I nternat. J. Math. & Math. Sci.
(1978)297-305
297
Vol.
EQUIVALENCE CLASSES OF THE 3RD GRASSMAN SPACE
OVER A 5-DIMENSIONAL VECTOR SPACE
KULDIP SlNGH
Department of Mathematics
The University of New Brunswick
Frederlcton, N.B., Canada E3B 5A3
(Received June i0, 1977)
ABSTRACT.
An equivalence relation is defined on
Arv,
the r
th
Grassman space
over V and the problem of the determnatlon of the equivalence classes defined by this relation is considered.
elements form an equivalence class.
For any r and V, the decomposable
For r
2, the length of the element
determines the equivalence class that it is in.
Elements of the same length
are equivalent, those of unequal lengths are inequivalent.
When r > 3, the
length is no longer a sufficient indicator, except when the length is one.
Besides these general questions, the equivalence classes of
dim V
A3V,
when
5 are determined.
KEY WORDS AND PHRASES. Grsman space, equivalent classes, representation of
AMS (MOS) SUBJECT CLASSIFICATION (1970) CODES.
14M15.
298
K. SINGH
Suppose V is a finite dimensional vector space over an arbitrary field F
and r is a positive integer.
Consider
define an equivalence relation on
write X
C (T)X
r
Arv
Arv,
We
the rth Grassman space over V.
If X and Y are in
as follows:
Arv,
we
a non-slngular linear transformation T:V-- V such that
Y iff
Y, where C r (T)
is the rth exterior product of
that if T and S are two linear transformations of V
and if T is non-singular, then
Cr(T-I)
Cr (T)-I’
T.
Using the facts,
then C
r
(T)Cr(S)
Cr(TS)
it follows that the above
relation is an equivalence relation.
We consider the problem of determining the number of equivalence classes,
Arv
into which the set
is decomposed, along with a system of distinct repre-
sentatives of these equivalence classes.
DEFINITIONS. i.
Arv
If X e
and X
x
I
^...^
Xr,
we say X is
decomposable.
Arv,
If X e
2.
we define its length, to be denoted
by (X) as (X)
posable elements
If X
3.
If X e
is a subspace of V and X E
Arv,
If X ,Y e
(+/-) P(X)
Arv
(i)
Let T:V
/
V be
a
s
Then Y
Y
dlm[X].
Y, then (i)
and X
P(Y).
PROOF.
If (X)
Aru}.
we define the rank of X to be denoted
by o(X) as p(X)
PROPOSITION 1.
is a sum of m decom-
we define a subspace [X] of V as
n{UIU
IX]
4.
Arv,
min{mlX
of Arv}.
s
X
C (T)X
r
i=ElXi,
where X
s
7. C (T)X
i.
i=l r
i
n.s.l.t, such that C
Arv
and (Xi)
This implies
(Y)
X implies (Y) < (X) and this proves (i).
r
(T)X- Y.
i.
< s
(X).
Similarly
299
EQUIVALENCE CLASSES OF THE 3RD GRASSMAN SPACE
implies Y e
ArT(U)
and hence P(X)
ArT-I(w),
implies X
If U and W are subspaces of V, then
PROOF.
Ar(u
Clearly
Xl,
YI’
Xl, x2,...,xk
^ xjll-<i<j-<k}
i
(Aru)
t
Ar(u
respectively.
If X e
(Aru)n (Arw),
then
X
aij xi ^ xj +
7.
A
AZ aijx
bij
i
dij
i and j.
xj +
A
cij
E
B
bij yi ^ YJ
CZ cijz
^ zj
i
+ E
D
+
n W),
Aru, Arw
y
and also
dij xi ^
EZ eijx
i
Arw Ar(u
n W).
j
Hence
^ zj.
,z t of W.
+ W.
l_<j_<t}, then the sets A, A u B u D, A
form bases of
n
,xk, z I
is a basis of U
E u F
X
T[X]
W and extend it to
be a basis of U
C u D
B
V is a n.s.l.t.
To prove the inclusion in
I
z
Aru
{Yi ^ Yj ll-<i<j-<s}’ C {z i ^
E-- {x ^ zjll_<i_<k l_<j_<t},
i
B
l_<j_<s},
i
(Arw).
n
of U and a basis x
Ys’ Zl’
^ yjll_<isk;
{Yi ^ zj ll-<i-<s;
D-- {x
n W) _=
Xk, yl,...,y s
Xl,...,Xk,
If A-- {x
/
P(Y).
PROPOSITION 2.
a basis
A
where T:V
From this remark, it follows easily that [Y]
the other direction, let
Then
Arw
and Y
C (T)X.
r
such that Y
F
Aru
We first remark that if U and W are subspaces of V, then X
(ii)
zjll-<i<j-<t},
C u E, and
and A r(U
aij
aij
+ W)
and
0 for all the appropriate values of the indices
eij
Thus X e Ar(U
W).
REMARK i.
The result of Proposition 2 holds for any number of subspaces
REMARK 2.
If X e
of V.
Ar[x]
Ar(ue Uen
ArV
and-=O
rU)"
{UIU
Thus X e
is a subspace of
Am[x]
V, X e
Aru},-
then
and IX] Isthe smallest such
subspace of V.
PROPOSITION 3.
Xl’’’’’Xk’ YI’’’’’Yk
PROOF.
Let X e
A2V,
(X)
k and X
k
x
i=l i
^ Yi
then
are linearly independent.
If not, then one of them (say)
Yk
is a linear combination of the
K. SINGH
300
remaining
Then x
k
X
i=II
Yk
alxl +
k-i
j--iIbjyj.
k-i
k
^’Yk
k
Let
Xl,...,Xk, yl,...,yk_ I.
I a
i--i
x
Ixk ^
i
+ I
b
j=l
J xk
A
Hence X can be written as
YJ
zi)’ where z i aix i + biy i, 1 _< i <_ k-l.
^ Yl + Xk ^ zl) i. If z i # 0, let a i # 0, then
-i
Xk ^ zi zi ^ (ai Yi- Xk)’ thus (x i ^ Yi + Xk ^ zi)
i=II (xi ^ Yi + Xk ^
then (x
i
xi ^ Yi +
If z
0,
i
< i.
Hence (X) < k-l, a contradiction
iX]
Z(X)
k
I x
k and X
^
i=l i
then
Yi
<Xl,...,Xk, yl,...,yk >.
PROOF.
3, dim U
X
A2V,
If X e
PdMARK 3.
Let U
k
i=17"xi ^ Yi’
dim IX] > 2k.
Thus [X]
[X]
U
k
Z x
Let X
formation of V
A2V,
^ Yi’
PROPOSITION 5.
Y
2k, P(Y)
2s.
x’i’ TYl
Yi’
Txi
Arv,
If X
Yk >"
P(Y), then X
P(X)
and [Y]
<Xl,...,Xk, yl,...,yk >
Proposition 3, P(X)
Again by Proposition 3,
<Xl,...,Xk, YI’
If X,Y e
By Proposition
let
e [X], 1 _< i -< k.
xi’ Yl
PROPOSITION 4.
PROOF
A2[X];
Also X e
2k.
then IX] c_ U.
<Xl,...,Xk, yl,...,yk>;
Z(X)
Y.
s
=ElXj ^ yj; then by Remark 3,
<x,...,x’s’ YI’’’’’Ys’> Also by
Thus k
Let T be a linear trans-
s.
i < i < k; then C (T)X
r
2, X
x
I
^ Xr
^
Y
+
Y
Thus X
Yl ^
^ Yr’
then X
<Xl,...,Xr, YI’’’’’Yr >"
PROOF. Let U
<Xl,...,Xr, yl,...,yr>;
at least one element
(say) xI
ing IX], i.e., U
yj
X
U
1
bjxI + wj, i
x ^ (Ar-Iw)
I
<Xl>
W =>
< j < r, where
and X
Aru
Let W be a complement of
W, [X] c_ W.
<Xl>
e
2
x
I
Let B be a basis of IX] and
is not in IX].
extend {x} u B to a basis of U.
Arw,
wi,w
(Xi)
^ (Ar-Iw) Arw.
and
alx I +
Let x i
W.
If IX] # U, then
then IX] c_ U.
1,2.
Also X e
in U, contain-
w l, 2 < i < r and
Then X--X
i, i
<Xl>
I +
X2,
where
But
Ar[x]"
C
Arw,
hence
EQUIVALENCE CLASSES OF THE 3RD GRASSMAN SPACE
X
X
X
I
2
Hence IX]
e
Arw.
U
<x
Thus X
Let X
PROOF.
If X,Y e
Arv,
Xl^...Axr
+
(Y)
(X)
U
Yl ^" ^Yr
basis of U
+
I n
r of U
s
and extend it to a basis
U2,
to a basis Zl,...,Zk,
I and
+ 2s.
k.
<x
I
Xr>
I + U 2.
Let
I
U
Zl,...,z k
Zk, Ul,...,Us,
Zl,
Vl’’’’’Vs
of
Y.
P(Y), then X
2, P(X)
<yl,...,yr >, then by Proposition 4, IX]
2
i, a contradiction.
,Xr, yl,...,yr >.
I
PROPOSITION 6.
k
(X)
=>
2
The above proposition is true also for (X)
Note:
U
X
0 and X
I
301
U2.
be a
where
Then
,Us are two bases of
Xl’’’’’Xr and Zl,...,Zk, uI
U I, hence
XlA.. AXr aZlA...AZkAUlA...AUs ZlA...AZkAU--IA...Au s, where
u--I auI. Similarly yl A...Ay r bZlA...AZkAVlA...AV S z 1 ^...AZkAVl A...Av s
where
i bVl Hence X ZlA AZk A (-IAU2A...^Us + IAV2A...AVs), where
Us’ V--l’V2’’’’’Vs is a basis of [X].
Zl’
Zk’ l’U2
Similarly Y
zl^...^z k ^ (-l^U2 ^. .^Us + Vl^V2^...^Vs) where
P(X)
k
Since
_
__w
zl’’’’’z’ Ul’U2’’’’’ U s’ Vl’V2’’’’’ Vs
Define T:V
Tz
--!
Y; hence X
REMARK 4.
U
1
+
U2,
Let X e
If X
YlA...^yr
aXlA...AXr,
P(X)
dim U
2r
THEOREM i.
E(2, s), s
PROOF.
I n U 2.
Arv,
2, then r + i
(X)
+ Yl ^
^Yr
2, 3,...,s.
for i
U
2
<
0(X) -< 2r.
then IX]
<yl,...,yr >.
where a is a scalar and
Hence r+l
P(X)
{XIX
Arv,
Let E(2, s)
r+l, r+2
all vectors of
Arv,
<Xl,...,Xr >,
I
vi,
Y.
XI^...AX r
where U
Vl, Tv i
ui, Tv I
Ul, Tu i
Then C (T)X
r
PROOF.
V, a linear transformation
>
zi, Tu I
i
is a basis of [Y].
(X)
<x ,...,X r, y
i
I
U
I #
U2,
Yr >
for otherwise
I.
2r.
(X)
2, P(X)
s}, then
,2r are all the equivalence classes on the set of
of length 2.
Follows from Proposition 6 and Remark 4.
K. SINGH
302
Arv
Let 0 # X e
PROPOSITION 7.
and x e V such that x
0
X
^
then
x IX].
PROOF.
Let x I, x 2,...,xm be a basis of IX].
Ar[x]
basis of
where
Qr ,m
X
^
basis of A
x
la
r+l<x ,IX]>
det X
m.
Thus x A X
> a
0
n
I a
ISeQr m is a
=0ee Qr, =>
m
IX], then {x
If x
^
is a
is a set of all the strictly decreasing sequences
of length r on the integers i, 2
then x
{sl SeQr,- m
Then
A
part of a
X
0, a
contradiction.
If 0
PROPOSITION 8.
^
Arv
e
By Proposition 7, x
PROOF.
x
# X
^
X # 0.
be a basis of Ix^X] and extend it to a basis x,
IX].
<x>
If U
Ar+l[xAX]
v e A
r+l
^ (Aru)
x
U.
Thus x^v
Ar+Iu.
Let xAX
Ar[x].
thus X- u
PROPOSITION 9.
ent vectors in IX].
Hence IX-u] c_ IX].
IX] and Ix^X]
A2V,
Suppose X
Then
YI’ Y2
xI ^
PROOF.
x
2
X e
E
l_<i<j_< 4
Xl,...,xk
of
,Xm
Aru
and
^
(X-u)
Now X e A r IX] and u e
c
U,
0.
Aru
If
_c
Ar[x]’
Thus x e IX-u] => x e IX], which is
u e
Aru.
Hence IX] _c U.
IX].
<x>
(X)
2, Xl, x 2 are linearly independ-
e IX] and % e F
x
I
^ Yl
3
+
X has one and only one
x2 ^ Y2
+
Yl ^ Y2"
A2V, (X)
2 => P(X)
x 2, x 3, x of IX].
4
Then X
Then x
xAu.
0 i.e., X
of the following representations: (i) X
(ii) X
Let x,
U, U c_ IX].
x^u+v, where u e
x e IX-u].
a contradiction and therefore X-u
Also U c_ IX], hence U
IX].
Xk, Xk+ 1
Xl,
<x>
0 and thus xAX
then by Proposition 7
IX].
<x>
If v # 0, then by Proposition 7, x e Iv]
0.
Hence v
a contradiction.
X-u # 0
then Ix^X]
<Xl,...,Xk >,
X]
A
Again by Proposition 7, since
Clearly Ix^X] _c <x>
0, hence x e Ix^X].
(x^X)
IX], then Ix
and x
aijx i ^
x
j’
a
ij
e F.
4.
Extend
Xl,
x
2
to a basis
Xl,
EQUIVALENCE CLASSES OF THE 3RD GRASSMAN SPACE
If
a12x2 + a13x3 + a14x4
x
If
a34 0, then
I ^ Yl + x2 ^ Y2"
+ a12)a34 a13a24 + a14a23 0
a34
X
(-%
0, take
Yl
(-%+a12)Xl^X2
Set Y
Then Y
-%xl^x2
+
+ X.
+
al3xl^x3
al4Xl^X4
and
a23x3 + a24x4’
Y2
Because of (i), (Y)
then
has a solution in F.
(i)
+
303
a23x2^x3
+
a24x2^x4
i; also Y e
+
a34x3^x4.
^
x
A2[X].
YI’ Y2 e IX] Y --Yl ^ Y2" Hence X-- %xl^x2 + yl^Y2
If X
xl^Y I + x2^Y 2 and also X %Xl^X2 + Zl^Z 2 then xI ^ X
and also x
I ^ X x I ^ z I ^ z 2. Thus 0 # Xl^X2^Y2
Xl^Zl^Z2
Thus
<xl,x2,Y2 >
z
alxl + a2x2 + a3Y2
Let z
<Xl,Zl,Z2 >.
x
I
2
^ Y2
and hence
and
blXI + b2x 2 + b3Y 2.
2
(alb2-a2bl)Xl^X2 + (alb3-a3bl)XlAY2 + (a2b3-a3b2)x2^Y2
Putting this expression for
Zl^Z 2 in X %Xl^X 2 + z I ^ z 2, we get two
different representations of X in the basis of A2[X], determined by the
Then
Zl^Z 2
Xl,X2,Yl,y 2
of
basis
IX]; thus X has precisely one of the two representations.
If X,Y e
PROPOSITION i0.
Arv
are decomposable, then X+Y is decomposable
iff dim[X] n [Y] -> r-l.
PROOF.
Let X
(=>) Let X+Y be decomposable, and X+Y
Xl^’’’^Xr’
any i, 1 -< i -< r
Y
Yl ^’’’^yr’
I
by Proposition 7, and [Z]
zi ^
(X+Y)
0 =>
Hence IX]
[Y].
If IX] # [Z], then for some i, z
zi^X -zi ^Y
IX].
i
---->
Let
Ul,...,Ur_ I
+
dim[Y]
(r+l)
be l.i. vectors in [X]
_< i.
[Z], then for
0, and thus z.i e [Y]
[Y], i.e., dim[X] n [Y]
dim
r-l.
<zi,[Y]>. Thus
space <zi,[X]>.
(<--) If dim[X]
[X], [Y]
Hence
[Y] > r-l.
[Y] and extend these to a basis
x, Ul,...,Ur_ 1 and a basis y, Ul,...,Ur_ 1 of IX] and [Y] respectively.
X
ax^ulA...^Ur_l,
Y
by
^ Ul^...^Ur_ I
r.
But
<zi,[X]>
are r-dimensional subspaces in an (r+l)
dim[X] n [Y] a dim[X]
If IX]
Zl^...^Zr.
0; but then z.^Y
i
z.^Z
i
z o^X
Z
Z, (Z)
for some a and b.
Thus
K. S INGH
304
Hence X+Y
(ax+by)^ul^... Ur_l,
THEOREM 2.
PROOF.
If dim V
X
a
l_<i<j<k_<5
(X)
_<
+
(a145xI
a134x4
(X)
then
XlAX2 ^ (a123x3
ijkXi ^xjAxk
Xl^X3^(a134x4
+
+
+
_<
_<
2.
Let
3.
Xl, x2, x3, x4,
x
5
+
a234x4 + a235x5"
a135x5’ Y2
+ (x I +
+
Hence (X)
_<
Let X
+ X2 +
124x4 + a125x5)
a135x5) + x2^x3^(a234x4 + a235x5)
yl,y 2
are i.+/-.; then
x4Ax5 %yl^Y2 % e F. Let a124x4 + a125x5
X
xlAx2^(a123x3 + blY I + b2Y2) + XlAX3^Yl
+ %(a145x I + a245x 2 + a345x3)Yl^Y2
a123XlAX2^X3
a
a245x 2 + a345x3)x4^x5
So we assume
3.
A3V,
Then
+
Yl
5, X e
We shall first prove that (X)
be a basis of V.
Let
i.e., X+Y is decomposable.
(b2x I
3
yl,y 2
<yl,Y2 >
blY I +
+
a345%Y2)AYl^(-blX2
x
If
are l.d., then
<x4,x5 >,
b2Y 2.
and thus
Then
x2^x3^Y 2
+
a145EY 2)
(a245 a345bl)%Yl)AX2^Y2
x
3
3.
X3, where XI, X2, X 3 are decomposable, X 1 Xl^X2^X3
X
yl^Y2AY3, X 3 ZlAZ2AZ3 Then 1 -< dim[XI] 0 IX2] _< 3.
2
CASE i. dim[XI] n IX 2]
%X for some
3. Then X
and thus (X) -< 2.
2
1
CASE 2. dim[XI] n IX 2]
2. Let Ul,U2,V and Ul,U2,W be bases of IX I]
and IX 2] respectively. Then X
%Ul^U2AV and X 2
Ul^U2AW. Then (X) _< 2.
1
CASE 3. dim[XI] n IX 2]
i. det Ul, u2, u and Ul, u4, u be bases of
5
3
IXI] and IX2] respectively. Then X 1 Ul^U2^U3 X 2 Ul^U4^U5; we have
X
1
assumed the co-effs, to be absorbed with the vectors
XI+X 2
u
I
^
Y, where Y
u2^u 3
+
u4Au5.
Also
ui’s
IX 1 + IX2
and
vi’s.
V.
Then
EQUIVALENCE CLASSES OF THE 3RD GRASSMAN SPACE
Since
n
dim<u2,u3,u4,u5 >
IX 3]
>
2, we can take X 3
By Proposition 9,
Wl,W 2
<u2,u3,u4,u5 >.
or Y
WlAV 1 + w2Av 2. If
-< 2. If Y Wl^V
1
WlAW2^W3
v I, v
%WlAW2 + VlAV2,
Y
305
2
Y
and %
where
%WlAW2 + VlAV 2
Ul^Y + WlAW2^W3
then X
+ w2^v2, then since Ul,Wl,W2,Vl,V 2 is also a
of V, let w
alu I + a2w I + a3w 2 + a4v I + a5v 2. Then
3
X
X + X + X
(u I
a4W2)AWlAVl + UI^W2AV2 + (a5v 2 + alUl)AWI^W2
1
2
3
length
length -< 2, since Z
dim<ul,w2,v2 >
REMARK.
n
uI^W2AV2
<a5v 2
+
+
+ alUl,Wl,W2 > -> 2
A3V
There exists X e
a basis of V and X
(a5v 2
Xl^X2AX3
+
with
alUl)AWlAW2
implies
(X)
then
xIAX4AX5
has
basis
has
and
f(Z) -<
i.
2; for if
(X)
Xl,X2,X3,X4,X 5
is
2, by Proposition i0.
A3V, dim V 5, (X) 2, then P(X) 5; for let
X
X + X2, where (XI)
(X2) i. Since X is not decomposable, then by
1
Proposition i0, dim[X I] n IX 2] < 2 and hence
dim[X] > dim[XI] + dim[X 2]
dim[XI] n IX 2]
5.
4, i.e., P(X)
It follows from Proposition 6 that if X,Y e A3V and (X)
(Y), then
Y. Hence all the equivalence classes of A3V are given by
X
S
o {XIX V, (X) 0} {0}
REMARK.
If X e
S
S
ACKNOWLEDGMENT.
1
2
{XlX
{XIX
e
e
A3V,
A3V,
(X)
I}
(X)
2}.
The author is grateful to Professor R. Westwick for his
invaluable help in the preparation of this manuscript.
REFERENCES
i.
Hodge, W. V. D. and D. Pedoe. Methods of Algebraic Geometry
University Press, Cambridge, 1953.
2.
Lim, M. J. S. L-2 Subspaces of Grassmann Product Spaces, Pacific J.
Math. 33 (1970) 167-182.
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Gurewich, G. B. Foundations of the Theory of. Algebraic Invariants,
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