Information Technology Reference
In-Depth Information
Lemma 7.
With the notation as above, define
a
:
V
⊕
V
→
F
2
by
a
(
x, z, y, w
)=
τ
(
x, z
)
·
(
y, w
)+
c
(
x, z
)
,
, y
∈
U, z, w
∈
W,
where
c
:
V
U
the
map
τ
x
is a permutation on
W
. Moreover, in this case, the partial dual of
a
with
respect to
W
→
F
2
.Then
a
is bent with respect to
W
⊕
W
if for every
x
∈
⊕
W
is given by
z
=
τ
−
x
(
v
)
.
a
W⊕W
(
x, u, y, v
)=
τ
z
(
x
)
·
y
+
u
·
z
+
c
(
x, z
)
,
where
Proof.
We have
H
W⊕W
(
a
)(
x, u, y, v
)=2
−k
1)
τ
(
x,z
)
·
(
y,w
)+
c
(
x,z
)+
u·z
+
v·w
(
−
z,w∈W
=2
−k
1)
τ
z
(
x
)
·y
+
τ
x
(
z
)
·w
+
c
(
x,z
)+
u·z
+
v·w
−
(
z,w
∈
W
=2
−k
1)
τ
z
(
x
)
·y
+
c
(
x,z
)+
u·z
1)
(
τ
x
(
z
)+
v
)
·w
.
(
−
(
−
z∈W
w∈W
The inner sum is zero unless
z
=
τ
−
x
(
v
), in which case the sum is equal to 2
k
.
Therefore
1)
a
W⊕W
(
x,u,y,v
)
,
H
W⊕W
(
a
)(
x, u, y, v
)=(
−
where
a
W⊕W
isgiveninthelemma.
{
1
,
2
,...,m
}
Now partition the set
into the two subsets
S
=
{
s
1
,...,s
k
}
and
T
=
{
t
1
,...,t
m−k
}
.
Given
x
V
, we shall write
x
S
=(
x
s
1
,...,x
s
k
)and
x
T
=(
x
t
1
,...,x
t
m−k
). As
before, let
U
and
W
be vector spaces over
∈
F
2
such that
V
=
U
⊕
W
and, if
(
x
1
,...,x
m
)
∈
V
,wehave
x
S
∈
W
and
x
T
∈
U
.
Theorem 8.
With the notation as above,
f
, given in (1), is bent with respect
to
U
⊕
U
and bent with respect to
W
⊕
W
. Moreover, the partial dual of
f
with
respect to
W
⊕
W
is given by
f
W⊕W
(
x
T
,x
S
,y
T
,y
S
)=
w
T
· y
T
+
x
S
· z
S
+
g
(
x
T
,z
S
)
,
where
z
j
=
x
j
if
j
∈
S
ψ
−
j
(
y
j
+
φ
j−
1
(
z
j−
1
))
if
j
∈
S,
and
w
j
=
φ
j−
1
(
z
j−
1
)+
ψ
j
(
x
j
)
for
j
∈
T.
By convention,
x
0
is the all-zero vector and
φ
0
is the identity map.
