# Why are the partial derivatives a basis of the tangent space?

Why are $$\frac{\partial }{\partial x_i}$$ , $$i=1,...,n$$, a basis for $$T_x\mathbb{R}^n$$?

My understanding is that the tangent space at $$x$$ is the set of all vectors beginning at $$x$$. I would be led to believe that all vectors beginning at $$x$$ would have the form $$\vec{v}-\vec{x}$$.

My question is three fold:
1. Why are the basis vectors partial derivatives and not the vector above?
2. How do they span all vectors originating at $$x$$?
3. Are elements of the tangents space thus linear functions?

• Recent related question Aug 21, 2019 at 15:21
• This is not an answer but just to ask for a little further clarification. The first answer above mentions an isomorphism $\phi: T_p(\mathbb{R}^n)\rightarrow D_p(\mathbb{R}^n)$. I just want to confirm the following: (1) May I assume the same can be said for a manifold $M$ (instead of $\mathbb{R}^n$), i.e.: $\phi: T_p(M)\rightarrow D_p(M)$? (2) The partials $\partial/\partial x^i$ form a basis of $D_p$, instead of the tangent space $T_p$ itself. When people still say the partials are the basis of the tangent space, is that because they don't distinguish between $T_p$ and $D_p$ simply because the
– Ruye
Apr 16, 2022 at 0:23

To distinguish between points and tangent vectors, let $$p=(p_1,...,p_n)\in \mathbb{R}^n$$ a point of $$\mathbb{R}^n$$ and $$v=(v_1,...,v_n)\in T_p(\mathbb{R}^n)$$ a point of the tangent space $$\mathbb{R}^n$$.

The line through $$p=(p_1,...,p_n)\in \mathbb{R}^n$$ with direction $$v=(v_1,...,v_n)\in T_p(\mathbb{R}^n)$$ has parametrization $$a(t)=(p_1+tv_1,...,p_n+tv_n)$$.

If f is $$C^\infty$$ in a neighborhood of $$p\in \mathbb{R}^n$$ and $$v$$ is a tangent vector at $$p$$, define the directional derivative of $$f$$ in the direction of $$v$$ at $$p$$ as $$D_vf=\lim\limits_{t \to 0} \frac{f(a(t))-f(p)}{t}.$$

By the multi-variable chain rule, we have $$D_vf=\sum_{i=1}^{n} \frac{da^i}{dt}(0)\frac{\partial f}{\partial x^i}(p)=\sum_{i=1}^{n} v_i\frac{\partial f}{\partial x^i}(p).$$

Of course in the above notation $$D_vf$$, the partial derivatives are evaluated at $$p$$, since v is a vector at $$p$$. Now, we can define a map $$D_v$$ (which assigns to every f which is $$C^\infty$$ the real number $$D_v(f)$$ ) with the natural way $$D_v=\sum_{i=1}^{n} v_i\frac{\partial }{\partial x^i}(p)=\sum_{i=1}^{n} v_i\frac{\partial }{\partial x^i}\Bigr\rvert_{p}.$$

This map $$D_v\in \mathcal{D}_p(\mathbb{R}^n)$$ is in fact a derivation at $$p$$. Finally, you can show that the map \begin{align} \phi :T_p(\mathbb{R}^n) &\to \mathcal{D}_p(\mathbb{R}^n) \\ v &\to D_v \end{align}

is a linear isomorphism of vector spaces (for surjectivity, you can use Taylor's theorem).

3) Υes, you can see every tangent vector $$v \in T_p(\mathbb{R}^n)$$ as a derivation $$D_v\in \mathcal{D}_p(\mathbb{R}^n)$$ using the isomorphism $$\phi$$.

2) Since $$e_1,...,e_n$$ is the canonical basis of $$T_p(\mathbb{R}^n)$$ and $$\phi$$ is an isomorphism, then $$\phi(e_1),...,\phi(e_n)$$ is a basis of $$D_v\in \mathcal{D}_p(\mathbb{R}^n)$$. But $$\phi(e_i)=\frac{\partial }{\partial x^i}\Bigr\rvert_{p}$$, hence $$\{\frac{\partial }{\partial x^i}\Bigr\rvert_{p}\}_{i=1}^n$$ is a basis of the tangent space $$\mathcal{D}_p(\mathbb{R}^n)\simeq T_p(\mathbb{R}^n)$$.

1) As a result, you can say that the basis of $$T_p(\mathbb{R}^n)$$ is $$\{\frac{\partial }{\partial x^i}\Bigr\rvert_{p}\}_{i=1}^n$$. You can say all that because of $$\phi$$.

• Great explanation ! I assume $e_i$ designates the canonical basis here, otherwise there would be some Jacobian in front of the $\dfrac{\partial}{\partial x^i}$ in case of a change of basis, is that correct ? May 7, 2020 at 15:03
• @Johann Yes, that is correct. Thank you for the remark and the clarification! $e_i$ is the canonical basis. Sections 2.1-2.3 of the book "An introduction to manifolds" by Loring Tu contain more information about the basis, tangent vectors and the isomorphism $\phi$ . May 8, 2020 at 16:38
• An statement/link to the multivariate chain rule would be a nice inclusion near the beginning, so that the instance applied at $D_{v}f=...$ in the second eq. is linked to that theorem, which is fundamental but a bit obscure.
– apg
Feb 9, 2022 at 15:09

The tangent space is not the set of all vectors beginning at $$x$$. It is a different vector space that is "attached" to the original vector space at $$x$$, and the vectors in this space are not elements of the original vector space. You have one of these tangent spaces attached at each point, and the tangent bundle for a vector space is the space together with all of the attached tangent spaces.

The tangent space is viewed as the space of infinitesimal displacements starting at $$x$$, since it does not bleed into other points but in a sense does contain $$x$$. That is why they are identified as directional derivative operators.

• So would $\frac{\partial}{\partial x_1}$ be an element of the tangent space which corresponds to an infinitesimal change at $x$ in the $x_1$ direction? If correct does this mean elements of the tangent space are only relevant when applied to something else? Aug 21, 2019 at 15:29
• @Matthew You can also embed the tangent bundle into another vector space to get actual tangent vectors. These would be tangent lines to the space. Aug 21, 2019 at 16:33
• The picture I now have in my head is that taking tangent vector from the tangent space and applying to a function is analogous to applying the directional derivative in that direction. Aug 22, 2019 at 8:13

You're right that on a plane, the tangent vectors at $$x$$ have the form $$v-x$$, but note that forming vectors by subtracting points only makes sense in Euclidean space. The points of an abstract manifold are not necessarily elements of a vector space, and if you embed the manifold and do subtraction in the ambient space, the vector between two points is not necessarily tangent to the surface (think of the sphere, say).

You could try to generalize the idea as follows: pick a point $$v\neq x$$ and draw the shortest curve possible that travels along the manifold from $$x$$ to $$v$$, and replace $$v-x$$ with the derivative of the curve at the starting point. This idea will lead to manifold exponential and logarithm maps. The issue is that defining "shortest possible" requires additional structure (e.g. a metric) beyond differential structure.

For this reason the tangent space is usually defined in terms of linear operators (as in Giuseppe's linked answer) or equivalence class of tangent vectors of curves through $$x$$.

About the basis of $$T_p M$$;

Let $$α:(-ε,ε)\to M$$ a curve in differential manifold $$M$$ such that $$α(0)=p\in M$$. If $$\mathscr D_p:=\{f:M\to \Bbb R |f$$ differentiable at $$p\}$$, then we define the tangent space to $$M$$ at $$p$$ to be the set

$$T_p M:=\{α'(0):\mathscr D_p\to \Bbb R|f\mapsto \frac {d}{dt}(f\circ α)(0)\},$$

which is a linear space (the proof is easy).

We get

$$α'(0)f=\frac {d}{dt}(f\circ α)(0)=\nabla f(α(0))\cdot α'(0)=(\sum_i x_i'(0)\frac {\partial}{\partial x_i}|_0)f,$$ where $$α(t)=(x_1(t), ...,x_n(t))$$ and $$x:U\to M, p\in U,$$ a parametrization of $$M$$.

We just proved that function $$α'(0)$$ is a linear combination of $$\{\frac {\partial}{\partial x_i}|_0\}_{i=1,...n}$$, which means that $$span\{\frac {\partial}{\partial x_i}|_0\}_{i=1,...n}=T_p M.$$

We need to show that $$\{\frac {\partial}{\partial x_i}|_0\}_{i=1,...n}$$ is linear independent; thus for $$λ_i\in \Bbb R$$ and for $$π_j:M\to \Bbb R:x\mapsto x_j$$ we have :

$$\sum_iλ_i \frac {\partial}{\partial x_i}|_0=0\implies (\sum_iλ_i \frac {\partial}{\partial x_i}|_0)π_j=0\implies \sum_iλ_i δ_{ji}=0\implies λ_j=0.$$