Thom–Boardman singularities

Remarks

Since $\Sigma^{i_1, \ldots, i_k}(f) \subseteq \Sigma^{i_1, \ldots, i_{k-1}}(f)$, the kernel of $f$ restricted to the smaller set lies in the kernel of $f$ restricted to the larger one. So the symbols are non-increasing: $i_1 \ge i_2 \ge \cdots \ge i_k$. Boardman constructed submanifolds $\Sigma^I$ of the jet space $J^k(M, N)$, depending only on $I$. Suppose the jet extension $j^k f$ is transverse to all of them. Then each $\Sigma^I(f) = (j^k f)^{-1}(\Sigma^I)$ is a submanifold and agrees with the inductive definition. By Thom's transversality theorem, this holds for a dense set of maps. If $\dim M = m$ and $\dim N = n$, the subset $\Sigma^i$ of $J^1(M, N)$ has codimension $i(n - m + i)$. Example: for $f(x, y) = (x, y^3 + xy)$ on $\R^2$, $\Sigma^1(f)$ is the curve $x = -3y^2$. Parametrising it by $y \mapsto (-3y^2, y)$, the restriction of $f$ is $y \mapsto (-3y^2, -2y^3)$, whose derivative vanishes only at $y = 0$. So the origin lies in $\Sigma^{1,1}(f)$, and the other points of the curve lie in $\Sigma^{1,0}(f)$. Non-example: for $f(x, y) = (x, y^3 - x^2 y)$, $\Sigma^1(f)$ is the set $3y^2 = x^2$, a pair of crossing lines. It is not a submanifold at the origin, so the inductive definition breaks down there; the jet of $f$ is not transverse to $\Sigma^1$.