TITLE: Conjugation is not expressible in terms of polynomials
QUESTION [5 upvotes]: In order to convince myself that the set $U(n)$ of unitary matrices (matrices with columns that are orthonormal under the complex inner product) is not an affine variety in $\mathbb{C}^{n^2}$, I need that the conjugate of an entry $\overline{x}_{ij}$ cannot be expressed using polynomials. If we say that complex polynomials have to be finite sums of monomials of the form $cx_1^{e_1}\dots x_n^{e_n}$ where $c \in \mathbb{C}$, $e_i \geq 0$, then is this really enough? It seems a little slippery.

REPLY [4 votes]: Here is a proof that $U(n)$ is not an affine variety in $\mathbb{C}^{n^2}$: If $X \subset \mathbb{C}^N$ is a smooth affine subvariety of $\mathbb{C}^N$, then, at every point of $X$, the tangent space to $X$ will be a complex sub-vector-space of $\mathbb{C}^N$. (Proof: The tangent space to a polynomial hypersurface is a complex sub-vector-space; the tangent space to an intersection is the intersection of tangent spaces. This argument also works for singular varieties if you use the Zariski tangent space.)
At the identity, the tangent plane to $U(n)$ is the Hermitian matrices. Mutiplyling a Hermitian matrix by $i$ does not give you a Hermitian matrix, so $U(n)$ is not an algebraic subvariety.