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In my differential geometry class, my teacher defined the co-tangent space as follows. Let $M$ be a smooth manifold and $p$ is a point on $M$. Now define two sets of $C^\infty$ real-valued functions defined on $M$.

  1. $\mathcal I_p := \left\{f\in C^\infty(M)\bigr|f(p)=0 \right\}.$ That is, the set of functions that vanish at $p$. This is not hard to imagine.
  2. $\mathcal I_p^2 := \left\{\displaystyle\sum_{i=i}^n f_ig_i\bigr|f_i, g_i\in \mathcal I_p \right\}.$ I could not visualise what this set looks like. Is it a subset of $\mathcal I_p$ or a bigger set and why?

After the definition of these two sets comes the even harder concept of quotient as follows. My teacher defined the co-tangent space $T_p^*M$ as the quotient space $\mathcal I_p/\mathcal I_p^2$. I have difficulty in understanding quotient thing in general, especially I have no idea of what this quotient looks like. Is there an intuitive way to understand this concept, please? Thank you!

LaTeXFan
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  • I have already came across something like that and I was confused as well (indeed I'm still confused=) Where did your teacher get the definition from? – PtF May 10 '14 at 00:50
  • I do not know. He wrote it down from his own notes. – LaTeXFan May 10 '14 at 00:52
  • Just a vague hint: $I_p$ is an ideal of $C^\infty(M)$ and $I_p^2$ is the product of this ideal with itself. You may find something about this kind of ideals in "Abstract Algebra" by D. S. Dummit and R. M. Foote.. – PtF May 10 '14 at 00:55
  • Thank you for the suggestion. – LaTeXFan May 10 '14 at 00:57
  • Since he defined $\mathcal{I}p / \mathcal{I}{p}^{2}$, certainly $\mathcal{I}_{p}^{2} \subset \mathcal{I}_p$. – Ivo Terek May 10 '14 at 00:57
  • The cotangent space not a class but the quotient, that is the set of equivalence classes. Each element of this quotient looks like an element of $\mathcal I_p$ but we deem two such elements equal if their difference lies in $\mathcal I_p^2$. Alternatively, you may view elements of the quotients as sets $f + \mathcal I_p^2$ for $f \in \mathcal I_p$. – Yuri Vyatkin May 12 '14 at 04:21

1 Answers1

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It is obvious that $\mathcal I_p^2 \subset \mathcal I_p$. Well, it is an easy exercise: simply use the definition of $\mathcal I_p$ to check that each element of $\mathcal I_p^2$ is in $\mathcal I_p$. Hint: 1) trivial: $a \cdot 0 = 0$ $\forall a \in \mathbb{R}$; 2) less trivial: products of smooth functions are smooth.

Straightforward verifications show the further necessary algebraic properties (being subspaces, ideals...) as the commenters have pointed out.

An important property of the space $\mathcal I_p^2$ is that each element of it has the vanishing differential at point $p$: $\forall f \in \mathcal I_p^2 \; \mathrm{d}f(p)=0$. Hint: use the product rule.

And the crucial fact: the quotient $\mathcal I_p / \mathcal I_p^2$ is a finite dimensional vector space. For a proof one uses the Taylor theorem (in the form that is also known as Hadamard's lemma).

The beauty of this construction is that we obtain a naturally defined finite dimensional vector space at each point of a smooth manifold without using coordinates!

It turns out that elements of this space can act on tangent vectors viewed at an appropriate angle (e.g. as derivations). The duality between vectors and covectors emerges immediately.

What is more important to get is that there is the notion of a cotangent space (of a smooth manifold at a point), and there are various definitions. The notion is something to visualize, the definitions being tools that allow us to work efficiently with the problems. Without a context everything will look unjustified and clumsy.

I hope that my remarks can help. For more details please see this and this notes. Actually, Wikipedia's articles on the tangent and the cotangent spaces are quite helpful too.

Yuri Vyatkin
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