The Ideal Tangent Space

The tangent space to a manifold \(M\) is usually defined in terms of derivations of smooth functions on \(M\). This definition works most of the time, but some applications benefit from a more ring-theoretic characterization. In this note we present this ring-theoretic characterization and prove its equivalence to the standard definition.

Tangent space as derivations

Suppose we are given a \(C^\infty\)-manifold \(M\) together with a point \(m \in M\). A linear function \(\delta \colon C^\infty(M) \to \mathbb{R}\) is called a derivation at \(m\) if it satisfies the following “product rule”:

\begin{equation*} \delta(fg) = \delta(f) g(m) + f(m) \delta(g) \end{equation*}

It is easy to see that the set of derivations at \(m\) is closed under addition and scalar multiplication, and hence that it is a vector space. The usual definition of the tangent space \(T_mM\) is precisely this vector space of derivations.

Ring theoretic tangent space

Now let \(\mathcal{O}\) denote the ring of germs of smooth functions at \(m\). Let \(\mathrm{ev}_m \colon \mathcal{O} \rightarrow \mathbb{R}\) be the evaluation map, ie.

\begin{equation*} \mathrm{ev}_m(f) = f(m) \end{equation*}

This evaluation map is a ring homomorphism, meaning that \(\mathfrak{m} = \mathrm{ker}(\mathrm{ev}_m)\) is an ideal of \(\mathcal{O}\). Moreover, \(\mathbb{R} \cong \mathcal{O}/\mathfrak{m}\) implies \(\mathfrak{m}\) is maximal. We show that the tangent space \(T_mM\) can be recharacterized as the dual space \((\mathfrak{m}/\mathfrak{m}^2)^\ast\). In other words, we exhibit an isomorphism of the form

\begin{equation*} T_mM \cong (\mathfrak{m}/\mathfrak{m}^2)^\ast \end{equation*}

Let \(\delta\) be any derivation at \(m\), and suppose we are given \(F \in \mathfrak{m}^2\). By definition we can write \(F = fg\) for some \(f, g \in \mathcal{O}\). Applying \(\delta\) we obtain

\begin{equation*}\delta(F) = \delta(f) g(m) + f(m) \delta(g) = 0\end{equation*}

Thus \(\delta\) descends to a linear function \(\mathfrak{m}/\mathfrak{m}^2 \to \mathbb{R}\). Clearly unique derivations have unique images in \((\mathfrak{m}/\mathfrak{m}^2)^\ast\), and so we have the inclusion

\begin{equation*}T_mM \subseteq \left(\mathfrak{m}/\mathfrak{m}^2\right)^\ast\end{equation*}

To obtain the reverse inclusion, suppose we have a linear functional \(\lambda \in (\mathfrak{m}/\mathfrak{m}^2)^\ast\) and an arbitrary \(f \in \mathcal{O}\). The modified function \(f - f(m)\) vanishes at \(m\), so that \(f - f(m) \in \mathfrak{m}\). We can therefore define a derivation \(\delta_\lambda\) according to the formula

\begin{equation*}\delta_\lambda(f) = \lambda(f - f(m) + \mathfrak{m}^2)\end{equation*}

The correspondence \(\lambda \mapsto \delta_\lambda\) is clearly injective, whence we may conclude

\begin{equation*}T_mM = \mathfrak{m}/\mathfrak{m}^2\end{equation*}

as claimed.