In elementary undergraduate multivariable calculus, when you have for instance a smooth function $f: \mathbb{R}^2 \to \mathbb{R}$, you can plot it as a graph in $\mathbb{R}^3$. At any point $(x,y)$ on the graph, you can take a vector, plot it, and see from inspection that it is a tangent vector to the graph at that point.

You also learn a way for identifying vectors in the tangent space: for instance, take $F(x,y,z) = f(x,y) - z$; its gradient at a point $(p,f(p))$ is normal to our graph $\Gamma(f)$ there, so we take the tangent space to be set of elements orthogonal to $\nabla F(x,y,f(x,y))$.

In Lee's Introduction to Smooth Manifolds, he adopts a notion of the tangent space $T_pM$ as comprising derivations at $p$.

Are these obviously reconcilable for embedded submanifolds of $\mathbb{R}^n$? (Say for instance that we have a level set of a smooth function on $\mathbb{R}^n$ (I know this isn't always an embedded submanifold!), or the graph of a smooth function $\mathbb{R}^{n-1} \to \mathbb{R}$. Is there an obvious way to go from a tangent vector in the multivariable calculus sense to a tangent vector in the derivations sense?)

My trouble essentially boils down to the formalization for tangent vectors seeming terribly abstract and being not at all obviously related to the naive notion.

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    $\begingroup$ You should consider reading Lee's book a bit more closely. Answers to your question are there. See also math.stackexchange.com/questions/1594618/… $\endgroup$ Commented Apr 4 at 0:44
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    $\begingroup$ You should take a look at the definition of tangent vectors as velocity vectors of curves. It's equivalent to the derivations version. All this is stated in Lee. That's how the two notions are related. $\endgroup$
    – random0620
    Commented Apr 4 at 0:45

1 Answer 1


Fix a curve $\gamma$ such that $\gamma(t_0)=p$. This defines a map $\gamma':C^\infty(M)\to \mathbb{R}$ as follows

$$\gamma'(t_0)f := \frac{d}{dt}f\circ \gamma(t)|_{t_0}$$

where the $\frac{d}{dt}$ is the normal derivative from calculus. This is a derivation because of the standard product rule

$$\gamma'(fg)(p) =\frac{d}{dt}fg \circ \left.\gamma(t)\right|_{t_0}= f(\gamma(t_0))\frac{d}{dt}\left.g(\gamma(t))\right|_{t_0} + \frac{d}{dt}g(\gamma(t))|_{t_0}f(\gamma(t_0)) = \\f(\gamma(t_0))\gamma'(g)(p) + g(\gamma(t_0))\gamma'(f)(p)\\=f(p)\gamma'(g)(p) + g(p)\gamma'(f)(p).$$

In this way, you can think of derivations as velocity vectors of curves, or little arrows attached tangent to your manifold.

You can also prove the converse that given a derivation, there is a unique equivalence class of curves with this velocity vector at the point.

Example: Let's compute the velocity vectors to curves in coordinates.

Parameterize $S^1$ using its angle in radians. Now we need to do this in coordinates. Consider the curve

$\gamma(t) = 2\pi t^2, \text{ where }t\in (0,1)$. Then

$$\gamma'\left(\frac{1}{2}\right) = \left.\frac{d\gamma}{dt}\left(\frac{1}{2}\right)\frac{d}{d\theta}\right|_{\gamma\left(\frac{1}{2}\right)}=\left.2\pi\frac{d}{d\theta}\right|_{\frac{\pi}{2}}.$$

$$\gamma'\left(\frac{3}{4}\right) = \left.\frac{d\gamma}{dt}\left(\frac{3}{4}\right)\frac{d}{d\theta}\right|_{\gamma\left(\frac{3}{4}\right)}=\left.3\pi\frac{d}{d\theta}\right|_{\frac{9\pi}{8}}.$$

which gives us a vector in the tangent space of $S^1$. See Lee page 69 for this.

Now consider the curve $\varphi(t) = 2\pi t$ where $t\in (0,1)$.

This curve has constant velocity, and as we would expect,

$$ \varphi'\left(\frac{1}{2}\right) = \left.\frac{d\varphi}{dt}\left(\frac{1}{2}\right)\frac{d}{d\theta}\right|_{\varphi\left(\frac{1}{2}\right)}=\left.2\pi\frac{d}{d\theta}\right|_{\frac{\pi}{2}}.$$

$$ \varphi'\left(\frac{3}{4}\right) = \left.\frac{d\varphi}{dt}\left(\frac{3}{4}\right)\frac{d}{d\theta}\right|_{\varphi\left(\frac{3}{4}\right)}=\left.2\pi\frac{d}{d\theta}\right|_{\frac{3\pi}{4}}.$$


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