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 $$g(x)=\Delta(x^{-1}-x^\dagger) \ \text{if} \Delta(x)\neq 0, 1 \ \text{else} \ $$ 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 $$\mathbb{V}(f)=j(U(n)) \ $$.

So, let $$x \ $$ be a matrix representation of a point in $$\mathbb{C}^{n^2} \ $$.


 * Flaws in this next part:
 * '' $$g \ $$ is not defined for all of $$\mathbb{C}^{n^2} \ $$, so $$f=g\circ j^{-1} \ $$ cannot be a function from $$\mathbb{R}^{2n^2} \ $$.
 * $$f \ $$ is a rational function, not a polynomial.
 * Demonstrates that $$U(n)\subseteq \mathbb{V}(f) \ $$, but does not demonstrate equality.

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) Suppose $$X_1, X_2 \in \mathbb{A}^k \ $$ are affine algebraic varieties. Then there exist polynomials $$f_1, f_2:\mathbb{A}^k \to \mathbb{C} \ $$ such that $$X_j=\mathbb{V}(f_j) \ $$.  Consider the polynomial $$g:\mathbb{A}^k \to \mathbb{C} \ $$ defined $$g(z)=f_1(z)f_2(z) \ $$, where $$z\in\mathbb{A}^k \ $$.  Then this polynomial will be zero precisely when $$f_1 \ $$ is zero, or $$f_2 \ $$ is zero, or both are.

Hence $$X_1 \cup X_2 = \mathbb{V}(g) \ $$, and so $$X_1 \cup X_2 \ $$ is an affine algebraic variety.

1.2.3) The twisted cubic curve is defined in Figure 1.5 as $$V=\mathbb{V}(x^2-y,x^3-z)=\mathbb{V}(x^2-y)\cap\mathbb{V}(x^3-z) \ $$ in $$\mathbb{R}^3 \ $$. Therefore, for any point $$\vec{r}=(x,y,z)^t \in V \ $$, we have $$x^2-y=x^3-z=0 \ $$.  Therefore, $$x^2=y, x^3=z \ $$, and so we can write any point $$\vec{r}\in V \ $$ as $$(x,x^2,x^3) \ $$.

2.1.2, 2.1.3,

2.1.5,

2.2.2,

2.2.3