U(n) is compact and algebraic, but not abelian—why not a contradiction?

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the subgroup of unitary matrices $\text{U}(n) \subset GL(n, \mathbb{C})$ is compact and definitely algebraic, with an algebraic group law; on the other hand, it's not abelian. why is this not a contradiction, in light of the following facts?

  • every complete variety with an algebraic group law is abelian (see e.g. [Milne, Cor. 1.4])
  • completeness for varieties over $\mathbb{C}$ amounts to compactness in the complex-analytic topology (see e.g. here);
  • $\text{U}(n)$ is algebraic (determined by a determinant relation), with an algebraic group law (matrix multiplication), and is compact (see e.g. here).

these together should imply that $\text{U}(n)$ is abelian, which is clearly false. my best guess is that something is going wrong with completeness, and that this thing isn't actually complete.

what gives?