Kac-Moody algebra from WZW model via Poisson brackets

In 'Non-abelian Bosonization in Two Dimensions', Witten shows that the Poisson brackets of the currents that generate the $$G\times G$$ symmetry of the WZW model give rise to a Kac-Moody algebra upon canonical quantization.

The Poisson brackets are calculated on page 465 to be \begin{equation} \begin{aligned} \left[X,Y\right]_{PB}&=-\frac{\hbar}{2}\delta'(\sigma-\sigma')\textrm{Tr}~g^{-1}(\sigma)Ag(\sigma)g^{-1}(\sigma')Bg(\sigma') \\&=-\frac{\hbar}{2}\delta(\sigma-\sigma')\textrm{Tr}\bigg([A,B]\frac{\partial g}{\partial\sigma}g^{-1}\bigg)-\frac{\hbar}{2}\delta'(\sigma-\sigma')\textrm{Tr}~AB. \end{aligned}\tag{29} \end{equation} for $$X=\textrm{Tr} A \frac{\partial g}{\partial \sigma}g^{-1}(\sigma)$$ and $$Y=\textrm{Tr} B \frac{\partial g}{\partial \sigma'}g^{-1}(\sigma')$$, where $$g$$ is a map $$g:\mathbb{R}\rightarrow G$$, and where $$A$$ and $$B$$ are arbitrary generators of $$G$$.

My question is, how does one go from the first line to the second line?

In my attempts so far, I am unable to obtain the second term of the second line. To be precise, starting from the first line above, writing $$\delta'(\sigma-\sigma')$$ as $$\frac{\partial}{\partial \sigma}\delta(\sigma-\sigma')$$, and using partial differentiation (and dropping the boundary term), one arrives at \begin{equation} \begin{aligned} \left[X,Y\right]_{PB}=&-\frac{\hbar}{2}\delta(\sigma-\sigma')\textrm{Tr}~g^{-1}(\sigma)\frac{\partial g}{\partial\sigma}(\sigma)g^{-1}(\sigma)Ag(\sigma)g^{-1}(\sigma')Bg(\sigma')\\&+\frac{\hbar}{2}\delta(\sigma-\sigma')\textrm{Tr}~g^{-1}(\sigma)A\frac{\partial g}{\partial\sigma}(\sigma)g^{-1}(\sigma')Bg(\sigma')\\=&-\frac{\hbar}{2}\delta(\sigma-\sigma')\textrm{Tr}\bigg([A,B]\frac{\partial g}{\partial\sigma}(\sigma)g^{-1}(\sigma)\bigg), \end{aligned} \end{equation} which is only the first term of the expression we want. It seems like allowing $$A$$ and $$B$$ to have dependence on $$\sigma$$ may give us the remaining term, but this is not correct, as Witten states explicitly that $$A$$ and $$B$$ are matrices on page 462.

I think he may be using the distributional identity $$f(x)\delta'(x-a)=f(a)\delta'(x-a) -f'(a)\delta(x-a)$$ which comes from differentiating $$f(x)\delta(x-a)=f(a)\delta(x-a)$$ with respect to $$x$$.

• This doesn't really make sense from the pov of distribution theory. Your second identity is more formally denoted by $\delta_a[f]=\delta_a[c_{f(a)}]$, where $c_y$ denotes the constant function $c_y(x)=y$. This identity is correct. But just because $\delta_a$ agrees on the functions $f$ and $c_{f(a)}$ does not mean that $\delta'_a$ must agree on those functions too (just like $f(a)=g(a)$ at some $a$ does not imply that $f'(a)=g'(a)$). So you cannot just "differentiate" your identity (it does hold for functions such that $f'(a)=0$, but not generally). Something is fishy about Witten's formula... – AccidentalFourierTransform Feb 21 at 19:46
• @Accidental . Mike's identity certainly looks fine integrated with a test function g(x), and also fine utilizing the narrow Gaussian limit for the δ - fctn, no? – Cosmas Zachos Feb 21 at 20:48
• @CosmasZachos Eh, maybe you're right. I didn't think too much before I posted my previous comment. On a second thought, the identity does seem correct... – AccidentalFourierTransform Feb 21 at 21:38
• Yes. It's a standard identity. You do need to use test functions for a proper proof though. – mike stone Feb 22 at 13:46

When you did the integration by parts you neglect the fact that this expression should be thought as being integrated together with a test function $$f(\sigma)$$ (or a test function $$f(\sigma')$$), so you need to be careful about throwing away the total derivatives since this derivatives will act on the test function as well.

Suppose you have the distribution $$\delta'(x)g(x)$$, then acting with a test function gives

$$\int dx\,\delta'(x)g(x)f(x) = - \int dx\,\delta(x)(g(x)f(x))'= - \int dx\,\delta(x)(g'(x)f(x)+g(x)f'(x))$$

so you have $$\delta'(x)g(x)\rightarrow -\delta(x)g'(x)-\delta(x)g(x)\frac{d}{dx}$$. The distribution that does this is

$$-\delta(x)g'(0)+\delta'(x)g(0)$$

as you may check by applying a test function on it.