I wouldn't say it depresses me, but it makes me somewhat nervous (and I truly dislike the masses of people on a shopping spree). I'm always looking forward to January 2 when all's over again.

Nervous about which part of it?

People's attitudes, expectations and behavior, I'd say.

Aw, this food is so spicy... :( // I have to live on this for the next 1.5 days...

What are you eating?

@tb I got some "Muhammara" on the way back since all shops will be closed tomorrow. I hope they keep Starbucks open at least.

This answer is soo hilarious...

Logarithmic differentiation, product rule and whatnot

@tb I think I know what you mean. Even though I'm indifferent towards my family.

@Srivatsan That looks really spicy. But it's just a dip, what are you having it with?

Pita bread.

Bah, I thought of buying some bread and bagel.

@Matt I think I understand. Let's not belabor this further here.

@tb I wasn't going to. : )

@Matt I was also telling myself :)

I don't quite see how this answers the query posed.

@tb phew : )

@JM Hi JM.

Good morning.

Hello @JM : ) How was the lasagne?

@JM Hi, at least it's a creative approach :)

JM: Also, did our use of camel case torment you enough?

Good morning, y'all. I'm still too tired to rail about camel case... :D

@Srivatsan Actually, I like that quite a bit. There was an Anatolian restaurant in Göttingen (about the only place that could produce edible food) that had an excellent acuka.

Ah, acuka = muhammara?

Yes.

@tb I do not know what acuka is. I like many of these dishes. [But this one is spicy, still good though.]

See Matt's comment above.

Aw, right.

Hi, I'm trying to relearn adjoints and I'm having a small problem.

Shoot.

Suppose we write vector space V as the direct sum of U and W, and consider the projection onto W. What is the adjoint of this operator.

Give me a second to collect and type my thoughts.

tb: If you don't mind, I'll see if I can have a go at it before you. =)

/me keeps silent

Ok, so first let's make sure I am getting all the definitions right. S is a map from V to V that especially "squashes" the part of the vector in U to zero, right?

Err, it's a map from V to W.

yes?

off-topic: Is there more to say than that the projective space is the space of one-dimensional subspaces of V?

@Potato Yes, that's right. So $S(u + w) = w$ if $u$ is from $U$ and $w$ is from $W$.

You just described the projection onto $U$, right?

Right, I fixed it now.

So we have a map $S$ from $V$ to $W$. The map $S^*$ is going to be map from $W^*$ to $V^*$ that sends $w^*$ to $w^* S$, yes?

@Potato How are you saying this?

That is the definition of adjoint, correct?

I assumed that the operator must be from $V$ to itself to define adjoints. // Wait, let me check Wikipedia.

@t.b. Could you clarify here? My book (A terse introduction to linear algebra) defines the adjoint of any linear transformation, between any two vector spaces (that are not necessarily the same vector space).

@Srivatsan I tried wikipedia and it was not helpful.

This page defines only for a transformation $H \to H$.

yes. The adjoint is a map $S^\ast: W^\ast \to V^\ast$ given by pre-composition

@tb But what forces one to consider $W$ as the co-domain? $S$ also happens to be an map from $V$ to itself.

@Srivatsan we don't have scalar products here and (parts of) the page is crap anyway (as I found out in this thread).

@tb I see. =)

So the problem is to interpret $S^*$. I'm not sure what the desired answer is. It appears to extend any linear functional on $W$ to one on $V$ by using the projection map?

@Potato BTW I am also a learner. I just realised that I pretty much forced myself on you. We can either figure things out together (with help from tb and others), or I will let you ask them directly...

That seems very simple, but I guess it is correct.

@Srivatsan It is no problem!

It is better to think about this invariantly first.

@Matt BTW: no leftovers. Everybody had seconds...

@t.b. What do you mean?

@tb Why don't you post this as a comment? That would be more useful there than here in chat : )

Yes, what does invariantly mean?

I need to leave for a bit; I'll read the transcript when I get back.

@tb Then the question boils down to: why don't we define it as the space of subspaces of $V$ of at most one dimensions?

@JM That sounds like a success! Congratulations on the cooking! : )

@Srivatsan I don't understand what you're saying.

@JM You cooked lasagne?

I meant: we have a linear map $S: V \to W$. Now the dual spaces $W^\ast$ and $V^\ast$ are spaces of linear maps $W^\ast \to k$ and $V^\ast \to k$ (where $k$ is the ground field)

@Srivatsan Sure. I like doing pasta. :)

@Srivatsan because.

(Of course, lasagne is one of the more complicated things to do, but it's the holidays after all...)

@Srivatsan I don't like isolated points in my manifolds. :)

@tb I said something along those lines in that thread.

In particular I think the variant projective space is disconnected.

@Srivatsan Anyway, what I was trying to say is: the only way we can produce a map between $V^\ast$ and $W^\ast$ using $S$ is to take $w^\ast$, precompose it with $S$ and get something in $V^\ast$ that is $w^\ast \mapsto (w^\ast \circ S : V \to k)$. Now this gives you the map $S^\ast : W^\ast \to V^\ast$.

I am not sure I got it. I have to think for a bit.

Can you explain what happens when $V$ is finite dimensional (say)?

I made no assumption on dimensions whatsoever.

Right, but does the map have a simpler description then?

@t.b. Would it best to think of it as giving maps that are invariant on $U$?

@t.b. I don't understand your interpretation.

I just said that the adjoint $S^\ast$ sends $w^\ast \in W^\ast$ to $S^\ast w^\ast$ which is the composition $V \; \xrightarrow{S} \; W \; \xrightarrow{w^\ast} \; k$

@t.b. Sorry, I was referring to my original question.

$S$ in this case is the projection onto W.

I remember that, I wasn't there, yet.

Now if you write $v = (u,w)$ then $Sv = w$, right?

Yes.

(It's bed time here, good night folks!)

Good night, Matt!

On the other hand, every functional $v^\ast : V \to k$ gives you a functional on $W$ and $U$ simply by restriction. These restrictions determine $v^\ast$.

In other words: $v^\ast = (u^\ast, w^\ast)$ where $u^\ast = v^\ast|_U$ and $w^\ast = v^\ast|_W$

Yes.

Above we (almost) computed the adjoint: $(S^\ast w^\ast) (u,w) = w^\ast(S(u,w)) = w^\ast (w)$.

This corresponds to the linear functional $(0,w^\ast)$ on $V^\ast$, right?

Yes.

So with these notations we have $S^\ast(w^\ast) = (0,w^\ast)$, and...

...the decomposition $v^\ast = (u^\ast,w^\ast)$ yields a decomposition $V^\ast = U^\ast \oplus W^\ast$ and $S^\ast: W^\ast \to V^\ast$ is simply the inclusion.

Yes, ok, that's what I was getting at above.

Now you should think about what the adjoint of the inclusion $T: U \to V$ is.

Ok. So we take some $v*$ and $T^*$ will be $v^* T$.

And that will be a map $V^\ast \to U^\ast$.

Can you describe it explicitly?

I think it's a sort of quotient map.

Exactly. It's the projection of $V^\ast$ onto $U^\ast$.

Yes, ok. This has been quite helpful, thank you.

You're welcome, of course. The take-away is: the adjoint of an inclusion is a projection and the adjoint of a projection is an inclusion.

Ok, so in the construction of elliptic functions, an initial attempt is the sum $\sum \frac{1}{{(z+\omega)}^2}$ for $\omega$ in some lattice. Why doesn't this converge absolutely?

Sorry, the sum is correct now.

"Gerry if you give an example then everyone will understand the question including you ."

What nerve, seriously?

When has he not been cheeky?

@JM I think the abstract-algebra tag is good for that question. The OP is confused about convergence in $\mathbf C[[x]]$ which is really some analysis/algebra.

Alright, I'm restoring it...

Thanks, JM.

Gerrys comment was terrible

it was typed out in caps lock before I flagged it for moderation

As yet it's still completely worthless, contains no information. But I can't downvote it

People seem to think it's cool to talk down to people though, hence the upvotes for that and his next - even more patronizing - comment

I need to be out for now. See you later.

Top of the not, to you what morrow.

Hey Asaf.

What's up?

Hey Asaf and Dylan.

@Srivatsan What's up Srivatsan?

Not much. Dry day.

Edited an awful lot of posts =)

It just started!

@Srivatsan I edited a question, answered it, then edited my answer, then edited the other answer too :-)

Nice answer.

I don't get it: "every countable partially ordered set can be embedded into cardinalities below the continuum." What does this mean really?

@Asaf, BTW, did you mean "slip through the cracks"?

(last line)

That given a countable partially ordered set, (for example finite/cofinite subsets of $\mathbb N$) then you can replace each of those with a cardinal number which is below $\beth_2$.

I see. Thanks.

I meant what I meant. I meant sip through the cracks. Like the rain sips into your house in Mumbai through the cracks in the roof :-)

Good morning everyone : )

Overlooking the tautology in "I meant what I meant" :-), I didn't know "sip through cracks" is a commonly used phrase or idiom.

@Srivatsan I'm not sure how common it is. It is wunderbar not to be a native speaker sometimes ;-)

On the contrary, I know water seeps through cracks.

Then that was the typo :-)

I'll correct it in a moment, I want to add some link anyway.

@Matt Morning, Matt.

@Srivatsan What time is it there?

@Matt 9:50 am

@Matt around 3am.

Stop living in the past!

I am living 10 mins into the future. At 3.10, I'll living in the past though =)

Us here in the future are much more advanced than you guys. We have an amazing technology allowing us to communicate with the folks from the past ;-)

Here it's coffee time : )

Here it is shower time and then heading out to the university time to pump myself full of caffeine and beer and sit through this paper in commutative algebra.

Beer? Now I can't tell whether you're serious or not.

@Matt What seems to be the problem, officer?

A drinking one? : )

I don't have a drinking problem, in general.

Hi, everyone

Hi Dr. Nick!

Hi tb and robjohn.

Oh wait, that would be the proper answer to "Hi everybody!"

Can somebody explain what "Then that does it then" could possibly mean?

@tb: Omit one "then".

Or add a vegetative state coma.

If I do that I'm just as clueless.

The coma or the omitting?

It also applies to your last 4 messages (but I'm used to that by now), but the intention was about the omitting.

Yes, it is true. I am living in my own world sometimes. :-)

Tell me about it :)

The inner chain of associations and lack of filtration are both very apparent.

@tb I am telling you about it!

Morning, tb!

Morning, Matt. I'm surprised that the sun still exists :)

Why? (But so much for white Christmas : ) )

@AsafKaragila or a comma coma (the state where you fall asleep at the keyboard with your finger on the comma key).

@Matt It managed to hide itself well over the entire month. From the cold temperatures we had I deduced that it was gone.

Hi robjohn : )

@robjohn I don't think that coma is a state of sleep :-)

@JM good afternoon.

@tb Do you make your mathematical deductions like this, too? : )

@tb Once again, you have to come visit Israel sometimes. Either Murphy's law will induce a massive blizzard - and I'll be very happy about it; or you'll enjoy the sun and heat of our summer (which I have no problems with).

@AsafKaragila Or step on a mine...

Speaking of that, here is a song for you...

The Doors - Waiting for the Sun

@Matt There are no mines here, only in small parts of the Golan heights.

@AsafKaragila Oh, sure. It's my hibernation hymn

@AsafKaragila I was joking, obviously.

Let's see if my girlfriend finishes getting ready before the song ends.

@Matt judging from the reception of my comma coma joke, I wouldn't take it personally :-)

@robjohn You're just mean, square.

@robjohn They're all asleep still : ) btw, did you get my ping yesterday?

@Matt which ping?

Well, time to go! See you folks later.

@robjohn I wanted to know how the gaijin came to like genmai cha. : )

See you later @Asaf!

@Matt oh, that one. I saw it, but I was on my way out for the day. Besides, I don't see what my ethnicity would have to do with my tea preference.

I doesn't!

Can you guys what ML means? From here: math.stackexchange.com/a/92478/13425.

The answer links to AoPS.

@Srivatsan No idea.

@Srivatsan No hints in the link either. Ask in a comment?

@JM I looked around in the AoPS site as well. I will ask in a comment.

@Srivatsan Either one says "slip through cracks" or "seep through cracks". :)

What do you call the symbol used to denote the cardinalities here?

That's "beth", $\beth$. Second letter of the Hebrew alphabet.

@JM :) Not that you brought it up, "sip through the cracks" sounds really funny. I'm imaging a mug with cracks in the side...

Thanks JM! (Doh, I could've looked at the latex!)

Thanks for the clarification, Matt and JM.

@JM Apparently the earlier answer is a bit more funny: math.stackexchange.com/posts/92478/revisions.

@JM as an idiom, I've heard slip through the cracks, but not "seep".

@robjohn Yes, seep through the cracks sounds too factual.

(whatever that means)

@Srivatsan "seep" isn't an idiom, yes. :)

But they'd be the same if the thing going through's a fluid.

@Matt See also beth numbers and the gimel function. As far as I know the first three letters are the only ones you occasionally encounter in math.

Nice, thank you tb!

@tb Would you happen to know an epimorphism in the category of torsion-free abelian groups which isn't surjective?

@ZhenLin wouldn't multiplication by two in $\mathbb{Z}$ work?

Oh, that's epi? I knew its cokernel was 0.

I'm back!

With coffee!

@ZhenLin Well, this means that the difference of two homomorphisms coinciding on the even integers factors over zero, doesn't it?

@tb Yes, there is no set theoretical use of further letters. I think I will have to do something about that.

@AsafKaragila Asaf-dalet numbers?

@tb Perhaps!

@tb Ah, of course. Thanks.

Cheers!

@tb "Mathematics is too big to fail." – Indeed it is!

There are two axes waiting to drop.

@Matt: Hello Matt, merry Christmas. I saw your answer, thank you. I'm stuck, I don't know what the problem is but I cannot figure it out. Assuming 1, 2, 3 and 4 as keys I can draw only one max-heap fulfilling the property that it should be filled from left to right, and 1 is the 4th biggest element. What am I doing wrong? I'm sorry if it's going to be annoying.

@Gigili Not annoying at all, I'm quite enjoying myself : ) Gives me a chance to learn what I should've learned some years ago : ) Can you give an ascii art version of the heap you have?

@Gigili Don't worry about what it looks like. You can just post it like so:

@Matt Sure, wait a minute please.

@Matt I don't know how that works, the way you've written. But I can draw it.

@Gigili You write each level of your tree on one line and separate the keys by spaces.

: D Even better!

:D

@Gigili So, what happens if you swap 3 and 2. Would it still be a valid heap?

Coffee and beer = love.

@AsafKaragila Good luck sleeping later...

@AsafKaragila I quit drinking yesterday.

@JM Oh, I don't want to sleep. I want to work.

@Matt Oh yes, it's a valid max heap. But it can only look like that, right? only swapping the values is possible?

@Matt You always quit drinking. That's a bad sign. It is a sign that you have a problem with drinking. I don't have a problem.

@Gigili What happens if you swap 2 and 1 instead? : )

"I don't have a problem. I can quit. I've quit plenty of times!"

@JM Exactly : )

@Matt Yes, so 3 possibilities? I guess I'm getting it ..

@Gigili Yes, very good! 3 valid max heaps for 4 values. Now for more keys you'll get more possibilities.

The proof that I don't have a problem is that I never failed quitting!

@Matt Hihi, thank a lot. And how you said the index of 4th biggest element could be 2 or 3 ? shouldn't it be 3 or 4?

@<Mathematics Lounge> Merry Christmas!!

@Gigili That depends. In reasonable programming languages (for example C) the index of an array starts counting at zero : ) But there are some (for example Eiffel iirc) where it starts at 1.

A merry Christmas to you too.

Yes, a merry Christmas to you too, FreakEnum.

So you guyz got special plans today? wanna say about?

:)

@FreakEnum Of course: play around on SE and chat.SE all day : ) What about you?

@Matt cool, I'll do same :)

@Matt Wow, you're awesome. Yes I forgot that. Last question: How the maximum index is 15 and not more?

@ZhenLin were you referring to this revived MO thread?

These days I'm back to home ( my lovely Nepal :))

@tb Yes.

@FreakEnum Nice for you! Can you stay for a while or will it only be for a few days?

@FreakEnum Oh, great to hear.

@tb just few days :( but I'm happy right now :)

Have fun.

When not in Nepal, where do you stay?

@FreakEnum Then enjoy them all the more!

@tb The measure theory question you recently answered -- it needs a better title, but I can't think of any. :)

@Srivatsan This time I'll move to Bangalore (to my Uncle's house , and doing education there:)

@Gigili Assume your 4th largest element has index greater than 15. In the heap it would mean that it is on level 5 (if the root is level 1).

@Srivatsan I'll fix that a bit later... (I hate it when my posts are at the same time when others are on an editing spree :))

@Gigili But what does this mean? All the elements before it in the tree are bigger than it. So if you traverse from the root to reach it you pass 4 elements, including the root, before you reach it.

@tb Hmm... You had all the choice, and you picked the time to post an answer =)

@tb Uff, sorry. I figured a low-traffic day might be a good time for en masse edits...

@Srivatsan Sure, but let me bump the thread when it's a bit more quiet...

@FreakEnum Nice. I like Bangalore. That's pretty much the only place in India I've been for an extended time, outside my hometown.

@Gigili But this means it cannot be the 4th biggest because you have already passed 4 bigger ones (as all the elements are different).

@Srivatsan Moving from Nepal to India costs nothing :D

@Gigili Hence it can't be on level 5. Which means it has to be somewhere between the root (level 1) and level 4.

@Matt Great, got it completely. You're an awesome teacher Matt, thank you for your time and help.

@JM No problem, it's a good idea of course. It just happened the last few times I posted that a few minutes earlier the editing started, go ahead!

@FreakEnum It should cost something. How do you go from Nepal to Bangalore? Flight or train?

@Srivatsan Train :) (I've never been on airport before :D )

@FreakEnum Hm, makes sense.

@FreakEnum Well, the first time I've been on a flight is when I came to the US. Tell me this: what train or bus do you take immediately out of Kathmandu (will be leaving from Kathmandu)?

@Gigili Thank you! That's a nice thing to be told. Pop by any time you have questions, it was a pleasure : )

@Srivatsan He had all the axiom of choice, but pickled an answer?

@Srivatsan 1st bus from Nepal to Bihar then to Bangalore

@FreakEnum Ok. I guess that'll be a quite a long journey.

@Srivatsan What do you do in US though? (your whole family there?)

@JM, In the SVD-PCA answer, why does the matrix have so much spacing?

Is there a reason or can I edit it? (This answer: math.stackexchange.com/a/3871/13425)

@Srivatsan Which, the Läuchli example?

I don't quite know why; if you see the source, it's just a standard pmatrix...

@JM you have quadruple backslashes. Wouldn't double backslashes work nowadays?

@FreakEnum I am studying here.

Aha, yes, I remember now. This was at a time when backslashes were iffy.

I've edited.

FreakEnum: I am from Chennai. Family is there.

@Srivatsan Grad or 12th? (If my Q's pester you anytime , lemme know I'll stop , won't ruin your mood :))

Aw, I am doing PhD here in computer science. I didn't think of the high school possibility.

@FreakEnum That's sheer nonsense FreakEnum. What makes you think my mood is not alright?

(As you perhaps surmised yourself, I am also simultaneously checking the main page -- almost all the time. So those interludes will be common :). Sorry about that.)

@Srivatsan 0_o , ok I'm done , you're way too above then me :)

@FreakEnum =)

@tb Look, you can get your sunglasses ready : )

@JM Take a deep breath. You're about to find out what ML means.

(and your warmest coat)

@Srivatsan Clearly we don't hang with the cool kids... :D

@FreakEnum What about you What do you do?

@JM Oh, I knew this about myself a long time ago. Why dya think I chose this chat in the first place? =)

@Srivatsan looking for job :)

@FreakEnum Good luck!

@Srivatsan Thanks (but I don't think I'll get that easily :)

@Matt yes, everything's ready for a stroll :)

this must have come up before, no?

I hate such non-answers :(

@tb Sounds like it. I looked around for a bit and got nothing. I will continue to look.

Anyways ,Merry Christmas once again :) Enjoy today ( And please don't chat about Q's and A's today for god sake :D)

Sure :)

Have fun, FreakEnum. See you around.

@tb I think so; I'm trying to comb through Eric's stuff, as I distinctly recall him dealing with this.

@Srivatsan sure , you too , bye:)

Hello guys, I have what I think is a trivial question on connected sets I hoped to ask.

Shoot.

I wanted to show that the interior of a connected set isn't necessarily connected.

I took a set $E\subset R^2$ to be the union of the closed circles of radius 1 centered on (1,0) and (-1,0).

@Srivatsan same here

@yunone circles = only the boundary, right?

@Srivatsan I misspoke then, I guess I should say disk?

I want to include all the points up to and including the boundary. Is that called a disk?

@yunone Closed disk will be most accurate and unambiguous, but disk is fine.

Ok, then I should say closed disks, thanks.

Anyway, I think it's not hard to see the interior of that union of closed disks is not connected.

But my question is, is there an easy way to see that $E$ is connected in the first place, (if it actually is?)

@yunone Yes, it is connected (and visually it is clear in the first place).

I just find it hard to prove that something is not a union of nonempty separated sets, since negative proofs are always difficult for me.

HINT It's also path-connected.

I'm reading Rudin right now, and he hasn't mentioned anything about path-connectedness.

@yunone Oh no no. You shouldn't have to do that every time. It is perfectly natural that negative proofs are hard.

@yunone Ok, there are several ways to prove this. I can think of two, both of them useful I think.

Before I start, give me five minutes. I'ld like to freshen up a bit =)

Maybe it's a dumb question, but why is it visually clear that it can never be written as the union of nonempty separated sets? I'd be grateful to see either of your explanations.

@Srivatsan Yes of course, please take your time!

Any of you guys happened to have a ring A which is not Noetherian (as a ring) but Spec(A) is Noetherian (as a topological space)?

@Srivatsan Yeah, you can't even flag them.

@AsafKaragila there's this

@tb Thanks!

@AsafKaragila even better

@tb Actually the link helps me with two whole questions, not just this one part of a question :-)

@AsafKaragila Is this homework? If it is you might want to tag it as such.

And you also might want to improve your acceptance rate.

: )

:D

@Matt Within the chat? Besides I was asking about something general, it's not my fault tb found a link answering more than what I needed for this question and exactly what I needed for other questions.\

@AsafKaragila Well, I don't want to flag something unless I am convinced that it's spam or answer meant to be a comment.

@Srivatsan Yeah, and what the guy wrote is essentially a long comment on other answers. Not an answer per se.

However this is such a long comment that there is no way but allowing it to be an answer.

@yunone Perhaps we can go over here: chat.stackexchange.com/rooms/2073/…

Sure, see you there.

I like this: I am on a quest to find a definition of "class function" and then I come across this: "...Informally, we call any collection of the form $\{x \mid \phi(x) \}$ a class. However formally, classes do not exist, and expressions involving them must be thought of as abbreviations for expressions not involving them...."

Yes, what seems to be the problem?

That classes don't exist.

They are syntactical objects.

Classes are simply formulae.

I think they are a sort of set even though you can't call them that. Just like a collection or a family. Why do there have to be several names to mean the same thing?

In a sense, yes.

See this answer of mine

@AsafKaragila Thanks! I'll read it in a bit, I'm reading something else at the moment.

Ok, I have it. If $A$ and $B$ are classes and $R \subseteq A \times B$ then $R$ is a class function if it's a function.

I think lecturers should be punished every time they use a word they haven't defined.

I think everyone should be shot.

Right in the back of the neck.

With paintballs?

Spitballs?

Rubber bullets?

7.62 :-)

@Matt Functionality is a property of ordered pairs. You can express it with a nice formula. So if all the elements satisfying a certain formula are ordered pairs with the property of a function we can say that $R$ is a function.