# antisymmetric mapping

Let $U$ and $V$ be a vector spaces  over a field $K$. A bilinear mapping $B:U\times U\rightarrow V$ is said to be antisymmetric if

 $B(u,u)=0$ (1)

for all $u\in U$.

A multlinear mapping $M:U^{k}\rightarrow V$ is said to be totally antisymmetric, or simply antisymmetric, if for every $u_{1},\ldots,u_{k}\in U$ such that

 $u_{i+1}=u_{i}$

for some $i=1,\ldots,k-1$ we have

 $M(u_{1},\ldots,u_{k})=0.$
###### Proposition 1

Let $M:U^{k}\rightarrow V$ be a totally antisymmetric, multlinear mapping, and let $\pi$ be a permutation  of $\{1,\ldots,k\}$. Then, for every $u_{1},\ldots,u_{k}\in U$ we have

 $M(u_{\pi_{1}},\ldots,u_{\pi_{k}})=\mathrm{sgn}(\pi)M(u_{1},\ldots,u_{k}),$

where $\mathrm{sgn}(\pi)=\pm 1$ according to the parity of $\pi$.

Proof. Let $u_{1},\ldots,u_{k}\in U$ be given. multlinearity and anti-symmetry imply that

 $\displaystyle 0$ $\displaystyle=M(u_{1}+u_{2},u_{1}+u_{2},u_{3},\ldots,u_{k})$ $\displaystyle=M(u_{1},u_{2},u_{3},\ldots,u_{k})+M(u_{2},u_{1},u_{3},\ldots,u_{% k})$

Hence, the proposition   is valid for $\pi=(12)$ (see cycle notation). Similarly, one can show that the proposition holds for all transpositions  $\pi=(i,i+1),\quad i=1,\ldots,k-1.$

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