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 $x$.

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  • $\begingroup$ 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... $\endgroup$ – AccidentalFourierTransform Feb 21 '19 at 19:46
  • $\begingroup$ @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? $\endgroup$ – Cosmas Zachos Feb 21 '19 at 20:48
  • $\begingroup$ @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... $\endgroup$ – AccidentalFourierTransform Feb 21 '19 at 21:38
  • $\begingroup$ Yes. It's a standard identity. You do need to use test functions for a proper proof though. $\endgroup$ – mike stone Feb 22 '19 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.

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