How Can the Vector Space $\mathbb{R}^n$ be identified with the Column space

Question

Hi i am reading An introduction to manifolds by Loring and have some doubts in section 2.1 and 2.3. First i am writing some statements from the sections from which i have doubts and then i'll state my doubts.So the first statement that i have doubt in is:

By the correspondence between arrows and column vectors, the space $\mathbb {R}^n$ can be identified with this column space.

The second statement that i have doubt in is:

We usually denote the standard basis for $\mathbb {R}^n$ or $T_p(\mathbb R^n)$ by $e_1,...e_n$.

The third statement i have doubt in is:

This proves that $\mathcal D=\mathcal D_v$ for $v=\langle \mathcal Dx^1,...\mathcal xD^n \rangle$.This theorem shows that one may identify the tangent vectors at $p$ with the derivations at $p$. Under the vector space isomorphism $T_p\mathbb R^n\simeq\mathcal D_p\mathbb R^n$ the standard basis $e_1,...e_n$ for $T_p(\mathbb R^n)$ corresponds to the set $\left.\frac{\partial}{\partial x^1}\right|_p,...,\left.\frac{\partial}{\partial x^n}\right|_p$.

My doubts are as follows:

1. How can we identify the space $\mathbb R^n$ with the column space? Is this an isomorphism ?
2. What does it mean that we have identified the space $R^n$ with column space? Does this mean we can generate the space $R^n$ as linear combinations of the vectors from the column space? Also it is written in the book that "we visualize the tangent space $T_p \mathbb R^n$ at $p$ in $R^n$ as the vector space of all vectors emanating from $p$". So will this vector space include all the possible vectors that have $p$ as starting point,even the normal vectors? Or only the vectos that pass tangentially through $p$?
3. In the second statement(doubt) why we are denoting the standard basis $e_1,...,e_n$ both for $R^n$ and $T_p \mathbb R^n$? Aren't they different vector spaces? How can they have the same basis. For example, in $\mathbb R^3$ we have $i,j,k$ as basis vectors.Are these basis vectors in $R^3$ different from $\left.\frac{\partial}{\partial x^1}\right|_p,\left.\frac{\partial}{\partial x^2}\right|_p,\left.\frac{\partial}{\partial x^3}\right|_p$?
4. In the third(doubt) how does $\mathcal D=\mathcal D_v$ for $v=\langle \mathcal Dx^1,...\mathcal xD^n \rangle$ mean allow us to identify the tangent vectors at $p$ with the derivations at $p$? I mean i know that the tangent vector is $v$ , $\mathcal D$ is the derivation at $p$ and $\mathcal D_v=v^i \frac{\partial}{\partial x^i}$ is the directional derivative at $p$.

For reference i am attaching the screenshots of the statement where i have highlighted different statements from the book.

Answer 1

Parts of your question have been answered here.

$\mathbb R^n$ is the vector space of all $n$-tuples of real numbers. Usually one writes $x \in \mathbb R^n$ as $x = (x_1,\ldots,x_n)$ in row form, but you can use also the column form $$x = \left[ \begin{array}{rrr} x_1 \\ . \phantom{.} \\ . \phantom{.} \\ . \phantom{.} \\ x_n \\ \end{array}\right]$$ This is just a notational issue. If you like, you can regard column vectors as $n \times 1$-matrices. Tu also prefers to use for coordinates upper indices instead of lower (for example $x^i$ instead of $x_i$).

Tu introduces the tangent space $T_p(\mathbb R^n)$ at $p$ in $\mathbb R^n$ as the vector space of all arrows emanating from $p$. He illustrates this for $n = 3$:

This generalizes of course to arbitrary $n$. Thus, formally we should regard such an arrow as a pair $a = (p,v)$ with $v \in \mathbb R^n$, that is we have $T_p(\mathbb R^n) = \{p\} \times \mathbb R^n$. The vector space structure on $\{p\} \times \mathbb R^n$ is of course given by $(p,v) + (p,w) = (p,v+w)$ and $r(p,v)= (p,rv)$. This achieves that the canonical bijection $$\beta : \{p\} \times \mathbb R^n \to \mathbb R^n, \phi((p,v)) = v $$ becomes a linear isomorphism of vector spaces. Note that an arrow emanating at $p$ can geometrically be viewed as the directed line segment from $p$ to $p + v$, in other words as the pair of points $(p,p+v)$.

Tu uses the convention to write the starting point $p$ of an arrow $a = (p,v)$ in row form and the "vectorial part" $v$ of $a$ in column form. In Tu's own words, he does so to distinguish between points and vectors. Moreover, by an abuse of notation he drops $p$ from $a = (p,v)$, i.e. identifies $a$ with the column vector $v$. Thus in the sense of Tu there is difference between the column vector with components $v^i$ (which is an element of $T_p(\mathbb R^n))$ and the row vector $v = (v^1,\ldots,v^n) \in \mathbb R^n$. Personally I do not like this way of notational distinction of $T_p(\mathbb R^n)$ from $\mathbb R^n$, it does not elucidate anything and does not have any advantage over properly writing $a = (p,v)$ for the elements of $T_p(\mathbb R^n)$. In fact, Tu introduces an oversubtle and unnecessary distinction between row and columns vectors - but perhaps it is a matter of taste.

Anyway, this should answer your questions 1. and 2. In 2. you also ask "So will this vector space include all the possible vectors that have $p$ as starting point, even the normal vectors? Or only the vectors that pass tangentially through $p$?" Here you are misled by Fig. 2.2. A tangent vector $v$ to a surface at $p$. In fact, $T_p(\mathbb R^3)$ is not the set of tangent vectors to some surface $S \subset \mathbb R^3$ at the point $p \in S$, it is the set of all column vectors in the above sense. The set of tangent vectors to $S$ at $p$ is a two-dimensional linear subspace of $T_p(\mathbb R^3)$, as you shall see later when studying the book. You can write $T_p(S)$ for it.

Concerning your question 3.: Yes, it is an abuse of notation. Tu uses the same symbols for basis elements of $\mathbb R^n$ and of $T_p(\mathbb R^n)$. If $e_1, \ldots,e_n$ denote the standard basis vectors of $\mathbb R^n$, he should properly write $(p, e_i)$ for the standard basis vectors of $T_p(\mathbb R^n)$. But due to Tu's notational conventions we may omit $p$ and regard $e_i$ as a column vector. I think the danger of confusion is very little.

Of course $e_i$ does not live in $\mathcal D_p(\mathbb R^n)$ and thus is different from the derivation $D_{e_i} = \dfrac{\partial}{\partial x_i} \mid_p$. But for each $v \in \mathbb R^n$ we have the usual directional derivative $D_v = D_v \mid_p$ at $p$. This is the map which associates to any smooth $f : U \to \mathbb R$ defined on an open neigborhood $U$ of $p$

$$D_v \mid_p(f) = \lim_{t \to 0}\frac{f(p +tv)-f(p)}{t} .$$ Tu formalizes this and gets a map $$D_v = D_v \mid_p : C^\infty_p \to \mathbb R$$ defined on the set $C^\infty_p$ of germs of smooth functions at $p$. This set is an $\mathbb R$-algebra (i.e. a vector space with a compatible multiplication of its elements). Tu shows that each $D_v$ is a derivation and thus gets his linear map $$\phi : T_p(\mathbb R^n) \to \mathcal D_p(\mathbb R^n), \phi(v) = D_v.$$ In this formula $v \in T_p(\mathbb R^n)$ is a column vector in the sense of Tu's convention. It would be more precise to understand it as $\phi((p,v)) = D_v \mid_p$.

The formula $D_v = \sum v^i \dfrac{\partial}{\partial x_i} \mid_p$ follows from the fact that $D_{e_i} = \dfrac{\partial}{\partial x^i} \mid_p$: We have $v = \sum v^i e_i$ and therefore $D_v = \phi(v) = \phi(\sum v^i e_i) = \sum v^i \phi(e_i) = \sum v^i \dfrac{\partial}{\partial x^i} \mid_p$.

It is by no means trivial that $\phi$ is an isomorphism of vector spaces. This is proved in Theorem 2.2. It answers your question 4.: As Tu writes

This theorem shows that one may identify the tangent vectors at $p$ with the derivations at $p$. Under the vector space isomorphism $T_p(\mathbb R^) \simeq \mathcal D_p(\mathbb R^n)$, the standard basis $e_1,\ldots,e_n$ for $T_p(\mathbb R^n)$ corresponds to the set $\dfrac{\partial}{\partial x^1} \mid_p, \ldots, \dfrac{\partial}{\partial x^n} \mid_p$ of partial derivatives.

Answer 2

1. In linear algebra, by default, the elements of $\Bbb R^n$ are considered as column vectors, so that the linear maps can be written as $x\mapsto Ax$ with a matrix (after choosing coordinate systems), corresponding to the way we write functions in analysis: $x\mapsto A(x)$.
   (While if we chose row vectors to work with, we need to consider the action of functions on the right: $x\mapsto xA$.)
   On the other hand, to distinguish notation, the points of the space are written as row vectors in the text.

2. Geometrically, the tangent plane of a plane (embedded in the space) at any point is the plane itself. Likewise, the tangent space $T_p(\Bbb R^n)$ at any point $p$ of the full space $\Bbb R^n$ is the full space, and there is no surface or anything to talk about normal vectors, all vectors are tangent vectors of the space.
   You can visualize it by drawing the vector $v\in T_p(\Bbb R^n)$ that corresponds to $\tilde v\in\Bbb R^n$ from point $p$ to point $p+\tilde v$.
   
   But be careful here, the tangent vectors at $p$ are probably defined abstractly: as the equivalence classes of smooth curves through $p$, with the equivalence given by having identical derivative at $p$.
   The correspondence to $\Bbb R^n$ is simply given by taking the derivative (velocity vector) of the curve at $p$, all done in the standard coordinates.

3. Well, when we write 'identifying' two kinds of objects, then afterwards they get identified in the text.
   By (the above) definition, we primarily consider tangent vectors as equivalence classes of curves. Then we introduce the directional differentiation operators that map real numbers to real valued smooth functions and satisfy the Leibniz rule at $p$, and these can also be identified with the tangent vectors: for a curve $\gamma:(-\varepsilon,\varepsilon)\to M$ with $\gamma(0)=p$, the directional derivative is $f\mapsto (f\circ\gamma)'(0)$.
   In the specific case when $M=\Bbb R^n$ (or locally, using a chart [=local coordinate system] of the $n$ dimensional manifold), the tangent vectors at a point $p$ can further be identified with the actual elements of $\Bbb R^n$, because they form a vector space with basis given by the curves $t\mapsto p+t\cdot e_i$ or by the corresponding directional differentiations $\frac\partial{\partial x^i}$.

4. So, the column vector $v=\pmatrix{v_1\\v_2\\ \vdots \\ v_n}$ is the correspondent of the directional differentiation $\partial_v=\sum_i v_i\frac\partial{\partial x^i}$ and also of the equivalence class of the curve $t\mapsto p+tv$.