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12:05 AM
@geocalc33 seems likely to me that any explanation will be a matter of physics as much as math. (For instance, the index of refraction of the plastic and its thickness, as well as the curvature of the container)
@Semiclassical yeah
does it relate to conic sections at all
or projections
yeah it's probably refraction
1:01 AM
I'm trying to calculate $\int_{S} \mathbf{F} \cdot d\mathbf{S}$, with $\mathbf{F} = x\mathbf{i} + y\mathbf{j}$ and $S$ being the half-sphere $x^2 + y^2 + z^2 = 1$, $z \geq 0$ with the normal vector pointing up. I tried $d\mathbf{S} = \left(\frac{x}{\sqrt{1-x^2-y^2}}\mathbf{i} + \frac{y}{\sqrt{1-x^2-y^2}}\mathbf{j} + \mathbf{k}\right)dxdy$ but that didn't seem to work very well. Is there an easier way?
@psa the double integral should be simplest in spherical coordinates
change $\iint \frac{x^2+y^2}{\sqrt{1-x^2-y^2}} dxdy$ to spherical coordinates?
oh hm I wouldn't have thought that, OK
Set it up in spherical coordinates to start with. But here's a way to do it super easily.
$\vec F\cdot\vec n = x^2+y^2$.
By symmetry, the integral is 2/3 the integral of $x^2+y^2+z^2=1$ over the hemisphere.
1:10 AM
I feel like I must be missing something obvious there
yeah I don't quite get it
The only symmetry I see is rotational symmetry
Hmm, maybe I do sorta see it
If you were on the whole sphere, you could exchange variables. It's clear that the integral of $x^2$ is the same as the integral of $y^2$ is the same as the integral of $z^2$.
Ah, I see where I was failing.
Yeah, so I screwed up. I need the full sphere to get 2/3. And then I take 1/2 because it's just a hemisphere.
1:14 AM
I had the wrong vector field in my head
But, even so, in spherical coordinates $dS = \sin\phi\,d\phi\,d\theta$ (using the American convention), so I'm integrating $\sin^2\phi dS$ and this integral is easy peasy.
Sometimes it's easier to do $\vec F\cdot\vec n$ than to do the whole parametrization apparatus grinding it out.
My saying throughout multivariable calculus classes (particularly integrals and line/surface integrals) is EXPLOIT SYMMETRY (whenever possible).
welp, I'm gonna have to practice more cause that took me awhile to understand
Well, it's worth understanding ... but of course you should know how to do the standard methods, too.
And, when you get to Gauss's (or the Divergence) Theorem, there's yet another easy way to do it.
1:20 AM
I've heard that
OK, I'm shutting up now.
The bit that I find interesting: suppose we’d started from the (rotation-invariant) sphere integral. Then there are two ways to try to proceed. One is to symmetrize the integrand over the sphere, so that $x^2+y^2\to \frac{2}{3}(x^2+y^2+z^2)$. The other is to use reflection symmetry across $z=0$. The former approach gives an easy answer, while the latter only makes things harder to see.
So to use spherical symmetry requires you to first ‘give up’ the reflection symmetry. (I don’t like how I’m saying it but I can’t think of anything better)
Yeah, you want to bring in all three variables to take advantage of the full symmetry (i.e., the equation) of the sphere.
Let $X$ and $Y$ have joint density $f(x,y)=\pi^{-1}\exp(-1/2(x^2+y^2))$ for $xy>0$; $f=0$ otherwise. Show that $X$ and $Y$ are normally distributed.
1:33 AM
That said, the divergence theorem is definitely the approach I like the best
Since we need $xy>0$ to define $f(x,y)$, we need $x$ and $y$ be positive and negative at the same time.
But I still don't know how to do it
If you condition on a particular $Y=y$, then either $X=0$ identically (if $y\leq 0$) or $X$ is distributed like a standard normal rv @simple
I wonder if that can be leveraged
@Semiclassic: But the standard error is to forget to correct by the flux coming in the bottom disk.
No, @Semiclassic. If $y\le 0$, then you still have $x\le 0$ in play.
But it's symmetric, so you might as well deal with them both positive.
Ah, you’re right. So X would be half-normal in those cases
(My intro physics students were simulating half-binomial random variables using coins this week, lol)
1:43 AM
I guess my comment above just amounts to the standard integrals for the marginal distributions
So what is $P(X\le a)$?
@TedShifrin hmm. Is there any flux through the bottom in the case under consideration? The vector field has no z-component
@Semiclassic: Correct. So no correction in this problem. But I would still take off if the student didn't show me that.
Which would seem to make this one particular case rather simple
Yeah, fair
@TedShifrin i don't know ):
1:50 AM
How do you get it from the joint density, @Simple?
As written, the S is just the upper hemisphere and not actually the boundary of a volume
Gotta establish that that isn’t a problem
That was precisely my point, @Semiclassic.
The $y$ integral should be from $-\infty$ to $\infty$, right?
1:52 AM
OK. So now do two cases: $a\le 0$ and $a>0$.
(I don’t know if I’d dock for missing that detail on a physics exam problem involving Gauss’s law, but the conditions are different there)
I don't remember ever having a non-closed surface when I took E&M.
It was always a sphere or a closed cylinder.
Let $F_{k+1}\subseteq F_k$ be a sequence of closed ,non empty sets in a metric space $X$ Show that if the diameter of $F_n$ tends to 0 then the itersection is non-empty
How do I start?
@TedShifrin I can think of two, actually. First one is to find the current passing through a loop by integrating the volume current density (which is a vector field)
1:58 AM
if $a\leq0$, $P(X\leq a)=\int_{a}^{0}\int_{-\infty}^{0}f(x,y)dydx$; if $a>0$, then $P(X\leq a)=\int_{0}^{a}\int_{0}^{\infty}f(x,y)dydx$
@topologicalmagician: I think the diameter tending to 0 shows that there is a unique point in the intersection. It's not needed for non-emptyness.
The other would be to find net power transmission through a surface (Poynting vector)
@Simple, no, both are wrong.
@Semiclassic: But then you don't use Gauss's law, do you? You do some Fubini trick or something.
Or you just calculate the flux directly. I don't see how Gauss can be rigged here.
@TedShifrin I'm sure I have to use the sequence characterisation of closed sets, but I'm not sure how to start constructing a sequence
@Simple: If $a\le 0$, you have $\int_{-\infty}^a$, right?
Just pick a point in $F_k$.
2:00 AM
Sure. I just meant in regards to “only closed surfaces”
Ohhh, I was referring to applying Gauss, @Semiclassic. Context.
I see, you have closed but not necessarily compact, @topologicalmagician. So the diameter condition will get you convergence.
fair enough. I think one could cook up an example regardless, but it’s hardly obvious
Yeah, otherwise, I could push the intersection to infinity.
right, then $P(X\leq a)=\int_{-\infty}^{a}\int_{-\infty}^{0}f(x,y)dydx$
OK, @Simple. So you can simplify that somewhat knowing what $f(x,y)$ is.
2:03 AM
@TedShifrin so idea is to choose $x_n\in F_n$ then hit that sequence of the condition that $diam(F_n)\right arrow 0$ and that will give me a point in the intersection?
Yup, you have to prove your sequence converges.
Not sure what why that condition gives me convergence though
If I were to build up an example, I’d probably make use of the continuity equation for the volume charge distribution: $\nabla\cdot\vec{J}=\partial \rho/\partial t$ where $\rho$ is the volume charge density
Well, what's a good way to prove a sequence converges? Think of a good name.
@Semiclassic: This sounds like my proof for the gravitational field inside the earth :P Still, using spheres.
2:07 AM
@TedShifrin Cauchy?
There you go.
but not $X$ nor $F$ are assumed complete
Then the result is false.
Take $X=\Bbb R-\{0\}$, and let $X_k = [-1/k,1/k]-\{0\}$.
Ok. I'll try to do it for the complete case.
Just so Im able to solve these type of problems
Do you see what happens with my counterexample?
You should be able to think about examples like that yourself.
2:12 AM
the result is still wrong even if you assume completeness
that's heavily number-theoretic, but a standard example is $\Bbb C_p$
if you assume localyl compact, then it's true
What's wrong with my proof, @Lukas? :P
@LukasHeger If $(X,d)$ is a complete metric space and $F_{k+1}\subseteq F_k$ is closed and non-empty for each k, then the intersection is non-empty, is false?
Intersection is always closed.
You wanted non-empty.
@TedShifrin sorry, I meant non empot
@TedShifrin how do you guarantee that the limit is in all the $F_n$? I don't see it
note that there's the notion of a spherically complete field, if this were true, we would just call them complete fields
In mathematics, a field K with an absolute value is called spherically complete if the intersection of every decreasing sequence of balls (in the sense of the metric induced by the absolute value) is nonempty: B 1 ⊇ B 2 ⊇ ⋯ ⇒ ⋂ n ∈ N B n...
2:14 AM
Because it's a limit of a sequence in a closed subset.
@LukasHeger if $(x_n)_n$ in closed subset converges to some limit then limit must be in subset
The decreasing sequence needn't have diameters going to 0?
its called sequence characterisation for closed sets
You wrote the wrong thing, @topologicalmagician.
It's the definition of closed sets I used in my book, actually.
@TedShifrin yeah, I saw it in your book
2:17 AM
@Lukas: The tail of the sequence is in any particular $F_k$, so its limit is in $F_k$.
hmm, you're right
that was nonsense
I got confused over spherical completeness
I should better sleep
sorry for the confusion
LOL, no need to apologize too profusely :P
@TedShifrin I'm unable to show convergence, still
@TedShifrin I got it
You haven't been trying very hard. You said the magic word, @topologicalmagician. Get to work on it.
Great, @Simple.
2:21 AM
Hey everyone!
oh, rats, it's Demonark
A silent Demonark.
I think I got it
Explain the idea.
2:28 AM
Let $\epsilon>0$ Since $diam(F_n)\rightarrow 0$, there exists $N$ such that $n\geq N$, $diam(F_n)<\epsilon$. Hence whenever $n>m\geq N$, we have $F_n \subseteq F_m$ and so $d(x_n,x_m)\leq diam(F_m)<\epsilon$.
ok, good. I thought it would be harder
And so you have a Cauchy sequence in $X$, so it converges (by completeness). Why is the limit in every $F_n$, as Mathein asked?
@TedShifrin otherwise, if $x\not F_n$ for some n, then since $x_n$ $\in F_n$, we may find $x_{n+1}, x_{n+2}....$ in $F_n$. This is also a subsequence which must converge to $x$
and since $F_n$ is closed $x\in F_n$
Just by closedness.
2:36 AM
You don't need contradiction.
I meant closed
Any subsequence (in particular, the tail) of a convergent sequence converges to the same limit.
@TedShifrin I don't know what a tail is, but its also a subsequence right? Do you mean $x_{n+I}$, the ones I mentioned above?
2:38 AM
Like that but for a sequence
Demonark to the rescue.
You need to elongate it, Demonark.
I will prove that as well, and it doesn't seem hard. Thank you once again. I should stop worrying that the problems are too difficult, it gets to me lol and I start to kinda get over anxious
Hahaha, yeah
And if you need to use a hypothesis they forgot to give, you should try to give a counterexample when it's missing.
Otherwise maybe it's really true without that extra hypothesis.
@TedShifrin yeah. Do you have any advice on the worrying part?
2:41 AM
If you worry so much, math isn't a good thing to study.
@TedShifrin no, I don't worry much. Only when there are much external factors involved ( and there are in this case)
but anyways
Thanks for you help @TedShifrin, I always appreciate it
@TedShifrin Have a wonderful day! Got to go!
5 hours later…
7:34 AM
> simple_wojowuToday at 5:37 PM
> What are the simple numbers tho?
> Even the integers are complex
> Or are they simple complex numbers
2 hours later…
9:51 AM
Let f be a smooth map X->Y, and D a distribution acting on smooth functions on Y. Further let g be a smooth function on Y. Is it true that (f* D)(f* g) = D(g) , where the first f* denotes the pullback of distributions, and the second is the pullback on functions (i.e. precomposition)? Looking at the obvious commutative diagram this seems to make sense, but I can't see how this can be justified precisely.
So factorial (n!) has precedence above multiplication. Why is this since it's defined as repeated multiplications?
repeated addition (i.e. multiplication) has precedence above addition
repeated multiplication (i.e. exponent) has precedence above multiplication
10:22 AM
Can a mapping be a contraction mapping as well as an expansive mapping?
@geocalc33 sounds like a number which is strictly less than 1 and at the same time strictly greater than 1 IRCC the definitions of contraction and expansive mappings
What's IRCC
Let $f(x)=1/x.$ What's an example of a dilation acting on $f$ that contracts one branch of $f$ and expands the other branch of $f.$
11:21 AM
i'm sure this is simple maths, but i'm just not very good at it. I have a worker that does jobs... each job takes somewhere between 5 and 10 seconds to do... let's say 7 seconds is the median. i have 900,000 of these jobs (for now, let's assume the jobs never grow) meaning the worker would take 72~ days to do all jobs.
how many workers would i need to clear them in 8 hours?
would it literally be 900,000/200?
11:54 AM
@TedShifrin Thanks, that does work.
@geocalc33 if I recall correctly with a typo
@TedShifrin Any idea why the same doesn't work for AsymptoticDSolveValue? I computed the first few coefficients for the Maclaurin series by hand and wanted to use Mathematica to see if there is some pattern
@Ilya Oh I see!
@TedShifrin nevermind, I fixed it
AsymptoticDSolveValue[{x'[t] == -y[t]^3, y'[t] == x[t]^3, x[0] == 1,
y[0] == 0}, {x[t], y[t]}, {t, 0, 5}]

this works correctly, however for n>5, I get "indeterminate expression ComplexInfinity + ComplexInfinity encountered" - any idea why?
12:11 PM
This the only way we know to go. Squad up never roll alone. And we gon' ride on forever. We, ride out together. Pull up right in your zone. Take over the street. That's how we roll. And we gon' ride on forever. We, ride out together.
If you love perpetual loxodromic transformations give me one clap! If you love isochronous monodromic isoclines give me two claps! If you enjoy torsive non-commuting expansive/anticontractive maps shake your hands in the air and say “totally complete isogenous metric space!”
12:26 PM
Suppose r
00 < Ez. A central problem in applied numerical probability
is the description of sub-naturally Noetherian subrings. We show that
there exists a stochastic u-totally bounded, partially semi-algebraic ring.
In [17], the main result was the characterization of domains. Recent
interest in functions has centered on describing homeomorphisms.
"The goal of the present article is to describe quasi-covariant hulls. In [10],
it is shown that the Riemann hypothesis holds. In future work, we plan to
address questions of minimality as well as positivit"
@courge9 I challenge you to show this is false... Definition 3.1. Let $l$ be a holomorphic monoid. A Gauss topos equipped with
a Hamilton–Landau monodromy is a functor if it is discretely infinite
I'll wait.
A definition cannot be right or wrong. One can only ask if the definition makes sense or is meaningful. Which here is obviously not the case.
Why isn't it the case?
Honestly, why are you asking this? You know that this is an automatically generated pseudo-math "paper", containing plain nonsense and gibberish.
if it's so obvious..
Everyone knows this kind of auto-generators. They are fun for a while, but nothing new.
12:40 PM
Okay yeah, I'll admit it.. I was being a drangus for several minutes
my b
Whoa, what a surprise. Never expected that.
Who are you????
Read my name.
How do you know the meaning of "drangus"?
Why would I not? And even if I didn't, there's a nice thing called the internet. Check it out. urbandictionary.com/define.php?term=drangus
12:42 PM
oh wow you're good
@Huy Wow hey how are you
I just got 0 hours of sleep
I got 0 hours of sleep last night
That's why I feel like I'm on crack
not good
and last night I got 9
but the night before that I got 0!
0+0+9=9 yikes
should be 27
it's not bad though
9/3 = 3
12:49 PM
one time I didn't sleep for 3 days
in 2012
and things got crazy
what happened on the fourth day?
I thought demons were chasing me
all day.
and time was really slow
like 1 hour felt like 10
so you normally sleep 9 hrs?/24
yeah I usually sleep 9
but every now and then I just can't sleep for about a week
actually now it's pretty much healed
but it happened a lot from age 11-21
I see.
1:00 PM
I tracked everything mathematically in a journal
How do you find the inverse for a complex number without using the notation with a + ib .. if you only have the definition of multiplication of (a,b), (a°,b°) where a° and b° should be inverses such that (a,b).(a°,b°)=(1,0). I tried to do this with only the definition of the multiplication and i do not get a correct result
the definition is (a,b).(a°,b°) = (a.a°-b.b°, a.b°- b.a°)
@MadSpaceMemer I don't know sorry
@MadSpaceMemer well, how do you translate between (a,b) and a+bi? if you know what the inverse of a+bi looks like, just translate this back into ordered pairs and you're done
Yes but that would be just guessing, i was trying to do systimatically.
Nvm i got it, i was making a mistake
Just solve two equations and you get correct results
1:12 PM
actually, are you sure that's what you want? should be ab°+ba° in the second component if you want the complex numbers
That is the mistake
I copied the definition wrong and was working on a wrong definition
that's good then
Can a group be 'almost' a lie group?
That reminds me of an advice one professor once gave: if you want to be a good scientist, make sure to copy correctly! ;-)
@geocalc33 "almost" in which sense?
@courge9 well take the 2-sphere for example. an infintesimal rotation leaves the sphere unchanged. but what if you start with a small number of points as subsets of the sphere, and have an iterative rotation process that increases that subset of points until the subset converges to satisfy the equation of a 2-sphere
1:26 PM
well that is a terrible advice, Copying is by no means a good meassurment of what a good scientist is. But i guess he was trying to make apoint, so okay
like for example, take the frame of a cube
with a fixed center at which it rotates about
actually I want to consider a square for simplicity
take a square, copy it, and rotate the copy by 45 degrees. then take the union of the two copies
initially starting with the one square, you have 4 points on some sphere
but after making another copy, rotating by 45 deg. and taking the union of the two copies you now have 8 points on that sphere
the next iteration should give 16 points on that sphere, I believe
and you get closer and closer to a circle (meant to say circle all this time)
so the rotation of the union of $n$ copies of the square after a certain number of iterations should be infintesimal right?
certainly, there is nothing infinitesimal after a finite number of steps
but even in the "limit", I think you only get rational multiples of 2pi
oh okay so in the limit you get pi/2
oh wait nvm, the sum of all the angles converges to like pi or something
because the angle you rotate after each step decreases like 1/2^n
so it would just be an ordinary group
I think I understand better now
1:47 PM
@MadSpaceMemer Of course he meant copying from your own notes or from the text of an exercise
@MadSpaceMemer NOT copying in the sense of plagiarism
Because if you start with a wrong assumption which stems from the fact that you made a mistake when writing off some text, exercise, theorem - whatever - then everything gets wrong
And of course, if you constantly fail at reading and writing things properly, you cannot be a good scientist.
That was the point he wanted to make - half serious, half tongue-in-cheek
2:05 PM
If you have a vector field $X$ partitioned into integral curves
A flow on a set $X$ is a group action of the additive group of real numbers on $X$. More explicitly, a flow is a mapping $\phi:X\times \Bbb R \to X$ s.t.
$\phi(x,0)=x$ and
Given x in X, the set ${\displaystyle \{\varphi (x,t):t\in \mathbb {R} \}}$ is called the orbit of x under φ. Informally, it may be regarded as the trajectory of a particle that was initially positioned at x. "If the flow is generated by a vector field, then its orbits are the images of its integral curves."
I don't understand that part in quotes
I thought the orbits would be exactly the integral curves
Do you know what it means for a vector field to be the infinitesimal generator of a flow?
Not completely.
The point is that a flow is just an action of the kind you described above. If the flow is smooth then it is generated by a vector field and follows its integral curves, but in principle that's not necessarily the case
oh okay that's what i was missing. I was just thinking continuity for everything
So a vector flow infintesimally generates a smooth flow, say for example on $(0,1)\times(0,1)$
and the source of the flow is at $(1,0)$
and the flow vanishes at $(0,1)$
is it possible for a particle initially configured at the source, to take an infinite amount of time to reach the sink?
like the velocity of the flow is so slow or something
2:46 PM
@BalarkaSen tired and you?
3:14 PM
Suppose we have a finite dimensional vector space $V$ and we're proving some statement about linear functionals, $\{f_1, \ldots, f_k \}$. We want to induct on the number of functionals, $k$. Now consider a subspace $W \subset V$ and suppose $\{g_1, \ldots, g_{k-1} \}$ satisfy whatever conditions we need, can we apply the induction hypothesis to the $g_{i}$? The main difference is that we assumed the hypothesis for the space V.
depends on the specifics
The $g_i$ are restrictions of the $f_i$ to the subspace. I can post the full statement of the theorem if you want.
It's in a textbook so we most likely can do that.
But for some reason it's bothering me.
the devil often is in the detail when it comes to induction
the crux here is that the statement you are inducting on is "the property holds for all vector spaces $V$", so your hypothesis is that the statement with $r-1$ functionals is true for all vector spaces $V$, so in particular for the functionals when restricted to $N_k$ and the induction works
the order is important in cases like these, as this would not necessarily work if you had fixed $V$ beforehand
3:27 PM
Mhm makes sense, that's what I was expecting, thanks!
@Huy Not great but not bad.
@BalarkaSen: The correct sentence would be "Not great, not terrible.".
Haha guess so
you must be like 14 now, right?
uhm 20 would be more accurate
3:35 PM
I try to ask my question again:
Let f be a smooth map X->Y, and D a distribution acting on smooth functions on Y. Further let g be a smooth function on Y. Is it true that (f* D)(f* g) = D(g) , where the first f* denotes the pullback of distributions, and the second is the pullback on functions (i.e. precomposition)? Looking at the obvious commutative diagram this seems to make sense, but I can't see how this can be justified precisely.
@BalarkaSen stop making me feel old
what are you up to, being all grown up now?
uh nothing very different really, pursuing an undergrad in math
being disenchanted by undergrad curriculum, and college life, among other things
where are you studying?
3:39 PM
in my country; indian statistical institute
what is the 1-quantile? As in $z_1$ in this answer math.stackexchange.com/a/3555116/72724
4:09 PM
Hi all.
What's up, @Thorgott? What have you been up to?
finally had my last exam today, so I'm on break now
How are you doing?
Not too horrible. Just doing an analysis assignment.
4:30 PM
an old asymptotic question which always gives me a headache whenever I've need to compute it: $\binom{2n}{n+x}\Big/ \binom{2n}{n}$ for large $n$ and small $x$
should be a Gaussian distribution but I've always been bad at showing it
Do you have one of those tex docs where you type up solutions to the most annoying things you always seem to encounter?
no, but it's a good idea
I had one of those, inspired by some professor I had.
the frustration on my part is that it's always seemed to me like there should be a better way to do that than just "hit it hard enough with Stirling's approximation for $n!$"
Unfortunately, most manipulations with binomial coefficients are some kind of Stirling's formula madness
4:38 PM
i mean, you can write it as $$\frac{n!}{(n+x)!}\frac{n!}{(n-x)!} = \frac{n(n-1)\cdots(n-x+1)}{(n+x)(n+x-1)\cdots(n+1)}$$
Maybe a combinatorist will help you out one day. B)
hope springs eternal
You could just find the kindest person working in combinatorics (in my experience, they are all kind) in your department, then just ask them if they had any idea.
One of the nice things about math is that there is enough people doing it that you don't have to do it alone.
i guess one can further write it as $$\left(1+\frac{x}{n}\right)^{-1}\left(1+\frac{x}{n-1}\right)^{-1}\cdots \left(1+\frac{x}{n-x+1}\right)^{-1}$$
4:54 PM
In a Hilbert space, finite dimensional subspaces are closed: I feel like it's not enough to use an orthonormal basis $(e_n)$ and then to express each member $x_i$ of a convergent sequence as $x_i = \sum_n \langle x_i,e_n\rangle e_n$...what am I missing?
Since convergent implies weakly convergent, the right hand side there converges to $\sum_n\langle x,e_n\rangle e_n$ (where $x_i\to x$).
3 messages moved from Basic Mathematics
@JackOhara Oh I forgot that moving a chat message doesn't put it at the end of the target room. But anyway, your question is about mathematica or at least certain primality test algorithms and so off-topic for the room you originally posted in.
what is the 1-quantile? As in $z_1$ in this answer math.stackexchange.com/a/3555116/744874
Nevermind, forget I asked, resolved!
ah.. that's not what it says
5:21 PM
@anakhro do you really need an Hilbert space here?
@AlessandroCodenotti no. :)
5:34 PM
I think you just need a TVS but I never think about spaces which are not Banach so I don't really know
5:47 PM
Or rather some completeness condition on the ground field
@AlessandroCodenotti Surely you need Hausdorff and that the ground field is R
I want to argue that a Cauchy sequence lying in an fd subspace must be bounded in some norm on that subspace, and apply compactness of the unit ball
6:21 PM
Makes sense
I think you just need TVS, but completeness makes it a jiffy.
6:36 PM
If V is a vector space, isn't $V^*\otimes V^*$ by definition the vector space of bilinear forms on V ?
Does this require finite dimensionality
Or is that only for naturality?
Oh I misread, I thought you were doing the Hom(V,W) thing.
Disregard me. :P
In this case V is finite dimensional :)
@Astyx I think you still need that.
You definitely have $(V\otimes V)^*$ is the space of bilinear forms on $V$.
But then to go from $(V\otimes V)^*\cong V^*\otimes V^*$ needs finite dimensional.
Ok that makes sense
Thank you
Make sure to test out that you understand the first isomorphism there (with bilinear forms).
It was my understanding we were only dealing with vector spaces, not topological vector spaces.
I do not know what wacky stuff occurs when you change $V^*$ to be the continuous dual rather than the algebraic.
In the finite dimensional case, not much.
But in the infinite dimensional, I trust there might be some hiccups.
6:43 PM
You clearly need more than TVS. R is a TVS over Q!
The reason why $(V\otimes V)^*$ is not always $V^* \otimes V^*$ is because we want forms to be continous which is not always the case in infinite dimension
I think
The ground topological field needs to be locally compact I think.
@MikeMiller I exclusively deal with TVSs over R or C. B)
What do you mean by "want forms to be continuous", @Astyx
Then "completeness makes it a jiffy" doesn't seem to parse since I was talking completeness of the ground field
I still am not entirely sure of the argument in general
I've always seen the definition of linear applications in infinite dimensional spaces to actually be bounded linear applications, because otherwise weird things happen
6:47 PM
If you want continuity you should not just say vector space
Hilbert space V?
Ah ok
it's in finite dimension anyways
Thank you for the insight
@Astyx What is your argument via universal property
@MikeMiller I meant Banach space.
It seems to me your argument produces a map $V^* \otimes V^* \to (V \otimes V)^*$
Should be straightforward to show this is injective, and you would get surjectivity by a dimension count
Uh let me think
I proved it in a non abstract way using a basis since I was dealing with finite dimension
ie $B = \sum B(e_i, e_j)e_i^*\otimes e_j^*$
6:53 PM
$V^\ast \otimes V^\ast = \operatorname{Hom}(V,V^\ast) = \operatorname{Hom}(V, \operatorname{Hom}(V,k)) = \operatorname{Hom}(V \otimes V, k) = (V \otimes V)^\ast$
How do you justify the second to last equality ?
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