Definition Suppose $C \subseteq \mathbb{R}^2$ is a curve and $p \in C$. The tangent space to $C$ at $p$, $T_pC$ is the set of all vectors tangent to $C$ at $p$.
Example
\[T_pC = \text{span} \{ \langle 1, f^{\prime} \rangle \} = \{ \langle c, c f^{\prime}(a) | c \in \mathbb{R} \}.\]
Let us say we have a curve \(y = f(x)\), with a point \(p = (a, f(a))\). A tangent vector at that point can be given by \(\vec v = \langle 1, f^{\prime}(a) \rangle\) where \(\langle \cdot, \cdot \rangle\) is an inner product. Then we can formulate the tangent space
The following code made the figure in the previous example.
How do we differentiate between points on $C \subseteq \mathbb{R}^2$ and vectors in $T_p C \subseteq \mathbb{R}^2$?
We can create a coordinate system on $C$
\[(x,y): C \mapsto \mathbb{R}^2\] \[(x,y)(p) = (x(p), y(p))\]and a coordinate system on $T_p C$:
\[\langle dx, dy \rangle : T_p C \mapsto \mathbb{R}^r\] \[\langle dx, dy \rangle = \langle dx(v), dy(v) \rangle\]Thus $x: C \mapsto \mathbb{R}$ and $y: C \mapsto \mathbb{R}$ are combined to get a coordinate vector in the ambient space.
Similarly, $dx: T_pC \mapsto \mathbb{R}$ and $dy: T_pC \mapsto \mathbb{R}$ are combined to get a coordinate vector in the tangent space.
Example Let $y = x^2$, then the coordinate vector is $(x,y)(p) = (a, a^2) \in C$. Similarly $\langle dx, dy \rangle (v) = \langle 1, 2a \rangle \in T_pC$ where $v$ is a tangent vector to the function.
The equation $\langle dx, dy \rangle (v) = \langle 1, 2a \rangle$ implies familiar result that $dy = 2a \cdot 1 = 2 a dx \implies \frac{dy}{dx} = 2a = f^{\prime}(a).$
The following code made the figure in the previous example.
Example Note that \(\mathbb{R}^2 = \text{span} \{ (1,0), (0,1) \} = \{ (x,y) | x,y \in \mathbb{R} \}.\)
That means that \(T_p \mathbb{R}^2 = \text{span} \{ \langle 1, 0 \rangle, \langle 0, 1 \rangle \} = \{ \langle dx, dy \rangle | dx, dy \in \mathbb{R} \}.\)
If you want to keep track of the tangent space at different points, you can denote $\langle dx, dy \rangle_p$ for the tangent space at point $p$.
Example Suppose we have $\langle 1, 2 \rangle_{(1,1)}$, we can think of the differentials as another set of axes as visualized below.
The following code made the figure in the previous example.
Definition A 1-form is a linear function $\omega : T_p \mathbb{R}^n \mapsto \mathbb{R}$.
Proposition For a 1-form $\omega : T_p \mathbb{R}^n \mapsto \mathbb{R}$ it holds that $\omega \in (T_p \mathbb{R})^{*}$ where $(T_p \mathbb{R})^{*}$ is the dual space of the tangent space $T_p \mathbb{R}$.
Example Given $\mathbb{R}^2$ and $T_p \mathbb{R}^2$ and $\omega : T_p \mathbb{R}^2 \mapsto \mathbb{R}$ be linear:
\[\implies \omega (\langle dx, dy \rangle) = adx + bdx = \langle a, b \rangle \cdot \langle dx, dy \rangle\]which also equals
\[\vert| \langle a, b \rangle \vert| \operatorname{scalar\_projection}_{\langle a,b \rangle} \langle dx, dy \rangle .\]This leads to an intuition that a 1-form is a multiple of the scalar projection on to the same line.
Example Let $\omega : T_p \mathbb{R}^n \mapsto \mathbb{R}$ then
\[\omega ( \langle dx_1, \ldots, dx_n \rangle ) = \sum_{i=1}^n a_i dx_i\]
Example Define $\omega (\langle dx, dy \rangle) = 3dx + 2dy.$ What line does $\omega$ project vectors onto?
- The line is in the direction $\langle 3, 2 \rangle$ b/c $\omega ( \langle dx, dy \rangle ) = \langle 3,2 \rangle \cdot \langle dx, dy \rangle$
Notice that $\langle 3,2 \rangle$ is parallel to $\langle 1, \frac{2}{3} \rangle$ which entails that $dy = \frac{2}{3} x.$
Example Suppose $\omega$ scalar projects onto the line $dy = 2dx$ with length of 3. Find $\omega .$
We know that $\omega (\langle dx, dy \rangle) = \langle a,b \rangle \cdot \langle dx, dy \rangle .$ We need $\langle a, b \rangle$ to be parallel to $\langle 1, 2 \rangle$ because $\langle 1, 2 \rangle$ is the vector in the direction of the line $dy = 2dx$.
So $\langle a, b \rangle = \langle a, 2a \rangle$ which needs that have \(\vert| \langle a, 2a \rangle \vert| = 3\) .
So \(\vert| \langle a, 2a \rangle \vert| = \sqrt{a^2 + 4 a^2} = 3 \implies a \sqrt{5} = 3 \implies a = \frac{3}{\sqrt{5}}\) . So $b=\frac{6}{\sqrt{5}}$
\[\implies \omega (\langle dx, dy \rangle) = \frac{3}{\sqrt{5}} dx + \frac{6}{\sqrt{5}} dy\]
Example Supposing $n=2$ and $\omega_1 \wedge \omega_2 : T_p \mathbb{R}^2 \times T_p \mathbb{R}^2 \mapsto \mathbb{R}$ then what is $\omega_1 \wedge \omega_2 (v_1, v_2)$ where $v_1, v_2 \in \mathbb{R}^2$ ?