# How does the definition of tangent space in terms of equivalence classes of curves relate with the intuitive notion of tangent vectors?

Let $M$ be an $n$-dimensional smooth manifold. One definition of the tangent space of $M$ at a point $p$ is the set of equivalence classes of curves $(\alpha,c)$, where $\alpha$ is a smooth map from a real open interval $J$ to $M$ such that $\alpha(c) = p$, and $(\alpha,c) \sim (\beta,d)$ if there exists a chart $(U,\phi)$ containing $p$ such that

$$(\phi \circ \alpha)'(c) = (\phi \circ \beta)'(d) \in \mathbb{R}^n$$

where the derivative of $\phi \circ \alpha$ at $c$ is identified with a vector in $\mathbb{R}^n$, and similarly for $\phi \circ \beta$.

I don't understand how this definition relates to the intuitive notion of what a tangent space should be. Imagine $M$ is a two dimensional surface. Then it seems like these equivalence classes of curves are arrows passing through $p$, tangent to the surface $M$, and pointing in a particular direction. How are the magnitude of these tangent vectors being taken into account in the definition of equivalent curves?

• Two curves are equivalent if their velocities coincide, and velocities encode both direction and magnitude. If you reparametrized a curve to go twice as fast, then it would double the magnitude of the corresponding tangent vector. Dec 15 '17 at 1:56
• I can't tell if you're confused because your "intuitive notion" of a tangent space doesn't allow for scaling, your "intuitive notion" of a curve is ignoring its parametrization, or you haven't noticed that these things go into the result of the derivative. (or something else!)
– user14972
Dec 15 '17 at 2:52

Oh, I think I maybe see your problem.

Are you thinking of something resembling a curve $\alpha : [0,1] \to M$ with $\alpha(0) = p$ as being an actual shape drawn on the manifold, starting from the point $\alpha(0)$ and ending at the point $\alpha(1)$?

The problem with such a picture is that the arrows are too long; generally, on the domain $t>0$, $\alpha(t)$ is telling us about things that are happening away from the point $p$, which is a problem when we're trying to describe what's happening at $p$.

There are a number of ways one might try to solve this problem. Take a moment to think about some of them yourself!

A number of ways of thinking about the problem lead to the solution of compressing the notion of the "extent" of an arrow to one that is entirely determined by the rate of change at $0$; the derivative is telling you "how fast" you're moving through the point. E.g. if one imagines continuing to draw a hypothetical tangent arrow at a fixed rate, the derivative would control how long the arrow would go.

Maybe you can think of $\alpha : [0,1] \to M$ as being the result of trying to do so over one unit of time, except that as soon as the curve gets "away" from the endpoint $\alpha(0)$ it's no longer governed by the tangent vector and so you can't take the rest of its trace seriously.

(Tangent vectors in $$\mathbb R^n$$ are intuitive) The fact that (individual) curves can be used to characterise tangent vectors at a point is relatively intuitive: if you picture a smooth manifold $$M$$ as some sort of curve or surface, taking a smooth curve on it passing through some $$p\in M$$, the direction said curve is going towards at $$p$$ is naturally tangent to $$M$$ at that point.

(We generalise to $$M$$ following the same intuition) Of course, the above naive picture only works as-is for a manifold embedded in $$\mathbb R^n$$. If we are not in some $$\mathbb R^n$$, then it doesn't really make sense to talk about the "direction the curve is facing at a given point $$p$$". Indeed, in such cases, we cannot talk of "directions" that are outside of the manifold itself.

We can fix this by identifying tangent vectors with the way curves behave locally around $$p\in M$$. This matches our intuitive notion of tangent vectors in $$\mathbb R^n$$, and can be seen to also work more generally.

(Precise definition of slope wrt a chart) More precisely, let $$M$$ be a smooth manifold, $$p\in M$$, and let $$(U,\phi)$$ be a local chart with $$p\in U$$. The idea is to leverage the chart, which identifies locally $$M$$ with $$\mathbb R^n$$, to talk of tangent vectors in the usual way. Given a smooth curve $$\gamma:(-1,1)\to M$$ witih $$\gamma(0)=p$$, its "slope" at $$p$$ with respect to the chart $$\phi$$ is then $$(\phi\circ\gamma)'(0) \equiv \lim_{\epsilon\to0} \frac{\phi(\gamma(\epsilon))-\phi(p)}{\epsilon}\in \mathbb R^n.$$ But wait, there can be several different curves passing through $$p$$ which have the same slope with respect to the same chart! To fix this, we consider equivalence classes of curves identified by their slope. More precisely, we define a set $$\mathrm{C}(p)\equiv \{ A\subset\mathrm{SmoothCurves}(p): \forall\gamma,\eta\in A,\,\, (\phi\circ\gamma)'(0)=(\phi\circ\eta)'(0) \},$$ where I defined $$\mathrm{SmoothCurves}(p)$$ as the set of smooth curves $$(-\epsilon,\epsilon)\to M$$ for some $$\epsilon>0$$ such that $$\gamma(0)=p$$, and $$\phi$$ denotes a(ny) coordinate chart defined around $$p$$. One can show that this is well-defined, and that we can use any chart in the definition, as also discussed here.

The elements of $$\mathrm C(p)$$ are often denoted with $$[\gamma'(0)]$$, which represents the set of curves with slope $$\gamma'(0)$$ when measures using some chart $$\phi$$.

One can then observe that $$\mathrm C(p)$$ has a vector space structure etc, and show that it is isomorphic to the space of derivations, which is an alternative equivalent way to discuss tangent vectors to a manifold.

It's not clear what your confusion is (I can think of at least three different ways to interpret your confusion), so I'll instead offer an additional interesting fact that may help demonstrate how everything comes together, to let you you sort out for yourself what's going on.

In your definition, $J$ is itself a smooth manifold, together with extra bells and whistles that encode features of our notion of the "real line".

In particular, $J$ has a tangent space at $c$, and this vector space comes with a notion of "orientation" as well as a notion of "length". Conventionally, we pick out a specific vector $\vec{v}$ as "special": the unique vector that is a unit vector pointing in the positive direction.

(and, in fact, in the usual way to identify the tangent space with $\mathbb{R}$, $\vec{v}$ is simply the vector $1$)

The smooth map $\alpha : J \to M$ induces a linear transformation $\alpha_* :T_cJ \to T_pM$ between the tangent spaces. The tangent vector corresponding to the curve $(\alpha, c)$ is precisely the vector $\alpha_*(\vec{v})$.

If $\vec{w}$ is the special vector for the curve $(\beta, d)$, then two curves are equivalent if and only if $\alpha_*(\vec{v}) = \beta_*(\vec{w})$.

The given definition is using the coordinate chart to understand this equation. $\alpha_*(\vec{v}) = \beta_*(\vec{w})$ is true if and only if $\phi_*(\alpha_*(\vec{v})) = \phi_*(\beta_*(\vec{w}))$, since at each point in $U$, $\phi_*$ is an isomorphism of tangent spaces.

But since $\phi \circ \alpha$ is a map between usual Euclidean spaces, the pushforwards reduce to ordinary multivariable calculus: $\phi_* \circ \alpha_*$ is the operation of multiplication by the usual derivative $(\phi \circ \alpha)'$ .

• Incidentally, there's an analogous picture of how differential forms work, but using functions $M \to \mathbb{R}$ rather than $J \to M$. And these two descriptions play well together; evaluating a differential form at a tangent vector produces the same result as (the derivative of) the real-valued function you get by composing function $J \to M$ with $M \to \mathbb{R}$.
– user14972
Dec 15 '17 at 3:09