An interesting idea about Big O

I am reviewing my course and preparing for the final exam. I observed something interesting.

Definition of big $O$$O$ I

Let $f,g$$f,g$ be functions defined on an interval of the form $(a,\infty )$$(a,∞)$.

We shall say that

$$\begin{array}{}\text{(1)}& f(x)=O(g(x))\text{ as }(x⟶\mathrm{\infty })\end{array}$$

If there exists $M>0$$M>0$ and $x_0>a$$x_0> a$ such that for all $x>x_0$$x>x_0$, $|f(x)|\le M|g(x)|$$|f(x)|\le M|g(x)|$.

We could also define a big $O$$O$ for $x⟶b$$x\longrightarrow b$.

Definition of big $O$$O$ II

if there exist positive numbers $\delta$$\delta$ and $M$$M$, such that $\mathrm{\forall }x\in 0<|x-b|<\delta$$\forall x\in 0<|x-b|<\delta$, $|f(x)|\le M|g(x)|$$|f(x)|\le M|g(x)|$.

One interesting thing is this definition looks like Stalk, i.e. $\begin{array}{c}{\mathcal{F}}_x:=\underset{U\ni x}{\underrightarrow{\lim }}\mathcal{F}(U)\end{array}$${\mathcal {F}}_{x}:=\varinjlim _{{U\ni x}}{\mathcal {F}}(U)\\$.

But change the relation from $f{|}_U=g{|}_U|$$f|_U=g|_U|$ to $f{|}_U(x)|\le M|g{|}_U(x)|$$f|_U(x)|\le M|g|_U(x)|$

 

Proposition

Let $\mathcal{F}(a,\mathrm{\infty })$$\mathscr F(a,\infty)$ be the space of all the functions defined on $(a,\mathrm{\infty })$$(a,\infty)$.

Then $O_f=\{f(x)\in \mathcal{F}(a,\mathrm{\infty })|f(x)=O(g(x))\}$$O_f=\{f(x)\in\mathscr F(a,\infty)| f(x)=O(g(x))\}$ is a subspace of $\mathcal{F}(a,\mathrm{\infty })$$\mathscr F(a,\infty)$.

Proof

We only need to prove that

$$\begin{array}{}\text{(2)}& f(x)-h(x)\in O_g,\lambda f(x)\in O_g\end{array}$$

Since $|f-h|\le |f|+|h|$$|f-h|\le|f|+|h|$, thus $|f-g|\le (C_f+C_h)g(x)$$|f-g|\le (C_f+C_h)g(x)$ for an open interval $(k,\mathrm{\infty })$$(k,\infty)$.

$$\begin{array}{}\text{(3)}& |\lambda f(x)|=|\lambda ||f(x)|\le |\lambda |C_f|g(x)|◻\end{array}$$

Moreover, if $f(x)=\mathrm{\Theta }(g(x))$$f(x)=\Theta(g(x))$, then $O_f=O_g$$O_f=O_g$.

 

Indeed, $f\sim g:=f(x)=O(g(x))$$f\sim g:=f(x)=O(g(x))$ is a preorder relation. Thus it can form a category in the usual sense.

The element of of preorder set can be the object and the preorder relation can be the morphism.

Let us call this preorder category $\mathcal{O}$$\mathscr O$

$$\begin{array}{}\text{(4)}& {\mathrm{Hom}}_{\mathcal{O}}(f,g)\ne \mathrm{\varnothing }⟺f(x)=O(g(x))\end{array}$$

Then $O_{(-)}$$O_{(-)}$ form a functor.

$$\begin{array}{}\text{(5)}& O_{(-)}:\mathcal{O}⟶{\mathrm{Vect}}_{\mathbb{R}}\end{array}$$

It maps $f$$f$ to $O_f$$O_f$, and it maps the preorder relation $f(x)=O(g(x))$$f(x)=O(g(x))$, to inclusion map $i_g^f:O_f⟶O_g$$i^{f}_{g}:O_f\longrightarrow O_g$.

 

SL(n,K) and U(n,K): an Algebraic Geometry Approach

I considered the singular matrix as an algebraic variety before.

I observed another interesting example today.

Let us consider the special linear group over algebraic closed fields $K$$K$.

$$\begin{array}{}\text{(1)}& SL(n,K):=\{X\in GL(n,K)|detX=1\}\end{array}$$

Which is a variety since it is the zero of $detX-1=0$$\det X-1=0$

Remark

$$\begin{array}{}\text{(2)}& detX\in K[X_{1,1},...,X_{n,n}]\end{array}$$

and

$$\begin{array}{}\text{(3)}& SL(n,K)\in {\mathbb{A}}_K^{n^2}\end{array}$$

Math Cat told me that this is a kind of algebraic group, i.e. group object in the category of variety.

$SL(n,K)$$SL(n,K)$ has a sub-variety, $U(n,K)$$U(n,K)$.

Which is the intersection of $X_{1j}^2+X_{2,j}^2+...+X_{n,j}^2-1=0$$X_{1j}^2+X_{2,j}^2+...+X_{n,j}^2-1=0$, $j$$j$ from $1$$1$ to $n$$n$.

One interesting question is what is the property of the coordinate ring of $SL(n,K)$$SL(n,K)$ and $U(n,K)$$U(n,K)$?