The level-one e=2, one-strand Hecke and KLR blocks are both scalar

In $\mathbb{C}$, $\xi =-1$ has order two, so $e=2$ and $\xi \mathrm{\ne }1$ is the nondegenerate branch [F1,F2]. By [F2], the $q$ in the cyclotomic presentation [F3] is $\xi =-1$. Since $⟨h_0,{\mathrm{\Lambda }}_0⟩=1$ and the other level-one pairing is zero, the height-one cyclotomic relation is $X_1-1=0$. At $d=1$ the only affine generators are $X_1^{\pm 1}$ [F3], so both become $1$ in the quotient; the scalar map sending its unit to $1$ is inverse to the scalar inclusion, hence $H_1^{{\mathrm{\Lambda }}_0}\cong \mathbb{C}$. The residue values ${\xi }^0=1$ and ${\xi }^1=-1$ are distinct; thus [F7] gives sole projector $e(0)=1$ and $e_{{\alpha }_0}=1\mathrm{\ne }0$.

Since ${\alpha }_0$ has height one, its only sequence is $(0)$; [F4] gives the idempotent $e(0)=1$, the dot $x_1$, and no crossing. The sole dot commutes with itself, so the presented algebra is $\mathbb{C}[x_1]e(0)$. The standard first-strand relation [F5] is $x_{1}e(0)=0$, and evaluation at zero maps $e(0)$ to $1$, so the quotient is nonzero and equals $\mathbb{C}e(0)\cong \mathbb{C}$. Its unit has degree zero and its dot is zero [F6], so the quotient is concentrated in degree zero. Every nonzero simple over this field algebra is one-dimensional because any nonzero vector spans a nonzero submodule; a graded simple is concentrated in one degree because the span of a homogeneous vector is a graded submodule. Thus the degree-zero representative and its grading shifts are exactly the graded simples.

In this rank-one data, [F3] sets $X_1=1$ and the distinct spectral values $1,-1$ make [F7] give $e_{{\alpha }_0}=e(0)=1\mathrm{\ne }0$. Also [F4] and [F5] give the nonzero quotient $\mathbb{C}e(0)$ at ${\alpha }_0$. Thus the hypotheses of [F8] hold over $\mathbb{C}$ with $e=2$, and its block isomorphism identifies the two scalar algebras.