 & 0 \\ 0 & t^{−1} I_{8} & 0 & 0 \\ 0 & 0 & t^{−1} I_{8} & 0 \\ 0 & 0 & 0 & t I_{8} \end{matrix} )  where t > 0,

which is not a subgroup of Spin(10, 1). Let 𝔯′ denote its Lie algebra. Calculation shows that

𝔯′ ⊕ [𝔰𝔭𝔦𝔫(10, 1), 𝔯′] = \{( \begin{matrix}w I_{8} & C R_{w} & u I_{8} & 0 \\ C L_{w} & −w I_{8} & 0 & u I_{8} \\ v I_{8} & 0 & −w I_{8} & −C R_{w} \\ 0 & v I_{8} & −C L_{w} & w I_{8} \end{matrix} ) \begin{matrix}u, v, w ∈ ℝ, \\ w ∈ 𝕆 \end{matrix} \} .

and that the sum 𝔰𝔭𝔦𝔫(10, 1) ⊕ 𝔯′ ⊕ [𝔰𝔭𝔦𝔫(10, 1), 𝔯′] is closed under Lie bracket. Thus, this defines a Lie algebra of dimension 66 that lies in the stabilizer of p.

The details of further analysis will be omitted, but by using arguments similar to those used in previous sections, one sees that this algebra is isomorphic to 𝔰𝔬(10, 2) and that the connected Lie subgroup of GL_{ℝ}(𝕆^{4}) whose Lie algebra is this one is simply connected, so that it this group is Spin(10, 2). Henceforth, this algebra will be denoted 𝔰𝔭𝔦𝔫(10, 2). Thus,

𝔰𝔭𝔦𝔫(10, 2) = \{( \begin{matrix}a_{1} + x I_{8} & C R_{w} & y I_{8} & C R_{y} \\ C L_{x} & a_{3} + w I_{8} & C L_{y} & u I_{8} \\ z I_{8} & C R_{z} & a_{1} − x I_{8} & −C R_{x} \\ C L_{z} & v I_{8} & −C L_{w} & a_{3} − w I_{8} \end{matrix} ) \begin{matrix}u, v, w, x, y, z ∈ ℝ, \\ w, x, y, z ∈ 𝕆, \\ a ∈ 𝔰𝔭𝔦𝔫(8) \end{matrix} \} .

Moreover, representation theoretic methods show that the only connected proper subgroup of SL_{ℝ}(𝕆^{4}) that properly contains Spin(10, 2) is Sp(16, ℝ), the symplectic group preserving the symplectic 2-form Ω defined by

Ω = dx_{1} ˆ·dx_{2} + dy_{1} ˆ·dy_{2} .

Of course, Sp(16, ℝ) does not stabilize any nonzero polynomials. It follows that Spin(10, 2) is the identity component of the stabilizer of p, and hence that the stabilizer of p must lie in the normalizer of Spin(10, 2) in GL_{ℝ}(𝕆^{4}). However, this normalizer is just ℝ^{+}·I_{32} × Spin(10, 2) and the only element in ℝ^{+}·I_{32} that stabilizes p is the identity element. It follows that Spin(10, 2) is the stabilizer of p.

6. Spin(9, 1). As a final note, inspection reveals that the subalgebra

𝔰𝔭𝔦𝔫(9, 1) = \{( \begin{matrix}a_{1} + x I_{8} & C R_{w} \\ C L_{x} & a_{3} − x I_{8} \end{matrix} ) \begin{matrix}x ∈ ℝ, \\ w, x ∈ 𝕆, \\ a ∈ 𝔰𝔭𝔦𝔫(8) \end{matrix} \} ⊂ 𝔰𝔩(16, ℝ) ,

which contains 𝔰𝔭𝔦𝔫(9), is actually the Lie algebra of a faithful representation of Spin(9, 1) on ℝ^{16} ≃ 𝕆^{2}. This action of Spin(9, 1) has the interesting feature that it has only two orbits: The origin and the set of all non-zero vectors. This follows because the compact group Spin(9) ⊂ Spin(9, 1) already acts transitively on the unit spheres, but the larger group does not even preserve the quadratic form.