Mathematics

@anon, yes i meant isomorphism. can we not say something in general, however? for any A and B? can't it be given as the quotient of a direct sum somehow?

anon

12:43 AM

How is it not already "for any submodules A,B"?

Peter Tamaroff

$$\eqalign{
& S = \left\{ {x:{x_2} - {x_3} - {x_4} = 0} \right\} \cr
& T = \left\{ {x:{x_1} + {x_2} + {x_4} = 0} \right\} \cr
& H = \left\{ {x:{x_1} + 2{x_2} - {x_3} + 2{x_4} = 0} \right\} \cr} $$

All in $\mathbb R^4$.

Thelonius

@anon, i want an expression that is valid for all modules A and B. you seemed to imply that A+B only made sense if they were submodules of a bigger ring. obviously they both have to be R-modules for some fixed ring R, but i don't see what you mean about being submodules

Peter Tamaroff

Now, I need to find a subspace $W$ such taht $S^\perp+T^\perp\supset W$ and such that $W\oplus (S\cap T)=H$.

I have done this.

anon

@Thelonius If $[A]$ and $[B]$ are two equivalence classes of modules, surely there are modules $M$ with the property that $X,Y\in[A]$ and $U,V\in[B]$ are submodules of $M$ but $X+Y$ and $U+V$ are neither equal nor isomorphic...

Peter Tamaroff

$$\eqalign{
& S = \left\langle {\left( {1,0,0,0} \right)\left( {0,1,1,0} \right)\left( {0,1,0,1} \right)} \right\rangle \cr
& T = \left\langle {\left( {0,0,1,0} \right)\left( {1,0,0, - 1} \right)\left( {0,1,0, - 1} \right)} \right\rangle \cr
& H = \left\langle {\left( {0, - 1,0,1} \right)\left( {1,0,1,0} \right)\left( {0,1,2,0} \right)} \right\rangle \cr} $$

And

Thelonius

12:47 AM

@anon what if $A$ and $B$ are finitely generated?

anon

forgetting some \perps I believe

Peter Tamaroff

$$\eqalign{
& S \cap T = \left\langle {\left( { - 1,1,1,0} \right)\left( { - 1,0, - 1,1} \right)} \right\rangle \cr
& S^\perp = \left\langle {\left( {0,1, - 1, - 1} \right)} \right\rangle \cr
& T^\perp= \left\langle {\left( {1,1,0,1} \right)} \right\rangle \cr
& S^\perp+ T^\perp = \left\langle {\left( {1,1,0,1} \right)\left( {0,1, - 1, - 1} \right)} \right\rangle \cr} $$

anon

@Thelonius What difference does it make? The intersection of two groups (e.g. $\Bbb Z$-modules) is (first of all empty unless they are both subgroups of a bigger group, but also) not determined at all by the isomorphism class of said subgroups. I don't see how f.g. has anything to do with the issue...

Thelonius

@anon your sight is better than mine

anon

You might be vaguely interested in something called amalgams of groups, after all. (A research topic of MSE's Arturo Magidin, as it happens.) It has to do with how different groups may be "joined at the hip" in an overgroup in different ways.

Thelonius

12:50 AM

i want to justify something in Coutinho, which says basically that the finite sum of holonomic modules is holonomic because submodules and quotients of holonomic modules are holonomic and the direct sum of holonomic modules are holonomic

i don't get the because

anon

What does holonomic mean, and is the sum internal and not necessarily direct?

But the implication would follow since, as you note, internal summations of submodules are isomorphic to direct sums quotiented by intersections.

Thelonius

he just says "finite sum". the definition of holonomic is not so important, his argument is. holonomic means a module over the Weyl algebra A_n which is of dimension n

if the sum were direct my question would be trivial

@anon just trying to understand what a reasonable expression of $\sum_1^N \text{holonomic}$ would be in terms of direct sums and quotients

anon

In general, $$\sum_{i=1}^k A_i\cong \underbrace{\left(\bigoplus_{i=1}^k A_i\right)}_{E}~\big/~\mathrm{ker}\left(E\to M:(a_i)_i\mapsto \sum_{i=1}^k a_i\right).$$

Thelonius

ahh, and this kernel must be a submodule!

M is the internal direct sum?

anon

no, $M$ is the overmodule that everything takes place in

the $A_i$ are each submodules of $M$

Thelonius

12:59 AM

@anon weird. he makes no reference to an overmodule...

i guess it doesn't matter which one...

still holonomic

anon

@Thelonius How do you add submodules together internally if they are not all submodules of something? That doesn't even make sense!

Thelonius

i see what you mean

anon

as an example in action: think of 1d subspaces (lines through the origin) of R^2. the external direct sum of n lines will be iso to R^n, but the internal sum of distinct subspaces will just be R^2.

sorry @Peter, I like this algebra more and my time is limited :-)

Peter Tamaroff

@anon Yeah. No problem.I think I got it.

Thelonius

thanks, @anon

Peter Tamaroff

1:15 AM

@anon If you have any keen observation, just tell me

Thelonius

1:29 AM

@anon given $I\subset J$ two left ideals of some ring $R$ (Noetherian if you like), can you express $I/J$ as a combo of sums, submodules or quotients of $R/K$, where $K$ is any left ideal you like

if you know this, even a hint would be fine

still working through same chapter

Peter Tamaroff

2:22 AM

@JayeshBadwaik Could you help me with some linear algebra?

Nevermind.

Peter Tamaroff

2:57 AM

@Thelonius I have a silly question

(ormaybe not)

Suppose $(5;-1;7)$ is the unique solution to $$Ax = \left( {3;4;1} \right)$$ where $A$ is $3\times 3$ and (the $;$ denote column vectors)

Let $B$ be the matrix

1 ,2,-3,-1
-1 ,0, 4, 2
1, 3, -1, 1

What are all the solutions to $(AB)x=(3;4;1)$?

Now since $Ax=b$, $b\neq \bf 0$, has unique solution, $A$ is inversible.

Jayesh Badwaik

3:15 AM

@PeterTamaroff Sorry, was AFK. if you still are having the problem, maybe I can help.

Peter Tamaroff

@JayeshBadwaik That above

@JayeshBadwaik On the matrix eqns

Jayesh Badwaik

Okay.

@PeterTamaroff B is a 4x4 matrix. How can it be multiplied with a 3x3 matrix?

Peter Tamaroff

@JayeshBadwaik $3\times 4$!

=)

Jayesh Badwaik

Sorry Sorry

Hmm.

You can find $A$.

Peter Tamaroff

@JayeshBadwaik What I can think of is considering $Bx=A^{-1}(3;4;1)$

@JayeshBadwaik I can?

Jayesh Badwaik

3:19 AM

@PeterTamaroff Yes that is what I was thinking.

Peter Tamaroff

@JayeshBadwaik Since $A$ is inversible.

Jayesh Badwaik

You can determine $A$ upto similarity I guess.

But another thing is check the rank of $B$, it might come out to be $4$, and probably no solution exists.

Peter Tamaroff

@JayeshBadwaik Oh! That is it.

@JayeshBadwaik I think that does it.

Jayesh Badwaik

But again, $Bx$ should not be possible, since $B$ is a $3 \times 4$ while $x$ is $3 \times 1$.

Unless this is a different $x$ from the one in $Ax = (3;4;1)$

Peter Tamaroff

@JayeshBadwaik Hmmm.... I'll sleep now. I'll ask my prof tomorrow.

Jayesh Badwaik

3:28 AM

@PeterTamaroff Good night.

Peter Tamaroff

@JayeshBadwaik Night

Jayesh Badwaik

Now, I will go away, and this room will be an empty room in JaX Shell.

user19161

@JayeshBadwaik Hi!

Jayesh Badwaik

@JasperLoy Good morning!! I guess you woke up, or not yet slept?

user19161

@JayeshBadwaik I woke up. I am trying to regulate my sleep to normal hours but keep failing.

user19161

3:32 AM

I see that the past few starred messages only have one star each mostly.

Jayesh Badwaik

@JasperLoy Yup. As I said, no real line punchers currently.

user19161

@JayeshBadwaik That definitely looks like KDE.

Jayesh Badwaik

@JasperLoy It is. It is.

Wait a min. Will repost that image.

user19161

@JayeshBadwaik Hey, you are not considering other countries for your grad school application?

Jayesh Badwaik

@JasperLoy I am considering, but most european countries demand a maths bachelor, and US countries demand a math GRE. There is very less chance otherwise (applying without a math subject GRE score for a non-math major). I am giving math GRE in November (that is the only time it is offered here.)

@JasperLoy what do you think?

user19161

3:45 AM

@JayeshBadwaik Hmm, I agree with your above paragraph.

Jayesh Badwaik

@JasperLoy okay.

user19161

@JayeshBadwaik Hmm, maybe I will tell you more about me in email some time. I feel very bad these days.

Jayesh Badwaik

@JasperLoy You are welcome. I would do my best to help.

Joe

4:50 AM

Any help on this integral please?
$$\int \frac {\sqrt{\tan \theta}} {\sin 2\theta} \ d \theta$$

I tried rewriting it as $$\int {\sqrt{\tan \theta}} \cdot \csc(2\theta) \ d\theta$$

Supposedly letting $u = \sqrt{\tan \theta}$ cleans up the integral to just $1$, but I don't see how. $$du = \frac{\sec^2 \theta}{2\sqrt{\tan(\theta)}} d\theta \implies 2u \ du = \sec^2 \theta \ d\theta$$

Here's where I'm not sure how expressing $\csc(2\theta)$ as a double angle and in terms of $u$ cleans it up so nicely.

user19161

@Joe Try posting on the main site.

Joe

I was considering it, but I figured it was a subtle thing I was missing that would be easier here.

Jayesh Badwaik

5:05 AM

Chat went wrong

@Joe Try putting $\sin 2\theta = 2 \sin \theta \cos \theta$ and then use the beta functions. Else substitute $(\sin \theta)^{-1/2} = t$ and try to solve the resulting integral (if possible in closed form.)

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$Mathematics$ Mathematics