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1:36 AM
Hello. I would like to discuss a small question in algebraic geometry.
 
no one relevant in the house at the moment.
 
@copper.hat Alright. Thanks!
 
are there any relevant people? i'm kinda surprised there isn't more algebraic geometry in the chat. it was everywhere when i was in grad school.
 
1:59 AM
\o
 
OK, great. let A be an abelian variety over a field of characteristic 2.
that's all i've got. i just want us to let A be that.
 
2:17 AM
summarily dismisses characteristic 2
then dismisses all finite characteristic
 
2:34 AM
let A be an elliptic curve over Q?
 
 
1 hour later…
3:51 AM
would an algebraic geometer be called an aggie?
maybe algae?
 
No, only if from Texas A&M
 
If I have $u_x + u_y = g(y - x)$, the homogeneous solution is $u = f(y - x)$. My understanding is that a particular solution can either be $x\int g(w)dw$ or $y\int g(w)dw$, where $w = y - x$, but when I plug in the guess I'm getting $\int g(w)dw = g(w)$. Should this type of guess be working?
 
4:20 AM
Does anyone know of any programs that do 3D graphing? Where you can just type in the equations and get the result?
 
wolfram alpha can do it. even google can do it for simple graphs, although i haven't figured out how to force this behavior. google.com/…
 
The fact that I didn't know Google possessed this ability -_-
 
i dunno what the best free, non-web thing would be. are you trying to generate figures for something, or just explore graphs, or what?
 
google knows all
 
4:24 AM
algebraic geometry is nothing compared to the us tax code.
 
Mostly just explore graphs. I'm doing some problems and I'm tired of working off of my ugly drawings
 
wolfram alpha can probably plot most of what a multivar calculus book would throw at you.
 
i use macsyma, but i am old school
 
Alright, I'll check out all of these and see what works best for me.
 
you use it to graph designs of the wheel (prototype expected 400BC)
 
4:26 AM
Thanks everyone
@leslietownes lol what
 
i was making fun of copper's age, or macsyma's age. take your pick. maybe both
 
Ooh, I see
 
i remember people using macsyma in college but we had the wheel by then.
maxima is still in development, i think.
 
I've never heard of macsyma
 
the sheer number of punch cards required to get it to graph z = x^2 + y^2...
 
4:33 AM
me, i still use switches to program my pdp-8
@UnderMathUate it is an old school symbolic analysis program. when i say old, i mean last century.
i like it mainly because i am used to the interface.
i worked with a professor who was involved in its dev
 
it's very efficient, too. you can get ten or twelve computations done on a single tank of kerosene
 
methylated spirits works too.
the best thing is that it is lisp based.
 
Hi everyone :))
 
hi rover
 
@copper.hat I just took a look at the Wiki page. Did you get involved with the development at all?
 
4:38 AM
Why we need to write proof , when the statement is logically true, I mean why to prove simple thing by weird things...?
 
@UnderMathUate no, i was just using it for some project
@Rover it is like putting up scaffolding when building
 
rover, one reason people prove things is because they don't know in advance or via some other route that it is true. if you do, i can see many instances where you might not care about a proof. but often you don't.
and sometimes (not always) a proof can shed light on "why" something is true.
the starting assumptions in many math classes seem 'strange' because they are often designed to be small, if not as small as possible, so you don't have to take large amounts of things for granted. in many applications it is more common to take large amounts of things for granted.
 
Ah, ok. Either way, that's pretty cool to have been able to be around when big stuff like that was happening. :D
 
Hm ok. Any suggestions for understanding and writing proofs or any book , videos ?
 
any recommendations i might have would depend on the subject matter you are interested in exploring. i personally think it's very hard to give meaningful advice on proof-writing without remaining specific to one topic, although there are books that try.
start with 'abstract algebra' and not 'analysis' is maybe a good first piece of advice. whether or not it's conceptually simpler, most people find it an easier start.
 
4:44 AM
I know about a book, " How to prove it" which I have started, but at present at the start lectures have started by proving things...
@leslietownes okay
 
what's the subject of the lectures?
 
@Rover the point of proof is to establish that something is true. you may have a hunch about something and a proof is a way of determining truth.
 
@leslietownes for now, it's sequences, convergence and divergence. I understand them, but writing it in way of writing proof..seems like writing in other language
 
4:59 AM
mm, i think sequences and series is probably the first time i saw anything resembling proof-like argument, too. it's an adjustment. there you usually eventually have access to a library of 'tests' you can take for granted and the trick is applying them.
tests involving inequalities tend to be harder for beginning students to apply than the tests that don't.
 
@copper.hat okay, so it's for times when can't say whether it's true just by statement, so we proceed by what we know is true?
 
the first proof i remember was that root 2 was irrational.
@Rover well, the idea is that there are some starting facts (axioms) and some rules of inference. the axioms are presumed true and the rules let us determine some 'new' truths and we can keep repeating.
the reality is that we skip most of the stuff and concentrate on the bits we are not sure about.
sometimes proofs are a bit like lock combinations, if you know the combo then it is easy, otherwise you can spent a lot of time scratching your head trying to figure stuff out.
 
@copper.hat Thanks, it makes sense 👍🏻
 
5:50 AM
just hit one more convex psq before bedtime...
@TedShifrin Happy H!
 
@Rover "How to Read and Do Proofs" by Daniel Solow has been a life saver for me, but you do need to take the time to understand the ideas and logic.
 
6:18 AM
So, a lot of mathematicians nowadays will name things by simply taking old names and adding an intensifier to them. Like, I would not be surprised to find someone had written some paper on something called "ultra-hyperbolic geometry" or "ideals of very strong finite type."
I'm wondering if this is a modern trend, or if people "back in the day" would name things in the same way, but we just don't recognize it as such because it's using older language than we don't immediately understand
 
interesting thought. my guess would be that it's at least somewhat modern, because if you go back too far, there isn't anything resembling sufficient standardization of core terminology for qualifiers or intensifiers to make sense.
there sort of still isn't, in some absolute sense. but mass publishing, popularity of key books, etc. helped a lot
even in the 1800s people would just say 'function' and mean different things by it, for example. continuity vs. uniform continuity took a while to be figured out, let alone nailed down in those words. maybe by 1900 there was hope of some standardization in some areas.
throwing a guess at the dartboard, i'll declare that mathematicians started doing this in earnest after WW2.
or maybe 1930. WW2 seems a little late for some stuff.
 
6:34 AM
i've been working on ultra convex analysis for a while now.
 
this is something you see a lot in obscure journals (usually with some pay-to-play element). introducing the theory of adjective X, where X is some known thing and adjective is your special sauce. you weaken or strengthen a hypothesis of X and run through what standard textbook stuff still goes through (perhaps with qualifications) and what doesn't. no applications except invented examples for this purpose.
there are real reasons people do that, too, but that's what it made me think of.
good examples of adjective are 'fuzzy' (more popular in 70s-80s) and 'quantum' (works great right now)
 
7:03 AM
these are all registered trademarks of Lesliecoin LLC, incidentally
 
except for the fuzzy extra standard analysis.
 
fuzzy extra standard, fuzzy platinum class, fuzzy VSOP, fuzzy premium, and fuzzy with techron
also fuzzy 21 year single malt
 
guaranteed by 4 field's medalists
seems like an obvious match up between mathematics and track & field
nike's axiom of choice - zorn ahead of your competitors
 
7:47 AM
Good morninrg.

i have read and understood the proof of Fubinis theorem multiple times now (about multidimensional integration) what i seem to fail to deveolop is an intuitive understanding, of how the functions defined in the proof connect with the multidimensional integration.. is there some good illustration of this or intutive way of understanding how these functions defined in the proof of fubinis theorem connect to the subject?
 
7:59 AM
I found some interesting discussion topic about AI in Facebook which is 'can AI have mental illness?'
 
 
6 hours later…
1:47 PM
anyone have a counterexample for this when $f$ does not have finite order ? let $f$ be analytic in the whole complex plane, such that $f f' f''$ never vanishes. Then $f(z) = \exp(az+b)$ for some constants $a,b$
(its true when $f$ has finite order by the hadamard factorization theorem)
 
2:00 PM
order?
 
2:10 PM
basically it means $f$ cannot grow too quickly at $\infty$, precisely $f$ has finite order if there is some number $a > 0 $ such that $|f(z)| \leq \exp(a|z|)$ for all $z$ sufficiently large
if $f$ has finite order and is entire, there is a result called hadamard factorization theorem that will tell us it has a specific form, and if we also know $f$ has no zeros, the form is exactly $\exp(p(z))$ where $p$ is a polynomial, then knowing in advance that $f'$ has nozeros will tell us that $p$ is necessarily linear
but in the absence of this additional information (namely that $f$ has finite order), I am unsure if this result still holds
what strikes me that this could still be true is that using the hadamard result, i dont think we need to know $f''$ never vanishes, it seems like all we needed is that $f f'$ never vanishes
 
Hey guys. Can someone tell me how we got from that integral to the lower integral with the given substution? My professor just wrote it and down and said (i dont have time to show you how you do this)
i am guessing it is about Polar coordinates. ( i know this from physics) but the problem is , we never learned them in math, so i am guessing maybe they used something else?
 
2:32 PM
given a partial differential equation what can you ask about it aside from the solutions to it?
 
oh
I think they meant like 1-x^2 to be constant... right? then we just dt they are meaning dy... right?
 
yeah he's substituting $c = 1 - x^2$ and the latter is a constant in $y$, just like $c$ is a constant in $t$
 
Yea i see that now, its written badly i think. but i get it.. lol
@geocalc33 order, linearity, etc but i am guessing thats not what you mean
@JoeShmo btw shes a she :D
 
Jam
@copper.hat im trying to prove that my function is also harmonic by proving it has the mean value property on every ball inside $B_1$ .Now when my ball is entirely on the upper ball or the lower ball obviously satifies the mean value property since $u$ does. Now if my ball is supposingly 3/4 on the upper ball and 1/4 on the lower some terms are killed because $v$ has opposite signs on lower and upper part. But still im left with some terms.
@copper.hat maybe i could just show is satisfies laplaces equation?
 
2:50 PM
@Jam definitely try to show that it satisfies Laplace's equation first
 
3:05 PM
@Jam why can't you just use that your thing is harmonic if and only if it satisfies the mean value property locally everywhere, this is clear in the upper and lower (open) half balls, then on the middle lower dimensional ball, you can show that the surface integral evaluates to zero as long as you choose a small enough ball centered at a point in the lower dimensional center ball
because your surface integral is just the integral of $u$ over the upper hemisphere and the integral over the lower hemisphere is the same as the integral of $-u$ over the upper hemisphere by construction
 
Jam
$v(x)=\[ \begin{cases}
u(x) & x_n \geq 0 \\
-u(x',-x_n) & x_n\leq 0
\end{cases}
\]$
 
so basically exactly how you would do it in the two dimensional case
 
Jam
this is my v
 
yeah, so in the two dimensional case, your $v$ is $-u(\overline{z})$ in the lower half part
 
Jam
my $u$ is harmonic. I proved v is $C^2$
 
3:07 PM
its just the analog of this
 
Jam
now if i consider a ball thats 3/4 on the upper and 1/4 on the lower why does it evalluate to the value at the centre?
 
it doesn't
 
Jam
i want to prove v is harmonic
 
choose it to be 1/2 in the upper and lower
 
Jam
to satisfy the meanvalue property Evans says for every ball inside your U
 
3:08 PM
since your $v$ is continuous, its necessarily $0$ at $x_n = 0$, no?
 
Jam
yes
 
oh sorry, i misspoke earlier, it is true that this needs to hold for every ball
but you can show that it will hold for every ball by showing that it holds for some ball centered at any point
these turn out to be equivalent conditions
 
Jam
ohh is that true? then i just make the ball small enough round the centre
 
yes
it does turn out to be true
the standard approach in the 2d case is to show that the mean value property (only holding locally) suffices for harmonicity
and the converse is immediate
(by greens theorem, say)
to show the local mean value property implies harmonicity you will basically need to consider the poisson kernel
 
Jam
yeah but i dont think we did that in class that the local condition is enough
 
3:11 PM
ah, cant help you then
 
Jam
so using it would be awkward but thanks
that sounds a good way to approach it
and for balls at the boundary is just zero
 
3:43 PM
why do i sometimes get an inbox notification even when the person who replied to my comment forgot to @name me?
(no the comment wasn't under my Answer/Question)
 
how do we do this step

$ \int_{[-R,R]\times [-R,R] } e^{-x^2-y^2}dxdy = [\int_{-R}^{+R} e^{-t^2} dt]^2$

with Fubinis theorem?
 
3:58 PM
@MadSpaces this is a famous integral. $\int_{[-R, R] \times [-R, R]} e^{-x^2} e^{-y^2} dy dx = \int_{-R}^R \int_{-R}^R e^{-x^2} e^{-y^2} dy dx = \int_{-R}^R e^{-x^2} (\int_{-R}^R e^{-y^2} dy) dx = (\int_{-R}^R e^{-y^2} dy) (\int_{-R}^R e^{-x^2} dx) = (\int_{-R}^R e^{-t^2} dt)^2$ where the interchange of the integrals is permissible due to Fubini
 
The first step you made is fubini right, but when you pull the dx infront of the other integral... whats the rigor behind that?
 
since observe that $\int_{[-R, R] \times [-R, R]} |e^{-(x^2 + y^2)}| dy dx < \infty$
 
@MadSpaces the other integral does not depend on $x$, you are pulling a constant out of an integral
 
what @AlessandroCodenotti said
 
not x.. but dx...
it seems to me a double integration has turned to multiplication. my spidey senses is telling me something is wrong
 
4:02 PM
$\int^R_{-R} e^{-y^2}dy$ evaluates to some real number $A$, you are pulling this number out of an integral
 
you have an integral inside that is not dependent on $x$. it integrates to a constant. therefore you may pull it out
 
Oh yea yea now is see it
Thanks very much :3
 
It's the same as $\int^R_{-R}Ae^{-x^2}dx$
 
4:47 PM
Let V be an inner product space then an operator $T\in L(V)$ is said to be a positive operator if $T$ is self adjoint and for all $v\in V$ the following holds: $\langle Tv, v\rangle \ge 0$. I think that the definition remains unchanged if I remove the condition that "T is self- adjoint". And I think so because if $\langle Tv,v\rangle \ge 0$ for all $v\in V$ then $\langle Tv, v\rangle= \langle v, Tv\rangle$ for all $v$. So by definition of adjoint of $T$, it follows that T is self adjoint.
Is my understanding correct? Thanks.
 
Huh?
 
the conclusion is correct. i'm not sure about your proof. all you would get from the definition of the adjoint is that if <Tw,v> = <w, Tv> held for all w, then T*v = Tv. which isn't what you have.
you'll could snake through a polarization identity or something else that relates a quadratic form to its 'diagonal' to fill in that gap.
 
Oh, this is not going to work over $\Bbb R$ at all.
 
i forgot that there is such a thing as a real vector space. :D
for a real vector space something like $\begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix}$ is a counterexample but if you toss in complex scalars this operator ceases to satisfy the hypothesis.
 
The argument — such as it is — doesn’t use positivity, just reality. I’m totally not a believer.
 
5:07 PM
Ah, I understood. I was concluding only on the basis of $\langle Tv, v\rangle =\langle v, Tv\rangle$ completely disregarding $\langle Tw, v\rangle=\langle w, T^*v\rangle$.
wait... I think my proof is correct. Instead of $\langle ,\rangle$ , I'll use dot .
 
realness is enough to deduce self adjointness in the complex case, but i do agree that the current argument does not explain why. a polarization formula would complete the picture.
koro as far as i know physicists are the main people who are likely to exclude 'self-adjoint' from the explicit definition of positivity (and also to use positivity of the form). there are twenty zillion characterizations of positivity, some of them quite baroque. it's just simpler to include it in the definition.
 
hi guys
im confused about how to prove that $u \cdot (v \times w) \neq \vec{0}$ for 3 linearly independent vectors $u, v, w$ of $\mathbb{R}^3$
my intuition would be that if it was $\vec{0}$, $u$ and $v \times w$ would be perpendicular
 
I understood now Leslie. On complex field, I can exclude "self -adjoint".
Because of complex field: T is self -adjoint if and only if $Tv.v\in \mathbb R$.
 
which doesnt make sense but im not sure how to put it into a proof
 
allie: that's true and more or less the definition of 'perpendicular'. can you deduce something interesting from u being perpendicular to v x w ?
 
5:19 PM
you again beat up my question to death, Leslie.
 
koro, another place where this almost comes up is in the notion of ordering on operators. one often writes A <= B if A and B are self adjoint and B-A is positive. it is possible for B-A to be positive even if A and B aren't self-adjoint (e.g. A = P + iQ and B = R + iQ where P, Q, R are self adjoint and Q is nonzero) but people tend not to write <= in that case because <= is just not very interesting on such operators.
lots of nooks and crannies in this area where the definitions seem to build in more hypotheses than necessary, and in some cases this is formally even the case, but it is nevertheless a standard thing to do.
you can tell when an author is in the thrall of some other field of mathematics when they state and prove that a long list of minimal sets of hypotheses are equivalent, and then say "an operator satisfying any of the conditions (a)-(z) is said to be positive." save that kind of goofy definition for some other kind of book.
this has been today's editorial.
 
5:38 PM
@Jam you showed it was $C^2$ and Laplace's equation follows. There is no need to do more. In fact, since it matches $u$ on an open set you can say more.
Leslie the abuser.
 
5:56 PM
lol
Sounds like a yu gi oh card :)
 
I met a question today: find cardinality of the set of continuous functions $f:[0,1]\to \mathbb R$ such that $\int_0^1 f(x)(1-f(x))\,dx=\frac 14$.
I think that only one such function is possible and that function is $f:x\mapsto \frac 12$ for all $x\in [0,1]$.
Proof: I claim that $f(x)(1-f(x))\le \frac 14$ for every $x$. This is so because maximum value of $t\mapsto t-t^2$ is $\frac 14$. It follows that $\int_0^1f(x)(1-f(x))\, dx\le \frac 14$.
Since integrand is continuous, the equality can hold iff $f(x)(1-f(x))=\frac 14$ for every $x\in [0,1]$. This gives $f(x)=\frac 12$ for every $x\in [0,1]$. So the cardinality of the set is 1.
QED.
Is my proof correct? Thanks.
 
6:12 PM
@leslietownes thats where im at rn and idk what to do next :/
another approach i might try is the fact that $[\vec{u} \. \vec{v} \. \vec{w}]$
 
allie: this may involve dipping into the specifics of how you've introduced the cross product but i'll try to speak to some ideas. because v and w are linearly independent, v x w is a nonzero vector, geometrically, it is a normal to the plane spanned by v and w. if u is perpendicular to that normal, then u is in the plane spanned by v and w.
 
i have the intuition for why it may be true but i dont know how to prove it
i understand that $v \times w$ is gonna be perpendiculr to the plane spanned by $v$ and $w$ and so $u$ and $v \times w$ cant be perpendicular to eachother since then $\mathbb{R}^3$ would have a dimension of 4 which is obv wrong
hmm wait
 
Hi Thor!!
 
how was the cross product defined? what properties do you know about it?
 
i know that $v \times v = \vec{0}$
i know that the cross product of 2 vectors is always perpendicular to both
with a magnitude determined by the sine of the angle between them
(hence the way that it is anticommutative, since its based on the sine not the cosine)
and the way it was defined was the "determinant"
 
6:19 PM
oh, the symbolic determinant with i, j, k and then the inputs in it?
 
@Koro looks good. were you concerned about something?
 
yup
or its also just $|u||v|\sin(\theta)\hat{n}$
right
 
so can you convince yourself that u . (v x w) is the determinant with u and then the inputs in it? it looked like you wrote a block matrix of this type (almost?) up above. that's one way to do this.
 
thanks a lot @copper. Actually, I was trying it in my head for a long time. But everytime I thought I'd solved it -I realized a mistake in the train of thought -For example: my subconscious mind was plotting $t(1-t^2)$ as concave upward. After fixing it, I decided to share the proof.
 
@leslietownes ya im gonna try the block matrix thing
 
6:23 PM
@Koro $t(1-t)$ presumably.
 
@leslietownes i probably could by just the brute force algebra
 
allie: yeah. when you do the dot product, you just replace i, j, and k with u_1, u_2, and u_3. which symbolically is the same thing as if you'd determinanted the matrix with u in it instead of the ijk thing.
 
yeah copper, right!!
 
my other idea was to somehow show that $u \perp (v \times w)$ suggests that $\mathbb{R}^3$ has dimension 4
 
which if i'm not mistaken is today's first use of 'determinant' as a verb.
 
6:25 PM
since it maybe implied that $(v \times w), u, v, w$ are all linearly independent
if you know what i mean
 
well the "maybe" is what we're getting at with this 3x3 matrix with u, v, w. if its determinant is zero, that tells you something about its rows/columns.
 
ohh
well i know the determinant being 0 suggests the columns are linearly dependent
but what does it suggest about the rows?
 
same thing
 
Allie, it's easy. Can you show that v, w and $v\times w$ are LI?
 
determinant is unchanged by transposing.
if the stacked matrix with u, v, w as rows has det zero, then (u | v | w) does too
 
6:30 PM
gotcha
@Koro ill try this approach in a second
@leslietownes so yea thats pretty much QED then because u v and w are assumed to be linearly independent
 
sure, you'll note that it answers your question in one line.
 
allie: yeah. u . (v x w) = det (stacked u, v, w) = det (u | v | w) = nonzero because u, v, w are LI
 
But Leslie's approach makes you ready for utilizing properties of matrices, which is very helpful in linear algebra. :)
 
@Koro ok what do i do to show that they are LI
being LI suggests there exists $c_i$ such that $c_1w + c_2v + c_3(v \times w) = 0$ has only the trivial solution
 
yes.
typo there, i think $c_1$ should be multiplied with w
 
6:35 PM
yup
saw that too
what next
yea that was confusing
 
Dot product both sides by $v\times w$ to get $c_3=0$
 
6:48 PM
$c_1w.(v\times w)+c_2v.(v\times w)+c_3 ||v\times w||^2=0$ and the first two terms on LHS are zero, whence one gets $c_3=0$ (note that $||v\times w||^2=(v\times w).(v\times w)\ne 0$ because if it is zero then v is parallel to w hence u,v and w can't be LI).
 
thank you
i apppreciate it
for the next problem i need to show $u \times v$, $v \times w$, $w \times v$ are LI
so what i have so far is i wrote oiut
$c_1(u \times v) + c_2(v \times w) + c_3(w \times u) = 0$
which i made into
$(c_1u-c_2w) \times v = c_3(u \times w)$
by some simple cross product calculationw
 
Now, one last step is: since $\mathbb R^3$ has dimension $3$ hence every vector $z\in \mathbb R^3$ can be written as linear combination of $v,w$ and $v\times w$.
 
allie: you could also just dot the original equation alternately with u, v, w and use the earlier exercise
 
What's the difference between strong duality and duality in convex problems? Is it that the first one guarantees the duality for concave functions too?
 
oh thats smart
 
6:51 PM
In particular, there exist some scalars p,q and r such that $u=pw+qv+r(v\times w)$. If $u.(v\times w)=0$ then r=0. This implies that $u=pw+qv$ which is a contradiction as $w,v$ and $u$ are LI. So $u.(v\times w)\ne 0$. QED.
 
is my approach salvagable at all
or should i just do it like that
hi balarka
 
hi allie
 
cuz idk what to do from here with my way
fuck it ill move to tyhe next one
 
in your way, dot product both sides by $v$ to get $c_3=0$ (noting that $v.(u\times w)\ne 0$ from the earlier exercise you did just now).
 
ahh gotcha
and since theyre the same vector, dot producting on both sides keeps it equal
this stuff is hard
the next question is, there are $v_1, ..., v_k$ such that when $i \neq j$, $v_i \cdot v_j =\vec{0}$
i assume this is gonna be one of those ones where i can do $v_1 \cdot (c_1v_1, ....)$
and then from there deduce that $c_1v_1 = 0$ since the RHS stays at $0$
 
7:28 PM
@Koro @leslie I don't know if your positivity discussion got resolved, but there's no way to get from what you had to what you need (polarization does not work). Moreover, you have to be suspicious since it fails over $\Bbb R$ but you can certainly consider matrices on $\Bbb C^n$ with all real entries.
 
I am trying to prove that $K(x,y) = \cos(x-y)$ is a PSD kernel on $[0,2 \pi] \times [0,2 \pi]$. The hint I am given is to consider $\phi : [0,2 \pi] \to \Bbb{R}^2$ given by $\phi(x) = (\cos (x), \sin (x))$. But I'm not sure how to proceed from this hint.
 
Hi @Allie @Balarka
 
this might not be the right identity (looks antilinear in the first coordinate :(() but i had in mind something similar to math.stackexchange.com/questions/561636/…
fischer's answer
 
Hmm ... While I was at PT I certainly convinced myself (working in my head, no paper) that the polarization cannot work over $\Bbb R$. I am still suspicious what happens to Daniel's argument if you start with a real non-symmetric positive matrix.
Oh, that came up in the comments.
This might be one of the only linear algebra results I know of that works $/\Bbb C$ but fails $/\Bbb R$.
 
it's just wrapped in something that only physicists care about. in the guise of a statement about invariant subspaces and eigenvalues it is less surprising.
 
7:35 PM
I think I once figured out that positive issue in the comment, many years ago when this discussion came up. I only define positivity in the self-adjoint setting.
 
every math book i have ever seen includes it in the definition. i think for some physics folks the quadratic form comes first and so they come at it differently.
 
professor Ted, I realized earlier during discussion with leslie that the proof (Over C, an operator is self adjoint iff $\langle Tv, v\rangle$ is real for all v) that I'm aware of (from Axler's) works when we are in field of complex numbers.
 
Comment to @allie: In my book, we only find out in Chapter 7 that linear independence of $n$ vectors in $\Bbb R^n$ is determined by nonzero determinant. In Chapter 1 that is established for $2$ vectors in $\Bbb R^2$. The right argument is what was given above. If $u$ is orthogonal to $v\times w$, then $u$ lies in the plane spanned by $v$ and $w$.
@Koro I guess that's a good tricky exercise, but as I've said, to me positivity is only relevant for self-adjoint operators in the first place. :P
 
philosophically i guess the vibe is that in R, <Tv, v> being real is simply a consequence of what < , > is, as opposed to a hypothesis with some bite.
 
smacks leslie for <'s and >'s
 
7:41 PM
^_^
 
How many people know how to interpret the hermitian inner product explicitly geometrically?
 
koro what pit of hell are you pulling these problems from anyway? is it still axler?
 
The first time, I learnt about positive semi definite was like this: A is symmetric and $x^TAx\ge 0$ for all x.
 
That's the definition you should remember :P
 
Leslie, that's quadratic form. Right?
I always wondered then why A had to be symmetric.
But things are getting clear now.
 
7:43 PM
this is one reason why wikipedia is a disaster for learning about this stuff. tons of related notions, with definitions that sometimes are and sometimes are not equivalent, scattered across pages written and edited by different people at different times.
 
yes, Leslie. I've been reading Axler.
 
@TedShifrin zero? or, i guess.. one [you]?
 
In past, I've also read Strang's linear algebra so I try to relate with Axler what I learned in Strang's. :)
 
Lots more than that, but it's a question you — for example — should contemplate. It comes up in complex geometry because the real part of a hermitian metric gives the associated Riemannian metric and the imaginary part is the associated Kähler form (i.e., is alternating).
Strang has various books. Which one?
 
Linear algebra and its applications
4th ed
 
7:48 PM
Oh, OK, "big" Strang.
He has lower-level books, too. The main virtue of Strang is the good treatment of interesting applications. It's not the place to learn proof-based linear algebra.
 
koro, one thing you sometimes see in complex * -algebras is the definition that T is positive if there's some B in the algebra with T = B* B. this is equivalent to the usual definition if the algebra is complex matrices, or bounded operators on a complex hilbert space. but if the elements your algebra isn't given as necessarily 'acting' on anything (let alone a hilbert space) it's a bit of work to relate this condition to other forms of positivity.
for example i don't think it's immediate that a sum of positive elements is positive, under this definition.
 
Hi @TedShifrin
 
( ͡° ͜ʖ ͡°)
 
@leslie I forgot to ping you with my hermitian metric comment above.
I bet Balarka has figured out the meaning of a hermitian inner product before :)
 
some books in this area basically apologize for introducing the complex conjugate in the hermitian inner product, saying stuff like 'just trust us, it makes <x,x> nonnegative' or 'we will come back to this later' or 'physics demands this'
the best approach is just to push those symbols onto the page, and then into readers' heads
 
7:57 PM
do we need the optimization problem to be convex for KKT conditions to hold? I'm trying to thing of a non-convex problem where KKT are still guaranteeing optimality, but I guess it's the h(x) = 0 condition that would be violated then?
 
Hermitian inner products are much deeper than people say it is in an intro course
 
Yeah, as leslie says, in an intro course the bar is there just to make the length of a vector nonzero.
 
$O(2n) \cap Sp(2n) = Sp(2n) \cap GL(n, \Bbb C) = GL(n, \Bbb C) \cap O(2n)$
 
Hmmm ... That seems orthogonal to my comments above, however.
 
I wish I could understand even 1% of it :P
 
8:01 PM
You can understand my comments, @Koro, if you wish to. Look at the real part of the hermitian inner product of two vectors $z$ and $w\in\Bbb C^n$. See that it is the real inner product if you view those as vectors in $\Bbb R^{2n}$. So, in particular, hermitian orthogonal implies real orthogonal. But there's more, obviously. What does the imaginary part tell you?
Oh, actually, maybe Balarka's equalities aren't orthogonal to my comments. I'd need to think, and that takes work.
 
I meant Balarka's comment. I recognize GL(n, C) though.
 
@TedShifrin :)
 
Well, Sp is the Lie group of symplectic matrices. The symplectic form has to do with the skew-symmetric comments I was making.
I think that's Koro's way of saying he's going to ignore the exercise(s) I gave him.
@leslie What are you sending me for lunch today?
 
I'll try the exercise you gave :). But not now :(. I'll be making some notes and will fall asleep soon (it's 1:30 AM here).
 
That's normal day hours for Balarka :)
 
8:11 PM
Very loosely and informally speaking, is it appropriate to call the parallel transport vector the "after image" of a tangent vector along a curve?
 
Where did those words come from?
It suggests contrast with pre-image, which means you're looking at push-forward, maybe, and that is not what parallel transport is, in general.
 
ted: i have some leftover turkey on a baguette. your choice of cheese and/or mustard.
ted: alternatively, three martinis.
 
Wait, you got actual turkey?
 
I came up with those names.
 
@TedShifrin Agreed
 
8:18 PM
ted: i got some on friday. my daughter ate a little bit of it.
 
@Hawk: The words parallel transport should not be deleted, as there are lots of other ways you could transport a vector along a curve.
Oh, OK, I'll take cheese (gruyere, not American), mustard, and two martinis.
 
maybe a little bit more precise, maybe I should've said: "after image" of a tangent vector parallel transported along a curve for a short period of time.
 
Just use short plain English. It's the parallel translate of the tangent vector along the curve.
 
In the complex plane, is the square of the line segment from $a$ to $b$ the quadratic Bézier curve with control points $a^2$, $ab$, and $b^2$?
 
I have no idea what the Bézier curve would be.
 
8:25 PM
As in, $(a(1-t)+bt)^2=(a^2(1-t)+abt)(1-t)+(ab(1-t)+b^2t)t$
That looks right to me
 
This is remarkably similar to the veronese/segre embedding of $\Bbb P^1$ into $\Bbb P^2$.
 
${\rm lerp}_t(a,b)^2={\rm lerp}_t({\rm lerp}(a^2,ab),{\rm lerp}_t(ab,b^2))$
 
Using $a^2,ab,b^2$ as a basis for the quadratic monomials.
 
@TedShifrin What is that?
 
WTH is lerp?
 
8:27 PM
lerp lerp lerp? are you having a stroke?
haha
 
@TedShifrin linear interpolation
${\rm lerp}_t(a,b)=a(1-t)+bt=a+(b-a)t$
 
i sort of figured it out. it's your notation for convex combinations
 
DogAteMy: You can embed $\Bbb P^m$ into $\Bbb P^N$ (for suitable $N$) by using homogeneous polynomials of degree $k$.
 
Relatively standard notation in computer science. Seems to be less common among mathematicians
A Bézier curve is easily described as an "iterated lerp"
 
As in I've never seen it in 50+ years as a mathematician.
But I never took numerical analysis, unfortunately.
 
8:30 PM
that's one heck of an abbreviation. L doing the work of linear, and erp doing the work of intERPolation. i guess nobody wanted to write lint although there the 'in' could apply to both words.
 
Make that three martinis, after all, @leslie.
 
I would just write $[a,b]$ or something like that
 
@AkivaWeinberger yes
 
YES got it to work
 
8:33 PM
Nice :)
 
Yeah that's the standard picture for Bézier curves. Iterated lerps ^_^
 
burps from the lerps
 
Nice animation
 
$a(1-t)^3+3bt(1-t)^2+3ct^2(1-t)+dt^3$, I think
Someone check that
@AlessandroCodenotti so a quadratic Bézier curve is $[[a,b],[b,c]]$? Eh, I'd want a $t$ in there somewhere
 
yes I guess my notation only really works for a pair of points
 
8:47 PM
I am finding the sine of the acute angle between the vectors a = 2i + j + 2k, b = -3j + 4k using the cross product. The textbook says to substitute |nhat| = 1 into the formula a x b = |a||b|sin(theta)nhat. Firstly, the formula uses nhat which is a vector, not |nhat|. Secondly, why can you substite |ncap| = 1?
 
Realization: the quadratic Bézier curve with control points $a$, $b$, and $c$ is the same as the cubic Bézier curve with control points $a$, $(a+2b)/3$, $(2b+c)/3$, and $c$
 
*...substitute |nhat| = 1?
 
Take the magnitudes of both sides
then you can use the fact that |nhat|=1 by definition of nhat
 
@AkivaWeinberger Can you explain a bit more, please? I understand the first part.
 
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