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Let $M$ be a smooth manifold of dimension $(m+n)$. Two curves $\gamma_1, \gamma_2 \colon \mathbf{R} \to M$ with $\gamma(0) = p$ are said to have contact at $p$ of order $k$ if for all smooth maps $\varphi \colon U \to R$ where $U \subset M$ is open about $p$, we have the following equality of jets: $J_0^k(\varphi\gamma_1) = J_0^k(\varphi\gamma_2)$.

Consider now submanifolds $N_1$ and $N_2$ of codimension $m$. How one would define $k$-th order contact at $p \in N_1 \cap N_2$?

EDIT: While we wait for Ted's reply, I find this definition to be reasonable:

$N_1$ and $N_2$ have $k$-th order contact at $p \in N_1 \cap N_2$ if for all $[\gamma]$ in the jet space $J_0^k(\mathbf{R}, M)_p$ there are submanifoldcharts $(x, U)$ and $(y, V)$ for $N_1$ and $N_2$ respectively such that $J_0^k(x\gamma) = J_0^k(y\gamma)$.

Andrew Thompson
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    Andrew, I won't be at a desktop computer until the end of the week, but I'll be glad to write a complete answer later. You basically need to define the higher-order tangent space (osculating space), working in a chart of $M$. You consider the span of the partial derivatives of order up to $k$ of any parametrization of your submanifold. This is well-defined and interesting when $m$ is quite large. Of course, if you vary $p$, you rarely get a vector bundle because the dimension of these spans may change dimension. This is basically taking a mapping of $J^k(TN)$ to $TM$ and looking at its image. – Ted Shifrin May 31 '15 at 21:25
  • Yes, for jet spaces $J_0^k(\mathbf{R}, M)_p$ I have been able to define a smooth fiber bundle which I find is simply $TM$ for $k = 1$, I'm assuming this is what you're getting at? – Andrew Thompson Jun 01 '15 at 09:22
  • Andrew, yes, but you want the "embedded" realization, rather than the abstract. Take a look at the beginning of my paper on surfaces in $\Bbb P^5$. Full disclosure: Far into that paper there is a mistake. – Ted Shifrin Jun 01 '15 at 16:45
  • Intuitively I find that the definition provided in the edit makes a lot of sense; is it nonsense? – Andrew Thompson Jun 01 '15 at 18:37
  • Andrew, it's intuitively appealing, but I worry about reparametrizations. Can you convince me that if the second-order tangent spaces, in my spanning sense, agree, that you can rig your definition? I'll think about that after I run errands today ... – Ted Shifrin Jun 01 '15 at 18:51
  • Hm, I guess I'll actually try to write a few elementary properties out to see where the definition might go wrong so that I can answer your question properly, but is the problem somewhat the same as with transversality? (I.e., given a picture they may look transverse, however the parametrization may be given in such a way that they slow down at the point of intersection.) – Andrew Thompson Jun 02 '15 at 06:45

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OK, so I want to define the $k$th order osculating space at $p$ of the submanifold $N$ to be $$T^{(k)}_p(N) = \text{Span}\big(\frac{\partial f}{\partial x_i}(0),\frac{\partial^2 f}{\partial x_i\partial x_j}(0), \dots, \frac{\partial^k f}{\partial x_{i_1}\dots\partial x_{i_k}}(0)\big),$$ where I'm assuming (having chosen a chart for $M$) that $f\colon U\to\Bbb R^{n+m}$ is a local parametrization of $N$ ($0\in U\subset\Bbb R^n$, $f(0)=p$). You can check, using the chain rule, that this is a well-defined subspace of $T_pM$.

Indeed, that calculation shows precisely that the definition you wished to use is equivalent. If $g=f\circ\phi$ for some diffeomorphism $\phi\colon U\to V$ with $\phi(0)=0$, then you can show inductively that $$D^ig(0) = \sum_{j=1}^i c_{jA} D^jf(0)(D^{\alpha_i}\phi(0),\dots, D^{\alpha_j}\phi(0))$$ for appropriate constants $c_{jA}$, where we sum over increasing multiindices $A$ with $\alpha_1+\dots+\alpha_j = i$. (Here I'm treating $D^jf$ as a symmetric $j$-linear map.)

Conversely, because we know that $Df(0)$ has maximal rank, we can also deduce that if the $k$th order osculating spaces according to $f$ and $g$ at $p$ are equal, then we can construct a polynomial $\phi$ of degree $k$ so that $g=f\circ\phi$ (on appropriate domains).

Ted Shifrin
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