278: Equivalence Between Derivation at Point of C^1 Functions and Directional Derivative

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A description/proof of equivalence between derivation at point of $C^1$ functions and directional derivative

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Topics

About: $C^{\mathrm{\infty }}$ manifold

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof
• 3: Note

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Starting Context

• The reader knows a definition of $C^{\mathrm{\infty }}$ manifold.
• The reader knows a definition of derivation at point of $C^k$ functions.
• The reader knows a definition of directional derivative.

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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.

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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.

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Main Body

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1: Description

For any $C^{\mathrm{\infty }}$ manifold, M, any point, $p\in M$, the set, ${\mathfrak{D}}_p^1(M)$, of derivations at p of $C^1$ functions, and the set, $T_{p}M$, of directional derivatives, the map, $\varphi :T_{p}M\to {\mathfrak{D}}_p^1(M)$, $\frac{df(v^{i}t)}{dt}\mapsto v^i\frac{\partial f}{\partial x_i}$ 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.

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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 $v^i\frac{\partial fg}{\partial x_i}=v^i\frac{\partial f}{\partial x^i}g+f{v}^i\frac{\partial g}{\partial x^i}$, 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 $\frac{df(v^{i}t)}{dt}=\frac{\partial f(x)}{\partial x^i}{v}_i$ 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.

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3: Note

Derivation at point of $C^{\mathrm{\infty }}$ functions is equivalent with directional derivative with the domain restricted to $C^{\mathrm{\infty }}$ functions, because as any $C^{\mathrm{\infty }}$ function is a $C^1$ function, the map, $\frac{df(v^{i}t)}{dt}\mapsto v^i\frac{\partial f}{\partial x_i}$, works for any $C^{\mathrm{\infty }}$ 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 $C^{\mathrm{\infty }}$ as well as $C^1$ and the Tailor theorem works for $C^{\mathrm{\infty }}$ 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 $C^{\mathrm{\infty }}$ functions", because $C_p^{\mathrm{\infty }}(M)\to \mathbb{R}$ cannot be said to be exactly equivalent with $C_p^1(M)\to \mathbb{R}$.

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References

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