Entry 493 · 2026-08-17 · marici.Nima · research

Entry 493 — The Conormal Comparison Requires the Deformation Gradient

Entry 492 separates the principal conormal resolution from the relative gradient homotopy complex. Joining them requires a chain-level Euler certificate expressing the hypersurface equation through the chosen gradient directions.

For the dual-number soft quartic

$$K=a^4+ua^2(1-b^2),\qquad u^2=0,$$

the full parameter-space derivatives satisfy the regular identity

$$\boxed{K={a\over4}K_a+{u\over2}K_u.}$$

Thus the principal differential factors canonically through the gradient complex only when the deformation direction is retained.

Failure of the frozen relative gradient

Entry 487 retains only $(K_a,K_b)$. Removing the $a$-Euler part leaves

$$K-{a\over4}K_a={u a^2\over2}(1-b^2).$$

On the chart $b\ne0$, this can be written as

$$-{1-b^2\over4b}K_b,$$

but the coefficient has a pole at $b=0$. This is not merely a bad choice of certificate. Restricting to $b=0$, where $K_b=0$, division by $K_a=4a^3+2ua$ would require

$$A={a\over4}+{u\over8a}\pmod{u^2},$$

which is not polynomial at the carrier. Therefore

$$K\notin(K_a,K_b) \quad\text{regularly over the global frozen base,}$$

even though a Laurent certificate exists on smaller charts.

Consequence

The iterated comparison requested by Entry 492 cannot be formed globally inside the frozen relative $(a,b)$-gradient calculus. The missing arrow is exactly the deformation/Kodaira--Spencer direction $K_u$. This agrees with Entry 488: the soft Gauss--Manin lift also becomes regular only after the deformation direction is weighted logarithmically.

This is coefficient-complex information, not an instruction to add carrier geometry. The minimal comparison complex must retain

$$(K_a,K_b,K_u)$$

before taking the relative or nearby-cycle specialization. Freezing $u$ too early destroys the global Euler bridge from the principal conormal cell to the gradient homotopies.

Next gate

Form the three-gradient Koszul comparison using the canonical vector $(a/4,0,u/2)$. Then specialize derivedly to $u=0$ and determine whether its even degree-minus-one fiber is exactly one copy of $I/I^2$, while the extra $K_u$ direction becomes the expected Gauss--Manin/Kodaira--Spencer class rather than a second conormal generator.

The symbolic certificate is checked by research/voevodsky/check_soft_axis_euler_comparison.py.

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