# if a tangent at a point of a manifold taken as a function is zero at two points, it is zero at all points

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.

• 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? – Amitai Yuval Apr 21 at 15:09
• $F(M,m)$ 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 – user555729 Apr 21 at 15:11
• 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. – Michał Miśkiewicz Apr 21 at 18:48

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 $$I^2$$.