Jump to content

Wedge product/Natural duality/Finite-dimensional/Fact/Proof

From Wikiversity
Proof

We consider the mapping (with factors)

with

For fixed , the mapping on the right is multilinear and alternating, as a direct verification using the determinant rules shows. Therefore, according to fact, we obtain an element in . Hence, we get altogether a mapping

A direct inspection shows that this assignment is also multilinear and alternating. Due to the universal property, there exists a linear mapping

We have to show that this mapping is an isomorphism. To show this, let be a basis of , with the corresponding dual basis . Because of fact, the family

is a basis of . Moreover, the family

is a basis of , with corresponding dual basis . We show that is mapped under to . For , we have

If , then there exists an that is different from all . Therefore, the -th row of the matrix is ; hence, its determinant is . If the index sets coincide, then we obtain the identity matrix with determinant . This effect coincides with the effect of .