Proposition 4

The homomorphism $\phi$ is an isomorphism of V onto $U^0$.

There exists a homomorphism $\phi'$ of U into $V^0$ mapping 1 to 1 and such that $\sigma_0 = \phi' \circ \sigma$. $\phi'$ can be considered as a homomorphism of $U^0$ into V. Then $\sigma_0 = \phi' \circ \phi \circ \sigma_0$ and $\sigma = \phi \circ \phi' \circ \sigma$ and hence $\phi' \circ \phi$ and $\phi \circ \phi'$ are the identity mappings of V and U. Hence the proposition.

V is identified with $U^0$ under the isomorphism $\phi$. Then $\sigma_0$ is identified with $\sigma$.

With this identification, the isomorphism $\theta : x \mapsto -x$ of $g$ onto $g^0$ defines an isomorphism $\tilde{\theta}$ of U onto $V = U^0$. This isomorphism can be considered as an antiautomorphism of U. It is called the principal antiautomorphism of U. If $x_1, \ldots, x_n$ are in $g$, then: