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9:02 PM
Goodnight @MikeM, @Tobias.
 
hello.. Can someone suggest good resource about Hochschild cohomology of banach algebras
 
@Ted: There was a good question earlier. Can you decompose $\Bbb R^2$ into subspaces homeomorphic to the circle?
 
I assume it's easy to see there's no such fibration?
 
Yes, or foliation (if you could foliated the disc with the boundary as a leaf, double it to get a foliation of the 2-sphere by circles, contradicting hairy ball)
But can you foliate $\Bbb R^3$ by circles? :)
 
9:12 PM
Whoa, now you're assuming smooth, not just homeomorphic ... and you're assuming the foliation extends to the boundary.
I meant fibration as opposed to locally trivial fiber bundle (which of course there can't be).
 
I am well aware of why I'm saying. When I say foliation here, yes, that's strictly weaker. And no, there's no such boundary assumption. A circle bounds a disc by Schoenflies which is then foliated.
 
Right, but why can't there be Reeb behavior? Why does the boundary component fit nicely into the local product structure of the foliation?
 
Fibration by circles have a simple local gorm which still forces it to be impossible. A neighborhood of a circle in a surface fibered by circles is either a cylinder or a möbius bNd.
 
Do you know the answer to topological fibration? I don't.
Your neighborhood statement applies to foliations on manifolds with boundary?
That begs the question, anyhow. I'm still not sure it's a foliation when you close it up.
 
@Ted: These are all questions that you could certainly answer. In any case, the answer I wrote to this assumes no more than the abiliry to decompose into disjoint subspaces homeomorphic to the circle, no local triviality assumed like in a foliation.
 
9:16 PM
Ah, ok.
I'm too busy keeping Balarka busy on email learning some single-variable analysis :P
 
But you can foliate $\Bbb R^3$ by circles :)
And here I really mean foliate.
 
@TedShifrin I am still thinking about your last email. I think I know what you're hinting at.
 
Oh, mentioning his name made him appear. It's worse than a genie.
The moral of the story, @Balarka, is that there is no converse to Taylor's Theorem.
 
I was here all the time. I have minimized my chatting to learn mult. calc.
 
Well, now we've regressed to sing. calc. :P
 
9:21 PM
But your problem is hard :(
 
@MikeM: I guess I don't know this. I don't see how to get it from the Hopf fibration.
 
You can also foliate $S^3$ by lines, which I believe is possible on $S^2$, but I still find it less absurd.
 
@MikeM: what do you mean by "lines"?
 
@TedShifrin Well, but I did prove that if $f$ is $C^1$ on a small interval around $0$ then $f(x) = o(x^2)$ does imply $f$ is twice differentiable. So there should be some sort of partial converse.
 
9:22 PM
@balarka have you come up with an example yet? :)
 
Nope, @SemiC. But Ted gave a hint.
 
Your proof is still suspect, @Balarka. You assumed differentiability of the error, and I think that's begging the question.
heya DogAteMy :)
hi @Semiclassic
 
Differentiability of the error term on that small interval follows from differentiability of $f$ on that small interval.
 
9:25 PM
@TedShifrin In the e-mail, are you hinting at something of the sort $\sin(1/x)$ where $x$ is rational and $0$ otherwise or something?
 
Why does $e'(h)/h\to 0$? (You're right about differentiability. I thought an $h$ had been factored out.)
No, @Balarka. My question with the rational/irrational answer is unrelated to the "right" sort of counterexample.
 
Ah, OK.
 
the example i came up with is of the $\sin(1/x)$ kind
 
What's the problem, if I may ask?
 
though i'll hold off on spoiling it for now
 
9:28 PM
Example of a function $f(x)=o(x^2)$ that is not $C^1$ at the origin, @Semiclassic?
That ^^^, DogAteMy.
 
it was intended as such. though that doesn't sound quite like how i heard the problem
 
That's why I asked
 
way i heard it was whether $o(x^2)$ implied existence of $f''(x)$ at zero
but that's $C^2$ not $C^1$, isn't it?
 
So, we want an $f$ where $\lim_{x\to0}\frac{f(x)}{x^2}=0$ but $f'(x)$ is not continuous (EDIT: at zero)?
 
9:30 PM
No, differentiable implies continuous, so $f'$ must be continuous in order to be differentiable.
Right, DogAteMy. Not continuous at $0$.
 
@Ted: What could I possibly mean by lines?
Oops, I meant to say "Which I believe is im possible on S^2"
Since it is impossible.
 
@MikeM: We don't need to play these irritating games. You mean foliating $S^3$ by noncompact $1$-manifolds?
 
@TedShifrin By the way, Shifrin, I watched one of your videos
 
then i must've misunderstood. the example i came up had $f(x)/x^2\to 0$ and $f'(x)\to 0$ but $f''(x)$ not continuous at zero i.e. not $C^2$
 
You're a pretty nice teacher
 
9:32 PM
@Semiclassical A ctrexample to Ted's question would be a ctrexample to that.
 
@Ted: Of course I do. I really and truly do not know what else I could have meant. That's why I asked.
 
Thanks, I think, DogAteMy ...
 
A ctrexample to Ted's question would be a ctrexample to that.
Ugh, internet.
 
@MikeM: Well, I had been thinking about filling up projective space with actual projective lines, so I wanted it clarified. Topologist-speak is not always universal.
 
$x^3\sin(1/x)$ is $C^1$, right?
 
9:34 PM
Gotcha, thank you for the clarification.
 
Yes, it is $C^1$, DogAteMy.
 
i might as well show my cards if my example isn't an example
 
@AkivaWeinberger It is.
@Semiclassical I feel better now that your example isn't an example for some reason! :P
 
@Balarka: How is it that your homework turns into homework for the whole chatroom? :D
 
I couldn't get sin(1/x) to work any way I could think of.
 
9:35 PM
Wait
I think I might have it
 
$f(x)=x^3/(1+x \sin(1/x))=o(x^2)$ is continuous and differentiable but not second-differentiable at the origin.
 
But I'm probably missing something
 
that was my example.
 
@Ted: Nothing else here is both interesting and accessible.
 
This is accessible. It's questionable if it's interesting. Although misconceptions about Taylor are actually interesting.
 
9:36 PM
@TedShifrin Well, it was a pretty public homework.
 
LOL .. I'm teasing.
I'm trying to grade topology homework — generalized tube lemma.
 
both $f(x)$ and $f'(x)$ squeeze to zero at zero, but that breaks down for $f''(x)$
 
I have little intuition for that quotient, @Semiclassic.
I would recommend sticking to products.
 
nor I, if i'm honest. i found it by playing around with mathematica
 
@TedShifrin How about this:
k phorq gvzrf fvar bs bar bire k fdhnerq
 
9:38 PM
WTF?
 
mathematica is cheating.
 
monoalphabetic cipher?
 
Copy/paste that into the link, because possible spoilers
 
pfff, plotting isn't cheating
 
Oh, good grief, just type the function.
 
9:38 PM
Yes it is.
 
$\displaystyle x^3\sin\left(\frac1{x^2}\right)$?
 
Applause to DogAteMy.
 
nicely done.
 
(and $0$ at $0$, of course).
 
9:39 PM
That makes sense.
 
Note that this phenomenon generalizes as much as you want to ruin arbitrary Taylor polynomials.
 
Wait, plotting is cheating?
 
@Ted: Just saw your previous ping. The construction of the foliation by circles was in 1989.
 
Oh wow, @MikeM. Can you hint to me how to visualize it?
 
9:40 PM
No, I haven't read it. :D
 
I don't think plotting is cheating. However, I think you should have come up with this by tweaking the usual $x^2\sin(1/x)$ example yourselves.
 
Yeah, I didn't think of tweaking the $1/x$.
 
A lot of deep mathematics (even differential geometry stuff with minimal surfaces) was proved only after computers drew pictures to suggest certain esoteric surfaces existed.
 
nor I
 
Anyhow, @Balarka, this makes you reassess the multivariable situation, I presume.
 
9:42 PM
yeah. regardless of whether computers are good at proving things, they're great for discovering things
 
@TedShifrin Yep.
 
I think computers actually proved things from Russell and Whitehead's Principia back in the early days of computing
 
DogAteMy: Did you at least pick a lecture on stuff so you'd learn something?
 
@Semiclassical: One topologist I have a lot of respect for is Nathan Dunfield, who does a lot of computer work. Part of the reason I believe the L-space conjecture so strongly is because he checked it for, like, a million manifolds.
 
i don't suppose you've got a lecture on bundles, @ted?
 
9:43 PM
Yeah, I chose that one where you introduced $dx_{i_1i_2\dots i_k}$
 
@mike nice
 
Differential forms
Still no idea what the connection is to calc
 
LOL, no, @Semiclassic. I didn't think to make an industry out of having students tape my lectures.
 
fair enough
 
Hello people.Is anyone interessted in open-source project with goal of creating mathematical free textbooks that are on the par with standard books?
 
9:45 PM
if i were to make a stab at the definition of a bundle, it'd be
 
@TheCoolDrop Hi again
 
@AkivaWeinberger What was your intuition behind choosing that function?
 
Thi is a good example.
 
The connection, DogAteMay, is that you need determinants to do $n$-dimensional volume, and hence to do $n$-dimensional integrals.
 
9:45 PM
@AkivaWeinberger heloooo :D
@aki
@AkivaWeinberger hoped I would meet some new people :D
 
$\pi:E\to M$ with $\pi$ being a sufficiently nice function (i should probably be invoking something about $\pi^{-1}$, but um)
 
@BalarkaSen I knew the function needed to get really small, but I wanted the derivative to stay pretty large
 
@TheCoolDrop o/
 
The graph explains it pretty well, I think
 
@Ted: Thwres a line in the Besse book which is like "Yeah so here's a guess about what Ricci curvature means. It's the trace of curvature, and trace is the derivative of detwrminant, and determinant is volume, so the Ricci curbsture is probably related to volume."
 
9:46 PM
@Hippalectryon what does that mean?
 
I found this to be a bit of a stretch but I guess it worked out.
 
Salut, M le Méchant.
 
@AkivaWeinberger But that happens with $\sin(1/x) x^3 $ too.
 
@TheCoolDrop Just a guy saying hi :-) like \o/ but with just one arm
 
No
Graph it, @BalarkaSen
 
9:47 PM
I probably wouldn't have recognized $\sin(1/x^2)x^3$ works at first glance.
 
The slope of the tangent mellows out
 
Actually, I think that's a good heuristic, @MikeM. I would have said "probably related to variation of volume."
 
@TedShifrin o/
 
@Akiva OK, you're right. Fair enough.
 
We should ask Hippa the question of the day.
 
9:48 PM
Wait, "trace is the derivative of determinant"? I thought it was the sum of the diagonal entries! @MikeMiller
 
@TedShifrin ?
 
@Hippalectryon oh well nice .Do you happen to take interest in taking part in the project ?I could link you with pdf of the work so far,its only 10 pages but I would appreciate input
 
it can be more than one thing
 
@Hippa: If $f(x)=o(x^2)$ can you conclude that $f$ is $C^1$ at the origin?
 
@TheCoolDrop What's the project ? I just arrived
 
9:48 PM
@Akiva: Yes.
 
@TedShifrin No. I wouldn't even be surprised if you could make a function nowhere derivable and still $o(x^2)$, koch style
 
Derivative of determinant must look horrible.
 
DogAteMy: Take the derivative of $\det (I+tA)$ at $t=0$ and you get the trace of $A$.
 
not always @balarka
 
@Ted: Fair enough. I was amused by a lot in that introduction. Another line is something like "Ricci curvature was defined by Ricci, but he doesn't seem to have done anything with it. We can't really blame him, since it's not obvious what to do with it."
 
9:50 PM
@Hippalectryon basicaly I would like to make series of mathematical book,created by community,driven by open source to develop.They should be real books,unlike wikibooks which are basically lists of facts with no monologue of author.Personally I look up to "homotopy type theory" and "stacks project",but this project would focus on teaching and not research
 
LOL, I like books with humor, @MikeM, although I've been criticized for it.
 
@TedShifrin Ah, I see
 
there's a formula i've used a lot in the past: $\text{tr}\ln M = \ln \det M$
 
DogAteMy, this is not unrelated to our differential forms discussion, btw.
 
Interesting.
I should watch the sequel video.
 
9:51 PM
@TheCoolDrop Well I'm happy to help if I can :-) I'm just a student though
 
@Semiclassical That's a cool formula.
 
That's just the algebraic property of log, @Semiclassic. Less interesting, somehow.
 
eh, it's pretty durn useful
 
$\det e^A = e^{\text{tr}\,A}$.
 
I believe it
 
9:52 PM
@Ted: I'm enjoying the book. :) Maybe this is how I'll actually get a good grasp on Riemannian.
 
that's easy to prove if $A$ is diagonalizable
 
Easy for diagonalizable ...
Yup ... then need density of diagonalizable/$\Bbb C$ for the easiest proof.
 
@Hippalectryon I am a student too,and not even student of mathematics at that.I believe that experts became experts by writing their work,even though most people believe that one has to be expert to write books.Of course I can link you up to the project so far,I will compile my work and give you a link in a second and you can tell me what you think.Its about set theory,introductory level, ZFC
 
Do you get that with all functions? That ${\rm det}\ f(A)=f({\rm tr}\ A)$?
 
9:53 PM
Speaking of, @Ted, I saw your proof of symmetric matrices having at least two eigenvalues.
 
Noooo, DogAteMy.
Huh? @Balarka
 
i mean, it definitely doesn't work for $f$ being the identity map
trace isn't determinant
 
Right
Yeah
 
@Balarka: How about 1: R -> R?
 
9:54 PM
@TedShifrin Well anyway for any continuous function there exists a nowhere derivable function that is as close as desired to the original function
@TheCoolDrop Link me :D
 
reason brought it up, though: if $M$ is $t$-dependent, then you get $$\text{tr}\frac{d}{dt}\ln M = \frac{(\det M)'(t)}{\det(M)}$$
 
That's not too surprising, though.
 
@Hippa: You can't perturb functions with $f(x)=o(x^2)$ and make them nowhere derivable! They have to have derivative $0$ at $0$.
 
@TedShifrin Well, nearly nowhere derivable. That's good enough. You just need it to be densely non derivable
 
ahem.
 
9:55 PM
which gives the derivative of $\det(M)$ in terms of $\det M$ and the trace of the derivative of $\ln M$
 
@Hippalectryon github.com/VanioBegic/Micromathics should you want to contribute,its open source,I welcome all input,authors will not be given credit individually,but as one community
 
LOL... yes, but all I wanted was to make Balarka et al give me a concrete example. They have.
 
@Ted: ??? Why not find an arbitrarily close smooth function and add an arbitrarily small non-differentiable function??
Oh, without breaking O(x^2)
 
Right ... without breaking ...
 
9:56 PM
@MikeMiller Ok, symmetric n x n matrices with n > 1.
 
Are you referring to the proof of the spectral theorem, @Balarka?
 
@TedShifrin Well if $f$ needn't be continuous it's likewise clear, just take $f(x)=x^3,x\in \Bbb{Q}$ and $f(x)=x^3+x^4,x\in\Bbb{R}-\Bbb{Q}$
 
@Balarka: How about 1: R^2 -> R^2?
 
combining that formula with the expression for the adjugate matrix gives fun things as well
 
@Hippa: I think you and I may have discussed this in the past. This arose because you can have something that looks like a Taylor polynomial for a function that doesn't have one :)
 
9:58 PM
We can even make the derivative unbounded, by the way — consider $x\mapsto x^3\sin(1/x^3)$
 
Right, @Hippa.
 
Still $o(x^2)$
 
Yup, DogAteMy
And you can jazz this up for $k$th derivatives, too.
 
@TedShifrin No, here's the proof: $Q$ be the quadratic form corresponding to $A$. Maximizing Q subject to $g(x) = ||x||^2 = 1$ gives eigenvalues of $A$. There must be two of them, because a sphere is compact.
 
I think we can make the derivative be equal to any fixed $g$ almost anywhere, in fact (though we'd need to break continuity)
 
9:59 PM
No, not as you said, @Balarka.
 
(i really want to say $\text{tr} \dfrac{d}{dt} M = \text{tr} (M^{-1} M'(t))$, but I forget if that's true)
 
You have to maximize and minimize ... or maximize and then maximize the restriction to the perp.
The correct version is an exercise in my book, @Semiclassic, but I'm too lazy to look it up.
Heya @PVAL
 
Yup, I am being sloppy. Critical points of $Q$ subject to $g = 1$ are precisely the eigenvectors of $A$.
 
@Ted: The Weitzenbock formula should hold for any vector bundle equipped with a metric & connection, yes?
 
yeah. i remember that the adjugate formula for $M^{-1}$ plays a role
 
10:00 PM
No, still wrong, @Balarka.
 
And there are two: the global maxima and the global minima.
 
@Balarka: If you note, I already gave you a counterexample.
 
oh, here
In matrix calculus, Jacobi's formula expresses the derivative of the determinant of a matrix A in terms of the adjugate of A and the derivative of A. If A is a differentiable map from the real numbers to n × n matrices, Equivalently, if dA stands for the differential of A, the formula is It is named after the mathematician C.G.J. Jacobi. == Derivation == We first prove a preliminary lemma: Lemma. Let A and B be a pair of square matrices of the same dimension n. Then Proof. The product AB of the pair of matrices has components Replacing the matrix A by its transpose AT is equivalent to permuting...
 
You need to restrict to orthogonal complements to get them all, @Balarka.
 
@TheCoolDrop What's the git command to create new repos from github ?
 
10:01 PM
Oh, all critical points ... that's right. You just don't know you get enough of them.
 
The adjugate is the determinant times the inverse, right? @Semiclassical
 
right @akiva
which, actually, justifies what I was saying above as what I wanted to say was true
 
OK, I'm going to finish grading point-set topology homework. Y'all misbehave without me.
 
only difference between the Jacobi formula and what I wrote is where the determinant sits
 
But that's what we did last time @TedShifrin
 
10:02 PM
@MikeMiller OK, let me get this sorted out.
 
@TheCoolDrop Nvm found it
 
The identity matrix does have at least two eigenvectors. Just not two eigenvalues.
 
main use i had for it was the case of $M(t)=1-t A$
 
@Hippalectryon I just checked it out and my father shut down my laptop during work,I lost a section and a half :D
 
since then the determinant is just (up to a factor of $t^{\dim M})$ the characteristic polynomial of $A$
 
10:04 PM
@TheCoolDrop What are the latex requirements ? \section{} outside of \begin{document} doesn't work in my editor for instance
 
@MikeMiller @TedShifrin The correct claim is "Every symmetric matrix have at least two linearly independent eigenvectors".
 
@TheCoolDrop Ah wait, do I need to just compile main.tex ?
 
Speaking of linear algebra, @Semiclassical, there's this neat proof of the fact that if $I+AB$ is invertible then so is $I+BA$
 
i always forget what the name of that identity is
 
@Hippalectryon yes but there is missing command in introduction ,the last line should be \end{axiom} it needs to be added to be compiled
 
10:04 PM
This follows from taking the quadratic form, extremizing it on the sphere (the extremums are precisely the eigenvectors of the matrix) and then use max value theorem to conclude there are at least 2.
 
@Hippalectryon please give me some way to contact you,I have some family business now,and we can continue the work tomorrow or later.Thanks for collaboration :D
 
Two linearly independent eigenvectors. :)
Or else you have not stated much.
 
@AkivaWeinberger That can be done by the power series thing, right?
 
10:06 PM
@MikeMiller Done. Apologies for my sloppiness, I think I am done with math for the day.
 
the proof on the Wiki page seems direct enough
 
@Semiclassical I didn't know they had the same determinants
 
Me neither.
That's a cool fact.
 
I just knew that they had the same invertibility status
 
yeah. though that definitely implies the above fact, since one determinant vanishes iff the other vanishes
 
10:09 PM
72
A: Jokes in the sense of Littlewood: examples?

Richard BorcherdsIf $1-ab$ is invertible for $a$, $b$ in a (noncommutative) ring then so is $1-ba$. Proof: $$(1-ba)^{-1} = 1+ba +baba+\cdots = 1+b(1+ab+abab+\cdots)a = 1+b(1-ab)^{-1}a,$$ The meaningless infinite series give the right answer (which is hard to guess).

 
@Balarka: Do you want an exercise anyway?
 
I'd like an exercise.
 
That version is slightly different, but just replace $b$ with $-b$
 
I do, @MikeMiller.
 
or multiply both of $a$ and $b$ by $i$
 
10:10 PM
But only if a solution would be publishable.
 
there's a version of that which I remember using a lot when doing linear algebra stuff in QFT
 
@PVAL What about the 4 dimensional smooth Poincare conjecture?
 
@Semiclassical Actually, this is a bit more general; it holds in any nonabelian ring!
 
@BalarkaSen Already thinking about that one.
 
Aw.
 
10:11 PM
Is that still open?
 
I suspect a solution to that would be publishable.
@Balarka: What on?
 
@MikeMiller Preferably calculus or analysis. Secondarily linear algebra or topology.
 
I'm certain some somewhat respectable journal like say the Annals of Mathematics might publish such a solution.
 
eh, that it works for nonabelian stuff doesn't surprise me. stuff not commuting is the main trouble in those kinds of calculations
 
@AkivaWeinberger Very much.
 
10:12 PM
Even after Perelman did his thing?
 
That's dimension 3.
This is 4.
 
@Balarka: Show that if $1/p+1/q=1$, then $L^p$ defines in an obvious way a subset of the dual space $(L^q)^*$. Show that $L^1$ is not all of $(L^\infty)^*$, but that $(L^2)^*=L^2$. You are free to replace these with sequence spaces instead of function spaces if you want.
 
@BalarkaSen Don't know if this is related to what you're learning, but can you prove that if $A$, $B$, and $C$ are all closed and cover $S^2$, then there exists $x\in S^2$ such that at least one of those contains $\{x,-x\}$ as a subset?
I.E. If three closed sets cover the sphere, at least one of them contains a pair of antipodal points.
 
I will crush him in a trash compactor if he can't prove that.
 
You need Borsuk–Ulam or something like that, though
 
10:16 PM
It seems strange to state a problem and then state the solution.
 
Oh
I didn't realize how simple the connection was
Well, I have a follow-up
 
@MikeMiller All the papers I've read must have been doing it wrong.
 
If $A$, $B$, $C$ are all closed and subsets of the sphere, define $D:=S^2\setminus(A\cup B\cup C)$ to be the rest of the sphere.
If $A\cap B\cap C=\varnothing$, prove that one of the four sets contains antipodal points.
 
@Balarka: Alternatively, find a 3-manifold that's prime under connected sum that covers a 3-manifold that's not prime under connected sum. However, show that if $M$ is prime under connected sum so is any cover of it.
Prime here not actually meaning prime. Instead I mean you cannot write something as M # N. Normally you would call that indecomposable but that word is taken.
 
10:19 PM
Now I have too many problems.
 
Me too.
 
@BalarkaSen but a bitch ain't one
 
@MikeMiller I think I will do this one first.
 
Got to go.
Meaning, I'm leaving my computer but I'm taking my phone, so I'll still be in the chat
Just, less
 
I don't know if that has an arrow pointing to something.
@PVAL: I'm more confused about what I'm working on than when I started this weekend.
 
10:22 PM
Yes, I was pointing to the prime problem.
 
@MikeMiller Does this mean you are no longer a "topologist"?
 
Sure, it's very believable that there is such a 3-manifold. I can come up with one.
 
@PVAL: Who's to say?
 
Ah so its an open question.
 
It's still 3-manifolds so I don't think I've changed.
Someone should go through the effort to update Kirby's list.
 
10:29 PM
I can't visualize $3$-manifolds :(
 
Spend enough time on it and you can.
 
When do I get to visualize $4$-manifolds?
Or do I just visualize $n$-manifolds and set $n=4$?
 
When you spend enough time on it.
 
I'm giving you a look
 
If I spent enough time trying to care I'm sure I would be able to.
 
10:36 PM
*throws linked spheres at Mike*
 
You think of $\Bbb R^2$ and set $\Bbb R=\Bbb C$.
 
*despairs*
By the way, an equation for a (Clifford) torus in $\Bbb R^4$ is $“x_1^2+x_2^2=x_3^2+x_4^2=1”$. Is there a similar thing for the projective plane?
 
Replacing R with C has historically been useful.
 
True
OK, torus in $\Bbb C^2$: $|x|=|y|=1$
 
I'm confused by what you mean "Is there a similar thing for the projective plane".
 
10:42 PM
Coordinate equations for something homeomorphic to the projective plane
Non-intersecting
 
are you asking if there is an embedding of the projective plane in $\Bbb R^4? Or are you specificaly interested in it being cut out by polynomials?
 
Well certainly theres both
 
The second is probably a good deal harder than the first.
If one takes an immersion RP^2 \to R^3, you just perturb it near the intersections to make it into an embedding into R^4.
 
Wouldn't be very elegant, though.
Also, not sure how to do that with polynomials.
 
10:47 PM
Look at Wikipedia. The algebraic embedding is easy to write down.
 
Where in Wikipedia are you looking?
 
In mathematics, the real projective plane is an example of a compact non-orientable two-dimensional manifold, that is, a one-sided surface. It cannot be embedded in standard three-dimensional space without intersecting itself. It has basic applications to geometry, since the common construction of the real projective plane is as the space of lines in R3 passing through the origin. The plane is also often described topologically, in terms of a construction based on the Möbius strip: if one could glue the (single) edge of the Möbius strip to itself in the correct direction, one would obtain t...
 
Oh, I didn't search the "real" part
I was looking at the "projective plane" article
 
I don't think I would have come up with that embedding.
 
hi @BalarkaSen
haha I just computed homology of real project plane
 
10:55 PM
@PVAL Doesn't look unique
 
@MikeMiller I have some incomplete ideas on the second question (i.e., covering space of prime manifold is prime), but I am sleepy. I think I'd rather think about this problem tomorrow.
 
Um of course it's not unique
 
hey @BalarkaSen
 
Hi.
 
It isn't "obvious". I'd like to say it isn't "tautological" but that has a precise technical meaning.
 
10:57 PM
No, I mean, the third coordinate looks really arbitrary
And the $2$ in the last coordinate seems unnecessary
 
The first idea is to note that if $M$ is prime, then $\pi_1(M)$ cannot be written as free product. But if some cover is prime, then this has a subgroup which is a free product. I am not sure if that is possible, but again, no proof.
 
There's a more obvious emvedding in (n+1) x (n+1) matrices: it's the space of orthogonal projections of rank 1
 
So that's of dimension $9$ in our case
 
The second idea is to recall that if $M$ is orientable prime, then every embedded sphere bounds a ball. So given a cover $\tilde{M} \to M$, an embedded sphere in $\tilde{M}$ can be pushed down to $M$ where it bounds a ball. But that doesn't necessarily mean it bounds an embedded ball in the cover, although it is nullhomotopic. Nothing final here either.
 
Would that embedding be polynomial?
 

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