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12:18 AM
@Obliv So this shows you why you have a very important misunderstanding. Saying $x=-1$ is not equivalent to saying $x^2=1$. You have to be very careful with algebra.
 
the answer to math.stackexchange.com/questions/3206959/… explains why gamma has a local minimum around there. (note there's often a shift-by-one connecting factorial and gamma, which is why the answer talks about a local min near 1.462 instead of 0.462)
there are other MSE posts about the value of x for which gamma(x) has that local minimum. apparently there's no nice expression for it.
 
According to wiki negative numbers weren't studied in mainstream math until the 1600s not sure how true that is
I wonder what was the motivation in doing so. Probably economics
 
If I scold a function, will it be well behaved or not?
 
Depends. If it's a young enough function, then possibly. If it is in its teen years, doubtful.
 
Its really badly behaved
Newton's algorithm keeps giving me stupid answers
 
12:24 AM
what is the function
 
That is what I'm trying to figure out
 
if $\frac{dy}{dx} = y^{2/3}$ asked to give a region in the xy plane that guarantees a unique solution for this d.e. that passes through the point $(x_0,y_0)$do i just find the bounds of which $y^{2/3}$ and $\frac{\partial (y^{2/3})}{\partial y}$ are continuous
or is it the partial w.r.t. the independent variable $x$
I guess since the partial is $\frac{2}{3\sqrt[3]{y}}$ it's all reals except $y=0$ ?
correct
for $\frac{dy}{dx} = \sqrt{xy}$ we have $\sqrt{xy} > 0$ then $\frac{\partial (\sqrt{xy})}{\partial y} = \frac{\sqrt{x}}{2\sqrt{y}}$
defined on $x\geq0,y>0$
 
12:55 AM
why does the partial derivative of a function $\frac{dy}{dx} = f(x,y)$ test a solutions uniqueness
 
That’s a matter of understanding the proof and also seeing counterexamples when the condition fails.
 
To calculate how many full houses (3 of a kind with another pair) are possible from a deck of 13 ranks and 4 suits (52 cards as usual), why is it $13\cdot 12{4 \choose 3}{4 \choose 2}$ and not ${13 \choose 2}{4 \choose 3}{4 \choose 2}$? I get we have $13$ possibilities for the "3 of a kind" rank (say aces) and we then have $12$ options for the pair's rank. But how come we're not choosing $2$ ranks from $13$?
 
@TedShifrin How to find a Sylow $3$-subgroup of $PSL(2,5)$ using the characteristic polynomial and conjugacy classes?
 
1:21 AM
@CottonHeadedNinnymuggins combinatorics gives me anxiety, but I think your proposed solution ignores that a six-king full house is different from a king-six full house
 
@Novice That was my suspicion, 13 choose 2 is picking 2 ranks without regard to the order (K,6 vs 6,K full house) as you said
 
Yeah, you shouldn’t think about $\binom{13}2$ here. Order matters.
 
So many ways to mess up counting
I'd calculate how many ways but I'm afraid I'd mess that up too
 
That may be uncountable!
 
1:32 AM
$\frac{dy}{dx} - y = x$ is very plainly obvious to exist and be unique for all values in the xy plane right
just making double sure i'm doing this right
 
A terrible mathematical sentence, but yes. You need to say things that make sense.
 
Right, I will try to be more accurate. Also would you say $(1+y^3)y' = x^2$ is implicit and $\frac{dy}{dx} = \frac{x^2}{1+y^3}$ is explicit?
or is it called the "normal" form? for the one on the right
Also what is preferred, this notation $I = (-\infty,-1) \cup (-1,\infty)$ or $x,y \in \mathbb{R} \cap -1$
-1 in brackets*
idk why both of those are duplicated
it fixed itself somehow that was odd
 
1:54 AM
@Obliv Maybe $x,y\in \mathbb{R}$ without $-1$. I don't know how to get the backslash "\" in there to indicate without $-1$
 
$\text{\}$ i think double backslash
yeah idk
 
I would say the first one us implicit but I'm not qualified to tell you lol.
I don't think there's such thing as "normal" either, but again, not qualified
 
I think you're right. Also do I have to carry around $\pm$ from something like $3x^2 - y^2 = C$ as a one parameter family of solutions for $y\frac{dy}{dx} = 3x$ so like $y = \pm\sqrt{3x^2-C}$ then plugging it into original as well as taking derivative. Do I have to treat it as two families of solutions $y_1 = \sqrt{3x^2 - C}$ and $y_2 = -\sqrt{3x^2 - C}$
it's like a two for one problem, i have to verify both :\ ?
 
good morning everyone, I need some clarification in the case of overdetermined system (i.e. $ m > n$) why it is always $Au \neq 0$ when $u\neq 0$. See the following picture
I need to understand the proof in the above mentioned picture.
I can see why $A^T A$ and $A A^T$ are positive semidefinite but I need some help why in the case of full rank, they are positive definite. I'm also aware of $rank(A^T A)=rank(A)$.
 
2:10 AM
Suppose we have an ordered set $S$. The usual definition of an upper bound of subset $B \subset S$ seems to be an element $\beta \in S$ s.t. for all $x \in B x \leq \beta$. Is an equivalent definition that $\beta$ is an upper bound of $B$ if for all $x' \notin B x' \geq \beta$?
 
@Croco what does full column rank mean
 
i feel like the answer is no
 
every row element in the column is non-zero?
 
@SillyGoose no, take a subset of the real line and choose a number outside this subset, to its left
 
my linear algebra is shaky, but $A$ having full column rank means $A$ as a map is injective
 
2:14 AM
okay i see; i am having trouble then seeing how rudin comes to the conclusion that 1+x itself is the upper bound in this proof (the last paragraph of the following image)
 
@Obliv full column rank means here $rank(A) = n$ in case $ m > n$.
since rank(A) = min(m,n)
 
ah okay
 
I feel liek after reading the last line surely any $ t > 1+x$ is an upper bound of the set, but I do not see how $1+x$ itself is an upper bound
 
linear maps send zero to zero, so note injectivity and then bob is your relation
 
How do I show that $PSL(2,5)$ has $10$ Sylow $3$-subgroups, and $5$
Sylow $2$-subgroups?
 
2:25 AM
@SillyGoose neither $t$ nor $t^n$ will be in the set if $t > 1+x$
@SillyGoose this designates all numbers to the right of $1+x$ on the real line
 
silly: maybe forget the specific setting of this argument and convince yourself that if b is any real number and E is any subset of real numbers, knowing that "if t > b then t is not in E" implies that b is an upper bound for E
 
so if no number to the right of $1+x$ is in the set, then $1+x$ is an upper bound
 
so in this special case, $\beta$ is an upper bound of $B$ if for all $x' \notin B x' \geq \beta$?
 
strictly greater
yes, in this case the statement works because the set includes everything on the left
it would not be true if the set didn't go to infinity
 
 
1 hour later…
3:56 AM
In words (combinatorial proof) how would I explain why there’s as many odd-sized subsets of $[n]$ as even-sized subsets of $[n]$? Any hints?
 
what is [n]?
 
$[n]=\{1,2,3,…,n\}$
 
there are lots of ways to do this, and i'm a little worried about the "in words" and "combinatorial" requirements, but a rule like "add 1 if it's not in there and remove 1 if it is" defines a bijection between the odd- and even-sized subsets of [n]
 
I guess I don’t know what you mean by “ add 1 if it's not in there and remove 1 if it is”
 
i'm not sure that counts as an 'explanation' or a 'proof', but it is a true statement
ok, can you say more about what you mean by a 'combinatorial proof'? are you familiar with proofs that construct a map and show it is a bijection?
 
4:06 AM
I am familiar, I just can’t tell what you meant
 
if $E$ denotes the set of even subsets of $[n]$, and $O$ denotes the set of odd subsets of $n$, then you can check that the rule $f(S) = \begin{cases} S \cup \{1\}, & 1 \not \in S \\ S \setminus \{1\}, & 1 \in S \end{cases}$ defines a function from $O$ to $E$ (or vice versa) and that this map is a bijection
 
Combinatorial proof being if we wanted to know why ${n \choose k}={n \choose n-k}$ we would say “counting the number of ways to pick $k$ people from $n$ people is the same as counting the number of ways to pick who isn’t in that group”
@leslietownes I see, and since there’s a bijection, the sets are the same size, right?
 
if $n$ is odd, complementation (the rule sending $S$ to $[n] \setminus S$) defines a map from $E$ to $O$ or vice versa that's maybe even a little easier to conceptualize as a bijection
yeah
 
@leslietownes good morning, would you please take a look at my above question if your time permits?
 
croco: Ax = 0 iff x_1 A_1 + x_2 A_2 + ... + x_n A_n = 0 (where here x = (x_1, ..., x_n) and A_k denotes the kth column of A). saying that A has "full" column rank is the same thing as saying that the columns A_k are linearly independent, so that Ax = 0 implies x = 0 in that case
 
4:22 AM
@leslietownes O my god, I can't believe I couldn't see that way. you're absolutely right and lifesaver.
 
I’m still a bit confused about the function you’ve defined. If we take $f(O)$ we remove $\{1\}$ from the set. How does that map to the set of even-sized subsets $E$? $\{1,2\}$ is in $E$ for example?
 
@leslietownes I'm interested in the properties of $A^T A$ or $AA^T$ where $A \in \mathbb{R}^{m\times n}$. Do you know if these matrices have specific name in the mathematical literature?
 
4:38 AM
I'm studying CT-DS markov process. Suppose $\tau_n$ be the holding time and $T_n$ be the time when transition happened. Then how to prove that $P(\tau_n>t+\delta|\tau_n>t,X_{T_n})=P(\tau_n>\delta|X_{T_n})$?
 
@CroCo positive semi-definite symmetric
 
He then used the definition of conditional probability to prove the exponential distribution of the holding time. So I guess he didn't use the definition of conditional probability to prove this result. I remember he used a picture to argue this but Idon't remember the argument.
 
@TedShifrin thank you so much.
 
@CroCo It’s easy to see that your matrices are that, once you learn definitions. The converse will require the spectral theorem, unless leslie knows a trickier way.
 
@TedShifrin I know they must be at least positive semi-definite but I thought there is a proper searchable term so I can read further.
 
4:50 AM
@CottonHeadedNinnymuggins f is a function defined on elements of the set O. "f(O)" doesn't make sense
 
Do you know about bases and diagonalizing and the spectral theorem?
 
for example if n = 5, then f({1,2,3}) = {2,3}, f({1,2,3,4,5}) = {2,3,4,5}, and f({2,3,4}) = {1,2,3,4}, f({1}) = emptyset, f({4}) = {1,4}, etc. put in an element of O, get out an element of E. f is a function from O to E
 
@TedShifrin yes.
 
@leslietownes Makes sense, thank you. I was picturing the entire set $E$ going in there at once lol
 
Then prove that every positive semidefinite symmetric matrix has a positive semidefinite symmetric square root.
 
4:55 AM
@TedShifrin Didn't know you could square root a matrix
 
Not always.
 
and sometimes, quite a lot.
 
I know you can when the matrix is semi-definite symmetric.
Wanna know how I know?
Because you said
 
ted we went to the little duck pond today and munchkin got to yell at ducks
 
Did they yell back?
 
4:58 AM
I am not sure of a proof for this problem on inequalities
0
Q: Proof of an implications involving inequalities

BAYMAXSuppose $$\lambda_{1} = \frac{1}{2} (x^2z - \sqrt{4wy^2 + x^4z^2}) , \lambda_2 =\frac{1}{2} (x^2z + \sqrt{4wy^2 + x^4z^2})$$ $$\lambda_3 = \frac{1}{2} (w^2 + y+xz^2 - \sqrt{-4w^2y + (-w^2-y-xz^2)^2}) $$ $$\lambda_4 = \frac{1}{2} (w^2 + y+xz^2 + \sqrt{-4w^2y + (-w^2-y-xz^2)^2})$$ $S=\{(x,y,z,w)| |...

 
@CottonHeadedNinnymuggins But I lie.
 
some of them quacked or honked back
 
@TedShifrin
For $AA^T$, it is clearly symmetric matrix that is
$$
(AA^T)^T= (A^T)^TA^T = AA^T.
$$
An $n\times n$ real symmetric matrix $M$ is positive semidefinite if
$$
x^T M x \geq 0 \text{ for all } x \in \mathbb{R}^n
$$
For $AA^T$, we have
$$
x^T(AA^T)x = (A^Tx)^T(A^Tx) = \lVert A^Tx \lVert^2 \geq 0.
$$
right?
 
Good job.
 
@TedShifrin Oh
We're all screwed
 
5:00 AM
Nah.
 
@PNDas Stochastic process people hang out here infrequently, if at all
In linear algebra, the Gram matrix (or Gramian matrix, Gramian) of a set of vectors v 1 , … , v n {\displaystyle v_{1},\dots ,v_{n}} in an inner product space is the Hermitian matrix of inner products, whose entries are given by the inner product G i j = ⟨ v...
 
Yeah, not sure who knows stochastic …
Good, Novice. I should have thought to give that name.
 
@TedShifrin thank you Sir. My problem was with positive definite proof and @leslietownes showed me why by considering the fact that $x_1 A_1 + x_2A_2+\cdots+x_nA_n = 0 \implies Ax\neq 0
$ when $x\neq 0$
@TedShifrin @Novice I'm not sure if I can extend the results for complex matrices. In the link posted, it says "The Gram matrix is symmetric in the case the real product is real-valued; it is Hermitian in the general, complex case by definition of an inner product."
 
In the complex case, you need the conjugate transpose.
 
How to reply to your own message here?
Anyway, the answer is quite easy. it follows by the time homogeneity and markov property.
 
5:10 AM
@PNDas click on the arrow of your comment. there is an option.
their chat is extremely limited though.
 
@CroCo Are you talking about the triangle on the left side? I don't see the reply option.
On others message I get an arrow on the right but I don't get that arrow on my own message.
 
Oh yes. I didn't realize that. Anyways, the chat is extremely limited.
 
Almost right
 
I clicked and copied the "permalink" option; strange, one can flag for a moderator on your own message
 
5:39 AM
@user7269591
 
5:59 AM
Can anyone please explain what is mean by this:" foot of ordinate of P(x,y)"?
I am asking this in ref of coordinate geometry.
Is it the point Q(x,0) on x-axis?
Or is it the point A(0,y) on y-axis?
 
Very awkward and old-fashioned. Give some context.
I would guess on the $y$-axis, because the other doesn’t depend on the ordinate
 
@TedShifrin I actually got this terminology while reading this theorem on caculus : A curve is concave or convex at P wrt the foot of the ordinate of P according as $y\frac{d^2y}{dx^2}$ is negative or positive.
@TedShifrin @TedShifrin you mean the point A(0,y)(on y-axis) is the foot of the ordinate of the point P(x,y) , right?
 
6:20 AM
[This terminology was encountered in the regional book of calculus]
 
Book written when? In what language?
 
@TedShifrin It's an Indian book written by RK Ghosh and KC Maity in 1960!
 
It seems wrong. Stilted language and wrong. Ordinarily, with $x$ the independent variable, I thought $y$ was the ordinate, bit maybe I’m wrong. But we talk about convexity of the function of $x$ with the second derivative.
Stilted and old-fashioned. Convex with respect to a point? Seems horrid to me. Forget you ever read this.
 
Second question, why are you using such an old textbook :P
 
Hmmm....but I wanted to know specifically the meaning of that terminology....
 
6:27 AM
Skim ahead...
 
@user7269591 in my university they suggested it. I live in India 🇮🇳 and here it's followed popularly
 
Ask your prof.
 
Off topic: Maybe in US the education system is much more cool and advanced .
@user7269591 nice joke! But my prof is not helpful at all
😂😂
That's the problem in India, in general, lack of cooperation.....Anyway sorry for being off -topic.
 
Have you tried the index and glossary?
Perhaps, other students?
Even other books.
 
@user7269591 Index is of no help here, I tried. Your second suggestion is not correct to me as other students literally dont care and ignorant neither are they so much of a mathematics enthusiast. Your third suggestion is the most appropriate but still......if I would've found the meaning of this weird "foot of ordinate" , it would've helped me. 😄 What do you think about this? A bit of clarifaction: in Mathematical text do we use this the ordinate x=2? I think this might solve this dilemma
 
6:39 AM
In common usage, the abscissa refers to the (x) coordinate and the ordinate refers to the (y) coordinate of a standard two-dimensional graph. The distance of a point from the y-axis, scaled with the x-axis, is called abscissa or x coordinate of the point. The distance of a point from x-axis scaled with the y-axis is called ordinate. For example, if (x, y) is an ordered pair in the Cartesian plane, then the first coordinate in the plane (x) is called the abscissa and the second coordinate (y) is the ordinate. In mathematics, the abscissa (; plural abscissae or abscissas) and the ordinate are...
But as Ted said, they are old-fashioned terms, and Western books haven't used them much in the last century.
 
@PM2Ring Thanks a lot! So the foot of the ordinate of P(x,y) is literally A(x,0) ? Thanks again. That helped me. It's true that western books haven't used them much but in Asia it's probably common( like in China, India atleast) , I maybe wrong. But in India it's followed here , those old terms.
Off topic: I am a big fan of German, Russian, American texts !
 
@porridgemathematics Ohh, I meant that I like your explanations; and I also liked the thought process behind 'how he got two spheres touching each other.' Thank you very much for taking the time to explain it. :-). It's something that I also like to do -how one got this answer? What mistake did he make? etc.
@Arthur not really.
I haven't seen 'abscissa' and 'ordinate' in a long time. These days, we just say -x coordinate or y-coordinate etc.
If you see Hindi translation of SL Loney's trigonometry, you'll find a chapter on 'Kshitiz niti', which is like 'Dip horizon'. During my highschool, I asked my teachers what 'Kshitiz niti' meant, they didn't know. So the terminology in a book also depends upon the time the book was written.
 
7:00 AM
@Koro Your conversations about circles on the torus reminded me of:
May 8, 2022 at 5:57, by PM 2Ring
I like how the stripes swap when you evert a punctured torus.
 
@PM2Ring wow!! it's cool!!
 
:)
 
it's amazing (to me) how the concepts like 'twisting', 'folding' etc. have been given mathematical notions. These are just quotienting a space by some equivalence relation. :)
I don't know who invented quotient topology.
Wikipedia page en.wikipedia.org/wiki/Quotient_space_(topology) doesn't mention who invented it.
hmm, also strange now that I think about it: in Topology I haven't yet come across people's name. Like in groups, you see Lagrange's theorem, Cauchy's theorem; in linear algebra you see Gram Schmidt process, Schur's lemma etc.
ohh nvm, there is Urysohnn's lemma.
and some compactifications.
 
the extended reals still have the least upper bound property right?
Since any set that is not bounded above in the reals is certainly bounded above in the extended reals and is only bounded above by one upper bound, namely $+\infty$
 
@Koro @Koro Just for curiosity, are you an Indian? Or it's just your gk knowledge? Not to mention, we Indians have contributed a lot lot lot lot in the field of mathematics 😉... from the Vedic era(if you know,from where I think ancient mathematics developed)
 
7:12 AM
Indians contributed zero! :D
May 25, 2021 at 15:16, by PM 2Ring
Medieval Indian mathematicians did some amazing work in trigonometry, and almost invented calculus. But their mathematical culture was very family-oriented, and they only shared results, not details, with outsiders. And they didn't really have our modern attitude towards proofs. The proof thing was still a bit of an issue in the late 19th / early 20th century. Hardy had great difficulties getting Ramanujan to understand the importance of proofs.
 
@PM2Ring Yeah, Aryabhatta discovered zero rather. Bhaskaracharya, Sridharaachrya, are some ancient names if I were to name a few. And of course the legendary S Ramanujan, PC Mahalanobis(a statistician) and the list continues:D....Of course other nations has many undeniable contributions...
 
@Arthur yes, I'm from India.
 
Indians discovered some really important trig identities a couple of centuries before Europeans re-discovered them, or learned about the Indian results via the Arabs.
 
@Koro cool me too! Which university a ur studying ,mind if u tell me? 🙃
I am from Calcutta
I am sorry if this goes off topic, couldn't control my excitement brother
Bye for now!
 
@Arthur I'm at ISI K.
 
7:21 AM
@Koro You are at my dream destination. I am at JU. But the faculty, I dint like. I plan to go for ISI again for the bstat course. I am preparing for the entrance ! I don't know whether I will be able to crack it...
 
wish you all the best for the entrance exam :).
 
Are u a bstat student ? Or mstat
Can you suggest me something?
If u dont mind of course...
I am solving IA Maron for calculus.
 
I'm an mmath student.
 
And 10+2 question papers
@Koro cool
 
@Arthur what suggestion?
 
7:28 AM
@Koro About cracking the exam ?
 
Ok. So I don't know much about bachelors' courses here as I come from engineering background. But I have heard a lot about 'TOMATO Book' amazon.in/Test-Mathematics-10-Level-ISI/dp/8176711004. You may try this one as well. I hope this helps.
 
You mean you had a btech vefore doing mstat
 
yes, before joining mmath.
 
Oo...so you have a btech and then opted for mmath , right?
 
yes
 
7:39 AM
Mmath is in bangalore if I am not mistaken, ok, so it might be in Kolkata now. Maybe u did btech from IIT?
You are really a good student and you are older than me
 
mmath is offered alternatively at ISI K and at ISI B.
 
Cool
 
yes, I did my BTech from an IIT.
 
Cool!
 
then I had industry experience. Since I am interested in maths also, I wrote exams, resigned from my job and joined ISI K for mmath.
 
7:42 AM
Nice
 
7:54 AM
Indians also did important work in combinatorics. stackoverflow.com/a/31678111/4014959 "the algorithm of the 14th century Indian mathematician Narayana Pandita [...] is still one of the fastest known ways to generate permutations in order, and it is quite robust, in that it properly handles permutations that contain repeated elements".
2
 
8:05 AM
@PM2Ring True indeed!
Highly Off topic:If I am not wrong, maybe you are an aussie ....looking for the upcoming test match between India and Australia cricket teams! 😂😂😂(I am a big cricket fan and I like listening to Australian bands say beegees, Men at Work). Have a great day our friendly southern neighbor 😀
 
I wish I was as hardworking as average Indian ;_;
 
8:21 AM
I like tunak tunak
 
 
2 hours later…
9:59 AM
0
Q: Product of topological groups is a topological group?

KoroLet $G_i,i=1,2$ be two topological groups. Is it true that the product space $G_1\times G_2$ is a topological group too? Suppose that $G:=G_1\times G_2$. Let $o_i$ denote the group operation of $G_i$ and let $i_j$ denote the map $x\to x^{-1}$ on $G_j$. $G$ is Hausdorff as product of two Hausdorff...

Can anyone please help me with my question?
My problem is in showing continuity of the group operation in the product.
here is a solution on page no. 44 greggrant.org/armstrong.pdf
But it is not clear to me: I think there is an issue with that which I have detailed in my post above.
 
10:24 AM
nvm, I got it. :-)
 
@TedShifrin Thanks Ted, was a typo from the professor :)
 
11:05 AM
Given a topological group G and a subset A of G, and an element c in G, can we say that A is homeomorphic to cA?
 
@AlessandroCodenotti: I observe that this boils down to showing that U if open in G iff cU is open in G.
I can show this in R, but I'm confused in case of G here.
 
Show that $g\mapsto cg$ is a homeomorphism of $G$
 
11:23 AM
hmm, I showed the following: given that the group operation m is continuous. So the restriction: m:{c}$\times U\to cU$ is continuous. If cU is open then by continuity $m^{-1}(cU)= ${c}$\times U$ is open, whence U is open.
$cU$ is open $\implies U$ is open. Conversely, U is open $\iff c^{-1}(cU)$ is open, hence by the previous part cU is open.
Now, I take $f: A\to cA: a\mapsto ca$. f is one-one, onto. I claim that f is continuous: Let $U_1\in cA$ be open. There exists open set U in G, $U_1=U\cap (cA)$. $f^{-1}(U_1)=\{x\in A: x\in c^{-1}U_1\}= c^{-1}U_1=c^{-1}(U\cap (cA))= c^{-1}U\cap A$, which is open in A as $c^{-1}U$ is open in G.
So f is continuous. $f^{-1}: cA\to A: ca\mapsto a$ is continuous using exactly the same proof as above. It follows that $A\simeq cA$.
@AlessandroCodenotti proved it for any subset A of G :).
Using this, I get the following: $\bar H$ (closure of a subgroup H of G) is normal if H is normal. Indeed, for any g in G, $H\subset gHg^{-1}= g\bar H g^{-1}$, the set on RHS is closed as it is homeomorphic to $\bar H$. So the smallest closed set containing $H$, i.e., $\bar H$ is also contained in $g\bar H g^{-1}$, i.e., $\bar H\subset g\bar H g^{-1}$
i.e., $\bar H g\subset g\bar H$. Changing g to $g^{-1}$, we get: $g\bar H \subset \bar H g$, whence it follows that $\bar H g=g\bar H$, i.e., $\bar H$ is a normal subgroup.
 
11:39 AM
@Koro I don't understand why $gHg^{-1}=g\bar Hg^{-1}$. Moreover if $A,B\subseteq X$ are subsets of some topological space which are homeomorphic and $A$ is closed in $X$, this doesn't imply that $B$ is closed in $X$. You have to use that left and right translations are homeomorphisms
 
Ohh. I thought that if A is closed, then by homeomorphism B is closed too.
can you please give me an example where it doesn't happen?
actually I am quite surprised by this fact.
ohh right, $\mathbb R\simeq (-\frac {\pi}2,\frac{\pi}2)$ but ...
@AlessandroCodenotti ohh, that was $gHg^{-1}\subset g\bar H g^{-1}$. = was a typo there.
but still, the proof is wrong as you said. Thank you so much.
 
12:09 PM
Here is a fix to the above: I have already showed that U is open in G iff cU is open in G. So we have $H\subset g\bar H g^{-1}, $ whence $\overline{g^{-1}Hg}\subset \bar H$. I claim that $g^{-1} \bar H g\subset \overline{g^{-1}Hg}$.
Take any $g^{-1}\bar h g\subset g^{-1} \bar H g$. Let $U$ be any open nbd. of $g^{-1}\bar h g$. It follows that $\bar h \in gU g^{-1}$. $gUg^{-1}$ is open because it has been proven already that left multiplication doesn't change openness and using the similar proof one can show that neither does right multiplication. Since $\bar h$ is in closure of H, $gUg^{-1}\cap H\ne \emptyset$
Take any p in this non empty intersection. It follows that $g^{-1}pg\in U\cap g^{-1} Hg$, i.e., the set on RHS is non empty for any arbitrary $U$. It follows that $g^{-1}\bar h g\subset \overline {g^{-1} H g}$.
This proves the claim.
So we have indeed: $g^{-1}\bar H g\subset \bar H$. Changing g to $g^{-1}$, we get the desired equality hence showing that $\bar H$ is normal in G. :-)
(disclaimer: I have erroneously mixed up $\in $ and $\subset$ at some places above.)
 
12:37 PM
@PM2Ring I wasn't aware of this. Will check more details.
 
Yeah, combinatorics is huge in India.
 
 
2 hours later…
2:40 PM
For self-studying, is it advised to follow a course syllabus that was put up on the internet, doing the recommended readings, problems, ... in the time they were intended to be done?
2
 
3
Q: Common mistakes in probability

Botnakov N.$\DeclareMathOperator\Var{Var}\DeclareMathOperator\Bern{Bern}\DeclareMathOperator\Pois{Pois}$Question: What not-trivial mistakes do students often make when solving problems in probability theory, mathematical statistics and random processes? Some examples of wrong solutions: Problem 1: Find dist...

 
2:56 PM
Every non trivial discrete subgroup of R is cyclic.
Proof: It is known that every additive subgroup of R is either cyclic or dense in R, and that a non trivial subgroup H is cyclic when inf ($R_{\gt 0}\cap H)>0$. Let H be a non trivial subgroup of R. Suppose on the contrary that it is not cyclic. Hence it is dense in R. There exists h in H, $h\in (0,1)$. Since H is discrete, there exists an open interval (a,b), a>0, b<1, a>b such that ${h}=(a,b)\cap H$. By density (a,(a+h)/2) should contain an element of H but that's not possible, which is a contradiction.
This proves the theorem. Q.E.D.
 
3:30 PM
5
A: every proper compact subgroup of the circle group is finite

José Carlos SantosConsider the map$$\begin{array}{rccc}\Pi\colon&\Bbb R&\longrightarrow&\Bbb T\\&x&\mapsto&\bigl(\cos(x),\sin(x)\bigr),\end{array}$$which is a surjective continuous group homomorphism. If $G$ is a subgroup of $\Bbb T$, $\Pi^{-1}(G)$ is a subgroup of $(\Bbb R,+)$. But every subgroup of $(\Bbb R,+)$ ...

I don't understand why $\Pi^{-1}(G)$ is a subgroup here. Can anyone please explain that?
 
Just check the definition.
 
The problem is the following: $\Pi^{-1}(G)$ should contain the identity. $\Pi^{-1}(1+i0)=\{2n\pi:n\in \mathbb Z\}$.
ohh I don't know what I was thinking. I was expecting that identity should be pulled back to identity. But that's not always true we want only containment of the identity.
which happens here. My doubt is clear now.
 
4:09 PM
the preimage of a subgroup is a subgroup
 
4:35 PM
yes :-).
 
 
1 hour later…
6:04 PM
@ILikeMathematics i don't see any reason why the specific timing or pacing suggested by the syllabus would necessarily be useful for self study, but the ordering of the subject matter and the materials relied upon could be useful. at least if the syllabus has been put together with care by someone who selected a decent textbook or set of notes.
 
6:57 PM
It is maybe a stupid question, but what is the advantage to consider an affine scheme instead of a sheaf. So I mean an affine scheme is per definition a locally ringed space of the form $(X,O_X)$ where X=Spec(A) so it contains the notion of a sheaf of rings O_X but I when we only consider O_X then we also have the information about the underlying topological space. so why exactly do we need the notion of a scheme?
 
7:17 PM
not all schemes are affine though
 
Yes sure I could also write scheme instead of affine scheme in my question sorry
 
let me also just say, for the record, that algebraic geometry is only part of mathematics, and many people within mathematics do not "need" the notion of a scheme any more than alexander grothendieck "needed" strichartz estimates for the kdv equation
:D
 
okey but what is the advantage of the notion of a scheme instead of only working with sheaves?
 
7:57 PM
A scheme has more structure
 
8:27 PM
@Astyx in which sense?
 
8:50 PM
Is there latex formatting in the chat?
 
9:08 PM
yes, there should be a link in a sidebar to enable tex rendering (the second tinyurl link)
 
9:20 PM
If I tell you the ring (or sheaf, if you prefer) of germs of smooth functions, do you know what the underlying manifold is? I'm not sure how I tell you that ring, anyhow. I can certainly describe a manifold in more explicit ways.
 
No we have not spoken about manifolds.
 
I don't remember getting so many lazy questions like this a few years ago.
A scheme is a generalization of a manifold. I'm speaking with an analogy. Give me a break.
 
I certainly have no idea how to talk about the sheaf $\mathscr O_{\Bbb P^n}(-1)$ without first understanding $\Bbb P^n$.
Nor $\mathscr O_{\Bbb P^n}(k)$ for any integer $k$, for that matter.
 
Ah so I know projective n space, we defined it by saying that $\Bbb P^n=K^{n+1}\setminus \{0\} / \sim$ where $(x_0,...x_n)\sim (y_0,...,y_n)$ iff there exists $a\in K$ s.t. $ax_i=y_i$ for all $i$
 
9:28 PM
Yes, and there are various geometric ways to picture this and work with it. I don't know how you begin to describe its structure sheaf without knowing what the space is.
 
Does anyone know some python here?
 
Monty Python.
 
or is there separate room for it?
Hello Ted!
 
Some
 
I would assume that belongs more on the computer programming site.
Hi, @Jack.
 
9:30 PM
Yes I thought so too
But for the life of me I could not find any
all must be frozen
 
@TedShifrin but why are we now speaking about the projektive n space?
 
Because I wanted to make a point. It is the most common complex manifold/nonsingular algebraic variety/nonsingular reduced scheme over $\Bbb C$ I know.
 
but I only wanted to know why we use schemes instead of sheaves, so what is the advantage.
 
@user123234 A scheme is a sheaf on a top space with extra constraints
 
@Astyx thanks!
 
9:33 PM
Why do we use polynomials instead of continuous functions? Why use continuous functions instead of general ones? Why look at graphs instead of general subset of product sets?
 
Salut, @Astyx.
 
Hello
 
10:10 PM
@Astyx Curious whether you know that there was a rock group named styx.
@Ted Surely you remember Styx, and @robjohn
 
And there's a French comic called Astérix :P
@amWhy Vaguely. Acid rock was never my thing. But I have visited the river Styx :)
 
@TedShifrin Never mine either, but I was force to hear it, given my brother was into all that. But I remember the songs played that I could relate to. Remember, at the time, I was a John Denver Fan!! :D
 
And Joni Mitchell and Tom Rush and Joan Baez ...
 
@TedShifrin I loved folk rock from that time, though I felt I was born too late, for a lot of it!
I loved Crosby Stills Nash and Young, and Neil Young's "Live Rust" tour, presented on HBO when 8 years old.
 
I'm guessing you're at least 10 years younger than I.
 
10:23 PM
@TedShifrin Perhaps. I don't know your age, so I wouldn't know. No need to reveal, though.
 
LOL ... I already announced that a week from today is Lincoln/Darwin/Ted day. :)
 
@TedShifrin Birthday?! Yippee.
 
Lincoln and Darwin were actually born on the precise same day! I wasn't quite around then. I will be one less than one third their virtual age (allowing obvious rounding).
 
@TedShifrin Clever approach!
@TedShifrin I was born roughly six months after a former president's death. President who died younger than any other president.
That should be a give away, @Ted!
Given your puzzle, your guess was pretty much spot on, give or take a year or so, @Ted ! You shall now be known as Ted The Sleuth!
 
10:52 PM
I'm not sure I follow the question that was posed
it is impossible to talk about a sheaf on $X$ without talking about the topological space $X$ to begin with
@TedShifrin a fact I'm very fond of is that the diffeomorphism type of a smooth manifold $M$ is determined by the ring isomorphism type of $C^{\infty}(M)$
of course, you know how badly the analogue fails in complex/algebraic categories
 
@TedShifrin you're turning 70 aren't you?
You look far younger in videos
 

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