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Let $v_a:=v|_a:=(a,v)\in\mathbb{R}^n_a:=\{a\}\times\mathbb{R}^n$ be a geometric tangent vector and $D_v|_a\in\mathcal{D}(\mathbb{R}^n)$ be a derivation, to be precise, a directional derivative. For $f\in\mathcal{C}^\infty(\mathbb{R}^n)$ the latter is defined as

$$D_v|_af:=D_vf(a):=\frac{d}{dt}|_{t=0}f(a+tv).$$

I want to show that the map $v_a\mapsto D_v|_a$ is linear, i.e. for $v,w\in\mathbb{R}^n$ and $\lambda\in\mathbb{R}$: $$D_{v+\lambda w}|_a=D_v|_a+\lambda D_w|_a.$$

How does one do this?

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  • $\begingroup$ It seems that you already had the answer... $\endgroup$ – John B May 25 '18 at 12:26
  • $\begingroup$ It came to me as I wrote the question. $\endgroup$ – Thomas Wening May 25 '18 at 12:27
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Let $v_a=v^ie_i|_a$ and $w_a=w^je_j|_a$ for some basis $(e_k|_a)_{k\leq n}$ of $\mathbb{R}^n_a$. Then, using the chain rule, we have

$$D_v|af=\frac{d}{dt}|_{t=0}f(a+tv)=v^i\frac{\partial f}{\partial x^i}(a).$$

Now we can evaluate $D_{v+\lambda w}|_af$ for some arbitrary $f$ using that equation. This leaves us with

$$D_{v+\lambda w}|_af=(v^i+\lambda w^i)\frac{\partial f}{\partial x^i}(a)=v^i\frac{\partial f}{\partial x^i}(a)+\lambda w^i\frac{\partial f}{\partial x^i}(a)=D_v|_af+\lambda D_w|_af.\qquad\square$$

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