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Let $C^{\infty}_M(TM)$ be the module over $C^{\infty}(M)$ of vector fields on a smooth (real) manifold $M$. Let $I_p=\{f\in C^{\infty}(M): f(p)=0\}$ be the ideal of smooth functions on $M$ that vanish at a point $p$. I want to show that $T_pM\simeq C^{\infty}_M(TM)/I_pC^{\infty}_M(TM)$ (I believe this to be true, a professor has said this in class, but I haven't seen a reference anywhere).

My thought process: We can define $e_p: C^{\infty}_M(TM)\to T_pM$ by $e_p(X)=X_p$. This is surjective, as we can extend any tangent vector $v_p$ to a vector field on $M$ using bump functions. Moreover, $e_p$ is $(C^{\infty}(M),\mathbb{R})$ linear. I want to show that $\ker e_p=I_pC^{\infty}_M(TM)$, then we can use the isomorphism theorem to deduce this.

That $I_pC^{\infty}_M(TM)\subset \ker e_p$ is immediate. The other inclusion is a bit more difficult for me to prove. Let $X\in C^{\infty}_M(TM)$ with $X_p=0$. How would we show that $X$ can be written as $fY$ for $f\in I_p$, $Y\in C^{\infty}_M(TM)$? I suspect it has something to do with the fact that $I_p\subset C^{\infty}(M)$ is a maximal ideal, but I'm not immediately seeing the connection.

Edit: I suspect it stems from the idea that $C^{\infty}_M(TM)\setminus I_p$ consists of units since $I_p$ is maximal.

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How would we show that $X$ can be written as $fY$ for $f \in I_p$, $Y \in C^\infty _M (TM)$?

This is not true! Fix a coordinate system $x^i$ with $x^i(p)=0$ and define the "radial" vector field $X^i = x^i$ (which can be smoothly extended to all of $M$) so that $X^i|_p = 0$ and $\partial_j X^i|_p = \delta^i_j$. If $X=fY$ for $f$ vanishing at $p$, then differentiating with the product rule we get $$\delta^i_j = \partial_j X^i|_p = (\partial_j f)(p) Y^i_p.$$ But the LHS has full rank while the RHS has rank at most one (since it is the simple tensor $Df \otimes Y$), giving us a contradiction in dimensions $\ge 2$.

However, I believe the proposition you started with is true, but you're interpreting it incorrectly. My algebra is rusty, but I'm pretty sure that in order for $I_p C^\infty_M(TM)$ to be something nice enough to quotient out by, you need to define it using an additive closure; i.e. $$I_p C^\infty_M(TM) := \{ \sum_{i=1}^n f_i Y_i | n \in \Bbb N, f_i \in I_p, Y_i \in C^\infty_M(TM) \}.$$

Using this definition the proof is easy: let $n$ be the dimension and choose $f_i = X^i$ and $Y_i = \partial_i$ in any coordinate system you please.

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  • $\begingroup$ Yes, now I feel silly. That is the definition of $I_pC_M^{\infty}(TM)$. $\endgroup$ – Moya Oct 15 '15 at 5:53

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