L.S.,
Denote $GO(2n,\mathbb{C})$ the group of matrices that leave an inner product invariant up to a scalar. We have for all $A \in GO(2n,\mathbb{C}), v,w$ vectors, that: $\langle A(v),A(w)\rangle = \lambda_A \langle v,w \rangle$.
It is known that the function $f$ that maps $A$ to $\lambda_A$ is a homomorphism from $GO(2n,\mathbb{C})$ to $\mathbb{C}^*$. However I find it somewhat unfortunate that we lose too much information, since for instance $f(-I) = f(I)$. Is there maybe a way to construct a homomorphism from $GO(2n,\mathbb{C})$ to $\mathbb{C}^*$ that doesn't lose this particular information? So that we have $f(-A) \neq f(A)$ for all $A \in GO(2n,\mathbb{C})$? I already found out that we can factor $A$ as $A = ab$, with $b \in O(2n,\mathbb{C})$ and $a^2 = \lambda_AI$.
Many thanks!
If $n>1$, I don't think it is possible.
Denote $f:GO(2n,\mathbb{C})\rightarrow \mathbb{C}^*$ a group morphism. You might already know that $SO(2n,\mathbb{C})$ is a simple Lie group. In particular, $SO(2n,\mathbb{C})$ contains no non-trivial normal connected subgroups.
It means that $Ker(f)\cap SO(2n,\mathbb{C})$ is either $SO(2n,\mathbb{C})$ or discrete. If your group morphism $f$ were not $SO(2n,\mathbb{C})$ then $SO(2n,\mathbb{C})$ quotiented by a discrete subgroup would be isomorphic to a subgroup of $\mathbb{C}^*$ which is not possible if $n>1$ (for dimensional reason).
Hence a group morphism $f:GO(2n,\mathbb{C})\rightarrow \mathbb{C}^*$ not only needs to trivialize $-I$ but also any element of $SO(2n,\mathbb{C})$.