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17:00
what is the best english word for: "give" all subsets of A. specify, state or simply give?
@robjohn compute it
@Null context?
@DHMO $262537412640768743.9999999999992500725972$
@DHMO i have to say all subsets of A, that are linearly independent.
@Null more context.
@robjohn lol, that was a joke, ok?
@DHMO it's all fun and games until someone puts a $\sqrt{-1}$ out.
17:03
@robjohn ?
@DHMO If $A_i$ is linearly dependent: give all subsets $A'$ of $A_i$, that are linearly independent and for which $<A'>_{\mathbb{R}}=<A_i>_{\mathbb{R}}$ holds.
@DHMO There's an old adage: "It's all fun and games until someone puts an eye out."
thats the excercise
@robjohn oh lol
@Null you're making exercises now?
but poorly translated^^
17:05
Given that $A_i$ is linearly dependent, find all subsets $A'$ of $A_i$, that are linearly independent and for which $<A'>_{\mathbb{R}}=<A_i>_{\mathbb{R}}$ holds.
@DHMO yes :)
(How can $A_i$ itself be linearly dependent?)
if some vectors are expressable as a linear combination of the others
What the hell is $A_i$? A matrix?
It reminds me this:
69
A: Mathematical "urban legends"

Asaf KaragilaThis is a story that I heard from one of the postdocs from my university, which in turn heard it from one of the professor at the university (I didn't bother to verify with him as the source seems relatively reliable). The said professor was a postdoc in some university in the USA a few decades ...

ah, it's a collection of vectors.
17:07
> Question: "Show that G1 and G2 are isomorphic."
Answer: "We will show that G1 is isomorphic..." and some nonsense, followed by "Now we'll show that G2 is isomorphic..." and more nonsense.
an example for $A_i$: $A_1=\{(0,0,0)\}$
alright, got it.
mmh, $A_1=\{(1,1,1),(0,0,0)\}$. would then $\{(1,1,1)\},\emptyset$ be all linearly independent subsets?
94
A: What are some examples of colorful language in serious mathematics papers?

Bruno MartelliA paper of David Bachman-Cooper-White describes a proof that a hyperbolic 3-manifold containing large embedded balls has large Heegaard genus. As they say at the end of the introduction, a proper subset of the authors wish to subtitle this paper “Big balls imply big genus” whch is indeed t...

Gold
17:17
180
A: What are some examples of colorful language in serious mathematics papers?

Gil KalaiFrom the ground-breaking paper: On the complexity of omega-automata by Muli Safra Acknowledgements The author thanks his advisor, Amir Pnueli, for his encouragement and many fruitful discussions on this research. Moshe Vardi initiated this research by a most illuminating mini-course on ω-au...

@Astyx
I am definitely keeping these
69
A: Mathematical "urban legends"

Asaf KaragilaThis is a story that I heard from one of the postdocs from my university, which in turn heard it from one of the professor at the university (I didn't bother to verify with him as the source seems relatively reliable). The said professor was a postdoc in some university in the USA a few decades ...

@Astyx
simple: the author's contribution is to wrote the paper itself so that these contributors received their acknowledgements
what means $<A>_{\mathbb{R}}$? The span of A in R?
51
A: What are some examples of colorful language in serious mathematics papers?

PandoraFrom Vector Calculus, Linear Algebra, And Differential Forms. A Unified Approach. by Hubbard: When a matrix is described, height is given first, then width: an m x n matrix is m high and n wide. After struggling for years to remember which goes first, one of the authors hit on a mne...

17:28
Colorful language? You mean like swear words?
@JessyCat not really.
Just light-hearted phrases/sentences.
Samuel L. Jackson talking about how he's tired of these motherflklsding vectors on this motherflklsding plane.
3
@JessyCat where?
In many an internet meme.
good morning
17:31
Unfortunately, no
@meow-mix hi =)
@meow hi
One time I said something along those lines here, and I got put in time out.
@JessyCat people are stupid
I didn't even say the real word. I used an abbreviation.
17:32
@DHMO not much haha
Yes you are
@Null go into the corner.
You just don't know it yet
abbreviations are exactly as bad as the real words
at least put some effort in
LOL! <--abbreviation
easy verifieable or easily verifieable?
> where if you just say "f" in a sentence, it means you're implying ... "fools!" You're saying they are very foolish, all the people who don't understand hexadecimal.
context: the joke is "there are 10 types of people in the world; those who understand hexadecimal, and F the rest"
@DHMO this is actually a nice variation
there is some bada$$ery going on here today.
^ go cast your $-magic someplace else
17:39
13
Q: What would base $0$ be? How would/could it work?

Jamie SanbornIf I was trying to take the number $123$ in base $10$ and try and convert it into base zero I would do something like this: $123 = 100 + 20 + 3$ $10^{\log_0(100)} + 10^{\log_0(20)} + 10^{\log_0(3)}$ But $\log_0(x)$ is the same thing as $\dfrac{\log(x)}{\log(0)}$ and the log of zero is undef...

You have no power here! Begone! Before somebody drops a house on you, too.
::facepalm::
:P
Don't hurt yourself.
Anyway, this is the real reason I came in here.
0
Q: Can this problem be solved with eigenfunction expansion? Thought it could, then things got weird.

Jessy CatI am being asked to determine whether the following problem $\begin{align} u_{xyy}(x,y) + u_{xxy}(x,y)=0, && 0<x,\, y<1 \\ u(x,0)=u(x,1)=0, && 0<x<1 \\ u(0,y)=f(y),\,u(1,y) = 0, && 0<y<1 \end{align}$ can be solved by eigenfunction expansion. Although it doesn't explicitly state them as suf...

A guy already gave me an awesome hint that I am going to go try to work through now, but just in case it doesn't work out, thought I'd post here to see if anybody could help :)
climbs back up into her tree
@DHMO Srry I had to do something, but I am back now and is the number of middle digits b/2 -3?
Hey Guys with 2 dices what is the chance to get atleast 1 six (two 6s is fine aswell) and what is the calculation
17:53
$1 - {5^2\over 6^2}$
The complement of getting no 6s at all
hi @Semiclassical
why do some people prefer \lange instead of < for span?
Hi @semi
\langle and \rangle look nicer
i.e. $\langle a\rangle$ versus $<a>$
The latter looks ugly to me.
17:58
haha yeah
You also see it when doing quantum mechanics because of Dirac notation e.g. $\langle \psi | \phi\rangle$
Though if I'm going to be doing a lot of that I'd define Latex functions \bra{}, \ket{} to save room
@DHMO I think I found a bijection for $C:=\{y\in\mathbb{R}:\exists a,b,c\in\mathbb{Q}:ay^2+by+c=0\}$ from $C\to\mathbb{N}$. $\mathbb{Q}\times\mathbb{Q}\times\mathbb{Q}$ represents the solution, but not uniquely. Can't we say $\mathbb{Q}\times\mathbb{Q}\times\mathbb{Q}\times\{0,1\}$ represents exactly one solution of the quadratic equation? (since any quadratic equation can have at most two solutions, we cover all with this)
So then I get a bijective mapping from $\mathbb{Q}\times\mathbb{Q}\times\mathbb{Q}\times\{0,1\}\to\mathbb{N}$, don't I?
@robjohn haven't thought about it, if I"m honest.
a hint would be nice, though.
Actually, I just noticed something nice @robjohn
@Ted Hi
Is there a specific characteristic of a trapezoid inscribed in a circle? Besides that the opposite angles sum up $180$.
18:06
Hi @Astyx
At least one of the diagonals of the trapezoid is the diameter of the circle
@Pichi: Doesn't it have to be isosceles?
If I define $b_n=1+a_n/2^n,$ then I can reorganize the recursion relation to $b_n=\sqrt{b_{n-1}}$
@EternusVia I doubt that.
@TedShifrin I'm not sure
18:08
so therefore $b_n=b_0^{1/(2n)}$
Think about reflection perpendicular to the parallel bases of the trapezoid, @Pichi.
Hi @Semiclassic
And probably one can deduce it from that.
hi @ted
Hi @Ted
@TedShifrin What do you mean with reflection perpendicular?
Hi @Alessandro
18:10
I'm not familiar with that term
I mean: Take the perpendicular bisector of the parallel bases, and it is a line of symmetry of the trapezoid.
(Not entirely sure I've transcribed the algebra right, but I think the main thrust is sound)
Hi all!
hi @danu
Hi one @Danu.
18:14
should've been $b_n=b_0^{1/2^{n}}$, woops
Hi @Danu
@TedShifrin Oh and that bisector goes through the center of the circle right?
Yup, @Pichi.
I'm studying the Euler class from MS, and I'm having problem deriving an equation. Let $\eta$ be a vector bundle with total space $E$, fiber $F$ and base $B$.

They (MS) define the Euler class as follows: $\iota:(E,\varnothing)\hookrightarrow (E,E_0)$ induces a restriction homomorphism on cohomology. The class $u$, used in the Thom isomorphism theorem (Thom class? Fundamental cohomology class?) is then mapped to $\iota^*u$. There is an isomorphism $\pi^*:H^n(E)\to H^n(B)$. Then we define $e(\eta)$ as the class such that $\pi^*e(\eta)=\iota^*u$.
@TedShifrin Oooh right, that's the cases of all quadrilaterals right?
18:17
I don't know what you mean, @Pichi.
Now, they claim that if I have two bundles ($\eta$ with rank $m$, $\xi$ with rank $n$), the Cartesian product $\eta\times \xi$ has Thom/fundamental class $u(\eta\times \xi)=(-1)^{mn}u_\eta\times u_\xi$. I don't see where the signs come from---they must come from commuting the tow $u$'s past each other but i don't see where I need it. This is what I did so far [to be continued]
@TedShifrin I mean that the center of a circle that inscribes a quadrilateral is the meeting of the perpendicular bisectors.
@Danu: You should be addressing this stuff to Mike. I don't think about things like this at all.
@Pichi: Inscribes or circumscribes?
@TedShifrin circumscribes*
Not every quadrilateral can be circumscribed by a circle, though
18:20
So you can only circumscribe a circle around a quadrilateral when the opposite angles add up to 180º.
But, yes, if there's a circumscribed circle, its center is equidistant from all the vertices, hence lies on perpendicular bisectors of the edges.
I want to see if $u|_{F_1(b_1)\times F_1(b_2)}$ is the ''positive generator'', given that $u_\eta|_{F_1(b_1)}$ and $u_\xi|_{F_2(b_2)}$ are the "positive" ones. So I wrote out the following: $$(u_\eta\times u_\xi)_{b_1\times b_2}=(\pi^*_1u_\eta\smile \pi^*_2 u_\xi)|_{b_1\times b_2}=(\pi_1\circ \iota_{b_1\times b_2})^*u_\eta\smile (\pi_2\circ \iota_{b_1\times b_2})^* u_\xi$$
Now, the composition $F(b_1)\times F(b_2)\hookrightarrow E_1\times E_2\to E_j$ is given by $\pi_j|_{b_1\times b_2}$, as far as I can tell. So I have
$(u_\eta\times u_\xi)_{b_1\times b_2}=(\pi_1|_{b_1\times b_2})^*u_\eta \smile (\pi_2|_{b_1\times b_2})^*u_\xi$.
Now I should be seeing some signs, I guess... But I don't see any: $(\pi_j|_{b_1\times b_2})^* u_j$ should be the positive generators. To me, it looks like if the first part and second are both "positive" then the total thing should be too.
I guess this was just me brainstorming :P
@TedShifrin Thank you very much
@Danu: Once again, you should talk to Mike. But I bet something's fishy with your pullbacks. And I remember some issue with comparing cross product and cup product (for example, cup product of cohomology classes skew-commutes, but cross product does not).
@Pichi: You're welcome.
@TedShifrin Alright, thanks for your comment.
The proof of the Grothendieck lemma worked out now, by the way Ted :) Do you wanna hear how to complete it or is it already clear to you? WHat was left was to split the SES we had.
hi @TedShifrin
I feel so tired mentally lately.
18:31
@Danu: So what makes it work on $\Bbb P^1$ but not on $\Bbb P^2$?
Karim: Graduate school is hard, hard work.
Our algebra professor gives us too much work..
yeah
I always assigned lots of work, both undergraduate and graduate. Other faculty are too lazy to bother. So I'm probably on the side of your algebra prof. But I don't know.
There's @MikeM ...
we are supposed to hand in a project + 2 assignment for algebra. However, we have other stuff other like marking and studying for other classes.
Yeah I better get used to it. Graduate school is indeed a lot of work.
hi @TedShifrin
A project? What sort of project? ...
hi @meow
18:34
@TedShifrin a 20 page write up about something in algebra.
I am doing mine on elliptic curves and elliptic curve cryptography.
@Ted I know nothing.
I guess I haven't heard of a first-year graduate course requiring a project. Often this shows up for a more advanced course.
@TedShifrin So in the induction step, @Ted, some isomorphisms are used that use the fact that we're on $\Bbb P^1$, as well as the Riemann-Roch formula for curves.
@MikeM: Bulls*** :P Did you see Danu's Euler class question for you above?
I'd have to think about which steps really rely on $n=1$.
18:35
Yeah @TedShifrin that is what everybody says, but she mentions that she would like us to get prepared and stuff. However, why do we have to do an extra assignment in the same week :S.
and the assignment isn't trivial.
That seems excessive, Karim. How many problems in an assignment?
7 problems but they are hard.
@Danu Most definitely.
Well, Karim, I don't entirely trust you on what is hard and what isn't :P
because it is 7 problems and each problem has parts to it.
But most likely this is ridiculously excessive.
@AndrewT :) Heya.
18:36
Hellu!
@TedShifrin she is a new professor so she is probably out of experience.
Hi @Andrew, long time no see
No, Karim, she's just not lazy yet.
@Ted I know nothing unless you can prove it in a court of law with my lawyer present.
2
Yup. How's life, @TedShifrin and @Danu?
18:38
I will just power through it. The semester is almost done then I will get a month off.
@MikeM: I'll use that line on you in the future, then.
@TedShifrin also that stuff we discussed about the map $\mathcal O(a_1)\to E$ being of constant rank. You see, only on $\Bbb P^1$ do sections of $\mathcal O(1)$ vanish only at a single point. This might actually be not-so-easy to fix.
We're alive, @AndrewT. :)
I may neglect my marking duties a little bit but I am sure professors who I am marking for understand that now is the last month of grad school.
@AndrewThompson Pretty good---essentially done with that Huybrechts book at last (about 2-3 "propositions" left, one being Kodaira embedding).
18:39
Karim: Don't assume. Go talk with them.
@Danu Yeey. I'll be working through part of Huybrechts book on Fourier-Mukai transforms next semester.
Ah, so in higher dimensions, the zero locus will be a local complete intersection but not necessarily a global one. True enough @Danu.
@Ted the images of linearly indendent vectors under a linear transformation are linearly independent, right?
Yeah, I will go talk to them.
NOOO @meow
Only if you know something about the rank of the linear transformation.
But if the linear transformation is nonsingular (invertible), then of course it's true.
18:42
but the linear transformation is from the general linear group
Yes, then it's true :)
@meow mapping everything to $0$ is a linear transformation, right?
ok, thanks
@Alessandro yes, but this exercise requires that the linear transformation is an element of the general linear group
One of my standard exam questions in linear algebra to determine whether a student deserves an A in the course is to prove that if a linear map $\Bbb R^n\to\Bbb R^m$ has rank $n$, then it maps linearly independent vectors to linearly independent vectors.
@Ted is it ok if i did a proof via contradiction? or would you like me to write another one that just proves that the $c$ is constant for every vector
18:44
The logic of this is a hard proof for someone first learning proofs. (And they don't really think very well about matrices as linear maps in the introductory course.)
Oh, that problem. I think your contradictory proof won't really need the contradiction, if you write it out. I think you'll see it can be made direct (to prove that the two $c$'s are equal).
But do whatever you're comfortable with.
I've become interested in documents about (attempts at) doing "more interesting than usual" mathematics with school children (of ages all the way down to 0 years old). Does any of you have any recommendations?
@Ted Danu's using my least favorite definition of Euler class. So I'm a little apathetic, unfortunately.
@TedShifrin i probably could have proven that the basis vectors of $\mathbb{R}^n$ have the same $c$, then shown that any linear combination of the basis vectors have the same $c$, and thus, all of $\mathbb{R}^n$ has the same $c$
By the way, is there any way af getting around this terrible Thom isomorphism stuff?
I hate it.
@meow: I think you can choose $x$ and $y$ arbitrarily and show that $c_x = c_y$ by considering $x+y$.
18:46
@meow however images of dependent vectors are dependent, which can be read as independence works well with preimages, not images (like a lot of other stuff actually)
@Danu: I get around it all the time by using curvature :P But Pfaffian is not a cake-walk.
The Euler class is the dual class to the zero set of a generic section.
Then your question is trivial and obvious with this definition.
I usually prove that ^^^ as a theorem. ;) Well, and the generalization for Chern classes.
@TedShifrin Pfaffian?
You still have some signs to worry about with orientations, maybe, @MikeM?
18:48
Oh crap.
Pfaffians are weird
@Danu: Uh huh. For matrices in $\mathfrak{so}(2n)$, there is one invariant besides the usual symmetric functions of the eigenvalues (trace,...,det). There's also the pfaffian, which is a square root of the determinant.
For yesterday's question: Let H be a hyperbola with center Z. Points A and B are selected on H. Suppose that the tangents to H at points A and B intersect at a point C distinct from A, B, Z. Prove that line ZC passes through a point X in the interior of segment AB and determine the ratio AX/AB. <--- Why do we need to use affine transformations?
They do show up in physics, interestingly enough.
albeit in a very specific context.
Probably for the same reason they show up in math, @Semiclassic, but that's interesting to hear.
18:49
@TedShifrin I am interested how did you manage your time in grad school ?
@Hiro I don't think you -need- to, it just makes it a lot easier.
@Semiclassical How so?
It lets you reduce it to a special case, I think?
i.e. one where the calculation is as simple as possible
The signs are the last thing I worry about on every case.
I worked a lot, Karim. But I did have an NSF graduate fellowship for 3 years, so I TAed only one quarter my first year and then taught my fourth year and fifth year (sort of, as there was no further departmental support then, so I taught for my adviser when he needed me to).
18:50
Incl the paper I'm not writing.
LOL @MikeM
and then you can argue that said property will be preserved by affine transformations, so therefore it works for -all- hyperbolas.
to be honest, though, i wouldn't go through that
I'd probably just do it in a more or less brute force way. Not the slickest approach, but eh
@Hiro i'd assume because affine transformations preserve ratios of side lengths
I'm a big fan of affine geometry to prove yucky things. And projective geometry, too.
See, @meow, you learned something :P
I think you can assume, without loss of generality, that your hyperbola is of the form $\frac{y^2}{b^2}-\frac{x^2}{a^2}=1$
18:53
@meow-mix Define side lengths
You can assume $a=b$, even, @Semiclassic, i'm pretty sure.
@TedShifrin I see.
I'm not thinking about it.
Maybe, but you probably need to do some arguing for that
Well, if you can rotate it, I can then stretch appropriately, and the composition of affine transformations is again an affine transformation.
18:54
And if there's a straightforward route with more algebra, that seems fine as a first approach.
@Semiclassical are you talking about this kind of affine transformation: imgur.com/a/jdTg7
@Hiro affine transformations consist of translations and linear transformations
You can represent said transformations like that, but you don't need to
compositions, @meow
well yes
18:54
invertible linear transformations
it's important to say things correctly :)
Just need to understand what they do.
i.e. of the form $A\boldsymbol{x} + \boldsymbol{c} : A \in GL(n,\mathbb{R}), \; \boldsymbol{c} \in \mathbb{R}^n$
The point is that there's ways to change a drawing which, while the lengths change, the proportions don't (appropriately defined)
Suppose I have someone whose head is as far from their waist as their waist is to their feet
what does transforming the hyperbola have to do with proving that a line ZC passes through a point X in the interior segment of AB?
iirc you can assume the hyperbola is of the form $x^2-y^2-1=0$ if you're interested in it up to affinities or $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ if you're working with isometries
18:57
It's with an eye towards that last bit of the question, @Hiro, about the ratio
@meow: You'll find that the exercises get way more involved and interesting. Projective duality is cool, and then there are interesting ones like 7, 11-17 (watch out: 18 is false), 19-20, 25-27.
@Alessandro: You are, as usual, correct.
If I now take a picture of that person and stretch it vertically, then their height will change but that property---distance of head to waist = distance of waist to feet---will stay the same.
So if someone changes the picture in that kind of way, I'll know that that property will stay the same
I can't read everything :P
Gone for now ... perhaps until evening.
Just at the 9/11 memorial. They built a mall next to it. Of course they'd like to make profits off of their dead.
later @ted
18:59
Whoa, you're in NY, @MikeM?
New York, New York, it's a hell of a town...
Have been since Wed.
I didn't have any idea.

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