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Thinking of the tangent space to a manifold as derivations is a concept which just kind of eludes me. I am comfortable thinking about tangent vectors as equivalence classes of curves and with the physicists' tangent space. Something about the derivation itself just does not lend itself in my head to the idea of a tangent vector, and as a result, I can't really work with this definition. In particular the following exercise is giving me a hard time.

For any $(U,x)$, consider the tangent lift $Tx: TU \rightarrow TV$ where $V=x(U)$. Since $V \subset \mathbb R^n$ we will identify $T_{x(p)}V$ with $\{ x(p) \} \times \mathbb R^n$. Using this identification show that $T_px \cdot v_p = (x(p), \frac{d}{dt}|_{t=0}(x \circ \gamma))$, interpreting both sides as derivations. (presumably $\gamma$ is a parameterized curve)

The point of this exercise is going to be to use this result to produce natural charts on the tangent bundle. This part I'm okay with, since it's mostly just manipulating the equations. I just feel stuck when trying to think about the tangent space as derivations - like in this question I don't even see what the Leibniz law has to do with this problem.

I'd appreciate it if someone could break down this idea for me in some amount of detail - I've been trying to read other explanations and I know what the definition of a derivation is, I just don't at all see what it has to do with tangent vectors, or how to use that idea to solve problems.

Thanks for the help.

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The reason this stuff is hard is because there are a lot of identifications going on behind the scenes when we write down equations like they one you're meant to prove. It can be tricky to keep track of them all until you gain some intuition. In the meantime, make everything explicit.

Let's start with $v_p \in T_p U$. It's a derivation of $C^\infty(U)$ at $p$, i.e. it sends each function $f \in C^\infty(U)$ to some number $v_p(f) \in \mathbb R$, and for any $f, g \in C^\infty(U)$, we have that $v_p(fg) = v_p(f)g(p) + f(p)v_p(g)$.

What about $T_px(v_p)$? It's an element of $T_x(p)V$, so it is a derivation of $C^\infty(V)$. Explicitly, we have for any $f\in C^\infty(V)$ that $T_px(v_p)\cdot f = v_p(f \circ x)$. This makes sense since $f\circ x \in C^\infty(U)$. It's easy to check that, under this definition, $T_px(v_p)$ really is a derivation of $C^\infty(V)$ and $x(p)$.

We've figured out what the left-hand side means as a derivation, so let's do the same thing for the right-hand side. First, it should be said explicitly that $\gamma: (-\varepsilon, \varepsilon) \to U$ is a smooth curve with $\gamma(0) = p$ and $\gamma'(0) = v_p$. To figure out what $\gamma'(0) = v_p$ actually means, we have to say how $\gamma'(0)$ is a derivation. By definition, for any $f \in C^{\infty}(U)$, $$\left.\gamma'(0) \cdot f = \frac{d}{dt}\right|_{t=0} f(\gamma(t))$$ So what we're saying is that, understanding $\gamma'(0)$ as a derivation, it is equal to the derivation $v_p$. It is a nontrivial fact (which you should have already seen a proof of) that every derivation $v_p$ is equal to $\gamma'(0)$ for some smooth curve $\gamma$.

With this in mind, let's try to understand $\frac{d}{dt}|_{t=0}(x \circ \gamma)(t)$ as a derivation of $C^\infty(V)$ at $x(p)$. For any $f \in C^\infty(V)$, we have that $$\left. \frac{d}{dt}\right|_{t=0} (x \circ \gamma)(t) \cdot f = \left. \frac{d}{dt}\right|_{t=0}(f \circ x \circ \gamma)(t) = \gamma'(0) \cdot (f\circ x) = v_p(f \circ x) = T_p x (v_p) \cdot f$$ This proves the claim.

Throughout the argument above, we thought of tangent vectors as derivations. We figured out what they were by asking how they acted on smooth functions. This is the same process you should always go through if you don't understand something having to do with tangent vectors.

By the way, there's very good reason to think of tangent vectors as derivations instead of as equivalence classes of curves. If we think of them as derivations, the vector space structure of $T_pM$ becomes obvious. Whenever you're adding tangent vectors or multiplying them by scalars, you should think of them as derivations. On the other hand, if you need some geometric intuition about them, thinking of them as coming from smooth curves is extremely useful. Try to get used to both ways of thinking, since they're both useful in their own right.

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  • $\begingroup$ This is exactly what I needed. I'm still not 100% confident, but I now feel like I follow the process. I imagine just more practice and exposure will seal the deal. Thanks so much!. $\endgroup$ – Alfred Yerger Jun 27 '15 at 20:59

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