antisymmetric mapping
Let U and V be a vector spaces over a field K. A bilinear mapping
B:U×U→V
is said to be antisymmetric if
B(u,u)=0 | (1) |
for all u∈U.
If B is antisymmetric, then the polarization of the anti-symmetry
relation gives the condition:
B(u,v)+B(v,u)=0 | (2) |
for all u,v∈U. If the characteristic of K is not 2, then
the two conditions are equivalent.
A multlinear mapping M:Uk→V is said to be totally antisymmetric, or simply antisymmetric, if for every u1,…,uk∈U such that
ui+1=ui |
for some i=1,…,k-1 we have
M(u1,…,uk)=0. |
Proposition 1
Let M:Uk→V be a totally antisymmetric, multlinear
mapping, and let π be a permutation of {1,…,k}. Then,
for every u1,…,uk∈U we have
M(uπ1,…,uπk)=sgn(π)M(u1,…,uk), |
where sgn(π)=±1 according to the parity of π.
Proof. Let u1,…,uk∈U be given. multlinearity and anti-symmetry imply that
0 | =M(u1+u2,u1+u2,u3,…,uk) | ||
=M(u1,u2,u3,…,uk)+M(u2,u1,u3,…,uk) |
Hence, the proposition is valid for π=(12) (see cycle notation).
Similarly, one can show that the proposition holds for all
transpositions
π=(i,i+1),i=1,…,k-1. |
However, such transpositions generate the group of permutations, and hence the proposition holds in full generality.
Note.
The determinant is an excellent example of a totally
antisymmetric, multlinear mapping.
Title | antisymmetric mapping |
Canonical name | AntisymmetricMapping |
Date of creation | 2013-03-22 12:34:39 |
Last modified on | 2013-03-22 12:34:39 |
Owner | rmilson (146) |
Last modified by | rmilson (146) |
Numerical id | 10 |
Author | rmilson (146) |
Entry type | Definition |
Classification | msc 15A69 |
Classification | msc 15A63 |
Synonym | skew-symmetric |
Synonym | anti-symmetric |
Synonym | antisymmetric |
Synonym | skew-symmetric mapping |
Related topic | SkewSymmetricMatrix |
Related topic | SymmetricBilinearForm |
Related topic | ExteriorAlgebra |