Every compact connected Lie group is actually a real algebraic group

Question

Every compact connected Lie group is actually a real algebraic group.

For example $U(n)$ fits the requests. I can't understand in which sense $U(n)$ is real, since its matrices are not real and its defining equations are not real.

Thanks!

Answer 1

$U(n)$ is a real algebraic group, and not a complex algebraic group, because the equations which define it are algebraic over $\mathbb{R}$, and not over $\mathbb{C}$.

Consider $$ U(1) = \left\{z \in \mathbb{C} \mid z \bar z = 1\right\} $$ In this form $U(1)$ is a subgroup of $\mathbb{C}^\times$, so is a group of complex numbers. And the equation defining $U(1)$ has complex variables in it. But that equation is not a polynomial equation in $z$, because of the $\bar z$ term.

On the other hand, $U(1)$ is isomorphic to the group $$ S^1 = \left\{(x,y) \in \mathbb{R}^2 \mid x^2 + y^2 = 1 \right\} $$ with multiplication rule given by $$ (x_1,y_1)\cdot(x_2,y_2) = (x_1 x_2 - y_1 y_2, x_1 y_2 + y_1 x_2) $$ Now the group's equation is a polynomial in the real variables $x$ and $y$.