# Basis for the Tangent space and derivations at a point

So i have been reading Tu's , "An Introduction to Manifolds" , and i have a question that is really bugging my mind.

So we have the tangent space at a point $$p$$, $$T_p(\mathbb{R}^n)$$ with basis $$e_1,...,e_n$$ is going to be isomorphic to the vector space of derivations at a point $$p$$, $$D_p$$. To do this the author creates the said isomorphism and proves it is in fact one. My question is the following, in this isomorphism basically is sending every tangent vector $$v$$ of the tangent to the directional derivative of the said vector.So far so good, my problem is that later he writes any tangent vector as a linear combination of partial derivatives and then latter when talking about differential forms and when we make computations in the proofs he uses this partial derivatives notation, so is this ok because the spaces are isomorphic ??Because they are not tangent vectors , they can just be identified as one, and can the partial derivatives in the sum really be interpreted as such??

Thanks.

• Are you only concerned with $\mathbb{R}^n$ as a manifold? Commented Sep 9, 2019 at 22:31
• I mean i would like to understand this for any manifold, but im really new to the theory so it might be easier to understand $\mathbb{R}^n$. Commented Sep 9, 2019 at 22:33
• In a word, yes. That's what isomorphic means: the two vector spaces can be used interchangeably. Commented Sep 9, 2019 at 22:38
• Or are you concerned about using derivations and partial derivatives interchangeably? Commented Sep 9, 2019 at 22:39
• I see, but it is ! I will type a few sentence about this in an answer because this can be confusing. Commented Sep 9, 2019 at 22:42

$$\newcommand{\R}{\mathbb{R}}$$ So because we are in $$\mathbb{R}^n$$ we can all imagine (or at least were told) how we should visualize what a tangent space is at a certain point $$p$$ in $$\mathbb{R}^n$$, i.e. some arrows that are tangent. The key point here is that this concept only holds because one can think of the $$\mathbb{R}^n$$ as vector space (and usually does it). Then a point in $$\R^n$$ is really nothing else then a vector in $$\R^n$$. But this is something that will not be true for manifolds. So when you think of $$\R^n$$ as being a manifold you should not think of any point in $$\mathbb{\R}^n$$ as vector but really just as point. Now, you can attach a vector space, that is the tangent space, at any point in $$\mathbb{R}^n$$, and then from there you have a vector space $$T_p(\mathbb{\R}^n)$$ associated with that point $$p$$ of the manifold $$\R^n$$.

Of course now that seems like just a lot of words and really for the $$\R^n$$ it probably is, but as we get more and more abstract those difference are key in the theory of manifolds.

Now, think of a donut or a sphere in $$\R^n$$, then again you can attach to any point in that manifold a tangent space. Intuitively you will know how to do it, because that object is imbedded into a euclidian space. But later on in your studies of manifolds this will not be the case! (At least not in any practical way).

So, we need to find a concept of tangent vectors that only rely on the local properties of a point of a manifold. And this is were the (partial) derivative notation comes into play. Because these are concepts that can be also defined for manifolds in general.

To the question at hand: As it was pointed out, isomorphic means it is the same thing! Both are $$n$$ dimensional linear vector space which are isomorphic to each other and now the only question is which kind of "names" you want to give these vectors. But that is all that is different: the symbols.

Edit: Maybe to make it more precise: Take the basis $$e_1,\dots,e_n$$ of $$T_p(\R^n)$$ and let $$\varphi : T_p(\R^n) \to D_p(\R^n)$$ be the isomorphism. Then $$\varphi(e_i) = \partial/\partial_{x^i}|_p$$ for all $$i\in\{1,\dots,n\}$$. Now, you do the calculations for this basis in $$D_p(\R^n)$$. So, whenever you now have a element $$v \in D_p(\R^n)$$ and you had enough of this derivative notation you simply do $$\varphi^{-1}(v)$$ or even more concrete: you have $$v = \sum_{i=1}^n v^i \partial/\partial_{x^i}|_p$$, so $$\varphi^{-1}(v) = \sum_{i=1}^n v^i e_i$$.

• Yes, Thank you , i am now getting a good idea of what is going on , also discussing it makes it easier to start understanding the theory! Commented Sep 10, 2019 at 9:44

I think what you want follows from the definition of the "partials" in an arbitrary manifold. If I misunderstood your question, please advise, and I will delete my answer.

Let $$x\in \mathbb R^n$$, and $$r_i$$ the projections from $$\mathbb R^n$$ to the $$i^{th}$$ coordinate. As you point out, Tu proves that the derivations $$\{\partial/\partial r_i\}_x$$ form a basis for $$T_x\mathbb R^n$$.

Now let $$p\in M$$, an $$n$$-dimensional manifold and consider a chart $$(\phi,U)$$ about $$p$$. Then, since $$\phi$$ is a diffeomorphism onto $$\phi(U),\ \phi_*^{-1}: T_{\phi(p)}R^n\to T_pU\cong T_p M$$ is an isomorphism.

Now, if we $$\textit{define}\ \left(\partial/\partial x_i\right)_p$$ to be $$\phi_*^{-1}\left(\partial/\partial r_i\right)_p,$$ we get a basis for $$T_pM$$.

From this definition, the rest follows: the covectors $$dx^i$$ are simply the (dual) basis for $$T_p^*M.$$

• The thing is , i have a "problem" proving that the covectors are the dual basis cause we use the partial derivative notation , at least the author did, and they are not really tangent vectors, they are just isomorphic. Commented Sep 9, 2019 at 23:00
• You have proved that $\left(\partial/\partial x_i\right)_p$ are a basis for $T_pM$ so there is a dual basis for the cotangent space $T^*_pM$ and these are denoted by $\{dx^i\}$. It's as simple as that. Or what am I missing? Commented Sep 9, 2019 at 23:21
• Oh alright , i had a different definition of the $dx_i$ and then we proved that they are the dual basis for the cotagent space using the partial derivatives notation. Commented Sep 9, 2019 at 23:24
• If you defined them like this: $dx_p^i(\left(\partial/\partial x^j\right)_p)=\left(\partial x^i/\partial x^j\right)_p=\delta^i_j$ then that's just another way of saying that the $dx_p^i$ are the dual basis for$\left(\partial/\partial x^i\right)_p$ Commented Sep 9, 2019 at 23:27
• Yes, that's how I look at it too. Have a look at this for a more detailed answer. Commented Sep 9, 2019 at 23:35