- - Define the projective space $ℙ(V)$ attached to a finite-dimensional vector space $V$ over a field $F$. Briefly explain how linear subspaces $U ⊂ V$ give rise to projective linear subspaces of $ℙ(V)$. Define the dimension of such a projective linear subspace. Define the notions of line and plane in $ℙ(V)$.
    Projective space $ℙ(V)$ is the set of lines through the origin in a finite-dimensional vector space $V$ over a field $F$.
    (Or: $ℙ(V)$ is the set of nonzero vectors in $V$ modulo the equivalence relation $\mathbf{v} ∼ λ \mathbf{v}$ for $λ ∈ F^*.$)
    A projective linear subspace $S$ of $ℙ(V)$ is the set of lines contained in a linear subspace $U ⊂ V$.
    (Or: A projective linear subspace $S$ of $ℙ(V)$ consists of equivalence classes represented by nonzero vectors contained in a linear subspace $U ⊂ V.$)
    The dimension of $S$ is $\dim S=\dim U-1$ for such an $S$. Lines and planes are $1,2$-dimensional projective linear subspaces of $ℙ(V)$ respectively.
  - Given two projective linear subspaces $L, M$ of $ℙ(V)$, define their span $⟨L, M⟩$ and briefly explain the geometric meaning of this notion. State and prove a dimension formula for $⟨L, M⟩$ in terms of data attached to $L, M$ and their intersection.
    Given two projective linear subspaces $L, M$ of $ℙ(V)$, let $U, W$ be the corresponding linear subspaces of $V$. Then $⟨L, M⟩$ is defined to be the projective linear subspace corresponding to the vector subspace $U+W$ of $V$.Geometrically, this is the union of lines $P Q$ with $P ∈ L, Q ∈ M$. The formula is $$ \dim⟨L, M⟩=\dim L+\dim M-\dim(L ∩ M) $$ with the convention that if $L ∩ M=∅$ then $\dim(L ∩ M)=-1$.
    To prove this, recall from Linear Algebra that for subspaces $U, W$ of $V$ as before, we have $$ \dim(U+W)=\dim U+\dim W-\dim(U ∩ W) $$ Now simply subtract one from each term on each side to get the required equality, noting that $L ∩ M=∅$ if and only if $U ∩ W=\{0\}$.
- Show that if $L, M$ are two lines in $ℙ(V)$, then $1 ⩽ \dim⟨L, M⟩ ⩽ 3$. Describe each of the three possible cases geometrically.
  If $L, M$ are two lines in $ℙ(V)$, then using the above formula, we obtain that $$ \dim⟨L, M⟩=2-\dim(L ∩ M) $$ Also $L ∩ M ⊂ L$ so its dimension is at most one. So there are three cases:
  1. $\dim(L ∩ M)=1$ so $L ∩ M=L=M$ so the two lines coincide, and $\dim⟨L, M⟩=1$.
  2. $\dim(L ∩ M)=0$ so $L ∩ M=P$ is the unique intersection point of the two lines, and $\dim⟨L, M⟩=2$ the plane they span.
  3. $\dim(L ∩ M)=-1$ so $L ∩ M=∅$, the lines are disjoint, and $\dim⟨L, M⟩=3$.
- Let $L, M, N$ be three distinct lines in $ℙ(V)$ that have non-empty pairwise intersection. Show that the lines $L, M, N$ either share a common point, or are all contained in the same plane.
  Let $S=⟨L, M⟩$ and $T=⟨L, N⟩$; by the discussion in (b), these are planes. Then by the dimension of intersection fornula, $$ \dim⟨S, T⟩=4-\dim(S ∩ T) $$ On the other hand, $L ⊂ S ∩ T ⊂ S$ so either $S ∩ T=S=T$ or $S ∩ T=L$. In the first case, all of $L, M, N$ are contained in the plane $S=T$ so we are done. In the second case, $\dim⟨S, T⟩=3$. Notice that $⟨S, T⟩=⟨S, N⟩$. Using the dimension of intersection formula again, $$ 3=\dim⟨S, N⟩=3-\dim(S ∩ N) $$ Thus $\dim S ∩ N=0$ so $S ∩ N$ is a point $Q$. This is the only intersection point of $N$ with the plane spanned by $L$ and $M$ but $L∩N,L∩M≠∅$ so this point $Q$ must also be on $L$ and $M$. Thus $Q$ is a common intersection point of all the three lines.
- - Suppose now that $\dim V=4$. Let $L, M$ be two disjoint lines in $ℙ(V)$, and let $P ∈ ℙ(V)$ be a point not lying on $L, M$. Prove that there exists a line $N$ through $P$ meeting both $L$ and $M$. [You may wish to consider the projective linear subspaces $S=⟨P, L⟩$ and $T=⟨P, M⟩$.]
    Let $S=⟨P, L⟩$ and $T=⟨P, M⟩$. Since $P$ is not contained in either of the lines, the dimension of intersection formula shows these are planes. Also they must be different planes, since otherwise $L, M$ would be coplanar and hence should meet. Thus $\dim(S ∩ T)<2$. On the other hand, $$ \dim⟨S, T⟩=4-\dim(S ∩ T) $$ The left hand side must be at most $3=\dim ℙ(V)$, so $\dim(S ∩ T)>0$. Thus $S ∩ T$ is a line $N$ which goes through $P$. It is also coplanar with $L, M$ so it must intersect them.
    (Or: By the discussion in (b), $⟨L, M⟩$ is three-dimensional since $L, M$ do not intersect, hence it must be the whole of $ℙ(V)$. Since this is the locus covered by lines through a point each of $L$ and $M$, there must exist a line $N$ incident with $L, M$ and through $P$.)
  - Let $P_1=[1: 0: 0: 0], P_2=[0: 1: 0: 0], P_3=[0: 0: 1: 0], P_4=[0: 0: 0: 1]$ and $P=[1: 1: 1: 1]$ be points of $ℙ^3$. Consider the lines $L=P_1 P_2$ and $M=P_3 P_4$. Find the coordinates of a point $Q$ different from $P$ such that the line $P Q$ meets both $L$ and $M$.
    Use coordinates $\left[x_0, x_1, x_2, x_3\right]$ on $ℙ^3$. The plane $S$ above, containing the points $P_1=[1,0,0,0], P_2=[0,1,0,0]$ and $P=[1,1,1,1]$, has equation $x_2-x_3=0$. Similarly, the plane $T$ containing the points $P_3=[0,0,0,1], P_4=[0,0,1,0]$ and $P=[1,1,1,1]$, has equation $x_0-x_1=0$. The intersection $N$ of these two planes satisfies both these equations. We simply need to choose another point satisfying both equations, for example $Q=[1,1,0,0]$.