Different definitions of the tangent space at a point of a smooth manifold: Biduals

differential-geometrytangent-spaces

There are two different definitions of the tangent space $T_pM$ at a point $p$ of a smooth manifold $M$.

  1. Define $T_pM$ as the set of all equivalence classes of curves (smooth functions) $u : (\mathbb R, 0) \to (M,p)$, where $u \sim v$ if $u, v$ have the same derivative at $0$.

  2. Define $T_pM$ as the set of all 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 $d(f\cdot g) = df \cdot g(p) + f(p)\cdot dg$.

Concerning 1. it is worth to mention that having the same derivative at some point $q$ of a smooth manifold $N$ is an equivalence relation on the set of smooth maps $f : N \to M$ which can be defined without previously introducing the concepts of tangent spaces and differentials $d_qf : T_qN \to T_{f(q)}M$. It works via charts around $q$ and $f(q)$.

It is well-known that both definitions are equivalent.

Definition 1. is certainly more intuitive, but its disadvantage is that there is no "intrinsic" definition of a vector space structure on $T_pM$. Only scalar multiplication has an obvious interpretation on the level of curves, addition has to be defined via using charts and observing that each curve with range in an open subset of $\mathbb R^n$ is equivalent to a "locally linear curve" determined by a vector $x \in \mathbb R^n$.

Definition 2. is less intuitive, but its advantage is that the vector space structure on $T_pM$ is provided for free.

My question:

It seems to me that the tangent space in the sense of 2. is something like the bidual of the tangent space in the sense of 1. My intuition says that $(T_pM)^*$ is something like equivalence classes of smooth maps $M \to \mathbb R$ (i.e. maps in $C^\infty(M)$, the equivalence relation again being "same derivative at $p$"), thus $(T_pM)^{**}$ would be something like suitable equivalence classes of maps in $(C^\infty(M))^*$ which could be related to derivations.
I have no idea how this could be made precice. Perhaps somebody can do this?

Best Answer

We're talking about finite-dimensional real vector spaces, so a posteriori $T_p M$ has to be the dual of $T^*_p M$ and $T^*_p M$ has to be the dual of $T_p M$. I'm just mentioning it to emphasize that the question here is rather meta-mathematical.

Anyway, one could introduce the cotangent space $T^*_p M$ first, and then define $T_p M := (T^*_p M)^*$ abstractly. The resulting definition would indeed be naturally consistent with Definition 2, see the explanation below.


So first introduce the (maximal) ideal ${\frak m}_p$ of all smooth functions vanishing at $p$: $$ {\frak m}_p := \{ f \in C^\infty(M) : f(p)=0 \}. $$ Next, consider the ideal ${\frak m}_p^2$ spanned by all possible products $f \cdot g$ for $f, g \in {\frak m}_p$.

Side note: a posteriori, one can see that ${\frak m}_p^k$ consists exactly of those functions for which all derivatives up to order $k-1$ vanish at $p$ (which justifies the definition below). For example, in $\mathbb{R}^n$ this observation can be verified by the Taylor expansion with integral remainder.

Then the cotangent space $T^*_p M$ can be defined as the quotient $$ T^*_p M := {\frak m}_p / {\frak m}^2_p. $$ Given any function $f \in C^\infty(M)$, one can define its differential $d_p f \in T^*_p M$ as follows: first taking the function $f-f(p) \in {\frak m}_p$, and then taking its equivalence class in the quotient space ${\frak m}_p / {\frak m}^2_p$.

Just by definition, one can check that the Leibniz rule holds: $$ d_p (fg) = d_p f \cdot g(p) + f(p) \cdot d_p g. $$ To see this, note that the two functions $fg-(fg)(p)$ and $(f-f(p)) \cdot g(p) + f(p) \cdot (g-g(p))$ differ by $(f-f(p)) \cdot (g-g(p))$, which is an element of ${\frak m}_p^2$.


Now, what if we define a tangent vector $V \in T_p M$ simply as an element of $(T^*_p M)^*$, i.e. as a linear functional on ${\frak m}_p / {\frak m}^2_p$?

Then we can also define the corresponding derivation $\bar{V} \colon C^\infty(M) \to \mathbb{R}$: $$ \bar{V}(f) = V(d_p f). $$ The Lebniz rule for $\bar{V}$ follows from the Lebniz rule for $d_p f$, so $\bar{V}$ is indeed a derivation.


I hope this answers the question how Definition 2 can be seen as dual to the definition of the cotangent space? However, Definition 1 is less algebraic and more directly related to the structure of euclidean spaces, so I don't see how it can be seen as predual to the definition of the cotangent space.