(i) The ordered pair ( F , t ) (F, t) ( F , t ) A A A .
(ii) Let j = ( 0 , e ) j = (0, e) j = ( 0 , e ) ( x , y ) = x e + y j (x, y) = xe + yj ( x , y ) = x e + y j x ∈ E , y ∈ E x \in E, y \in E x ∈ E , y ∈ E T F T_F T F N F N_F N F F F F
(24)
T F ( x e + y j ) = T ( x ) , N F ( x e + y j ) = N ( x ) − γ N ( y ) .
T_F(xe + yj) = T(x), \quad N_F(xe + yj) = N(x) - \gamma N(y).
T F ( x e + y j ) = T ( x ) , N F ( x e + y j ) = N ( x ) − γ N ( y ) . (iii) For F F F E E E
For ( x , y ) ∈ F (x, y) \in \mathbf{F} ( x , y ) ∈ F
( x , y ) + t ( ( x , y ) ) = ( x + x ‾ , 0 ) = T ( x ) e
(x, y) + t((x, y)) = (x + \overline{x}, 0) = T(x)e
( x , y ) + t (( x , y )) = ( x + x , 0 ) = T ( x ) e (25)
( x , y ) t ( ( x , y ) ) = ( x , y ) ( x ‾ , − y ) = ( x x − γ y ‾ y , y x ‾ − y x ) = ( N ( x ) − γ N ( y ) , 0 ) = ( N ( x ) − γ N ( y ) ) e .
(x, y)t((x, y)) = (x, y)(\overline{x}, -y) = (xx - \gamma \overline{y}y, y\overline{x} - yx)
= (\mathrm{N}(x) - \gamma \mathrm{N}(y), 0) = (\mathrm{N}(x) - \gamma \mathrm{N}(y))e.
( x , y ) t (( x , y )) = ( x , y ) ( x , − y ) = ( xx − γ y y , y x − y x ) = ( N ( x ) − γ N ( y ) , 0 ) = ( N ( x ) − γ N ( y )) e . (26)
To prove both (i) and (ii), it therefore suffices to show that t t t F \mathbf{F} F t t t
t ( ( x , y ) ⋅ ( x ′ , y ′ ) ) = t ( ( x x ′ + γ y ‾ ′ y , y x ‾ ′ + y ′ x ) ) = ( x ‾ ′ x ‾ + γ y ‾ y ′ , − y x ‾ − y ′ x ) = ( x ‾ ′ , − y ′ ) ( x ‾ , − y ) = t ( ( x ′ , y ′ ) ) t ( ( x , y ) )
t((x, y) \cdot (x', y')) = t((xx' + \gamma \overline{y}'y, y\overline{x}' + y'x)) = (\overline{x}'\overline{x} + \gamma \overline{y}y', -y\overline{x} - y'x)
= (\overline{x}', -y')(\overline{x}, -y) = t((x', y'))t((x, y))
t (( x , y ) ⋅ ( x ′ , y ′ )) = t (( x x ′ + γ y ′ y , y x ′ + y ′ x )) = ( x ′ x + γ y y ′ , − y x − y ′ x ) = ( x ′ , − y ′ ) ( x , − y ) = t (( x ′ , y ′ )) t (( x , y )) and hence
t t t is an antiautomorphism.
It remains to prove (iii). As E is identified with a subalgebra of F, E may be assumed to be associative. Let u = ( x , y ) u = (x, y) u = ( x , y ) u ′ = ( x ′ , y ′ ) u' = (x', y') u ′ = ( x ′ , y ′ ) u ′ ′ = ( x ′ ′ , y ′ ′ ) u'' = (x'', y'') u ′′ = ( x ′′ , y ′′ )
{ ( u u ′ ) u ′ ′ = ( ( x x ′ + γ y ‾ ′ y ) x ′ ′ + γ y ‾ ′ ′ ( y ‾ x ′ + y ′ x ) , ( u u ′ ) u ′ ′ = ( y x ‾ ′ + y ′ x ) x ‾ ′ ′ + y ′ ′ ( x x ′ + γ y ‾ ′ y ) ) u ( u ′ u ′ ′ ) = ( x ( x ′ x ′ ′ + γ y ‾ ′ ′ y ′ ) + ( γ ( x ′ ′ y ‾ ′ + x ‾ ′ y ‾ ′ ′ ) y , u ( u ′ u ′ ′ ) = y ( x ‾ ′ ′ x ‾ ′ + γ y ‾ ′ y ′ ′ ) + ( y ′ x ‾ ′ ′ + y ′ ′ x ′ ) x ) .
\begin{cases}
(uu')u'' = ((xx' + \gamma \overline{y}'y)x'' + \gamma \overline{y}''(\overline{y}x' + y'x), \\
\phantom{(uu')u'' = } (y\overline{x}' + y'x)\overline{x}'' + y''(xx' + \gamma \overline{y}'y)) \\
u(u'u'') = (x(x'x'' + \gamma \overline{y}''y') + (\gamma(x''\overline{y}' + \overline{x}'\overline{y}'')y, \\
\phantom{u(u'u'') = } y(\overline{x}''\overline{x}' + \gamma \overline{y}'y'') + (y'\overline{x}'' + y''x')x).
\end{cases}
⎩ ⎨ ⎧ ( u u ′ ) u ′′ = (( x x ′ + γ y ′ y ) x ′′ + γ y ′′ ( y x ′ + y ′ x ) , ( u u ′ ) u ′′ = ( y x ′ + y ′ x ) x ′′ + y ′′ ( x x ′ + γ y ′ y )) u ( u ′ u ′′ ) = ( x ( x ′ x ′′ + γ y ′′ y ′ ) + ( γ ( x ′′ y ′ + x ′ y ′′ ) y , u ( u ′ u ′′ ) = y ( x ′′ x ′ + γ y ′ y ′′ ) + ( y ′ x ′′ + y ′′ x ′ ) x ) . (27)
Examining these formulae shows that the commutativity of E implies the associativity of F. Conversely, if F is associative, formulae (27) applied with y = y ′ = 0 y = y' = 0 y = y ′ = 0 x ′ ′ = 0 x'' = 0 x ′′ = 0 y ′ ′ = e y'' = e y ′′ = e ( 0 , x ′ x ) = ( 0 , x x ′ ) (0, x'x) = (0, xx') ( 0 , x ′ x ) = ( 0 , x x ′ ) x ′ x = x x ′ x'x = xx' x ′ x = x x ′ x , x ′ x, x' x , x ′
Note also that, in the above notation, for x , y x, y x , y
y j = j y ‾ , x ( y j ) = ( y x ) j , ( x j ) y = ( x y ‾ ) j , ( x j ) ( y j ) = y ‾ x e
yj = j\overline{y}, \quad x(yj) = (yx)j, \quad (xj)y = (x\overline{y})j, \quad (xj)(yj) = \overline{y}xe
y j = j y , x ( y j ) = ( y x ) j , ( x j ) y = ( x y ) j , ( x j ) ( y j ) = y x e (28)
(29)
The Cayley algebra ( F , t ) (\mathbf{F}, t) ( F , t ) Cayley extension of ( E , s ) (\mathbf{E}, s) ( E , s ) defined by γ \gamma γ
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
Nothing.