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12:00 AM
Another way to think about it is "you defined $d$ as some crazy alternating sum, so that all the terms in $d^2$ cancel out nicely" :P
 
@Ilya_Gazman sounds interesting
 
@MatheinBoulomenos It's not crazy at all. It's just a signed sum, where the sign takes care of orientation
Modulo that it's $\partial^2 = \emptyset$ like I said.
I guess I am alternating between boundary morphism and the coboundary morphism (i.e., homology/cohomology)
 
howdy
 
Hey @PVAL
 
Hi @PVAL
 
12:06 AM
@LeakyNun are you there?
 
@MrAP ?
 
Can you help me in deriving the expression for XOR
 
sure
 
Without truth table
 
why is there a constant objection towards truth table
I mean, they're perfectly valid
 
12:09 AM
@Kirill it is! I also tested it on big numbers like RSA-1024. I been able to find $f(x)$ that is only 494 bits size
 
@MrAP what is the expression then?
 
Well i know how to do it using truth table. I want to know how to derive them algebraically.
 
algebraically, XOR is addition in $\Bbb F_2$
 
p xor q = p.(~q)+(~p).q
 
that just means (p and not q) or (not p and q)
how would you define xor?
the most common way I've seen is (p or q) and not (p and q)
= (p or q) and (not p or not q)
= (p and not p) or (p and not q) or (q and not p) or (q and not p)
= (p and not q) or (q and not p)
= (p and not q) or (not p and q)
= p.(~q)+(~p).q
 
12:12 AM
I have read xor operator gives output 1 for odd number of 1s in input.
 
Logic is better than differential geometry because at least I can read the transcript and understand what is going on
 
then it follows trivially
 
@Ilya_Gazman but I am unable to look at this now - I am working with graphs now. Still, very interesting question!
 
p xor q is true iff (p is true but q isn't) or (q is true but p isn't)
the most beautiful diagram in existence
 
tnx Kirill
 
12:14 AM
the last row is the second isomorphism theorem / nine lemma
 
Can you explain this (p or q) and not (p and q)
 
@Ilya_Gazman I've bookmarked that :)
 
@MrAP it just depends on how you define xor. some people define it as "p or q is true, but not both"
@MatheinBoulomenos is it actually second iso?
 
what is second iso?
 
12:15 AM
I don't know
 
what were you asking?
 
I just randomly choose a number between 1 and 3
the last row in my diagram above
 
What diagram is that?
 
@MrAP relation between different groups of matrices
 
Is it related to xor
?
 
12:17 AM
hardly
$\mathbb{PGL}(V)/\mathbb{PSL}(V) \cong (\mathbb{GL}(V)/\mathbb{Z}(V))/(\mathbb{SL}(V)/\mathbb{SZ}(V)) ~??~ (\mathbb{GL}(V)/\mathbb{SL}(V))/(\mathbb{Z}(V)/\mathbb{SZ}(V)) \cong F^*/(F^*)^n$
@MatheinBoulomenos ^
 
If all theorems can be expressed in such clean fashion as categories, there will be no more headache on what it means
 
oh thanks
 
Our natural language is sometimes just too wordy to read
 
@MatheinBoulomenos are coequalizers and quotients the same?
 
@LeakyNun I don't see how the last row is any particular isomorphism theorem. But the fact that $PGL(V)/PSL(V) \cong F^*/(F^*)^n$ follows from the diagramm (this is an instance of the so-called "nine lemma")
 
12:22 AM
8 mins ago, by Leaky Nun
the last row is the second isomorphism theorem / nine lemma
:)
 
ah yeah
@LeakyNun I don't know a definition of "quotient" in a general category, so I'd say no
 
What does the isomorphism theorems look like in commutative diagrams?
 
@MatheinBoulomenos in groups?
@Secret first iso:
 
why do we need the two nodes of zeros at the top of the quotient group?
 
@LeakyNun the coequalizer of $f,g: A \to B$ is the quotient of $B$ by the normal subgroup generated by all elements of the form $f(a)g^{-1}(a)$, yes
 
12:27 AM
@Secret if $0 \to A \xrightarrow{f} B$ is exact then $f : A \hookrightarrow B$ is injective
if $A \xrightarrow{f} B \to 0$ is exact then $f : A \twoheadrightarrow B$ is surjective
third isom ^ (@MatheinBoulomenos amirite)
@MatheinBoulomenos but can I have $A/\ker(f)$?
 
What do you mean?
you can also make a diagram such that 3. iso follows from the nine-lemma
 
oh really
this is beautiful lol
@MatheinBoulomenos right, I think I see it, just need to put some $\operatorname{id}$ somewhere
 
yup
that's the thing
 
Hmm... I am missing something here...
 
@Secret what is it?
 
12:35 AM
f does not look really injective to me
 
the first colum is $1 \to N \to N \to 1 \to 1$
 
↪ is a cover, right?
I am not familiar with the use of $\twoheadrightarrow$
 
↪ is for monomorphisms and ↠ for epimorphisms
 
@LeakyNun yup, that's it
 
12:38 AM
@Secret no, the opposite (embedding)
@Secret exact means image = kernel, right
so the intersection in the middle has to be the same
i.e. the thing is continuous
(the thing = line above, cone below)
@MatheinBoulomenos but the previous one looks much more beautiful
6 mins ago, by Leaky Nun
user image
@MatheinBoulomenos I prefer $0$ because abelian categories
 
O I see, it's the image of the maps equal to the kernel of the next map, not every domain maps to the kernel of the next domain
 
right
@MatheinBoulomenos how would I do second iso though
 
no idea
 
Sometimes I do wonder whether we can visualise those maps directly without drawing the domains, but guess not since each map is uniquely specified with its domain in addition to its image. which mean for any diagram that is used to visually illustrate a given map, it always consists of 3 glyphs, corresponding to the domain, the map and the image
 
but no diagram compares to the butterfly lemma
$B(A \cap C)/B(A \cap D) \cong (A \cap C)D/(B \cap C)D$
 
12:51 AM
 
Actually... I might be able to do it this way...
so the maps to and from the trivial objects constraint the other maps
 
@LeakyNun that diagram I posted is actually super useful
 
@MatheinBoulomenos what is that theorem?
@Secret zero objects
 
it's not a theorem
 
what is the use?
@TedShifrin hi
@Secret I don't understand the diagram
 
12:55 AM
hi Leaky. If you're trying to understand $d$, $d^2=0$, and vector calculus, you should watch one (or more) of my relevant lectures. I did that all very carefully.
 
it allows you to reduce statements about exact sequences of the form $A \to B \to C$ to statements about exact sequences of the form $0 \to A \to B \to C \to 0$
rerehi @Ted
 
I've lost count, Mathei. I'm only here briefly before I retire to the kitchen.
 
Ye
Ugh cheating
 
Oh, good grief, Balarka is unsleeping badly.
 
I am going to fix my sleep schedule today
Going to keep awake all day
 
12:56 AM
Yeah, sure.
 
@MatheinBoulomenos how?
 
As the saying goes, promises, promises ...
 
I sense disbelief in your voice
 
In my voice and in my fingers.
 
Thy shall not be disappointed
 
12:57 AM
@LeakyNun the exactness of the middle sequence is equivalent to the exactness of the three diagonal sequences
@BalarkaSen you have sleep schedule? I don't believe you
 
Not that Mathei is doing much better.
 
@TedShifrin What do you usually cover in difftop after Mayer-Vietoris?
 
why coker(f) = im(g)? and did you spell coker as koker?
 
That was near the end of my G&P course, Kevin. Remember I wasn't teaching a grad course. ... I was in the hospital with anemia last time and someone covered for me and didn't even do what I wanted.
 
I'm German
 
12:59 AM
@MatheinBoulomenos but did you spell koker as coker?
 
It's not a category theory diagram, but a visual illustration. The idea is that cones represents the kernel of the map (since everything on the LHS will map to the zero object), parallel lines represents bijections, frustums represents a neither injective nor surjective map, and arrows represents mapping from zero object to zero object.

So if we are given the maps in blue, and the requirement that the maps form an exact sequence, then the black glyphs have no choice but to be constrained by the blue maps, which is why when you compose the two maps in the top row, you get two arrows thus gu
 
German doesn't use c like that.
Ich habe cein Idee. Um, no.
 
lol
 
Kokaine
 
stupid autocorrect: Infectivity->injectivity
 
12:59 AM
Ah ok, gotcha. It's near the end of our course as well, we just did it.
 

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