Displacement vector tangent to ambient space or manifold

Question

When addressing the difference between single and multivariable integration, Terence Tao mentions:

It turns out that there will be two dimensions which will be relevant: the dimension $n$ of the ambient space $\mathbf R^n,$ and the dimension $k$ of the path, oriented surface, or oriented manifold $S$ that one will be integrating over.

...

The displacement $\Delta x_i:= x_{i+1} − x_i \in \mathbf R^n$ from $x_i$ to $x_{i+1}$ is now a vector rather than a scalar. (Indeed, one should think of $\Delta x_i$ as an infinitesimal tangent vector to the ambient space $\mathbf R^n$ at the point $x_i.$)

The question is whether the vector is supposed to be tangential to the ambient space, or rather to the manifold $S$ (embedded in an ambient space) at a point $x_i$ .

Answer 1

From your quote:

Indeed, one should think of $Δx_i$ as an infinitesimal tangent vector to the ambient space $\mathbb R^n$ at the point $x_i$.

To see why this comment is relevant to your question, suppose that the ambient space is now the unit sphere $\mathbb S^2$ embedded in $\mathbb R^3$ and that your submanifold $S$ is the circular arc $\gamma$ parametrized by $(\cos t, \sin t, 0)$ for $t \in (0,\pi)$. Partition the arc into two smaller arcs, setting $t_0 = 0$, $t_1 = \pi/2$, and $t_2 = \pi$. Then the vector $\gamma(t_1) - \gamma(t_0)$ is $(-\sqrt 2/2,\sqrt 2 / 2)$, which is tangent to neither $\mathbb S^2$ nor the arc $\gamma(0,\pi)$. However, if we partition the arc even further, allowing the mesh size $|t_{i+1}-t_i|$ to go to zero, we see that, by approximating $\gamma$ to first order, $$\begin{split} \gamma(\tau) &= \gamma(0) + \gamma'(0)\cdot \tau + \mathtt o (\tau) \end{split}$$ as $\tau \to 0$, that is $$\frac{\gamma(\tau) - \gamma(0)}{\tau-0} \xrightarrow{\tau \to 0}\gamma'(0) = (0,1,0) $$ which is a vector pointing in the $y$-direction. If you imagine this vector anchored at the point $\gamma(0) = (1,0,0)$, you see that it is tangent to both $\mathbb S^2$ and the path $\gamma(0,\pi)$.

You may try the same procedure around $\pi/2$ instead of $0$ to see that you could approach $\gamma(\pi/2)$ from both sides (something you cannot do at $0$ or $\pi$ because the path ends there), or try a completely different path in $\mathbb S^2$. The vectors you'll find will always be tangent to both the path and the ambient space $\mathbb S^2$.

Beware, however: if a vector is tangent to the submanifold $S$ of the ambient space, it will surely be tangent to the ambient space too. The converse is not true. Think the vector $(0,0,1)$ anchored at $\gamma(\pi/2)$: it is surely tangent to $\mathbb S^2$, but orthogonal to every vector tangent to $\gamma$ at $t = \pi/2$.