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2:03 PM
@iwriteonbananas what is $X$, again?
solid torii pasted via boundary?
 
yes, via identity
 
the maps in the mayer-vietoris sequence are well-defined : they're not arbitrary
 
yeah, the map $H_n(A)\oplus H_n(B) \to H_n(X)$ is the sum of the maps induced by inclusions, right?
 
yes.
i.e., you add the images of the simplices pushed to $X$ via inclusion and then mod out appropriately via boundary
so what should it be in this case?
 
user147690
@Balarka I don't really get what they are doing in corollary 3 page 228 D&F
 
2:10 PM
I don't have it with me right now, @AlexClark. what's it about?
huh, @iwriteonbananas?
 
yeah nvm
 
user147690
Any finite integral domain is a field:

Let $R$ be a finite integral domain, and let $a$ be a nonzero element of $R$. By the cancellation law, the map $x\mapsto ax$ is an injective function. Since $R$ is finite this map is also surjective. In particular there is some $b\in R$ such that $ab=1$, i.e. $a$ is a unit in $R$. Since $a$ was arbitrary nonzero, $R$ is a field.
 
@BalarkaSen im not sure
 
user147690
Can you give me an alternative proof, since I have no idea what I don't understand here
 
user147690
Their goal was to show every element is a unit, hence all invertible, hence a field I imagine
 
2:14 PM
what do you not understand, in particular?
 
user147690
Well how do they show that all elements are units?
 
$x \mapsto ax$ is surjective : so for any element $b$ in $R$, $ax = b$.
take $b = 1$. thus, $ax = 1$.
does that make sense?
OK, @iwriteonbananas, let's get our hands dirty.
 
user147690
No for some reason, but give me a minute more
 
Let's look at this piece of the long exact sequence :
$0 \to H_2(X) \to H_1(U \cap V) \to H_1(U) \oplus H_1(V) \to H_1(X) \to 0$
Homology is reduced, but I am too lazy to put the tildes.
 
sure
in the last nonzero map
the generator of the circles $U$ and $V$ are both mapped to the same circle in $X$, right?
under the maps induced by inclusion
 
2:23 PM
Hello!!!
 
Now, the kernel of the 3rd map is $\Bbb Z$ : it takes the generators $(a, b)$ of the boundary torus to $a \oplus b$
 
user147690
How do we know $1$ is even in the ring?
 
A ring always has an identity.
It's in the definition of the ring.
 
user147690
Not in D&F
 
We have: $b(x)=\left\{\begin{matrix}
1-|x| &, |x| \leq 1 \\ \\
0 &, |x|>1
\end{matrix}\right.$

How can we find the integral $\int_0^x e^{2t} b(t) dt$ , where $x \in \mathbb{R}$ ?
 
user147690
2:25 PM
That is called a ring with identity in D&F
 
user147690
In D&F multiplication is just an associative magma not a monoid
 
Balarka: what nonsense
 
Look carefully : there must be a 1 in the definition of integral domain.
Otherwise the proof doesn't make sense.
And in any case, who cares about non-unital rings?
 
user147690
Oh indeed, an integral domain is only defined when we have identity and no zero divisors
 
told ya
 
user147690
2:28 PM
But a ring need not have 1 :P
 
OK, @iwriteonbananas, so since the kernel of the 3rd map is $\Bbb Z$, the image of the second map is $\Bbb Z$ by definition of an exact sequence.
@AlexClark there are no non-unital/non-commutative rings.
 
user147690
Proof?
 
it's called the Grothendieck-Uhlenback theorem
 
is "retraction" = "left-inverse"?
 
@AlexClark proof by intimidation.
 
user147690
2:30 PM
LOL
 
@SohamChowdhury a retraction is a left inverse to the inclusion map
@BalarkaSen yeah, im just not sure why the kernel in $\Bbb{Z}$. can you elaborate on what the map is?
 
user147690
@BalarkaSen Why does where we mapped to have $1$?
 
Unfortunately, I have to go once again. Sorry for being unhelpful, @iwriteonbananas. We'll continue the discussion on the algebraic topology chat next time.
@AlexClark ?
 
i know it's $(i_*, j_*)$ where $i:U\cap V \to X$ is the inclusion
and j analogously
 
user147690
Is this map $R\to R$
 
2:33 PM
but
 
it's the map $H_\bullet(A \cap B) \to H_\bullet(A) \oplus H_\bullet(B)$ I am talking of
given by including the generators into each copy of $A$, $B$.
 
you meant to write intersection?
 
this map, in this case, is $(a, b) \mapsto a \oplus a$
 
So what's the kernel?
 
2:34 PM
exactly
$\Bbb{Z}$
 
right. thus, the image of the snake map $H_2 \to H_1$ is $\Bbb Z$
 
user147690
@SohamChowdhury How many pages of aluffi have you done?
 
~80
Why?
 
yeah, so $H_2(X)$ is $\Bbb{Z}$
 
You planning to start too?
 
user147690
2:35 PM
@SohamChowdhury How many pages in does it get fun?
 
i dont wanna hold you any longer, we'll chat later
 
@AlexClark wait.
 
user147690
@SohamChowdhury I would like to if I get time. I think a bit of category theory perspective would be interesting.
 
right, bu-byes. I believe you can do most of the exercise by yourself from this point.
 
user147690
Cya @balarka
 
2:36 PM
yeah, i think so, too
take care
 
@AlexClark it starts with the usual "set theory basics"
 
@AlexClark yes, 'cause multiplication by $a$ sends elements of $R$ to elements of $R$
now I am really gone
 
@AlexClark category theory perspective on what?
 
user147690
@iwriteonbananas Algebra in general
 
oh, yeah it's kewl (not that i've studied it much)
 
2:38 PM
@AlexClark the first diagram is on page 10, which got me all excited
 
what course are you studying for?
 
user147690
@iwriteonbananas The rest of my life I suppose.
 
user147690
But really courses next semester haha
 
the category theory discussion starts on 18. it's a fun ride from there!
 
2:39 PM
do you have winter break?
 
user147690
@SohamChowdhury Alright, I'll give it a shot tomorrow probably. I doubt I'll go through it as fast as you have
 
user147690
@iwriteonbananas Yep, and I'll be doing a research project on tropical geometry
 
user147690
E.g min-plus algebra
 
and everything possible is introduced with a universal property.
what that does is essentially show that, for example, if you have any two ways to define disjoint unions, set products, group homomorphisms, free groups or any other sort of construction, they are *bound to be* isomorphic
this is great imo
 
cool (i have no idea what that is)
 
2:40 PM
you have the book, or a semi-legal version?
 
user147690
@SohamChowdhury Yep got it entirely illegally I imagine
 
user147690
But I will buy the books after I start tutoring\
 
user147690
Nice expensive hardcover copies
 
those are around 90 USD here :'(
 
user147690
@SohamChowdhury Sounds good. I have dealt only once with the universal property
 
user147690
2:42 PM
@SohamChowdhury Probably 150USD here - but I will be on about $350USD a week
 
user147690
I'll likely get Munkres, D&F, Aluffi, Hatcher(since I will be using that for next sem)
 
it helped me get rid of the nagging feeling I had about "how do i know that this is the only possible def. of a disjoint union?" because, you know, the universal proves that they're defined uniquely "up to isomorphism"
 
user147690
I think I'll get Rudin's functional, but not sure
 
you don't have DF?
 
user147690
I have an illegal copy
 
2:43 PM
ooooooh
 
user147690
I am illegal until I can buy these things next semester. I don't feel bad at all on the basis that I will be buying them
 
user147690
Otherwise I would feel bad solely due to the fact they are so good xD
 
i remember watching Sherlock in Germany, and a teacher popped his head into my room saying "that's illegal over here bro" or some equivalent
@AlexClark any book with non-unital rings is bad
@AlexClark when?
 
user147690
@SohamChowdhury In class, for showing $\oplus,\times$ were equivalent in the category $\bf{Vect}$
 
oh. i just did that yesterday for $\sf{Ab}$
 
user147690
2:45 PM
I didn't understand it very thoroughly though, so don't ask me to explain it
 
category?
you work with them?
 
user147690
Almost not at all, that was my only encounter
 
intro algebra class?
 
user147690
Nah Australia is a little weird, we do a really weak algebra class, then a medium one and then you can take 3 hard ones
 
and I've begun looking at Munkres (my "first pass", in my lingo)
 
user147690
2:47 PM
This was in the medium one
 
the first chapter is all set theory stuff, with a little choice and well-ordering thrown in
 
user147690
We are more analysis heavy for some reason
 
i'm afraid of turning into a logician for some reason
 
user147690
@SohamChowdhury Yes the first chapter is rather boring
 
user147690
It always is though.
 
2:48 PM
every effing book starts with a set theory review
at least at undergrad level
 
user147690
Actually no, the first chapter of Lee's Topo manifolds I think was nice.
 
dunno what that is
 
user147690
@SohamChowdhury It is important haha.
 
user147690
I think I'll sleep now
 
yes, but it gets boring after you've done it once
@AlexClark yeah, must be late
 
user147690
2:49 PM
@SohamChowdhury Yep I do it just for completeness and it is very quick to do
 
start Aluffi tomorrow
and we can do exercises together like Mr. Sen and bananas
 
user147690
Yeah it's semi-late. I normally go to bed 3 hours ago, but I woke up super late. Well ta-ta da-da
 
if you wish
 
user147690
@SohamChowdhury I'll try to keep up, you seem like a prodigy :P
 
sarcastic tone yeah, right
more like really want to get to the sweet stuff soon
good night
 
3:14 PM
@TedShifrin Morning. I think of Heegaard diagrams as handlebody decompositions rather than in terms of Dehn twists.
 
For future generations to copy and paste: [Chat guidelines](http://meta.math.stackexchange.com/questions/3890/main-chatroom-etiq‌​uette-rules) | [$\LaTeX$ in chat](http://www.math.ucla.edu/~robjohn/math/mathjax.html)
 
3:31 PM
@MikeMiller I saw a post of yours mentioning Hatcher's K-theory book, and it looks quite interesting
 
The book, or the post?
The only thing in the post that's in the K-theory book is the clutching function business at the end.
 
All of this topology stuff you do that I don't understand, but I intended the book, there
 
Hatcher's book is incomplete but otherwise rather good. The standard book to learn about the vector bundle/characteristic classes story is Milnor and Stasheff's "Characteristic classes", but I hate that book.
To do some of it you need to know cohomology, because that's where characteristic classes live. But I remember the first chapter being readable without really much background at all.
 
Yeah, I was just skimming the first chapter. I'm definitely not quite at the level of the first chapter (at least, I imagine all of the definitions will seem mysterious)
 
I'd love to pick up some stuff from VBKT sometime.
 
3:40 PM
@SohamChowdhury To me the real value of the set theory section is familiarizing yourself with the conventions of the author. For instance, Munkres deals with collections of sets in a particular way that becomes important in later chapters.
 
@Fargle Yeah, that's actually a very good point.
 
4:06 PM
What is K-theory
 
I have no idea, really.
 
study of the letter K
 
the earliest and simplest (and prettiest) notion of K-theory is the study of "vector bundles" over a space. one is basically adding a vector space above each point so that these vary continuously; you can "twist" the vector spaces as you move around your space in interesting ways
 
Heya @Fargle: How's geometry going?
 
4:08 PM
there were soon after important algebraic and operator algebraic generalizations
 
oh I see
interesting
 
that left me completely unsatisfied
 
Where did $K(\pi,n)$ notation come from, @Mike? I never knew.
 
@TedShifrin: No idea; it was around before K-theory. I bet it's in Dieudonne.
 
all I know about K-theory is that you can define a group K_0(X) made out of doing grothendieck construction on vector bundles over X w/ whitney sum, and defining K^-n(X) = K(S^nX). I guess {K^*} defines a coherent cohomology theory, but I have no idea, of course.
 
4:11 PM
Why do people write $T$ for kinetic energy?
Susskind: "Dunno."
 
its same as like why you write $L$ for angular momentum
very weird letter
 
And $V$ for potential energy.
 
this is almost certainly known, just not by me
 
I actually was solving a question related to angular momentum earlier
I thought was cool
 
@Karim: You should do my favorite pair of physics questions.
 
4:14 PM
Determine what height should the edge of billard table be in order for no reaction to occur between ball and the table and it the height was 7/10 the diameter of the ball.
very nice question @TedShifrin
which is your favorite pair of physics questions?
 
(1) A point mass starts at height $h<R$ above the center of a sphere and slides along the sphere. No friction, but gravity downward. At what height does it fly off the sphere? [Standard problem]
 
Hello@TedShifrin
 
(2) A ladder with mass leans against a wall. It starts to slide. At what fraction of its original height does it fall away from the wall?
hi @Remember
 
Welcome the new remember....
 
What do you mean by "no reaction to occur between ball and table"? @Karim
 
4:16 PM
that is no slipping occur for the ball
 
Hmm, I'll have to ponder that.
 
@TedShifrin: just looked at the original paper; they define $K(\Pi)$ to be what we would now call $K(\pi,1)$. Probably $K$ was used for complex. Likely the $n$ was later inserted when it was realized that it was interesting for all $n$!
 
pity that there is no notion of long exact sequences in $\mathsf{Top}$
 
@Mike: Was it Whitehead?
 
No, it was Eilenberg and MacLane.
 
4:19 PM
Oh ...
Oh, of course, I knew that.
 
@TedShifrin I will solve your questions tomorrow today I am on a break from work :D
 
Any time you're interested ... No obligation. On a break from work but here? :)
 
yeah :D my gf is dragging me today to buy some stuff
for the apartment
 
Be a good boy.
 
lol
 
4:27 PM
lol
 
Anyone ever run across equations like $f(x) = 12\cos(x) + \sqrt{144\cos^2(x) + 756}$? It looks like it should be expressible as a transformed cosine, graphically, but that square root seems to resist simplification.
 
@TedShifrin: Any ideas on this? We can't do it in dimension 2, because we can just pick a particularly nice symplectic form with given area to work with.
 
Huh? I don't follow.
Why can't we take $z\to\bar z$ on $\Bbb C/\Bbb Z^2$?
 
@TedShifrin: The only invariant of symplectic forms (up to isotopy!) in dimension 2 is area. So pick an unusually symmetric representative that you know does support a diffeomorphism $f$ with $f^*\omega = -\omega$. This is essentially obvious for the sphere, say.
Sorry, that was terribly worded of me. I meant "We can't find a counterexample in dimension 2".
I find the claim hard to believe, so my focus is on counterexamples.
 
What claim? He asked if there is an $f$?
 
4:37 PM
Yes.
 
So what do you mean by counterex?
 
I'm sorry, I don't understand what's hard to parse about this. I am saying I believe there must be some $(M,\omega)$ that doesn't support a diffeomorphism $f$ with $f^* \omega = -\omega$.
 
He just asked for an example, I thought. Let me reread.
Oh, I see, you interpreted it as $M$ general, and I thought he wanted an example.
I get it.
 
It's not far from trivial if he just wants an example, as it's easy to do such a thing on the sphere or torus. (Or the point with the identity map. :) )
 
so I see we want a $(4n+2)$-dim example with orientation-reversing diffeo.
 
4:42 PM
without* orientation reversing diffeo... that's how we get a contradiction
Maybe one can find a better example somehow in low dimensions, though?
 
I guess I don't know a nec and suff condition for there to be an orient-reversing diffeo.
 
The thing with Nash just reminded me of something. Don Knuth is going to die one of these days, too. :(
 
@TedShifrin: Me neither. I found a source that says every 6-mfld has one, as seen in the comments.
@SohamChowdhury: So're you. No need to obsess.
 
Yeah, really.
Is Munkres sufficient background for Hatcher?
I've started going through it today, in my spare (read: non-algebra) time.
 
Just do what you find fun.
 
4:47 PM
So when is $\text{Diff}(M)$ connected? Is this not in Smale or Hirsch or something?
 
I don't find being exposed to a huge amount of unfamiliar terminology fun, so I need a few prereqs, @Mike.
 
@TedShifrin: This is not really what you want to ask. There is never an isotopy between orientation reversing and preserving diffeos, but there are plenty of non-isotopic diffeomorphisms in most cases...
I would be astonished if you could find a manifold $M$ with connected $\text{Diff}(M)$.
And no, that is emphatically not anywhere. You're asking about when the mapping class group is trivial; the mapping class group is barely understood in dimensions 2 and 3 (where we have a presentation of it for most manifolds). In dimension 4 it's incredibly mysterious - most work on the subject is by Danny Ruberman, who can say a few algebraic facts about $\text{MCG}(M)$ for a few 4-manifolds, and all in about 4 papers, at most. I have no idea what's known for $n>4$.
Understanding $\text{Diff}(M^4)$ better is one of the projects I'm most interested in.
(We actually know a finite-dimensional approximation of the homotopy type of $\text{Diff}(M)$ for a wide class of prime 3-manifolds. I wrote up a list of references at some point but I'd have to check back to see what's still open. I think there's a way to patch these together for connected sums via a fibration but I never looked at the details.)
 
They're called irreversible manifolds, @MikeM. We need examples.
Danny's an old friend. I Could email him :)
 
My comment about Danny was a non-sequitur to the problem at hand, just talking about diffeomorphism groups in the broad.
I've never heard the term irreversible manifold (you're referring to the ones with no orientation-reversing diffeomorphism?) I've heard the term chiral.
Try this paper for examples. I found one in there of dimension 14 before, I think.
 
@Mike: Try this.
 
4:59 PM
If that's what I think it is, that's where I found my 14-manifold.
 
Ah, funny.
 
It is what I think it is indeed.
His 10- and 14-dimensional examples don't support symplectic forms for cohomological reasons.
I think it's probably folly to try to solve it this way.
 
Zero signature is necessary, for example.
 
What does signature mean in singly even dimensions?
 
Ah, duh.
 
5:04 PM
Hi everyone! Just a quick question: Let $\mathbf{x}\in\Bbb{R}^n$ be an $n$-dimensional real vector, $U\in\Bbb{R}^{n \times n}$ be an orthonormal matrix, and $\Lambda$ be a diagonal matrix with strictly positive elements. We define the $n$-dimensional vector $\mathbf{y}\in\Bbb{R}^n$ as follows
$$
\mathbf{y} = U\Lambda\mathbf{x}.
$$
What could we say about the squared Euclidean norm of $\mathbf{y}$ w.r.t. the squared Euclidean norm of $\mathbf{x}$?
Thank you very much for your time!
Apologies for spamming you @DanielFischer, @TedShifrin :)
 
To quote @Pedro, I'm derping too much.
 
Hello again
 
@TedShifrin: Let's stop working so hard. Is there even a diffeomorphism of $\Bbb{CP}^2$ which reverses the sign of $H^2$?
 
You can give bounds in terms of entries of $\Lambda$, @nullgeppetto, but that's it.
 
@TedShifrin, thank you very much, well, that's what I was afraid of... I guess that the norm is bounded by the minimum an maximum element of $\Lambda$, right?
 
5:07 PM
What happenss if you conjugate, @Mike?
yup, @Nullg.
 
@TedShifrin, thanks again!
 
Could you maybe take a look at my question?
0
Q: Prove that Euler's equation can be written in a specific form

evindaAccording to my notes, the following theorem holds: If $y$ is a local extremum for the functional $J(y)= \int_a^b L(x,y,y') dx$ with $y \in C^2([a,b]), \ y(a)=y_0, \ y(b)=y_1$ then the extremum $y$ satisfies the ordinary differential equation of second order $L_y(x,y,y')- \frac{d}{dx}L_{y'}(x,y,...

 
@TedShifrin: I think I'm confusing myself. Looking at the Kahler form it seems it doesn't change anything, because of that $\log |z|^2$ in there, but this seems weird!
 
Looks wrong, @evinda.
 
So is there a typo? I found the exercise in my book... @TedShifrin
 
5:14 PM
You still have to pullback $\partial\bar\partial$. Pullback commutes with $d$, but ...@Mike
 
oh, that's probably where the problem lies. I've never thought about the complex $d$s...
 
Do it locally on $\Bbb C^2$, Mike.
 
OK, I agree now.
Yeah, I just did.
Darn, so $\Bbb{CP}^2$ is out, and that's the best symplectic 4-manifold I know. :)
 
Sowwee ...
 
@TedShifrin have you an idea about this please: math.stackexchange.com/questions/1298321/no-compact-embedding
 
5:21 PM
No, @Vrouvrou. I haven't thought about such things in 40 years.
 
@TedShifrin i have this definition
in my case T is the injection $i$
how to do the negation please
 
What do you think?
 
Morning Math SErs
 
G'night, @Clarinetist :)
 
there exist a bounded sequence $(x_n)$ in $X$ and $\forall (Tx_{n_k})\subset (Tx_n)$ $T(x_{n_k})$ do not converge in $Y$ ?
@TedShifrin
 
5:28 PM
@Clarinetist Morning, Mr. Actuary
 
@SohamChowdhury Oh boy :P
 
Right @Vrouvrou .... Or it is not continuous.
 
@TedShifrin: I've got a couple more hours to think about this before I really really should get to work, so I'll let you know if I think of anything.
 
@TedShifrin It is given that $a>0$, $b$ continuous on $[0,+\infty)$ and $\lim_{x \to +\infty} b(x)= l \in \mathbb{R}$.
How could we calculate the limit $\lim_{x \to +\infty} e^{-ax} \int_0^x e^{at}b(t)dt$ ?
 
5:30 PM
@TedShifrin to say that $\forall (Tx_{n_k})\subset (Tx_n)$ $T(x_{n_k})$ do not converge in $Y$ , means that $(Tx_n) $ do not converge in $Y$ no ?
 
So I learned about nonnegative definite matrices for the first time yesterday. Fascinating stuff.
 
Happy Mem Day, Mike. I won't be around much once I fly to SD Wed.
 
@SohamChowdhury Surprised I've never heard of that before today. Only one I know on that list is Edward Elgar
 
Mmh, Cello Concerto
 
Gone now.
 
5:31 PM
Thanks, @Ted. I need to get back to work on my talk for Friday, but I'd rather not for now... :)
 
@TedShifrin ?
 
@SohamChowdhury A classic. I hope you've heard the Enigma Variations as well
 
@Clarinetist Handel is well-known too, I believe
@Clarinetist Yes!
And you're really a clarinetist?
 
@SohamChowdhury Ah yeah, I missed Handel on that list. Yep, 'been playing for about... 13 years? but on and off
@TedShifrin You're leaving us?!?!
 
@Clarinetist Behold. You might have respiratory distress laughing.
Wait for the solo.
 
5:35 PM
@TedShifrin are you there ?
 
@SohamChowdhury How I looked at the video: O_O . Some really absurd multiple tonguing in there too that I would not be able to do.
 
Saw all of it?
 
@SohamChowdhury Just finishing it up now
Oh that solo...
 
I play guitar. Necrophagist is sadness for us because it's hard. And then this guy . . .
 
Fast notes are EASY on clarinet... they just require some practice
Oh boy
I have some classical guitar things for you to listen to @SohamChowdhury
 
5:39 PM
Yeah, I guess, all winds/horns generally. The guitar technique of sweeping was invented so that Coltrane and friends weren't the only ones playing fast.
@Clarinetist Please.
 
Some parts of the guitar solo remind me of the Toccata and Fugue.
 
@SohamChowdhury That I can see.

I'm always jealous of classical guitarists. Baffles me how they can play that quickly :P
@SohamChowdhury After some Googling, this is probably the most agreed on most difficult piece for guitar
 
please tel me when we say : $\forall (Tx_{n_k})\subset (Tx_n)$ $T(x_{n_k})$ do not converge in $Y$ , it means that $T(x_{n})$ don not converge in $Y$ right ?
 
@Clarinetist So you're jealous of me? ;)
 
5:45 PM
@SohamChowdhury Haha
Guitar and piano are two instruments I wish I had picked up in my youth
Piano I know decently well, but hey, it would be nice to play a Rachmaninoff prelude some day :P
Not looking into any concerto playing anytime soon :P
 
Is Prof. @TedShifrin still around? :)
 
@nTuply I think he left as of 10-20 minutes ago?
 
Okay.
 
Or this! @Clarinetist
 
Ah, Jeux d'eau! @SohamChowdhury I know one person who just graduated with a B.M. in Piano Performance whose dream is to play that piece. DANG it's difficult. Especially some of those "arpeggios" :P
Ravel is probably my most favorite composer
 
5:49 PM
@Clarinetist Relevant
 
His piano stuff is great, but his orchestration is precise and... just beautiful. I don't know how any human can write the scores Ravel wrote
@SohamChowdhury Lol
 
@anon hello, have you an idea about this : math.stackexchange.com/questions/1298321/no-compact-embedding , do you know a bouded sequence in W^{1,p}_0 but every subsequence of it do not converge in $L^{p^*}_1$ ?
 
I'm not good enough to play Ondine yet :P @SohamChowdhury I did play a similar one from Ravel's Miroirs
 
@Clarinetist you have not heard this.
 
5:52 PM
Triple-guitar chords <3
And here's a little Xenakis for you.
 
@SohamChowdhury Yay for 20th Century music :)
 
You into Ferneyhough?
I like his scores.
 
@SohamChowdhury I know of him, don't know much about him
 
They're handwritten art.
 
I've mainly kept myself to Ravel/Debussy/Stravinsky-type music
 
Wow, REALLY strange rhythms
 
19:26-ish polyrhythms too
 
Nice to get to know a musician on this site @SohamChowdhury :) Out of curiosity, what's your musical training? Just wondering
 

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