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The formal delta function is $\,\,\displaystyle\delta(x):=\sum_{n\in\mathbb Z}x^n. $
If we agree that expressions $(x+y)^n$ for $n\in\mathbb Z$ are always expanded in non-negative powers of the second variable: $\,\,\displaystyle (x+y)^n:=\sum_{k\in\mathbb N}(\begin{smallmatrix} n\\ k\end {smallmatrix})x^{n-k}y^k $,
then we have the following well known formulas:

$$ z^{-1}\delta\Big(\frac{y-x}z\Big) \,\,\,=\,\,\, y^{-1}\delta\Big(\frac{z+x}y\Big), $$ $$ x^{-1}\delta\Big(\frac{y-z}x\Big)+ x^{-1}\delta\Big(\frac{z-y}{x}\Big) \,\,\,=\,\,\, z^{-1}\delta\Big(\frac{y-x}{z}\Big) $$

which are important in the theory of vertex algebras.

Q1: Is there a way of proving these formulas without writing down lots of binomial coefficients?

Added later:
Q2: Are there identities for the non-formal delta function (an element of the dual of some appropriate function space) that parallel and imply the above two formulas?

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I don't think there's many binomial coefficients in the expansion as in Frenkel-Lepowsky-Meurman's book (Prop 8.8.5). However, you can use the formal Taylor expansion [(8.8.10) in FLM] $e^{-x \partial y} \delta(y/z) = \delta\left( \frac{y-x}{z} \right)$

from where your first identity becomes

(*) $\frac{y}{z} e^{-x \partial_y} \delta(y/z) = e^{x \partial_z} \delta(z/y)$

Now note that $\delta(y/z) = \delta(z/y)$ and both sides are power series in $x$. At $x = 0$ the RHS is simply $\delta(z/y)$ while the LHS is $\frac{y}{z} \delta(y/z) = \delta(y/z)$ by the first property of the $\delta$ function.

Taking derivatives with respect to $x$ on both sides $n$ times and putting $x=0$ you simply get $\partial^n_t \frac{y^{n+1}}{z^{n+1}} \delta (y/z) = \delta^{(n)} (z/y)$ where $t=y/z$ which is true for the same reasons.

To clean up a little and not use the issue of differentiating several times let us write the identity (*) more as the one you wrote (commuting some variables past its derivatives) $e^{-x\partial_y} \frac{1}{z} \delta(y/z) = e^{x \partial_z} \frac{1}{y} \delta(z/y)$

Now note that we have the straightforward identities

$\frac{1}{z} \delta(y/z) = \frac{1}{y} \delta(y/z) = \frac{1}{y} \delta(z/y) = \frac{1}{z} \delta(z/y)$

And by direct computation (no binomial coefficients)

$\partial_y \frac{1}{z} \delta(y/z) = \sum_{n \in \mathbb{Z}} n y^{n-1} z^{-n-1} = - \sum_{n \in \mathbb{Z}} n y^{-n-1} z^{n-1}= -\partial_z \frac{1}{y} \delta(z/y)$

Finally passing $e^{x \partial_z}$ to the LHS your identity reads

$ e^{-x (\partial_y + \partial_z)} \frac{1}{z} \delta(y/z) = \frac{1}{y} \delta(z/y)$

and by the identity we just proved we have that $\partial_y + \partial_z$ acts by zero, hence the LHS is simply $\frac{1}{z} \delta(y/z)$ which trivially equals the RHS.

I am myself more used to the notation for $\delta(x-y) := \sum_{n \in \mathbb{Z}} x^n y^{-1-n}$ in which case the analogous formulas can be proved by using arguments as in Frenkel-Ben-Zvi's book by noting that $\delta$ is the difference of the expansions of $(x-y)^{-1}$ in different fields of the form $k( (x))((y))$ and $k((y))((x))$. From where identities like $(x-y) \delta(x-y)=0$ and $(\partial_x + \partial_y) \delta(x-y) = 0$ follow trivially by noting that both expansions of $1 = (x-y)/(x-y)$ agree.

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  • $\begingroup$ Ooops, thanks Scott, edited accordingly. $\endgroup$ – Reimundo Heluani Nov 1 '14 at 8:55
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A comment (to the first part of Q.2) disguised as an answer since the system doesn't let me. The equation at the start of the question can be given a precise verification (rather than being purely formal). This is done in terms of Köthe's theory of distributions on the unit circle as generalised boundary values of analytic functions on the complement of the latter in the complex plane. In the case of the delta function (nota bene concentrated at the point $1$, not at zero as is usually intended by the notation $\delta$), the corresponding functions are ${\dfrac{1}{1-x}}$ in the interior of the unit circle and ${\dfrac{1}{x-1}}$ on the exterior. The infinite series in the question is obtained by combining the Taylor series of the former with the Laurent series of the latter. Whether this can be used to answer the query is another matter. Reference: Köthe, Math. Z. 57 (1952) 13-33.

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  • $\begingroup$ Yes, this is interesting. Do you have anything similar to say concerning the second (more interesting) equation $ x^{-1}\delta\Big(\frac{y-z}x\Big)+ x^{-1}\delta\Big(\frac{z-y}{x}\Big) \,\,\,=\,\,\, z^{-1}\delta\Big(\frac{y-x}{z}\Big) $? $\endgroup$ – André Henriques Nov 9 '14 at 22:42
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Your second formula seems incorrect. The correct statement is:

$$x^{-1} \delta \left(\frac{y-z}{x}\right)-x^{-1}\delta\left(\frac{z-y}{-x}\right)=y^{-1} \delta \left(\frac{z+x}{y}\right).$$

Observe also that your first formula follows from the second. This is because the left hand-side of the (corrected) second formula is invariant under $y \leftrightarrow z$ and $x \leftrightarrow -x$.

For the second identity, you only have to recall that $$\frac{1}{z-y}+\frac{1}{y-z}$$ (expanded by using your conventions) is the delta function. Then you differentiate the formula $n$-times, write the $x$-generating series over $n$, and finally apply the formal Taylor theorem (see Reimundo's answer). This proof doesn't require "lots of binomial coefficients".

Here's my favorite one-line "proof" of the second identity: one-dimensional vector space $\mathbb{C}$ (spanned by the vacuum) has a vertex algebra structure.

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  • $\begingroup$ You differentiate the formula n-times with respect to which variable? $\endgroup$ – André Henriques Jan 17 '15 at 13:07
  • $\begingroup$ It doesn't matter matter, you can take $z$. If you go with $y$ your final answer will be the LHS in formula (1). $\endgroup$ – user45130 Jan 17 '15 at 13:16
  • $\begingroup$ Observe also that the LHS in formula (2) has no negative powers of $x$ (because of $(y-z)^n-(-1)^n(z-y)^n=0$ for $n \in \mathbb{N}$). $\endgroup$ – user45130 Jan 17 '15 at 13:18
  • $\begingroup$ I see. I had taken my formula from eq. (2.7) on page 48 of the paper arxiv.org/abs/1012.4193 without actually checking it. Do you have anything to say concerning my second question: is there an identity for the non-formal delta function that somehow mirrors the above formal identity? $\endgroup$ – André Henriques Jan 17 '15 at 23:03
  • $\begingroup$ I'm not sure. If viewed as non-formal delta functions $\delta((y-z)/x)=\delta((z-y)/(-x))$, so it seems unlikely. $\endgroup$ – user45130 Jan 17 '15 at 23:38

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