In a short talk, I had to explain, to an audience with little knowledge in geometry or algebra, the three different ways one can define the tangent space $T_x M$ of a smooth manifold $M$ at a point $x \in M$ and more generally the tangent bundle $T M$:

  • Using equivalent classes of smooth curves through $x$
  • Using derivations near $x$
  • Using cotangent vectors at $x$

Just by looking at the definition, it is not at all clear why they should all define the same object. I went through the proof, but judging from their reaction, it was not very meaningful. I wonder if there is any way I can let them "see", with just intuition, that the three definitions are, in certain sense, the same.

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    $\begingroup$ Maybe part of the conceptual issue is that you can define the "tangent space" of a set $K \subseteq {\mathbb R}^n$ (not a submanifold), but that different notions aren't equivalent. For example, you can try to think of smooth functions on ${\mathbb Q}^n$ as those which are restrictions of smooth functions on ${\mathbb R}^n$, then you will be able to discuss differential operators (as limits of difference quotients), but you won't have smooth curves into the space. On a related note, you might think that the $C^k$ tangent space is different from the $C^{k+1}$ tangent space. $\endgroup$
    – Phil Isett
    Feb 19, 2012 at 3:19
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    $\begingroup$ Why, in a short talk to those without much background, would you be feel compelled to explain these equivalences instead of just using one of them for whatever the main point of the talk was? $\endgroup$
    – KConrad
    Feb 19, 2012 at 17:17
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    $\begingroup$ I was talking about homotopy continuation on an algebraic variety. So eventually I must use the last two definitions. But those, I thought, are too abstract, so I wanted to start with something more intuitive, i.e., the first definition. I didn't have to show that they are all equivalent, but that'd be an obvious question to ask. $\endgroup$
    – ssquidd
    Feb 20, 2012 at 16:03

4 Answers 4


What all three definitions have in common is that they each try to capture the first order behavior of a smooth function on $M$.

  1. The derivative of a smooth function $f$ along a curve $\gamma$ with $\gamma(0) = p$ depends on $\gamma$ only insofar as it depends on $\gamma'(0)$, and indeed it recovers the directional derivative of $f$ at $p$ in the direction $\gamma'(0)$. The directional derivatives of $f$ determine the total derivative of $f$ which in turn determines the first order behavior of $f$ (more or less by the definition of the total derivative).

  2. Since a derivation $D$ at $p$ sees only the values of a function $f$ and its derivatives at $p$ (not near $p$), we can replace $f$ by a polynomial by Taylor's theorem. By the Leibniz rule, $D(P)$ depends only on the linear part of a polynomial $P$ and hence $D(f)$ depends only on the first order part of $f$.

  3. Recall that the cotangent bundle of $M$ at $p$ is the space $I/I^2$ where $I$ is the ideal in $C^\infty(M)$ consisting of functions $f$ such that $f(p) = 0$. If we imagine replacing $C^\infty(M)$ by a polynomial ring then $I$ represents the ideal of polynomials whose lowest order part has degree $1$ and $I^2$ is the ideal of polynomials whose lowest order part has degree $2$. In this case $I/I^2$ is naturally identified with the space of linear polynomials. Thus the cotangent bundle at $p$ is in a sense the space of "first order parts" of smooth functions on $M$.

This intuition acutally allows us to be a little more explicit about how the relevant identifications are made.

It's very easy to go from 1 to 2: if $\gamma$ is a curve in $M$ with $\gamma(0) = p$ then $D(f) = (f \circ \gamma)'(0)$ is a point derivation at $p$ which depends only on the equivalence class of $\gamma$ in $T_p M$.

To go from 2 back to 1, let $f$ be a smooth function and let $f(x) \sim \sum_\alpha c_\alpha x^\alpha$ (multi-index notation) be its Taylor series in a coordinate system centered at $p$. Then for any derivation $D$ at $p$ we have $D(f) = c_1 D(x_1) + \ldots + c_n D(x_n)$ by the Leibniz rule, so $D$ corresponds to the tangent vector $(D(x_1), \ldots, D(x_n))$.

To go from 2 to 3, let $D$ be a derivation at $p$ and observe that $D(f) = 0$ for any $f \in I/I^2$ by Taylor's theorem and the Leibniz rule. Thus $D$ determines a linear functional in $(I/I^2)^*$.

Finally, to go from 3 back to 2, let $\ell \in (I/I^2)^*$ and define a point derivation by $D(f) = \ell(f - f(p) + I^2)$.

  • $\begingroup$ This is very good, I did intend to use it in the setting of algebraic varieties, so the use of degree one polynomial feels quite natural. $\endgroup$
    – ssquidd
    Feb 20, 2012 at 16:06

In each of the three cases, your definition is capturing the intuition of "directions near x" --- an equivalence class of curves defines a "direction" in which the curves head out from x. A derivation or a linear functional on the cotangent vectors is a directional derivative, hence determined by a choice of direction. (Of course it takes proof that these account for all the derivations, etc. but the intuition is pretty clear.)


Klaus Jänich's undergraduate-level book "Vector Analysis" includes a section showing the equivalence between the three descriptions of the tangent space that you mention. He gives rigorous proofs of everything, and also provides a fair amount of motivation.


Here is an explanation based on the category-theoretical formulation of tangent spaces.

In short: To give a definition of tangent spaces, one actually define a functor from the category of (pointed)manifolds to the category of vector spaces that is a natural extension of the tangent space of Euclidean spaces satisfying the principle of localization(see below). It can be shown that the tangent space functor is uniquely determined by these two conditions, up to a natural isomorphism.

Hence to check that a concrete definition of tangent space is reasonable, one only need to show that this definition generalizes the concept of tangent spaces of Euclidean spaces, and satisfies the principle of localization. If you have two such definitions, then they are equivalent in the sense that there exists a natural isomorphism between the functors representing the two definitions.

In detail: The category of pointed manifold $\mathcal M$ has objects in the form $(x,M),$ where $M$ is any smooth manifold and $x$ an element of $M.$ A morphism from $(x,M)$ to $(y,N)$ is an ordered quadruple $(x,f,M,N),$ in which $f$ is a functional relation on $M\times N$ with $\mathrm{dom}(f)$ a (not necessarily open)neighborhood of $x,$ $f$ differentiable at $x$ and $f(x)=y.$

More exactly, $f$ is a subset of $M\times N$ such that if $(x,y),(x,y')\in f$ then $y=y'.$ By definition, if $O$ is an open neighborhood of $x$ in $M$ such that $\mathrm{dom}(f)\subset O,$ and $O'$ is an open subset of $N$ with $\mathrm{Im}(f)\subset O',$ then $(x,f,O,O')$ is a morphism from $(x,O)$ to $(y,O').$

The composition of morphisms is given by \begin{equation} \boxed{(y,g,N,Q)\circ (x,f,M,N):=(x,g\circ f,M,Q)} \end{equation} where $g\circ f:=\{(x_1,x_3)|\ \exists \ x_2\in N,\ s.t. \ (x_1,x_2)\in f,\ (x_2,x_3)\in g \}$ is the composition of relations.

If we consider Euclidean spaces only, then we get a full subcategory $\mathcal M(\mathbb E)$ of $\mathcal M.$ Now we define the canonical tangent space of $(x,\mathbb R^n)$ to be simply $\mathbb R^n,$ and the canonical tangent map $T_{\rm canonical}(x,f,\mathbb R^n,\mathbb R^m):=D_xf,$ then we get a functor $T_{\rm canonical}:\mathcal M(\mathbb E)\to \mathsf{Vec}(\mathbb R).$ \begin{equation} \boxed{T_{\rm canonical}\big[(x,\mathbb R^n)\stackrel{(x,f,\mathbb R^n,\mathbb R^m)}{\longrightarrow} (y,\mathbb R^m)\big]:=\mathbb R^n \stackrel{D_xf}{\longrightarrow}\mathbb R^m} \end{equation}

A tangent space functor defined on the full category $\mathcal M$ must generalize the concept of tangent spaces of Euclidean spaces, that is, it must be an extension of $T_{\rm canonical}.$ In fact this extension should be natural, that is to say, the restriction of $T$ to $\mathcal M(\mathbb E)$ is nutural isomorphic to $T_{\rm canonical}.$ In other words, for any object $(x,\mathbb R^n)$ there is an isomorphism $\alpha_{(x,\mathbb R^m)}:T(x,\mathbb R^n)\to T_{\rm canonical}(x,\mathbb R^n)=\mathbb R^n$ such that if $(x,f,\mathbb R^n,\mathbb R^m):(x,\mathbb R^n)\to (y,\mathbb R^m)$ is a morphism, then the following diagram commutes: enter image description here We say that a functor $T:\mathcal M\to\mathsf{Vec}(\mathbb R)$ gives a definition of tangent spaces, if it satisfies the followng two conditions:

  1. $T$ is a natural extension of $T_{\rm canonical}.$
  2. (Principle of localization)If $(x,f,M,N),(x,g,M,N)$ are morphisms and $f,g$ coincide on a neighborhood of $x$ then $T(x,f,M,N)=T(x,g,M,N).$

Of course, if $T$ and $\widetilde T$ are two such functors, then their restriction to $\mathcal M(\mathbb E)$ are naturally isomorphic through the transformation $\beta:=\widetilde\alpha^{-1}\circ\alpha,$ since we have the commutative diagram enter image description here

To show the uniqueness of tangent space functor, we need to find a natural isomorphism between $T$ and $\widetilde T.$

For any object $(x,M)$ of $\mathcal M,$ we define a vector space isomorphism $\gamma_{(x,M)}:T(x,M)\to \widetilde T(x,M)$ by \begin{equation}\boxed{\gamma_{(x,M)}:=\widetilde T(\varphi(x),\varphi^{-1},\mathbb R^n,M)\circ \beta_{(\varphi(x),\mathbb R^n)}\circ T(x,\varphi,M,\mathbb R^n)}\end{equation} in which $(U,\varphi)$ is a coordinate chart for $x$ in $M.$ To show that $\gamma_{(x,M)}$ is well-defined, we have to verify that if $(V,\psi)$ is another coordinate chart for $x$ in $M$ then \begin{equation}\widetilde T(\varphi(x),\varphi^{-1},\mathbb R^n,M)\circ \beta_{(\varphi(x),\mathbb R^n)}\circ T(x,\varphi,M,\mathbb R^n)=\widetilde T(\psi(x),\psi^{-1},\mathbb R^n,M)\circ \beta_{(\psi(x),\mathbb R^n)}\circ T(x,\psi,M,\mathbb R^n)\end{equation}

In fact, principle of localization yields the commutative diagram enter image description here

A similar digram commutes for the functor $\widetilde T.$ On the other hand, since $\psi\circ\varphi^{-1}$ is a functional relation on $\mathbb R^n\times\mathbb R^n,$ it follows that the following diagram commutes: enter image description here

Thus we have a combined commutative diagram

enter image description here

which is desired.

It remains to show that $\gamma$ is a natural isomorphism between $T$ and $\widetilde T.$ Let $(x,f,M,N)$ be a morphism from $(x,M)$ to $(y,N)$ and let $(U,\varphi)$ and $(V,\psi)$ be a coordinate chart for $x$ and $y,$ respectively, then $(\varphi(x),\psi\circ f\circ\varphi,\mathbb R^n,\mathbb R^n)$ is a morphism from $(\varphi(x),\mathbb R^n)$ to $(\psi(y),\mathbb R^n),$ hence \begin{equation}\label{key}\widetilde T(\varphi(x),\psi\circ f\circ\varphi^{-1},\mathbb R^n,\mathbb R^n)\circ\beta_{\varphi(x),\mathbb R^n}=\beta_{\psi(y),\mathbb R^n}\circ T(\varphi(x),\psi\circ f\circ\varphi^{-1},\mathbb R^n,\mathbb R^n)\end{equation} Plugging \begin{equation}(\varphi(x),\psi\circ f\circ\varphi^{-1},\mathbb R^n,\mathbb R^n)=(y,\psi,N,\mathbb R^n)\circ (x,f,M,N)\circ(\varphi(x),\varphi^{-1},\mathbb R^n,M)\end{equation} into the former equation gives \begin{equation}\widetilde T(x,f,M,N)\circ \widetilde T(\varphi(x),\varphi^{-1},\mathbb R^n,M)\circ \beta_{(\varphi(x),\mathbb R^n)}\circ T(x,\varphi,M,\mathbb R^n)=\widetilde T(\psi(y),\psi^{-1},\mathbb R^n,N)\circ\beta_{(\psi(y),\mathbb R^n)}\circ T(y,\psi,N,\mathbb R^n)\circ T(x,f,M,N)\end{equation} hence we have the commutative diagram enter image description here as expected.

The principle of localization is equivalent to the following statement: for any two open subsets $U,U'$ of any manifold $M,$ if $x\in U\cap U'$ then there is an isomorphism $\Delta^{(x,M)}_{U,U'}$(abbreviated as $\Delta_{U,U'}$) satisfying the cocycle condition $\Delta_{U',U''}\circ \Delta_{U,U'}=\Delta_{U,U''},\ \Delta_{U,U}=\mathrm{id}$ and if $(x,f,M,N)$ is a morphism from $(x,M)$ to $(y,N),$ then the following diagram commutes: enter image description here Here $f|_{U}^V:=f\cap (U\times V)$ and it's easy to see that $(x,f|_U^V,U,V)$ is a morphism from $(x,U)$ to $(y,V).$

In fact, assume that the principle of localization hold, then defining \begin{equation}\Delta_{U,U'}^{(x,M)}:=T(x,\mathrm{id}_{U\cap U'},U,U')\end{equation} will make the cocycle diagram commute.

Conversely,assume that the cocycle diagram commute, and assume $f$ coincide $g$ on a neighborhood $U$ of $x,$ then we have the commutative diagram enter image description here Since $\Delta_{U,M}\circ\Delta_{M,U}=\mathrm{id},\ \Delta_{N,N}=\mathrm{id},$ it follows that $T(x,f,M,N)=T(x,g,M,N),$ which establishes the equivalence of the two statements.


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