Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

In Aaron's answer here...

"Given a manifold $M$, and a point $p\in M$, we have a vector space $T_pM$ of the tangent vectors to $M$ at $p$. For example, if you take the hollow sphere sitting inside $R^3$, you can look at the plane that sits tangent to a point, and turn it into a vector space. These tangent vectors act on functions by taking the directional derivative of a function at a point."

....this is from Aaron's answer in the link mentioned in the question. Here he says that any vector in the tangent plane to the sphere at a given point belongs to the tangent space at that point. fine.. but again he says that "These tangent vectors act on functions by taking the directional derivative of a function at a point." I do not understand the last statement and I'd appreciate some help. Thank you

share|improve this question
    
thank you for all the answers, they were very helpful. –  Rajesh D Jul 13 '11 at 12:41

4 Answers 4

up vote 5 down vote accepted

If $\gamma: [0, 1]\to \mathbb R^n$ is a curve passing through the point $p\in \mathbb R^n$, then the tangent vector of $\gamma$ in $p$, usually, is defined as the vector $$ \dot \gamma(t_0) = \left.\frac {\operatorname d} {\operatorname d t}\right\vert_{t_0} \gamma(t)\in \mathbb R^n \tag 1 $$

We can associate to $\gamma$ the differential operator $$ X_p :f\in C^\infty(\mathbb R^n) \to \left.\frac {\operatorname d} {\operatorname d t}\right\vert_{t_0} f(\gamma(t)) \in \mathbb R \tag 2 $$

The following equality holds $$ X_p f = \dot\gamma(t_0)\cdot\nabla f $$ It shows the operator $X_p$ is completely determined just by $\dot\gamma(t_0)$, therefore calling it the tangent vector of $\gamma$ in $p$ is as natural as the definition (1).

The former definition can't be directly used for curves on manifolds: since they aren't generally vector spaces we can't take sum of points or products of points with real numbers, so we can't form the derivative in (1).

The definition (2), instead, can be extended without modifications to manifolds.

Now, a tangent vector to a manifold at a point $p$, $X_p$, is the tangent vector at $p$ to some curve $\gamma$, so it is an operator that associates to smooth functions their derivatives along the path $\gamma$.

share|improve this answer

For any path on the manifold $M$ passing through the point $p$, it is natural to ask what a given function's rate of change is along that path specifically at the point $p$. In Euclidean space, studied in vector calculus, this quantity is represented by the dot product of the gradient with the curve's tangent vector, i.e. $\nabla f \cdot \vec{n}$. In the more general case we get basically the same thing; the gradient is a vector field which, when poised with the tangent vector $v \in T_pM$ in tensor contraction, tells us the rate of change of the function at the point $p$ along any path passing through it tangent to $v$.

On a more semantic note, when mathematicians say $X$'s "act" on $Y$'s they mean there is a rule of applying the $X$'s to the $Y$'s so that, when the $X$ is held fixed it amounts to a mapping of the $Y$'s to some other structure. Linear maps "act" on vector spaces by taking their elements to some other space, permutations "act" on multivariable functions by permuting their input variables and thus creating a (most of the time) new and distinct function, etc.

share|improve this answer
    
this answer is not helping me. It is not able to clarify the confusion mentioned in the questions. –  Rajesh D Jul 13 '11 at 8:30
    
Would I be correct in presuming your confusion is with the word "act"? –  anon Jul 13 '11 at 8:39
    
In his answer Aaron mentions that a tangent vector is any vector in the tangent plane to the sphere at that point. If this definition of tangent vector is fixed they how and why it would act on functions. Why would a tangent in a 2-D plane act on functions. Finally what is the message...what is the final definition of a tangent vector ? this is not clear to me. –  Rajesh D Jul 13 '11 at 8:44
1  
thank you for the comment in your edit –  Rajesh D Jul 13 '11 at 8:50
1  
You're taking the word "action" too literally I think. It's somewhat figurative language; tangent vectors aren't wild animals in function space jungle searching for functions to differentiate and frothing at the mouth. Rather, there is a set of ways to take a directional derivative of an arbitrary scalar function at a point, and this set is identified with (put in one-to-one correspondence with) the set of tangent vectors at the point. The Wikipedia article on Tangent space has a good graphic on how tangent vectors are intuitively understood as the "directions" curves can have through points. –  anon Jul 13 '11 at 8:53

The following may help you (though it is specifically for $\mathbb{R}^n$):

The linear map $\phi$ defined by

$\phi : T_p (\mathbb{R}^n) \to D_p(\mathbb{R}^n)$

Given explicitly by

$v \rightarrow D_v = \sum v^i \frac{\partial}{\partial x^i} \Bigg|_p$

is an isomorphism between vector spaces, where $D_p$ is the vector spaces of all directional derivatives at a point $p$ in $\mathbb{R}^n$. The symbol $D_v$ means the directional derivative in the direction of a tangent vector at $p$.

share|improve this answer

Supopse we have local coordinates $x_1,\ldots,x_n$. Consider differential operator $L=a_1\frac{\partial}{\partial x_1}+a_2\frac{\partial}{\partial x_2}+\ldots+a_n\frac{\partial}{\partial x_n}\ $. The coordinates of vector $a=(a_1,\ldots,a_n)$ changes in the same way as coeffitients of $L$, so there is a natural correspondence between them (actually vectors are defined as such operators with the help of curves as described on the wiki page in your post.) Applying $L$ to a smooth function $f$ we get $Lf(x)=a_1\frac{\partial f(x)}{\partial x_1}+a_2\frac{\partial f(x)}{\partial x_2}+\ldots+a_n\frac{\partial f(x)}{\partial x_n}\ $. As mentioned above in Euclidian space it it the directional derivative of $f$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.