Lattice and Boolean Algebra over vector space (1)

Consider a vector space $V$$V$, all the subspaces of $V$$V$ with $\subseteq$$\subseteq$ can be a lattice, denoted as $L_V$$L_V$

Observe that $A\cap B$$A\cap B$ is the greatest lower bound, and $A+B$$A+B$ is the least upper bound.

Remark. $A+B:=\{a+b|a\in A,b\in B\}$$A+B:= \{a+b|a\in A,b\in B\}$

And we have

$A\subseteq B$$A\subseteq B$ iff $A+B=B$$A+B= B$, $A\subseteq B$$A\subseteq B$ iff $A\cap B=A$$A\cap B=A$

Thus the absorb law holds. $A\cap (A+B)=A,A+(A\cap B)=A$$A\cap (A+B)=A,A+(A\cap B)=A$

And easy to check that for $\cap ,+$$\cap,+$, close, associative law, and commutative law holds

$0$$0$ is the $inf{L}_V$$\inf L_V$, $V$$V$ is the $sup{L}_V$$\sup L_V$, and $0\cap A=0,V+A=V$$0\cap A=0,V+A=V$

Thus $L_V$$L_V$ is a Lattice.

And if we consider an inner product space $(V,⟨-,-⟩)$$(V,\langle-,-\rangle)$ and the lattice $L_V$$L_V$

We can define orthogonal complement as complement or duality

Because

$†.0^{\mathrm{\perp }}=V,V^{\mathrm{\perp }}=0,A\cap A^{\mathrm{\perp }}=0,A+A^{\mathrm{\perp }}=V$$\dagger. 0^{\bot}=V,V^{\bot}=0,A\cap A^{\bot}=0,A+A^{\bot}=V$

$†.A\subseteq B\iff A^{\mathrm{\perp }}\supseteq B^{\mathrm{\perp }}$$\dagger. A\subseteq B\Leftrightarrow A^{\bot}\supseteq B^{\bot}$

$†.(A^{\mathrm{\perp }}{)}^{\mathrm{\perp }}=A$$\dagger. (A^{\bot})^{\bot}=A$

Thus we have De Morgen Law

$(A\cap B{)}^{\mathrm{\perp }}=A^{\mathrm{\perp }}+B^{\mathrm{\perp }}$$(A\cap B)^{\bot}=A^{\bot}+B^{\bot}$

$(A+B{)}^{\mathrm{\perp }}=A^{\mathrm{\perp }}\cap B^{\mathrm{\perp }}$$(A+B)^\bot=A^\bot\cap B^\bot$

Because $\mathrm{\perp }:(L_V,\subseteq )\to (L_V,\supseteq )$$\bot:(L_V,\subseteq)\to (L_V,\supseteq)$ is a partial order isomorphism

And partial order isomorphism preserves the greatest lower bound and least upper bound

In general, for ${}^{\prime }$$’$ have $a\le b\iff a^{\prime }\ge b^{\prime },(a^{\prime }{)}^{\prime }=a$$a\le b\Leftrightarrow a'\ge b',(a')'=a$, We will have De Morgan Law

$(a{\vee }_{\le }b{)}^{\prime }=a^{\prime }{\vee }_{\ge }{b}^{\prime }=a^{\prime }{\wedge }_{\le }{b}^{\prime }...$$(a\vee_\le b)'=a'\vee_\ge b'=a'\wedge_\le b'...$

But the distributive law does not hold; therefore it can not be a Boolean Algebra.

But, if we consider the standard orthogonal basis of $(V,⟨-,-⟩)$$(V,\langle-,-\rangle)$

$S=\{e_1,e_2,e_3,...,e_{n-1},e_n\}$$S=\{e_1,e_2,e_3,...,e_{n-1},e_n\}$,

We know that $(P(S),\cup ,\cap ,\mathrm{\varnothing },S{,}^c)$$(P(S),\cup,\cap,\empty,S,^c)$ is an Boolean Algebra,

And $\mathrm{Span}:P(S)\hookrightarrow L_V$$\mathrm{Span}:P(S)\hookrightarrow L_V$ is a Boolean homomorphism

That is the $\mathrm{im}\mathrm{Span}\subseteq L_V$$\mathrm{im\,Span}\subseteq L_V$ is a Boolean Algebra

$\mathrm{Span}(A\cap B)=\mathrm{Span}(A)\cap \mathrm{Span}(B)$$\mathrm{Span}(A\cap B)=\mathrm{Span}(A)\cap \mathrm{Span}(B)$

$\mathrm{Span}(A\cup B)=\mathrm{Span}(A)+\mathrm{Span}(B)$$\mathrm{Span}(A\cup B)=\mathrm{Span}(A)+\mathrm{Span}(B)$

$\mathrm{Span}(\mathrm{\varnothing })=0$$\mathrm{Span}(\empty)=0$ (Recall another definition of $\mathrm{Span}(S):=$$\mathrm{Span}(S):=$ $\begin{array}{c}\bigcap _{S\subseteq V_i}{V}_i\end{array}$$\bigcap_{S\subseteq V_i} V_i\\$ )

$\mathrm{Span}(S)=V$$\mathrm{Span}(S)= V$

$\mathrm{Span}(A^c)=A^{\mathrm{\perp }}$$\mathrm{Span}(A^c)= A^{\bot}$

Then we can define Boolean Ring over $\mathrm{im}\mathrm{Span}$$\mathrm{im \,Span}$

$A\mathrm{\Delta }B=(A+B)\cap (A\cap B{)}^{\mathrm{\perp }}$$A\Delta B=(A+B)\cap(A\cap B)^{\bot}$

Another interesting thing is, in $\mathrm{A}\mathrm{b}$$\rm Ab$, we have the second isomorphism theorem

$HK/H\cong K/H\cap K$$HK/H\cong K/H\cap K$

This shows an interesting property of the glb and lub

If we consider in ${\mathrm{Vect}}_{\mathbb{F}}$$\mathrm {Vect}_\mathbb F$, then we will have $\dim (H+K)-\dim H=\dim K-\dim (H\cap K)$$\dim(H+K)-\dim H=\dim K-\dim(H\cap K)$

That is, $\dim (H+K)+\dim (H\cap K)=\dim H+\dim K$$\dim(H+K)+\dim(H\cap K)=\dim H+\dim K$