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I have a question maybe more relevant to an English language section of StackExchange, but I doubt that anybody but a Mathematician could properly answer my question.

Let $\mathcal M$ be a smooth manifold, let $X$ be a smooth vector field on $\mathcal M$ and let $\Sigma$ be a smooth hypersurface of $\mathcal M$. Let $t\mapsto \gamma(t,x)$ be the integral curve of $X$ originating at $x\in \Sigma$. The vector field $X$ is tranverse to $\Sigma$ at $x\in \Sigma$ whenever $\dot \gamma(0,x)\notin T_x(\Sigma)$: we may say as well in that case that the curve $\gamma$ is transverse to $\Sigma$.

When $\gamma$ is not transverse to $\Sigma$, I suppose that for some $k\ge 2$ $$ \frac{d^l\gamma}{dt^l}(0,x)\in T_x(\Sigma)\quad\text{for $1\le l\le k-1$}, \quad \frac{d^k\gamma}{dt^k}(0,x)\notin T_x(\Sigma), $$ and I want to say that [the order of contact of $\gamma$ with $\Sigma$ is exactly $k$] or that [the curve $\gamma$ is tangential to $\Sigma$ with a contact of order $k$]. Are these formulations correct?

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  • $\begingroup$ I like "the curve gamma is tangent to Sigma with a contact of order k" $\endgroup$
    – user44143
    Commented Feb 15, 2017 at 15:55
  • $\begingroup$ I like the first one. It seems pretty standard: for example the Wikipedia page en.wikipedia.org/wiki/Contact_(mathematics) uses more or less this formulation. $\endgroup$ Commented Feb 15, 2017 at 16:50
  • $\begingroup$ Both of your examples sound like idiomatic mathematical English to me. Possibly other ways of saying it are more established in the literature, but the fact that you are asking this question puts an upper bound on just how well established these hypothetical other ways are. $\endgroup$ Commented Feb 15, 2017 at 16:55
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    $\begingroup$ I would be more concerned about specifying whether this counts as order $k$ or order $k-1$. $\endgroup$ Commented Feb 15, 2017 at 17:57
  • $\begingroup$ Sigma has a tangential curve gamma with an order of contact k $\endgroup$
    – JMP
    Commented Feb 16, 2017 at 7:08

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