Tangent vectors of smooth manifolds as "modified" derivations?

Question

It is a well-known approach to define the tangent space $T_pM$ of a smooth manifold $M$ at $p \in M $ as the vector space of derivations $d : C^\infty(M) \to \mathbb R$ at $p$. Here $C^\infty(M)$ denotes the set of all smooth functions $M \to \mathbb R$ and a derivation at $p$ has the property $d(f\cdot g) = df \cdot g(p) + f(p)\cdot dg$.

In Hitchin's definition of tangent space and tangent vectors we see that $T_pM$ can be identified with the dual space of $C^\infty(M)/Z_p(M)$, where $Z_p(M)$ is the subspace of all $f \in C^\infty(M)$ having derivative $0$ at $p$. That is, a linear map $C^\infty(M) \to \mathbb R$ is a derivative iff it annihilates $Z_p(M)$.

In Paul Frost's answer to Different definitions of the tangent space at a point of a smooth manifold: Biduals? we see that $C^\infty(M)/Z_p(M)$ can be identified with $$C^\infty((M,p),(\mathbb R^m,0))/Z_{(p,0)}M $$ where $C^\infty((M,p),(\mathbb R^m,0))$ is the linear subspace all $f \in C^\infty(M)$ with $f(p) = 0$ and $Z_{(p,0)}M = Z_p(M) \cap C^\infty((M,p),(\mathbb R^m,0))$.

We conclude that $T_pM$ can be identified with the dual space $\tilde T_pM$ of $C^\infty((M,p),(\mathbb R^m,0))/Z_{(p,0)}M$, thus with the set of linear maps $C^\infty((M,p),(\mathbb R^m,0)) \to \mathbb R$ annihilating $Z_{(p,0)}M$.

Question: Is there a nice alternative characterization of linear maps $C^\infty((M,p),(\mathbb R^m,0)) \to \mathbb R$ annihilating $Z_{(p,0)}M$ as "modified derivations"?

It is clear that $d(f\cdot g) = df \cdot g(p) + f(p)\cdot dg$ is inadequate here because $g(p) = f(p) = 0$ in this case, but perhaps there is a bypass?

Answer 1

You are wrong when you say "It is clear that $d(f\cdot g) = df \cdot g(p) + f(p)\cdot dg$ is inadequate here because $g(p) = f(p) = 0$ in this case". In fact, the Leibniz rule reduces to $$d(f \cdot g) = 0 \text{ for } f,g \in C^\infty((M,p),(\mathbb R^m,0)) . \tag{1} $$ But $(1)$ is a restriction, not every linear map $d : C^\infty((M,p),(\mathbb R^m,0)) \to \mathbb R$ will satisfy it. It reflects the well-known fact from elementary calculus that the product of two functions having a common zero at some point has derivative zero at this point (apply the product rule).

As an answer to your question let us prove

Lemma. A linear map $d : C^\infty((M,p),(\mathbb R^m,0)) \to \mathbb R$ annihilates $Z_{(p,0)}M$ if and only if it satisfies $(1)$.

Proof. We extend $d$ to a map $\bar d: C^\infty(M) \to \mathbb R$ by $$\bar d(f) = d(f - f(p)).$$ Note that we regard $f(p)$ as the constant function with value $f(p)$. Clearly $\bar f = f - f(p) \in C^\infty((M,p),(\mathbb R^m,0))$. We have $$\bar d(f + g) = d(f+g -(f(p + g(p)) = d((f - f(p)) + (g - g(p)) = d(\bar f +\bar g) = d(\bar f) + d(\bar g) \\=\bar d(f) + \bar d(g) ,$$ $$\bar d(af) = d(af-af(p)) = d(a(f - f(p))) = d(a\bar f) = a\bar d(f) .$$ Therefore $\bar d$ is a linear map.

1. Assume that $d$ satisfies $(1)$. For $f,g \in C^\infty(M)$ we can easily verify that $$\overline{ f \cdot g} = f \cdot g - f(p)\cdot g(p) = \bar f \cdot \bar g + g(p)\bar f + f(p) \bar g$$ which implies $$\bar d(f \cdot g) = d(\overline{ f \cdot g}) = d(\bar f \cdot \bar g) + g(p)d(\bar f) + f(p) d(\bar g) = g(p)d(\bar f) + f(p) d(\bar g) \\= g(p)\bar d(f) + f(p) \bar d(g) .$$ Therefore $\bar d$ is a derivation, i.e. annihilates $Z_p(M)$. Since $\bar d$ extends $d$, wed see that $d$ annihilates $Z_{(p,0)}M$.

2. Assume that $d$ annihilates $Z_{(p,0)}M$. Then $\bar d$ annihilates all $f \in C^\infty(M)$ such that $\bar f \in Z_{(p,0)}M$. But clearly $f$ has derivative $0$ at $p$ iff $\bar f$ has. Thus $\bar d$ annihilates $Z_p(M)$. It is therefore a derivation. Hence $d$ satisfies $(1)$.

The above considerations also show that restriction $$\rho : \mathcal L(C^\infty(M),\mathbb R) \to \mathcal L(C^\infty((M,p),(\mathbb R^m,0)),\mathbb R)$$ and extension $$\bar {\phantom d} : \mathcal L(C^\infty((M,p),(\mathbb R^m,0)),\mathbb R) \to \mathcal L(C^\infty(M),\mathbb R)$$ establish inverse isomorphisms between the subspaces of derivations.