# Tangent vectors as derivatives

I am reading "An introduction to manifolds" by Loring Tu and I struggle understanding a concept. In the book the directional derivative is given a tangent vector $v$ at a point $p$ the map : $$D_v : \mathcal{C}_p^\infty \rightarrow \mathbb{R}$$ The follows by explaining that a map satisfying the leibniz rule is a "Derivation at p". The we can define the map : $$\phi : T_p(\mathbb{R}^n) \rightarrow \mathcal{D}_p(\mathbb{R}^n)\\ v \rightarrow D_v = \sum v^i \frac{d}{dx^i} |_0$$ where : the "d" are partials evaluated at $0$.

So far so good, then the book proceed to the proof that this map is an isomorphisme of vector spaces. And by this decide to write every tangent vectors $v = (v_1,...,v_n) = \sum v^i e_i$ as $v = \sum v^i \frac{d}{dx^i}|_p$

I have trouble understanding how a vector could be equal to a function. For me the RHS takes a function and the LHS is just a vector. I would like to know how is that intuitively correct and also what is the proof of the last equation.

• You only need to review Linear Algebra: every linear function is determined at the basis, $f(\sum a_ie_i) = \sum^n a_i f(e_i)$, and as such there is unique matrix A such that f(v) = Av = y. – Weaam Oct 5 '15 at 21:46
• I am sorry but I failed to understand how this is related with my problem. Here I am trying to understand how different objects can be equal to each other. The object on the LHS is a vector and the object on the RHS is an operator that takes a function. also How can the standar basis of the vector space $(e_1, ..., e_n)$ could be equal to the derivatives $(\frac{d}{dx^1}, ... ,\frac{d}{dx^n} )$. Thanks ! – user149705 Oct 5 '15 at 21:56

A real vector space is any set $V$ endowed with an addition operator $+ : V \times V \to V$ and a scalar multiplication operator $\cdot : \Bbb R \times V \to V$ such that $+$ is commutative, associative, has an identity $0$ and an inverse $-v$ for each vector $v\in V$, and such that $a(v + w) = av + aw$ for $a\in \Bbb R, v,w \in V$.

It doesn't matter what the elements of the vector space $V$ are. As long as you can define the two operations, it is a vector space, and we can refer to its elements as vectors, and apply the whole theory of vector spaces to it, including the concept of bases. A very VERY common source of vector spaces is spaces of maps. If $S$ is any set, and $W$ is a vector space itself, consider the set V of all maps from $S \to W$. For any two functions $f, g \in V$ and any scalar $a \in \Bbb R$, define $(f+g)(s) = f(s) + g(s)$ and $(af)(x) = af(x)\;\; \forall s \in S$. These make $V$ a vector space. You can choose smaller spaces of maps, provided they are closed under addition and scalar multiplication.

$T_p(\Bbb R^n)$ is just such a vector space. The set $S$ is the set $\mathcal F(\Bbb R^n)$ of smooth functions real-valued functions on $\Bbb R^n$ (or some more local equivalent - it depends on the author which they prefer). The vector space $W = \Bbb R$. And in this case, since $\mathcal F(\Bbb R^n)$ we can, and do, restrict $T_p(\Bbb R^n)$ to linear maps from $\mathcal F(\Bbb R^n) \to \Bbb R$, and further to those that satisfy the Liebnitz condition $v(fg) = f(p)v(g) + g(p)v(f)$. The resulting set of maps $v$ is still closed under addition and scalar multiplication, and so still forms a vector space.

Now consider what this means: an element $v \in T_p(\Bbb R^n)$ is a function that carries functions in $\mathcal F(\Bbb R^n)$ to real numbers. That is, it is an operator on $\mathcal F(\Bbb R^n)$. But because it is also a member of the vector space $T_p(\Bbb R^n)$ it is a vector.

Anything can be a vector, as long as you can define addition and scalar multiplication. That is why vector spaces are so useful. They apply to so many things.

Edit: I did not realize that you had only seen the tangent space defined for $\Bbb R^n$, and where they used the canonical vector space structure of $\Bbb R^n$ to be its own tangent space. In general, manifolds are not vector spaces. When thinking about manifolds, I usually view them as being an undulating surface. But a more concrete example may be helpful to you: the sphere. Consider a point on the sphere and all the various vectors tangent to the sphere at that point. These vectors form a plane that is tangent to the sphere. That plane is the tangent space. (This picture is slightly misleading, as the tangent planes at two different points intersect in $\Bbb R^3$, but we need the actual tangent spaces to be completely disjoint, so we define them a unique abstract objects.) For any curve in the sphere that passes through the point, its derivative is a vector in this plane (all mappings mentioned here are assumed to be smooth). If I have a function $f$ on the sphere and a curve $\phi$ with $\phi(0) = p$ and let $v$ be the tangent vector $\phi'(0)$, then $f \circ \phi$ is from $\Bbb R$ to $\Bbb R$. Further if $\psi$ is a second curve with $\psi(0) = p$ and $\psi'(0) = v$, then you can show that $$\left.\frac{d(f \circ \phi)}{dt}\right |_0 = \left. \frac{d(f \circ \psi)}{dt} \right |_0$$ In other words, the value of this derivative is dependent only on $v$, not on the curve chosen. Thus we can define the directional derivative $$D_vf(p) = \left.\frac{d(f \circ \phi)}{dt}\right |_0.$$ Note that since $v$ is a tangent vector at $p$, the $p$ in the definition is actually redundant, but I left it in to make it apparent that this is a value is at $p$.

The directional derivative $D_v$ is an operator on the space of smooth functions at $p$. It is linear, and Liebnitzian. In addition to being linear on the functions, it is also linear in $v$. Further, you can show that for any real-valued linear Liebnitzian operator on the space of smooth functions at $p$, there is some vector $v$ such that the operator is $D_v$. That is, the tangent space $T_p(S^2)$ is isomorphic as a vector space to the space of linear Liebnitzian operators at $p$.

A significant problem with the above development is that it depends on how the sphere is imbedded in $\Bbb R^3$. The tangent vectors are vectors in $\Bbb R^3$. But in general what we are interested in with manifolds are their own properties, not the properties of how they sit in another space. So we need a more abstract development of a manifold. But then where do we get the tangent vectors from, if they are not vectors in some containing vector space? While the development of the tangent space as a tangent plane is no longer available, the vector space of linear Liebnitzian operators still is. And since is was isomorphic to the imbedded tangent space, it has exactly the properties we need. That is why in general, the tangent space is defined to be the space of linear Liebnitzian operators at $p$. We just identify each vector $v$ with its directional derivative operator.

• I guess the missing part of my understanding is why $T_p(\mathbb{R}^n)$ is the vector space that send smooth functions to reals. In the book $T_p(\mathbb{R}^n)$ is defined as the set of vectors starting at $p$ and all elements from it can be written according to the standard basis in $\mathbb{R}^n$, $(e_1, ..., e_n)$. – user149705 Oct 5 '15 at 23:54
• See the edit above. I didn't realize you had not seen tangent spaces for general manifolds. – Paul Sinclair Oct 6 '15 at 3:31
• Thank you for the edit ! So If I understand correclty the $v = \sum v^i \frac{d}{dx^i}|_p$ is actually not the same as the $v = \sum v^i e_i$. It is just since the two vector space are isomorphic we can relate to one another with a known bijection. So instead of considering the vector in $\mathbb{R}^n$ we consider the vectors in the derivation space ? The book made it really confusing if this is the idea. – user149705 Oct 6 '15 at 17:37
• Not quite. In general, this space of operators is the only canonical vector space available that we can define the tangent space to be. We could choose some coordinate patch and use that to define a vector space, but then the actual vector space is dependent on our choice of coordinate patch (even though all the patches give rise to isomorphic vector spaces). So we would have endless complications this way. But the space of linear Liebnitzian operators at $p$ is not coordinate system dependent, and has all the properties we need, so we define $T_p(M)$ to be it. – Paul Sinclair Oct 6 '15 at 19:45