A linear representation of the group of jets at 0 under composition

Question

Let $G$ be the set of sequences $(f'(0), f''(0), f'''(0), \ldots)$ of derivatives at zero of functions $f : \mathbb{R} \to \mathbb{R}$ with $f(0) = 0$ and $f'(0) \ne 0 $. The set is a group under composition and, setting $a_k = f^{(k)} (0)$, the operation corresponds to multiplying (infinite) triangular matrices of the form

$$ \begin{pmatrix} a_1 & & & & & \\ a_2 & a_1^2 & & & & \\ a_3 & 3 a_1 a_2 & a_1^3 & & & \\ a_4 & 4 a_1 a_3 + 4 a_2^2 & 6 a_1^2 a_2 & a_1^4 & & \\ a_5 & 5 a_1 a_4 + 10 a_2 a_3& 10 a_1^2 a_3 + 15 a_1 a_2^2 & 10 a_1^3 a_2 & a_1^5 \\ & & \vdots & & & \end{pmatrix} $$

Could you provide references for this fact? I am especially interested in the many properties of such matrix, it looks as fascinating as a Pascal triangle.

Answer 1

I think you want to look at Faà di Bruno's formula, and the description of composition of formal power series, particularly the historical remarks.

The linear representation of this group that allows you to write them as infinite matrices corresponds to the fact that composition defines a natural right action of this group on the vector space of formal power series with vanishing term of order $0$.

Of course, these formulae and their multivariable generalizations are of critical importance in applying Lie's method of prolongation for computing differential invariants. There was a lot of work on these structures in the late 19th century by Drach, Vessiot, Medolaghi, and ultimately, Cartan.

Answer 2

These are the incomplete Bell polynomials.

You need $4a_1a_3 + 3a_2^2$ where you have $4a_1a_3 + 4a_2^2.$

They work as follows:

The number of partitions with only one part of a set of six elements is $1.$ Hence $a_6.$ (You'll see the pattern as we proceed.)

The list of partitions with two parts contains $6$ partitions of the form $1+5$ and $15$ of the form $2+4$ and $10$ of the form $3+3.$ Hence $6a_5a_1+ 15a_4a_2 + 10a_3^2.$

The list of partitions with three parts contains $15$ of the form $4+1+1$ and $60$ of the form $3+2+1$ and $15$ of the form $2+2+2.$ Hence $15a_4a_1^2 + 60a_3a_2a_1 + 15 a_2^3.$

The list of partitions with four parts contains $20$ of the form $3+1+1+1$ and $45$ of the form $2+2+1+1.$ Hence $20a_3a_1^3+ 45a_2^2a_1^2.$

The list of partitions with five parts contains $15,$ all of the form $2+1+1+1+1.$ Hence $15a_2a_1^4.$

The list of partitions with six parts contains only one, of the form $1+1+1+1+1+1.$ Hence $a_6.$

So the sixth row of your matrix is: $$ \begin{array}{ccrccccc} a_6, & 6a_5a_1 + 15a_4a_2 + 10a_3^2, & 15a_4a_1^2 + 60a_6a_2a_1 + 15a_2^3, \\ & & 20a_3a_1^3+ 45a_2^2a_1^2, & 15a_2a_1^4, & a_1^6 \end{array} $$

Faà di Bruno's formula is easiest to understand when one uses functions of multiple variables: \begin{align} & \frac{\partial^3}{\partial x_1\,\partial x_2\,\partial x_3} f(u) \\[8pt] = {} & \phantom{{}+{}} f'(u) \frac{\partial^3 u}{\partial x_1\,\partial x_2\,\partial x_3} \\[8pt] & + f''(u) \left( \frac{\partial u}{\partial x_1} \cdot \frac{\partial^2 u}{\partial x_2\,\partial x_3} + \frac{\partial u}{\partial x_2} \cdot \frac{\partial^2 u}{\partial x_3\,\partial x_1} + \frac{\partial u}{\partial x_3} \cdot \frac{\partial^2 u}{\partial x_1\,\partial x_2} \right) \\[8pt] & + f'''(u) \cdot \frac{\partial u}{\partial x_1} \cdot \frac{\partial u}{\partial x_2} \cdot \frac{\partial u}{\partial x_3} \end{align}

There is one term for each partition of the set of $x$ variables. The order of the derivative of $f$ is the number of parts in the partition.

Then let the several $x$ variables become indistinguishable from each other, and the left side becomes $(d^3/dx^3) f(u)$ and the terms that become indistinguishable from each other on the right side are collected into one term that counts set-partitions of the respective form.

Looking again at the incomplete Bell polynomials from $a_6$ to $a_1^6$ listed above, suppose we make them coefficients in a polynomial in $y,$ thus \begin{align} p_6(y) = {} & a_6 y + (6a_5a_1 + 15a_4a_2 + 10a_3^2)y^2 \\ & + (15a_4a_1^2 + 60a_3a_2a_1 + 15a_2^2)y^3 \\ & + (20a_3a_1^3+ 45a_2^2a_1^2)y^4 \\ & + 15(a_2a_1^4) y^5 + a_1^6 y^6 \end{align}

Then we have a polynomial sequence of "binomial type": $$ p_n(y+z) = \sum_{k=0}^n \binom n k p_k(y) p_{n-k}(z). $$

Every polynomial sequence of binomial type is formed in this way from incomplete Bell polynomials. The set of all such sequences is a non-Abelian group under "umbral composition":

If $$ p_n(y) = \sum_{k=0}^n p_{n,k} y^k $$ then the $n$-th degree term in the umbral composition of two sequences $(p_n(y))_{n=0}^\infty$ and $(q_n(y))_{n=0}^\infty$ is $$ (p_n \circ q)(y) = \sum_{k=0}^n p_{n,k} q_k(y). $$

This is isomorphic to the group of which you wrote. (Note that in $\text{“}(p_n\circ q)\text{”},$ $p_n$ has the subscript $n$ and $q$ has none.)

• Roman, Steven M.; Rota, Gian-Carlo (1978), "The umbral calculus", Advances in Mathematics, 27 (2): 95–188, doi:10.1016/0001-8708(78)90087-7, ISSN 0001-8708, MR 0485417

• G.-C. Rota, D. Kahaner, and A. Odlyzko, "Finite Operator Calculus," Journal of Mathematical Analysis and its Applications, vol. 42, no. 3, June 1973. Reprinted in the book with the same title, Academic Press, New York, 1975.