Mathematics

bjb568

12:00 AM

And the Calculus Question of the Day award goes to…

-3

Q: what is 2×=(360÷4)=

2×=(360÷4)= 2×(360÷4)= 2×=(360÷4)=

calculus

ᴇʏᴇs

The parade outside is too noisy to study

bjb568

Oh, I like your eyes btw.

Mike Miller

the fundamental question is something I think would be solved by seeing some computations, so I'm just looking for a reference... rather than bugging people to solve my question

ᴇʏᴇs

I like your cat

bjb568

:D

Ted Shifrin

12:01 AM

Retag it @bjb568

bjb568

Meh, it's going to burn in flames and I dunno what it's even asking. Also, don't have 2k.

Ted Shifrin

I'll look.

Mike Miller

You've seen all that's there

Ted Shifrin

Closed.

Which spectral sequence, @Mike?

Mike Miller

@TedShifrin: say for convenience $c_n \in H^n(K(\Bbb Q,n);\Bbb Q)$ is the canonical element. Let $X$ fit into the fibration $K(\Bbb Q,4n-1) \to X \to K(\Bbb Q, 2n)$, classified by principal obstruction $c_{2n}^2 \in H^{4n}(K(\Bbb Q,2n);\Bbb Q)$. I'd like to compute the rational cohomology of this.

(so Serre)

this reduces to showing that $H^{4n-1}(K(\Bbb Q,4n-1);\Bbb Q) \to H^{4n}(K(\Bbb Q,2n);\Bbb Q)$ (the transgression map) is an isomorphism

I don't understand the transgression map, and thus my predicament

Ted Shifrin

12:12 AM

I'm worthless ...

ccorn

Hi folks. I have not been active here for some time, but for this question I had to work out the answer.

Mike Miller

Me too, @TedShifrin

Clarinetist

Suppose $M^2 = M$ and $v \in C(M)$. Why does $v = Mb$? (it's unknown to me what $b$ is, but my guess is that $b \in \mathbf{R}^n$).

$C(M)$ is the column space of $M$

Ted Shifrin

Definition of column space.

Clarinetist

Column space = span of columns of $M$

Ted Shifrin

12:23 AM

Remember that multiplying a matrix by a vector takes a linear comb of the columns.

Work an example.

Clarinetist

$$Mb = \begin{bmatrix}
a_{11} & \cdots & a_{1n} \\
\vdots & \ddots & \vdots \\
a_{n1} & \cdots & a_{nn}
\end{bmatrix}\begin{bmatrix}
b_1 \\
b_2 \\
\vdots \\
b_n\end{bmatrix} = \begin{bmatrix}
\sum_{j=1}^{n}a_{1j}b_j \\
\sum_{j=1}^{n}a_{2j}b_j \\
\vdots \\
\sum_{j=1}^{n}a_{nj}b_j
\end{bmatrix} = \sum\limits_{j=1}^{n}b_j\begin{bmatrix}
a_{1j} \\
a_{2j} \\
\vdots \\
a_{nj}\end{bmatrix}$$

ccorn

@Clarinetist Note that this does not require $M^2=M$.

Clarinetist

Yep noted

Suppose $M^{\prime} = M$ and $w \in C(M)^{\perp}$. Then I know that for all $x \in C(M)$, $x \cdot w = 0$. Then $Mw = 0$ should follow immediately, right? Why is symmetry necessary? The book says $Mw = M^{\prime}w = 0$.

Nvm I figured it out

Ilham

1:07 AM

Can you all see the LaTeX? All I see are dollar signs in chat.

N3buchadnezzar

Heya

@TedShifrin Hey ted =)

Ilham

Oh never mind, I see the guidelines to the side for LaTeX.

N3buchadnezzar

@Ilham I usually never bother activating it, I do not see LaTeX anymore. All I see are blondes, brunettes and redheads.

Ilham

@N3buchadnezzar All I have are blackheads and whiteheads on my face.

N3buchadnezzar

That was a refference to the matrix.

Samuel Yusim

1:17 AM

apparently the product of quotient maps isn't always a quotient map? what's a counterexample?

Mike Miller

hell if I know

think I saw an MSE question about that once

Samuel Yusim

I was thinking about asking if it was the case here but I figured I'd look it up first but people kept using the phrase "locally compact" and I got scared

N3buchadnezzar

@SamuelYusim Whats so scary about that definition?

Samuel Yusim

I just don't know it yet

Mike Miller

locally compact should intuitively just mean "finite dimensional"

N3buchadnezzar

1:24 AM

I am desperately trying to figure out why $$ K_0 = \{ (x_1 , x_2) \mid x_1 \in \mathbb{R} , x_2 \in \mathbb{R} , -1 \leq x_1 \leq 1 , -1 \leq x_2 \leq 1\} $$ can not be approximated uniformly on compact subsets (eg Runge).

Fargle

@SamuelYusim Let $p : \Bbb Q \rightarrow \Bbb Q / \Bbb Z$ be the natural quotient map, let $id : \Bbb Q \rightarrow \Bbb Q$ be the identity. Then $p \times id$ is not a quotient map.

N3buchadnezzar

First I thought this represented a square, but now I am not so sure. It says that $(z_1 + i z_2 , z_1 - i z_2)$ maps it onto the polydisc $\Delta^2$.

Ted Shifrin

You want to be in $\Bbb C$ or in $\Bbb C^2$, @N3B?

Mike Miller

oh, I guess locally compacr shouldn't be intuited as finite dimensional, since $\Bbb Q$ isn't locally compact. gross

Ted Shifrin

No, it won't be the polydisk.

You're too used to manifolds, @Mike.

Mike Miller

1:28 AM

for manifolds and things that are reasonably like manifolds that's what it means

N3buchadnezzar

@TedShifrin $\mathbb{C}^2$

@TedShifrin Or the bidisk / clifford torus. I just really can not wrap my head around this simple paper.

Mike Miller

@TedShifrin: if I do Chern-Weil theory for $SU(2)$ I recover te first two Chern classes. What should I get if I worked with $SO(3)$?

or maybe, say, $Sp(n)$

Ted Shifrin

You're doing complex rank $2$-vector bundles in the first case, @Mike. An oriented rank $3$ vector bundle has only its Euler class. No Pontryagin.

$\text{Sp}(n)$ embeds as a subgroup of $SO(2n)$?

Fargle

@SamuelYusim (Where $\Bbb Q / \Bbb Z$ is the space given by identifying every integer to every other integer and leaving non-integers alone; i.e. $x \sim y \iff (x = y) \vee (\{x,y\} \subset Z)$)

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