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There is a very basic invariant of $\mathcal{C}^{\infty}$-germs $f: (\mathbb{R}^n, 0) \to (\mathbb{R}^k, 0) $ ($n \leq k$), which I call ''maximal order along curves''. I would like to interested whether it appeares in the literature and how does it relate to other invariants.

Definition: Consider a germ of curve $\gamma: (\mathbb{R}, 0) \to (\mathbb{R}^n, 0)$, $\gamma'(0) \neq 0$. The order of $f$ along $\gamma$ is $$ \mbox{ord}_{\gamma} (f)= \mbox{ord}_0 (f \circ \gamma)=\min_{i=1}^k (f_i \circ \gamma). $$ The maximal order of $f$ along regular curves is $$\mbox{Ord} (f)=\max \{\mbox{ord}_{\gamma} (f) \ | \ \gamma: (\mathbb{R}, 0) \to (\mathbb{R}^n, 0) , \ \gamma'(0) \neq 0 \}.$$

$\mbox{Ord} (f) $ is invariant under contact equivalence of germs, i.e. if $g \sim_{\mathcal{K} }f$, then $\mbox{Ord} (f) =\mbox{Ord} (g) $.

Question: How does it relate to other invariants? Especially, how can it be expressed from the local algebra $Q(f)$ of $f$?

Examples and additional information:

  1. For corank-1 germs (i.e. $\mbox{rk} (\mbox{Jac}_0(f))=n-1$), $\mbox{Ord} (f)$ is equivalent with the Thom-Boardman symbol of $f$. In fact, $f$ belongs to the Thom-Boardmann class $A_d=\Sigma^{1_d}$ if and only if $ \mbox{Ord} (f)=d+1$. In this case the local algebra $Q(f)$ of $f$ is isomorphic with $\mathbb{R}[t]/(t^{d+1})$. This case is described in Bérczi-Szenes.
  2. $\mbox{Ord}(f)$ cannot be read directly from the Taylor series of $f$. E.g. consider $$ f(x, y)=(x-y^2, xy).$$ Although both component has degree 2, our invariant is $\mbox{Ord}(f)=3$. In fact, $\gamma(t)=(t^2, t)$ is a regular curve with $f (\gamma (t))=(0, t^3)$. Moreover $f$ is right equivalent with the Whitney cusp $g(x, y)=(x, y^3+xy)$. The fold curve $\Sigma^1(g)$ is a parabola, but the order of $g$ along $\Sigma^1(g)$ is only 2. Hence the curves along the maximal order realizes do not coinside with the Thom-Boardman strata.
  3. Maybe this question somehow relates to the Porteous probe model of the Thom-Boardman classes, but I cannot see the exact relation. See Porteous: Probing singularities, PhD Thesis of Balázs Kőműves.
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  • $\begingroup$ It seems that $\mbox{Ord}(f)=d$ if $\mathfrak{m}^d \nsubseteq I(f) $ but $\mathfrak{m}^{d+1} \subset I(f) $. (Where $I(f) $ is the ideal generated by the components of $f$, $\mathfrak{m}$ is the maximal ideal in the algebra of $\mathcal{C}^{\infty}$ germs). It's called the ''depth'' of $I(f)$, or $Q(f)=\mathcal{O}_n/I(f)$, I don't know this terminology. And it seems that the Bérczi-Szenes proof in the corank-1 case can be generalized to prove this - I cannot see every details yet. $\endgroup$ Commented Jul 6, 2022 at 7:47
  • $\begingroup$ The previous comment is not true. I can prove only an inequality, and $(x^2, y^2)$ is a counterexample for the equality. $\endgroup$ Commented Feb 16 at 12:23

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