For a smooth manifold $\mathscr M$ I have seen following definition for the tangent space at a point $m\in\mathscr M$.

Define it to be $(F_m/F_m^2)^*$, where $F_m$ denotes the set of germs of smooth functions vanishing at $m$. My question is, why not replace $F_m$ by $\tilde F_m$ ($\tilde F_m$ denotes the set of germs at $m$) in the "numerator"? I don't see why $(\tilde F_m/F_m^2)^*$ wouldn't work. Thank you in advance fo any response.

  • $\begingroup$ You mean, replace set of germs of functions vanishing at $m$ by set of germs of all functions? $\endgroup$
    – xyzzyz
    Apr 19, 2013 at 17:39
  • $\begingroup$ yes, exactly. I don't understand why it is done like this. $\endgroup$ Apr 19, 2013 at 18:04

2 Answers 2


We have $\tilde F_m= F_m\oplus \mathbb R $, so that $\tilde F_m/F_m^2= F_m/F_m^2\oplus \mathbb R$ and thus $(\tilde F_m/F_m^2)^*= (F_m/F_m^2)^*\oplus \mathbb R^*$, which has dimension one more than the dimension of the genuine tangent space.

So, no, $(\tilde F_m/F_m^2)^*$ is not a correct substitute for the tangent space $(F_m/F_m^2)^*$.

  • $\begingroup$ Thank you for the response, Very helpful. The mistake I made was that I forgot to add the constant functions to the space where the derivations vanish. If I instead chose $\tilde F_m/( F_m^2\oplus C)$ where C is the subalgebra of germs of constant functions would it then work? Thanks $\endgroup$ Apr 20, 2013 at 18:19
  • $\begingroup$ Hello, how can i see $\mathbb{R}$ as a set of germs of $m$ in the manifold? $\endgroup$ Mar 23, 2018 at 22:04
  • 1
    $\begingroup$ @Eduardo: identify a real number $r$ with the germ at $m$ of the constant function $\mathscr M\to \mathbb R:x\mapsto r$. $\endgroup$ Mar 23, 2018 at 23:28

This definition is motivated by the fact that any finite dimensional vector space is isomorphic to its dual space: $V \cong V^*$. As you probably know, the dual space $V^*$ consists of all linear function on the vector space $V$, i.e. all linear $\operatorname{f} : V \to \mathbb{R}$. One key point to note is that $\operatorname{f}(\vec{0}) = 0$, i.e. all linear maps send the zero vector in $V$ to zero in $\mathbb{R}$.

The isomorphism $V \stackrel{\sim}{\longrightarrow} V^*$ is canonical. We send the vector $\vec{v}$ to the linear function $\vec{w} \mapsto \langle \vec{v},\vec{w}\rangle.$

Now, consider a smooth manifold $M$. The tangent space at each point $p \in M$ is a vector space, often denoted $T_pM$. The tangent space consists of all vectors tangent to the manifold $M$ and based at the point $p \in M$. The dual vector space is the cotangent space, written $T^*_pM$. This consists of all linear functions (in this case differential one-forms) defined on the tangent space $T_pM$. In other words, $\omega \in T^*_pM$ is a linear map $\omega : T_pM \to \mathbb{R}$. As we saw above: $T^*_pM = (T_pM)^* \cong T_pM$.

All linear functions on the tangent space $T_pM$, i.e. all one-forms $\omega \in T^*_pM$, must vanish at $p$ since the point $p$ plays the role of the origin in the vector space $T_pM$. In your algebraic notation, $F_p$ are the functions vanishing at $p$. However, these need not be linear functions. Taking the quotient by $F_p^2$, leaves only the linear functions.

As an example, let $F_p$ be the set of analytic functions in $x$ and $y$ which vanish at $x=y=0$. Quotienting by $F^2_p$ means we quotient out by $x^2$, $xy$ and $y^2$. All that is left are the terms generated by $x$ and $y$, i.e. the linear terms vanishing at the origin.

What you end up with is, $F_p/F^2_p \cong T^*_pM \cong T_pM$.

Be careful using $\mathscr{M}$. This is usually reserved for maximal ideals, and not manifolds.

  • 7
    $\begingroup$ Dear Fly by Night, the isomorphism $V \stackrel{\sim}{\longrightarrow} V^*$ is the prototype of a non canonical isomorphism for a vector space not endowed with a euclidean structure (and there is no euclidean structure in Nicolas's question). It is a very, very bad idea to identify $ T^*_p M$ and $ T_pM$ : do not do it! $\endgroup$ Apr 19, 2013 at 20:02
  • $\begingroup$ Thank you very much. I realize now what mistake I made in thinking this. $\endgroup$ Apr 20, 2013 at 18:14

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