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There is an identity $e^x=\lim_{n\to \infty} (1+x/n)^n$, and I always thought it is a purely analytic statement. But then I discovered its curious interpretation in pure algebra:

Consider the ring of formal infinite sums of monomials in infinitely many variables $\varepsilon_1, \varepsilon_2,\ldots$ satisfying $\varepsilon_i^2=0$.

$$ R=\mathbb{Q}[\![\varepsilon_1, \varepsilon_2, \ldots]\!]/(\varepsilon_i^2: i=1,2,\ldots). $$

Then the sum $x=\sum_{i=1}^\infty \varepsilon_i$ makes sense and is not infinitesimally small, in fact we have $$ x^n = n! \sum_{1\leq i_1<i_2<\ldots<i_n} \varepsilon_{i_1} \cdots \varepsilon_{i_n}. $$ So the ring of polynomials $\mathbb{Q}[x]$ embeds into $R$. Moreover, in $R$ we have the identity $$ \prod_{i=1}^\infty(1+\varepsilon_i) = \sum_{n=0}^\infty \frac{x^n}{n!}. $$ So somehow we multiplied infinitely many elements infinitely close to $1$ and managed to get away from $1$ and obtain the right answer.

I was wondering if this is well-known and if there are applications of this idea. For instance, one can probably use it to recover the formal neighborhood of $1$ in an algebraic group from the Lie algebra.

In positive characteristic the right hand side doesn't make sense, but the left hand side still does. In fact, symmetric functions in $\{\varepsilon_i\}$ form a ring with divided power structure. Can one build $p$-adic cohomology theories based on this idea instead of divided power structures?

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    $\begingroup$ Just a trivial remark, ur x is a topological nilpotent element of R and so determines an embedding of power series into R, not just polynomials. $\endgroup$ – EBz Jun 20 at 14:24
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    $\begingroup$ Related to the Trotter product formula (or Lie-Trotter-Kato) of use to physicists, I believe. $\endgroup$ – Tom Copeland Jun 21 at 14:47
  • $\begingroup$ You start with an identity and leave it hanging. You could delete it without changing the title or rest of post. But since you put it first, it sounds like you want it addressed. You could make sense of that identity in $\mathbb Q[\![x]\!]$. It is false there, because the coefficient of $x^2$ is $(n-1)/2n$, which does not converge in $\mathbb Q$. As @EBz says, $\mathbb Q[\![x]\!]$ embeds in your ring, so it does not change the notion of convergence, so it is still false. $\endgroup$ – Ben Wieland Jun 22 at 17:24
  • $\begingroup$ @BenWieland I don't understand your comment. Did you expect $e^x=\lim_{n\to\infty} (1+x/n)^n$ to hold in $\mathbb{Q}[[x]]$? Of course not, as you explain the limit doesn't even exist. This limit exists in the usual topology on $\mathbb{R}[[x]]$ or $\mathbb{R}$ for fixed $x$. The point of the first identity, because it is well known, is a motivation for the rest. Are you asking if there is a direct connection between this formula and the formula with epsilons? $\endgroup$ – Anton Mellit Jun 22 at 18:05
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    $\begingroup$ @TomCopeland The analogue of the Trotter formula is $\prod_i (1+(A+B)\varepsilon_i) = \prod_i (1+ A \varepsilon_i)(1+ B \varepsilon_i)$ for some elements $A$,$B$ of a Lie algebra, which in this setting holds trivially because $\varepsilon_i^2=0$. Note that the products are not commutative. It is natural to ask for a proof of Baker-Campbell-Hausdorff, but in this setting another formula, Zassenhaus formula (en.wikipedia.org/wiki/…) follows more easily, by reordering the product. $\endgroup$ – Anton Mellit Jun 22 at 22:05
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$\DeclareMathOperator{\ex}{ex}$ One way to look at this is through symmetric functions. To be consistent with standard notation I'll take infinitely many variables $x_1, x_2,\dots$ (instead of $\varepsilon_1, \varepsilon_2,\ldots$). I will follow the presentation in Richard Stanley's Enumerative Combinatorics, Vol. 2, page 304.

There is a homomorphism ex from the ring of (unbounded degree) symmetric functions to power series in $t$ that can be defined by $\ex(p_1) = t$, $\ex(p_n)=0$ for $n>1$, where $p_n$ is the power sum symmetric function $x_1^n+x_2^n+\cdots$. Then $\ex$ is the restriction to symmetric functions of the homomorphism on all formal power series in $x_1,x_2,\dots$ that takes each $x_i^2$ to 0 (where $t$ is the image of $p_1$). It has the property that for any symmetric function $f$, $$\ex(f) = \sum_{n=0}^\infty [x_1x_2\cdots x_n] f \frac{t^n}{n!},$$ where $[x_1x_2\cdots x_n] f$ denotes the coefficient of $x_1x_2\cdots x_n$ in $f$. In particular, $\ex(h_n) = \ex(e_n) = t^n/n!$ where $h_n$ and $e_n$ are the complete and elementary symmetric functions.

This idea is well known and is very useful in enumerative combinatorics. It allows one to derive exponential generating functions for objects with distinct labels (e.g., permutations or standard Young tableaux) from symmetric function generating functions for objects with repeated labels (e.g., words or semistandard tableaux). There are related homomorphisms that preserve more information; see, for example, section 7.8 of Stanley.

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  • $\begingroup$ So when you say "ring of (unbounded degree) symmetric functions", this is bigger than what most people use "ring of symmetric functions" to mean, right? E.g., $\sum_{i=0}^{\infty}p_i$ is in the ring you are considering? $\endgroup$ – Sam Hopkins Jun 20 at 17:17
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    $\begingroup$ @SamHopkins. Yes, this is the ring usually denoted $\hat\Lambda$. $\endgroup$ – Ira Gessel Jun 20 at 17:32
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    $\begingroup$ Yeah, in combinatorics this is quite natural because if you interpret the ring of symmetric functions as the ring of representations of symmetric groups, then $\operatorname{ex}$ sends a representation of $S_n$ to its dimension times $t^n/n!$. So this is precisely the map that allows you to "forget" the $S_n$ action when you don't care about it. $\endgroup$ – Anton Mellit Jun 20 at 21:27
  • $\begingroup$ Nice answer! Going the other way, thinking of the exponential specialization as evaluation in the quotient ring where $x_i^2 = 0$ makes me feel I understand it much better. $\endgroup$ – Mark Wildon Jun 21 at 10:34

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