Given a vector x in the n=6-dimensional Euclidian space Rn, do there exist n−1 continuous functions f1 to fn−1 such that the matrix (x,f1(x),…,fn−1(x)) is invertible for all x in Rn. I am aware of the results of T. Wazewski, Sur les matrices dont les elements sont des fonctions continues. Composito Mathematica, tome 2 (1935), p. 63-68 B Eckmann, Mathematical survey lectures 1943-2004. Springer 2006. There is no topological obstruction for n=6 whereas there is for n odd. I know the solution for n=2 and 4. It is based on permutations with appropriate signs in order to make fi(x) orthogonal to x. This does not work for n=6. Hence the question for n=6 and even larger n=8, …

pilinyir1

pilinyir1

Answered question

2022-09-25

Given a vector x in the n=6-dimensional Euclidian space R n , do there exist n−1 continuous functions f 1 to f n 1 such that the matrix
( x , f 1 ( x ) , , f n 1 ( x ) )
is invertible for all x in R n .
I am aware of the results of
T. Wazewski, Sur les matrices dont les elements sont des fonctions continues. Composito Mathematica, tome 2 (1935), p. 63-68
B Eckmann, Mathematical survey lectures 1943-2004. Springer 2006.
There may be no topological obstruction for n=6 while there may be for n unusual. I know the answer for n=2 and four. it is based totally on diversifications with appropriate signs and symptoms in an effort to make f i ( x ) orthogonal to x. this doesn't paintings for n=6. as a result the question for n=6 or even large n=8, …

Answer & Explanation

ruinsraidy4

ruinsraidy4

Beginner2022-09-26Added 17 answers

restricting x to the unit sphere and projecting all the vectors f i ( x ) onto the orthogonal supplement of x, that is equal to asking whether or not the sector S n 1 is parallelizable (because the f i ( x ) might be a basis for the tangent space at x for each x S n 1 ). via a famous deep theorem (see, as an instance, phase 2.three of those notes with the aid of Allen Hatcher), the simplest parallelizable spheres are S 0 , S 1 , S 3 , and S 7 . especially, S 5 is not parallelizable, so this is not possible on your case.

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