I am looking for a quick definition of the tangent space $T_p M$, where $p$ is a point of a smooth manifold $M$. I mean a definition that allows easily to prove that: (1) $T_p M$ is a vector space of the same dimension of $M$; (2) if $x^1, x^2, \ldots,x^d$ are local coordinates functions then $\partial / \partial x^i|_p$ is a basis of $T_p M$; (3) The elements of $T_p M$ are the same thing of derivations of functions definited on a neighborhood of $p$ (W. Warner, Foundations of Differentiable Manifolds, called them "germs" and denoted their set by $\tilde{F}_p$).

Surely the old-style definition of $T_p M$ by equivalence classes of curves which passes through $p$ is inadequate.

The definition of $T_p M$ like the vector space of derivations of germs is, obviously, OK for the point (3) but seems to me that points (2) and (1) require to look at the cotangent space $T_p^* M$. I do not like to talk about the cotangent space before I have finished to talk about the tangent space.

I saw that someone defines the cotangent space first and then the tangent space, however the Warner definition of cotangent space like $\tilde{F}_p / \tilde{F}_p^2$, where $\tilde{F}_p^2$ is the ideal of finite linear combinations of product of two germs, I think it is rather artificial, and justifiable only in retrospect.

In conclusion I really appreciate any advice on how to define $T_p M$ so that (1), (2) and (3) are easy to prove. Thanks.

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    $\begingroup$ You can prove (1) and (2) in one go using the Taylor theorem using the definition in terms of derivtions, if you write that theorem in the form: every smooth function $f$ can be written locally as $f(x)+\sum_i\frac{\partial f}{\partial x_i}x_i+\sum_{i,j}f_{i,j}x_ix_j$ for some smooth functions $r_{i,j}$. $\endgroup$ – Mariano Suárez-Álvarez May 21 '12 at 17:02
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    $\begingroup$ Why is the old-style definition inadequate? $\endgroup$ – Angelo May 21 '12 at 17:04
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    $\begingroup$ @Angelo: I think it is inadequate because it is not immediately clear that $T_p M$, viewed as equivalence classes of curves, is a vector space. $\endgroup$ – Johannes Ebert May 21 '12 at 17:31
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    $\begingroup$ @Johannes: on any chart, it is immediate to prove that $T_pM$ is a vector space with the old-style definition (and that the vector structure does not depend on the chart). It is immediate to prove (1) and (2). $\endgroup$ – Bruno Martelli May 21 '12 at 21:00
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    $\begingroup$ @Micheal-grade83: you can easily show that if $v$ is a derivation at $x$ and $f$ and $g$ are functions which vanish at $x$ then $v(fg)=0$. Use that. $\endgroup$ – Mariano Suárez-Álvarez May 22 '12 at 2:36

I think the description of tangent space $T_pM$ you're looking for is one defined entirely locally by "pushing forward" tangent spaces of $\mathbb{R}^n$ via local coordinate charts. ie. the tangent space $T_pM$ at a point $p$ in a trivial neighborhood $U$ of the manifold $M$ is described via a coordinate chart $\phi: (\mathbb{R}^n, 0) \to (U,x)$ as simply the image of $T_0 \mathbb{R}^n$ under the differential $d\phi$. This yields your (1), (2) immediately, while (3) follows once one understands why its true on $\mathbb{R}^n$.

The basic point of smooth manifolds, I think, is that there is no calculus on manifolds, but only calculus on $\mathbb{R}^n$ and pushforwards and pullbacks.

When I was first learning the basics of smooth manifolds I found Warner's book to be very helpful. He demonstrates very clearly the formalism of (co)tangent spaces and how to actually perform differential computations.

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    $\begingroup$ But we need to place some equivalence relation on these things, right? Or is there a way of avoiding this? $\endgroup$ – Dylan Moreland May 21 '12 at 20:32
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    $\begingroup$ More formally, if $\mathcal{A}_x$ is the local atlas of $M$ at $x$, we define $\left( T _ x M, \{ d\phi(x) \} _ {\phi\in\mathcal{A} _ x}\right)$ as the limit cone of the diagram of the differentials of the transition maps at $x$, i.e. the linear isomorphisms $d(\psi\phi^{-1})(\phi(x))$. This notation is consistent with the chain rule of differentiation, for $d(\psi\phi^{-1})(\phi(x))d\phi(x)= d\psi(x)$. $\endgroup$ – Pietro Majer May 21 '12 at 21:34

Let $p\in M$ and $q\in N$ be points. Given two smooth maps $M\to N$ both taking $p$ to $q$ (each maybe only defined in a neighborhood of $p$), call them first-order equivalent if when expressed in terms of coordinates in both manifolds they have the same first derivatives at $p$. Call the equivalence classes $1$-jets $(M,p)\to (N,q)$.

Define cotangent vectors of $M$ at $p$ to be $1$-jets $(M,p)\to(\mathbb R,0)$. These clearly form a vector space. Show that it is isomorphic to $\mathfrak m/\mathfrak m^2$ where $\mathfrak m$ is the ideal of germs of functions on $M$ vanishing at $p$, and that its dimension is that of $M$.

Define tangent vectors of $M$ at $p$ to be $1$-jets $(\mathbb R,0)\to (M,p)$. These do not obviously form a vector space until you note that there is a canonical bijection between them and the linear functionals on the cotangent space (given by composing $\mathbb R\to M\to\mathbb R$).


I think what you want is something like this: Define a vector bundle whose transition maps are the ones that we know to be the transition maps of the tangent bundle. Then define that vector bundle to be the tangent bundle. Once you've shown that this is all well-defined, the properties that you want follow easily. I like to think of this as the physicists' definition: A vector is an object that transforms like a vector.

(Btw, you're free to prefer one definition over another, but remember that it's just your own opinion. It's a bit much to say that "surely" the equivalence class definition is "inadequate.")


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