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12:00 AM
But yeah so continuing the problem, that's our setup, we want to extend by $0$ to get a sheaf on $X$. Define $\mathcal{G}(V) = \mathcal{F}(V)$ if $V\subset U$, and $0$ otherwise. Then sheafify. Okay well do we need to sheafify? Turns out yes, and the example I thought of was to take a tiny bump function within $(0,1)$, and then extend by $0$ way outside
The usual proof if the deRham theorem is sheafical.
Why can't I edit? Grrr
Oh, that I didn't know. And that's weird, trying to edit the most recent message?
If the group is discrete, what you're saying doesn't work?
Well, chat on iPad sucks.
Or not that it's necessarily not a sheaf, just that in general it isn't automatically a sheaf
So I just took that as a single counterexample of, take the sheaf of smooth functions etc
Working with an open set is troublesome, anyhow, because things can blow up as you go to the boundary. So you get the sheaf of discontinuous sections or something.
Anyhow, if you look at Bott/Tu (e.g.) you see plenty of sheaf stuff in the non-complex setting.
12:08 AM
I see
I should check it out soon for sure
Yeah, the presheaf of bounded holo fns (working on $\Bbb C$) is interesting, too.
12:28 AM
Hmmm, what would be the general approach to show that if $\text{char}(K)\neq 2$ and $[F:K]=2$ then $F$ is Galois over $K$? (In particular, where does the $\text{char}(K)\neq 2$ come into play?)
Well, what's the most obvious example of a degree 2 field extension?
$\mathbb{Q}(\sqrt{d})$ where $d$ is squarefree
That works, so what are the Galois automorphisms? Explicitly
Can I say that the degree of the minimal polynomial must be 2? Not sure where to go with that, though
Well, the automorphisms would be the trivial automorphism and the automorphism that takes $\sqrt{d}$ to $-\sqrt{d}$ and vice versa.
Does something here sound like it could go awry in char 2?
12:32 AM
Ah, right. $-1=1$
Good point
Play around and see what you can get
12:47 AM
Every degree 2 extension in characteristic=/=2 is generated by a square root (omplete the square)
2 hours later…
3:12 AM
in Probability and Statistics, 37 mins ago, by Simple
Let $\{S_n\,:\,n\geq0\}$ be a simple random walk with $p$ be the probability of stepping up and $q=1-p$ be the probability of stepping down. Let $T_k=\min\{n\geq1\,;\,S_n=k\}$. If $p<1/2$, is $E(T_1\,|\,T_1<\infty)<\infty$?
2 hours later…
5:29 AM
@Rithaniel $\chi(K) \nmid [F:K]$ so $F/K$ is separable; take a normal closure $N/K$, then $\operatorname{Gal}(N/F)$ is a subgroup of $\operatorname{Gal}(N/K)$ of index $2$, so it is normal, so $F/K$ is normal
6:13 AM
Hello. Anyone know how to search the OEIS. I am looking for non-primes which do no have perfect squares...
@deostroll $n!+4$ is always a non-square
I want the full sequence...
i have a few terms: 6, 8, 10, 12, 14, 18, 20, 21, 24, etc...
6:42 AM
@deostroll what do you mean by non-primes which do not have perfect squares?
ok I guess I know what you mean now
I don't see why you would find this in OEIS
@LeakyNun apparently $n!+4$ is never a perfect power
Do you know why?
I am not able to prove it either
Hi I have a small question: in theoretical computer science, sometimes a problem is a hard to solve, but sometimes when we reduce the problem to metric space like in TSP, then it becomes hard but still we can approximate it. Now my question is: what is the space of graph? I mean in general a graph has vertices and edges but it did not say anything about space so what is the space of this graph.
So, since every graph can be considered as a groups/rings, then does this mean that graph is on complex space since ring C is the highest level of number and cannot go beyond that.
TSP is the traveling salseman problem
if my question is not clear, then let me know
2 hours later…
8:30 AM
@AlessandroCodenotti halp why is $(a_1, \cdots) \mapsto (a_1/1, a_2/2, \cdots): \ell^2 \to \ell^2$ compact
8:54 AM
@LeakyNun What is your definition of compactness? Do you know that compact operators are exactly the ones which are limit of finite rank operators under the operator norm?
i want to die, my riemann surfaces course is filled with functional analysis
ya lol its too much for me
@BalarkaSen image of bounded contains convergent subsequence
my text says I can choose a subsequence such that each component is convergent
and I'm like, huh?
yeah that should be workable
I see how to make finitely many components convergent
8:58 AM
yeah idk man im not going to try to do it this way altho it shouldnt be too hard
too annoying
can i draw affine vectors in geogebra
Ah I can
@BalarkaSen also my text claims that identity plus compact operator is fredholm of index 0; they spent 3 paragraphs proving the 3 conditions of fredholm; but they didn't prove index 0 ;_;
also why is a hilbert space isomorphic to infinite direct sum of itself?
(R^n reacc only)
Tryna prove that a lattice $\Gamma \subseteq \Bbb R^n$ is complete iff $\Bbb R^n/\Gamma$ is compact. I got one direction (take the closure of a fundamental parallelepiped for $\Gamma$ and take the image of this guy under the canonical projection). The other direction seems harder (I think I just lack characterisations of compactness). I guess it would be easier to show that if $\Gamma$ is not complete then the quotient is not compact, but I can't draw any conclusions!
@EdwardEvans what is a lattice and when is it complete?
(I've seen 100 definitions of lattice and complete lattice)
a lattice is a discrete subgroup of R^n and complete means it has full rank
or alternatively just $\Gamma = \Bbb Zv_1 + \cdots +\Bbb Zv_n$
well if it doesn't have full rank then the quotient is just (R/Z)^rank x R^unrank right
9:09 AM
unrank rofl
I guess so
and then the R/Z factors are compact but the R's are not
conclusion: always choose basis
it just seems almost too easy
@LeakyNun That also becomes clearer if you use the fact about compactness I mentioned. If T is finite rank I + T has index zero because ker(I + T) = im T, and dim im (I + T) = dim ker (I + T)^* = im T^*, and those match
@EdwardEvans why must a discrete subgroup of R^n have a finite set of generators?
and then you just take a limit
the hard part as Leaky points out is classifying lattices in R^n
but also that's not too hard
you project to every coordinate and classify discrete subgroups of R instead
9:16 AM
ker(I+T) = im T?
well, it's a theorem that a subgroup of an n-dimensional R-vector space is a lattice iff it is discrete. The original definition in Neukirch is just that it is a subgroup of the form Zv_1 + ... + Zv_m with m <= n
@Leaky Yikes, I am not paying attention
Let me fix that
Ok, dim im (I + T) also computes the dim of the eigenspace of -1 of T
isn't that dim ker I+T
Wait lol what was index again
dim ker - codim im right
codim im(I + T) = dim ker (I + T)^* = dim ker (I + T^*)
so that's dim of eigenspace of -1 of T^*
Everything's finite dimensional so these agree
why dim ker I+T* = dim ker I+T?
9:32 AM
@LeakyNun Balarka's explanation of a limit of finite rank operators is the fastest way, I agree
dimension of eigenspace of lambda for an operator on a finite dimensional inner product space and it's adjoint are the same
facts from linear algebra yo
fun fact: it was open for a long time whether "every compact operator is a limit of finite rank operators" holds in any Banach space, until Enflo constructed a counterexample in 1973 and won a goose from Mazur because of it
There's also a photo of the prize being awarded en.wikipedia.org/wiki/Approximation_property
Mazur looks salty
9:35 AM
the fuh
lol Mazur is Barry and not Stanislaw in my mind
Barry is a good guy
he was a topologist before he went full nuts
9:37 AM
nuts is an acronym for NUmber TheoriSt right
Functional analysis is not very pogchamp
@LeakyNun Because Hilbert spaces are classified by the cardinality of an orthonormal basis
@BalarkaSen Unnecessary argument
Everything about my life is unnecessary so why not
You should talk to Leaky I am gonna duck out
9:47 AM
If you know that Fredholm + compact = Fredholm and you know homotopy invariance, then i(F+K) = i(F) because F+tK are all Fredholm
O yeah I never was able to prove homotopy invariance
Someday I will try it again
@BalarkaSen Homotopy invariance for surjective/injective Fredholm operators is clear; let's take injectivity for simplicity. Then we may think of the cokernel as a subspace $J$ of the codomain, so that the map $F \oplus I: V \oplus J \to W$ is an isomorphism, and hence of index zero. Now for small $t$ we also have that $F_t \oplus I$ is an isomorphism, because isomorphisms are open; and this $i(F_t) = i(F_t \oplus I) - i(I) = - \dim J = i(F)$, as desired.
10:04 AM
What do you do for the general case
Now suppose $F$ is a general Fredholm operator --- again let's say negative index for simplicity. Pick $J_F \subset V$ as the kernel. Then using projection we have a map $\tilde F: V \to W \oplus J_F$ which is injective, and $i(\tilde F) = i(F) - \dim J_F$. Now for $F'$ near to $F$, we have that $\tilde F'$ is near to $\tilde F$ (using the same $J_F$ --- don't replace it with $J_{F'}$)
So $\tilde{F}'$ is also injective --- and by the first argument, because it is near to $\tilde F$, we have $i(\tilde F) = i(\tilde F')$. Putting this together, we get $$i(F) - \dim J_F = i(\tilde F) = i(\tilde F') = i(F') - \dim J_F,$$ so that as desired $i(F) = i(F')$.
Thus index is locally constant and hence constant on path components
Instead of analyzing dimensions of kernels and cokernels everything is reduced in the end to stability of isomorphisms
my text says the space of fredholm operators is weakly homotopy equivalent to Z x BU lol
oh and it used the Mazur swindle to show that U(H) is contractible
the adjective weakly is unnecessary, but sure
the statement just means that maps to the space of fredholm operators give rise to vector bundles up to iso.
Q: Why is $\frac{987654321}{123456789} = 8.0000000729?!$

marty cohenMany years ago, I noticed that $987654321/123456789 = 8.0000000729\ldots$. I sent it in to Martin Gardner at Scientific American and he published it in his column!!! My life has gone downhill since then:) My questions are: Why is this so? What happens beyond the "$729$"? What happens in bas...

Well looks like I learned a new way to investigate the cause of mathematical coincidences
by rewriting any funny number in base n
and see what pops out in the resulting algebraic expression
10:21 AM
so suppose $X$ is compact, and $X \to \text{Fred}(H)$ a continuous map. You can find some large finite-dimensional subspace $J \subset \text{Fred}(H)$ so that $\text{Im}(F_x) + J = H$ for all $x$. Thus the map $\tilde F_x = F_x \oplus I: H \oplus J \to H$ is surjective for all $x$, and hence its kernel gives a vector bundle $V$. The cokernel is zero, but we added on this stupid extra term $J$, so that we call the "virtual index bundle" by the difference $\text{ker}(\tilde F_x) - J$
This is a virtual vector bundle, which is what $\Bbb Z \times BU$ classifies
$$ab = \sum_{k=0}^{\infty} \frac{a_k}{n^k} \sum_{\ell=0}^{\infty} \frac{b_\ell}{n^{\ell}} = \sum_{k=0}^{\infty}\sum_{\ell=0}^{\infty} \frac{a_kb_{\ell}}{n^{k+\ell}}$$ this said little about how multiplication propagates
unless I need to take a mod somewhere...
the point of the above construction is that you'd really like to just say $I(F) = \text{ker}(F) - \text{coker}(F)$ is a virtual vector bundle because both ker and coker are vector bundles, but the rank of the "kernel bundle" can jump, as can the rank of the "cokernel bundle". So you instead just add a factor you'll formally subtract later so that the cokernel bundle is just zero, and ker(F) is locally constant rank.
10:47 AM
This all sounds like the precursor to trying to jump into the Atiyah-Singer index formula (via heat asymptotic stuff)
Or at least that's the context in which I learned stuff like this
This is just indices of Fredholm operators in general, of course knowing that is a pre-requisite to knowing how to calculate indices of differential operators ;)
If you're doing heat kernels you won't need to know so much of the topology, like that $\Bbb Z \times BU \simeq \text{Fred}(H)$
I didn't know you learned that stuff
Yea that was the last bit of maths I did at university!
'Pseudodifferential Operators and the Atiyah-Singer Index Formula' was the name of my dissertation at the end of it all
It was actually really fun despite the lack of supervision for most of the time I had to do the project
Also yea the topology stuff above is super new to me!
Impressive that you were able to do it without much supervision --- pretty difficult material! What have you moved on to since then?
It eventually reduced to calculating stuff on a Riemann surface after I did some of the general stuff
I have done very little since then!
I took a break for a year to try and weigh up my options and realised I didn't want to study anymore so I've been learning how to program for the last month :)
iirc robjohn did his thesis on Pseudodifferential Operators
11:00 AM
Needless to say I don't remember very much of stuff like that ↑
@skullpatrol Long time no see!
Hi pal @Khallil
Wow I haven't robjohn in ages
I tried to tag @skullpatrol and realised 'skillpatrol' was another option
Have you been roaming these parts with more than one account? xD
yup :D
i picked up the habit from anon
@Khallil Good for you!! I have good friends who decided they were at the right spot to stop studying math, and made the same move, and they're very happy now.
I hope it's rewarding for you too
Thank you! Programming doesn't come too naturally for me but it's good fun
I've actually been doing some math related problems on Project Euler which has been interesting
Never been a massive fan of number theory but it seems quite important for programming (at least on an elementary level)
11:10 AM
Really, outside of project Euler? How odd
Yea it's quite strange! It seems like optimisation (for the practice problems I'm doing) usually boils down to using number relations to reduce the number of checks/conditionals
I don't think Project Euler is meant to be representative of skills necessary for programming in general, just that one needs to get better at programming (and specific programming skills) to do the problems as they get harder, but I might have misinterpreted you
Ahhh I meant to write 'important for programming similar problems to Project Euler'
Haha, fair enough
Apart from complex geometry, have there been applications of algebraic geometry that's yielded results for manifolds?
11:21 AM
It's hard for me to interpret that question, since I'm not sure what exactly qualifies as results for manifolds
Okay yeah true that question is a bit vague, I guess like just one aspect I'm thinking about was if there been applications of alg geom to construct new invariants of manifolds
But I think I do have an answer for that, moduli spaces of manifolds seem to be a growing area of research currently
moduli space isn't a term exclusive to algebraic geometry, it just means "space of objects"
the space of all squares in Euclidean space up to isometry could be called a moduli space; it is homeomorphic to $(0,\infty)$, the homeomorphism given by sidelength
Oh I see, I just assumed moduli spaces were from algebraic geometry
Thanks for linking that paper btw
Here is a much older nice paper
you could look at walter neumann's work, who has studied 3-manifolds arising from singularities of complex polynomials and their associated invariants
or you could start by looking into the Casson invariant, which is defined by a (perturbed count of) intersection points in a certain algebraic variety, the representation variety
11:38 AM
Thanks for those suggestions!
Though I do need a bit more background before I could tackle some of those things I guess
probably not for the milnor paper
anyway, the point is just that there are things to do, keep going with your studies
This kinda stuff has really piqued my interest lol
Michael David Spivak (born May 25, 1940) is an American mathematician specializing in differential geometry, an expositor of mathematics, and the founder of Publish-or-Perish Press. Spivak is the author of the five-volume A Comprehensive Introduction to Differential Geometry. In 1964 Spivak received a Ph.D. from Princeton University under the supervision of John Milnor. In 1985 Spivak received the Leroy P. Steele Prize. Spivak was born in Queens, New York.Spivak has lectured on elementary physics. Spivak's most recent book, Physics for Mathematicians: Mechanics I, which contains the material...
Can somebody explain this picture?
Sanity check : are the irreducible representations of $SO_3$ in bijections with the irreducible representations of $SU_2$ of odd dimension ?
11:54 AM
Ted can explain that, I think
> I think it's a picture of him doing a stretching exercise, which some Wikipedia vandal has given a funny caption.
from tweeter
I think he just came back from Australia and had a hard time readjusting to not being upside down
he does look flexible
What does "math is a reflection of the universe mean"?
12:13 PM
just that you can find math possibly everywhere if you search for it, and also apply it
so math is everywhere. Is that why they call it the field of mathematics?
"field" means "subject area" in that sentence
"field" is to denote the particular domain of study, i am not sure why exactly that word is "used", but work is done in fields (and was more in ages before, i mean like agriculture and similar), so that could be a clue
that´s more a question about genesis and evolution of particular languages
@geocalc33 here
12:25 PM
np, pal
Sai ak un borstan, pero marque zelarico moi tambose paco, nosga saime a toi to yenne
Sai ak un soire, moise nocto pranagarte duelo crisee normale, eso tris poino ben
okay that's enough out of me.
that's the language i'm making lol^^
12:41 PM
Is it some kinda cypher of the Latin alphabet?
no it seems to be some sort of mix of Romance languages
@Thorgott When you say "anything that is improperly Riemann-integrable is also HK-integrable", does that mean there are no restrictions at all? So the function could be non-negative (I've seen this as a condition both in Tonelli's theorem and the corollary of Fubini's theorem for improper integrals)?
1:00 PM
@Khallil Ted has explained it before in chat, stretching exercise despite the wiki caption
Jan 16 at 0:18, by Ted Shifrin
It was about limberness (he did all sorts of marshall arts stuff) ... not about shoe-sniffing.
what is usually considered to be a neighborhood of a point in metric spaces? some open set containing that point?
I've seen neighbourhood denote both an open set containing the point and just a set containing the point.
I've also seen the term open neighbourhood to mean an open set containing a point.
I think it just depends on how your lecturer/text defines the convention to begin with!
@AlessandroCodenotti I guess only the highest up can practice marshal arts xD
this is in the context of neighborhoods homeomorphic to $\mathbb R^n$ so i think they need to be just open sets containing that some point
@Masterphile Either an open set containing the point or a set containing an open set containing the point, depending on definitions (but it makes no real difference)
ahhh I forgot the open set in the set that contains the point
Thanks, @AlessandroCodenotti!
1:14 PM
am i observing something is imprecise in the following definition of the manifold:

a manifold is a metric space $M$ with the following property:

if $x \in M$ then there is some neighborhood $U$ of $x$ and some integer $n \geq 0$ such that $U$ is homeomorphic to $\mathbb R^n$
i think this could be better phrased
Why? It's quite clear
but this allows the option that $U=\{x\}$, just choose $n=0$ for every point
do you know what a neighbourhood is?
yes, i almost always take it to mean an open set U containing some point
$\{x\}$ is not open in most metric spaces though
1:20 PM
If $\{x\}$ is an open subset of $M$, then sure, this is true
yes, but then $n \geq 0$ can be changed into $n>0$
@VJ123 well, I'm not sure which restrictions you mean. Surely such a function can be non-negative
@Masterphile since you usually want the same $n$ for each point $x\in M$, this is just noting that discrete spaces are $0$-manifolds
@Masterphile the reason we allow $n=0$ is because a discrete metric space is also a manifold
how is discrete metric space defined? what it means discrete in that context?
a metric space such that every subset is open
1:23 PM
A metric space in whose topology is discrete, meaning that $\{x\}$ is open for all points of the space
So you can take $d(x,y)=1$ for all $x\neq 1$ as metric for example
why is the choice of metric not specified in the definition? does that means that if $U$ is homeomorphic with $\mathbb R^n$ with some metric $d$ then it is homeomorphic with some other metric $w \neq d$?
or that´s irrelevant for homeomorphisms?
In the definition you start with a metric space, so a metric has been given
yes, but in this context all metrics are, sort of, equivalent in some precise sense, so i can think in terms of "Pythagorean distance", or, "the usual metric" as it is sometimes called
(The definition of a manifold is something more general than a metric space whose open neighbourhoods are homeomorphic to a copy of $\mathbb{R}^{n}$)
@Thorgott for example, on the following link: math.libretexts.org/Courses/Montana_State_University/…
If you scroll down to equation 21, it says underneath that "It is very important to note that we required that the function be nonnegative on D for the theorem to work" (dealing with improper integrals).
This was the kind of restriction I was talking about, though I'm not sure why this is
1:34 PM
That just seems like a weaker version of Tonelli and I genuinely don't understand what about those integrals is "improper"
@Masterphile both the notion of a discrete space and of a continuous function are topological notions, so it doesn't really matter which metric the space carries, only what topology this metric induces
yes, that seems understandable
@Khallil Is it? I'm not convinced
If you define a manifold to be a locally euclidean hausdorff second countable space, then all manifolds are metrizable, but Masterphile's definition also considers as manifolds uncountable discrete spaces, so it seems to be the more general one
and that one can be made even more general if one replaces metric space with topological space as in that case, not all manifolds are metrizable, no?
Depends on what you ask, specifically in whether you ask either paracompact or second countable or only locally euclidean
locally euclidean topological space would be the most general definition
and then you can take stuff like the long line
1:44 PM
@Thorgott Hey, weißt du ob der Begriff "F invariant" das Gleiche bedeutet wie "commutes with F" auf Englisch? Bzw wird es normal so benutzt auf Deutsch? :)
Yea I think my definition always omits second countability so I always allowed for manifolds whose topology isn't metrisable @AlessandroCodenotti
Oha zwei mal bedeutet geschrieben lol
Yea precisely what @Thorgott said. Thanks!
@Thorgott I think everyone requires Hausdorff, but apart from that I agree
Otherwise you get the line with two origins and that kind of weird stuff

"[If you know anything about topological spaces, you can replace “metric
space” by “topological space” in our definition; this new definition allows some
pathological creatures which are not metrizable and which fail to have other
properties one might carelessly assume must be possessed by spaces which are
locally so nice."
1:46 PM
@Edward In welchem Kontext?
@EdwardEvans Is F a group acting on something?
Was your original question just that you wanted to relax the dimension from $n \geqslant 0$ to $n > 0$?, @Masterphile?
Ja also vom Kontext her ist es irgendwie offensichtlich: Ich krieg Tr(Fz) = FTr(z) lol, wollt nur fragen ob invariant normal so benutzt wird lol
one must be mad to consider non-Hausdorff spaces to begin with
@Thorgott D:
1:48 PM
@AlessandroCodenotti F is just an involution on a C-vector space
Doing Minkowski theorx hehe
ich würde unter F-invariant eher Tr(Fz)=Tr(z) verstehen
Okey ich auch, aber irgendwie benutzt der Neukirch invariant für kommutiert mit
@Khallil i am just considering some variations of the definition, i started to "do" differential geometry with only some basic knowledge of calculus and linear algebra, so i could "run" into some difficulties with understanding of the material
Zb <.,.> ist ein hermitesches Skalarproduct das unter F auch invariant ist, und in Neukirch steht <Fx,Fy>=F<x,y>
Lol ah well
F-invariant to me sounds like applying F or not makes no difference, not that it commutes with F
1:53 PM
Yeah exactly
W e i r d
if $F$ is an involution on the vector space, what even is $F\langle x,y\rangle$?
Loooolmhe then goes on to describe points that are invariant under F as points that are unaffected by the action of F
Oh sweet. I really enjoyed diffgeo when I studied it a while back. I hope you enjoy it too :) @Masterphile
@Khallil haha, i am not a student, could be mine age is strictly larger than yours , i have much free time and in it i much do math :)
@Thorgott Good question
1:58 PM
@Thorgott Hmmm. Even for Tonelli's theorem, the condition is that the function must be non-negative and measurable (en.wikipedia.org/wiki/Fubini%27s_theorem). Why can the function not be negative?
maybe he fixed an embedding of the base field into the space
This is starting to sound like you have $G=\Bbb Z/(2)$ acting on the vector space and the base field and you want to talk about $G$-equivariant things
Hang on I'll give you more context: we're looking at the space $K_\Bbb C = \prod_\tau \Bbb C$ where $\tau$ are embeddings of a number field $K$ into $\Bbb C$. There's a hermitian scalar product on this C-vector space called $\langle \cdot, \cdot \rangle$. Then we define $F : z_\tau \to \overline{z}_{\overline{\tau}}$.
The notation is weird
Where's Mathei when we need him
2:01 PM
This is getting too algebraic for me and I need to study for my exam, I'm out for a while
See you later
No worries, cya
Best of luck with your exam, Alessandro!
Now Neukirch says that $\langle \cdot, \cdot \rangle$ is $F$-invariant and writes "so $\langle Fx, Fy\rangle = F\langle x, y\rangle$, which works out, and also that the trace I described is F-invariant, which also works out, and then on the next line he talks about F-invariant points of $K_\Bbb C$, which are those such that $\overline{z}_\tau = z_{\overline{\tau}}$ lol
It's the last one I need to finish my Masters and it's going to be on skype which doesn't sound great
You could check what the English translator of Neukirch's book decided to use
2:04 PM
yeah good idea lol
@VJ123 Well, this has something to do with a kind of fundamental asymmetry we often see in measure theory. If $f$ is a non-negative, measurable function, the integral $\int f$ always makes sense, but it may be infinite. To deal with not necessarily non-negative measurable functions, we split them up into a positive part, $f^+$, and a negative part, $f^-$, for which the integrals are defined as previously and then define $\int f=\int f^+-\int f^-$.
For this to work out nicely, we require both integrals on the RHS to be finite (we don't want something like $\infty-\infty$ to stand there) and
2:21 PM
I can vouch for this ^
Seriously well-written, @Thorgott!
thanks a lot :)
the books starts with a theorem without a proof of it! :)
it seems the theorem is needed for a nice start
Which book are you reading through, @Masterphile?
2:37 PM
one professor sent me his book, i am reading Spivak´s book
he said he will endorse me if needed
on arxiv
now i only need to prove something new! :P
or something old with new methods
anyway, it´s a challenge
3:39 PM
That sounds pretty cool!
@Thorgott Wow that's a really nice explanation. Really helped me understand where all of the stuff about Lebesgue integrability comes from. Thank you very much.
@Khallil yeah, i am already trying to prove something, but am distracted by NT stuff i also do
@Thorgott So with Fubini's theorem and improper integrals - say you have a function that becomes unbounded at some point in the region, then you don't have to worry about the function being non-negative as long as the integral of the absolute value of the function is finite?
mostly elementary NT
4:29 PM
How do I show that the adjoint representation of $SU_2$ has $SO_3$ as its image ? I've shown that the image is included into $SO_3$ but I don't see why it has to be all of $SO_3$
@VJ123 yeah
@Thorgott And just a final clarification, for a function to be Lebesgue integrable, does it also have to be measurable? Is that also part of the definition?
Yea it's part of the definition!
@Khallil Thanks :)
Is there any easy to way define what it means to be measurable?
(I don't have much knowledge of measure theory at all)
Have you constructed a measure before?
Oh right, then it's probably best to consult an introductory text
That way you'll be able to follow the whole development of integration, starting with sigma algebras, measures, measurable functions etc.
4:38 PM
Okay cool. And I read somewhere that practically any function that can be described is measurable. How accurate of a statement is that?
what is the possessive form of "Tom and I"?
Tom's and my?
@VJ123 honestly it's probably best for you to learn about the preliminary stuff like sigma algebras, measures and measurable spaces before you move onto measurable functions
That way you can build your intuition up properly!
@Khallil All right cheers. I appreciate your help.
if I have a non-measurable set, I can easily describe a non-measurable function
but the point really is that essentially all functions you will naturally encounter will be measurable
e.g. a Vitali set ^
4:46 PM
(in the context of euclidean spaces)
the folklore is that if you construct a function without thinking about the axiom of choice, it will be measurable
@AlessandroCodenotti you sent this ten thousand years ago but the paper is called "hyperbolic groups" lol
I know, Balarka told me in the meantime
But thanks
That image is amazing, @LeakyNun
5:03 PM
If I have a morphism between two Lie groups $\phi:G\to G'$, such that its derivative is a Lie algebra isomorphism, how can I show that the image of the morphism is the connected component of the neutral element in G' ?
@Astyx consider $G = G' = C_2$?
What's $C_2$ ?
Cyclic group of order 2?
I only wanted it to contain the neutral component
5:18 PM
@Astyx I guess it's because the component of the identity is locally homeomorphic to the Lie algebra via exponential
I don't know
Yes that's what I'm thinking as well
I can't convince myself though
I did some digging and found this
It's been so long since I studied Lie stuff that I genuinely don't remember the basics haha
5:33 PM
My professor wrote "Define $\mathcal{L}$ to be the set of vectors that can be obtained as linear combinations of the columns of $X \in \mathbb{R}^{n\times p}$
does that just mean the column space?
5:54 PM
That's literally the definition
The 'set of vectors that can be obtained by linear combinations' is also called the span of those vectors
(Replace columns with rows to get an equivalent definition for the row space!)
@Khallil super! i thought so, but i had no idea why he was calling it L
@TedShifrin hey Ted!
6:22 PM
Hi Stan
L for linear span, I s’ppose.
@Astyx: presumably an open/closed argument.
@Astyx Yes, because the map has $df_e$ an isomorphism (would be sufficient for it to be a surjection), by the inverse (implicit) function theorem you find that the image is open. Then you just need to know that a subgroup of a Lie group that contains a neighborhood of the identity is also closed (equivalently, contains the identity component).
(Of course, because it's a subgroup and multiplication is continuous, containing an open neighborhood of the identity is equivalent to being open.)
And the rest just follows because, if $g_n \to g$, then consider $g_n g^{-1}$; for large $n$ this is contained in your subgroup $H$ because it contains a neighborhood of the identity; say $g_n g^{-1} = h_n$. Then in particular we have that $g = h_n^{-1} g_n$ for all sufficiently large $n$, and in particular $g$ lies in your subgroup, so it's closed.
@TedShifrin his notation just generally sucks so i'm not sure its intentional
hey chat
Hi Lucas
@TedShifrin what have u been up to lately?
6:35 PM
Nothing fun ... trying not to get sick.
Let $\mathbb R^{\mathbb R}$ be the set of all real-valued functions. What's the probability that a random element of this set is continuous?
(I'm not sure if my question is well-defined)
@TedShifrin that's what I guessed, but I'm not sure of how I'd prove that
6:37 PM
Of course, you didn't tell us a topology or measure ....
Measure theory looks cool
for every continuous one f choose an infinite number of those that are the same as f but the difference is only that those are discontinuous at one point, and you´re done
Just by cardinality reason it should be 0 for any reasonable probability measure
Still 0, I bet, if discontinuous just somewhere.
yeah, continuity is rather rare :D
but if you asked for a probability that it is continuous at at least one point or more, then...hm?
6:49 PM
Yeah, I typed the wrong thing, most likely. Not paying attention.
What do people mean when talking about "differential groups"? What's the differential part of this group?
I'm sorry about the sporadic, random questions. I guess my low effort questions are welcome for people who are also wasting time :p
hmm, what does the product-$\sigma$-algebra on $\mathbb{R}^{\mathbb{R}}$ look like
Thank you @TedShifrin and @MikeMiller
Hi all
Q: Fixed point question for branches

mickLet $A(z)$ be a real-entire function. $$ A(z) = a_0 + a_1 z + a_2 z^2 + ... $$ Also for real $x$ we have $$ A’(x) > 0 $$ Let $f$ be The functional inverse : $ f(A(z)) = A(f(z)) = z $. More specific $f$ is the main branch inverse. An example could be $A(z) = exp(z) $ and $f(z) = ln(z) $ ( no...

7:32 PM
@TedShifrin Are there old practice problems for your Linear Algebra course online? Lately, I've been thinking about purchasing the textbook.
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