Monday, October 3, 2016

Invariance under inversion


Commenter M left a comment that made me look at the symmetry principles that leave the information equilibrium (IE) condition:

$$
\frac{dA}{dB} = k \; \frac{A}{B}
$$

invariant. This may seem like an esoteric point, but in physics it is an incredibly powerful principle that in fact leads to everything we know. I could say special relativity ... or I could say Poincare invariance.  Quantum electrodynamics is basically a Poincare invariant theory with a local $U(1)$ gauge group symmetry. The standard model (core theory) is the gauge group $SU(3) \times SU(2) \times U(1)$ plus general covariance (general relativity).

Anyway, I've previously noted the scale invariance of the IE condition where $A \rightarrow \alpha A$ leaves the equation unchanged. This manifests in the real life invariance to changing the number of zeros on the price of something in different currencies (e.g. about 100 Yen versus 1 dollar versus about 6-7 Francs when they existed). We could suddenly talk about everything in "cents" in the US and it shouldn't change the economic theory. In a sense, this scale invariance is related to the homogeneity of degree zero of supply and demand functions in economics (which Bennett McCallum relates to long run neutrality and the quantity theory of money).

I have a bad habit of referring to this scale invariance as a conformal symmetry (due to dealing with approximately conformal field theories like QCD in the large N expansion where quantum corrections are suppressed [see here]). Scale invariance is just a subset of conformal symmetry. But M's comment made me see if the IE condition was invariant under more of the conformal symmetry group.

It turns out it is, and as you could probably tell from the title, it is invariant under inversion where $A \rightarrow 1/A$ and $B \rightarrow 1/B$:

$$
\begin{align}
\frac{d(1/A)}{d(1/B)} & = k \; \frac{1/A}{1/B}\\
\frac{-1/A^{2}}{-1/B^{2}}\; \frac{dA}{dB} & = k \; \frac{B}{A}\\
\frac{dA}{dB} & = k \; \frac{B}{A} \;\frac{A^{2}}{B^{2}}\\
\frac{dA}{dB} & = k \; \frac{A}{B}
\end{align}
$$

This may end up being more consequential than the scale invariance, but other than T-duality, I can't think of an immediate example of relevance in physics. In fact, see this Wikipedia paragraph on inversion symmetry. What this symmetry says is that we could talk about 1/NGDP (one over aggregate demand) or 1/L where L is the labor supply and that would be the same theory (with the same, not inverted, price). Also, zero and infinity are essentially the same thing. The equation isn't invariant under translations ($A \rightarrow A + c$), so it won't be invariant under special conformal transformations.

In any case, this give me something to ponder.

...

Update 4 October 2016

Commenter M expanded this to transformations of the form

$$
\begin{align}
A & \rightarrow \alpha A^{\gamma}\\
B & \rightarrow \beta B^{\gamma}
\end{align}
$$

with $\alpha$, $\beta$, and $\gamma$ nonzero. Inversion is the special case of $\gamma = -1$.

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