My Math Forum  

Go Back   My Math Forum > College Math Forum > Topology

Topology Topology Math Forum


Reply
 
LinkBack Thread Tools Display Modes
April 22nd, 2016, 06:37 AM   #1
Newbie
 
Joined: Oct 2015
From: algeria

Posts: 5
Thanks: 0

Hello friends,
I have a question. Given a submanifold Mⁿ of codimension p in a Riemannian manifold N^{n+p}, let (e₁,...,e_{p}) be an orthonormal basis of the normal vector space of Mⁿ in M^{n+p}.
For every e_{i}, we can define the shape operator A_{e_{i}} of the second fundamental form corresponding to the normal vector e_{i}.
We know that each shape operator A_{e_{i}} is diagonalized.
My question is: Is there a basis that diagonalized all the shape operators A_{e_{i}} simultaneously (at the same time)? And if the response is yes, is there any method, theorem or idea to find this basis? If it possible.
Thank you.

Last edited by skipjack; April 22nd, 2016 at 08:05 PM.
zamzar1992 is offline  
 
April 22nd, 2016, 01:14 PM   #2
Global Moderator
 
Joined: May 2007

Posts: 6,704
Thanks: 669

Wrong forum - this is not a high school algebra question!

Last edited by skipjack; April 22nd, 2016 at 08:05 PM.
mathman is offline  
April 22nd, 2016, 06:29 PM   #3
Global Moderator
 
greg1313's Avatar
 
Joined: Oct 2008
From: London, Ontario, Canada - The Forest City

Posts: 7,923
Thanks: 1122

Math Focus: Elementary mathematics and beyond
Moved to Topology.
greg1313 is online now  
Reply

  My Math Forum > College Math Forum > Topology

Tags
diagonalisation, operators, shape



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Operators on inner product spaces Luiz Linear Algebra 2 October 7th, 2015 05:08 PM
Rational operators jamesuminator Number Theory 10 December 12th, 2010 06:41 PM
Example of operators karkusha Real Analysis 1 November 4th, 2010 12:23 AM
New Operators brangelito Number Theory 7 April 26th, 2010 02:08 PM
Compact Operators Nusc Real Analysis 1 April 11th, 2009 06:14 PM





Copyright © 2019 My Math Forum. All rights reserved.