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12:08 AM
After they answered, the argument continuous
*continues. They're really not good at explaining type theory to a newbie
 
12:26 AM
all greek to me
there oughta be a button on math.SE where it displays the list of comments in an animated crawl like the beginning of a star wars movie
 
12:56 AM
wait wait wait
wikipedia the set of arithmetic functions with usual addition and Dirichlet convolution is a commutative ring
but why not a field? The elements have mutliplicative inverses
 
I have no clue, but mult inverse under convolution?
 
right, you don't have that for all nonzero elements
 
I truly have no idea what you’re talking about, but it was the obvious question.
 
it's nice how mathematicians don't know anything about each other's areas
like what have I gotten into? A bottomless pit of endless math?
 
You did not ask a remotely self-contained question. In general, I have a decent acquaintance with lots of fields of mathematics.
I can throw lots of words around, too.
 
1:09 AM
i do not know number theory but do have the vibe that inverses under convolution-like operations are subtle. they pop up all over the place.
but yes, there's a bottomless pit of endless math.
or a limitless field of beautiful growing math that is always in bloom. it's a matter of perspective, man.
 
Happy ducks to the munchkin.
 
one of my friends sent her a gift today. it is a tshirt with a graphic of a dinosaur that has a leg in a cast, and the caption "can't you see my legissaur?" ha, ha, ha. and a black cat puzzle.
 
A bit too punny for her age, but cute.
 
she found out about halloween from the big kids at day care and can't stop talking about it. she can't decide if she wants to be a 'scary pumpkin' or a ghost.
 
How trite. I expect more creativity.
 
1:21 AM
that's what i thought, although they really don't make scary costumes for toddlers. so maybe she has identified an unmet market need. once you get to the 5-10 age range is when you start to see scary stuff and it is usually branded 'boys' (i don't know why childrens costumes need gendering). all the pumpkin stuff for toddlers is 'cute,' which does not seem to be what she's going for.
i'm hoping she settles on ghost because i can make that using stuff from around the house. it appeals to my sense of thrift.
 
She and Olivia should team up on claws.
 
you can get little bat wings to put on a cat's back. i thought about getting some for olivia, but most the ones online are either low quality or very expensive.
 
A clawing (cloying?) ghost!
 
a ghost cat. there is something to this.
 
Yup!
I get partial credit.
 
 
1 hour later…
2:47 AM
@TedShifrin i'll note that the original setting of the problem sorta had the "group action" part baked in.
suppose you have a set of orthonormal basis vectors $\{v_k\}$. then basis vectors for the $N$-fold tensor product is just various multisets of those vectors
then the question (coming from physics stuff) was to count how many permutations fix a particular basis vector
 
3:04 AM
something something symmetrization
 
So you’re splitting this $n^N$ element set into orbits.
 
right
ultimately it amounts to checking a normalization condition
 
Huh?
 
yeah, that was not at all meaningful
i meant that the original problem is that of checking whether certain tensor products are normalized, but it's equivalent to the counting problem
 
If you say so :)
 
3:21 AM
well, for instance, if you had $e_1\otimes e_1\otimes e_2$ then in our construction that (iirc) gets converted to $$v=\frac{1}{\sqrt{3!}\sqrt{2!}\sqrt{1!}}(2e_1\otimes e_1\otimes e_2+2e_1\otimes e_2\otimes e_1+2e_2\otimes e_1\otimes e_1)$$
and the problem is to check that $\|v\|=1$
which...hrm. doesn't look like i've done it right
 
some people normalize those operators differently, depending on what they like. it's like h vs. h-bar.
 
spivak has some of these calculations in his diff geo book. or maybe an appendix to same.
 
physicists want normalized basis vectors
which is sometimes a bit silly, b/c measurement is easier to understand imo when you don't bother to renormalize
 
I do not understand.
 
3:25 AM
in QM, a projective measurement has you start with a state vector $v$ and end up with a state $Pv$
but that's not normalized, so we'd usually take the new state to be $Pv/\|Pv\|$
 
Converted to? These are not scalar multiples. Are you applying a symmetrization? Yes, that’s what you’re doing, appropriately scaled.
Your 2s just muddy the water unnecessarily. Factor out the scalar.
 
yeah, that was a bit disconnected. i was talking about why physicists like normalized vectors, even when it makes things tedious
i was writing them b/c the way it's defined is literally "take $e_1\otimes e_1\otimes e_2$, apply every possible permutation, and add those up"
 
Not sure what your inner product is on the tensor product.
Oh, ok, it’s just $1/\sqrt3$ times the sum of three orthogonal unit vectors.
Duh.
 
yeah. but i've screwed up the normalization somehow
 
There is all sorts of standard representation theory going on here that I once knew decades ago.
 
3:30 AM
definitely
 
No, your constant is right.
The 2 cancels two $\sqrt2$ factors.
 
really? looking at it right now i thought i'd have $1/(3!2!1!)(4+4+4)=2$
i'm probably doing something silly
 
Factor out the damn 2s.
Your arithmetic is 1.
 
oh dear
i was doing 3! = 3 mentally
so yeeeah
 
3! is a 3 that really means it
 
3:34 AM
I always taught my students to keep factors factored to minimize issues.
Right!
 
while 3!! is a 3 that really doesn't
anyways. the course i'm grading for has a HW problem for which one way to compute is to show that the version of that with a generic $\bigotimes_{k=1}^n e_{i_k}$
the book positions them to do it in another way with a different formalism
but i wanted to see how it worked directly from the definition
*show that the version of that ... is properly normalized
 
4:39 AM
trusting websites because of sources
trusting websites becuase of css
 
that would make a good drake meme
 
indeed
 
4:55 AM
If V and W are finite dimensional and $T\in L(V,W)$ then rank T=1 if and only if the matrix of T with respect to basis of V and W has all entries equal to $1$. By matrix of T w.r.t. V and W, I mean that: jth column of the matrix is $[c_{1j},c_{2j},...,c_{mj}]^t$ such that $Tu_j=\sum_{i=1}^m c_{ij}w_i$, where m= dim W, and $u_i$'s are basis of $V$ and $w_i$'s basis of W. My confusion is: since $u_i$'s will contain some $u$'s corresponding to null T so we'll always have $Tu=0$ for such $u$'.
Hence $c$'s (scalers) are zero corresponding to these. How can 1 come in such columns? What am I missing?
 
This sounds implausible.
Take $V=W=\mathbb{R}^2$. Then any multiple of $e_1e_1^t$ is certainly rank one, but its matrix doesn't have any 1s at all in the canonical basis.
 
Please consider this "if and only if there exist basis of V and W, with respect to which matrix of T..." I erroneously forgot to incorporate this in the question statement.
 
oh. so some basis
that sounds more plausible
So you want to show that you can always find one
 
yes, but the problem is this question is not intuitively clear to me unless T is one -one. :(
 
to be clear, which direction are you dealing with right now? showing that such a representation means T is rank 1, or showing that such bases always exist for rank-1 T?
 
5:06 AM
a rank one operator isn't going to one-to-one except in the case where dim V = 1
 
@Semiclassical The direction to show existence of such bases of V and W.
 
thought so. the first one is pretty straightforward
to get intuition, it might help to start with the case $e_1e_1^t$ i said earlier. that's obviously not T=[ [1,1], [1,1]] with respect to the canonical basis
so you're looking for vectors such that $T u_1=T u_2=w_1+w_2$.
simplest choice is probably something like $w_1=(1,1)$, $w_2=(1,-1)$, $u_1=(2,2),$ $u_2=(2,-2)$.(there's obviously a lot of freedom in that choice).
 
@Semiclassical Yes, that makes sense. I think now I can take it from here. Thanks a lot :)
 
np
there's probably some clever way to choose the bases for T in general but i don't know what it'd be
 
@Semiclassical that's the famous rank 1 matrix :)
 
5:19 AM
indeed
 
:)
 
5:33 AM
cute problem.
 
5:50 AM
Let $\displaystyle u_{1} ,u_{2} ,...,u_{k}$ be a basis for $\displaystyle V$.

$\displaystyle Tu_{1} ,\ Tu_{2} ,...,Tu_{k}$ span range T.

Now, if $\displaystyle Tu_{1} \neq Tu_{2}$ then rank T is atleast 2, which is a contradiction.



Hence all $\displaystyle Tu_{i}$'s should be the same.

Let $\displaystyle Tu_{i} =c_{1} w_{1} +c_{2} w_{2} +...+c_{n} w_{n}$, where $\displaystyle w_{i} '$s are basis of W.

The above is true for all $\displaystyle u_{i}$.

So matrix of $\displaystyle T=\begin{bmatrix}
How do I show now that all $c_i$'s are $1$? @Semiclassical
and @Leslie.
 
6:03 AM
Why would a rank of 2 be a contradiction?
 
mm, you will need to be more strategic in your choice of basis than that.
note for example that T u_1 and T u_2 being different doesn't imply that they are linearly independent.
consider the matrix M with rows (1,0) and (0,0) and the standard basis on R^2. M(e_1) and M(e_2) are different but M does not have rank 2.
copper at the outset of all of this T was assumed to be rank one.
 
@leslietownes thanks! i missed the pretext
 
we are having the wooden siding of our house power washed tomorrow. it is going to be noisy. probably zero chance of me getting any work done.
 
@leslietownes ahh, you're right. I should have said non-zero. I'll think more about this. Thanks a lot Leslie :)
 
play loud raucous music in your headphones?
 
6:18 AM
most definitely. i just don't normally work that way.
maybe i'll hide in the closet with our cat. i'm sure she's going to love it.
koro, one thing that it might help to keep in the back of the mind, which i don't think ruins the problem for me to mention, is that there is nothing resembling uniqueness here, and yet completely arbitrary bases won't work either. it's in that annoying middle area of problems, where a solution has to make choices that are kinda arbitrary but cannot be made totally freely.
 
6:53 AM
Let $\displaystyle u_{i} '$s be a basis of null T. Then extending it to basis of V gives

$\displaystyle u_{1} ,u_{2} ,...,u_{m} ,\ v_{1} ,v_{2} ,...,v_{n}$

So $\displaystyle Tv_{1} ,\ Tv_{2} ,...,Tv_{m}$ span range T

$\displaystyle \sum c_{i} Tv_{i} =0\Longrightarrow T\sum c_{i} v_{i} =0\Longrightarrow \sum c_{i} v_{i} =\sum d_{i} u_{i} \Longrightarrow $all $\displaystyle c_{i}$'s are zero.



So $\displaystyle Tv_{i}$'s are LI. Hence $\displaystyle n=1$.

So basis of V is $\displaystyle u_{1} ,u_{2} ,...,\ u_{m} ,v$, where $\displaystyle Tv$ spans range T.
 
7:04 AM
i'm a little confused by the first chain of implications. particularly the middle \implies. and the appearance of d's and u's at the end.
maybe also confusing to let m denote the dimension of dim null(T). isn't m already in use as the dimension of W? i agree that you extend a basis for null(T) to a basis of V by adding only one vector. that's the right conclusion, but i'm not sure i follow your proof.
you do seem to have hit upon the idea of a construction that works. if i were writing it up for an audience i would include more detail about why the {w, w_1, ..., w_k} are linearly independent. the c's and that equation come out of nowhere although somebody familiar with the definition of linear independence could see where they are coming from.
 
7:20 AM
@leslietownes I shared the working that I did skipping details so it is confusing. Sorry about that. I'll clarify: d's and u's: So I had $T\sum c_i v_i=0$ which means that $\sum c_iv_i$ is in null T (which has a basis $u_1,u_2,...,u_m$). But this means that there exist $d_i$'s (scalars) such that $\sum c_iv_i=\sum d_i u_i$. Now taking $u$'s to LHS, we get a linear combination of $u$'s and $v$'s which must imply that all scalars $c_i$'s and $d_i$'s are 0 because $u$'s and $v$'s are basis of V.
Then I claimed that $w, w_i$'s are LI. Proof: suppose that for any scalars $c_i$'s $c_1w+c_2w_1+...+c_k(w_k)=0$,writing $w=Tv-\sum w_i$ results in what I had written
and all $c_i$'s (here) turn out to be $0$ as $Tv, w_1,...,w_k$ are LI
I'll share the complete working with all details (in one piece) shortly.
 
you've convinced me :) just offering unsolicited advice for presentation to others
 
@leslietownes you have told me this earlier also (my presentation issue!). I'm just trying to improve that. Thanks a lot Leslie :)
 
7:38 AM
This is a silly question, but does a regular C^0 curve make any sense to you?
I am reading a paper that has such an implication not sure it can be true!
 
koro, you might think about generalizing. say S, T are in L(V,W) and (v_j), (w_k) are fixed bases for V and W respectively. when can you find bases (v_j'), (w_k') for V and W so that the matrix of T with respect to (v_j', w_k') is equal to the matrix of S with respect to (v_j), (w_j)
 
Let $\displaystyle u_{i} '$s be a basis of null T. Then extending it to basis of V gives


\begin{equation}
u_{1} ,u_{2} ,...,u_{m} ,\ v_{1} ,v_{2} ,...,v_{n} \tag{1}
\end{equation}


So $\displaystyle Tv_{1} ,\ Tv_{2} ,...,Tv_{m}$ span range T. (Because for any for $\displaystyle x\in V$, there exist some scalars $\displaystyle x_{i} '$s and $\displaystyle x_{i} '$'s \ such that $\displaystyle x=x_{1} u_{1} +...+x_{m} u_{m} +x_{1} 'v_{1} +...+x_{n} 'v_{n}$ hence $\displaystyle Tx=\sum _{i=1}^{n} x_{i} 'Tv_{i} \in $range T. So $\displaystyle \{Tx:\ x\in V\} =span( Tv_{1} ,Tv_{2} ,...,Tv_{n}
the complete solution.
@leslietownes ok. I'll think about it. Thanks a lot. :)
 
 
3 hours later…
10:23 AM
Just came here to say: I feel like I am transforming myself into a god by learning additional 2 more language Chinese and French. Now I got 7 language. I feel like I am gonna levitate on air soon and I my brain is going to generate electricity for whole Earth for free.
Cringe*
 
10:34 AM
Please remove that post I regret writing that.
 
 
1 hour later…
11:46 AM
Today my physics teachers gave a small intro on calculus... he explained with riemann's sum and showed us that for $f(x) = x^2$ The area under the curve between 0 and 1 approaches 0.333... when he increases the number of rectangles. Again he did this for 0 and 2 and it approaches 2.6666..., so this common observation led to the idea of anti-derivatives ? Or is there any logical approach ?
 
@Thorgott Yes the assumption is for excision. In my diagram, I indirectly showed $H_n(M^{\prime},M^{\prime}-B)\rightarrow H_n(M,M-B)$ is an isomorphism by showing all other maps in the diagram are isomorphisms.
 
ok, if you wanna do it that way, why is $H_n(M^{\prime},M^{\prime}-B)\rightarrow H_n(M^{\prime},M^{\prime}-y)$ an isomorphism?
 
12:08 PM
Ok for the isomorphism $H_n(M,M-B)\to H_n(M,M-x)$ for general manifold $M$, I use excision. I thought replacing $M$ by $M'$ is not a problem
 
that's not a statement that follows from excision
it couldn't possibly
when you excise, you excise from both the ambient space and the subspace
 
I thought the isomorphism is by the above diagram
And I think the lower horizontal isomorphism is by excision
maybe not
$U$ is a local chart of $M$ that contains $x$. $U\cong \Bbb B$
 
12:36 PM
Oh the excision make sense in the lower horizontal map if $cl(B)\subset U$
Return to the original question, once $cl(B)\subset M'$, then three isomorphisms in my diagram hold so $H_n(M^{\prime},M^{\prime}-B)\rightarrow H_n(M,M-B)$ right @Thorgott ?
I should always assume $B$ to be small enough so that $cl(B)$ is contained in the coordinate chart at least
 
@love_sodam this is correct
but you also need $\mathrm{cl}(B)\subseteq U$ for the left vertical map to be an iso
 
@Thorgott even if I don't refer the commutative diagram?
 
how would the commutative diagram help
 
I mean using the commutative diagram, I conclude the left vertical map is an iso since all other maps are iso.
 
you've told me the same thing about the right vertical map though
so that's a circular argument
 
12:52 PM
@Thorgott Oh I was talking about this diagram
sorry about the confusion
 
how does this help?
tell me which map you want to conclude is an iso
 
1:09 PM
@Ishwaran I think the answer is “not quite”, and the name is a clue: Riemann sums are named after the mathematician Bernhard Riemann, whose contributions to math were mid-19th century. By contrast, Newton/Liebniz’s invention of calculus is generally dated to the late 17th century
That’s not definitive, and apparently Riemann sums go at least as far back as Cauchy in the 1820s. But there’s still a large gap between that and Newton/Leibniz
This doesn’t mean the concept of integration and its relation to differentiation was unknown to the inventors of calculus, but the concept wasn’t entirely formalized at that point
For more discussion on the history see for instance digitalcommons.ursinus.edu/cgi/…
In particular, the following line sticks out: Most mathematicians before Cauchy’s time preferred to think of integration as the inverse of differ- entiation: to evaluate $\int_a^b f (x) dx$ you found an antiderivative $F$ of $f$ and evaluated $F (b) − F (a)$.”
Which is great if you can find F explicitly. What’s useful about the Riemann sum calculation is that you only need the original function
There’s also issues that arise when f fails to be continuous
 
1:27 PM
I am about ready to murder one of my students.
 
Do tell.
 
In all of his vast experience as a student, he has developed certain notions about how a class should be run, and tells me so ever time I violate his rules.
 
Yyyikes
 
When I assigned the first exam, he told me that all of his other instructors have always given him two hours to complete an exam, and so I should give them two hours, instead of the 50 minutes I gave them.
(My exams are short, and written to be completed in a 50 minute class.)
I got another email from him this morning explaining how all of his previous teachers have handed out study guides, so I should hand out a study guide.
 
1:29 PM
DUDE! That is what the HOMEWORK is.
I mean, I have taught classes where I have handed out study guides, but I have determined that they are a lot of work, and they don't actually seem to help students very much (they either include all of the topics which I expect student to know, and are therefore as broad as saying "do all the homework", or they are too narrow, and students end up not actually studying "that one problem" which shows up on the exam, and complain about that).
 
I mean, for a final exam it might be different, but for the first one? Different instructors do different things
 
I mean, my teaching practice is relatively intentional... I've thought about the points this student raises, and made decisions.
 
@Semiclassical Sounds interesting
 
But the f'king attitude... "My other teachers do this. You should do it, too."
ARG!
Not even "Why have you chosen to do it this way?" Nope. "You are doing it wrong."
 
1:32 PM
@Ishwaran typical real analysis stuff
 
And the worst part is that this student is currently the highest scoring student in the class. Like... dude... you are doing fine. Stop complaining.
 
@XanderHenderson not a small amount of Dunning-Kruger in effect there
 
Okay, I have vented. Back to your regularly scheduled programming.
 
@XanderHenderson: really interesting. I wonder if the student motivates others in class to complain in similar manner :)
 
Geyser eruption has ceased for today
Entirely unrelated question
Is the Hopf fibration also a bundle? If so, I don’t get why it’s called the Hopf fibration and not the Hopf bundle
 
1:38 PM
@XanderHenderson In India, No one would dare complaining about these to teachers
 
Unless, say, the S^3 version is a fiber bundle but not for S^7 or S^15
 
they're all fiber bundles
 
Hmm
Then why call them fibrations? Seems odd
 
I'm not sure why the term Hopf fibration is more prevalent. To an algebraic topologist, the specificity of being a fiber bundle rather than just a fibration is usually not relevant.
 
Hmm, ok
 
1:42 PM
The term Hopf bundle is used too at times, though. On the other hand, the Hopf fibrations arise as the sphere bundles of tautological line bundles over certain projective spaces and I've seen those line bundles themselves alternatively referred to as Hopf bundles too.
 
language, what're you gonna do
 
Say I have a matrix exponential $e^{A}$ for some real symmetric matrix $A$, is there an easy way to know what the value of $\log \big( e^{A}_{i,j} \big)$ is?
oh wait thats not exactly what im looking for
 
1:58 PM
@Flows. a typical too is tr(log M) = ln(det M)
which for M=e^A gives tr(A) = ln det(e^A)
which is useful but not fine-grained enough for your purpose
@Flows. to check, do you know both e^A and A itself, or just e^A?
 
I know $A$
well im reading something
 
then i'm not sure what's stopping you from just saying $\log e^A=A$
 
so basically
the author of what im reading seems to know what $\log (e^{A})_{i,j}$ is
like I couldnt tell you what $e^{A}$ is ...
Only $A$
 
what are you reading?
 
some paper
 
2:03 PM
how very specific
 
hold on
here $c\in \mathbb{R}^{N\times N}_+$ is the matrix
then he seems to be able to easily calculate ^ ... $\pi$ is just some other matrix.
There must be something simple im missing.
reading
: Entropic Approximation of Wasserstein Gradient Flows
 
where does that log show up?
 
basically it seems he can calculate what $\log (\xi_{i,j})$ is very easily
 
well, there is stuff you can do to find matrix logs computationally
 
isn't $\xi_{i,j}$ just a number
 
2:13 PM
yeah
the i,j component of $\xi$,
 
@Thorgott Just a number? JUST a number?!
Every number is a special little snowflake, and you need to ACKNOWLEDGE that!
 
Man... "just" a number. That is some colonialist BS, right there.
:P
 
if $\pi_{i,j}$ and $\xi_{i,j}$ are just matrix elements of their respective matrices
then i don't know why computing $\log(\pi_{i,j}/\xi_{i,j})$ would be an issue
pick the two matrices, find the ratio of those matrix elements, and compute the log
 
im not explaining myself well. dw guys :) thanks though !
 
2:16 PM
the point is that $\log(\pi_{i,j}/\xi_{i,j})$ isn't a matrix
 
yes
 
the K-L divergence of $\pi$ on $\xi$ is defined in terms of the matrix elements of both
so I'm not seeing why you couldn't just list out all the matrix elements and compute, however tedious that would be
 
2:30 PM
@XanderHenderson :P
 
@XanderHenderson What is "number"? natural?integer?rational?real?complex?quarternion?transendental?algebraic?contructable?computable? idk
 
how about grassmann numbers, those are fun
(calling them numbers is physicist silliness but so it goes)
 
Or numbers are the elements of any algebraic structure like groups , rings , fields?
 
that's a way to look at it, i guess. ultimately the notion of "number" proves pretty useless
if you define it narrowly, you exclude a lot of useful stuff
if you define it widely, then it's so broad as to be pointless
also, rings/groups/fields have axioms
there's no axioms for what it means to be a number
 
2:51 PM
@Prithubiswas $p$-adic.
 
0
Q: Finding continuous functions $f$ with property that $f(x)\in \mathbb Q$ if and only if $f(x+1)\in \mathbb R\setminus \mathbb Q$?

KoroThe question is to find such continuous functions $f:\mathbb R\to \mathbb R$ which have the property that $f(x)\in \mathbb Q$ if and only if $f(x+1)\in \mathbb R\setminus \mathbb Q$? I tried to solve this question as follows: I'll prove that such a function does not exist.Suppose on the contrary ...

Is my proof correct? Thanks.
I think that I skipped only one detail there but I think that that doesn't change the final conclusion. The detail being the case when $h(x)=f(x)f(x+1)=0$ for all $x$.
 
@Thorgott I mean the left vertical map $H_n(M',M'-B)\to H_n(M,M-B)$ is an iso as all other maps are iso.
 
for that argument to work, you have to tell me why $H_n(M^{\prime},M^{\prime}-B)\rightarrow H_n(M^{\prime},M^{\prime}-y)$ is an iso
 
3:09 PM
koro it's such a long argument i lost interest in checking the details after a while but it looks right. :) note you could do something simpler by noticing not only that your g(x) is constant but that k(x) = f(x+1) - f(x) is constant for essentially the same reason. so g(x) - k(x) is constant...
 
Isn't it also followed by excision? Taking $B$ small enough.
 
again, this cannot possibly follow from excision
excision is about excising a subset from both the ambient and the subspace
 
@leslietownes yeah, I somehow had that (that k(x) is also a constant) in the back of my mind but didn't write for some reason unknown to me. Also, an answer has been posted now and as I suspected that $h(x)$ could be zero, seems to be one issue with my proof.:)
 
3:32 PM
@Thorgott From excision, $H_n(U,U-B)\simeq H_n(M',M'-B)$ and $H_n(U,U-y)\simeq H_n(M',M'-y)$ where $U$ is a coordinate chart with $y\in cl(B)\subset U$. Since $H_n(U,U-B)\simeq H_n(U,U-y)$, $H_n(M',M'-B)\simeq H_n(M',M'-y)$ I think.
 
my point was and still is that you still have to argue why $H_n(U,U-B)\cong H_n(U,U-y)$ is an iso
this, too, requires $\mathrm{cl}(B)\subseteq U$ and once you figure out why this is the case, you will realize that restricting yourself to a chart does actually not change the argument
 
@Thorgott $H_n(U,U-B)\simeq H_n(U,U-y)$ since $(U,U-B)\simeq (U,U-y)$?
Homotopy equivalence
 
@Thorgott one thing which is confusing me. On one hand, most of the references I’m seeing talk about just the four Hopf fibrations per the four normed division algebras
But one set of slides extends that further for the real/complex/quaternionic cases
 
What is the extension?
 
when someone is talking about the canonical inner product between matrices in $\mathbb{R}^{N\times N}$ what do they mean? what is $\langle A , B \rangle$
 
3:43 PM
Eg $S^1\hookrightarrow S^{2n+1}\to \mathbb{C}P^n$ for complex
 
Ah yeah
it depends on what is to be considered as a Hopf fibration
 
@love_sodam this is always false. in fact, any map $(U,U-y)\mapsto(U,U-B)$ induces $0$ on homology. don't make guesses, think
 
I figured it’s a matter of definition
 
yeah, these are all the sphere bundles of tautological bundles over real/complex/quaternionic projective spaces
the total spaces of these bundles are spheres as well
the base is a sphere iff n=1
 
Works for the octonion too.
 
3:46 PM
flows i would discourage the use of 'canonical' in this context but if the reference is identifying matrices with tuples of entries it is probably <A,B> = sum_{i,j} a_{ij} b_{ij}, i.e., what you'd get if you thought of A and B as lists of entries and used the euclidean inner product from R^(n^2)
flows note this <A,B> can be written slightly more "matrixy" as trace(B^T A)
 
Maybe I will take back my octonion comment.
It's not clear that octonionic lines are the right things
 
Yeah, the slides I’m thinking of set that case aside
Anyways. So the difference is that the usual four have spheres as base space?
 
yup
they're the only fiber bundles with fiber, total space and base all spheres by a famous theorem
 
Whereas the rest are just various projective planes for base space. Neat
 
I don't think that's true. There are topologically exotic bundles $S^3 \to S^7 \to S^4$.
 
3:56 PM
Isn’t that just the quaternionic case?
 
I think distinct from the quaternionic Hopf bundle.
That's how I think one constructs exotic $S^7$'s, but I don't know much about that.
You can have two bundles with total, base, fiber all homeomorphic but still the bundles are not isomorphic, of course.
 
Yeah, you’re right. From Wikipedia’s page on Hopf fibrations: “ As a consequence of Adams's theorem, fiber bundles with spheres as total space, base space, and fiber can occur only in these dimensions. Fiber bundles with similar properties, but different from the Hopf fibrations, were used by John Milnor to construct exotic spheres.”
So same dimensions but different constructions
 
Right.
 
what confuses me a bit is what it's getting to at the end, re: the connection to QM
i think i might see what they're getting at but i'm not sure
to wit: the state vectors for a two-level system in QM are just points in $\mathbb{C}^2$
you typically normalize them to have $$\|(z_1,z_2\|^2=|z_1|^2+|z_2|^2=x_1^2+y_1^2+x_2^2+y_2^2=1$$ so really that's $S^4$
you also only really care about those states up to a phase, so a $S^1$ degree of freedom
and there's a way to map those states to $S^2$, which is really just the Hopf map
so the usual pretty story about qubits in QM is just the Hopf fibration in action
i get that much
where i get confused is when they apply it to the case of two-qubit systems
which i've mostly seen described in relation to the quaternionic Hopf fibration $S^3\hookrightarrow S^6\to S^4$
 
@Thorgott You mean $(U,U-y)$ and $(U,U-B)$ are not homotopy equivalent?
 
4:06 PM
and I don't quite get why it's the quaternionic case rather than just the complex case with $n=4$
 
yes, they never are
 
Does a regular C^0 curve make any sense? since regularity is defined for differentiable curves I think it does not, but I am reading a paper on Bicycle model that has such implication!
 
@leslietownes thanks I believe this is the case for what im reading too :) May I also ask how would you define a gradient of a function of a matrix ?
see the start of section 3.2 of Entropic Approximation of Wasserstein Gradient Flows
 
i guess this maybe comes down to wanting to respect the two-qubit structure rather than just thinking of a generic 4-level system
aaaanyways
 
@Semiclassical You mean $S^3$. What does $S^3 \to S^2$, $(z_1, z_2) \mapsto z_2/z_1$, physically mean?
Oh you're just quotienting by the phase, OK
It's really $(z_1, z_2) \mapsto [z_1 : z_2]$. Nothing particularly groundbreaking.
 
4:49 PM
@Thorgott Is it follows from some commutative diagram or just a simple argument?
 
it follows since every map $(U,U-y)\rightarrow(U,U-B)$ factors through $(U-B,U-B)$
 
5:28 PM
@RozaTh I would not think so, but it could mean topologically embedded (homeomorphism to its image).
 
@TedShifrin Thanks! There is this lemma that says: If a bicycle's rear wheel's path is regular C^k, K >= 1, then the path of the front wheel is regular C^{k−1}. The proof is easy for k>=2 but for k=1 it would imply a regular C^0 curve for the front wheel!
Its from this paper: arxiv.org/pdf/0801.4396.pdf
 
5:45 PM
I am quite familiar with the bicycle question. It appears in two of my books, including the differential geometry book. Yes, you certainly can get cusps. Lemma 2.1 does not include the word regular. Where are you getting that?
 
@TedShifrin Wow! you are right. My advisor gave me the old version (2009 Experimental Mathematics), I searched to post the link here and did not recognize they have changed the lemma! in the old version there is the word regular!
 
LOL ... so there! :)
 
Yeah, they really mean $k\ge 2$ there, of course, since $\gamma''$ appears.
Yeah, there would certainly be no justification for topological embedding from that computation when $k=1$. In fact, the computation is not valid.
 
Hey guys, can someone help me? I need to solve this question:

Assume that the duration X, in minutes, abut a certain call conversation is a random variable with probability density function:

$$f(x) = \left\{ \begin{array}{ll} \frac{1}{5}e^{\frac{-x}{5} & , if x > 0 \\ 0 & , otherwise \end{array} \right.$$

(a) Determine the mean duration E(X) of this call conversation type.

My think: I don't know how I do this because the call can last over the person's entire life if he/she wants.
 
5:54 PM
@TedShifrin Thanks a lot I got much more than a answer. I will continue reading from the new version!
 
Great :)
Why is that a problem, @Matheus?
It lasts infinitely long with probability $0$, of course.
 
matheus: if the probability of long calls decays fast enough with the length of the call the mean can still be finite. just use the formula.
this is a calculus problem of evaluating the integral from 0 to infinity of x * 1/5 e^(-x/5) dx. it's an 'improper' integral but it converges.
 
@Ted and @leslie Ok, I was thinking it was a discrete random variable
because you can count the call seconds (No one will do this but it's possible)
Thanks for help
 
Not when they give it to you as a function of $x\in\Bbb R$. :P
I don't see why that makes a difference, anyhow.
 
you could have the same phenomenon with a discrete variable, although the formula for the pdf would be different and you'd be testing convergence of a series.
 
6:01 PM
Ah ok, I understood now. Thanks @Ted and @leslie
 
I think @leslie was more explicitly helpful. I was just asking questions, as usual :P
 
if it was helpful it was only by accident. they are power washing the exterior of our house right now and i am very distracted by the noise.
 
You always have some distraction to complain about. You'd complain about the munchkin if she were home.
 
yes. it's not me, it's the distractions.
 
6:32 PM
Why is epsilon afraid of zeta?
(Because zeta eta theta.)
 
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