Here ρ is the half sum of the roots in Δ , and τ is dominant with respect to Δ .
In order to see that π has nonzero Dirac cohomology, we need the following lemma.
Lemma 2.4. The restriction of the g-module E(ρ) to m is isomorphic to F (ρm)⊗

n, where F (ρm) denotes the irreducible m-module with highest weight
ρm and ρn denotes the half sum of roots in Δ(n).
Proof. Since g and m have the same rank, we can use Lemma 2.2 to replace
E(ρ) and F (ρm) by the corresponding spin modules. Recall that the spin module
Spinm can be constructed as

is a maximal isotropic subspace of
m. We can choose m+ so that it contains all the positive root subspaces for m, as
well as a maximal isotropic subspace h+ of the Cartan subalgebra h. To construct
Sping, we can use the maximal isotropic subspace g+ = m+ ⊕n of g. It follows that
Sping = Spinm C−ρn

n. The ρ-shift comes from the fact that the highest
weight of Spinm is ρm and the highest weight of Sping is ρ, while the highest weight
of Spinm

n is ρm + 2ρn = ρ + ρn.
Since π is unitary, the computation for its Dirac cohomology is
π E(ρ) : E(τ ) = πm E(−τ ) |m : E(ρ) |m =
πm E(−τ ) |m : F (ρm) C−ρn

n =
Cξ+ρn πm F (ρm) E(−τ ) |m :

n .
Here the first equality used Frobenius reciprocity, while the second equality used
Lemma 2.4. Note that the dual of E(τ ) is the module E(−τ ) which has lowest
weight −τ with respect to Δ .
Using (2.12) and (2.9), we can write
(2.14) −τ = −2λ + ρ = −ξ μm νm + ρ = −ξ τm ρm + ρ .
We have assumed ξ to be dominant for Δ(n), and 2λm is dominant for Δm. Thus
Δm Δ, Δ . Because of (2.10), the LHS of the last line of (2.13) contains the
Cξ+ρn F (τm) E(−τ ) |m⊇ Cξ+ρn F (τm τ ).
Namely, F (τm −τ ) is the PRV component of F (τm)⊗F (−τ ) F (τm)⊗E(−τ ) |m.
By (2.12) and (2.9), τm τ = −ξ ρm + ρ , so
Cξ+ρn F (τm τ ) F (ρn ρm + ρ ) = F (wmρ + ρ ),
where wm is the longest element of the Weyl group of m. Namely, wm sends all
roots in Δm to negative roots for m, while permuting the roots in Δ(n), so wmρ =
−ρm + ρn.
So we see that the LHS of the last line of (2.13) contains the m-module F (wmρ+
ρ ) = F (wmρ +ρ). Namely, both wmρ+ρ and wmρ = wm(wmρ+ρ ) are extremal
weights for the same module.
We will show that
(2.15) F (wmρ + ρ) :

n = 0.
This will prove that (2.13) is nonzero, and consequently that π has nonzero Dirac
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