definition of gradient of \(C^\infty\) function over Riemannian manifold with boundary
Topics
About: Riemannian manifold
The table of contents of this article
Starting Context
- The reader knows a definition of Riemannian manifold with boundary.
Target Context
- The reader will have a definition of gradient of \(C^\infty\) function over Riemannian 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:
\( (M, g)\): \(\in \{\text{ the Riemannian manifolds with boundary }\}\)
\( f\): \(: M \to \mathbb{R}\), \(\in \{\text{ the } C^\infty \text{ functions }\}\)
\(*grad f\): \(: M \to TM\), \(\in \{\text{ the } C^\infty \text{ vectors fields }\}\)
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Conditions:
\(\forall m \in M (\forall v \in T_mM (g (grad f, v) = v f))\)
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2: Note
So, "gradient" is defined only over any Riemannian manifold with boundary, not over a \(C^\infty\) manifold with boundary.
Let us see that \(grad f\) is indeed well-defined.
Does it really exist?
Let \((U_m \subseteq M, \phi_m)\) be any chart.
Let us see that \(grad f := \widetilde{g}^{j, l} \partial f / \partial x^l \partial / \partial x^j\), where \(\widetilde{g}\) is the \(C^\infty\) \((2, 0)\)-tensors field induced by \(g\), satisfies the condition.
\(g (grad f, v) = g_{m, n} d x^m \otimes d x^n (\widetilde{g}^{j, l} \partial f / \partial x^l \partial / \partial x^j, v^a \partial / \partial x^a) = g_{m, n} d x^m (\widetilde{g}^{j, l} \partial f / \partial x^l \partial / \partial x^j) d x^n (v^a \partial / \partial x^a) = g_{m, n} \widetilde{g}^{j, l} \partial f / \partial x^l v^a d x^m (\partial / \partial x^j) d x^n (\partial / \partial x^a) = g_{m, n} \widetilde{g}^{j, l} \partial f / \partial x^l v^a \delta^m_j \delta^n_a = g_{m, n} \widetilde{g}^{m, l} \partial f / \partial x^l v^n = \delta^l_n \partial f / \partial x^l v^n = \partial f / \partial x^l v^l = v^l \partial / \partial x^l f = v f\).
Let us see that it is consistent with the choices of charts.
Let \((U'_m \subseteq M, \phi'_m)\) be any other chart around \(m\).
Hereafter, we profusely use the proposition that for any \(C^\infty\) manifold with boundary and the \((p, q)\)-tensors space at any point, the transition of the standard bases with respect to any charts is this and the proposition that for any \(C^\infty\) manifold with boundary and the \((p, q)\)-tensors space at any point, the transition of the components of any tensor with respect to the standard bases with respect to any charts is this.
On \(U_m \cap U'_m\), \(\widetilde{g}'^{m, n} = \partial x'^m / \partial x^j \partial x'^n / \partial x^l \widetilde{g}^{j, l}\).
On \(U_m \cap U'_m\), \(\partial / \partial x'^n = \partial x^o / \partial x'^n \partial / \partial x^o\).
So, \(\widetilde{g}'^{m, n} \partial f / \partial x'^n \partial / \partial x'^m = \partial x'^m / \partial x^j \partial x'^n / \partial x^l \widetilde{g}^{j, l} \partial x^o / \partial x'^n \partial f / \partial x^o \partial x^s / \partial x'^m \partial / \partial x^s = \partial x'^m / \partial x^j \partial x^s / \partial x'^m \partial x'^n / \partial x^l \partial x^o / \partial x'^n \widetilde{g}^{j, l} \partial f / \partial x^o \partial / \partial x^s = \delta^s_j \delta^o_l \widetilde{g}^{j, l} \partial f / \partial x^o \partial / \partial x^s\), by the proposition that for any \(C^\infty\) manifold with boundary, on the intersection of any 2 charts, the looks-like-chain-rule-for-partial-derivative-of-composition-of-transitions-of-coordinates rule holds, \(= \widetilde{g}^{j, l} \partial f / \partial x^l \partial / \partial x^j\).
So, \(grad f := \widetilde{g}^{j, l} \partial f / \partial x^l \partial / \partial x^j\) is consistently defined all over \(M\).
\(grad f\) is \(C^\infty\), because its components are \(\{\widetilde{g}^{j, l} \partial f / \partial x^l\}\), which are \(C^\infty\) with respect to \(\{x^l\}\).
So, there is at least 1 \(grad f\).
Let us see that \(grad f\) is uniquely defined.
Let \(V': M \to TM\) be another \(C^\infty\) vectors field that satisfies \(g (V', v) = v f\).
\(g (grad f, v) - g (V', v) = v f - v f = 0\), so, \(g (grad f - V', v) = 0\). Especially, \(v\) can be taken to be \(grad f - V'\) and \(g (grad f - V', grad f - V') = 0\), which implies that \(grad f - V' = 0\), so, \(V' = grad f\).
So, \(grad f\) is well-defined.
