Hochschild homology of the rank-one Soergel bimodule

Compute the differential in the displayed basis. [L1, L2, given, algebra] In the left $R$-basis $\{1\otimes 1,1\otimes y\}$ one has $y\otimes 1=y(1\otimes 1)$ and $1\otimes y$ as the second basis vector, so $\delta (1\otimes 1)=y\otimes 1-1\otimes y$ has coordinates $(y,-1)$. Using the balancing relation $1\otimes y^2=y^2\otimes 1=y^2(1\otimes 1)$ and $y\otimes y=y(1\otimes y)$, $\delta (1\otimes y)=y\otimes y-1\otimes y^2$ has coordinates $(-y^2,y)$. The matrix is therefore $\left(\begin{array}{cc}y& -y^2\\ -1& y\end{array}\right)$, with columns the images of the two basis vectors, and both columns are homogeneous of degree $2$ since $y$ has degree $2$.

Compute kernel and cokernel. [step 1.1, algebra] An element $a(1\otimes 1)+b(1\otimes y)$ lies in $\ker \delta$ exactly when $ay-b{y}^2=0$ and $-a+by=0$; since $R$ is a domain the second equation gives $a=by$, and the first is then automatic. Hence $\ker \delta =\mathbb{Q}[y]\cdot (y\otimes 1+1\otimes y)$, free of rank one with generator of degree $2$. The first column $(y,-1)$ generates the image, because the second column $(-y^2,y)$ equals $-y$ times the first; the cokernel is therefore free of rank one and the class of $1\otimes 1$ is a generator of degree $0$, so $\mathrm{coker}\delta \cong R$ with the class of $1\otimes 1$ as basis element.

Apply the diagonal theorem and read the Hochschild groups. [L1, L2, L3, L5, step 2.1, algebra] By [L3] applied to $M=B_1$, $H{H}_0(R,B_1)\cong H_0=\mathrm{coker}\delta \cong R$ in degree $0$, and $H{H}_1(R,B_1)\cong H_1=\ker (\delta :B_1\{2\}\to B_1)$. The kernel of $\delta$ has its generator in degree $2$ inside $B_1$; the shifted term $B_1\{2\}$ represents the coefficient together with its single Koszul symbol $\theta$, so its degree is $2+2=4$. Thus the class $[(y\otimes 1+1\otimes y)\theta ]$ has total internal degree $4$ and $H_1\cong R\{4\}$. For $h\ge 2$ the complex has no terms, so $H{H}_h=0$. AC is used only here, in [L3].

Record the normalization. [L1, L4, step 3.1] The generator of $H{H}_1$ is the class of $r{b}_1(1)$ times the symbol, so it is the degree-$2$ element $r{b}_1(1)$ shifted by the degree-$2$ symbol; the resulting degree is $4$, and no other normalization is asserted. The two other boundary cases are immediate: for $h=0$ there is no symbol and the degree is the degree of the class of $1\otimes 1$, namely $0$; the complex is empty above $h=1$.