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Let $M$ be a manifold and $m\in M$. We call $t$ a tangent at $m$ if for every pair $(f,g)$ of smooth real functions defined in a neighborhood of $m$ and for every pair $(a,b)$ of real numbers, we have $t(af+bg)=at(f)+bt(g)$ and $t(fg)=t(f)g(m)+f(m)t(g)$. (The domain of $t$ is the the set of all smooth real maps defined in a neighborhood of $m$ and its codomain is $\mathbb R$.)

Now suppose that $M,N$ are manifolds and $p,q$ are respectively projection maps of $M\times N$ onto $M$ and $N$.

I want to show that if for all real smooth map $f$ defined in a neighborhood of $m\in M$ and for all real smooth $g$ defined in a neighborhood of $n\in N$ we have $t(f\circ p)=0, t(g\circ q)=0$ where $t$ is a tangent at $(m,n)$, then $t\equiv 0$. What is the idea here?

I already know a solution but not the solution I am looking for. This solution is easy. It follows from the answer to Maps between tangent space of product manifold and sum of tangent spaces.

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  • $\begingroup$ It would be helpful if you added the type of a tangent. It seems to be some kind of map, but what are its domain and codomain? $\endgroup$ – Amitai Yuval Apr 21 at 15:09
  • $\begingroup$ $F(M,m)$ is defined to be the set of smooth real maps $f$ defined in a neighborhood of $m\in M$. This set is the domain, and the set of real numbers is the codomain. @AmitaiYuval $\endgroup$ – user555729 Apr 21 at 15:11
  • $\begingroup$ So what kind of solution are you looking for? The problem is trivial once you have other equivalent definitions, and obviously some analysis is needed here. $\endgroup$ – Michał Miśkiewicz Apr 21 at 18:48
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Consider any tangent $t$ at $(m,n)$. It follows from the definition that $t$ vanishes on constants and on $I^2$, where $$ I = \{ f : f(m,n) = 0 \} $$ is the ideal of functions vanishing at $(m,n)$. For a function $f$ defined on a neighborhood of $(m,n)$, let us define
$$ \overline{f}(x,y) := f(x,y) + f(m,n) - f(x,n) - f(m,y). $$ We claim that $\overline{f} \in I^2$ for any $f$.

Taking this for granted, let us take $t$ as described. Since $t$ vanishes on $f(x,n)$ and $f(m,y)$ (by assumption), on $f(p,q)$ (as a constant), and on $\overline{f}(x,y)$ (as an element of $I^2$), it also vanishes on any $f(x,y)$, thus $t \equiv 0$.


To see that $\overline{f} \in I^2$, one needs to refer to mathematical analysis (say, the Taylor expansion or the fundamental theorem of calculus). Adopting some product coordinates around $(m,n)$, we have $$ \overline{f}(x,y) = (x-m) \otimes (y-n) \cdot \int_0^1 \int_0^1 D^2 f(m + t(x-m), n + s(y-n)) dt ds, $$ where the coordinates of $(x-m) \otimes (y-n)$ are products of two functions vanishing at $(m,n)$ and the integral is a smooth function of $(x,y)$ around $(m,n)$. In consequence, $\overline{f}$ lies in $I^2$.

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  • $\begingroup$ Thanks, I was just looking for a solution without resort to analysis but apparently that is not possible $\endgroup$ – user555729 Apr 21 at 19:58

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