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Consider a line $k$ in the projective space $P(\mathbb{R}^4)=\mathbb{R}P^3$ and 3 points $A_0,A_1,A_2\in k$. Let $B$ be an arbitrary point in $P(\mathbb{R}^4)$ but not on $k$ and $l$ be a "changeable" line such that $l\cap k=\emptyset$ and $B\notin l$. Suppose now that the cross-ratio $(\alpha_0,\alpha_1,\alpha_2,\beta)=1981$, where $\alpha_i=A_i+l$ for $i=0,1,2$ and $\beta=B+l.$

Show that there exists a line that intersects with all such lines $l$.

Now, we can readily conclude that $\alpha_i$ and $\beta$ are hyperplanes of $P(\mathbb{R}^4)$. Furthermore, they belong to a hyperplane bundle with axis $l$ (i.e. $\bigcap_i\alpha_i\cap\beta$=l).

By transversality properties of projective spaces, every line, say $h$, that does not intersect with $l$ crosses every hyperplane $\alpha_i,\beta$ in exactly one point, denoted by $P_i(h),P_B(h)$, and $(\alpha_0,\alpha_1,\alpha_2,\beta)=(P_0(h),P_1(h),P_2(h),P_B(h))$.

I can't figure out how to construct such a line that intersects with every line $l$. How is the transversality property useful here? Or am I following a wrong path?

Any help is welcome!

TheHunter
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  • @JeanMarie Why wouldn't that be possible? I would just say that both lines have no projective points in common, otherwise I don't think I understand your question... – TheHunter May 15 '21 at 22:51
  • But lines can be non intersecting and non parallel? – TheHunter May 16 '21 at 07:06
  • Connected: https://math.stackexchange.com/q/607348 – Jean Marie May 16 '21 at 07:39
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    $P(\mathbb{R}^4)=\mathbb{R}P^3$ is the projective space associated to the vector space $\mathbb{R}^4$, i.e. the quotient set $\mathbb{R}_0^4/\sim$, where $\sim$ identifies all the points that lie on the same line through the origin in $\mathbb{R}^4$. I hope this is clear because I didn't know what notation was conventional. – TheHunter May 16 '21 at 08:14

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