# A mysterious Heisenberg algebra identity from Sylvester, 1867

I am trying to understand two papers by James Joseph Sylvester:

P92: "Note on the properties of the test operators which occur in the calculus of invariants, their derivatives, analogues, and laws of combination; with an incidental application to the development in a Maclaurinian series of any power of the logarithm of an augmented variable."

and

P95: "On the multiplication of partial differential operators."

[The numbering is from volume 2 of Sylvester's Collected Works. These, incidentally, are spread over four volumes: volume 1, volume 2, volume 3, volume 4 (first two courtesy of anonymous book scanners), with some duplicates on the Internet Archive. All of them are out of copyright.]

On the first page of P95 (aka page 11 of the linked two-paper PDF), Sylvester states that

"If $$\phi$$ be any such function [i. e., a polynomial or power series in infinitely many commuting variables $$x$$, $$y$$, $$z$$, ..., $$\delta_x$$, $$\delta_y$$, $$\delta_z$$, ... (here, $$\delta_x$$, $$\delta_y$$, $$\delta_z$$, ... are just symbols, not differential operators!) which is multilinear with respect to $$\left(\delta_x,\delta_y,\delta_z,...\right)$$], [we have]

$$e^{\displaystyle t\phi\star} = \left[e^{\displaystyle \left(e^{\displaystyle t\phi\star}-1\right)\phi}\right]\star$$."

Here, as far as I understand, the $$\star$$ operation is defined as follows (see page 1 of P92, aka page 1 of the linked PDF): If $$\psi$$ is any polynomial or power series in infinitely many variables $$x$$, $$y$$, $$z$$, ..., $$\delta_x$$, $$\delta_y$$, $$\delta_z$$, ..., then $$\psi\star$$ means the differential operator we obtain if we collect all the $$\delta_x$$, $$\delta_y$$, $$\delta_z$$, ... variables at the right end of every monomial and replace them by the partial derivative operators $$\frac{\delta}{\delta x}$$, $$\frac{\delta}{\delta y}$$, $$\frac{\delta}{\delta z}$$, .... I cannot say that I am sure about this, though, because no matter how I try to obtain a small, verifiable example for the formula, I get some nonsense which is either wrong or I am not able to check.

Sylvester studied these in the context of classical invariant theory, but nowadays quantum field theorists are interested in these differential operators as elements of the Heisenberg algebra. Is there any modern (readable) reformulation of the above identity? Has anyone else tried to comprehend its meaning? Is it related to the identity $$\left(\exp a\right)\left(\exp b\right)\left(\exp a\right)^{-1} = \exp\left(\left(\exp\left(\mathrm{ad} a\right)\right)\left(b\right)\right)$$ which holds for any two elements $$a$$ and $$b$$ of a ring for which these exponentials make sense? (This is speculation based on nothing more than the appearance of nested exponentials in both identities.)

• I don't have it with me now but it seems to me that the book of Ben-Zvi and Frenkel on vertex algebras might help you. – DamienC Jul 15 '12 at 11:53
• Hmm, thanks. This is certainly related to the normally ordered product in the universal enveloping algebra of the Heisenberg algebra; but I don't yet see much connection to vertex algebras. (That said, I barely know the definition of the latter.) – darij grinberg Jul 15 '12 at 12:48
• Interesting work by Sylvester. On pages 570 and 571 of the first paper he explains the roles of Hermite and Cayley in the birth of highest weight theory. His $E_1$ and $H_1$ are the modern $e$ and $f$ for $\mathfrak{s}\mathfrak{l}_2$, acting on an irreducible representation. – Abdelmalek Abdesselam Apr 2 '15 at 22:47
• @AbdelmalekAbdesselam: Nice! How much of the surrounding text do you understand? I hate to say, but I have made about no progress on this 1866 paper since 2012. – darij grinberg Apr 3 '15 at 0:13
• @Darij: Sylvester's classical invariant theory for studying these differential operators is about irreducible reps of $SL(V)$ with $V$ two-dimensional INSIDE $S^m(S^n(V))$. These are identified via highest wieght vectors, i.e., objects annihilated by his $E_1$ or $H_1$ depending on one's convention. This is explained in Section 2 of my paper arxiv.org/abs/1203.4761 – Abdelmalek Abdesselam Apr 3 '15 at 14:49

The following instance of your identity is well known in physics (and is sometimes called an "operator disentangling" identity)

$$:\exp\left[\left(e^W-1\right)_{ij}a_i^\dagger a_j\right]:\;=\exp\left(W_{ij}a^\dagger_i a_j\right)$$

where $$::$$ denotes normal ordering, $$a_i$$ and $$a^\dagger_j$$ are canonical Bose annhilation and creation operators satisfying $$\left[a_i,a^\dagger_j\right]=\delta_{ij}$$, and W is an arbitrary matrix (summation implied).

For general (rather than just quadratic) $$\phi$$ the formula is completely new (indeed remarkable) to me.

Frustratingly, it's hard to track down the origins of the above formula. Here's a recent discussion that includes the above version for a single boson mode (Eq. 30):

Combinatorics and Boson normal ordering: A gentle introduction American Journal of Physics, 75 (7), pp. 639 (2007)

The authors' comments after Eq. 30 seem to imply that the formula doesn't generalize simply.

EDIT: I realized that Sylvester initially states the quadratic form above, and then limits his generalization to functions $$\phi$$ "linear quantic in $$\delta_x$$, $$\delta_y$$, $$\delta_z$$,...". Still, this generalization appears to contradict Eq. 31 of the above article.

• According to Harold Davis in The Theory of Linear Operators (Principia Press, 1936, pg. 199), the commutator [q,p]=1 "was apparently first studied by Charles Graves as early as 1857." Davis goes on to use the commutator to get some "normal ordering" results obtained by Graves and to expand on them. – Tom Copeland Mar 7 '14 at 10:32
• Very interesting! Graves essentially had the form of the star product. – Austen Mar 7 '14 at 16:08
• You were referring to the formulas on p.194 of Davis and the star product as discussed by Coutinho on p. 602 in The Many Avatars of a Simple Algebra? (maa.org/programs/maa-awards/writing-awards/…) – Tom Copeland Feb 27 '16 at 21:44

Thanks, Austen; I didn't think a particular case could be that strong. (I can't say it's an immediate particular case, though. It took me some transformations to get the $$e^W-1$$ term.)

Meanwhile I have understood the claim (and found a proof; more about it later today or tomorrow).

The formula I quoted above is slightly wrong, at least as I understand it. It should be $$e^{\displaystyle t\phi\star} = \left[e^{\displaystyle \left(\left(e^{\displaystyle t\phi\star}-1\right) / \left(\phi\star\right)\right) \phi}\right]\star$$, where $$\left(e^{\displaystyle t\phi\star}-1\right) / \left(\phi\star\right)$$ is to be understood "formally" (as in, "plug in $$\phi\star$$ as $$x$$ in the power series $$\left(e^{tx}-1\right)/x$$"). Sylvester gets this right in his first paper (P92) even though he misses an assumption (multilinearity with respect to the $$\delta$$ variables) there. Apparently, the wrong formula is due to an off-by-$$1$$ error.

Let me rewrite the correct formula in some more modern notations.

Theorem 1 (Sylvester). Let $$K$$ be a commutative ring.

Let $$K\left[a,b,c,...\right]$$ be the ring of polynomials in the commuting indeterminates $$a$$, $$b$$, $$c$$, ..., and let $$K\left[\delta_a,\delta_b,\delta_c,...\right]$$ be the ring of polynomials in the commuting indeterminates $$\delta_a$$, $$\delta_b$$, $$\delta_c$$, ... (which, as for now, have nothing to do with $$a$$, $$b$$, $$c$$, ... except being similarly labelled). Let $$\mathrm{Diff}\left(a,b,c,...\right)$$ be the ring of polynomial differential operators on $$K\left[a,b,c,...\right]$$. Then, we can define a $$K$$-module isomorphism

$$M : K\left[a,b,c,...\right] \otimes K\left[\delta_a,\delta_b,\delta_c,...\right] \to \mathrm{Diff}\left(a,b,c,...\right),$$

$$P \otimes Q \mapsto P \cdot Q\left(\dfrac{\partial}{\partial a},\dfrac{\partial}{\partial b},\dfrac{\partial}{\partial c},...\right)$$.

(Only $$Q$$, not $$P\cdot Q$$, is being evaluated at $$\left(\dfrac{\partial}{\partial a},\dfrac{\partial}{\partial b},\dfrac{\partial}{\partial c},...\right)$$ here.)

Let $$\left(K\left[\delta_a,\delta_b,\delta_c,...\right]\right)_1$$ be the $$K$$-submodule of $$K\left[\delta_a,\delta_b,\delta_c,...\right]$$ spanned by $$\delta_a$$, $$\delta_b$$, $$\delta_c$$, ... (that is, the degree-$$1$$ part of $$K\left[\delta_a,\delta_b,\delta_c,...\right]$$).

The ring $$\mathrm{Diff}\left(a,b,c,...\right)$$ acts on the tensor product $$K\left[a,b,c,...\right] \otimes K\left[\delta_a,\delta_b,\delta_c,...\right]$$ by acting on the first tensorand only. Denote this action by $$\rightharpoonup$$. In other words, for any differential operator $$R\in \mathrm{Diff}\left(a,b,c,...\right)$$, any $$P\in K\left[a,b,c,...\right]$$ and any $$d\in K\left[\delta_a,\delta_b,\delta_c,...\right]$$, set $$R\rightharpoonup \left(P\otimes d\right) = R\left(P\right)\otimes d$$, and extend this by linearity to an action of $$\mathrm{Diff}\left(a,b,c,...\right)$$ on the whole tensor product.

Let $$D \in K\left[a,b,c,...\right] \otimes \left(K\left[\delta_a,\delta_b,\delta_c,...\right]\right)_1$$ be arbitrary. Let $$T$$ be the power series

$$\sum\limits_{i\geq 1} \left(M\left(D\right)\right)^{i-1} \rightharpoonup D \dfrac{t^i}{i!} \in \left(K\left[a,b,c,...\right] \otimes K\left[\delta_a,\delta_b,\delta_c,...\right]\right)\left[\left[t\right]\right]$$.

Then, $$M\left(\exp T\right) = \exp\left(tM\left(D\right)\right)$$, where $$M\left(\exp T\right)$$ is shorthand for "the image of $$\exp T$$ under the canonical map $$M : \left(K\left[a,b,c,...\right] \otimes K\left[\delta_a,\delta_b,\delta_c,...\right]\right)\left[\left[t\right]\right] \to \left(\mathrm{Diff}\left(a,b,c,...\right)\right)\left[\left[t\right]\right]$$ induced by $$M$$".

Note that this is closely related to normal ordered products, but I find the $$:...:$$ notation for normal ordered products annoyingly restrictive (as it forces everything between the colons to be normal ordered, but the proof of Theorem 1 needs a normal ordered product with a non-normal ordered product inside). Maybe this is because I don't really understand the subtleties of this notation. I would personally just use some different symbol for the commutative multiplication map on $$\mathrm{Diff}\left(a,b,c,...\right)$$ transferred from $$\left[a,b,c,...\right] \otimes K\left[\delta_a,\delta_b,\delta_c,...\right]$$ by the isomorphism $$M$$. What about $$\boxdot$$?

Theorem 1 is by no means the whole content of the two papers I've linked, and I'd welcome any readable "translations" of the other results. (I'm going to try that myself, too.)

Note that there are two typos in the second equation on page 568 of P92. It says

$$\phi_1\star\phi_1\star\phi_1\star = \left(\phi_1^s + 2 \phi_1\phi_2 + \phi_3\right)\star$$.

First, the $$s$$ should be a $$3$$ (this is probably because whoever made the djvu set the threshold for identity of shapes too liberally; the original paper had it right); second, the $$2$$ coefficient should be a $$3$$.

• So you've ended up with a rep of the Bell partition polynomials oeis.org/A036040 as Sylvester's equation $e^{t\phi _1*}=(e^{T})*$ suggests, with $T=e^{t\phi_{.} }$ umbrally? – Tom Copeland Jul 17 '12 at 14:07
• I interpret Sylvester's $e^{tx\frac{\mathrm{d} }{\mathrm{d} x}*}=\left ( e^{\left ( e^t-1 \right )x\frac{\mathrm{d} }{\mathrm{d} x}} \right )*$ as $e^{tx\frac{\mathrm{d} }{\mathrm{d} x}}=e^{tB_{.}(\widehat{x\frac{\mathrm{d} }{\mathrm{d} x}})}=e^{\left ( e^t-1 \right )\widehat{x\frac{\mathrm{d} }{\mathrm{d} x}}}$ where $(B_{.}(x))^n=B_{n}(x)$ are the regular Bell polynomials and $(\widehat{x\frac{\mathrm{d} }{\mathrm{d} x}})^n=x^{n}\frac{\mathrm{d^n} }{\mathrm{d} x^n}$. – Tom Copeland Jul 17 '12 at 21:47
• Your phrase "by acting on the first tensor only" reminds me of a pre-Lie algebra. – Tom Copeland Mar 4 '14 at 7:10
• Tom, thanks for reviving this thread. Unfortunately I'm lacking time these days, but I hopefully will come back to it. Which links are broken? (NB: I really don't understand umbral notation and anything remotely similar to it -- Young's raising operators, the symbolic method etc.; sorry.) – darij grinberg Mar 5 '14 at 5:50

This can be interpreted cleanly using the notion of Pre-Lie algebra.

Indeed, vector fields on an affine space form a Pre-Lie algebra. To prove the desired identity, it is enough to consider the free Pre-Lie algebra on one generator.

This identity is known to be related to a pre-Lie version of the Baker-Campbell-Hausdorff formula. This is written in several places in the literature.

Here are a few references:

• Agracev, A. A. and Gamkrelidze, R. V., Chronological algebras and nonstationary vector fields

• Dominique Manchon, A short survey on pre-Lie algebras, E. Schrödinger Institut Lectures in Math. Phys., Eur. Math. Soc.

• Pierre Cartier, Vinberg algebras, Lie groups and combinatorics. (English summary) Quanta of maths, 107–126, Clay Math. Proc., 11,

A formula similar to Sylvester's:

$$exp(t\;g(z) D)=exp[:[<exp(t\;g(z) D)z>-z]\; D:]=exp[:<z^{-1}exp(t\;g(z) D)z-1>\;z D:].$$

Derivation:

First, given a fct. $\omega=h(z)$, its compostional inverse $z=h^{-1}(\omega)$, and $g(z)=\frac{1}{h^\prime(z)},$ then

$$exp(t\;g(z) \frac{d}{dz})f(z)=exp(t\; \frac{d}{dh(x)})f(x)=exp(t\; \frac{d}{d\omega})f(h^{-1}(\omega))=f(h^{-1}(t+\omega))=f(h^{-1}(t+h(z))).$$

(This expression for the Lie derivative was studied as early as 1857 by Charles Graves.)

In particular with $f(z)=z,$

$$exp(t\;g(z) \frac{d}{dz})z=h^{-1}(t+h(z)).$$

(Note that with $f(z)=z$, $h(0)=0$, $g(0)=1$ and $t=h(y)$, the operator generates a formal group law (FGL), so the operator can be used to simply explore the properties of the FGL in general. For an easy intro to FGLs, see Olver, Applications of Lie Groups to Differential Equations.)

But also using the generalized shift operator (generalized Taylor-Maclaurin series),

$$exp[:[h^{-1}(t+h(z))-z]\; \frac{d}{dz}:]f(z)=f(h^{-1}(t+h(z))),$$

and, in particular,

$$exp[:[h^{-1}(t+h(z))-z]\; \frac{d}{dz}:]z=h^{-1}(t+h(z)),$$

where $:ABC:^n=A^nB^nC^n,$ i.e., the power distributes over the operators maintaining their order. This notation allows more succinct and suggestive formulas and shouldn't be confused with normal ordering although it can give the same result. In this case, $:[h^{-1}(t+h(z))-z]\; \frac{d}{dz}:^n=[h^{-1}(t+h(z))-z]^n\frac{d^n}{dz^n}.$

So recursively, using $D=d/dz,$

$$exp(t\;g(z) D)f(z)=f(h^{-1}(t+h(z)))=exp[:[h^{-1}(t+h(z))-z]\; D:]f(z)=exp[:[<exp(t\;g(z) D)z>-z]\; D:]f(z),$$

where <...> denotes evaluation within the symbols.

Removing the intervening steps, we obtain a formula similar to Sylvester's:

$$exp(t\;g(z) D)=exp[:[<exp(t\;g(z) D)z>-z]\; D:]=exp[:<z^{-1}exp(t\;g(z) D)z-1>\;z D:].$$

Checks:

A) Let $g(z)=z$, $h(z)=ln(z)$, $h^{-1}(z)=e^z$.

Then

$$z^{-1}h^{-1}(t+h(z))-1=z^{-1}exp(t+ln(z))-1=e^t-1,$$

$$exp(t\;z D)=exp[(e^t -1) :zD:],$$

and

$$exp(t\;z D)f(z)=f((e^t-1)z+z)=f(e^tz).$$

B) With $g(z)=z^2$, $h(z)=-1/z$, $h^{-1}(z)=-1/z$,

$$z^{-1}h^{-1}(t+h(z))-1=\frac{1}{1-t \cdot z}-1,$$

$$exp(t\;z^2 D)=exp[:[\frac{1}{1-t \cdot z}-1] zD:],$$

and

$$exp(t\;z^2 D)f(z)=f([\frac{1}{1-t \cdot z}-1]z+z)=f[\frac{z}{1-t \cdot z}].$$

Explicit Expansion of LHS:

For expansion of the LHS of the formula in terms of rooted trees, see references in OEIS A139605.

Explicitly for the third order term with $g^j_i=[D^i g(z)]^j$,

$$(g(z)D)^3=(g^1_0 g^2_1+g^2_0g^1_2)D+3g^2_0g^1_1 D^2+g^3_0D^3.$$