Benefits of trying to learn homology theory from three different sources: I end up actually learning homology theory!

sure

@AkivaWeinberger Tell me what the homology groups of a surface of genus $g$ is.

i.e., tell me how you'd find them.

@BalarkaSen Uh, $H_1(M_g)$ (is that the right notation?) is $\sum_{2g}\Bbb Z$, right?

And $H_0=H_2=\Bbb Z$

Yes, but how would you prove that? Also, just write it as $\Bbb Z^{2g}$.

Yeah, I wasn't sure, because we've been using $\oplus$ instead of $\times$

But, yeah, that

Simplicial or singular?

Singular. Simplicial is dumb because you have to use singular to prove it's homotopy invariant.

(or, well, use the simplicial approx thm, but that's of no concern right now)

So whenever I talk to you I'd be talking about the singular homology.

Presumably I break it up into $M_{g-1}$ and $T$ (each minus a disk and overlapping each other a bit) and apply Mayer–Vietoris

$T:=M_1=S^1\times S^1$

Probably not actually what I should be calling it

Mayer–Vietoris would be a pain to type out

You should work out the details.

At some point

I'm actually in the middle of dinner right now

Everyone else keeps stealing my food 'cause I'm on my phone

Ah, sure. But note that even if you use M-V it won't be immediate what the homology groups are. You're going to face trouble, me thinks.

And the trouble is why I gave you this problem :)

Ah, OK

Oh, yeah, 'cause $H_1(U\cap V)\ne0$

So that'd be annoying

Mhm. $H_2$ would not be obvious.

I think you should just try out the long exact sequence. The trouble would be the same.

i.e., to figure out the snake map.

$H_{\color{Red}2}$ wouldn't be obvious?

Huh, I guess it seems obvious until I actually try to prove it

(Like Jordan)

@AkivaWeinberger Actually both $H_2$ and $H_1$ wouldn't be :)

In any case, pure algebra won't tell you what they are. The snake map needs to be dealt with.

@rorty consider the quadratic formula

Is what you call "snake map" what my book calls the "connecting map" (IIRC)?

The $\Delta$ thing

Yes.

The $\partial$ thing.

Huh, my (non-Hatcher) book uses $\Delta$.

Yes, I use Hatcher's notations.

I struggled with this for some time. When it clicked, it was great. There is actually something very geometric going on, if you already haven't read about it.

Ping me if you want to let me know your thoughts. I'll tell you a story afterwards.

I'm heading to bed. See ya.

Isn't it 7am where you are?

@BalarkaSen

hi

I am wondering what the word parallels means in geometry

here is an example: Let P be a point such that the parallels through A′,B′,C′ to AP, BP, CP respectively concur at a point Q.

@AkivaWeinberger: Just tell him it's past his bedtime when you see him. Don't worry about what time it is; it's probably true.

@BalarkaSen This is false.

Ah, you said so.

can someone help me understand what "parallels" means?

Parallel lines

@AkivaWeinberger I mean in this example. " Let P be a point such that the parallels through A′,B′,C′ to AP, BP, CP respectively concur at a point Q."

It doesn't mean parallel lines there

The line through A' parallel to AP, the line through B' parallel to BP, etc.

the drawing doesn't correspond to that

i am reading this paper

ijgeometry.com/wp-content/uploads/2015/10/1.pdf

Gib page number?

7 at the bottom

lemma 1.3

Are you looking at Figure 2?

'Cause that seems to refer to Sondat's theorem (Theorem 1.2), not Lemma 1.3.

@user19405892

oh i was

but in any case there is no Q in figure 1

or P

I think parallels here just means line connecting

I don't think Lemma 1.3 has a figure

I think you just need to draw it yourself

I don't know what a parallel through A' to AP is though

Oh is it just a parallel line through A' parallel to AP?

That's what I thought it was

Parallels isn't a noun so I think they should say "parallel lines"

Eh

My dictionary has "parallel" as a noun

@AkivaWeinberger what about parallels?

Definition 1a is "a parallel line, curve, or surface"

Um, plural of "parallel"?

I don't see that in my dictionary

Eh

i did see though in my dictionary that "parallels of latitude" is a phrase

Right

so maybe parallels can be used

"orbifold" is also not in my dictionary but mathematicians use it anyway

you'll survive

In a sense, trampolines only work 'cause there's no deformation retraction from the disk to its boundary

So, I've asked this before and I'm still not sure I get it: Let $\alpha$ be a generator of $H_3(S^3)$, and let $f:S^3\to S^2$ be the Hopf fibration. Why isn't $f_*(\alpha)$ a nonzero element of $H_3(S^2)$?

In other words, how is $f_\#(a)$ a boundary?

take $S^3$ as an $S^1$ bundle over $S^2$. fill in those circles with discs to get a $D^2$ bundle over $S^2$; this is a smooth 4-manifold with boundary $S^3$

since f takes each of the fiber circles to a point, it's obvious how to extend it to the larger 4-manifold: make it constant on each disc

this map bounds your map.

@AkivaWeinberger pinging in case you didn't see that

anyone seen Batman vs Superman? Supposedly horrible

@StanShunpike yes I thought of watching it today. But then the reviews made me decide otherwise

do

@Akiva @MikeMiller A geometric way to think about it is this. There's a map $D^4-{0} \to \Bbb CP^1$ given by taking the complex line through the point and the origin. The boundary of this map is the Hopf fibration and it doesn't extend over D^4 (or else $\pi_3(S^2)=0$. However if you cut out $0$ from $D^4$ and replace it by all the lines through $0$ it obviously extends (just by mapping the lines to themselves). This new space is the 4-ball blown up at 1 point.

@MikeMiller I knew you're going to say it's false, thus the addition.

@PVAL I am not sure if I understand the construction, can you please elaborate? You're looking at the quotient map $\Bbb C^2 - 0 \to \Bbb{CP}^1$. What do you mean when you say you cut out $0$ and replace it by all the lines through $0$?

Like, remove a neighborhood of $0$ homeomorphic to $S^3$ and glue in the manifold @Mike mentioned?

@MikeMiller Thanks! That makes sense. Though I'm not entirely sure how to define the topology on that.

@Akiva It's a disk bundle on S^2.

I'm specifically confused about the preimage of the North Pole

Preimage of any point is a disk. What do you mean?

You know how the preimage of a line of longitude is a torus?

At the poles, the tori become degenerate

You're talking about the Hopf map? The preimage of the longitudes are torii, yes.

And yes, at the poles, the tori become circles.

I'm visualizing $S^3$ as its projection onto $\Bbb R^3$ plus a point at infinity. The preimage of the North Pole is a line through that point

Yes, I know.

so I'm not sure exactly how to fill it in with a disk

You can't do it in $\Bbb R^3$ without self intersection obviously.

Right.

I mean, given any two points which are sufficiently away from each other, preimage of those two points is a Hopf link. You can't fill those in by disks without self intersection inside $S^3$, because Hopf link is a nontrivial link.

Yeah. So I'm not sure how exactly to create the actual $4$-manifold in which they don't intersect.

morning

@AkivaWeinberger This just means you can't do this inside $\Bbb R^3$, is all. If you want a rigorous definition, I guess you have to glue things of the form $D^2 \times U$ systematically, where $U$ are trivializing nbhds of the Hopf bundle.

I.e., use the transition functions of the Hopf bundle or something.

Right, I think that's exactly what you have to do.

@BalarkaSen "Trivializing neighborhoods"? "Transition functions"?

Define your topological space as $\coprod_i D^2 \times U_i/\sim$ where $\sim$ is defined as follows.

What does "trivializes" mean here? Something to do with looking like a product locally?

@AkivaWeinberger Have a look at the definition of a fiber bundle from somewhere. Yes, trivializes means the preimage of the neighborhood is a product.

Anyway, we define $\sim$ by gluing $D^2 \times (U_i \cap U_j)$ of $D^2 \times U_i$ to $D^2 \times (U_i \cap U_j)$ of $D^2 \times U_j$ by extending $\varphi_j \circ \varphi_i^{-1}$ above to a homeomorphism $D^2 \times (U_i \cap U_j) \to D^2 \times (U_i \cap U_j)$.

Note that I am using a crucial fact here that every self-homeomorphism of $S^1$ extends to one of $D^2$.

I think this should be your definition. @MikeMiller ^ is this right?

I don't know much about fiber bundles so it's better Mike looks at this and tells if it's ok.

Was that supposed to be $U_i\cap U_i$ there? (That is, $U_i$ twice)

Sorry, $U_i \cap U_j$.

@BalarkaSen: If you were doing this carefully you would want to use the fact that the transition functions of the circle bundle are linear, which have canonical extensions. You want this to be a smooth bundle hence a smooth manifold hence triangulated.

PVAL's construction = mine so works fine.

But, ignoring the details, the intuition is "fill in the circles"?

@MikeMiller What do you mean by linear transition functions when the fibers are circles?

Yes, @Akiva.

What could I possibly mean?

Multiplying by a matrix in O(2)?

Yes.

This is actually an $SO(2)$ bundle, which is even easier, but to get the canonical extension to a disc bundle all you need is $O(2)$.

OK, great. But why do we want to use the transition functions are linear?

Oh, I get it.

If my extensions vary fiberwise I'm doomed. I need a fixed extension.

This is all beyond me. I'll just trust that it works.

@AkivaWeinberger: What I would have said if you asked me is not to worry about the details.

Just understand the picture as you take circles and fill them in. Nbd.

@MikeMiller Now I understand what you meant when you said canonical extensions. That's good point, thanks.

I just realized you were implicitly using homeomorphisms instead of diffeomorphisms, in which case it's actually fine, there's a canonical way to extend a homeomorphism - you cone off.

Hmm, ok.

@AkivaWeinberger If you want to understand this filling construction, look at the Klein bottle. That's a circle bundle over the circle. Try to understand what happens when you fill in the circle fibers, maybe.

or the torus for something more eminently visualizable....

If you wanted to do this with diffeomorphism you'd run into trouble because there's no canonical way to extend

@MikeMiller Right. So my construction won't give me a smooth fiber bundle.

@akiva this whole conversation is going "woosh" over my head :/

in particular there's not actually any nontrivial homomorphism $\text{Diff}(S^1) \to \text{Diff}(D^2)$, and I'm only talking group-level here

That I didn't know.

Interesting.

Is anyone of you familiar with coding theory?

Right, if I recall correctly, the way one proves a self-diffeo of $S^1$ extends to one of $D^2$ is by isotoping the diffeo to the id/-id diffeo (you gave that to me as an exercise, I think). That's not canonical.

@MikeMiller how does one prove this?

well, you do it more or less canonically, since it actually gives a def. retraction $\text{Diff}(S^1) \to O(1)$

it just doesn't behave homomorphically

ultimately this doesn't matter; you just say "$S^1$ bundles are isomorphic to a unique $O(2)$-bundle, which I can fill in with discs in an obvious and way"

dumb terminology question: $S^1$ bundle = fibers are $S^1$ or base space is $S^1$?

the former

blah bundle over a blah

mmkay

@BalarkaSen: Previous statement is of course wrong. You need to restrict to the identity components of both (ie, orientation preserving diffeos of $S^1$)

oh, yes, I agree.

@Balarka a proof is here, though I didn't read it

@MikeMiller Thanks. I doubt I'll understand that, but I'll bookmark it for future.

why do you doubt you'll understand it?

Primarily (1) because I don't know much (as in, anything) about diffeomorphism groups. Secondarily (2) because I want to force myself to believe I won't understand anything except calculus right now to not get distracted.

(2) is an acceptable response. (1) is not. Stick with (2).

"Syndiffeonesis means difference-in-sameness. The idea is that no two things can ever be absolutely different, because to have the relation of difference they must be similar in that they both have this relation."

Syndiffeonesis: forums.philosophyforums.com/threads/…

@MatsGranvik Uh, what

@AkivaWeinberger: Anyway, the point is that whereas for maps to be null-homotopic they need to bound a (map from a) ball; for maps to be "null-bordant" they need to bound a manifold; and for them to be null-homologous is even weaker, roughly amounting to having to bound a singular manifold, with singularities in codimension 3.

So, the situation we're talking about is only the last of those? (I've got to go, will read your response later)

Why do you say "roughly"? Because what you said was right. It need to bound an image of a simplicial complex which is a manifold everywhere except singularities in codimension 3.

Roughly because I didn't define it.

Also didn't define "bounds".

@AkivaWeinberger: No, the thing we constructed was a manifold.

Fair enough.

In practice if you show some chain is null-homologous you'll either do it by algebra or by showing it bounds a manifold. You're probably not going to find some dumb singular thing.

Most of the time doing it by algebra is way more easier than looking for a thing it bounds.

ofc yes

clickhole.com/video/incredible-man-names-several-pokemon-2951

6? really? I probably can't name more than 2.

they had incredible in the title for a reason

I also like the sequel

lol!

This seems redundant, though. I'm repeating the (P - P * D) in two locations... Intuitively, I know there's a better way to express this, but I can't think of it

You could write that as (P-PD)(1+T)

the 1 says you're including the original price; adding T says you're adding tax. All I did is factor your expression

Mike, would you be kind (and patient) enough to walk me through the steps of how you got from mine to (P-PD)(1+T)?

Let's call P-PD "x". It's not a big deal, I just don't want the symbols to get in the way.

Your expression is x+xT. Agreed?

Yes

Now multiply out x(1+T).

Ok... Let me do it on paper

The rule you want to use is called "distributing".

Ahhhh! Of course!

A bit scary, though, that I couldn't see that!

You will get better at it with practice, as we all do.

Thank you

Calculating the tax first would of course be (P + PT)(1 - D)

Now can you factor that again?

Factoring it even more than you already have, I mean.

Hmm.. let me think about it for a minute

So, some naive question about the definition of the first cohomology group of a Riemann surfaces with coefficients in some sheaf $\mathscr F$: Why do we have to do this construction with taking the disjoint union of all $H^1$'s for different coverings and then take an equivalence relation?

P(1 + T) ?

Yup, you get P(1+T)(1-D).

wow! immediate improvement! haha

thanks!

Just a sec!

We had a minute ago that if we do discount first, we get (P-PD)(1+T).

What do you get when you factor that?

The opposite... P(1-D)(1+T)

Why can't I just do something more naive like... Define it to be the $H^1$ with respect to the open covering that consists of the full topology

@Mohamad: But does it matter what order you multiply in?

Addition and multiplication are commutative... so that means no...

Or in this case are we talking about the grouping? in which case it would be associative... and again no

I know we're going somewhere with this, but I'm not thinking about it the right way...

Well, you just decided that discount first is P(1-D)(1+T), and tax first is P(1+T)(1-D). What do you conclude?

they're equal

let me verify that with a number other than 100

You don't really need to verify. You just told me the key word: "Multiplication is commutative and associative". So if I want to multiply a bunch of numbers - such as, for example, P, 1-D, and 1+T - I can do it in whatever order I want.

So the original question was "Would you prefer to calculate tax first or discount first?" Can we come to a conclusion?

Yes, it doesn't really matter!

Ayup.

Learning math is becoming addictive and fun, but it's also humbling. I can't believe i stumbled this far in life without knowing the basics!

Godspeed to ya.

I grew up in Beirut during the civil war, and sometimes we missed our school days, other times we had no text books. Eventually, when we had them, they were in English, and we hardly knew any English.

Any place where some intuition is built for the definitions involved in sheaf cohomology? Something concrete?

@Danu: Give me a second.

When I got to England, I did my GCSE's, and somehow I passed them. Now it's all catching upto me.

Anyway, Mike, thank you so much for your help!

@Danu: The idea here is that if you include "large open sets", that might break what the sheaf cohomology group should be. Morally the sheaf cohomology group is what you get when you look at the values over little tiny open balls and see how they compare with values in other little tiny open balls.

@Mohamad You must have had a tough childhood. Great that you've decided to relearn math.

I suspect $H^1$ using the open cover that consists of every open set might actually just be zero, but I don't remember.

@Mohamad Gladly, and good luck.

@Danu $H^1(M,\mathcal O^*)$ is the space of holomorphic line bundles, and replacing that sheaf with the sheaf of continuous/smooth functions to $\Bbb C^\times$ is the set of (topological) complex line bundles. Understanding why this is true and how the correspondence goes should be the foundation of your intuition for sheaf cohomology.

@MikeMiller So the quotienting removes things that are too large

Yeah, that's exactly what it does. If you have a finer cover it says "OK, finer cover, you win! You win! I give!"

And instead of just finding the right cover to do the job, you can just say: "Well, let's just take the limit of covers as they get finer and finer and finer..."

Because if $A$ and $B$ both contain some $C$ then the quotienting will glue cocycles on $A$ and $B$ together

Maybe a good exercise would be for you to tell me what happens when the whole space is an element of the cover. I'd like to know.

maybe that wasn't so clear

@MikeMiller What happens where? Are you fixing one cover?

Yeah

Ah, you're looking at my proposal for taking just the full topology?

Just do $H^1$ for convenience of notation with some open cover, one of whose sets is the whole space

OK, I got the setup. So what are you asking about?

Tell me if it's zero

So if every cocycle splits?

@MikeMiller can I ask you a question related to algebra?

Sure

So I'm looking at some 1-cocycle $(f_{ij})_{i,j\in \cal A}$

I don't actually want to hear about it though ;D

Now I want to find a $0$-cochain $(g_k)_{k\in\cal A}$ such that $f_{ij}=g_i-g_j$ on $U_i\cap U_j$

what does it mean that $f$ is a cocycle, explicitly?

Kernel of $\delta$---equivalently $f_{ij}=f_{ik}+f_{kj}$ on $U_i\cap U_j\cap U_k$

There's a group $G$ which has order $6$. The element $a \in G$ has order $3$. There is another element $b \in G$ such that $b$ is not an element of $<a>$. We have to show that $e,a,a^2,b,ab, ab^2$ are distinct elements of $G$. Proving that $e,a,b$ are in $G$ is fine. And I've made the argument that $a^2 \in G$ as $<a> \in G$ and since $ord(a)=3$ $a^2 \in G$. How do I go about proving the same for $ab$ and $a^2 b$ though? Just a hint would suffice thanks @MikeMiller

Since $\delta$ sends $f_{ij}$ to $f_{ik}+f_{kj}-f_{ij}\in B^2$

maybe think about what that means in our setting.

I'm just trying to explicitly find a 0-cochain...

which is not what I just asked you to do. :)

You're asking me if $H^1$ is zero, i.e. if every 1-cocycle splits

So I'm trying to split it

dude I just said to think about what it means to be a cocycle in our setup

that will be a place to start

I have a hard time to seeing how having one particular open set (the full space) in my covering can make any difference for something like the cocycle condition which is for all triples of open sets

I guess I'll check what happens if I consider $U_i=X$

I see

Then $f_{ij}$ and $f_{ik}$ can be regarded as $0$-cochains

and I get my splitting

yeah, you define $g_j = f_{ij}$ if $U_i = X$.

trrrrricky

but this came from the cocycle condition, which is why you had to start there.

the point being that by having such a "big" element in the cover, you're restricting what all the terms in the cocycle can actually be.

Also, I feel super clumsy throwing these words "cochain" "cocycle" "coboundary" around all the time---do I come across as such, or is it just me?

they're ugly words but I haven't noticed you sounding awkward or whatever

@MikeMiller Is it correct to rephrase/expand that a bit as follows: If there is a very large open set, it makes the cocycle condition very restrictive by causing many things to automatically split, making $H^1$ trivial "too easily"

something like this

if you wanted you could extend the above to showing $H^j$ is also zero for all $j \geq 0$

Well, I'm only doing Riemann surfaces, and Forster takes a "minimalist approach", only defining what's really needed---so I don't even know how $\delta$ acts on higher cochains.

He only defined its action on the 0- and 1-cochains

should be easy to work it out with your teeth

get a solid bite in there, blood and guts and veins and all that

note to self: beware - mike is a vicious carnivorous.

Do Not Feed.

And I walked in and sat down and they gave me a piece of paper, said, "Kid, see the psychiatrist, room 604." And I went up there, I said, "Shrink, I want to kill. I mean, I wanna, I wanna kill. Kill. I wanna, I wanna see, I wanna see blood and gore and guts and veins in my teeth. Eat dead burnt bodies. I mean kill, Kill, KILL, KILL." And I started jumping up and down yelling, "KILL, KILL," and he started jumping up and down with me and we was both jumping up and down yelling, "KILL, KILL."

And the Sargent came over, pinned a medal on me, sent me down the hall, said, "You're our boy."

Woody Guthrie! I've recently started really appreciating country from the first half of the 20th century; I was previously completely unaware of its strong social themes.

arlo, not woody

D'oh

Son-of, though. Close enough!

for those who haven't heard it

Hey @AlexClark.

No idea what that is, as usual :)

reaaaal big

like moist?

a lot of people find that word uncomfortable, I think

somehow a generalized blah is not always a blah

similar to how a red herring is neither red nor a herring

?

@AlexClark yeah, at one point of time, I wanted to study algebraic geometry over an Artin stack, replacing the word "Artin stack" by "algebraic varieties" each time I saw them.

I don't know, because I never tried. Just wanted to do it.

I think it was just a failed joke @AlexClark

No, I mean Balarka's artin stack thing

Red herrings aren't fish... they're either something that distracts an investigation (points the wrong way) or possibly a character in "A Pup Named Scooby Doo" who acted as a red herring

@MikeMiller You'll never know now.

I have no idea what a kipper is

where are you learning your algebras?

@AlexClark What do you want to know?

@AlexClark: I remember finding this book useful

@AlexClark OK, so affine algebraic varieties you know about. Those are zero set of a bunch of polynomials in $n$ variables over field $k$, inside the affine space $\Bbb A^n$.

do you know singular cohomology?

But affine varieties are generally too big, in the sense that if you look at $k = \Bbb C$, and give your zero set the subspace topology from $\Bbb C^n$, you'll get noncompact things.

The compact analogue is what projective varieties are. This should be the key motivation.

then idk how helpful that book will be

there's a nice relationship in that given a compact Lie group $G$, $H^*_{dR}(G) \cong H^*(\mathfrak g)$, its Lie algebra using Lie algebra cohomology. this doesn't extend to the noncompact case. there's also the relationship for discrete groups, $H^*(BG) = H^*(G;\Bbb Z)$, where the latter is taken as group cohomology

so there's a natural question: is there a relationship between various flavors of group cohomology for a) noncompact b) positive-dimensional groups and the cohomology of BG?

and I remember ch3 of that book being helpful for that question.

Here's the definition. First, recall projective spaces. Given the affine space $\Bbb A^n$, the projective space is defined to be the collection of all lines through the origin inside $\Bbb A^n$. This is denote by $\Bbb P^n$. Rigorously, $\Bbb P^n = \Bbb A^n - 0/\sim$ where $\sim$ is defined by $(x_0, \cdots, x_n) \sim (\lambda x_0, \cdots, \lambda x_n)$ where $\lambda \in k^\times$.

Points of $\Bbb P^n$ thus correspond to lines through the origin in $\Bbb A^n$. Points in $\Bbb P^n$ are denoted by $[X_0: \cdots : X_n]$ (i.e., these are the eq classes of $\sim$). Note that due to the definition, $[X_0: \cdots : X_n] = [\lambda X_0 : \cdots : \lambda X_n]$ for $\lambda \in k^\times$.

A projective variety in $\Bbb P^n$ is a subset $X$ such that there are homogeneous polynoms $F_1, \cdots, F_m \in k[T_0, \cdots, T_n]$ for any $[X_0 : \cdots, X_n] \in X$, $F_0(X_0, \cdots, X_n) = \cdots = F_m(X_0, \cdots, X_n) = 0$. Note that this is well defined, as if a homogeneous polynomial vanishes on $(X_0, \cdots, X_n)$, it must also vanish on $(\lambda X_0, \cdots, \lambda X_n)$.

Sorry: there's a small typo I made above: replace all the $\Bbb A^n$'s above by $\Bbb A^{n+1}$ (but leave the $\Bbb P^n$ alone!) We're working in $n+1$ variables.

Example: consider $\Bbb P^2$ and the subset $\{[X:Y:Z] \in \Bbb P^2 | XY - Z^2 = 0\}$.

@AlexClark So now you know what a projective variety is. What's next?

no, the number of homogeneous polynomials need not be the same as the number of variables.

Look at the example I posted.

The number of variables is $n + 1$, the dimension of the affine space we started working with.

Oh and by the way, to relate to my previous statement of projective varieties being the compact analogue: let $k = \Bbb C$. $\Bbb C^{n+1}$ is a very noncompact object in the complex topology, whereas $\Bbb {P}^n$ (which inherits the quotient topology) is compact (you should prove this at some point of time).

@AlexClark You're reading about generalized Kac-Moody algebras?

@AlexClark Ah. I've encountered them in my physics studies.

In conformal field theory, they're used a lot

I lost my pen. Ugh.

I know, @AlexClark. What I wrote down could take you some minutes to digest. To be honest, I am still digesting it, even though I had stumbled upon projective spaces earlier in topology albeit from a different viewpoint.

i'm not working today so i've just been downloading books and putting them in my bookmarks bar

@AlexClark there are more solutions that just $[1 : 1 : 1]$.

Sure.

@MikeMiller All day?

we'll see how long I manage to keep doing this

@MikeMiller You must have a huge bookmarks bar.

@MikeMiller Ah, this is the end of my day ;)

@AlexClark: if you're working in P^n, which is n-dimensional, and cutting out solutions to k equations, you expect the solution space to n-k dimensional

Seems like I have to get off my bed and fetch another pen. Blah. Annoyed.

@BalarkaSen i haven't had a working laptop in a while so i have a new one. rebuilding the bar.