0
$\begingroup$

Let $X \subseteq \mathbb R^n $ be a smooth manifold i.e. $\exists k\in \mathbb N$ such that for every $x\in X , \exists$ an open set $W_x$ in $X$ containing $x$ such that $W$ is diffeomorphic with $\mathbb R^k$ . Now let $X \subseteq \mathbb R^K,Y\subseteq R^L$ be smooth manifolds , $dim X=k , dim Y=l$ and let $f: X \to Y$ be a smooth map i.e. for every $x \in X$ there is an open set $\hat W_x $ in $\mathbb R^K$ containing $x$ and a smooth map $F : \hat W_x \to \mathbb R^L$ such that $F|_{\hat W_x \cap X }=f$ . Now for $x \in X$ , if $\phi : U \to X$ is a local parametrization of $X$ at $x$ , where $(0 \in ) U $ is an open subset of $\mathbb R^k$ and $\phi (0)=x$ and $\psi : V \to Y$ is a local parametrization of $Y$ at $f(x)$ where $(0 \in) V$ is an open subset in $\mathbb R^l$ and $\psi(0)=f(x)$ , then the derivative of $f$ at $x \in X$ is defined as $df_x:=d\psi_0 \circ dh_0 \circ (d\phi_0)^{-1}$ ( noting that $d \phi_0 $ is an isomorphism ) , where $h:=\psi^{-1} \circ f \circ \phi : U \to V$ . It can be checked that this definition of $df_x$ is independent of the choice of parametrization . My question is : Is it true that $df_x = dF_x$ , where $F$ is the smooth extension of $f$ locally at $x \in X$ ?

$\endgroup$
0
1
+50
$\begingroup$

As differentials they are different. $df_x$ is a map from the $k$-dimensional tangent-space $T_xX$ to the $\ell$-dimensional tangent-space $T_yY$ (with $y=f(x)$) whereas $dF_x$ is from the $K$-dimensional space $T_x {\Bbb R}^K \sim {\Bbb R}^K$ to the $L$-dimensional $T_y {\Bbb R}^L \sim {\Bbb R}^L$.

In local coordinates (e.g. between your $U$ and $V$) the map $df_x$ is described by an $\ell\times k$ matrix, while $dF_x$ is given by an $L\times K$ matrix. The second contains a lot of useless information when considering the map between tangent-spaces (e.g. it prescribes what happens to normal vectors to the tangent-space, which you may not be interested in).

They agree on $T_x X$ in a certain sense which I think becomes more clear if you introduce the embeddings: $j_X : X \hookrightarrow {\Bbb R}^K$ and $j_Y : Y \hookrightarrow {\Bbb R}^L$. Then we have: $$ j_Y \circ f = F \circ j_X $$ and for the differentials $$ dj_Y \circ df = dF \circ dj_X $$ where $dj_Y$ in local coordinates is an $L\times \ell$ matrix and $dj_X$ a $K\times k$ matrix.

$\endgroup$

Your Answer

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