Twisted Vertex Operator Construction of Affine sl2

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This is the second part in a two part series.

Over at our informal SeanForScience YouTube channel, we’ve been running a technical series on Vertex Operator Algebras. It mostly follows the Frenkel, Lewposky, Meurman text, although I’ve included some extra physics material for context.

After months of producing episodes, we’re about to publish our first actual representation of an affine Lie algebra using Twisted Vertex Operators. Because of that, I included a whole bunch of review material - mostly in the form of exercises - just to remind everybody how most the structures involved work.

The twisted vertex operator construction of \(\hat{\mathfrak{sl}_{2}}\)

Our main result amounts to realizing \(\sigma_{\pm}(z)\) in terms of the twisted vertex operators \(X(\pm \sigma, z)\). In particular,

Proposition:

The twisted vertex operators form a representation of \(\hat{\mathfrak{sl}}_{2}[\sigma_{1}]\),

$$ [\sigma(r),X(\pm\sigma,z)] = \pm 2 z^{r} X(\pm\sigma,z),$$

$$[X(\sigma,z) , X(-\sigma ,w)] = \frac{1}{2}\sigma(w)\delta(z^{1/2}/w^{1/2}) - \frac{1}{2} D_{z}\delta(z^{1/2}/w^{1/2}$$

Some preliminaries

We begin with a few simple computations with formal operators.

Lemma 1:

Let \(X\) and \(Y\) be operators whose Lie bracket with \([X,Y]\) vanishes, then

$$e^{X}e^{Y} = e^{Y}e^{X} e^{[X,Y]}.$$

Proof:

Recall the Baker-Campbell-Hausdorff formula for such operators,

$$e^{X}e^{Y} = e^{X + Y + \frac{1}{2}[X,Y]}.$$

And compute as an exercise. \(\blacksquare\)

Lemma 2:

$$\sum_{r\in\mathcal{Z}_{+}} \frac{1}{n}X^{n} = -\log\left(\frac{1 - X^{1/2}}{1+X^{1/2}}\right).$$

Proof:

We start with the right hand side,

$$ \log\left(\frac{1 - X^{1/2}}{1+X^{1/2}}\right) = \log(1 - X^{1/2}) - log(1 + X^{1/2}),$$

and recall the formal expansion of the logarithm,

$$\log(1 - aX) = - \sum_{n\geq 1} \frac{a^{n}}{n}X^{n}.$$

This gives us

$$ \log\left(\frac{1 - X^{1/2}}{1+X^{1/2}}\right) = - \sum_{n\geq1} \left(1 - (-1)^{n}\right)\frac{1}{n}X^{n/2} $$

The even terms cancel,

$$ \log\left(\frac{1 - X^{1/2}}{1+X^{1/2}}\right) = -\sum_{n\geq1,\, \mathrm{odd}} \frac{2}{n}X^{n/2},$$

which is nothing more than

$$-\sum_{r\in\mathcal{Z}_{+}} \frac{1}{n}X^{n}. \quad\blacksquare$$

Lemma 3:

$$\left(\frac{1+X}{1-X}\right)^{2} = 1 + 4\sum_{n \geq 1} nX^{n}$$

Proof:

As operators the numerator and denominator commute, so we can expand,

$$\left(\frac{1+X}{1-X}\right) = (1 + X)\sum_{n\geq 0} X^{n}.$$

We now compute

$$\left(\frac{1+X}{1-X}\right) = \sum_{n\geq 0} \left(X^{n} + X^{n+1}\right).$$

Removing the zeroth term and shifting the second index,

$$\left(\frac{1+X}{1-X}\right) = 1 + \sum_{n\geq 1} \left(X^{n} + X^{n}\right) = 1 + 2\sum_{n\geq1}X^{n}.$$

Squaring, we find,

$$\left(\frac{1+X}{1-X}\right)^{2} = (1 + 2\sum_{n\geq1}X^{n})^2,$$

which expands to

$$\left(\frac{1+X}{1-X}\right)^{2} = 1 + 4\sum_{n\geq1}X^{n} + 4\sum_{m\geq 1}\sum_{n\geq1} X^{n+m}.$$

We can resum the double sum over terms of fixed \(\ell = n+m \),

$$\sum_{m\geq 1}\sum_{n\geq1} X^{n+m} = \sum_{\ell\geq1} \sum_{n=1}^{\ell -1} X^{\ell} = \sum_{\ell \geq 1}(\ell -1)X^{\ell}.$$

We therefore have

$$\left(\frac{1+X}{1-X}\right)^{2} = 1 + 4\sum_{n\geq1}X^{n} + 4\sum_{m\geq 1} (m-1)X^{m},$$

which simplifies to the desired result. \(\blacksquare\)

We now employ these two lemmas to prove a result about the operators \(E^{\pm}(\alpha,z)\)introduced in past discussions. Recall that

$$E^{\pm}(\alpha,r) = \exp\left( \sum_{r \in Z_{\pm}} \frac{\alpha(r)}{r}z^{-r}\right),$$

where \(\alpha\) is odd under the \(\theta\) involution, and we have in mind terms analogous to the Cartan subalgebra, which for \(\mathfrak{sl}_{2}\) unfortunately only includes terms in \(\mathbb{F}\sigma\). A generalization of this result will eventually apply.

Lemma 4: (FLM Prop 3.4.1)

$$E^{+}(\alpha,z)E^{-}(\beta,w) = E^{-}(\beta,w)E^{+}(\alpha,z) \left(\frac{1-w^{1/2}/z^{1/2}}{1+w^{1/2}/z^{1/2}}\right)^{\langle \alpha , \beta \rangle}$$

Proof:

Consider the commutator

$$\left[ \sum_{r\in\mathcal{Z}_{+}} \frac{\alpha(r)}{r}z^{-r} , \sum_{s\in\mathcal{Z}_{-}} \frac{\beta(s)}{r}w^{-s}\right].$$

Since \(\alpha,\beta \in \mathbb{F}\sigma\), only the centrally extended terms remain,

$$\left[ \sum_{r\in\mathcal{Z}_{+}} \frac{\alpha(r)}{r}z^{-r} , \sum_{s\in\mathcal{Z}_{-}} \frac{\beta(s)}{r}w^{-s}\right] = -c\langle \alpha , \beta \rangle \sum_{r\in\mathcal{Z}_{+}} \frac{1}{n}\left(\frac{w}{z}\right)^{n}$$

By Lemma 2 we have

$$\left[ \sum_{r\in\mathcal{Z}_{+}} \frac{\alpha(r)}{r}z^{-r} , \sum_{s\in\mathcal{Z}_{-}} \frac{\beta(s)}{r}w^{-s}\right] = \langle \alpha , \beta \rangle \log\left(\frac{1 - (w/z)^{1/2}}{1 + (w/z)^{1/2}}\right),$$

which is

$$\left[ \sum_{r\in\mathcal{Z}_{+}} \frac{\alpha(r)}{r}z^{-r} , \sum_{s\in\mathcal{Z}_{-}} \frac{\beta(s)}{r}w^{-s}\right] = \log\left(\frac{1 - (w/z)^{1/2}}{1 + (w/z)^{1/2}}\right)^{\langle \alpha , \beta \rangle}.$$

Appealing to Lemma 1, we find the desired result. \(\blacksquare\)

Corollary (of Lemma 4):

$$:X(\alpha, z )X(\beta, z ): \;= X(\alpha, z )X(\beta, z ) \left(\frac{1-w^{1/2}/z^{1/2}}{1+w^{1/2}/z^{1/2}}\right)^{\langle \alpha , \beta \rangle} $$

Proof:

Recall that the normally ordered product moves all negative powers of the formal variables to the left, and all positive powers of formal variables to the right. This was done as per our earlier discussion amount ensuring the various infinite combinations of operators were well-defined. Recalling that and the definition of the twisted vertex operator for \(\mathfrak{sl}_{2}\),

\begin{equation}\label{voam039:X}X(\alpha,z) = \frac{E^{-}(-\alpha,z)E^{+}(-\alpha,z)}{2^{\langle \alpha,\alpha \rangle}},\end{equation}

we find,

$$:X(\alpha, z )X(\beta, w ): = \frac{1}{2^{\langle\alpha,\alpha\rangle + \langle\beta,\beta\rangle}}E^{-}(-\alpha,z)E^{-}(-\beta,w)E^{+}(-\alpha,z)E^{+}(-\beta,w).$$

Issues of convergence and summability aside, the standard product of twisted vertex operators would be given by

$$:X(\alpha, z )X(\beta, w ): = \frac{1}{2^{\langle\alpha,\alpha\rangle + \langle\beta,\beta\rangle}}E^{-}(-\alpha,z)E^{+}(-\alpha,z)E^{-}(-\beta,w)E^{+}(-\beta,w).$$

So we see the difference amounts to the order of the middle two operators. Applying the proposition gives the desired result. \(\blacksquare\)

Proof of the Proposition

We now state the proof Proposition 1.

Proof:

Recall from earlier discussions that the twisted vertex operator is, up to an overall scalar, the unique solution to the operator equations,

$$[h(m) , U(\alpha,z)] = \langle h,\alpha \rangle z^{m} U(\alpha,z),\quad [d,U(\alpha,z)] = -DU(\alpha,z).$$

For \(\alpha = \sigma\) and \(h(r) = \sigma(r)\), - recalling that \(\langle \sigma , \sigma \rangle = 2\) - these are just the affine Lie algebra brackets,

$$[\sigma(r) , \sigma_{\pm}(z)] = 2 z^{r}\sigma_{\pm}(z). $$

Thus, by their very construction, the twisted vertex operators \(X(\sigma,z)\) satisfy that particular bracket. It remains to show that

$$[\sigma_{+}(z),\sigma_{-}(w)] \sim [X(\sigma, z), X(-\sigma,w)].$$

To that end, let us compute,

$$[X(\sigma, z), X(-\sigma,w)] = X(\sigma, z) X(-\sigma,w) - X(-\sigma, w) X(\sigma,z),$$

which we can now express in terms of normally ordered operators thanks to the Corollary,

$$[X(\sigma, z), X(-\sigma,w)] = :X(\sigma, z), X(-\sigma,w): \left( \left(\frac{1-w^{1/2}/z^{1/2}}{1+w^{1/2}/z^{1/2}}\right)^{-2} - \left(\frac{1-z^{1/2}/w^{1/2}}{1+z^{1/2}/w^{1/2}}\right)^{-2} \right).$$

Appealing to Lemma 3, we find a simplification,

$$[X(\sigma, z), X(-\sigma,w)] = 4 :X(\sigma, z), X(-\sigma,w): \left( \sum_{n\geq1} n\left(\frac{z}{w}\right)^{n/2} - \sum_{n\geq1} n\left(\frac{w}{z}\right)^{n/2} \right),$$

which neatly compresses to

$$[X(\sigma, z), X(-\sigma,w)] = 4 :X(\sigma, z), X(-\sigma,w): \sum_{n\in\mathbb{Z}} n\left(\frac{w}{z}\right)^{n/2}.$$

Given that

$$X(\sigma,z) = :X(\sigma,z): = \frac{1}{4}:e^{D^{-1}_{z} \sigma(z)}:,$$

where the inverse degree derivation is well-defined on formal sums over \(\mathcal{Z}\), we have

$$[X(\sigma, z), X(-\sigma,w)] = \frac{1}{4}:e^{D^{-1}_{z}\sigma(z) - D^{-1}_{w}\sigma(w)}: \sum_{n\in\mathbb{Z}} n\left(\frac{w}{z}\right)^{n/2} ,$$

where finally the sum may be expressed in terms of the delta function

$$[X(\sigma, z), X(-\sigma,w)] = \frac{1}{4}:e^{D^{-1}_{z}\sigma(z) - D^{-1}_{w}\sigma(w)}: 2D\delta\left((w/z)^{1/2}\right)$$

Now, in lecture 29, we explored how the delta function impacts expressions like these, notably, we proved a result:

$$W(w,z) D\delta(w/z) = W(w,w) D\delta(w/z) + (D_{z}W)(w,w)\delta(w/z).$$

This results holds only for integral powers of $z$ and $w$, so we need to re-express things in terms of the variables

$$x = z^{2},\quad y = w^{2},$$

to apply it. Careful treatment of $D$ through the change of variables gives us the expression

$$[X(\sigma, z), X(-\sigma,w)] = \frac{1}{4}:e^{2D^{-1}_{y}\sigma(y^2) - 2D^{-1}_{x}\sigma(x^{2})}: D\delta(x/y)$$

To that end, let

$$W(y,x) = : \exp\left( 2D^{-1}_{x}\sigma(x^{2}) - 2D^{-1}_{y}\sigma(y^{2})\right) :, $$

and note that

$$W(x,x) = 1.$$

We also have

$$D_{x}W(y,x) = :D_{x}\left(2D^{-1}_{x}\sigma(x^2)\right)\exp\left( 2D^{-1}_{x}\sigma(x^{2}) - 2D^{-1}_{y}\sigma(y^{2})\right):,$$

so that

$$D_{x}W(y,x) = 2:\sigma(x^2)\exp\left( 2D^{-1}_{x}\sigma(x^{2}) - 2D^{-1}_{y}\sigma(y^{2})\right):.$$

Therefore,

$$W(y,y) = 2\sigma(y^{2}).$$

Putting all this together, we find,

$$ W(y,x) D\delta(x/y) = D\delta(x/y) + 2\sigma(y^{2})\delta(x/y)$$

This then gives us

$$[X(\sigma, z), X(-\sigma,w)] = \frac{1}{4}\left( D\delta(x/y) + 2\sigma(y^{2})\delta(x/y) \right)$$

In terms of the original variables $z$ and $w$ we find,

$$ [X(\sigma, z), X(-\sigma,w)] = \frac{1}{2} D_{z}\delta(w^{1/2}/z^{1/2}) + \frac{1}{2}\sigma(w)\delta(w^{1/2}/z^{1/2}),$$

or

$$ [X(\sigma, z), X(-\sigma,w)] = \frac{1}{2}\sigma(w)\delta(z^{1/2}/w^{1/2}) - \frac{1}{2} D_{z}\delta(z^{1/2}/w^{1/2}).\quad \blacksquare$$