Mathematics

leslie townes

12:02 AM

oeis.org/A001065 might be helpful. OEIS often includes varying descriptions of a sequence.

Thorgott

@love_sodam if $\mathrm{cl}(B)\subseteq U$, yes

that's the point I've been trying to get to all along

do you see why this implies $(U,U-B)\subseteq(U,U-y)$ inducing an iso on homology

love_sodam

@Thorgott Then I would say from the l.e.s of pairs $(U,U-y)$ and $(U,U-B)$ and five lemma tell they are isomorphism

Derivative

@leslietownes oeis.org/A005114

I found it after running the python

Thorgott

oh yeah, that works too

leslie townes

derivative, math.dartmouth.edu/~carlp/rangeofs6.pdf might have interesting references.

Thorgott

12:04 AM

I was thinking LES of the triple, but either approach is fine

leslie townes

if carl pomerance has written a paper about it, it may be insanely hard. :)

Derivative

This is an introductory book :( why are authors like this

leslie townes

introductory book in analytic number theory. very different from many uses of the term 'introductory' :)

love_sodam

@Thorgott So in the diagram chat.stackexchange.com/transcript/message/59204619#59204619 two vertical maps are iso by excision and horizontal maps are iso by what we just said. So adding $\mathrm{cl}(B)\subset U$ makes the proof works

Thorgott

yes

Ted Shifrin

12:18 AM

@leslietownes Not necessarily.

Depends on the paper and journal. He did a lot of expository stuff.

Analytic number theory is NOT introductory number theory.

Check out Paul Pollack’s Springer book. That was basically his undergraduate honors thesis.

Koro

1:57 AM

Given a finite dimensional vector space V with a basis $v_1,v_2,…,v_k$, and a subspace U of V, is it possible to reduce the basis to a basis of U?

Ted Shifrin

What do you think?

Koro

It is clear to me that for any non zero subspace U, we have U= span of some $v_i$’s

Ted Shifrin

Clear?

How many subspaces does $\Bbb R^2$ have?

Koro

3 (zero subspace, straight lines through origin and $\mathbb R^2$ )

Ted Shifrin

How many lines are there?

Koro

2:03 AM

Infinitely many, a point on a straight line through origin is $(x,y)= x(1,m)$. So span {(1,m) } gives a straight line through origin for every real m.

Ted Shifrin

Reread your question.

Koro

Hmm, no. Because standard basis $e_1,e_2$ can’t be reduced to a basis of a straight line $y=5x$ 😀

Ted Shifrin

Bingo.

Koro

Yay! Thanks a lot professor Ted. So the original problem (which I have not stated) needs more work. :)

Ted Shifrin

The converse of the extension lemma is false:)

2

Don’t forget to think about concrete examples of things.

2

Koro

2:48 AM

I see that now professor Ted. Thank you very much. :)

copper.hat

4:06 AM

Draw pictures.

Koro

4:36 AM

Let V,W be finite dimensional vector spaces. Let $T\in L(V,W)$. Let basis of W be given as $w_1,...,w_m$. Then it is to be shown that there exists a basis of V such that first row of matrix of T w.r.t. V and W is either a zero row or has first entry equal to 1 and rest all entries zero.

The idea I have is to assume any basis of $V$ say, $v_1,v_2,...,v_n$. Then there exists scalars c's such that $Tv_i=\sum_r c_{ir}w_r$.

and then finding some linear combination of $v$'s such that the new set obtained say $v_1',v_2'...,v_n'$ is linearly independent and is such that $Tv_1'=w_1$ and $Tv_i'$(for all $i\ge 2$) is a linear combination of $w_j$'s $j\ge 2$.

But it seems difficult.

leslie townes

are you familiar with elementary row and column operations? one way of looking at this is that you can either zero out the first row or put a 1 in the (1,1) entry and zeros everywhere else using only elementary column operations.

the corresponding fact with row operations is maybe better known but it's the same principle

Koro

@leslietownes yes, I'm familiar with those.

leslie townes

you could do the exercise you were doing the other day about rank one operators this way too, although it's less explicitly constructive and maybe less illuminating than the direct proof you came up with

the idea would be, pick some basis v_1, ..., v_n for V and consider the matrix M of T with respect to the (v_i) and (w_k) basis. the thing about elementary column operations tells you that there's an invertible matrix A for which MA has either a first row of all zeros, or a 1 in the (1,1) entry and zeros everywhere else.

A or A^(-1) (i can never keep this straight and have to draw a diagram) implements a change of basis from the v_i that you started with to a basis v_i' that has the desired property

the effect on the matrix M of changing a basis for V is right-multiplying by an invertible nxn matrix (which you can think of as a product of elementary column operations), and the effect of the matrix M of changing a basis for W is left-multiplying by an invertible mxm matrix (a product of elementary row operations)

if you actually think about your construction from the rank one problem in these terms you'll see how the formulas you came up with can be thought of as row/column operations on what you started with

Koro

4:58 AM

We have $[Tv_1, Tv_2,..., Tv_n]$, wherein each $Tv_i$ is a column. I'll do column operations to get desired zeros in the first row. (Column operations because row operations will changes basis for W). Each column operation is equivalent to multiplying by some elementary matrix (just as it is in case of row operations).

Koro

5:50 AM

@leslietownes I think I understand your idea now.

no, I'm still working on it.

Koro

6:37 AM

Let $\displaystyle w_{1} ,w_{2} ,...,w_{m}$ be the given basis of W. \

Let $\displaystyle u_{1} ,u_{2} ,...,u_{n}$ be a basis of V.

If $\displaystyle Tu_{i} =\sum _{r} c_{ir} w_{r}$

$\displaystyle \begin{bmatrix}
Tu_{1}\\
Tu_{2}\\
Tu_{3}
\end{bmatrix} =\begin{bmatrix}
c_{11} & c_{12} & c_{13}\\
c_{21} & c_{22} & c_{23}\\
c_{31} & c_{32} & c_{33}
\end{bmatrix}\begin{bmatrix}
w_{1}\\
w_{2}\\
w_{3}
\end{bmatrix} =C\begin{bmatrix}
w_{1}\\
w_{2}\\
w_{3}
\end{bmatrix}$

If the first row is a zero row then we are done. Suppose that the first row has one non-zero entry.

I think I understand how to get the basis now @Leslie.

@leslietownes it seems to be a very powerful way. Thanks a lot :)

love_sodam

7:16 AM

Here's an interesting question from a graduate school interview : Let $f:S\to\Bbb R$ be a function which is continuous at $0$ where $S = \{\frac{1}{n}\mid n\in\Bbb N\}\cup\{0\}$. If $f$ uniformly continuous?

Sumanta

8:11 AM

Is the universal cover figure-eight embeddable in plane?

I mean from weak topology on universal cover onto subspace topology of plane?

Sumanta

8:39 AM

I hope I got the answer. It is on page 59 of Hatcher's book. The main point is placing the real lines with some lengths in the right positions so that an arbitrary collection of vertices should not converge to a point on the tree.

Sorry for inconvenience.

SmokenSieEinBitteChebaHitBits

9:20 AM

code meatrix

I'm working on the subgraph isomorphism search engine now for Abstract Spacecraft

It's an NP-complete problem, but for small enough graphs, you can do it speedily enough

monoidaltransform

11:24 AM

@Thorgott you once recommended a category theory book, I can't, for the love of me, remember what it's called

Thorgott

11:58 AM

Jun 14 '20 at 16:00, by Thorgott

must've been Abstract and Concrete Categories - The Joy of Cats by Adámek, Herrlich and Strecker or Handbook of Categorical Algebra by Borceux

Xander Henderson

1:07 PM

ARG! I left my keys on a shelf at home. So I get to sit outside in the cold for an hour until someone comes to let me into my office. :(

But at least I can grade out here!

love_sodam

2:03 PM

1

Q: Galois group of $x^n-p\in\Bbb Q[x]$

Compute the Galois group of $x^n-p$ over $\Bbb Q$. Here, $n\in\Bbb N$ and $p$ is a prime number. Let $\eta$ be the primitive $n$th root of unity. Then $\Bbb Q(\sqrt[n]{p},\eta)$ is a splitting field of $x^n-p$ over $\Bbb Q$. Let $H$ be a fixed field of $\Bbb Q(\eta)$ under the Galois correspond...

I have a question related to Galois group. I want to know if the s.e.s argument holds.

Xander Henderson

2:22 PM

Yay! I am inside now! My fingers are thawing!

Transcript for

$Mathematics$ Mathematics