I'm reading Frankel's The Geometry of Physics, a pretty cool book about differential geometry (at least from what I understand from the table of contents). In the first chapter, we are introduced to the notion of a tangent vector to a manifold:

A tangent vector, or contravariant vector, or simply a vector at $p_0 \in M^n$ [where $M^n$ is an n-dimensional "nice" enough manifold], call it $\mathbf{X}$, assigns to each coordinate patch $(U, x)$ [where $U \subset M^n$ and $x=(x^1,\dots,x^n)$ are the coordinates) holding $p_0$ an $n$-tuple of real numbers

$$(X^i_U) = (X^1_U,\dots,X^n_U)$$

such that if $p_0 \in U \cap V$, then

$$X^i_V = \sum_j \left(\frac{\partial x^i_V}{\partial x^j_U} \right)_{p_0}X^j_U $$

So far it's fine, this is a reasonable definition of a vector. Now if $f:M^n \to \mathbb{R}$, we define the derivative of $f$ with respect to $\mathbf{X}$:

$$\mathbf{X}_p(f) = D_{\mathbf{X}}(f) = \sum_j \left( \frac{\partial f}{\partial x^j} \right)_p X^j$$

Again, this makes sense. The author points out that there is a one-to-one correspondence between vector and differential operators of the form $\sum_j X^j\left( \frac{\partial}{\partial x^j}\right)_p$, which is not hard to picture. But then the book says that

we shall make no distinction between a vector and its associated differential operator.

What's the point of this? I understand that there's an association between vectors and operators, and that this might be useful. But why would we make no distinction? It seems that, while equivalent, the two have quite different interpretations.

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    $\begingroup$ Dear Javier, you are right : there is no reason to make no distinction! Actually there are three ways to define tangent vectors: the two you mention and a third by equivalence classes of curves going through $p$.You should elegantly choose between them according to context, but there is definitely no harm done by remembering that they are different beasts. $\endgroup$ Sep 23, 2013 at 22:53
  • $\begingroup$ I answered a similar question recently, though it seems you're more looking for motivation than explanation. In some sense it's just a matter of notational convenience - we can simply write $Vf$ instead of $D_V f$, and likewise the Lie bracket $[V,W]$ is literally the commutator of the differential operators $V,W$. When working with an abstract manifold you can no longer identify vectors with point displacements, so the other "interpretation" doesn't mean much any more. $\endgroup$ Sep 23, 2013 at 23:43
  • $\begingroup$ @GeorgesElencwajg: If you think that's the answer, then post it as so. But I still wonder: if there's no reason to make no distinction, why does the author make no distinction? $\endgroup$
    – Javier
    Sep 25, 2013 at 22:26
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    $\begingroup$ Dear Javier: I think one should maintain the distinction, but that's just an opinion and I don't want to make an answer of it. I don't know why the author wants to blur that distinction. $\endgroup$ Sep 26, 2013 at 6:40
  • $\begingroup$ @JavierBadia You may find this previous answer of mine useful: math.stackexchange.com/questions/65766/… $\endgroup$
    – Neal
    Sep 30, 2013 at 3:00

5 Answers 5


There are three common ways to define tangent vectors $v\in T_{p}M$ on abstract manifolds:

  • As linear operators $C^{\infty}\left(M\right)\to\mathbb{R}$ satisfying the Leibniz law $v\left(fg\right)=f\left(p\right)vg+g\left(p\right)vf$

  • As equivalence classes of curves satisfying $\gamma\left(0\right)=p$, where two curves are equivalent if their first derivatives at zero agree in some chart

  • As assignments of tuples $v_{\varphi}=\left(v_{\varphi}^{1},\ldots,v_{\varphi}^{n}\right) : \operatorname{dom}( \varphi )\to \mathbb{R}^{n}$ to charts $\varphi$ such that $v_{\varphi},v_{\psi}$ are related by the Jacobian of $\varphi\circ\psi^{-1}$.

The reason the first is so popular is that it does not require one to talk about coordinate charts - all the dependence on coordinates is encapsulated in the dependence on the ring of smooth functions, $C^{\infty}\left(M\right)$. Once one has established the well-definedness of $C^{\infty}\left(M\right)$, this definition is truly coordinate-free and self-evidently well defined. This "cleanliness" is just an aesthetic advantage (one can of course check that the other definitions are perfectly consistent and that all three are equivalent), but it's a significant one - it's much easier to handle the concepts in DG if you keep the definitions as clean as possible, even if computations sometimes require one to fix a chart and get their hands dirty.

Now, as to interpretation - I argue that in general, the first and second definitions above are much more natural interpretations of what a vector actually is than the third. My reasoning stems from thinking about what you can actually do with a vector in an abstract smooth manifold.

In an affine space, the natural interpretation of a tangent vector is as a displacement - one can take a vector $v$ at a point $p$ and translate over to $p+v$; and this addition is literally addition of components in cartesian coordinates, so we have a very close relationship with the vector components.

This does not work on a general smooth manifold - you need at least a metric for this to have any analog at all (the exponential map), and even then it's not true that $``p+v"^i=p^{i}+v^{i}$. In the abstract case, vectors indicate an "infinitesimal" direction, which can be formalised as one of the first two definitions I gave above:

  • vectors are the directions in which you can differentiate functions; or

  • vectors are the directions in which curves can be travelling.

The picture you have in your head should look the same as before - vectors are still directions with magnitudes attached to points. The difference is that since there is no longer a literal formalisation of this concept (displacements in affine spaces), we have to choose a new formal definition of what a "direction" is; and I believe that while the component definition is the most familiar in terms of computation, it is not a good conceptual model.

  • $\begingroup$ I think I like your answer the most, but I'll wait a bit longer just in case another one pops up. Thanks! $\endgroup$
    – Javier
    Sep 30, 2013 at 2:30

In this paper by David Hestenes, a modern proponent of applying clifford algebra in physics, Hestenes explores some of his own frustration with this identification. He attributes the "formula"

$$e_i = \frac{\partial x}{\partial x^i}$$

to Cartan. Hestenes concludes that Cartan did not mean this literally (when you write the derivative as a limit, a difference between points has no meaning) and merely used it as a "heuristic device." Since none of Cartan's later arguments relied on this particular point of logic, it didn't matter. Hestenes attributes it to later mathematicians that the subtle nonsense in the formula involved, useful though it may have been, drove people to abandon it by dropping the $x$, leaving

$$e_i = \frac{\partial}{\partial x^i}$$

and thus the identification of tangent vectors and partial derivatives was born. Is this accurate to history? Not sure. But it makes a nice and concise story to tell your children who are struggling with their differential geometry bedtime stories.

Anyway, Hestenes has reasons for attacking this identification. It's fundamentally incompatible with the greater structure of clifford algebra (partial derivatives commute, whereas clifford algebra demands a wedge product of vectors that anticommutes).

However, clifford algebra as applied to differential geometry can, in principle, neatly dodge some of the problems involved in having to consider a proliferation of charts to get anything useful done. By considering an isomorphism between a typical manifold and a so-called "vector manifold," Hestenes turns the points on a manifold into vectors in some vector space (in other words, he embeds everything; the arbitrary nature of embedding is dodged by embedding in an infinite dimensional vector space) and is able to identify $x$ as a vector and thus $\partial x/\partial x^i$ as a meaningful geometric quantity.

At any rate, what I wish to point out is that while the identification of tangent vectors and partial derivatives may have had good motivations, it does deprive one of being able to apply clifford algebra to differential geometry problems.

Given the power of clifford algebra in concisely describing geometric quantities beyond simple vectors, this is a nontrivial price to pay, in my opinion. Being able to algebraically handle the tangent space not as some abstract thing that holds vectors but as an oriented $k$-vector called the pseudoscalar of the manifold is particularly powerful, for instance. It makes questions of orientable manifolds trivial because the pseudoscalar is multi-valued for nonorientable manifolds. And when phrased that way, orientability becomes one of the most natural things in the world. Compared to the clumsy, clumsy definitions you usually see in differential geometry, involving determinants of transition maps and so on, a clifford algebra viewpoint of differential geometry puts much more emphasis on the geometry, in my opinion.


Well, there are numerous equivalent interpretations of tangent vectors and tangent spaces, where given a smooth manifold $M$, locally having a smooth chart $ \phi : U \rightarrow V\subset \mathbb{R}^n $ for $ U \subseteq M $, you can define tanjent vectors as vectors of the form $ J_\phi^{-1}(\phi(p))(a_1, a_2,...,a_n) $ for a $ p \in M $ where $ J_\phi $ is the Jacobian, so this is clearly a vector when $M$ is embedded in some Eucleadean space. But in general it is defined as duals, i.e.linear map $X_p : C^\infty_p(M) \rightarrow \mathbb{R} $ that satisfies Leibneiz's rule i.e. for all $ f , g \in C^\infty_p(M)$(this means you indentify smooth functions that are locally same in neighbourhood of $p$) $$ X_p(fg) = f(p)X_p(g) + g(p)X_p(f) $$ Now for $ \phi = (x^1,...,x^n)$ if you define linear maps $\partial/\partial x^i |_p $ as $$ \frac{\partial}{\partial x^i}\Bigg|_p(f) = \frac{\partial (f\circ \phi^{-1})}{\partial x_i}(\phi(p))\ \ \ \text{which is the usual partial derivative} $$ then it's easy to show that $\{\partial/\partial x^i|_p\} $ span $T_p(M) $ i.e. the vector space of all such $X_p$. Now you can identify $$ J_\phi^{-1}(\phi(p))(a_1, a_2,...,a_n) \sim \sum^n_{i= 1} a_i\frac{\partial}{\partial x^i}\Bigg|_p $$

Both these definitions are interchangeably used in different areas according to convenience, if you look for some analytic problem then the fisrt one is more useful. However the second definition has beautiful algebraic properties which are put to lot of use. This derivation algebra is closed under differential operator, and provide a rigorous foundation of differential forms which can just be defined as alternating subspaces of tensor products of the dual $ T_p(M)^* $. You can get to know lots of these as you go along. There are lots of such things studied in differential geometry, where such identifications are made. The point is to study the geometric properties of such objects without distinguishing whether they are operators, vectors, equivalence class of curves, or any other abstract algebraic elements.


I think the point that the author tries to make is that it is the only sensible way to define a vector on a manifold. When you have a manifold, vectors are elements of the tangent space, but the tangent space changes from point to point, so how would one describe a basis of vectors $\{ e_i \}$ at a given point?(try it!) The only sensible way to do it is by finding independent directions at each point. Because of the fact that we have a well defined coordinate system we can guarantee that the directions of change of the coordinate system at a point will be linearly independent. These directions may be represented by any symbol, but the differential operators $\frac{\partial}{\partial x^i}$ have the intention of providing the rate of change of any (differentiable) function defined on the manifold, while being a basis for the tangent space at each point. One can think of the $\frac{\partial}{\partial x^i}$ as directions (vectors) or as rates of change along directions (differential operators), whichever is more useful, but they are just that: a way to encode independent directions at each point, that depends on the chosen coordinate system.

  • $\begingroup$ Why is it the only sensible definition? If I have a point $p \in M$ and an open set $U \ni p$ with local coordinates $\phi: U \to \mathbb{R}^n$, then I can choose (for example) the standard basis of $\mathbb{R}^n$, and if I want to use another coordinate system, the vectors transform as dictated by the Jacobian, so my basis is well-defined regardless of the coordinates. Isn't this right? $\endgroup$
    – Javier
    Sep 29, 2013 at 23:23
  • $\begingroup$ Or we can define tangent vectors to be equivalence classes of smooth curves passing through a point. $\endgroup$
    – user14972
    Sep 30, 2013 at 4:40
  • $\begingroup$ The basis is well defined regardless of the coordinate system you use. If you do a coordinate transformation you will end up with a basis of the same type, given by differential operators $\frac{\partial }{\partial y^j}$. You can also define non-coordinate basis, these are basis formed by linear combinations of differential operators that cannot be transformed into a canonical basis of the type $\frac{\partial }{\partial y^j}$ for any coordinate system. Still, the only sensible way is to define linear independent directions, and these include also l.i. combinations of "coordinate directions". $\endgroup$ Sep 30, 2013 at 8:14

A vector is an element of a vector, or 'linear' space. that is it pure and simple.

A linear space is any ordered pair of sets $(X, F)$ with F a field and with an operator + defined on X such that if $x$, $y$ are in $X$, then so is $aX+bY$ where a b are in F. The operator $+$ is commutative in the sense that $a + b = b + a$, etc.

The author of the book is asserting an isomorphism between the two linear spaces he discusses.

In applying these ideas to smooth manifolds, I find it useful to think in terms of fibre bundles. A fibre bundle is a set X with a distinguished subset A called the base space and a projection $$\pi :X \to A % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqiWdaNaai % OoaiaadIfacqGHsgIRcaWGbbaaaa!3C01! $$satisfying a set of obvious rules. Now think of your manifold as the base space. the fibre at the point $$a \in A % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyyaiabgI % Giolaadgeaaaa!3926! $$is given by $${\pi ^{ - 1}}(a) % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqiWda3aaW % baaSqabeaacqGHsislcaaIXaaaaOGaaiikaiaadggacaGG$$ If each fibre is a linear space this forms a vector bundle. If we insist that a fibre is the set of all basis representations of the vector space, then we have a principal bundle together with the 'gauge group' $U(a)$ of unitary mappings between bases. See for example; https://www.omicsonline.org/open-access/unitary-representations-of-the-translational-group-acting-as-localdiffeomorphisms-of-spacetime-2090-0902-1000233.php?aid=90406

  • $\begingroup$ This site supports latex markup, eclose the latex formula between two \$ signs. $\endgroup$
    – peterh
    Jan 14, 2018 at 8:23
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    $\begingroup$ sorry i can do advanced math but at 69 latex is a foreign language $\endgroup$ Jan 14, 2018 at 8:31
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    $\begingroup$ Ok, no problem. It is better if you would mention this in your profil, so people around will help to improve your posts, particularly if they find advanced math in it. $\endgroup$
    – peterh
    Jan 14, 2018 at 8:42
  • $\begingroup$ Meanwhile I am eighteen with latex familiar and advanced math as foreign language LOL $\endgroup$ May 1, 2021 at 8:50

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