Why $U_n$?

hermitian pairings seem really un-natural algebraically at first;
why take the -opposite- structure on one factor?

B/c you're abstracting $\norm{z}^2 = \bar z \cdot z$, not $z^2$;
there's no difference over $\bR$, but there is over $\bC$.

- b/c then $\norm{\omega v} = \norm{v}$ for all $\omega \in S^1 < \bC$
  (instead of being $\omega^2\norm{v}$)
- b/c then it's positive definite!
- b/c $U_n$ behaves rather more like $O_n \bR$ than $O_n \bC$ does
- more deeply, b/c $\fu_n$ is a compact lie algebra,
  like $\so_n \bR$ but unlike $\so_n \bC$, I think...

essentially, it's -geometrically- much nicer

also, b/c of the magical:
$O_{2n} \bR, \GL_n \bC, \Sp_n < \GL_{2n} \bR$:
pick any 2, and their overall intersection is $U_n$;
more geometrically, any 2 of:
- riemannian structure
- almost complex
- symplectic
...determines a complex structure.
[an -integrable- almost complex structure, say]
