This is an exercise from An Invitation to Algebraic Geometry by Karen Smith. It asks to show that the set of unitary matrices $U_n$ is not an affine algebraic variety in complex space $C^{n^2}$. However it is a real algebraic variety.

The second part is pretty easy. I expanded the complex variables into $a+bi$ and make them two real variables. Using the equations $\Sigma_j^{n} x_{ij} \bar{x_{jk}}=1$ when $i=k$ and $\Sigma_j^{n} x_{ij} \bar{x_{jk}}=0$ when $i\neq k$, I got a bunch of polynomials in terms of the $2n^2$ variables. So the zero set must be an affine algebraic variety in reals.

For the first part, I think it has something to do with the fact that $f(z)=\bar{z}$ is not analytic. But how to show that the zero set cannot come from polynomials? Thanks for any help!

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    $\begingroup$ Hint : Unitary matrices are a compact subspace of $\mathbb{C}^{n^2}$ when equipped with the Euclidian topology and this cannot be the case of an algebraic variety $\endgroup$ – marwalix Jan 22 '15 at 20:23
  • $\begingroup$ I see it now! Thanks! $\endgroup$ – KittyL Jan 22 '15 at 20:34
  • $\begingroup$ @marwalix: Sorry I just made a mistake. I see how it is compact now. But why cannot it be compact? Isn't a set of finite points compact? $\endgroup$ – KittyL Jan 22 '15 at 20:49
  • $\begingroup$ Unless it is a finite set of points as a consequence of the Noether normalisation theorem it cannot be compact. $\endgroup$ – marwalix Jan 22 '15 at 21:46
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    $\begingroup$ Noether's normalization theorem says that if $X$ is a variety of dimension $n$ , there exists a finite surjective morphism $X\rightarrow\mathbb{A}^n$ . Since, in the euclidian topology over $\mathbb{C}$ , affine space $\mathbb{A}^n$ is not compact for $n\geq 1$ , $X$ is not compact either. $\endgroup$ – marwalix Jan 22 '15 at 21:58

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