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I have some questions about the definition of tangent space that arose after reading the book Differential Geometry of Curves and Surfaces of Manfredo do Carmo.

First, what's the best way to define tangent space? Using tangent vector to curves or derivations?

I've researched about the concept of derivation but it seems to require knowledge of abstract algebra, is is possible to understand derivations without knowing anything of abstract algebra?

Can someome recommend some material about this topic? And if it's really needed to understand derivations, can someone recommend also some material about derivations?

Thanks in advance.

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There is no best way to do it. Like with all useful things, what is best is to know many ways to construct it. You do not need to know any abstract algebra to know what a derivation is in the context of manifolds—as usual, the more you know the better, of course, but with nothing more than the definition of what a derivation is you are good to go. In fact, if you are studying manifolds, as you seem to be, then the best plan is probably to not further your knowledge of derivations and "their theory" at this point and instead concentrate on, well, manifolds! –  Mariano Suárez-Alvarez Aug 31 '12 at 19:12
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3 Answers

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For smooth manifolds the tangent space given by equivalence classes of smooth curves and the set of derivations on smooth functions is equivalent. You can give an isomorphism between these constructions. Each has its advantages.

  • For example, in the equivalence class of curves set-up the differential is really simple; $dF(\gamma) = F \circ \gamma$ the mapping $F: \mathcal{M} \rightarrow \mathcal{N}$ pushes the curve on $\gamma$ on $\mathcal{M}$ to the curve $F \circ \gamma$ on $\mathcal{N}$. To show $d(F \circ G)=dF \circ dG$ for manifolds we can argue $$ d(F \circ G)(\gamma) = F( G (\gamma)) = F( dG(\gamma)) = dF(dG(\gamma)) = (dF \circ dG)(\gamma) $$ I think the proof in the derivation view is less slick.

  • Derivations naturally arise from coordinates systems as the partial derivatives with respect to the manifold coordinates. The differential again pushes vectors from one manifolds tangent space to another. Or from the manifold to itself, but where the coordinate system has two overlapping charts. Believe it or not, you've already done this calculation in multivariate calculus. Changing coordinates from cartesian to polar for example: $$ \frac{\partial}{\partial x} = \frac{\partial r}{\partial x}\frac{\partial}{\partial r} +\frac{\partial \theta}{\partial x}\frac{\partial}{\partial \theta} = \cos(\theta)\frac{\partial}{\partial r}-\sin(\theta)\frac{\partial}{\partial \theta}$$ This can be understood as the vector $\frac{\partial}{\partial x}$ being pushed forward to the vector $\cos(\theta)\frac{\partial}{\partial r}-\sin(\theta)\frac{\partial}{\partial \theta}$ by the differential of the transition map. In other words, there are constructions for which the derivation formulation of tangent space nicely dovetails. Usually the derivation is sold as a sort of directional derivative like object. The formula $X(f) = \sum_{i=1}^n X^i \frac{\partial f}{\partial x^i}$ is like $D(f)(\vec{X}) = (\nabla f) \cdot \vec{X}$.

In either case, it is a bit unsettling to trade seemingly static objects like geometric vectors for classes of curves or differential operators. There is at least one other popular view, the contravariant vector formulation of tangent vectors.

  • a contravariant vector is typically described in physics by saying something like $v^{\mu}$ is a contravariant vector because under a change to barred coordinates we find $\bar{v}^{\mu'} = \sum_{\nu=1}^n\Lambda^{\mu'}_{\nu} v^{ \nu}$ where the coordinates transform via $\bar{x}^{\mu'} =\sum_{\nu=1}^n\Lambda^{\mu'}_{\nu} x^{ \nu}$. This is the rule underwhich $X^i$ of the derivation changes. That fact follows from the chain rule for manifolds. This viewpoint has great advantages for those folks who do detailed calculations in one particular system of coordinates; a.k.a. physicists.

It should be commented that when we consider $C^k$ manifolds this correspondence breaks down.See page 49 in Conlon's 2nd edition of Differentiable Manifolds. In his Lemma 2.2.20 he builds functions which are used to frame the isomorphism between curves and derivations. The construction fails for $C^k$ manifolds. Moreover, the reason for this is that the space of derivations on $C^k$-germs of functions is infinite dimensional.

I'd love to say you can pick one to understand. However, the nature of the subject of differential geometry requires you to be conversant in all these views.

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James S. Cook, thanks a lot for your answer, I think I understood the difference, but can you recommend good material where the author explains the construction using smooth curves and another material where the author explains using derivations? I'd like to go more deep into that to aquire a good knowledge in both views. I didn't studied any abstract algebra as I told you. This book Differentiable Manifolds that you recommended has both views? Thanks very much again. –  user1620696 Sep 1 '12 at 14:49
    
@user1620696 yes Conlon has both views and explains the connection between them. However, if you want more on the topic then many other books also work. For example, the text by Warner mentioned by Andrew is also very careful. The advice found here: mathoverflow.net/questions/7834/… is probably helpful to your inquiry. –  James S. Cook Sep 1 '12 at 15:39
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I think the best way depends on your goals. One reason derivations are so important is that they are general enough to be used in contexts other than manifolds (they are crucial for doing algebraic geometry).

There is a nice description on p.12 of Warner's text in the real manifold context.

For the algebraic theory you can look at chapter 16 of Eisenbud.

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I am just trying to add my understanding to the mix. As in one of the books I used, the author remarks that the definition of tangent vectors and ergo the tangent space is much more intuitive and "pleasant" when curves are employed.

However as the answers already given so neatly tell, it is much more beneficial in a general context to look at tangent vectors as derivations.

And though it helps to know abstract algebra to understand derivations better, a derivation is in essence just a linear map that satisfies the familiar Leibniz product rule seen in usual derivatives.

That it is defined on an Algebra over a ring or a field can be neglected if you are studying it from the point of view of Differential Geometry atleast at this juncture.

I think Loring Tu's book on manifolds and tensors has a nice way of introducing derivations without delving much into abstract algebra. He does mention modules , but explains the background himself and that suffices in that context.

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