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I just checked out this thread: help in understanding tangent vectors and I still have some problems understanding this.

The tangent vector on a manifold at point $t_0$ is intuivitely $$\dot \gamma(t_0) = \left.\frac {\operatorname d} {\operatorname d t}\right\vert_{t_0} \gamma(t)\in \mathbb R^n \tag 1$$ Apparently this expression doesn't make sense in some manifolds, so $X_p$, defined $$X_p :f\in C^\infty(\mathbb R^n) \to \left.\frac {\operatorname d} {\operatorname d t}\right\vert_{t_0} f(\gamma(t)) \in \mathbb R \tag 2$$ is used as the tangent vector instead. But this isn't equivalent to ($1$) (which is a real valued vector, not an operator), and I have no idea what effect the function $f$ has. I realise it's similar to a directional derivative on the function $f$, but don't see why it is relevant.

If anyone can explain with reference to the sphere, that would probably make it easiest to understand. It's easy to see that the tangent space defined by ($1$) gives the plane that best approximates a point on the sphere, but ($2$) doesn't make sense to me. Thanks.

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up vote 14 down vote accepted

The first expression doesn't make sense on an abstract manifold at all. You can't subtract points on a manifold. It makes sense on a manifold equipped with an embedding into $\mathbb{R}^n$, but then how do you know that the answer is independent of the embedding in some meaningful way? (How do you even know that the dimension of the resulting tangent space is independent of the embedding?)

One solution is not to talk about what the tangent vector is at all, but instead to talk about what it does: it's a thing you can take directional derivatives with respect to. Given a function $f$, we can intuitively measure the rate of change of $f$ in the direction of a tangent vector $v$ at a point $p$, and these rates of change capture everything there is to know about $v$, so we can safely identify the tangent vector $v$ with the directional derivative it defines at $p$. This definition is independent of an embedding as desired.

To really feel comfortable with this definition you should feel comfortable with a certain idea, coming from algebraic geometry and functional analysis, of studying a space $X$ by studying a suitable space of functions on $X$. For smooth manifolds $M$, that means studying the space $C^{\infty}(M)$ of smooth functions $M \to \mathbb{R}$. I believe it's known that $M$ can be completely recovered from $C^{\infty}(M)$, so any definition you want to make about an abstract smooth manifold you can phrase in terms of $C^{\infty}(M)$ instead.

For example, a point $p \in M$ is the same thing as a morphism $C^{\infty}(M) \to \mathbb{R}$ (given by $f \mapsto f(p)$). More generally, I believe it's known that a smooth map $N \to M$ is the same thing as a morphism $C^{\infty}(M) \to C^{\infty}(N)$ (postcomposition).

But since we're working algebraically, we can talk about more general things. A tangent vector is nothing more than a morphism $C^{\infty}(M) \to \mathbb{R}[\epsilon]/\epsilon^2$; it sends a function $f$ to $f(p) + \epsilon df_p(v)$ where $p \in M$ is a point and $df_p(v)$ is the directional derivative of $f$ at $p$ in the direction of $v$.

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Thank you for the answer. I think the second part is a bit above my current level, but I get the basic idea of it. – M.X Aug 17 '11 at 18:33
At least for compact manifolds, Milnor and Stasheff's characteristic classes book has an excercise for showing that $C^\infty(M)$ determines $M$ and that the maximal ideals in $C^\infty(M)$ are in natural 1-1 correspondence with the points of $M$ given by the map $p\rightarrow \ker(ev_p)$ where $ev_p:C^\infty(M)\rightarrow \mathbb{R}$ evaluates the function at $p$. In the noncompact case, it's easy to find other maximal ideals, so at least this proof won't show that $C^\infty(M)$ characterizes $M$. I'm not sure if $C^\infty(M)$ characterizes $M$ for noncompact manifolds or not. – Jason DeVito Aug 17 '11 at 18:37

There are quite a few definitions of the tangent space to a manifold. Happily, these are all equivalent in some sense.

One definition that you might like, if you like curves, is the following. Let $M$ be a manifold, and let $P \in M$. Then the tangent space $T_PM$ at $P$ consists of smooth curves $\gamma\colon J \to M$, where $J$ is an open interval in $\mathbf R$ containing $0$, such that $\gamma(0) = P$ under the following equivalence relation: $\gamma_1 \sim \gamma_2$ if there exists a chart $\varphi\colon U \to \mathbf R^n$ around $P$ such that $(\varphi \circ \gamma_1)'(0) = (\varphi \circ \gamma_2)'(0)$. (This will then hold for every chart around $P$, by the chain rule.)

This has the advantage of being nearly obviously the same as your (1) when $M$ is embedded in Euclidean space, yet working in great generality.

Now, how can we connect this to the operators in (2)? Call a linear map $X \colon C^\infty(M) \to \mathbf R$ a derivation at $P \in M$ if it satisfies the Leibniz rule $$ X(fg) = g(P)Xf + f(P)Xg $$ for all $f, g$. You can check that in $\mathbf R^n$, for $v \in \mathbf R^n$, the directional derivative gives a derivation at $P$: define $\gamma\colon (-\varepsilon, \varepsilon) \to \mathbf R^n$ by $\gamma(t) = P + tv$. Then $$ f \mapsto D_v|_P f = (f \circ \gamma)'(0) $$ is a derivation.

In general, we get a well-defined linear map $T_PM \to D_P$ by sending the class of $\gamma$ to the derivation $f \mapsto (f \circ \gamma)'(0)$. It should be straightforward to show that this association is injective, since we could take $f$ to be a coordinate function. It's less obvious (to me, at least) that it's surjective. But this does follow from the multivariable Taylor's formula. I won't write out a proof here, but you can find one in the case of $M = \mathbf R^n$ (so, proving that every derivation has the form $D_v|_P$) as Proposition 3.2 in Lee's book. From this the case of general $M$ follows.

What I like about the second definition is that it doesn't mention charts or equivalence relations, and it is obviously a vector space. It also generalizes, with only slight modifications, to more algebraic settings.

I apologize for the lack of mental pictures here. To me, the first definition I gave is the most intuitive, and you have this natural (!) map to another space which turns out to be an isomorphism.

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Thank you very much for the answer, very useful, particularly the connection. – M.X Aug 17 '11 at 18:34

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