There I was, innocently doing some category theory, when up popped a totally outlandish operation on polynomials. It seems outlandish to me, anyway. I'd like to know if anyone has seen this operation before, in any context.

The categorical background isn't relevant to the question, so I'll skip it. All I want to emphasize is that *a priori*, it has nothing to do with polynomials. It's just some universal property, which produces this in a special case. (For the curious, the categorical connection is that some functors $\mathbf{Set}^n \to \mathbf{Set}$ can be viewed as "polynomial", in the sense that they're built up from products, $\times$, and coproducts, $+$.)

By a **polynomial** I mean a polynomial in commuting variables $X_1, \ldots, X_n$, with coefficients in the natural numbers $\mathbb{N}$ (which include $0$).

Here's the operation. Given a polynomial $f = f(X_1, \ldots, X_n)$, define a new polynomial $f^*$ as follows.

Write $f$ as a sum of products of $X_i$'s.

Change every occurrence of $+$ to $\times$, and every occurrence of $\times$ to $+$. Call the resulting polynomial $f^*$.

**Examples:**

Take $f(X, Y) = (X + Y)^2$. Step 1 writes $f$ as $$ f(X, Y) = (X \times X) + (X \times Y) + (X \times Y) + (Y \times Y). $$ Step 2 then gives $$ f^*(X, Y) = (X + X) \times (X + Y) \times (X + Y) \times (Y + Y) = 4XY(X + Y)^2. $$ Now let's calculate $f^{**}$. Step 1: $$ f^*(X, Y) = 4X^3 Y + 8X^2 Y^2 + 4X Y^3. $$ Step 2: $$ f^{**}(X, Y) = (3X + Y)^4 (2X + 2Y)^8 (X + 3Y)^4. $$

Generally, if $$ f(X_1, \ldots, X_n) = A X_1^{a_1} \cdots X_n^{a_n} + B X_1^{b_1} \cdots X_n^{b_n} + \cdots $$ ($A, a_i, B, b_i, \ldots \in \mathbb{N}$) then $$ f^*(X_1, \ldots, X_n) = (a_1 X_1 + \cdots a_n X_n)^A (b_1 X_1 + \cdots + b_n X_n)^B \cdots. $$

By the previous example, $f^{**} = f$ if $f$ is a monomial ($X_1^{a_1} \cdots X_n^{a_n}$) or linear ($a_1 X_1 + \cdots + a_n X_n$).

Since the empty sum is 0 and the empty product is 1, it's meant to be implicit in (2) that 0s become 1s and 1s become 0s. E.g. if $f = 0$ then $f^* = 1$, and if $f(X) = X^2 + 1$ then $f^*(X) = 2X \times 0 = 0$.

**Edit**: Similarly, if $f$ has nonzero constant term then $f^* = 0$.

I'm interested to hear about anywhere that anyone has seen this operation.

Feel free to add tags as appropriate.