The n-Category Café

(To expand a little more on my silly answer: I had a hunch that the answer could be something to do with entropy just on sociological grounds, as a long-time reader of this blog; but I just stupidly misremembered the formula for entropy and hence got something which didn’t work)

Well, of course you’re right that $W = 0$ is a solution! More generally, it’s clear that the space of solutions is closed under taking scalar multiples.

And I suspected some people might guess it was basically entropy on sociological grounds, but there was nothing I could think of doing to avoid that…

BTW, here’s a way of guessing the answer - though perhaps not quite proving.

Suppose that instead of $W(cx_1, \dots, cx_n) = c W(x_1, \dots, x_n)$ we required $W(cx_1, \dots, cx_n) = c^\lambda W(x_1, \dots, x_n)$ for some $\lambda$. Clearly, the multiplicative group of positive reals acts on the space of solutions of the functional equation, so looking for other eigenfunctions seems like a reasonable thing to do.

Denote $w_n := W(1, \dots, 1)$ (with $n$ copies of $1$). By breaking an $mn$-tuple into an $n$-tuple of $m$-tuples we get

$w_{mn} = m w_n + n^\lambda w_m$

And by symmetry, also

$w_{mn} = n w_m + m^\lambda w_n$

Subtracting these, we get

$w_n = n^\lambda - n$, up to a multiplicative constant that doesn’t depend on $n$ that we can set to $1$.

Now, if our $x_i$’s are integer multiples of some small $\varepsilon$, we can represent $W(x_1, \dots, x_n)$ as an intermediate step in a calcuation of the sum of $\varepsilon^{-1} (\sum_i x_i)$ copies of $\varepsilon$, reducing everything to the sequence $(w_n)$:

$W(x_1, \dots, x_n) = \varepsilon^{\lambda} \left( w_{\varepsilon^{-1} \sum x_i} - \sum w_{\varepsilon^{-1} x_i} \right)$

so by a continuity argument:

$W(x_1, \dots, x_n) = (\sum x_i)^\lambda - \sum x_i^\lambda$

The limit as $\lambda \to 1$ is $0$, but normalizing it by $(\lambda-1)^{-1}$, we get our logarithms.

Yup!

That’s a nice argument. It’s an interesting fact that the various deformations of Shannon entropy are often easier to axiomatize than Shannon entropy itself, because you can use the kind of symmetry-breaking argument that you’ve used here. Some more examples of this technique are here.

You do need some sort of regularity assumption, otherwise you can replace $\log$ in your formula by $f \circ \log$ where $f : \mathbb{R} \to \mathbb{R}$ is some function that is additive but not $\mathbb{R}$-linear. (A classical argument using the axiom of choice shows that such functions exist.) Continuity is enough. Measurability is enough if you also assume that $W$ is symmetric in its arguments. I don’t know whether measurability is enough without that symmetry argument. If I had to guess, I’d say yes.

Nice!

There are various papers that take this homological approach to entropy. I think Jean-Louis Cathelineau may have been first to see the connection, e.g. there’s something on p.58-59 of his paper Sur l’homologie de $SL_2$ a coefficents dans l’action adjointe, and it’s also right there at the start of his paper Remarques sur les différentielles des polylogarithmes uniformes. Maxim Kontsevich also came upon it in a privately-circulated note he wrote for Hirzebruch’s retirement celebrations, later published as an appendix to an article by Elbaz-Vincent and Gangl. And Pierre Baudot and Daniel Bennequin have a whole 65-page paper called The homological nature of entropy.

To clear up a couple of small points from your comment:

You didn’t write down an equation for adding three things

I’m not sure I understand what you mean by that, but the big displayed equation just before the picture of mounds of rubble includes what you need. Take $n = 2$ and $k_1 = 2, k_2 = 1$ to get

$$W(x, y, z) = W(x, y) + W(x + y, z).$$

And similarly, you can get

$$W(x, y, z) = W(y, z) + W(x, y + z).$$

Then as you say, equating the two expressions for $W(x, y, z)$ gives the cocycle identity.

I figured you probably wanted $W(0,0)=0$

This follows from the homogeneity requirement:

$$W(0, 0) = W(0 \cdot 0, 0 \cdot 0) = 0 W(0, 0) = 0.$$

Yes, absolutely. And the explicit formula for that homogeneous extension is the formula at the start of Alexander’s first comment, up to a constant factor.

The theorem needed is more or less the classical Faddeev characterization of entropy. (At least, assuming $W$ is symmetric.)

I say “more or less” because he (like most other writers on these things) used a non-symmetric special case of the main equation in my post: in my notation above, it’s the case $k_1 = 2$, $k_2 = \cdots = k_n = 1$. This forced him to also assume a symmetry axiom. In the presence of the symmetry axiom, this special case is equivalent to the general case by a mundane induction. But if you assume the general case, you don’t need symmetry.

From numerical simulation it looks like the above system of equations truncated to $n$ has kernel with dimension equal to the number of prime factors in $n$. I’m guessing this corresponds to the choice of value of $\log(p)$ for prime $p$.

Also, I finally found a proof that $W(3,1)\lt W(2,2).$ It uses the two equations $$\begin{array}{c} W(2,1) + W(3,1) = W(1,1) + W(2,2) \\ W(6,2) + W(8,1) = W(2,1) + W(6,3). \end{array}$$

These simplify to $$\begin{array}{c} W(2,1) + W(3,1) = 3W(1,1)\\ 2W(3,1) + W(8,1) = 4W(2,1). \end{array}$$ Multiplying the first equation by 4 and then combining with the second equation we get $$6W(3,1) + W(8,1) = 12W(1,1) = 6W(2,2)$$ and because $W(a,b)\gt 0$ the result follows, as long as we are willing to cancel the factor of 6.

All of this makes me think that entropy can be defined taking values in a rig, or 2-rig, or traced monoidal category, or some other gadget like that. Working just over the natural numbers I would hope to see the “real” entropy show up as a colimit, similar to how Paolo Perrone explains here .