Graphics Reference
In-Depth Information
4.9.1. Proposition.
(1) L
0
(
V
) =
R
and L
1
(
V
) =
V
*.
(2) If a
1
, a
2
,..., and a
n
are a basis for
V
*, then the element a
1
Ÿa
2
Ÿ ...Ÿa
n
is a
basis for L
n
(
V
). In particular,
L
n
()
ª
VR
.
(3) The determinant map
n
det:
RR
Æ
is a basis for L
n
(
R
n
). More generally, the maps
v
v
Ê
ˆ
1
Á
Á
Á
˜
˜
˜
2
(
)
Æ
n
vv
,
,...
v
determinant of some fixed k
¥
k minor of
,
vR
Œ
,
12
k
i
M
Ë
¯
v
are a basis for L
k
(
R
n
).
(4) L
k
(
V
) =
0
, for k > n.
(5) aŸa = 0 for all aŒL
k
(
V
). If a, bŒL
1
(
V
), then aŸb = -bŸa. In particular, the
product Ÿ is not commutative.
Proof.
The listed facts are easy consequences of the definitions and Properties
(1)-(3) of Ÿ. We shall only prove fact (3). Specializing fact (2) to
R
n
implies that if we
pick any basis a
1
, a
2
,..., anda
n
for
R
n
*, then
det
=ŸŸŸ
c
aa
. . .
a
,
1
2
n
for some c Œ
R
. Now if we choose a
1
, a
2
,..., anda
n
to be the dual basis for a basis
v
1
,
v
2
,..., and
v
n
for
R
n
, then Property (3) of Ÿ implies that
(
)(
)
=
aa
ŸŸŸ
...
a
vv
,
,...,
v
1
.
1
2
n
1
2
n
Choosing
v
i
=
e
i
, shows that
(
)
=
(
)(
)
=◊
det
ee
,
,...,
e
c
aa
ŸŸŸ
...
a
ee
,
,...,
e
c
1
.
12
n
1
2
n
12
n
In other words, c = 1 and the first part of fact (3) is proved. The second part is proved
in a similar fashion.
Proposition 4.9.1(4) explains why the summation in (4.25a) stops at n, since all
the other spaces are
0
.
Returning to
R
n
, it is convenient to think of L(
R
n
) as the exterior form algebra
associated to the origin of
R
n
, because the coefficients of the elements are just reals.
We would like to extend the exterior algebra notion by allowing functions f :
R
n
Æ
R