- Commit
- 72cae0c32d16abae7e84645b02f2f6290069eb8d
- Parent
- bbf4cbb07b54e0656b791a4e4a7b9ce5bbcd2a40
- Author
- Pablo <pablo-escobar@riseup.net>
- Date
Fixed a typo
Riemannian Geometry course project on the manifold H¹(I, M) of class H¹ curves on a Riemannian manifold M and its applications to the geodesics problem
Fixed a typo
1 file changed, 4 insertions, 3 deletions
Status | File Name | N° Changes | Insertions | Deletions |
Modified | sections/structure.tex | 7 | 4 | 3 |
diff --git a/sections/structure.tex b/sections/structure.tex @@ -204,9 +204,10 @@ find\dots Similarly, if \(t_0 > t_1\) then by inverting the orientation of the curve \(\gamma\) we can see that \(\norm{\xi}_\infty \le \sqrt{2} \norm{\xi}_1\). More precisely, if we set \(\eta(t) = \gamma(1 - t)\) and \(\zeta \in - H^1(\eta^* TM)\) with \(\zeta_t = \xi_{1 - t}\) then \(\norm{\xi}_\infty = - \norm{\zeta}_\infty \le \sqrt{2} \norm{\zeta}_1 = \sqrt{2} \norm{\xi}_1\) - because of the inequality (\ref{eq:one-norm-le-sqrt-two-infty-norm}). + H^1(E)\) with \(\zeta_t = \xi_{1 - t}\) -- a section of \(E\) along \(\eta\) + -- then \(\norm{\xi}_\infty = \norm{\zeta}_\infty \le \sqrt{2} \norm{\zeta}_1 + = \sqrt{2} \norm{\xi}_1\) because of the inequality + (\ref{eq:one-norm-le-sqrt-two-infty-norm}). \end{proof} \begin{note}