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Let $M$ be a smooth manifold (finite dimensional, Hausdorff and second-countable) and $p\in M$ a point. The higher cotangent space at $p$ is defined to be quotient: $$ {T^*_p}^rM:= \eta_p/\eta_p^{r+1} $$ Where $\eta_p:=\{ f\in C^{\infty}_p(M) \; | \; f(p)=0 \}$ the germs of smooth functions that vanish at $p$. The $r$-tangent space at $p$ is the dual: $$ {T^{\square}_p}^rM:={({T^*_p}^rM)}^* $$ If $v\in {T^{\square}_p}^rM$, we have a linear map: $$ v: C^{\infty}_p(M) \longrightarrow \mathbb R$$ $$ v(f):=v[f-f(p)] $$

I'm looking for an abstract characterization of the $r$-tangent space in terms of linear maps: $$ v: C^{\infty}_p(M) \longrightarrow \mathbb R$$ If $r=1$, $v\in T^{\square}_pM$ if only if satisfy the Leibniz rule: $$ v(fg)=v(f)g(p)+f(p)v(g) $$

This is due to the comparison: $$ fg-f(p)g(p)=(f-f(p))g(p)+f(p)(g-g(p))+(f-f(p))(g-g(p))$$

This may be easy but I have difficulties having a higher identity: $$f_1...f_{r+1}-f_1(p)...f_{r+1}(p)=(f_1-f_1(p))...(f_{r+1}-f_{r+1}(p))+...$$

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    $\begingroup$ Based on your definition, your $v$ is an order $r$ partial differential operator. The best you can say is that there exists a finite set of pairs $\{(u_i, w_i)\} \subseteq ( T_p^{\Box^{r-1}})^2$ such that $$ v(fg) - v(f)g - f v(g) = \sum u(f)w(g) + u(g)w(f)$$ $\endgroup$
    – Willie Wong
    Commented Nov 11, 2020 at 17:27
  • $\begingroup$ Thanks. I was thinking that always appears factors of the form $(f-f(p))(g-g(p))...$. Then if $f(p)=g(p)=0$, $v(f\cdot g)=\delta (v)(f,g)$, with the coproduct $\delta: {T^{\square}_p}^rM \longrightarrow {T^{\square}}^{r-1}M \otimes {T^{\square}}^{r-1}M $. It gives us the set of vectors $\{(u_i,w_i)\}$. I think this result it's not very useful. $\endgroup$
    – alexpglez98
    Commented Nov 11, 2020 at 21:33

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