I'm a physics student with a very loose understanding of the mathematics I use. I'm trying to learn a little more about very basic topology, manifolds, and Riemannian geometry. I'm using Nakahara's Geometry, Topology, and Physics for self-study. I'm trying to read up about vectors on manifolds and the concept of a tangent vector has me thoroughly confused. Up until now I had always pictured the tangent space something like a plane tangent to a point on the surface of a manifold, however if I'm understanding my book correctly the elements of the tangent space seem to be something more like differential operators. I will lay out the definition from my book and then see if perhaps you guys might be able to help me understand exactly what a tangent vector, and what the tangent space is.

My book defines a tangent vector at $c(0)$ as the direction derivative of a function, $f:M \rightarrow \mathbb{R},$ along a curve $c:(a,b) \rightarrow M,$ or $c(t)$ where $a<0$, $b>0$ at $t=0$. The rate of change is

$$\frac{df(c(t))}{dt} \Bigr|_{t=0}.$$

Assuming we've defined all coordinates correctly and stuff this becomes

$$\frac{\partial f}{\partial x^\mu} \frac{d x^\mu(c(t))}{dt} \Bigr|_{t=0}.$$

Defining the differential operator $X$ as

$$X=X^\mu \left( \frac{\partial}{\partial x^\mu} \right),$$ $$X^\mu=\frac{dx^\mu(c(t))}{dt} \Bigr|_{t=0},$$

the directional derivative can be found by applying $X$ to $f$. Putting it all together, this means

$$\frac{df(c(t))}{dt} \Bigr|_{t=0} = X^\mu \left( \frac{\partial f}{\partial x^\mu} \right) = X[f].$$

The book then says, "the last equality defines $X[f]$. It is $X=X^\mu \partial / \partial x^\mu$ which we now define as the tangent vector to $M$ at $p=c(0)$ along the direction given by the curve. This makes the tangent vector a differential operator, it seems like. Intuitively I had thought that the tangent vector would be $X^\mu \partial f / \partial x^\mu$ along the direction of the curve.

The book goes on to say that instead of identifying a tangent vector with a curve you instead identify a tangent vector with the following equivalence class of curves

$$[c(t)] = \left\{ \tilde{c}(t) \, \Bigr| \, \tilde{c}(0)=c(0) \text{ and } \frac{dx^\mu (\tilde{c}(t))}{dt} \Bigr|_{t=0} = \frac{dx^\mu (c(t))}{dt} \Bigr|_{t=0} \right\},$$

and that the tangent space is the collection of all such equivalence classes at $c(0)=p.$ So I'm a little confused. The equation above seems to define a tangent vector as an equivalence class of curves. The previous definition seems to define a tangent vector as a differential operator. My intuition tells me something different altogether. Can someone help me sort through all this? Thanks.

  • $\begingroup$ The equivalence classes of curves define a differential operator, and in fact two curves define the same differential operator iff they're equivalent in this sense. $\endgroup$
    – user98602
    Aug 7, 2015 at 19:01
  • $\begingroup$ Could you possibly explain how the equivalence class defines a differential operator? That just looks like a set of curves to me. Also, this basically says that the tangent vector is independent of the function $f$ to me, which seems very counterintuitive. Shouldn't the magnitude of the tangent vector matter also, which would be dependent on f? $\endgroup$ Aug 7, 2015 at 19:10
  • $\begingroup$ Also, when we think of tangent vector in Euclidean space, is it a differential operator in the same sense? $\endgroup$ Aug 7, 2015 at 19:11
  • $\begingroup$ What happens if you let the basis differential operators operate on the coordinate curves in $E^3? What do you get? Do you think there exists an isomorphism between the differential operator basis and the standard basis? $\endgroup$
    – jeo15
    Aug 7, 2015 at 19:17
  • 1
    $\begingroup$ The function $f$ is arbitrary. The point is that we are defining a tangent vector by the set of directional derivatives of all functions. And the magnitude depends on the parametrisation of the curve(s), i.e. rescaling of the parameter $t$ rescales the vector. $\endgroup$
    – Rhys
    Aug 7, 2015 at 19:18

2 Answers 2


The answer, in my opinion, depends on your general philosophy about mathematics, that is, if you lean towards intuitionism, or logicism, or formalism, or other flavours. For these see e.g. the article on Stanford Encyclopedia, or the Introduction of Lindström et al. (2009), or even better read some of their history in Kline's (1982) fascinating book.

The answer I give here has a markedly strong intuitionist flavour.

There are many concepts in mathematics and geometry which we understand intuitively but are difficult to nail down with a definition. The very notion of number is one of these. What is the number "2", for example? Its Zermelo-Fraenkel set-theoretic definition is this: $$\{\{\},\{\{\}\}\}.$$ Well, this definition is quite far from what I intuit as "2". And I'm not alone; see Benacerraf's (1965) article for example. But this definition, in the whole theory it is part of, captures many important properties of what we understand as "2".

Another example is the notion of function. It's defined in terms of a subset of a Cartesian product, and this definition captures its properties. My intuition of it, though, is more in terms of a procedure or rule associating this with that; probably this is your intuition of it too.

Another example is the notion of (local) orientation of a manifold. It can be defined as an equivalence class of charts and coordinate orderings (Choquet-Bruhat et al.) or of $n$-forms (Marsden et al. 2007). Well, again, my intuition of orientation is somehow more immediate than such definitions, but they capture its properties well.

Still another example is the notion of outer-oriented vector, also called "twisted vector". If you've never heard about such an object, I warmly invite you to take a look at Burke's (1983, 1987, 1995, 1995b), Bossavit's (1991, 1998), Schouten & van Dantzig's (1940), and Schouten's (1989) brilliant works (the last two are more technical but have very neat pictures). It's a beautiful and geometrically very intuitive object: it's like a vector with an orientation not along its own line, but "around" it. A picture can give you a good intuition: within a two-dimensional Eclidean space it is something like this: enter image description here; within a three-dimensional one it is something like this: enter image description here. This object and its cousins can be added, subtracted, etc. just like vectors; they have many useful applications in mechanics, electromagnetism, and the theory of integration on manifolds. The definition of an outer-oriented vector involves equivalence classes of pairs of orientations and vectors (Bossavit 1991, 1998), and captures all the properties of this notion. But my intuition of this object is much more immediate than such a definition.

The point is that saying what something is is different from saying how something is defined. In mathematics and geometry we have many notions that we understand intuitively. When we axiomatize a theory, we aim for economy: we choose a small set of such intuitive notions as primitives – and therefore undefined – and define the remaining ones in terms of these. There are good reasons for proceeding this way: for example, we are less likely to create logical clashes in the properties of the primitives. But the price for this economy is that the definitions of other intuitive notions often end up being very convoluted.

The point of this long preamble is that I don't think that your question can be answered by a definition. Now imagine that someone asks you "what exactly is the number 2?". You could tell them "it's $\{\{\},\{\{\}\}\}$", but such a definition would probably leave them unsatisfied. What would be your answer? The problem, as you understand, is that they lack an intuitive understanding of "2" and its complex uses. They need to build such an understanding. It can only be built by considering many different definitions, examples, and especially by working with the concept.

So my (non-)answer to your question is this: you probably just need to be patient and build an intuitive understanding of it, by considering alternative definitions and especially by using it in practice.

I found the different definitions and perspectives by the following authors very helpful:

  • The great Cartan (1983) apparently thought of tangent vectors as traditional "arrows" in $R^n$, "tangent" to a manifold in the traditional sense of the word – just like you do! He imagined the manifold embedded in some $R^n$, a conceptual device which we can always use for any manifold (Hirsch 1961). (This view of Cartan's was discussed by a mathematician or a philosopher of mathematics, but I can't remember who and have been unsuccessful in finding this reference, sorry.)

  • Chevalley (1946), followed by Penrose & Rindler (1987), sees tangent vectors as differentiation operators; what's interesting is that his definition of manifold doesn't involve atlases of coordinates.

  • Burke's (1987) definition, § II.8, is similar to the traditional one in terms of an equivalence class of curves, but he offers a more pictorial understanding; see his discussion about "addition of curves".

  • Choquet-Bruhat et al. (1996), § III.B.1, discuss and compare several traditional definitions.

  • Kennington (2018) offers many, many insightful and well thought-out remarks and historical notes on the notion of tangent vectors. In particular, search for Remark: Curve-classes and differential operators are unsatisfactory as tangent vectors, for Styles of representation of tangent vectors, and for The true nature of tangent vectors (section numbers aren't reliable, as this book is very frequently updated). I recommend doing a text search for the string tangent vec to find other interesting passages.

If I may add a personal note, in my concrete experience I've always had to deal with parameterized curves – concrete, particular curves on particular manifolds – and therefore I've never needed to consider equivalence classes of curves. This has also been possible because many objects usually represented as tangent vectors, e.g. electric fields or forces, can in fact be better represented as forms, which in my opinion are much easier to understand intuitively and picture geometrically (see Burke's and Bossavit's works above).



The equivalence relation on the (parametrised) curves is exactly that they give the same differential operator at the point $p$; this reconciles those two definitions. See below for some more details.

As for intuition, that can be a little harder. The thing to recognise is that we need a definition of tangent vectors without reference to an ambient space. In other words, we may have a manifold defined abstractly, rather than embedded in some Euclidean space (in general relativity, this is often the case).

Let $\{y^i\}$ be coordinates on $n$-dimensional Euclidean space. Notice that if we take a 'vector' $\vec{v} = (v^1, \ldots, v^n)$, we can think of this as giving the derivatives of the coordinate functions along the straight line parametrised by $y^i(t) = y^i_0 + v^i t$. Indeed, the directional derivative of any function $f$ along this line is given by $$ \frac{df}{dt} = \sum_i v^i \frac{\partial f}{\partial y^i} $$ So we can associate $v$ with the differential operator $\sum_i v^i\partial/\partial y^i$. You can easily check that this gives a linear isomorphism between the vector space $\mathbb{R}^n$ and the space of operators spanned by $\{\partial/\partial y^i\}$.

Let me return briefly to the 'equivalence classes of curves' bit. Choose some other parametrised curve $\gamma : [0,1] \to \mathbb{R}^n$ passing through the same point, i.e. $y^i(\gamma(0)) = y^i_0$. Then the directional derivative of $f$ along the curve at the point $(y^1_0,\ldots, y^n_0)$ is $$ \left.\frac{df(\gamma(t))}{dt}\right\vert_{t=0} = \sum_i \left.\frac{\partial y^i(\gamma(t))}{\partial t}\frac{\partial f}{\partial y^i}\right\vert_{t=0} $$ We define two curves $\gamma, \gamma'$ to be equivalent iff they give the same directional derivative for every function $f$, i.e. the same differential operator. This gives our isomorphism between equivalence classes of parametrised curves passing through some point and first-order differential operators at the point. Note that rescaling the parameter $t$ will rescale the differential operator, which is why we talk about parametrised curves.

Now suppose we have some embedded manifold $X$, with local coordinates $x^\mu$, and embedding functions $y^i(x^\mu)$. A tangent vector (in the familiar sense) to $X$ just gives the infinitesimal change in the coordinates $y^i$ when we change the coordinates $x^\mu$ by an arbitrary infinitesimal amount $\delta x^\mu$. We have $$ \delta y^i = \sum_\mu \delta x^\mu \frac{\partial y^i}{\partial x^\mu} $$ So the corresponding tangent vector, using the notation we introduced before, is $$ \sum_{i, \mu} \delta x^\mu \frac{\partial y^i}{\partial x^\mu} \frac{\partial}{\partial y^i} $$ The $\delta x^\mu$ are arbitrary, and as we vary them, we map out a linear subspace spanned by the 'vectors' $$ \sum_i \frac{\partial y^i}{\partial x^\mu} \frac{\partial}{\partial y^i} = \frac{\partial}{\partial x^\mu} $$

Now we realise that the operators $\partial/\partial x^\mu$ don't actually depend on the embedding at all, and we have successfully defined tangent vectors in an intrinsic way on any differentiable manifold!

This has become a very long answer, but I hope it's helpful.

  • $\begingroup$ Yes, thank you, that does help. To be perfectly honest I was having a little trouble understanding some of what you were explaining because I was a little unfamiliar with the notation. However, it was basic vector calculus stuff, and felt rather stupid about asking follow-up questions regarding it. I figured I would figure that out on my own before following up. $\endgroup$ Aug 11, 2015 at 19:01
  • $\begingroup$ Now that I think I have understood it, let me ask a couple of follow-up questions. 1.) It seems in your typical vector calculus course the definition of the tangent vector is a bit different than what we're talking about here. From what I learned it seems that a tangent vector in the vector calculus sense is the operator you've defined above applied to a particular field, whereas here it is the operator itself. Is that correct? 2.) Can you explain a little bit about how the equivalence class if equivalent to a differential operator? Thanks, and regards $\endgroup$ Aug 11, 2015 at 19:03
  • $\begingroup$ I really don't understand the "applied to a particular field" bit; I've never seen anything like this as the definition of a tangent vector, and don't see how it could make sense. A single function can't possibly encode an entire tangent space; its gradient is just a single vector. Let me add a little more info to my answer regarding the equivalence relation. $\endgroup$
    – Rhys
    Aug 11, 2015 at 19:13
  • $\begingroup$ Okay, thanks regarding the equivalence class. I think I see my mistake regarding the other part now. $\endgroup$ Aug 11, 2015 at 19:22

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