Spin(8) tensor-product decompositions

Write the positive roots as ${\xi }_i-{\xi }_j$ and ${\xi }_i+{\xi }_j$ for $i<j$, with $\rho =(3,2,1,0)$ and ${\lambda }_1={\xi }_1$, ${\lambda }_2={\xi }_1+{\xi }_2$, ${\lambda }_3=({\xi }_1+{\xi }_2+{\xi }_3-{\xi }_4)\mathrm{/}2$, ${\lambda }_4=({\xi }_1+{\xi }_2+{\xi }_3+{\xi }_4)\mathrm{/}2$, as fixed in [F1]. Pairing the six roots for each $i<j$ in [F3] gives denominator $(3^2-2^2)(3^2-1^2)(3^2-0^2)(2^2-1^2)(2^2-0^2)(1^2-0^2)=5\cdot 8\cdot 9\cdot 3\cdot 4\cdot 1=4320$, so $d(\mu )=\frac{{\prod }_{i<j}(({\mu }_i+{\rho }_i{)}^2-({\mu }_j+{\rho }_j{)}^2)}{4320}\mathrm{.}$ For ${\lambda }_1,{\lambda }_3,{\lambda }_4$, the shifted-coordinate squares are respectively $(16,4,1,0)$, $(49,25,9,1)\mathrm{/}4$, and $(49,25,9,1)\mathrm{/}4$, giving numerator products $34560,34560,34560$ and dimensions $8,8,8$. For ${\lambda }_2$, the squares are $(16,9,1,0)$, giving $7\cdot 15\cdot 16\cdot 8\cdot 9\cdot 1=120960$ and dimension $28$. For $2{\lambda }_1,2{\lambda }_3,2{\lambda }_4$, the squares are $(25,4,1,0)$, $(16,9,4,1)$, $(16,9,4,1)$, giving numerator products $151200,151200,151200$ and dimensions $35,35,35$. For ${\lambda }_1+{\lambda }_3,{\lambda }_1+{\lambda }_4$, the squares are $(81,25,9,1)\mathrm{/}4$ in both cases, giving $14\cdot 18\cdot 20\cdot 4\cdot 6\cdot 2=241920$ and dimension $56$; for ${\lambda }_3+{\lambda }_4$, the squares are $(16,9,4,0)$, giving $7\cdot 12\cdot 16\cdot 5\cdot 9\cdot 4=241920$ and dimension $56$. Each weight is dominant integral because it is a nonnegative integral combination of the displayed fundamental weights.

Define $\beta :8_s\otimes 8_c\to 8_v$ by $⟨\beta (s,t),v⟩=⟨c(v)s,t⟩$. It is Spin-equivariant: for $g\in \mathrm{Spin}(8)$, Clifford conjugation sends $c(v)$ to $c({\rho }_{g}v)$ and the three inner products are invariant by [F1]. It is nonzero because for a unit $v$ and nonzero $s$, Clifford multiplication gives a nonzero $c(v)s\in 8_c$, and choosing $t=c(v)s$ makes $⟨\beta (s,t),v⟩=\mathrm{\parallel }c(v)s{\mathrm{\parallel }}^2>0$. Since $8_v$ is irreducible by [F1], $\beta$ is onto and its kernel has dimension $64-8=56$.

The invariant real inner products in [F1] complexify to invariant Hermitian inner products on each tensor representation. Taking orthogonal complements repeatedly therefore decomposes each complexified representation into irreducible Spin(8)-submodules. By [F1], the group is generated by plane-rotation exponentials and its central kernel. The vector action in Spin(8) vector and half-spin modules is conjugation, so $-1$ acts trivially on $8_v$; the spin action is left Clifford multiplication, so $-1$ acts as $-I$ on $8_s,8_c$. It therefore acts scalarly on every tensor product here. Hence a subspace is Spin(8)-invariant exactly when it is invariant under the complexified rotation Lie algebra. The highest-weight vector of weight $\lambda$ in one factor tensored with the highest-weight vector of weight $\mu$ in another is nonzero, has weight $\lambda +\mu$, and is killed by every positive-root operator because such an operator acts on a tensor as the sum of its actions on the factors. Every tensor-product weight is at most $\lambda +\mu$. Projections onto the irreducible summands commute with the Lie algebra action, so a nonzero projection of this vector is a highest-weight vector of weight $\lambda +\mu$ in one summand. Its highest weight is dominant integral, and [F2] identifies that summand as $L(\lambda +\mu )$; its dimension is the value computed in step 1.1 by [F3].

The trace map ${\mathrm{Sym}}^2(8_v)\to \mathbb{R}$ is onto because the invariant metric tensor has trace $8$, and its real kernel has dimension $8\cdot 9\mathrm{/}2-1=35$. The square of a highest-weight vector of weight ${\lambda }_1$ is highest of weight $2{\lambda }_1$; its irreducible summand from step 2.1 has dimension $35$. The trace map vanishes on that summand: a nonzero equivariant map from a $35$-dimensional irreducible module to the one-dimensional trivial module would have zero kernel and hence impossible image dimension. Thus the complexified summand equals the complexification of the real trace-free kernel, proving that kernel irreducible; the invariant metric splits off the real trivial line. Also ${\mathrm{\Lambda }}^2(8_v)$ identifies with $\mathfrak{s}\mathfrak{o}(8)$ by $u\wedge v\mapsto (x\mapsto ⟨v,x⟩u-⟨u,x⟩v)$; this is an equivariant isomorphism onto the $28$-dimensional adjoint module, irreducible by the simplicity in [F1].

For $X=8_s$ or $8_c$, the invariant metric gives an onto trace map ${\mathrm{Sym}}^2(X)\to \mathbb{R}$ with real kernel dimension $8\cdot 9\mathrm{/}2-1=35$. The square of a highest-weight vector has highest weight $2{\lambda }_4$ for $X=8_s$ and $2{\lambda }_3$ for $X=8_c$; step 2.1 supplies an irreducible $35$-dimensional complex summand. Its trace is zero because a nonzero equivariant map from this irreducible $35$-dimensional summand to the trivial line would be injective and exceed the target dimension. Hence it equals the complexification of the real trace-free kernel, proving that kernel irreducible; the metric tensor splits off the real trivial line. For either chirality, call the real trace-free kernel $X_0$ and the corresponding real Hodge eigenspace from [F4] $Y_0$. Their complexifications are irreducible of the same highest weight, so [F2] gives a complex Lie-algebra intertwiner $\varphi :(X_0{)}_{\mathbb{C}}\to (Y_0{)}_{\mathbb{C}}$ that is an isomorphism. For each coordinate-plane infinitesimal rotation $T$, the identity $\varphi T_X=T_Y\varphi$ gives $\varphi \exp (t{T}_X)=\exp (t{T}_Y)\varphi$ by the convergent matrix-exponential series. These exponentials are the actions of the rotation lifts from [F1]. The kernel element $-1$ acts trivially on $X_0$, as a subspace of a half-spin tensor square, and on $Y_0$, whose action factors through $SO(8)$. Rotation lifts and this kernel generate Spin(8) by [F1], so $\varphi$ intertwines the full group actions. Let ${\sigma }_X,{\sigma }_Y$ be the conjugations whose fixed spaces are $X_0,Y_0$. The map $d={\varphi }^{-1}{\sigma }_Y\varphi$ is a conjugate-linear equivariant involution on $(X_0{)}_{\mathbb{C}}$, and $d{\sigma }_X$ is a complex-linear equivariant endomorphism; by Schur's lemma it is $aI$ (a complex endomorphism has an eigenvalue, whose nonzero eigenspace is invariant, hence the whole irreducible module). Since $d=a{\sigma }_X$ and $d^2=I$, $\mathrm{\mid }a\mathrm{\mid }=1$. Choose $c\in \mathbb{C}$ with $c\mathrm{/}\bar{c}=a$; then ${\sigma }_Y(c\varphi )=\bar{c}a\varphi {\sigma }_X=c\varphi {\sigma }_X$, so $c\varphi$ restricts to a real equivariant isomorphism $X_0\cong Y_0$. Thus ${35}_s\cong {\mathrm{\Lambda }}_{+}^{4}V$ and ${35}_c\cong {\mathrm{\Lambda }}_{-}^{4}V$ in these orientation and chirality conventions.

The tensor of highest-weight vectors in the complexification of $8_s\otimes 8_c$ has highest weight ${\lambda }_4+{\lambda }_3$; step 2.1 gives an irreducible $56$-dimensional summand of that weight. Its restriction under the complexified $\beta$ is zero: a nonzero map from an irreducible $56$-dimensional module to the irreducible $8$-dimensional module would be injective and have image dimension $56$, impossible. Thus this summand equals the complexified kernel, since the real kernel has dimension $56$ by step 1.2 and complexification preserves dimension. Hence the real kernel is irreducible, because any real invariant subspace complexifies to an invariant subspace of this irreducible complexification. The invariant real orthogonal complement of $\ker \beta$ maps isomorphically to $8_v$: it has dimension $8$, intersects the kernel trivially, and $\beta$ is onto. Therefore $8_s\otimes 8_c\cong 8_v\oplus {56}_v$ as real modules.

The map ${\gamma }_{-}:8_v\otimes 8_c\to 8_s$, $v\otimes t\mapsto c(v)t$, is equivariant and onto because $c(v_0{)}^2=I$ for a fixed unit vector $v_0$; its real kernel has dimension $56$. The complexified highest-weight tensor has weight ${\lambda }_1+{\lambda }_3$, whose irreducible summand has dimension $56$ by step 2.1. Equivariance and the unequal target dimension make its image zero, so it is the full complexified kernel; therefore the real kernel is irreducible. Its invariant real orthogonal complement has dimension $8$, intersects the kernel trivially, and maps onto $8_s$ because ${\gamma }_{-}$ is surjective. Hence $8_v\otimes 8_c\cong 8_s\oplus {56}_s$.

For either real half-spin module $X$, the invariant inner product identifies ${\mathrm{\Lambda }}^{2}X$ equivariantly with the skew-adjoint endomorphisms $\mathfrak{s}\mathfrak{o}(X)$. The plane-rotation infinitesimal action from [F1] maps $\mathfrak{s}\mathfrak{o}(8)\to \mathfrak{s}\mathfrak{o}(X)$, and its complexification is nonzero because $X_{\mathbb{C}}$ has highest weight ${\lambda }_3$ or ${\lambda }_4$. Its complexified kernel is an ideal in the simple algebra $\mathfrak{s}\mathfrak{o}(8,\mathbb{C})$, so it is zero; hence the real map is injective. Both real Lie algebras have dimension $8\cdot 7\mathrm{/}2=28$, so it is onto and intertwines their adjoint actions. A real invariant subspace of this adjoint module complexifies to an ideal in $\mathfrak{s}\mathfrak{o}(8,\mathbb{C})$, so simplicity makes it zero or the whole space. Therefore ${\mathrm{\Lambda }}^{2}X\cong 28$ irreducibly.

Clifford multiplication defines an equivariant real map ${\gamma }_{+}:8_v\otimes 8_s\to 8_c$, $v\otimes s\mapsto c(v)s$. It is onto because for any fixed unit $v_0$, $c(v_0{)}^2=I$ maps $8_s$ bijectively to $8_c$, so its kernel has real dimension $64-8=56$. In the complexification, the tensor of highest-weight vectors has weight ${\lambda }_1+{\lambda }_4$; step 2.1 supplies an irreducible $56$-dimensional summand. Its map under the complexified ${\gamma }_{+}$ is zero by the same injectivity-and-dimension argument as in step 3.3, so it equals the complexified real kernel. Hence the real kernel is irreducible. Its invariant real orthogonal complement has dimension $8$, intersects the kernel trivially, and maps onto $8_c$; this gives $8_v\otimes 8_s\cong 8_c\oplus {56}_c$ over $\mathbb{R}$.

Steps 3.1, 3.2, and 4.1 prove all symmetric- and exterior-square decompositions, and steps 1.2, 3.3, 3.4, and 4.2 prove all mixed products with the kernels named in the Statement. Reversing chirality exchanges ${\lambda }_3$ and ${\lambda }_4$ by [F1] and therefore exchanges the corresponding $s,c$ labels without altering any dimension.