User:J D Bowen/Math899 HW1

1.1.2) In order for $$U(n) \ $$ to be a subvariety, it must be a variety in its own right. It is not, and we can prove it.

Assume to the contrary $$\exists \ $$ a polynomial $$f:\mathbb{C}^{n^2}\to\mathbb{C} \ $$ such that $$U(n)= \mathbb{V}(f) \ $$.

If we let $$x \ $$ be an $$n\times n \ $$ matrix representing a point in $$\mathbb{C}^{n^2} \ $$, we know that setting $$g(x)=\Delta(x^{-1}-x^\dagger) \ $$ gives us a function whose zero set includes $$U(n) \ $$. Therefore, $$x\in U(n) \implies f(x)=g(x) \ $$. But as a polynomial, $$f \ $$ is holomorphic everywhere, and as a function involving complex conjugates, $$g \ $$ is holomorphic nowhere. A holomorphic function is defined globally by its values on a connected subset of its domain. Since this means that $$f|_S=g|_S \implies f=g \ $$, and we know $$f\neq g \ $$, we cannot have $$f|_S = g|_S \ $$ on any connected set $$S\in\mathbb{C}^{n^2} \ $$. Since $$U(n) \ $$ is connected, no such polynomial $$f \ $$ exists.

Now let's show that $$U(n) \ $$  is  a variety in $$\mathbb{R}^{2n^2} \ $$. Let $$j:\mathbb{C}^{n^2}\to\mathbb{R}^{2n^2} \ $$ be the map $$f(z_1, \dots, z_{n^2}) = (\text{Re}(z_1), \text{Im}(z_1), \dots, \text{Re}(z_{n^2}), \text{Im}(z_{n^2}) ) \ $$.

To demonstrate that $$j(U(n)) \subset \mathbb{R}^{2n^2} \ $$ is an algebraic variety, we must find $$f:\mathbb{R}^{2n^2}\to\mathbb{R} \ $$ such that $$\mathbb{V}(f)=j(U(n)) \ $$.

This is fairly simple. Let $$x \ $$ be a matrix representation of a point in $$\mathbb{C}^{n^2} \ $$. Then if we set $$g(x)= \Delta(x^{-1}-x^\dagger)\overline{ \Delta(x^{-1}-x^\dagger)} \ $$, we can form the map $$f=g\circ j^{-1}:\mathbb{R}^{2n^2}\to\mathbb{R} \ $$.

By construction we have $$U(n)\subseteq \mathbb{V}(f) \ $$.

1.2.1,

1.2.3,

2.1.2, 2.1.3,

2.1.5,

2.2.2,

2.2.3