2:20 AM
@user10478 The only rational values of cos(θ) for rational θ (in degrees) are 0, ±½, and ±1
Same for sin(θ)
If you want cos(θ) and sin(θ) to be rational at the same time, then, you'd need one to be 0 and the other to be ±1.
@user10478 $\theta/\pi$ you mean
I have a proof of this, if you want
or you can try to find one yourself

2 hours later…
4:36 AM
$0 \to H_i(X; \mathbf{Z})\otimes A \, \overset{\mu}\to \, H_i(X;A) \to \operatorname{Tor}(H_{i-1}(X; \mathbf{Z}),A)\to 0$
@BalarkaSen I don't get the diagram in P.12 of pi.math.cornell.edu/~hatcher/AT/ATch5.pdf
is it wrong?

What do you not get / why do you think it's wrong

5:42 AM
@MikeMiller why isn't the first row Z2, 0, Z2, 0, Z2, ...

the (0,0) term is H_0( ..., H_0(...)) right
why would it be Z2 lol

5:57 AM
oh I mean the 1-st row

Oh - then H_i(RP^\infty, Z2) = Z2 for all i! (think cellular chain complex)

I thought it's 0 for even i
is there a fibration $S^0 \to S^\infty \to \Bbb RP^\infty$?
$S^\infty$ is contractible right

yes / yes

so let's run SSS!
well this isn't looking great
$\begin{array}{c} H_0(\Bbb RP^\infty; H_0(S^0)) & H_1(\Bbb RP^\infty; H_0(S^0)) & \cdots \\ 0 & 0 & \cdots \\ \vdots \end{array}$
all differentials are zero
$H_0(S^0) = \Bbb Z^2$
so $H_n(\Bbb RP^\infty)^2 = \Bbb Z, 0, 0, \cdots$
????
@loch something must be wrong

i think when your base is not simply connected you have to be careful

6:11 AM
oh no
I see

but i never learnt what happens in that case so i'll let others rescue you
:p

Hatcher said that $\pi_1(B)$ needs to act trivially on $H(F)$
which is not true in this case
let's just use cellular homology
$\Bbb RP^\infty = \bigcup_{n \in \omega} (\Bbb D^n)^\circ$
so it's the homology of $\cdots \to \Bbb Z \to \Bbb Z \to \Bbb Z$

6:41 AM
@AkivaWeinberger

How in the world does one even begin to solve for $k$ when given $\int_k^1 \left(\sqrt{1-x^2}-\frac{1}{3}\right) dx=\int_{-k}^k \left(\sqrt{1-x^2}-\frac{1}{3}\right) dx$ ?
I mean I know the solution, I just don't remember how to do an integral with a variable in the boundary things.

7:03 AM
@JennaSloan well integrate it first
$x = \sin \theta$

I forgot why I'm doing this in the first place

7:49 AM
Oh yeah, it was because I wanted to cut a circular cake into 9 equally sized slices using 2 horizontal and 2 vertical cuts, with each cut being equidistant from the center of the cake.
So far I've determined that it's not possible.

@LeakyNun @loch If $\pi_1(B)$ doesn't act trivially on $H(F)$ you still have a spectral sequence but now with $E^2_{p, q} = H_p(B; \underline{H_q(F)})$ where $\underline{H(F)}$ is the local system on $B$ coming from the representation $\pi_1(B) \to \text{Aut}(H(F))$.
@LeakyNun Which is alternately degree $2$ and degree $0$. For example, the 2-cell of $\Bbb{RP}^2$ is attached to 1-skeleton/0-skeleton = $S^1$ by a degree 2 map $S^1 \to S^1$. The 3-cell of $\Bbb{RP}^3$ is attached to 2-skeleton/1-skeleton = $S^2$ by a degree $0$ map $S^2 \to S^2$.
It's tricky to see that it has degree 0. The 3-cell is attached to the 2-skeleton by the double cover $S^2 \to \Bbb{RP}^2$. So that might lead you to think $S^2 \to \Bbb{RP}^2 \to \Bbb{RP}^2/\Bbb{RP}^1 = S^2$ is degree 2, but $\Bbb{RP}^2$ is nonorientable so degree doesn't work that way. Think instead of the map $S^2 \setminus S^1 \to \Bbb{RP}^2 \setminus \Bbb{RP}^1$ by throwing away the equator. This is a map $D^2 \sqcup D^2 \to D^2$ which is a homeomorphism on each factor.
The first factor is degree $1$, the second is degree $-1$! One is obtained from another by the antipodal map.
So $1 + (-1) = 0$ is the degree.
This in particular implies $\Bbb{RP}^3/\Bbb{RP}^1 \cong S^3 \vee S^2$ if I'm not mistaken.
In general $\Bbb{RP}^n/\Bbb{RP}^m$ are called stunted projective spaces.
@MikeMiller Are the homotopy types of these spaces completely understood?

8:48 AM
Let $H$ be a separable Hilbert space and let $K(H)$ be the algebra of compact operators on $H$. Is there an easy way to see that $K(H)$ has no nontrivial characters (*-morphisms into $\Bbb C$)? I can show this but taking a huge detour

What is a *-morphism

A morphism of algebras $\varphi:A\to B$ with $\varphi(a^\ast)=\varphi(a)^\ast$, where $A$ and $B$ are *-algebras
The involution (the *) in K(H) is taking adjoints and the involution in C is the conjugate, in particular
This condition is surprisingly (to me) strong, a *-morphism is automatically contractive

Ah I see.
So if K(H) -> C is a *-morphism, it's determined upto multiplication by reals by the image of the diagonal operators, yes?
Eh, no, what am I saying. On the subset of normal operators.
That doesn't help

9:03 AM
Because the normal ones are unitarily diagonalizable?

Diagonalizable guys are dense in K(H). Is there any continuity this character needs to satisfy?
@Alessandro That was my thinking, yeah

Because it's an algebra homomorphism, so in particular linear, I suppose
So it is determined on the subset of diagonalizable operators in K(H), which means it's determined upto real scalars by the image of diagonal operators...

@BalarkaSen wait why?

$\varphi(TDT^*) = \varphi(T)\varphi(D)\overline{\varphi(T)}$ and $\varphi(T)\overline{\varphi(T)}$ is real, right?
Am I talking nonsense

9:09 AM
Actually I see that a character of a unital C*-algebra is continuous, I'm not sure about the general case right now
@BalarkaSen I agree, but why is this enough to determine the whole morphism?
Oh, density, right

It seems like a natural question is where the diagonal guys get sent to, but I am not sure I can answer this

Glad to be of $\varepsilon$ help

10:06 AM
@BalarkaSen I doubt it. A postdoc explained to me once how you can cook up the first Morava K-theory with these guys
Sounds like a complicated homotopy type to me

10:25 AM
@MikeMiller Huh I see
Tell me the Morava K-theory story sometime

why not lol

It's too hard and I don't know it

2 hours later…
1:04 PM
@BalarkaSen can't we have a CW-complex with one cell for each dimension with each attaching map having degree 2?
I was thinking, this would break the cellular complex
but I don't see why we can't have such a CW-complex

1:16 PM
@LeakyNun What do you mean "having degree 2"? Degree is only a notion I know for maps between closed oriented manifolds

I mean the maps in the cellular homology
I'm wondering why is the cellular complex a complex to begin with

Do you mean that the corresponding cellular chain complex has $C_k = \Bbb Z$ and the boundary map $C_k \to C_{k-1}$ is always multiplication by 2?
Oh
I mean read a proof of it

sure

1:49 PM
$\cdots \to H_n(X^{n-1}) \to H_n(X^n) \to H_n(X^n,X^{n-1}) \color{red}\to H_{n-1}(X^{n-1}) \to \cdots$
$\cdots \to H_{n-1}(X^{n-2}) \to H_{n-1}(X^{n-1}) \color{red}\to H_{n-1}(X^{n-1},X^{n-2}) \to H_{n-2}(X^{n-2}) \to \cdots$
@MikeMiller the map in the cellular chain complex is the composition of these two maps right
if so then this is a proof that it is a complex
although not a very enlightening one

you would have to explain more carefully why that's a proof, normally I expect to see a more complicated diagram showing how they fit together
but yeah pretty much that, the proof should be about exactly as enlightening as the definition
if this proof is unenlightening then that probably means you're unsatisfied with the definition of the cellular chain complex using differential given by the composite of those two maps
another argument is that $C_*(X)$ is a filtered chain complex, with $F_k C_*(X) = C_*(X^k)$. Filtrations give spectral sequences, and the cellular chain complex is the $E^2$ page of this spectral sequence, and the $E^2$ page has its own differential; because the $E^2$ page here is concentrated in a single line, there are no further differentials

2:20 PM
@LeakyNun It is an enlightening proof if you spend time understanding it

ok thanks

The cellular boundary maps are closely related to the snake maps in the homology long exact sequence

If I know that $\Lambda_1, \Lambda_2 \subset A$ are two orders in a $\mathbb{Q}$-algebra $A$ with $\Lambda_1 \subset \Lambda_2$, does it follow that $\Lambda_1^\times$ has finite index in $\Lambda_2^\times$?
Also does the same hold for, say, the subgroup $SL_n(\Lambda_1)$ of $SL_n(\Lambda_2)$?

To be precise the snake map $\partial : H_n(X, A) \to H_{n-1}(A)$ comes from taking a relative class $[\xi]$ and sending it to $[\partial \xi]$ since by definition $\xi \in Z_n(X, A)$ means $\partial \xi \in Z_{n-1}(A)$
In light of this, interpret the cellular boundary maps and why they are "obviously" the right maps

So I have an upcoming course on Global Analysis, in the winter semester. The book that they are going to follow is S Ramanan's Global Calculus. Considering I have functional analysis now and that I am attending this school on Operator theory and NC Geometry, what else is necessary for Global Analysis?

2:27 PM
Lol
Global Calculus is sheaf theory
Rather, differential geometry from the point of view of sheaf theory

I know horse shit bout sheafs

It's self-contained if I remember correctly.

That's better. There are talks that Ramanan might come for guest lectures which will be awesome

Nice!
I saw him a few days ago

How does he move around, he's 80 man

2:30 PM
He's in great shape.

So his PhD student is taking the course, Kapil Hari Paranjape so lucky lol
@BalarkaSen thats cool

Oh I know Kapil Paranjape
Nice

@BalarkaSen He's pretty cool. You know him through conferences or did you meet him?

No, no, just know his name. He's famous.
I didn't know he was at IISER Mohali

Yea he's like fulltime here. I didn't know he was famous lol, I just knew him as the guy who knows Ramanan
I see now why my recco's work so nicely.
Damn
Do I even deserve his recco now

2:47 PM
I'm looking for an example of a bounded operator on an Hilbert space that has a closed invariant subspace $\mathcal N$ such that $\mathcal N^\perp$ is not invariant, any idea?

3:16 PM
@AlessandroCodenotti how is $\bot$ defined?

$\mathcal N^\perp$ is $\{x\in H\mid \langle x,n\rangle=0\text{ for all }n\in\mathcal N\}$

3:35 PM
@AlessandroCodenotti let your Hilbert space be $\Bbb R^2$ and your operator be $\begin{pmatrix}1 & 1\\0 & 1\end{pmatrix}$ and your invariant subspace be $\langle (1,0)\rangle$

In general if the operator doesn't preserve the inner product there's no reason to believe that it's going to preserve complements.

Oh, neat, I was thinking about $\ell^2$ and way more complicated examples

Hi chat

@BalarkaSen right, you want a self-adjoint operator. Or more generally a set of bounded operators closed under taking adjoints

@MikeMiller I'm taking Geometry and Algebra at the graduate level this fall semester

5 hours later…
8:20 PM
$\cdots \xrightarrow 2 \Bbb Z \xrightarrow 0 \Bbb Z \xrightarrow 2 \Bbb Z \xrightarrow 0 \Bbb Z$
$H_n(\Bbb RP^\infty;\Bbb Z) = \Bbb Z, \Bbb Z/2\Bbb Z, 0, \Bbb Z/2\Bbb Z, 0, \cdots$
$0 \to H_i(X; \mathbf{Z})\otimes A \, \overset{\mu}\to \, H_i(X;A) \to \operatorname{Tor}(H_{i-1}(X; \mathbf{Z}),A)\to 0$
$0 \to H_i(\Bbb RP^\infty; \Bbb Z)\otimes_\Bbb Z (\Bbb Z/2\Bbb Z) \to H_i(\Bbb RP^\infty; \Bbb Z/2\Bbb Z) \to \operatorname{Tor}(H_{i-1}(\Bbb RP^\infty; \Bbb Z), \Bbb Z/2\Bbb Z)\to 0$

Too hard my dude

how do you compute $H_n(\Bbb RP^\infty, \Bbb Z/2\Bbb Z)$ then?

Just compute the cellular homology with $\Bbb Z/2$ coefficients. You get $\cdots \to \Bbb Z/2 \to \Bbb Z/2 \to \Bbb Z/2$ with all zero homomorphisms between them

oh is that how it works
oh yeah this is just tensoring the complex with $\Bbb Z/2\Bbb Z$
ah nice

It's literally tensoring your cellular chain complex with $\Bbb Z/2$

8:24 PM
sniped

Yes, exactly
The universal coefficient theorem tells you how tensoring a chain complex with a module changes the cohomology in general
That's the homological algebra content, at least

I see
$0 \to \begin{array}{c} \Bbb Z/2\Bbb Z \\ \Bbb Z/2\Bbb Z \\ 0 \\ \Bbb Z/2\Bbb Z \\ 0 \\ \vdots \end{array} \to H_i(\Bbb RP^\infty; \Bbb Z/2\Bbb Z) \to \operatorname{Tor}(\begin{array}{c} 0 \\ \Bbb Z/2\Bbb Z \\ \Bbb Z/2\Bbb Z \\ 0 \\ \Bbb Z/2\Bbb Z \\ \vdots \end{array}, \Bbb Z/2\Bbb Z)\to 0$

Nice TeX shenanigans

$0 \to \Bbb Z/2\Bbb Z \xrightarrow{\frac12} \Bbb Q/\Bbb Z \xrightarrow 2 \Bbb Q/\Bbb Z \to 0$

What the fuck take projective resolution you absolute madman

8:31 PM
Tor is injective right
and this is the shortest injective resolution I think

You can take a projective resolution as well
That's how normal people define it

lmao

$0 \to \Bbb Z \to \Bbb Z \to \Bbb Z_2 \to 0$

oh no I always get it wrong
$0 \to \Bbb Z \xrightarrow 2 \Bbb Z \to \Bbb Z/2\Bbb Z \to 0$
$0 \to \operatorname{Tor}(\Bbb Z/2\Bbb Z, \Bbb Z/2\Bbb Z) \to \Bbb Z/2\Bbb Z \xrightarrow 0 \Bbb Z/2\Bbb Z \to \Bbb Z/2\Bbb Z \to 0$

@BalarkaSen lol

8:34 PM
I never understood this obscene fascination with injective resolutions
What exactly is simpler with those? Injective modules are fucking complicated!

so $\operatorname{Tor}(\Bbb Z/2\Bbb Z, \Bbb Z/2\Bbb Z) = \Bbb Z/2\Bbb Z$
oh and the big diagram above is wrong, it should be:
$0 \to \begin{array}{c} \Bbb Z/2\Bbb Z \\ \Bbb Z/2\Bbb Z \\ 0 \\ \Bbb Z/2\Bbb Z \\ 0 \\ \vdots \end{array} \to H_i(\Bbb RP^\infty; \Bbb Z/2\Bbb Z) \to \operatorname{Tor}(\begin{array}{c} 0 \\ \Bbb Z \\ \Bbb Z/2\Bbb Z \\ 0 \\ \Bbb Z/2\Bbb Z \\ \vdots \end{array}, \Bbb Z/2\Bbb Z)\to 0$
which becomes:
$0 \to \begin{array}{c} \Bbb Z/2\Bbb Z \\ \Bbb Z/2\Bbb Z \\ 0 \\ \Bbb Z/2\Bbb Z \\ 0 \\ \vdots \end{array} \to H_i(\Bbb RP^\infty; \Bbb Z/2\Bbb Z) \to \begin{array}{c} 0 \\ 0 \\ \Bbb Z/2\Bbb Z \\ 0 \\ \Bbb Z/2\Bbb Z \\ \vdots \end{array}\to 0$
hey they complement each other lol
therefore the middle term is always $\Bbb Z/2\Bbb Z$

Right.

if A is an R-algebra and M an R-module, is Tor_n(A,M) always an A-module?
I think it is, by unfolding the definition via projective resolution
but that doesn't give me well-definedness

You're tensoring an R-module resolution by A, so there's a A-module structure on the resulting chain complex

and if I take another resolution?

8:39 PM
Namely, extension of scalars
@LeakyNun Good point. You'd have to check that the chain-homotopy gives an A-module isomorphism
It should, I think, but I haven't checked.

@loch

Just check it man. If it's well defined the proof should make it apparent
The chain homotopy equivalence between the two projective resolution is given by iteratively extending the morphisms by projectivity

and tensor the homotopy with A...?
using the fact that tensor is additive

Tensor the equivalence with A, yeah.

i don't want to check this

8:44 PM
lol
@LeakyNun Learn what goes wrong when you do universal coefficient with your coefficient ring not a PID and then teach it to me. The point should be that if $M$ is an $R$-module and $R$ is a PID then there's a 3-term resolution $0 \to F_1 \to F_0 \to M \to 0$ so the whole abelian group proof pushes through. There's a universal coefficient spectral sequence which says there's a spectral sequence with $E^2_{p, q} = \text{Tor}^q_R(H_p(X;R),M)$ that converges to $H_*(X;M)$ or some such.
Namely, the higher Tor terms won't vanish
I guess it makes sense. If you have a $R$-module chain complex $C_*$ and you take a projective resolution of $C*$, tensor with $M$, that's a double complex. You filter the totalization in the obvious fashion and take it's cohomology. The $E^1$ page takes cohomology in vertical direction giving rise to resolutions for $H_p(X; R)$ tensored with $M$. Then the $E^2$ page takes cohomology in the horizontal direction giving rise to $\text{Tor}^q_R(H_p(X; R), M)$.
Simple enough.
I don't suppose this spectral sequence is computable without effort in general. Who knows what the higher differentials do

9:07 PM
If $\gamma : [a.b] \to \Bbb{C}$ is a function of bounded variation and $f : [a,b] \to \Bbb{C}$ is continuous, how does one define $\int_{a}^{b} f(t) d \gamma(t)$? Is it defined as $\sup \{\sum_{k=1}^m f(\tau_k)(\gamma (t_k)-\gamma(t_{k-1})) \mid \text {etc.} \}$?

@user193319 Riemann-Stieltjes integral

Yeah, I guess so. Is that how the Riemann-Stieltjes integral is defined?

In mathematics, the Riemann–Stieltjes integral is a generalization of the Riemann integral, named after Bernhard Riemann and Thomas Joannes Stieltjes. The definition of this integral was first published in 1894 by Stieltjes. It serves as an instructive and useful precursor of the Lebesgue integral, and an invaluable tool in unifying equivalent forms of statistical theorems that apply to discrete and continuous probability. == Formal definition == The Riemann–Stieltjes integral of a real-valued function f {\displaystyle f} of a real variable with respect...
@BalarkaSen Hatcher says that if $0 \to A \to B \to C \to 0$ is exact then there is a corresponding fibration $K(A,1) \to K(B,1) \to K(C,1)$

Yep

so there should be a fibration $S^1 \to S^1 \to \Bbb RP^\infty$?

9:19 PM
Sure. These are homotopy fiber sequences, mind.

what's the second map...?
$\Bbb RP^1$?
here it says that the composition needs to be a constant...?

Take any contractible space $X$ on which $\Bbb Z$ acts freely properly discontinuouslyand any contractible space $Y$ on which $\Bbb Z_2$ acts freely properly discontinuously. Consider $(X \times Y)/\Bbb Z$ where $\Bbb Z$ acts on the first factor usually and on the second factor by factoring through $\Bbb Z \to \Bbb Z_2$.
There's a map $(X \times Y)/\Bbb Z \to Y/\Bbb Z = Y/\Bbb Z_2$ by forgetting about the first factor.
$(X \times Y)/\Bbb Z$ is weak homotopy equivalent to $S^1$ and $Y/\Bbb Z_2$ is weak homotopy equivalent to $\Bbb{RP}^\infty$ :)
"Weak" can be removed if the spaces in question are CW complexes and the action is cellular etc
So for an explicit example, consider $(\Bbb R \times S^\infty)/\Bbb Z \to S^\infty/\Bbb Z_2$.
This is your homotopy fibration map $S^1 \to \Bbb{RP}^\infty$
@LeakyNun $S^1 \to \Bbb{RP}^1 \subset \Bbb{RP}^\infty$ is not a fibration, but it can be made into one. See this.
Meh nevermind the last comment. It is a double cover, so whatever.
Anyway the point is you shouldn't think of $K(A, 1)$ simply by it's homotopy type. They are by construction massive spaces. They just shrink a lot sometimes.
OK, no, my comment was correct, actually. If you take $S^1 \to \Bbb{RP}^\infty$ by just naively sending $S^1$ to the 1-skeleton $\Bbb{RP}^1$, then make it a homotopy fiber sequence using the wikipedia page I linked, it's homotopy fiber is going to be $S^1$.
Because suppose it is $F$. Then you have a homotopy fiber sequence $F \to S^1 \to \Bbb{RP}^\infty$, where the middle $S^1$ is not really $S^1$ but some homotopy-enlarged space.
The map $S^1 \to \Bbb{RP}^\infty$ induces $\Bbb Z \to \Bbb Z_2$ in $\pi_1$. So by running the homotopy fibration sequence, you get $1 \to \pi_1 F \to \Bbb Z \to \Bbb Z_2 \to 0$ and $\pi_n F \cong 0$ for all $n \geq 2$.
That means $\pi_1 F = \Bbb Z$ and $F$ is a $K(\Bbb Z, 1)$ space. A circle.
Alright, I gotta go now. It's 3 AM

2 hours later…
11:50 PM
write $S^\infty$ for the unit sphere in $\Bbb C^\infty$, and let $S^1$ act on itself with weight two; it is convenient to rename this second circle, maybe call it $C$. then the projection of your fiber sequence can be modeled as $$S^\infty \times C \to (S^\infty \times C)/S^1 = \Bbb{RP}^\infty$$