description/proof of visualization of covector of tangent vectors space at point on \(C^\infty\) manifold With boundary
Topics
About: \(C^\infty\) manifold
The table of contents of this article
Starting Context
- The reader knows a definition of tangent vectors space at point on \(C^\infty\) manifold with boundary.
- The reader knows a definition of covectors (dual) space of vectors space.
Target Context
- The reader will have a description and a proof of a visualization of any covector of the tangent vectors space at any point on any \(C^\infty\) manifold With boundary.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(M\): \(\in \{\text{ the } d \text{ -dimensional } C^\infty \text{ manifolds with boundary }\}\)
\(m\): \(\in M\)
\(t\): \(\in {T_mM}^*\) such that \(t \neq 0\)
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Statements:
\(t\) can be visualized as the trace of the tangent vectors that \(t\) operates to be 1, \(t^{-1} (1) \subset T_mM\), which is a hyperplane on \(T_mM\)
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2: Proof
Whole Strategy: Step 1: understand the visualization of tangent vectors: Step 2: see that no arrow visualizes \(t\); Step 3: see that the trace of the tangent vectors that \(t\) operates to be 1 characterizes \(t\); Step 4: see that \(t^{-1} (1)\) is a hyperplane on \(T_mM\).
Step 1:
Usually, a tangent vector at \(m\), \(v\), is visualized as an arrow that starts at \(m\).
What does that mean?
\(v\) is represented by a \(C^\infty\) curve that passes through \(m\), and the arrow is tangent to the curve at \(m\) and the length is the speed of the curve at \(m\).
So, the arrow characterizes \(v\).
Step 2:
Can \(t\) be visualized by an arrow likewise?
The issue is that \(t\) is not represented by any curve, and so, the logic for \(v\) does not apply to \(t\) at all.
So, we do not see any justification to represent \(t\) by an arrow.
Step 3:
Our goal is to characterize \(t\) in a graphical manner.
According to the definition of covector, \(t\) is characterized by how \(t\) operates on the tangent vectors.
As the operation is linear, if \(t (v) = 1\), \(t (r v) = r\), and how \(t\) operates on \(r v\) is naturally imagined.
So, let us think of \(S_t := t^{-1} (1) \subset T_mM\).
\(S_t\) is a set of some tangent vectors at \(m\), which can be visualized as the set of the arrows.
So, the trace of the arrows visually characterize \(t\).
Step 4:
Let us see that \(t^{-1} (1)\) is a hyperplane on \(T_mM\), which means that it is a '\(d - 1\)'-dimensional affine subspace of \(T_mM\), which means that for a \(v_0 \in t^{-1} (1)\), \(V := \{v - v_0 \vert v \in t^{-1} (1)\}\) is a '\(d - 1\)'-dimensional vectors subspace of \(T_mM\): \(t^{-1} (1)\) is not any vectors subspace, because \(0 \notin t^{-1} (1)\).
Let us see that \(V = t^{-1} (0)\).
For each \(v' \in V\), \(v' = v - v_0\), and \(t (v') = t (v - v_0) = t (v) - t (v_0) = 1 - 1 = 0\). So, \(v \in t^{-1} (0)\).
For each \(v' \in t^{-1} (0)\), \(v' = v' + v_0 - v_0\), and \(t (v') = t (v' + v_0 - v_0) = t (v' + v_0) - t (v_0) = t (v' + v_0) - 1 = 0\), so, \(t (v' + v_0) = 1\), so, \(v := v' + v_0 \in t^{-1} (1)\), and so, \(v' = v - v_0 \in V\).
Let us see that \(V\) is a vectors subspace of \(T_mM\).
For each \(v_1, v_2 \in V\) and each \(r_1, r_2 \in \mathbb{R}\), \(t (r_1 v_1 + r_2 v_2) = r_1 t (v_1) + r_2 t (v_2) = r_1 0 + r_2 0 = 0\): \(V = t^{-1} (0)\).
Let us see that \(V\) is '\(d - 1\)'-dimensional.
For any chart, \((U_m \subseteq M, \phi_m)\), for each \(v \in V\), \(t (v) = t_j v^j = 0\).
As \(t \neq 0\), \(t_l \neq 0\) for an \(l\).
So, \(v^l = (- \sum_{j \neq l} t_j v^j) / t_l\) where \(\{v^j \vert j \neq l\}\) can be taken arbitrarily, with \(d - 1\) free variables.
So, \(V\) is '\(d - 1\)'-dimensional.
3: Note
For example, for \(M = S^2\), \(t^{-1} (1)\) does not become any circle, because then, \(V\) would not be any vectors subspace of \(T_mM\).
