Constructing the Tangent and Cotangent Space

This one confused me for a long time, so I figured I should write this down before I forgot again.

Let ${M}$ be an abstract smooth manifold. We want to define the notion of a tangent vector to ${M}$ at a point ${p \in M}$ . With that, we can define the tangent space ${T_p(M)}$ , which will just be the (real) vector space of tangent vectors at ${p}$ .

Geometrically, we know what this should look like for our usual examples. For example, if ${M = S^1}$ is a circle embedded in ${\mathbb R^2}$ , then the tangent vector at a point ${p}$ should just look like a vector running off tangent to the circle.
Similarly, given a sphere ${M = S^2}$ , the tangent space at a point ${p}$ along the sphere would look like plane tangent to ${M}$ at ${p}$ .

However, the point of an abstract manifold is that we want to see the manifold as an intrinsic object, in its own right, rather than as embedded in ${\mathbb R^n}$ . This can be thought of as analogous to the way that we think of a group as an abstract object in its own right, even though Cayley’s Theorem tells us that any group is a subgroup of the permutation group. (This wasn’t always the case! During the 19th century, a group was literally defined as a subset of ${\text{GL}(n)}$ or of ${S_n}$ . In fact Sylow developed his theorems without the word “group” Only much later did the abstract definition of a group was given, an abstract set ${G}$ which was independent of any embedding into ${S_n}$ , and an object in its own right.) So, we would like our notion of a tangent vector to not refer to an ambient space, but only to intrinsic properties of the manifold ${M}$ in question.

So how do we capture the notion of the tangent to a manifold referring just to the manifold itself? Well, the smooth structure of the manifold lets us speak of smooth functions ${f : M \rightarrow \mathbb R}$ . In the embedded case, we can thus think of taking a directional derivative along ${\vec v}$ (i.e. some partial derivative). To give a concrete example, suppose we have a smooth function ${f : S^2 \rightarrow \mathbb R}$ and a point ${p}$ . By the structure of a manifold, near the point ${p}$ , ${f}$ looks like a function on some neighborhood of the origin in ${\mathbb R^2 = T_p(M)}$ So we are allowed to take the partial derivative of ${f}$ with respect to any of the vectors in ${T_p(M)}$ .

For a fixed ${v}$ this partial derivative is a linear map ${D : C^\infty(M) \rightarrow \mathbb R}$ . It turns out this goes the other way: if you know what ${D}$ does to every smooth function, then you can figure out which vector it’s taking the partial derivative of. This is the trick we use in order to create the tangent space. Rather than trying to specify a vector ${\vec v}$ directly (which we can’t do because we don’t have an ambient space), we instead look at arbitrary derivative-like functions, and associate them with a vector. More formally, we have the following.

Definition 1

A derivation ${D}$ at ${p}$ is a linear map ${D : C^\infty(M) \rightarrow \mathbb R}$ (i.e. assigning a real number to every smooth ${f}$ ) satisfying the following Leibniz rule: for any ${f}$ , ${g}$ we have the equality

$\displaystyle D(fg) = f(p) \cdot D(g) + g(p) \cdot D(f) \in \mathbb R.$

This is just a “product rule”. Then the tangent space is easy to define:

Definition 2

A tangent vector is just a derivation at ${p}$ , and the tangent space ${T_p(M)}$ is simply the set of all these tangent vectors.

In fact, one can show that the product rule for ${D}$ is equivalent to the following three conditions:

${D}$ is linear, meaning ${D(af+bg) = a D(f) + b D(g)}$ .
${D(1_M) = 0}$ , where ${1_M}$ is the constant function on ${M}$ .
${D(fg) = 0}$ whenever ${f(p) = g(p) = 0}$ . Intuitively, this means that if a function ${h = fg}$ vanishes to second order at ${p}$ , then its derivative along ${D}$ should be zero.

This suggests a third equivalent definition: suppose we define

$\displaystyle \mathfrak m_p \overset{\mathrm{def}}{=} \left\{ f \in C^\infty M \mid f(p) = 0 \right\}$

to be the set of functions which vanish at ${p}$ (this is called the maximal ideal at ${p}$ ). In that case,

$\displaystyle \mathfrak m_p^2 = \left\{ \sum_i f_i \cdot g_i \mid f_i(p) = g_i(p) = 0 \right\}$

is the set of functions vanishing to second order at ${p}$ . Thus, a tangent vector is really just a linear map

$\displaystyle \mathfrak m_p / \mathfrak m_p^2 \rightarrow \mathbb R.$

In other words, the tangent space is actually the dual space of ${\mathfrak m_p / \mathfrak m_p^2}$ ; for this reason, the space ${\mathfrak m_p / \mathfrak m_p^2}$ is defined as the cotangent space (the dual of the tangent space). This definition is even more abstract than the one with derivations above, but it has the advantage (or so I’m told) that it can be transferred to other settings (like algebraic varieties).

EDIT (Oct 5 2015): Reproducing this Reddit comment by tactics, The beauty of the definition given in this blog post is stuffed away into an easily-forgotten sentence. This definition:

Does not rely on any kind of parametrization, and
It is defined only in terms of the ring of “regular functions” defined on the space.

The former is nice for philosophical reasons. The latter is nice because we can pull in a lot of intuition about manifolds into the study of algebraic varieties, and similarly, we can inject a lot of ring theory into the study of manifolds.

With all these equivalent definitions, the last thing I should do is check that this definition of tangent space actually gives a vector space of dimension ${n}$ . To do this it suffices to show verify this for open subsets of ${\mathbb R^n}$ , which will imply the result for general manifolds ${M}$ (which are locally open subsets of ${\mathbb R^n}$ ). Using some real analysis, one can prove the following result:

Theorem 3

Suppose ${M \subset \mathbb R^n}$ is open and ${0 \in M}$ . Then

$\displaystyle \begin{aligned} \mathfrak m_0 &= \{ \text{smooth functions } f : f(0) = 0 \} \\ \mathfrak m_0^2 &= \{ \text{smooth functions } f : f(0) = 0, (\nabla f)_0 = 0 \}. \end{aligned}$

In other words ${\mathfrak m_0^2}$ is the set of functions which vanish at ${0}$ and such that all first derivatives of ${f}$ vanish at zero.

Thus, it follows that there is an isomorphism

$\displaystyle \mathfrak m_0 / \mathfrak m_0^2 \cong \mathbb R^n \quad\text{by}\quad f \mapsto \left[ \frac{\partial f}{\partial x_1}(0), \dots, \frac{\partial f}{\partial x_n}(0) \right]$

and so the cotangent space, hence tangent space, indeed has dimension ${n}$ .

6 thoughts on “Constructing the Tangent and Cotangent Space”

JM October 5, 20155:40 AM Reply

The definition of m^2 should be the abelian group spanned by the products fg.

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1. Evan Chen (陳誼廷) October 5, 20157:08 AM Reply
  
  Yep, thanks for catching that.
  
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2. Kip Ingram April 27, 20187:30 PM Reply
  
  I recently watched a series of YouTube videos by Leonard Susskind on general relativity. In lecture one he said that we were just going to take it for granted that our coordinate systems were such that we could do all of the things we needed to do – take derivatives, etc. etc. Now I see why.
  
  A couple of days ago I waded into all of this for one reason: I wanted to achieve the same sort of intuition about cotangent spaces that I had about tangent spaces. I must have looked at 50 different documents online, and from what I can tell there is a veritable Mt. Everest of tedious work involved in carefully building all this stuff up, all just to be able to do something like take the partial derivative of something.
  
  I absolutely see why Susskind didn’t want to fret with it – he wanted to get down to the business at hand.
  
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timurvural September 9, 201612:48 PM Reply

Typo at definition 1 – it should be D(fg) = f(p)D(g) + D(f)g(p)

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1. Evan Chen (陳誼廷) September 9, 20161:59 PM Reply
  
  Yep. Thanks, I will fix this.
  
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Differential Geometry Formalism – timurvural September 10, 20163:16 PM Reply

[…] that . is the vector space of all derivations at , and we call members of tangent vectors. See here for a more thorough introduction to the tangent […]

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