# Tangent vectors as differential operators vs equivalence classes of curve in manifold

Suppose we have a curve $$\alpha: \mathbb{R} \rightarrow \mathbb{R}^2$$ through $$p \in \mathbb{R}^2$$ and $$f: \mathbb{R}^2 \rightarrow \mathbb{R}$$ is a smooth function. Then $$f \circ \alpha: \mathbb{R} \rightarrow \mathbb{R}$$ is a smooth function and we can differentiate it at $$0$$: $$(f \circ \alpha)^{\prime}(0)=a_1^{\prime}(0) \frac{\partial f}{\partial x_1}(\underbrace{p}_{\alpha(0)})+a_2^{\prime}(0) \frac{\partial f}{\partial x_2}(\underbrace{p}_{\alpha(0)})$$ by the Chain rule. We therefore have a map $$f \mapsto(f \circ \alpha)^{\prime}(0)$$ from functions to $$\mathbb{R}$$ given by $$f \mapsto\left(\left.a_1^{\prime}(0) \frac{\partial}{\partial x_1}\right|_p+\left.a_2^{\prime}(0) \frac{\partial}{\partial x_2}\right|_p\right) f,$$ which is a differential operator acting on functions. If we think of $$\left\{\left.\frac{\partial}{\partial x_1}\right|_p,\left.\frac{\partial}{\partial x_2}\right|_p\right\}$$ as a basis for a $$2$$ dimensional vector space, then we identify this map is the tangent vector to $$\alpha$$ at $$p$$.

The good thing about this is that we can replace $$\mathbb{R}^2$$ by any manifold $$M$$, since $$f \circ \alpha: \mathbb{R} \rightarrow \mathbb{R}$$ can be differentiated, so this definition still works. Explicitly, if $$\alpha: \mathbb{R} \rightarrow M$$ is a curve through $$p \in M$$ and $$f: M \rightarrow \mathbb{R}$$ is a smooth function then we let $$(U, \varphi)$$ be a coordinate chart at $$p$$ and write $$\varphi \circ \alpha(t)=$$ $$\left(a_1(t), \ldots, a_n(t)\right) \in \varphi(U) \subseteq \mathbb{R}^n$$. Then \begin{align} (f \circ \alpha)^{\prime}(0)&=\left.\frac{\mathrm{d}}{\mathrm{d} t}(f \circ \alpha)(t)\right|_{t=0}\\ &=\left.\frac{\mathrm{d}}{\mathrm{d} t}\left(f \circ \varphi^{-1} \circ \varphi \circ \alpha\right)(t)\right|_{t=0}\\ &=\left.\frac{\mathrm{d}}{\mathrm{d} t}\left(f \circ \varphi^{-1}\right)\left(a_1(t), \ldots, a_n(t)\right)\right|_{t=0}\tag1\\ &=\sum_{j=1}^n a_j^{\prime}(0) \frac{\partial\left(f \circ \varphi^{-1}\right)}{\partial x_j}(\varphi(p))\tag2\\ &=\left(\left.\sum_{j=1}^n a_j^{\prime}(0) \frac{\partial}{\partial x_j}\right|_{\varphi(p)}\right)\left(f \circ \varphi^{-1}\right) \end{align} Hence, using the $$\left.\frac{\partial}{\partial x_j}\right|_{\varphi(p)}$$ as a basis, we can identify the tangent vector to the curve $$\varphi \circ \alpha$$ in $$\mathbb{R}^n$$ at $$\varphi(p)$$ with the differential operator $$\left.\sum_{j=1}^n a_j^{\prime}(0) \frac{\partial}{\partial x_j}\right|_{\varphi(p)}$$ acting on the function $$f \circ \varphi^{-1}$$ (which is how we identify functions on $$M$$ locally with functions on $$\mathbb{R}^n$$ ).

In the first paragraph, considering $$\left\{\left.\frac{\partial}{\partial x_1}\right|_p,\left.\frac{\partial}{\partial x_2}\right|_p\right\}$$ as a basis make sense, as it can be thought of as the small increment/direction of each coordinate direction in $$\mathbb R^2$$. But that's not happening for second paragraph, specially $$\frac{\partial\left(f \circ \varphi^{-1}\right)}{\partial x_j}$$. Because there is no coordinate system, as we are considering in the abstract manifold space $$M$$.

• Then how to interpret $$\frac{\partial}{\partial x_j}$$ here? And I am unable to understand how $$(2)$$ come from $$(1)$$?
• Why the summation follow the dimension of the chart? And what's that mean?

It will be a great help if anyone can help me to understand the definition. TIA

## Updated

Definition 4.2.1: Given a $$C^k$$ manifold, $$M$$, of dimension $$n$$, for any $$p \in M$$, two $$C^1$$-curves, $$\left.\gamma_1:\right]-\epsilon_1, \epsilon_1[\rightarrow M$$ and $$\left.\gamma_2:\right]-\epsilon_2, \epsilon_2\left[\rightarrow M\right.$$, through $$p$$ (i.e., $$\gamma_1(0)=\gamma_2(0)=p$$ ) are equivalent iff there is some chart, $$(U, \varphi)$$, at $$p$$ so that $$\left(\varphi \circ \gamma_1\right)^{\prime}(0)=\left(\varphi \circ \gamma_2\right)^{\prime}(0) .$$ Definition 4.2.2 (Tangent Vectors, Version 1): Given any $$C^k$$-manifold, $$M$$, of dimension $$n$$, with $$k \geq 1$$, for any $$p \in M$$, a tangent vector to $$M$$ at $$p$$ is any equivalence class of $$C^1$$-curves through $$p$$ on $$M$$, modulo the equivalence relation defined in Definition 4.2.1. The set of all tangent vectors at $$p$$ is denoted by $$T_p(M)$$.

One of the defects of the above definition of a tangent vector is that it has no clear relation to the $$C^k$$-differential structure of $$M$$. There is another way to define tangent vectors that reveals this connection more clearly. As a first step, consider the following: Let $$(U, \varphi)$$ be a chart at $$p \in M$$ (where $$M$$ is a $$C^k$$-manifold of dimension $$n$$, with $$k \geq 1$$ ) and let $$x_i=p r_i \circ \varphi$$ (where $$pr_i: \mathbb R_n \rightarrow \mathbb R$$, are defined by $$pr_i(x_1, \cdots x_n) = x_i, 1\leq i\leq n)$$, the $$i$$ th local coordinate $$(1 \leq i \leq n)$$. For any function, $$f$$, defined on $$U \ni p$$, set $$\left(\frac{\partial}{\partial x_i}\right)_p f=\left.\frac{\partial\left(f \circ \varphi^{-1}\right)}{\partial X_i}\right|_{\varphi(p)}, \quad 1 \leq i \leq n .$$ (Here, $$\left.\left(\partial g / \partial X_i\right)\right|_y$$ denotes the partial derivative of a function $$g: \mathbb{R}^n \rightarrow \mathbb{R}$$ with respect to the $$i$$ th coordinate, evaluated at $$y$$.)

Here the partial derivatives are defined as $$\left(\frac{\partial}{\partial x_i}\right)_p f=\left.\frac{\partial\left(f \circ \varphi^{-1}\right)}{\partial X_i}\right|_{\varphi(p)}, \quad 1 \leq i \leq n .$$ which seems different from @J.V.Gaiter answer,

$$\frac{\partial}{\partial x_j}\vert_pf:= \frac{\partial (f\circ \phi)}{\partial x_j}(p)$$

I couldn't come up with their equivalence.

The $$\frac{\partial}{\partial x_j}\vert_p$$'s are objects that come with the coordinate charts we place on the manifold. We commit abuse of notation (or convenient identification) by saying $$\frac{\partial}{\partial x_j}\vert_p\in T_pM$$ defined by $$\frac{\partial}{\partial x_j}\vert_pf:= \frac{\partial (f\circ \phi)}{\partial x_j}(p)$$ for $$\phi$$ the chart for which $$x_j$$ is a coordinate.

If you think of the tangent vectors as "infinitesimal displacements" then the fact that for an $$n$$-dimensional manifold, $$T_pM$$ is $$n$$-dimensional should be manifest. If in a coordinate chart there are $$n$$-dimensions in which we can move, then the space of infinitesimal displacements should have that same dimension.

EDIT:

In order to get the fact that the space of tangent vectors is $$n$$-dimensional on an $$n$$-dimensional manifold, one needs to apply Taylor's theorem.

If $$f\in C^\infty(M)$$ and $$\phi: U\to \mathbb{R}^n$$ is a chart, then in a small neighborhood of $$p\in U$$ we can write $$f(x+p)=f(p)+df(x)+\sum_{i,j} A_{ij}(x)x^ix^j$$ for some smooth functions $$A_{ij}(x)$$. $$df$$ is a linear map $$\mathbb{R}^n\to \mathbb{R}$$. The Leibniz rule tells us that for a derivation based at $$p$$, $$\delta: C^\infty(M)\to \mathbb{R}$$, we have $$\delta(f(x+p))=\delta(f(p))+\delta(Df)(p)+\delta(\sum_{ij}A_{ij}(x)x^{i}x^j)=\delta(Df)(p)$$. This means that $$\delta$$ only depends on the value of $$Df(p)$$. As $$df(p)$$ is an element of $$(\mathbb{R}^n)^*$$ it lies in an $$n$$-dimensional vector space and we can construct functions $$f\in C^\infty(M)$$ with $$Df=\xi$$ for $$\xi\in (\mathbb{R}^n)^*$$with respect to any given compatible coordinate chart using bump functions. This means that the dimension of the space of derivations at $$p$$, $$\mathrm{Der}_p(M)$$ is at most $$n$$. We can show that the space is $$n$$ dimensional by using the coordinate curves, i.e. we can find $$n$$ linearly independent elements of $$\mathrm{Der}_p(M)$$ given by the maps $$f\mapsto \frac{d}{dt}\vert_{t=0}f(p+tx^i)$$.

(Note, here I am abusing notation and identifying $$f:M\to \mathbb{R}$$ with $$f\circ \phi: U\to \mathbb{R}$$)

• Thanks for your response @J.V.Gaiter Can you comment on how $(1)\implies(2)$? I guess it use the same logic of 1st paragraph. But I want to know your perspective. Again thanks. Aug 6, 2023 at 7:48
• Thanks for your edit @J.V.Gaiter. But I am still not clear how $$\left.\frac{\mathrm{d}}{\mathrm{d} t}\left(f \circ \varphi^{-1}\right)\left(a_1(t), \ldots, a_n(t)\right)\right|_{t=0}\rightarrow \sum_{j=1}^n a_j^{\prime}(0) \frac{\partial\left(f \circ \varphi^{-1}\right)}{\partial x_j}(\varphi(p))$$ Aug 9, 2023 at 4:22
• That follows from the chain rule from multivariable calculus. Aug 9, 2023 at 14:33
• I see. Actually, I see a lot of definition of tangent vector here and there. And each have its own distinct idea. I see there was another way to define $\frac{\partial}{\partial x_i}$, (added the definition as well as the source in the updated thread)$$\left(\frac{\partial}{\partial x_i}\right)_p f=\left.\frac{\partial\left(f \circ \varphi^{-1}\right)}{\partial X_i}\right|_{\varphi(p)}, \quad 1 \leq i \leq n .$$How they related each other? I hope this will be the last question for this thread. And I will accept your answer. Sorry to ask that much questions, Thanks for your help @J.V.Gaiter Aug 9, 2023 at 17:36
• That's the same definition I give, just instead using the convention that charts are maps $\varphi:U\to M$ where $U\subset \mathbb{R}^n$ is open rather than maps $\phi: V\to \mathbb{R}^n$ where $V\subset M$ is open. The exact details and conventions are not important, but rather the idea that we transport notions from multvariable calculus to differntial geometry by using the charts to convert objects on our manifold into objects living over open subsets of $\mathbb{R}^n$. Aug 9, 2023 at 22:34