# The Universal Property of Tangent Spaces

There are many different ways to define the tangent space $$T_pM$$ at some point $$p\in M$$, just like there are different ways to define the tensor product $$V_1\otimes\cdots\otimes V_n$$ or the cartesian product $$V_1\times\cdots\times V_n$$ of vector spaces. In the latter cases, all definitions have a universal property that guaranturees the existence of exactly one natural isomomorphism from one set to the other.

Does $$T_pM$$ have such a universal property, too?

By the way, I searched for this on the Internet and I found this post on Mathoverflow, but I understand neither the question, nor the answers, as they are very advanced (I am an undergraduate). I'm hoping that there is an easier answer like in the case of the tensor product. (For comparison, here is an explanation for the case of the cartesian product that I like.)

• One way of doing it is explained on pages 362-3 of Tammo tom Dieck's book "Algebraic Topology". bit.ly/3lgQ42J . Nov 29, 2020 at 17:05
• @MichaelCohen Thank you :) Are there different ways of doing it? Nov 29, 2020 at 18:19
• I did work out my own way some years ago but it's not that different from tom Dieck's. Nov 29, 2020 at 21:44
• @MichaelCohen: the page you mention isn't shown on Google books. Would you mind writing the property down as an answer? I'm really interested in it. Dec 5, 2020 at 8:14
• @Vercassivelaunos I added an answer (based on the book mentioned above), I hope it helps. Dec 5, 2020 at 8:42

$$\newcommand{\dif}{\mathrm{d}}$$ The following is inspired by the pages 362-3 of Tammo tom Dieck's book "Algebraic Topology". The definition of tangent spaces and the differential exploits the fact that we have already defined the differential of a differentiable function $$f\colon U\subset\mathbf{R}^n\to\mathbf{R^n}$$: The differential $$Df\colon U\to L(\mathbf{R}^n,\mathbf{R^n})$$ is given by the Jacobian: $$Df(a)(x)=\partial_if^j(a)x^ie_j=\begin{pmatrix}\partial_if^1(a)x^i\\ \vdots\\ \partial_if^n(a)x^i \end{pmatrix}=\begin{pmatrix}\partial_1f^1(a)&\cdots&\partial_nf^1(a)\\ \vdots&&\vdots\\ \partial_1f^n(a)&\cdots&\partial_nf^n(a) \end{pmatrix}\cdot\begin{pmatrix} x^1\\ \vdots\\ x^n \end{pmatrix}$$

## The universal property

Let $$M$$ be a differentiable, $$n$$-dimensional manifold - a manifold with an atlas such that all transition charts are differentiable. Consider the following subset of $$A$$ for some $$p\in M$$: $$\begin{equation*} A_p:=\left\{x\in A:p\in\text{Dom}(x)\right\} \end{equation*}$$ (According to the definition of an atlas, $$A_p\neq\emptyset$$ for all $$p\in M$$.)

A tangent space at $$p\in M$$ is a pair $$(V,\phi)$$, consisting of a $$n$$-dimensional real vector space $$V=:T_pM$$ and a map \begin{align*} \phi\colon A_p&\to L\left(T_pM,\mathbf R^n\right)\\ x&\mapsto\phi(x)=:\dif x_p \end{align*} with the following properties:

• $$\dif x_p$$ is an isomorphism for all $$x\in A_p$$. Thus, $$\begin{equation*} \left(\frac{\partial}{\partial x^1}|_p,\ldots,\frac{\partial}{\partial x^n}|_p\right):=(\mathrm{d}x_p^{-1}(e_1),\ldots,\mathrm{d}x_p^{-1}(e_n)) \end{equation*}$$ is a basis of $$T_pM$$ - the basis induced by $$x$$.
• We have $$$$\tag{1} D(y\circ x^{-1})_{x(p)}=\dif y_p\circ\dif x_p{}^{-1}$$$$ for all $$x,y\in A_p$$.

Lemma. The tangent space is defined up to a natural isomorphism:

Let $$(V,i)$$ and $$(V',i')$$ be tangent spaces at $$p$$. An isomorphism $$f\in L(V',V)$$ is called natural, if there is an $$x\in A_p$$ such that $$f=\mathrm{d}x_p^{-1}\circ\mathrm{d}x_p'$$. You can prove that there is exactly one natural isomorphism using the chain rule.

## Definition - The Differential

Let $$(M,A)$$ and $$(M',A')$$ be two differentiable manifolds.

• A function $$f\colon M\to M'$$ is called differentiable at $$p\in M$$ if $$y\circ f\circ x^{-1}$$ is differentiable at $$x(p)$$ for some pair $$(x,y)\in A_p\times A'{}_{f(p)}$$. In this case, $$y\circ f\circ x^{-1}$$ is differentiable at $$x(p)$$ for all $$(x,y)\in A_p\times A'{}_{f(p)}$$.
• The differential of a differentiable function $$f\colon M\to M'$$ is defined as $$\begin{equation*} \mathrm{d} f_p:=\mathrm{d}y_{f(p)}^{-1}\circ D(y\circ f\circ x^{-1})_{x(p)}\circ\mathrm{d}x_p\in L(T_pM,T_{f(p)}M') \end{equation*}$$ You can prove that $$\mathrm{d} f_p$$ doesn't depend on the choice of $$(x,y)\in A_p\times A'_{f(p)}$$ by using the chain rule.

Let me conclude with a useful equation: $$\begin{equation*} \mathrm{d}f_p\frac{\partial}{\partial x^i}|_p=\underbrace{(\mathrm{d}y_{f(p)}^{-1}\circ\underbrace{D(y\circ f\circ x^{-1})_{x(p)}\circ\underbrace{\mathrm{d}x_p)\frac{\partial}{\partial x^i}|_p}_{=e_i}}_{=\partial_i(y^j\circ f\circ x^{-1})_{x(p)}e_j}}_{=\partial_i(y^j\circ f\circ x^{-1})_{x(p)}\mathrm{d}y_{f(p)}^{-1}e_j}=\partial_i(y^j\circ f\circ x^{-1})_{x(p)}\frac{\partial}{\partial y^j}|_{f(p)} \end{equation*}$$

• Thanks! This looks very similar to a "physicists perspective" for the tangent space, where a tangent vector is defined by the coordinates it has in different coordinate systems/charts. Dec 5, 2020 at 9:30
• @Vercassivelaunos Thank you for appreciating my answer :) I just wanted to let you know that I made a huge edit, since I think that I have significantly improved my answer. Feb 5, 2021 at 13:32