In the tutorial we discussed doubly ruled quadrics, i.e., quadrics of signature (+,+,-,-) in $\mathbb{R}P^3$. The term “doubly ruled” expresses the fact that those quadrics contain two families of lines, where each family of lines generates the quadric. Two lines of those families are skew if they are contained in the same family and they intersect if they are contained in different families. Therefore, each point of the quadric can be described as the intersection point of two unique lines, which also span the tangent plane to the quadric at the intersection point.
It is an important fact that any three skew lines in $\mathbb{R}P^3$ determine a unique doubly ruled quadric that contains the three given lines. You may use this statement for the solution of Exercise 8.3.
Happy holidays,
Emanuel
Just to be sure, in exercise 1 we should assume that $q$ is given in such a way s.t. $Q_1$ is non-degenerate?
From the definition of $U_1$ it follows that $Q_1$ is non-degenerate. Why?
Well, of course. Somehow I got confused and thought an empty quadric is degenerate… So I rephrase my question:
Should we assume that $q$ is given in such a way s.t. $Q_1 \ne \varnothing$?
I see. Well, if $q\big|_{U_1}$ is definite, i.e., $Q_1 = \emptyset$, then of course $Q$ is just the k-plane $P(U_0)$ and you can’t write $Q$ as union of lines that connect points in this k-plane with points on $Q_1$. Therefore, assume that $Q_1 \ne \emptyset$.
I just realized that in exercise 8.3. one should not consider skew hexagons as they are defined, but instead talk about non-planar hexagons.
Hint: From the fact that a hexagon is non-planar, you can deduce that some of the extended edges are skew. Which ones? For some pairs of extended edges, this is not clear and the existence of further intersection points is related to the question, whether the hexagon is contained in a quadric of signature $(+,+,-,-)$…
The exercise sheet has been updated accordingly.