# Discretization of derivative of delta function and affine Kac-Moody algebra

In equation (4.16) of https://arxiv.org/abs/1506.06601, a discretization of the (classical) affine Kac-Moody algebra is presented:

$$\frac{1}{\gamma}\left\{J_{m}^{1}, J^2_{n}\right\}=J_{m}^{1} J_{n}^{2} \overline{\mathfrak{r}} \delta_{m n}-\overline{\mathfrak{r}}^{t} {J}_{m}^{1} J_{n}^{2} \delta_{m n}-J_{m}^{1} \overline{\mathfrak{r}} J_{n}^{2} \delta_{m, n+1}+J_{n}^{2} \overline{\mathfrak{r}}^{t} J_{m}^{1} \delta_{m+1, n},\tag{4.16}$$

where $$A^1=A\otimes 1$$ and $$B^2=1\otimes B$$. Here $$\overline{\mathfrak{r}}$$ is a solution of the classical Yang-Baxter equation that satisfies

$$\overline{\mathfrak{r}}+\overline{\mathfrak{r}}^{t}=2 \mathfrak{C},\tag{4.15}$$

where $$\mathfrak{C}$$ is the tensor Casimir of the associated Lie algebra, i.e., $$\sum_aT_a\otimes T_a$$.

The claim is that by expanding $$J_{n} \sim 1+\frac{2 \pi}{k} \Delta \mathscr{J}_{+}(x),\tag{4.14}$$

we obtain

$$\left\{\mathscr{J}_{+}^{1}(x), \mathscr{\mathscr { J }}_{+}^{2}(y)\right\}=\frac{1}{2}\left[\mathfrak{C}, \mathscr{J}_{+}^{1}(x)-\mathscr{\mathscr { J }}_{+}^2(x)\right] \delta(x-y) \pm \frac{k}{2 \pi} \mathfrak{C} \delta^{\prime}(x-y).$$

My question is, how does one show this? An important step seems to be the identification between the Kronecker and Dirac deltas as explained in Where does the delta of zero $\delta(0)$ come from? , $$\frac{1}{\Delta}\delta_{mn}\rightarrow \delta(x-y).$$

But it seems that we also need to discretize the derivative of the Dirac delta, $$\delta'(x-y)$$, and it ought to be expressible in terms of $$\Delta^2\delta_{m+1,n}$$. How does one do this?

• One way you can discretize a Dirac delta derivative is by applying a discretized derivative to the discretized Dirac delta. Depending on exactly how you discretize the derivative, you would get something like $\Delta^{-2} \left(-\delta_{m, n-1} + \delta_{m, n+1}\right)$. Commented May 3, 2022 at 14:28

Let $$\varphi$$ be a smooth compactly supported function. Then : \begin{align} \int \varphi(x)\delta'(x)\text dx &= -\varphi'(0) \\ &=\lim_{\Delta\to 0} -\frac{1}{\Delta}(\varphi(\Delta)-\varphi(0))\\ &=\lim_{\Delta\to 0} \Delta\sum_n \frac{1}{\Delta^2} (-\delta_{n,1}+\delta_{n,0})\varphi(n\Delta) \end{align}
Therefore, a correct discretization for $$\delta'(x)$$ is $$\frac{1}{\Delta^2}(\delta_{n,0} - \delta_{n,1})$$. For $$\delta'(x-y)$$, we get $$\frac 1{\Delta^2}(\delta_{n,m} - \delta_{n,m+1})$$