Mathematics - 2019-07-21

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.

Leaky Nun

7:03 AM

@JennaSloan well integrate it first

$x = \sin \theta$

Jenna Sloan

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

Jenna Sloan

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.

Balarka Sen

@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?

Alessandro Codenotti

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

Balarka Sen

What is a *-morphism

Alessandro Codenotti

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

Balarka Sen

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

Alessandro Codenotti

9:03 AM

Because the normal ones are unitarily diagonalizable?

Balarka Sen

Diagonalizable guys are dense in K(H). Is there any continuity this character needs to satisfy?

@Alessandro That was my thinking, yeah

Alessandro Codenotti

They are continuous for free

Balarka Sen

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...

Alessandro Codenotti

@BalarkaSen wait why?

Balarka Sen

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

Am I talking nonsense

Alessandro Codenotti

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

Balarka Sen

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

Alessandro Codenotti

Me neither, I'll think about it. But that was helpful, thanks!

Balarka Sen

Glad to be of $\varepsilon$ help

Mike Miller

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

Balarka Sen

10:25 AM

@MikeMiller Huh I see

Tell me the Morava K-theory story sometime

Mike Miller

No

Balarka Sen

why not lol

Mike Miller

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

Leaky Nun

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

Mike Miller

1:16 PM

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

Leaky Nun

I mean the maps in the cellular homology

I'm wondering why is the cellular complex a complex to begin with

Mike Miller

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

Leaky Nun

sure

Leaky Nun

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

Mike Miller

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

Balarka Sen

2:20 PM

@LeakyNun It is an enlightening proof if you spend time understanding it

Leaky Nun

ok thanks

Balarka Sen

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

abenthy

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 $SL_n(\Lambda_2)$?

Balarka Sen

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

Sayan Chattopadhyay

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?

Balarka Sen

2:27 PM

Lol

Global Calculus is sheaf theory

Rather, differential geometry from the point of view of sheaf theory

Sayan Chattopadhyay

I know horse shit bout sheafs

Balarka Sen

It's self-contained if I remember correctly.

Sayan Chattopadhyay

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

Balarka Sen

Nice!

I saw him a few days ago

Sayan Chattopadhyay

How does he move around, he's 80 man

Balarka Sen

2:30 PM

He's in great shape.

Sayan Chattopadhyay

So his PhD student is taking the course, Kapil Hari Paranjape so lucky lol

@BalarkaSen thats cool

Balarka Sen

Oh I know Kapil Paranjape

Nice

Sayan Chattopadhyay

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

Balarka Sen

No, no, just know his name. He's famous.

I didn't know he was at IISER Mohali

Sayan Chattopadhyay

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

Alessandro Codenotti

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?

Leaky Nun

3:16 PM

@AlessandroCodenotti how is $\bot$ defined?

Alessandro Codenotti

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

Mike Miller

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 $\langle (1,0)\rangle$

Balarka Sen

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

Alessandro Codenotti

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

Ultradark

Hi chat

Alessandro Codenotti

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

Ultradark

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

Leaky Nun

8:20 PM

$\cdots \xrightarrow 2 \Bbb Z \xrightarrow 0 \Bbb Z \xrightarrow 2 \Bbb Z \xrightarrow 0 \Bbb Z$

$H_n(\Bbb RP^\infty;\Bbb Z) = \Bbb Z, \Bbb Z/2\Bbb Z, 0, \Bbb Z/2\Bbb Z, 0, \cdots$

$0 \to H_i(X; \mathbf{Z})\otimes A \, \overset{\mu}\to \, H_i(X;A) \to \operatorname{Tor}(H_{i-1}(X; \mathbf{Z}),A)\to 0$

$0 \to H_i(\Bbb RP^\infty; \Bbb Z)\otimes_\Bbb Z (\Bbb Z/2\Bbb Z) \to H_i(\Bbb RP^\infty; \Bbb Z/2\Bbb Z) \to \operatorname{Tor}(H_{i-1}(\Bbb RP^\infty; \Bbb Z), \Bbb Z/2\Bbb Z)\to 0$

Balarka Sen

Too hard my dude

Leaky Nun

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

Balarka Sen

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

Leaky Nun

oh is that how it works

oh yeah this is just tensoring the complex with $\Bbb Z/2\Bbb Z$

ah nice

Balarka Sen

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

Leaky Nun

8:24 PM

sniped

Balarka Sen

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

Leaky Nun

I see

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

Balarka Sen

Nice TeX shenanigans

Leaky Nun

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

Balarka Sen

What the fuck take projective resolution you absolute madman

Leaky Nun

8:31 PM

Tor is injective right

and this is the shortest injective resolution I think

Balarka Sen

You can take a projective resolution as well

That's how normal people define it

loch

lmao

Balarka Sen

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

Leaky Nun

oh no I always get it wrong

$0 \to \Bbb Z \xrightarrow 2 \Bbb Z \to \Bbb Z/2\Bbb Z \to 0$

$0 \to \operatorname{Tor}(\Bbb Z/2\Bbb Z, \Bbb Z/2\Bbb Z) \to \Bbb Z/2\Bbb Z \xrightarrow 0 \Bbb Z/2\Bbb Z \to \Bbb Z/2\Bbb Z \to 0$

Alessandro Codenotti

@BalarkaSen lol

Balarka Sen

8:34 PM

I never understood this obscene fascination with injective resolutions

What exactly is simpler with those? Injective modules are fucking complicated!

Leaky Nun

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 $\Bbb Z/2\Bbb Z$

Balarka Sen

Right.

Leaky Nun

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

Balarka Sen

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

Leaky Nun

and if I take another resolution?

Balarka Sen

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.

Leaky Nun

@loch

Balarka Sen

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

Leaky Nun

and tensor the homotopy with A...?

using the fact that tensor is additive

Balarka Sen

Tensor the equivalence with A, yeah.

loch

i don't want to check this

Balarka Sen

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

user193319

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 $\sup \{\sum_{k=1}^m f(\tau_k)(\gamma (t_k)-\gamma(t_{k-1})) \mid \text {etc.} \}$?

Leaky Nun

@user193319 Riemann-Stieltjes integral

user193319

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

Leaky Nun

Riemann–Stieltjes integral

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 $K(A,1) \to K(B,1) \to K(C,1)$

Balarka Sen

Yep

Leaky Nun

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

Balarka Sen

9:19 PM

Sure. These are homotopy fiber sequences, mind.

Leaky Nun

what's the second map...?

$\Bbb RP^1$?

here it says that the composition needs to be a constant...?

Balarka Sen

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

Mike Miller

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 $$S^\infty \times C \to (S^\infty \times C)/S^1 = \Bbb{RP}^\infty$$

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