Let me quote Guillemin, Pollack here. "We can use derivatives to identify the linear space that best approximates a manifold $X$ at a point $x$. Suppose $X\subset R^n, \phi:U\rightarrow X$ is a local parametrization around $x\in X$, $U$ is open in $R^k$ and $\phi(0)= x$".

So I understand that the best linear approximation to $\phi$ at $0$ is the map $u\rightarrow \phi(0)+d\phi _0(u) = x+d\phi _0(u)$.

1) In the line "The parallel translate $x+T_{x}(X)$ is the closest flat approximation to $X$ through $x$", I don't understand what is $x+T_{x}(X)$ parallel to ($T_{x}(X)$??).
2)$\because T_{x}(X) = Im(d\phi _0)$, I get that $x+T_{x}(X)$ is a linear approximation to $\phi:U\rightarrow X$. Visually, is $x+T_{x}(X)$ like a tangent plane (subspace of $R^n$) to say a surface?. Also, can I visualize $T_{x}(X)$ by itself?

May sound rudimentary but appreciate your explanation to make this information tangible.

  • 1
    $\begingroup$ If you found tangent lines to curves and tangent planes to surfaces in $\Bbb R^3$ in your multivariable calculus class, this is what G&P are doing with $x+T_x(X)$. $T_x(X)$ is a subspace of $\Bbb R^n$, i.e., is a $k$-dimensional plane through the origin. $\endgroup$ – Ted Shifrin Feb 14 at 17:26

The idea is the following: If $U$ is an open subset of $\mathbb{R}^k$, then we know that $T_0U\cong \mathbb{R}^k$ as vector spaces. If we use the map $\phi:U\to X$ as our parametrization, then we can see that $d\phi_0$ acts on $T_0U$ as a linear isomorphism. In order to get the best linear approximation to $\phi$, we should use the first order Taylor expansion $\phi(u)\approx\phi(0)+d\phi_0(u).$The best linear approximation to $X$ as a submanifold of $\mathbb{R}^n$ is given by the image of the tangent plane $T_0U\cong \mathbb{R}^k$. This is exactly the set of points $\phi(0)+d\phi_0(u)=x+d\phi_0(u)$ for all $u\in T_0U$.

So, we apply a linear transformation $d\phi_0:T_0U\to T_xX\subseteq \mathbb{R}^n.$ Then we add $\phi(0)+x$ to shift this linear space $T_xX$ to be tangent to $X$ at $x$. As a very concrete example, take the manifold $S^1\subseteq \mathbb{R}^2$. Near $(0,1)$ it has graph coordinates $(x,\sqrt{1-x^2})$. That is, we can parametrize a neighborhood of $(0,1)\in S^1$ by $(-1,1)\to S^1$ given by $\phi:t\mapsto (t,\sqrt{1-t^2})$. If we visually inspect $S^1$ at $(0,1)$ we expect its best linear approximation line to be given by a horizontal line $y=1$ passing through $(0,1)$, call this $A$.

The recipe given in Guillemin and Pollack says that we can find this plane by calculating the Jacobian of the parametrization, $d\phi_0$, then writing $A=(0,1)+d\phi_0 T_0(-1,1).$ $d\phi_0$ is the $2\times 1$ matrix $$ d\phi_0=\begin{bmatrix} \frac{\partial x}{\partial t}\\ \frac{\partial y}{\partial t} \end{bmatrix}_{t=0}.$$ This is $$ d\phi_0=\begin{bmatrix} 1\\ 0 \end{bmatrix}.$$ The moral is that the image of $T_0(-1,1)=\mathbb{R}$ is the $x-$axis. $$ d\phi_0T_{0}(-1,1)=\{(x,y)\in \mathbb{R}^2:y=0\}.$$ Then if we add $(0,1)$ we get that $A$ is precisely the horizontal line passing through $(0,1)$. Indeed, here $T_x(X)$ is the $x-$axis, and the best linear approximation is the shifted $x-$axis.

In response to the second question, $x+T_x(X)$ is literally a shifted subspace tangent to the manifold $X$ at the point $x$. $T_x(X)$ can be visualized as $x+T_x(X)$ but shifted in an affine manner so it passes through the origin. The advantage to calling $T_x(X)$ the tangent space is that when it passes through the origin it is a bona fide linear subspace of $\mathbb{R}^n$.

  • $\begingroup$ Perfect, thanks for a detailed explanation and that clarified my doubts. $\endgroup$ – manifolded Feb 14 at 17:37

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