Show $T_p M$ is isomorphic to $D_p(M)$.

Question

Here is the problem:

Let $M$ be a smooth manifold and $p\in M$ a given point. Let $D_p(M)$ be the set of derivations of $C^{\infty}(M)$ at $p$ meaning that $D_p(M)$ is the set of linear maps $\zeta:C^{\infty}(M)\rightarrow \mathbb{R}$ satisfying $\zeta(fg)=f(p)\zeta(g)+g(p)\zeta(f)$.

Prove that $D_p(M)$ is a vector space, and that there is an isomorphism $A_p:T_p M\rightarrow D_p(M)$.

I had no trouble proving that $D_p(M)$ is a vector space and I feel that I shouldn't have too much trouble showing the isomorphism if I could get my hands on what exactly things in $T_p(M)$ look like.

Thanks for any help!

Answer 1

• The map is,

$$ v \mapsto D_v:=\sum_j v^i \frac{\partial}{\partial x^j}\Bigr|_p,\ v= \begin{pmatrix} v^1 \\ \vdots \\ v^n \end{pmatrix}$$

• First establish the result for $M= \mathbb{R}^n$ the result is true i.e $v$ is really that $n$-tuple above. Here $D_p(M)$ are just real-valued linear functions which obey the Leibniz Rule. Show first that this map is injective.

• Use Taylor's theorem which says that we can write any germ $f$ locally as,

$$ f(x) = f(p) + \sum_j \frac{\partial f}{\partial x^j}(p)(x^j-p^j)$$

• Let now $D$ be any derivation. You'll be done by applying $D$ to both sides (this will show $D = D_v$ i.e our map is surjective) if you first prove the following lemma: If $D$ is any point derivation at $p$, then $D(c) = 0$ for any constant.

• Now when $f: U \subset M \to \mathbb{R}$ and $M$ is not necessarily embedded in Euclidean space, what do you do? Maybe use a chart?

$\textbf{Comment}$: The above definition for $D_v$ was very natural. It came from looking at the direction derivative and just removing the function. In this case you now see $D_v$ a linear map which acts on smooth functions. Even more, it obeys the Leibniz i.e it is a derivation for all $v$. Since this is the only element of $\mathcal{D}_p\mathbb{R}^n$ that we know about, it makes sense to see if all elements have this form and they do (by the above)!

If your question is now about how do elements of $T_pM$ look, then first you have to make sense of a smooth map $f: U \subset M \to \mathbb{R}$. Take a chart $(U,\phi)$ then identify $f$ with the map,

$$\tilde{f} = f \circ \phi^{-1}: \phi(U) \subset \mathbb{R}^{\textbf{dim}(M)} \to \mathbb{R}$$

and now smoothness makes sense again. Now you have to make sense of partial derivatives. Well, it makes sense to define,

$$ \frac{\partial}{\partial x^j}\Bigr|_p f = \frac{\partial }{\partial r^j}\Bigr|_{\phi(p)} (f \circ \phi^{-1})$$

where $(r^1,...,r^n)$ are the coordinate variables on $\mathbb{R}^n$. Ah, but now $\partial/\partial x^j$ are linear maps and derivations (check this). These end up becoming a basis for $T_pM$. Why? You just refer to tangent vectors as derivations since by the above there is a natural isomorphism with,

$$ e^j \mapsto D_{e^j} = \frac{\partial}{\partial x^j}\Bigr|_p$$

where $e^j = (0,...,1=x^j,0,...,0)$.

Answer 2

You have to know first what exactly the elements of $T_pM$ are. If you define $T_pM$ as a set of equivalence class of curves which pass $p\in M$, i.e $[c]$, $c : I \rightarrow M$ s.t $c(0)=p$, we can make the isomorphism $$ T_pM \rightarrow D_pM $$ by define the map as $$ [c] \mapsto v_p \in D_pM \quad \text{such that} \quad v_p f := \frac{d}{dt}\Bigg|_{t=0} f \circ c $$ We can verify that it is a derivation and the map is isomorphism.