I.2.15c

3/9/2021

This exercise has, let's say, a twisted ending. First of all, we are asked to find curve(s) other than the lines I found
back in part (b) last time. So to show that P^{1} × P^{1} and Q aren't homeomorphic under ψ, the implicit suggestion
is that we have to show that closed sets in the former map only to lines in the latter. Hence, ψ will "miss" the
special nonline curve(s) in Q.

Lemme save finding such a "nonline" curve for later, and for now let me show that ψ does indeed only map closed
sets to lines.

I struggled quite a bit with this, for the same reason I struggled quite a bit all the way back in 1.4: The product
topology is oddly weird to think about.

Now what does an arbitrary open set in the product topology of P^{1} × P^{1} look like?

An arbitrary open set looks like ⋃
_{i}U_{i} ×V _{i}, where U_{i} and V _{i} are open in P^{1}. Now, let me define the closed sets
Z_{i} = P^{1} - U and C_{i} = P^{1} - V .

Then an arbitrary closed set in P^{1} × P^{1} can be written as

P^{1} × P^{1} -⋃
_{i}U_{i} × V _{i} | ||||||

= ⋂
_{i}P^{1} × P^{1} - (U_{i} × V _{i}) | ||||||

= ⋂
_{i}((P^{1} - U_{i}) × P^{1}) ∪ (P^{1} × (P^{1} - V _{i})) | (See here) | |||||

= ⋂
_{i}(Z_{i} × P^{1}) ∪ (P^{1} × C_{i}) |

Now, my friends, let's take a look at, say Z

Basically, Z

Now what does ψ do to an arbitrary element (a,b,c,d) ∈{P}×P

W | = Y | ||

X | = Z |

(on the constant multiple thing: note that if I used (ka,kb) instead, te ks cancel out)

Look familiar? Maybe not, because now I'm using W,X,Y,Z instead of A,B,C,D (because I realize that those are a more appropriate choice of coordinate names if our coordinate ring is k[w,x,y,z], lol). But if you look closely, this is just what I called a "TYPE K" line. Muahahaha. So our {P}× P

"But Mr. Holeinmyproof, you just divided by b! What if b = 0? What then?" Good question, but don't call me that again. What does happen if b = 0? Well, then our image point would instead be (W,X,Y,Z) = (ab,ac,bc,bd) = (ab,ac, 0, 0). I.e. we'd get

Y | = 0 | ||

Z | = 0 |

Look familiar? Again, I'm using W,X,Y,Z now instead, but this is just what I called a TYPE H line. Remember how types H and K ended up going together, as what I called "TYPE M" lines?

So, basically, Z

Similarly, you can guess that P

So in total, our arbitrary closed set gets mapped to intersections of lines in Q, which we showed back in part (b) are either empty, singletons, or lines themselves.

So ψ maps closed sets only to lines in Q (or more trivial things: singletons, empty sets, P

Now all I have to show is that there are in fact nonline curves in Q, and I'm done

Okay.

*Sigh*

*Sigh*

So, remember how in Club Penguin you used to dress up as a pink Penguin and *Sigh* in the corner of the pizza place so that other Penguins would come to you and ask you "what's wrong?" and you could get easy attention and pity that way, as well as possibly an e-bf with whom you could go on romantic dates and have e-sex with in your igloo?

What? Just me? Okay, well, anyway, that's how I feel right now. I feel like a lonely, sighing pink penguin in the corner.

The reason is.... Early on, in yesterday's post, I had a little mini freak out about how our xy - zw looks super familiar to one of the polynomials used to generated [CENSORED]? Well, in particular, it looks familiar to the xw - yz polynomial you can find on the Wikipedia page.

Plot twist: They're actually the fucking same. The reason is that the Wikipedia page (as well as my work in 2.12d) uses the coordinate ring k[x,y,z,w], whereas here we're using the coordinate ring k[w,x,y,z]. If you just relabel the coordinates, you'll see that they're actually the same surface. Yes, the quadric surface Q actually contains the t-t-t-tWISTED CUBIC CURVE.

SO, YOU WANT A NONLINE CURVE IN Q, MOTHERFUCKERS? YOU WANT IT? WELL, I'M GOING TO USE THE TWISTED MOTHERFUCKING CUBIC CURVE AS MY EXAMPLE (SINCE THE TWISTED CUBIC CURVE IS OBVIOUSLY NOT A LINE, ESPECIALLY NOT ONE OF THE TYPE M OR TYPE L LINES FROM Q), AND THIS EXERCISE IS FINISHED.

Yes, the twisted cubic curve actually saved me. Maybe it doesn't hate me after all?

Would it ask me *what's wrong* if I sighed in the corner?