A description/proof of equivalence between derivation at point of \(C^1\) functions and directional derivative
Topics
About: \(C^\infty\) manifold
The table of contents of this article
Starting Context
- The reader knows a definition of \(C^\infty\) manifold.
- The reader knows a definition of derivation at point of \(C^k\) functions.
- The reader knows a definition of directional derivative.
Target Context
- The reader will have a description and a proof of the equivalence between any derivation at point of \(C^1\) functions and the corresponding directional derivative.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Description
For any \(C^\infty\) manifold, M, any point, \(p \in M\), the set, \(\mathfrak{D}^1_p (M)\), of derivations at p of \(C^1\) functions, and the set, \(T_pM\), of directional derivatives, the map, \(\phi: T_pM \rightarrow \mathfrak{D}^1_p (M)\), \(\frac{d f ({v^i t})}{d t} \mapsto v^i \frac{\partial f}{\partial x_i}\) on any chart where \({v^i}\) is any combination of real numbers, is an isomorphism, which does not really depend on the choice of the chart, with the same result on any \(C^1\) function.
2: Proof
Any directional derivative can be represented by \(c (t) = {v^1 t, v^2 t, . . ., v^n t}\) with the unique \({v^1, v^2, . . ., v^n}\), because only the tangent of the curve at p matters. \(v^i \frac{\partial f}{\partial x_i}\) is a derivation at p, because \(v^i \frac{\partial f g}{\partial x_i} = v^i \frac{\partial f}{\partial x^i} g + f v^i \frac{\partial g}{\partial x^i}\), satisfying the Leibniz rule. The map is an injection, because for 2 distinct \({v_1^1, v_1^2, . . ., v_1^n}\) and \({v_2^1, v_2^2, . . ., v_2^n}\) with f as \(x^i\), the results are \(v_1^i\) and \(v_2^i\), which differ for an i. The map is a surjection, because by Tailor's theorem with reminder, any \(C^1\) function, f, satisfies \(f (x) = f (p) + (x^i - p^i) f_{r_i} (x)\) where \(f_{r_i} (p) = \frac{\partial f}{\partial x^i} (p)\). Applying any derivation at p on the both sides, \(D_v (f (x))|_p = (D_v ( (x^i - p^i)))|_p f_{r_i} (p) + (p^i - p^i) (D_v (f_{r_i} (x))|_p = (D_v (x^i - p^i))|_p f_{r_i} (p) + 0 = (D_v (x^i))|_p \frac{\partial f}{\partial x^i}|_p\). So, any derivation at p can be expressed as with \((v^1, v^2, . . ., v^n) = (D_v (x^1)|_p, D_v (x^2)|_p, . . ., D_v (x^n)|_p)\).
The result of any derivation is the same with that of the corresponding directional derivative, because \(\frac{d f ({v^i t})}{d t} = \frac{\partial f (x)}{\partial x^i} v_i\) by the chain rule.
The isomorphism does not depend on the choice of chart, because directional derivative does not depend on the choice and derivation equals the choice-independent directional derivative.
3: Note
Derivation at point of \(C^\infty\) functions is equivalent with directional derivative with the domain restricted to \(C^\infty\) functions, because as any \(C^\infty\) function is a \(C^1\) function, the map, \(\frac{d f ({v^i t})}{d t} \mapsto v^i \frac{\partial f}{\partial x_i}\), works for any \(C^\infty\) function, the injectivity and the surjectivity are intact (because the materials used in the proof for \(C^1\) functions can be used also for this case: \(x^i\) is \(C^\infty\) as well as \(C^1\) and the Tailor theorem works for \(C^\infty\) functions, which are \(C^1\) functions), and the both sides yields the same result with the domains just restricted. We have to say "directional derivative with the domain restricted to \(C^\infty\) functions", because \(C^\infty_p (M) \rightarrow \mathbb{R}\) cannot be said to be exactly equivalent with \(C^1_p (M) \rightarrow \mathbb{R}\).
