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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?

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  • $\begingroup$ Using coordinate charts we can transport all the discussion to Euclidean spaces, as derivatives at a point and derivations at a point are both defined locally. In Euclidean spaces derivations are exactly the directional derivatives, and the derivative of a curve at a point defines a direction, that at the same time can be assigned to a directional derivative in such direction (and viceversa). Thus all three concepts: directions in Euclidean spaces, derivations and derivatives of curves, are all equivalent in the sense that can be related by bijections. $\endgroup$
    – Masacroso
    Sep 6, 2021 at 1:37
  • $\begingroup$ (continuation) However I dont see any relation to dual or bidual maps. $\endgroup$
    – Masacroso
    Sep 6, 2021 at 1:38
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    $\begingroup$ There is a standard direct definition of the cotangent space as a quotient $I_p/I_p^2$, where $I_p$ is the ideal of all smooth functions that vanish at $p$. You can then show that its dual is the space of derivations at $p$. $\endgroup$
    – Deane
    Sep 24, 2021 at 17:10

2 Answers 2

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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 $$ T^*_p M := {\frak m}_p^2 / {\frak m}_p. $$ 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 ${\frak m}_p^2 / {\frak m}_p$.

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 ${\frak m}_p^2 / {\frak m}_p$?

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.

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This is an interesting question. It is in fact well-known that both definitions are equivalent, but I have never before considered the corresponding vector spaces as biduals.

I think your point of view is inspiring. Let $T^c_pM$ denote the variant of the tangent space based on curves and $T^d_pM$ the variant based on derivations. Certainly $T^d_pM$ is not the bidual of $T^c_pM$ in the formal sense $T^d_pM = (T^c_pM)^{**}$, where $V^*$ denotes the standard dual space of the vector space $V$, but it comes close to it in a natural way. This means that if somebody knows the construction of $T^c_pM$, he will probably come upon $T^d_pM$ by considering the bidual of $T^c_pM$.

As mentioned in the question, $T^c_pM$ is the vector space of equivalence classes of smooth maps $(\mathbb R,0) \to (M,p)$, where the equivalence relation is having the same derivative at $0$. The key is to explain that we can naturally identify the cotangent space $(T^c_pM)^*$ (formal dual!) with the vector space of equivalence classes of smooth maps $(M,p) \to (\mathbb R,0)$, where the equivalence relation is again having the same derivative at $p$. Roughly speaking, dualizing can be performed on the level of equivalence classes of smooth maps by exchanging domain and range. At first glance this may look surprising, but it is a consequence of the fact that this is (more or less trivially) true for $M = \mathbb R^n$.

  1. For manifolds $M, N$ we let $C^\infty(M,N)$ denote the set of smooth functions $M \to N$. For $p \in M , q \in N$ we let $C^\infty((M,p),(N,q))$ denote the subset of functions mapping $p$ to $q$.

  2. For $f, g \in C^\infty(M,\mathbb R^m)$ we define the relation of having the same derivative at $p \in M$, $f\sim_p g$, by requiring that for each chart $\phi : U \to U' \subset \mathbb R^n$ around $p$ we have $D_{\phi^{-1}(p)}f \circ \phi^{-1} = D_{\phi^{-1}(p)}g \circ \phi^{-1}$. Here $D_x u$ denotes the usual derivative of a smooth map $u : U' \to \mathbb R^m$ at $x \in U'$ which is linear map $\mathbb R^n \to \mathbb R^m$. Let us write $[M,\mathbb R^m]_p$ for the set of equivalence classes. $C^\infty(M,\mathbb R^m)$ is vector space and $\sim_p$ is compatible with the vector space operations. Hence $[M,\mathbb R^m]_p$ is again a vector space. It is easy to verify that $$[M,\mathbb R^m]_p = C^\infty(M,\mathbb R^m)/Z_pM$$ where $Z_pM$ denotes the subspace of all smooth maps have a zero-derivative at $p$. Clearly $\sim_p$ restricts to an equivalence relation on the vector space $C^\infty((M,p),(\mathbb R^m,0))$ which produces a quotient vector space $$[(M,p),(\mathbb R^m,0)] = C^\infty((M,p),(\mathbb R^m,0))/Z_{(p,0)}M .$$ Note that the inclusion $C^\infty((M,p),(\mathbb R^m,0)) \hookrightarrow C^\infty(M,\mathbb R^m)$ induces an isomorphism $[(M,p),(\mathbb R^m,0)] \to [M,\mathbb R^m]_p$. This is true because each $f \in C^\infty(M,\mathbb R^m)$ is equivalent with respect to $\sim_p$ to a smooth map $\bar f \in C^\infty((M,p),(\mathbb R^m,0))$ - compose $f$ with the translation from $f(p)$ to $0$. Abusing notation we shall sometimes write $$[M,\mathbb R^m]_p = [(M,p),(\mathbb R^m,0)] .$$ Note that for $M = \mathbb R^n$ each $[f] \in [\mathbb R^n,\mathbb R^m]_p$ has a unique linear representative, that is, we can identiy $$[\mathbb R^n,\mathbb R^m]_p \equiv \mathcal L(\mathbb R^n,\mathbb R^m) .$$ Moreover, each chart $\phi : U \to U' \subset \mathbb R^n$ around $p$ induces an isomorphism $$\phi^* : [\mathbb R^n,\mathbb R^m]_{\phi(p)} \to [M,\mathbb R^m]_p .$$

  3. In the definition of $[M,\mathbb R^m]_p$ we cannot replace $\mathbb R^m$ by an arbitrary $m$-manifold $N$. To define the relation of $f,g$ having the same derivative at $p \in M$ we need to consider charts around the image points of $p \in M$, but we cannot expect that $f(p)$ and $g(p)$ always lie in a common coordinate neigborhood. We therefore restrict to the set $C^\infty((M,p),(N,q))$ and can now define $f\sim_{(p,q)} g$ similarly as above via charts around $p$ and $q$. This gives us the set of equivalence classes $[(M,p),(N,q)]$. It does not have the structure of a vector space in an obvious way as in the special case $[(M,p),(\mathbb R^m,0)]$, but each chart $\psi : V \to V' \subset \mathbb R^m$ around $q$ with $\psi(p) = 0$ induces a bijection $$\psi_* : [(M,p),(N,q)] \to [(M,p),(\mathbb R^m,0)] = [M,\mathbb R^m]_p .$$ This induces a vector space structure on $[(M,p),(N,q)]$ which turns out to be independent of the choice of $\psi$.

  4. Each chart $\psi : V \to V' \subset \mathbb R^m$ around $q$ with $\psi(p) = 0$ indces an isomorphism $$T^c_pM = [(\mathbb R,0),(M,p)] \stackrel{\psi_*}{\to} [(\mathbb R,0),(\mathbb R^m,0)] = [\mathbb R,\mathbb R^m]_0 = \mathcal L(\mathbb R,\mathbb R^m) \equiv \mathbb R^m.$$ Here $\equiv$ denotes the obvious isomorphism. Therefore $$(T^c_pM)^* \approx (\mathbb R^m)^* = \mathcal L(\mathbb R^m,\mathbb R) \equiv [\mathbb R^m,\mathbb R]_0 \approx [M,\mathbb R]_p .$$
    Both isomorphisms $\approx$ are induced by $\psi$ and one can check that the isomorphism $$(T^c_pM)^* \to [M,\mathbb R]_p = C^\infty(M,\mathbb R)/Z_pM$$ does not depend on $\psi$. This fact is interesting (and well-known): The dual of $T^c_pM = [(\mathbb R,0),(M,p)]$ can be naturally identified with $[(M,p),(\mathbb R,0)] = [M,\mathbb R]_p$.

  5. We conclude that the bidual $(T^c_pM)^{**}$ can be identified with the dual of $[(M,p),(\mathbb R,0)] = [M,\mathbb R]_p$. But the latter "is" nothing else than $T^d_pM$. See Hitchin's definition of tangent space and tangent vectors. In that sense the bidual of $T^c_pM$ is $T^d_pM$.

Remark:

The dual space construction comes together with a natural bilinear form $$\beta : V^* \times V \to \mathbb R, \beta(f,v) = f(v) .$$ This is a special case of a dual system $(W,V,b)$ where $W,V$ are vector spaces and $b: W \times V \to \mathbb R$ is a non-degenerate bilinear form. For a finite-dimensional $V$ all dual systems $(W,V,b)$ are naturally isomorphic (in the obvious sense). In other words, they are essentially nothing else $(V^*,V,\beta)$.

For $V = T^c_pM = [(\mathbb R,0),(M,p)]$ we can define $$b : [(M,p)(\mathbb R,0)] \times [(\mathbb R,0),(M,p)] \to \mathbb R, b([f],[u]) = (f \circ u)'(0) .$$

It is a nice exercise (using charts) to show that this is a well-defined non-degenerate bilinear form. This provides an alternative proof for $(T^c_pM)^* \approx [(M,p)(\mathbb R,0)]$.

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