# Visualizing tangent space from its definition

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.

• 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. – 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 $$\mathbb{R}^n$$.