# Circular Morera Regularizer

- ID: 1796
- Canonical URL: https://synthcore.org/idea/1796/circular-morera-regularizer
- API JSON: https://synthcore.org/api/idea/1796.json
- API Markdown: https://synthcore.org/api/idea/1796.md
- Verification status: unverified
- Source: [arXiv:2608.04540](https://arxiv.org/abs/2608.04540)
- Category: regularization
- Solves: accuracy, generalization, stability
- ML areas: regularization, mlp, embedding
- Math tags: functional-analysis, pde, measure-theory
- Ratings: usefulness 5/10; difficulty 3/10; novelty 7/10

## Idea description

Add a multiscale circular-integral penalty to a complex-valued neural field f_theta: R^2 -> C. The penalty directly tests the local contour condition that characterizes holomorphic functions, providing a derivative-free alternative to explicitly penalizing the Cauchy-Riemann residual.

## Mathematical statement

The paper defines partial = 1/2(partial_x - i partial_y) and bar-partial = 1/2(partial_x + i partial_y), and proves that if D is a domain, f is continuous on D, and for every center a in D, the contour integral of f over the circle of radius r centered at a is o(r^2) as r tends to zero, then f is holomorphic, equivalently bar-partial f = 0. Here a is the complex circle center, r is the radius, and o(r^2) means that the magnitude of the integral divided by r^2 tends to zero. For a smooth field, the local expansion is integral over the circle of f(zeta) dzeta = 2 pi i r^2 bar-partial f(a) + O(r^3), so the normalized circular residual estimates the antiholomorphic derivative without automatic differentiation. The adapted loss is L_circ = E over centers a and radii r of the squared magnitude of the normalized contour integral, with the contour integral approximated by quadrature.

## Key formulas

- $$\partial=\frac{1}{2}(\partial_x-i\partial_y),\qquad\bar{\partial}=\frac{1}{2}(\partial_x+i\partial_y).$$
- $$\int_{|\zeta-a|=r}f(\zeta)\,d\zeta=o(r^{2})\qquad(r\to0^{+})\ \Longrightarrow\ \bar{\partial}f=0\ \Longrightarrow\ f\text{ is holomorphic}. $$
- $$\oint_{|\zeta-a|=r} f(\zeta)\,d\zeta=2\pi i r^2\bar{\partial}f(a)+O(r^3).$$
- $$L_{\mathrm{circ}}(\theta)=\mathbb{E}_{a\sim\mathcal{D},\ r\sim\mathcal{R}}\left[\left|\frac{1}{2\pi i r^2}\oint_{|\zeta-a|=r}f_\theta(\zeta)\,d\zeta\right|^2\right].$$

## Implementation notes

1. Integration point: use a complex-valued MLP f_theta(x,y) = u_theta(x,y) + i v_theta(x,y) as a coordinate-field decoder, image representation, or analytic latent embedding. Add lambda times L_circ to the reconstruction or supervised loss during training; inference is unchanged. Sample centers a = (x,y) from training coordinates and reject circles that leave the valid domain. 2. Pseudocode: `for each batch of centers a: sample radii r from logarithmically spaced values; sample t_j = 2*pi*j/K for j=0,...,K-1; z_j = a + r*exp(i*t_j); f_j = f_theta(z_j); dz_j = i*r*exp(i*t_j)*(2*pi/K); I = sum_j f_j*dz_j; q = I/(2*pi*i*r*r); L_circ += abs(q)^2; loss = L_task + lambda*mean(L_circ); backprop(loss)`. Start with K = 16 or 32 and two to four radii. Gradients must flow through every f_theta(z_j). 3. The theorem supplies the o(r^2) certificate and the 2*pi*i*r^2*bar-partial f(a) interpretation. Quadrature error, radius range, lambda, and the number of contour samples are estimated or selected empirically. Normalize by r^2 to prevent large circles from dominating, and optionally use the maximum residual across radii to prevent the model from satisfying only one scale. 4. First experiment: train a small complex MLP on analytic targets such as f(z) = z^3 + 0.2 z^2 and on nonanalytic targets containing a conjugate-z term. Compare ordinary MSE, MSE plus the circular penalty, and MSE plus an explicit Cauchy-Riemann penalty. Then evaluate a 64 by 64 coordinate decoder on a smooth image-fitting task. Measure task error, circular residual at unseen centers and radii, extrapolation outside the training region, and robustness to noisy coordinate samples. Success means lower unseen-region error or better noise robustness at matched parameter count, with normalized residual decreasing consistently as radii shrink.

## Disclaimer

AI-generated research hypothesis, automatically tested. Not peer-reviewed.