Is there a system to delete some tags?

(Just asking...)

@tb 4) prove or disprove P=NP.

hey, quick question: is there an efficient way to find a basis of the intersection of two subspaces given bases for the two?

I don't think so.

@robjohn no that's off-topic :) RH or BSD seems like a better fit...

Given a basis for the intersection, we can always extend which you already know, but the other way, I don't think so.

We might have a basis that does not intersect the intersection at all.

Incredible wall of text

Can some algebra expert say what do they think about David's quick question?

this is the simplest way to do it, and the fastest

for example, in mathematica, if a an b are bases of the two subspaces, you can say NullSpace[Join[NullSpace[a], NullSpace[b]]] to get a basis of the intersection

So, then, there is an algorithm that works!

@DylanMoreland In atiyah macdonald they mention very little of base extension. And the isomorphism I mentioned above was not trivial to see and there was a lot of working out to see the isomorphism.

@Dylan obviously, he should become an english major.

@DavidWheeler It has dawned on me how alienated high level mathematics is from the way people live their lives everyday.

@BenjaminLim, see people.maths.ox.ac.uk/bui/ideal.pdf

Story of my life.

hahahahah

You know Zeilberger fooled me ... :/

@MarianoSuárezAlvarez I often have great difficulty sustaining conversation with my peers

talk about something else :)

I can't connect with people...

don't go Jordan on me now

they seem to be talking about lots of movies, internet cultures, reddit, etc

@MarianoSuárezAlvarez hahahahahhahahahahahahahahhaah

@MarianoSuárezAlvarez I love you man :D

heh

But i'm just saying

do you have this trouble??

Because for me whenever my tutor and I meet

we suddenly both get very excited

that line sounds a bit weird :D

NO SEXUAL CONNOTATION INTENDED

Like that day he was trying to explain to me fibred products

I know, just kidding :P

We started talking really loudly

But I mean I am worried that I am not able to function normally in society

sure

I see that as I go deeper into say commutative algebra

it is always very nice to find someone who gets what you enjoy

I get further and further away from daily life

@MarianoSuárezAlvarez Well not exactly because he is an analyst/geometer more of that....

it is just a different aspect of your daily life

@BenjaminLim this sort of reminded me of this

one difference between being, say, a violinist and a mathematician

is that everyone can perceive what the violinist does, even if they hate it

but you cannot even start talking to someone who does knot know what a vector space

if what you do is commutative algebra, say

that is why mathematicians have so many meetings and congresses and so on :D

yeah what about AC

@tb @MarianoSuárezAlvarez Wow you guys are giving great support, thanks so much man

da bosses

with time, you'll develop standard lies to tell to people who ask you what you do

@MarianoSuárezAlvarez hahahahahhahahahahahahhaaaaaaaaaaaaaaaaaaaaaahaaaaaaaaaaaa

what a pro comment

true story :)

@MarianoSuárezAlvarez You've been there done that :D :D

@DavidWheeler Is everything Euclidean, or do you mean more general types of spaces?

Like that day I was trying to explain to my friend who does DG what the tensor product of vector spaces is

I said: Consider the Free vector space on the set $V \times W$

but if he's doing DG he should know what a tensor product is :)

Let us quotient out by the subspace generated by the relations bla bla bla

(if DG = differential geometry)

@tb They don't! They dump them the definition of the dual space and then tensor product

Embarrassingly not many of them know that $V \cong V^{\ast}$

knot know? reminds me of the knot knotes.

@MarianoSuárezAlvarez So true. How do you mingle at parties?

I wonder how those people do DG when they don't know that $V \cong W$ iff $\dim V = \dim W$...

well, you can do other things apart math

Writers like Eric Temple Bell, Ian Stewart and Martin Gardner have done superb jobs of producing layperson's explanations of what mathematicians do. I don't understand what stops a mathematician from explaining what he/she does to a layperson.

most of them are not Eric Temple Bell, Ian Stewart nor Martin Gardner !

I think that mathematicians should try to be more like these esteemed gentlemen.

and, really, none of these three have explained to a layperson anything about the moduli space of quasi-conformal structures on the moduli space of anything...

Maybe there should be an undergraduate course on it.

anything you tell the layperson about the work of voevodsky will have nothing to do with the work of voevodsky

to pick a simple example

Sorry, just had to go and chase a cat out of my yard.

in fact, most of you tell an actual average mathematician about, say, the Langlands program will have little to do with the Langlands program!

@DavidWallace say what?

I haven't heard of Vodvodsky, sorry. But I still think it should be possible for a sufficiently eloquent mathematician to share what they do with non-mathematicians?

@DavidWheeler Otherwise, I end up with poo all over my lawn.

that is a claim that only carries force when the claimer has tried :)

(I don't require having succeded...)

well, my girlfirend says things like: well, you're always fooling with your numbers.....frankly, i DETEST numbers.

the layperson does not even know about the objects involved

@MarianoSuárezAlvarez Does that give mathematicians a kind of sense of superiority?

a chemist can talk about what he does, because people think they know what atoms and such are

(even though they of course don't)

yeah

so can a physicist

mathematician?

I for one don't think I am superior

most of math is not difficult at all

once you get the hang of it

but it takes time, lots of time

trivial once you see it eh?

not trivial

provided you spend enough time

but accessible

Try explaining to a lay person why $M \otimes_A N \otimes_A k \cong (M \otimes_A k) \otimes_k (N \otimes_A k)$

you simply can't explain that

because you'd first have to explain what it means for that to be true

One does not simply explain tensor products to a lay person

but why would one want to?

one can explain pieces of mathematical knowledge to people

i want Benjy to explain tensor products to ME

I've had immense pleasure explaing (and proving!) the fact that there are five regular solids to young highschoolers

@DavidWheeler Do you see why $A/a \otimes_A M \cong M/aM$?

@MarianoSuárezAlvarez GROUP THEORY TIME. With orbit stabiliser?

last year I spend 3 hours sitting on the floor at the door of my university explaining kids about wallpaper groups

one can, for example, talk about symmetry without going into the group theory involved....for example, one can display pictures of examples of various frieze groups

with complete proofs

but I'd never attempt explaining Hochschild cohomology to them

@MarianoSuárezAlvarez Math is indeed exciting

People say that when I start talking about it I get excited and then I stammer a little from the excitement

:)

the only thing i recall about the platonic solids is something about euler characteristic, and they, um, wouldn't go out with me

@DavidWheeler There's a way to do it without mentioning euler characteristic

i mean the edges, faces and vertices thing

@DavidWheeler math.stackexchange.com/a/128606/5783

what you can do, with considerable work, is to try to have them feel the rush of what mathematical knowledge is

that is better than trying to explain "what one does"

@MarianoSuárezAlvarez Yeah, that's what I always try to do. To make them feel the adrenalin pumping through their blood the same way I do.

I guess if you want to explain something to laypeople you need something that they can connect to and experience by "seeing" it, for example. Geometry is very good for that, because you can see the stuff. However, abstract algebra is too abstract for many math majors, so it's completely hopeless for someone who has no experience with math whatsoever.

For example when I meet kids, I ask them what are they learning now at school

graph theory works great, too

i blame the schools

@tb Even worse commutative algebra

when people learn arithmetic, so much emphasis is put on the numbers, rather then the laws they have to obey, or go to jail if they can't make bail

@DavidWheeler Tensor the exact sequence $0 \rightarrow a \rightarrow A \rightarrow A/a \rightarrow 0$ with $M$ and use the fact that $A \otimes_A M \cong M$

Combinatorics at the most superficial level can be interesting.

@MarianoSuárezAlvarez I then try to explain to them something interesting about whatever they are learning at school.

For example there was this kid in year 5 learning about area and perimeter

I asked her how to measure the area of a circle

@Benjamin but don't i need to verify that the functor involved is right-exact, then?

Well prove it using the fact that $\operatorname{Hom}(M \otimes N, P) \cong \operatorname{Hom}(M, \operatorname{Hom}(N,P))$

ok, i see the $M$ part, but what about the $aM$ part?

i mean it seems as if you're claiming $a \otimes_A M \cong aM$

huh?

@MarianoSuárezAlvarez

you asked me to use the fact that $A \otimes_A M \cong M$ but your original result involves $M/aM$

use first isomorphism theorem

Verify that the kernel of the map from $A \otimes M$ to $A/a \otimes M$ is actually $aM$

well, when i tensor the first sequence i get $a \otimes_A M$ at the start, right?

then you will have your isomorphism because by exactness the map from $A \otimes M$ to $A/a \otimes M$ is surjective

yes

Now the image of $a \otimes M$ in $A \otimes M$ is just by the inclusion map

that's what i'm struggling with....by the exactness of the tensored sequence, that should be the kernel of the surjective map

yeah

Now look:

so how do i verify that this kernel is actually $aM$

Look at an element in $a \otimes M$

Send it into $M$

what does that element look like??

send it..how...by what mapping?

Do you know what is the explicit description of the isomorphism $f : A \otimes M \rightarrow M$?

It sends an elementary tensor $s \otimes m$ to $sm$ for $s \in A$, $m \in M$

that's one of the things i want to know

Sorry I sort of assumed you knew that

@DavidWheeler are you undergrad?

ok, so if $x \in a$, then $x \otimes m$ should go to $xm \in aM$

oh no wait

here it goes

an element in $a \otimes M$ looks like:

$\sum a_i \otimes m_i $

In $M$, it will be $\sum a_im_i$

which is an element of $aM$

yeah but the maps are additive, so i don't care about sums

Yeah but the sum part is important

it's like in linear algebra, just follow the basis elements

@DavidWheeler careful with that analogy.

I usually write out everything in full

unless when checking maps, I check on elementary tensors and then extend linearly

but here it is important to keep the sum in there

ok, why is that?

because what do elements of $aM$ look like?

the product of two ideals, one being a, the other M itself

explicit description please

$M$ is not an ideal

it's a module

we're not looking at algebras?

NONONONOONONONONONNONONONONONOONONONONONONONONO

TENSOR PRODUCT OF MODULES

ok, then we have linear combinations $\sum_i a_i m_i$

yeah

see why it's important to keep the sum in there?

so we get a submodule?

yes

$aM$ is a submodule of $M$

we have to have closure under module addition

yes

@DavidWheeler are you undergrad??

not anymore...i don't have to go to school anymore :P

grad student?

when i was undergrad, modules were just barely covered...that's why i'm looking into AM

well i'm first semester 2nd year long way to go for me :D :D :D

really, i want to understand the group ring F[G] better, because reps of G are F[G]-modules

So we already know that $A \otimes M/(a \otimes M) \cong A/a\otimes M$ by FIT

But then $A \otimes M/(a \otimes M) \cong M/aM$

so we are done

good :)

so now, go back to that "bigger" result with the field k

@DavidWheeler well the field $k$ thing was a lot more complicated

Not so easy to explain just like that

It would be 2 pages of a proof

@DavidWheeler Can I ask how old you are? You may delete the comment after I see it

Because if you are older than me you are probably wiser

well a representation is just a $G$ - module no?

well, we need to "extend" G to make it be a ring, since addition doesn't a priori exist

anyway

I want to ask you something about life

As I go deeper into mathematics

my age is no big secret...i am 50

I feel my life is becoming more and more isolated and separated from the conventional reality

Good, you have prob got a big lot of experience behind you :D

mathematics, especially, is largely mental...it lacks the physical interactions of a lot of many other fields

yeah, that's the thing

I feel like I am growing separate and different from real life if you want to put it that way

I mean my days are spent only doing maths

I don't do everything else

and mathematics...it has this internal perfection of structure, that life lacks

@DavidWheeler Hey joking young man!?!!

when you're proving a result....the rules are clear. when you're ordering pizza...well, who knows what might happen?

An old man's reaction would have been spontaneous. I caught you there. :P

lol, look at the boy wonder

@KannappanSampath Don't know if you feel the same way I do

I'm not doing a Jordan but just saying the way I feel

the isolation used to bother me more than it does now

@BenjaminLim I do feel that, but not very often.

And, very often I am in this chat room.

but...but...i want you to get back to your "non-trivial" tensor product result

i'm tired man

ah well

so tired

all those days of getting maggot

i swear on sat night that cake I ate was laced with marijuana

bye

$$test \tag{in$g$}$$

From Wikipedia I'm familiar with $\mathrm{Ind}_H^G\pi$ being defined as either (coset representatives in subscript) $$\bigoplus_{x\in G/H}xV$$ or as $k[G]\otimes_{k[H]}V$. I see how the above is well-defined and equivalent to the latter. The text I'm looking at (Lie Groups by Procesi) however defines the induced representations as $$\mathrm{Ind}_H^GM=\{f:G\to M: f(gh^{-1})=hf(g)~~\forall h\in H,g\in G\}$$ Hints on seeing the equivalence of this definition?

Morning everyone.

mroing

(allusion)

That's awful. : )

Ello teddy : )

morning

Cold feet?

Of course. Saint Peter went crazy. Too much Easter celebrating, I guess.

@anon They are equivalent provided $G$ is finite. Note that $k[G] \mathbin{\otimes_{k[H]}} M$ represents certain bilinear forms $k[G] \times M \to Z$ and so does $\operatorname{Ind}_{H}^G M$.

(where $Z$ is an arbitrary $k[G]$-module)

'Ello.

'ello

@anon: Are you a, umm, native English speaker or something?

Ello Gigili : )

@Gigili: I am. @tb Sorry, I'm looking up stuff on the bilinear map / universal property yadda yadda definition of tensor product. (I just knew what the elements looked like and how to add or scalar multiply them.)

(well, "scalar" being relative)

I have a question. when talking about, for instance, Respiratory tract and gastrointestinal tract or mouth, are they the most common "places" for viral diseases? I don't know an appropriate word to use there.

I have no idea about your medical question, but the word "places" is fine. I suppose you could be more specific and say "bodily area" or somesuch (was that your question?)

@Gigili, this reminds me: I saw my vet friend the other day and she confirmed that what I said about DNA strings being finite and stuff was right. Do you remember? : )

@anon Yeah, thank you. So "the most common bodily areas".

@MattN Ah yes. They're finite?

jiggily!

Yes. You don't have infinite space inside each of your cells. : )

'Ello David!

viruses like mucus membranes

The number is just so large that you can't imagine. So certainly large enough that you couldn't fit all the hoomins on planet earth if you implemented all the possibilities.

@MattN True that, but how each two DNA strings are different?

@MattN Umm, that must be the case. So large means finite or infinite?

because the sequence of the amino acids are different

It means finite. There are $N$ possibilities to make (genetically different) hoomins. For $N$ very large : )

Right, there are sequences as many as human beings.

during reproduction the DNA codon sequences are recombined, making the chromosomal expression for all intents and purposes unique

@Gigili That's because we are only $n$ hoomins today, for $n$ comparably small. So there are still $N-n$ (new) possibilites to make people that don't exist at the moment.

as far as numbers go, jig, 9 billion isn't that big

I think I got it.

Cool : )

@t.b. If I understand correctly, any bilinear form $A\times B\to C$ is equivalent to $A\times B \to A\otimes B \to C$ (first the tensoring followed by a linear map) - this is the definition of the tensor product. Intuitively I feel this means the tensor prod "prerepresents" bilinear forms, in the sense that realizing any bilinear map is equivalent to first moving everything onto the tensor product and then applying a linear map. Would this be accurate?

Thank you for pursuing the subject, Matt.

No problem : )

@anon yes, that's the idea: the tensor product allows you to replace bilinear maps on the product $A \times B$ uniquely by linear ones on $A \otimes B$. You can go back by composing the latter with $(a,b) \mapsto a \otimes b$. The universal property "yadda yadda" tells you that this property together with the map $A \times B \to A \otimes B$ determines the tensor product uniquely.

(up to unique isomorphism)

So: to check that $X$ represents a tensor product of $A$ and $B$, you check that you have a bilinear map $A \times B \to X$ and that every bilinear map $A \times B \to Z$ factors over a linear map $X \to Z$.

Ok. Going to do some stuff. See you later!

would it be fair to say that $A \otimes B$ is universal among bilinear maps from AxB?

@MattN Have fun.

@DavidWheeler Almost. It would be a bit fairer to say that the map $(a,b) \mapsto a\otimes b$ is universal among bilinear maps from $A \times B$.

It is important that the tensor product comes equipped with such a map. It is meaningless without it.

and i suppose we call this map "$\otimes$"?

Yes, it's the tensor product of elements of $A$ and $B$.

The point of the universal property is that it doesn't really matter how you implement the tensor product $T$ of $A$ and $B$. The important thing is that you get a bilinear map $A \times B \to T$ which satisfies the unique factorization property for bilinear maps out of $A \times B$.

so, concretely, what is $\Bbb{R} \otimes_{\Bbb{R}} \Bbb{R}$?

It is $\mathbb{R}$ itself. Multiplication $\mathbb{R} \times \mathbb{R} \to \mathbb{R}, (s,t) \mapsto st$ gives you a bilinear map and now you can check that this map has the universal property.

well, it seems analogous to the products and quotient groups we were looking at earlier...we have a guaranteed factorization of (yadda yadda thing) through our universal

yes.

ok, what is $\Bbb{R}^n \otimes_{\Bbb{R}} \Bbb{R}^n$?

i'm thinking $\Bbb{R}$ again, still, because of inner products

You can implement it explicitly by $n \times n$ matrices via $(x,y) \mapsto x y^T$ if you want.

(every bilinear form is determined by how it acts on the basis vectors, hence by an $n \times n$-matrix)

$(x,y)\mapsto xAy^T$

That's the bilinear map represented by $A$.

but that means $x \otimes y$ is always a real number...i mean we have a set, but we need the bilinear form to include it with our definition, right?

can we realize det as a tensor product?

I meant what I wrote. $(x,y) \mapsto x y^T$. This is the bilinear map $(a,b) \mapsto a^T (xy^T) b$

(thinking of $\mathbb{R}^n$ as column vectors).

Hmm, $f:G\to M$ as I defined earlier is determined by its values on the coset representatives $x_i\in G/H$ so I see the isomorphism between the definitions now. I don't think I'm at the right level to see the equivalence through tensor products (even if I know naively how the basic algebra works in them).

@anon, what are G and M?

G is a group, M the vector space it acts on, H the subgroup of G

@anon That's right. The relevant bilinear maps satisfy in addition $B(gu,v) = gB(u,v)$ and $B(uh,v) = B(u,hv)$.

(left $G$-linear in the first variable and "balanced over $H$").

Sorry to interrupt, what does "you will be good to go" mean?

what does "balanced over H" mean?

what I wrote: B(uh,v) = B(u,hv).

@Gigili You will be ready for whatever needs to be done.

@anon Thank you.

@Gigili: for example, if the mechanic is all finished with your car repairs, he may well say: "you're good to go"

@anon...what is that "Ind" thingy?

Induced representation.

We have $k[G]$ which we view as a left $k[G]$ module and as a right $k[H]$-module and we have a left $k[H]$-module $M$. When we have an element $u \otimes v$ of the tensor product $k[G] \mathbin{\otimes}_{k[H]} M$ then we can act as follows

@t.b. you're a fine one for saying "what what i said means is what i just said"

ah, thankee

we can act with $G$ on the left $(gu) \otimes v = g(u \otimes v)$

@DavidWheeler Meaning "it's ready to use"? Does it have the same meaning in this context: "Follow the instructions to get it set up on your computer and you will be good to go!"?

and we can act with $H$ in the middle $(uh) \otimes v = u \otimes (hv)$.

@Gigili yes, it means "you're all set!"

Thank you.

@tb is $k[G]$ a $(k[G],k[H])$-bimodule, then?

That's the way we think of it in this instance, yes.

lol, first time a chat convo has crashed my mathjax

i love this stuff...it's like tinker-toys for grown-ups :)

The idea of induction is to make a $k[G]$-module out of a $k[H]$-module.

is that "expansion of scalars"?

extension, yes.

Of course, $H$ is a subgroup of $G$.

i figured that when anon said cosets

We can always restrict a $G$-representation to a $H$-representation and induction is one way to go back from $H$-representations to $G$-representations.

it seems analogous to the group concept of extending a group G by H

I don't quite see the analogy.

you get a bigger thing that contains the smaller thing you start with that you got from the bigger thing before

i am fascinated by representations, and scared of them, too

but i want to learn more about them before fooling around with Lie schtuffs

If you want to pursue this, look up Frobenius reciprocity

oh, they're adjoints!

Yes, induction is left adjoint to restriction.

well, i know physicists do this because they can use it for quantum stuff like with gauge symmetries...but i just want to understand what people are talking about when they say cool things like: well, just look at the fiber bundle, because, y'know it's imprimitive

i'm afraid i don't have enough years left to satisfy my curiosity, but oh well

Then you should look at Mackey's work. Physicists like Barut, Rączka, but I only read small parts of it.