Consider smooth map $f:\mathbb{R}^{n}\to \mathbb{R}$, let $a \in \mathbb{R}^{n}$ be any point, $X \in T_{a}\mathbb{R}^{n}=\mathbb{R}^{n}$ be tangent vector at point $a$.

I have probably proven the following:

Claim: $df_{a}(X)=\frac{d}{dt}|_{t=0}f(a+Xt)$, where $d$ is the tangent map or derivation of $f$.

""Proof"": Let $h:\mathbb{R}\to \mathbb{R}$ be some smooth function. Let's evaluate $df_{a}(X)$ at $h$.

$df_{a}(X)h = X|_{a}(hf)$ (this means directional derivative by $X$ at point $a$. also this point follows from definition)

Now by chain rule, we obtain $X|_{a}(hf)=\frac{d}{dt}|_{t=f(a)}h \cdot X|_{a}f$.

I.e. $df_{a}(X) = X|_{a}f \cdot \frac{d}{dt}|_{t=f(a)}$.

Because we identify real numbers $x \in \mathbb{R}$ with one dimensional derivations $x \cdot \frac{d}{dt}|_{t=f(a)}$, we conclude:

$df_{a}(X) = X|_{a}f$

Which by the definition equals to $\frac{d}{dt}|_{t=0}f(a+Xt)$. Q.E.D.

The whole reasoning seem clumsy and far fetched to me. If not wrong. Is my proof correct? What would you suggest to change/improve? Thanks in advance.


You proof is correct. However, there are various equivalent definitions of the tangent space $T_a\mathbb{R}^{n}$. It is isomorphic to $\mathbb{R}^{n}$, but you should make precise which definition you use and how the isomorphism $T_a\mathbb{R}^{n} \to \mathbb{R}^{n}$ is given.

Obviously you use the description via derivations. In this case $X \in \mathbb{R}^{n}$ is identfied with the directional derivation $D_{a,X}$ at $a$ in direction $X$ which you write as $X \mid_a$. In the one-dimensional case you say correctly that $x$ is identified with $x \cdot \frac{d}{dt} \mid_{f(a}$ which is the directional derivation $D_{f(a),x} = x \mid_{f(a)}$. Your claim should then precisely be understood as follows:

Under the above identications of tangent spaces, $df_{a}(X)=\frac{d}{dt}|_{t=0}f(a+Xt) = D_{a,X}(f) = X \mid_a(f)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.