# Show that the set $U(n)$ of unitary matrices can be described as the zero locus of a collection of polynomials with real coefficients in $R^{2n^2}$

In Karen Smith's An Invitation to Algebraic Geometry, we are asked to show that "Show that the set $U(n)$ of unitary matrices can be described as the zero locus of a collection of polynomials with real coefficients in $R^{2n^2}$".

The part of question before this asked us to show that $U(n)$ is not an affine algebraic sub-variety of $C^{n^2}$. I think it's because $U(n)$ always have determinant of 1 so never vanishes in terms of determinant variety(?) I also saw one's past question whose answer was about Noether normalization and how the set of $U(n)$ is compact, but I didn't understand that solution...

The second part is not very obvious to me as well. I'd appreciate any help!

Regarding the second part: The unitary matrices are $U_n = \{S \in \mathbb{C}^{n \times n} \mid S^* S = I\}$, i.e. the zero locus of the equations $$0 = \sum_{j=1}^n (S^*)_{ij} S_{jk} = \sum_{j=1}^n \overline{S_{ji}} S_{jk} \qquad \text{for 1 \leq i,k \leq n}.$$ The main problem is that this the above are not complex polynomials in the entries $S_{ij}$ because we need to conjugate. (This does not show that $U_n$ is not the zero locus of some complex polynomials, but at least the obvious thing does not work. In constrast, for $O_n \subseteq \mathbb{R}^{n \times n}$ this would work.)
But if we identify $\mathbb{C}^{n \times n} \cong \mathbb{C}^{n^2}$ with $\mathbb{R}^{2n^2}$ then the above equations are real polynomials in the real entries $\Re(S_{ij})$, $\Im(S_{ij})$ for $1 \leq i,j \leq n$, so it is the zero locus of real polynomial.
• Thanks! But I'm sorry I'm lost here: what do you mean by "the above are not complex polynomials in the entries because we need to conjugate. " So $S_{ij}$ are matrices or the entries? – nekodesu Jan 9 '16 at 16:57
• The entries. Take for example the special case $n = 1$. Then $U_1 = \{z \in \mathbb{C} \mid \overline{z} z = 1\}$. But $f(z) = \overline{z} z$ is not a complex polynomial; the problem is, that $z \mapsto \overline{z}$ is not a complex polyonmial. If, on the other hand, we regard $\mathbb{C} = \mathbb{R}^2$ then the conjugation is given by $(x,y) \mapsto (x,-y)$, which is polynomial in each entry. So $f(x,y) = x^2 + y^2$ is a polynomial in the real entries $x,y$. – Jendrik Stelzner Jan 9 '16 at 17:00