Mathematics - 2023-12-18

Jakobian

00:06

@DannyuNDos What's that?

Dannyu NDos

The class of all decision problems. If the set of alphabet is fixed, that's a set with $\beth_1$ members.

robjohn

01:46

@TedShifrin Oh, bother...

Ted Shifrin

02:02

@robjohn oh, help and bother!

Koro

05:21

Let $g:A\to B$ be a diffeomorphism of open sets in $R^k$. Let $\beta: B\to R^n$ be a smooth map. Define a parametrized manifold $Y_\beta= \beta(B)$. $Y_\beta=Y_\alpha=Y$(say). Let $f:Y\to R$ be continuous. Suppose $\alpha=\beta \circ g$. Then it is to be shown that the integrals $\int_{Y_\beta} f dV= \int_{Y_\alpha} f dV$.

$\int_{Y_\beta} f dv:= \int_B f\circ \beta V(D\beta)$

Then change of variables gives: $\int_A f\circ \beta \circ g V(D(\beta \circ g))$, right?

But in Munkres, this last bit is written as $\int_A f\circ \beta \circ g V(D(\beta) \circ g))|det(Dg)|$

shouldn't it have been $V(D(\beta \circ g))|det(Dg)|$?

V here means k -volume function (for appropriate k above), i.e., if $(x_1,...,x_k)$ is a k tuple of vectors in R^n, then $V(x_1,...,x_k)= (det(X^{t} X))^{1/2}$, where X=$[x_1,..., x_k]$ is an n by k matrix.

oh nvm.

it seems they are considering the function $V(D\beta)(y)$ and change of variables is $y=g(x)$ etc.

Koro

07:56

can I have a 3 tensor on R^2?

Unknown x

08:19

0

Q: $\frac{\partial}{\partial t}\frac{\partial }{\partial s}=\frac{\partial}{\partial t}\frac{\partial s}{\partial t}+\kappa^2\frac{\partial}{\partial s}$

Prove that $$\frac{\partial}{\partial t}\frac{\partial }{\partial s}=\frac{\partial}{\partial t}\frac{\partial s}{\partial t}+\kappa^2\frac{\partial}{\partial s}.$$ My attempt:- $$\frac{\partial}{\partial t}\frac{\partial }{\partial s}=\frac{\partial}{\partial t}\Big(\frac{1}{v}\frac{\partial }{\...

PNDas

10:35

I am watching youtube.com/watch?v=ed3_j2Ss9b8. Can you give any references for the talk more specifically for what is happening after 4:56?

I don't understand his accent.

He is defining derivatives abstractly and asked is there any derivative of order 1 which is isotropic?

I don't know what is isotropic. He said something but I couldn't undertsand.

Thomas Finley

11:11

If $A,B$ are ideals of a commutative ring $R$ with unity then, why is $AB+BA\subseteq AB$?

Jakobian

11:23

@ThomasFinley Because $AB = BA$

Adding ideal to itself gives you that ideal back

$IJ$ is defined as the set of sums $\sum x_iy_i$ for $x_i\in I, y_i\in J$

If your ring $R$ is commutative and $z\in AB$, so that $z = \sum x_iy_i$ for some $x_i\in A, y_i\in B$, then clearly $\sum x_iy_i = \sum y_ix_i$ so that from definition, $z\in BA$

Thomas Finley

@Jakobian yeah, I get it now. Thanks!

Jakobian

Thus $AB\subseteq BA$ and similar argument shows $BA\subseteq AB$ so that $AB = BA$

and if $I$ is any ideal, then $z\in I+I$ if $z = x+y$ for some $x, y\in I$. But since ideals are closed under addition, $z\in I$.

Here, $I = AB$, is an ideal as a product of ideals

Thomas Finley

@Jakobian Thanks again! That looks rigorous.

If $P$ is a prime ideal of a commutative
ring $R$ all of whose elements satisfy $x^n = x$, then $R/P$ is an integral domain with the same property. This implies that $R/P$ is a finite integral domain.

Why is $R/P$ finite?

Koro

11:39

"$k$ forms are a generalization of scalars and vectors in R^3 to $R^n$". What does this mean?

why is it a generalisation?

Thomas Finley

@Koro Can you please help me with this: chat.stackexchange.com/transcript/message/64875202#64875202 ?

Jakobian

11:56

@ThomasFinley I'm assuming $n > 1$, otherwise this makes no sense. If $A$ is an integral domain with $x^n = x$ for all $x\in A$, then $x = 0$ or $x^{n-1} = 1$ for all $x\in A$. In particular, each non-zero element $x$ is invertible, so $A$ is a field. Since the polynomial $p(t) = t^n-t$ has at most $n$ different roots, $|A|\leq n$

Or you can go directly from the same theorem but about integral domains if you know that

Theorem. If $p(t)\in A[t]$ is a polynomial of degree $n$ over an integral domain $A$, then $p$ has at most $n$ roots.

$A$ is an integral domain, $p(t) = t^n-t$ is a polynomial of degree $n$, and all elements of $A$ are its roots, so $|A|\leq n$

Soumik Mukherjee

Anyone interested to do a study group over this playlist?

Complex Geometry by Pierre Albin

Pedro

@robjohn Wonder if you'd like to take a stab a this one: let $\{S_n\}$ be a symmetric random walk with $S_0=0$, and let $\tau_a = \min\{ n : S_n = a \}$.

Now consider $\tau = \min(\tau_a,\tau_{-b})$ for $-b< 0 < a$.

I know that $\mathbb E(\tau) = ab$, but I couldn't compute its variance.

Thomas Finley

13:06

@Jakobian is $A=R/P$ in my case?

Jakobian

yes, you're applying what this argument was for $A = R/P$

treat it like applying a theorem

Thomas Finley

@Jakobian There seems to be a problem with me: If $(x+P)\in R/P$ then, $$(x+P)^n=x+P\implies (x+P)^n-(x+P)=P\implies (x+P)((x+P)^{n-1}-(1+P))=P\implies x+P=P $$ or $$(x+P)^{n-1}=1+P\implies x^{n-1}+P=1+P\implies x^{n-1}-1\in P$$.

If $x\notin P$ then, $$x^{n-1}+P=1+P$$

But how is $(x+P)$ invertible?

John Zimmerman

13:24

Consider the $1$-parameter family of functional equations on the critical strip $$\zeta(1-s)\Gamma[\hat\Omega_r,1-s ]=\zeta(s)\Gamma[\Omega_r,s]$$

where the Gamma factor $$ \Gamma[\Omega_r,s]= 2\int_0^1e^{\frac{r}{\log(|x|)}}x^{s-1}dx=4\sqrt{\frac{r}{s}}\, K_1\left(2\, \sqrt{r s}\right)$$

satisfies the degenerate linear PDE

$$ r^2\frac{\partial^3}{\partial r^3}\Gamma[\Omega_r,s]=s^2\frac{\partial}{\partial s}\Gamma[\Omega_r,s]. $$

We can reduce this to second order through the Fourier transform

The main idea here is that the Gamma factor satisfies a somewhat nice PDE

which is absent in the case for the classical gamma factor of $\pi^{-s/2}\Gamma(s/2)$

I forgot the hats on the $\Gamma[]$ in complex heat equation

$\hat \Omega$ just notates the F.T. of $\Omega$

John Zimmerman

13:45

In 1976, Newman showed that the Riemann hypothesis failed
for sufficiently large negative t. Combining this with de Bruijn’s
results, we conclude that there exists a real number
−∞ < Λ ≤
1
2
, now called the de Bruijn-Newman constant, such
that the Riemann hypothesis at time t is true if and only if t ≥ t0.
The classical Riemann hypothesis is then equivalent to the
assertion Λ ≤ 0. Newman then made the opposite conjecture
Λ ≥ 0; intuitively, if the (classical) Riemann hypothesis is true,
then it is only “barely so”: any deformation of the Riemann xi

Jakobian

14:52

@ThomasFinley There is no reason to obscure notation in this way. Sorry but I won't read this

Thomas Finley

@Jakobian ok, I understand. But the takeaway is: "A polynomial of degree n over an integral domain has atmost n roots" is a standard theorem, correct? Anyways, you helped me a great deal, though.

Jakobian

15:23

@ThomasFinley Its a standard theorem. Even more known is the one for fields. However, you can prove the one for the integral domains from the one for fields, for example.

Jayaraman adithya

can someone please answer my question math.stackexchange.com/q/4825370/942050

Jakobian

@Jayaramanadithya $\mu_B(x)\leq \inf\{s : p(x) < s\} = p(x)$

or say, take $s_n = p(x)+1/n$ then $p(x) < s_n$ so $\mu_B(x) \leq s_n$, and now since $s_n\to p(x)$ you obtain $\mu_B(x)\leq p(x)$

Jayaraman adithya

15:42

@Jakobian Thank you very much

Thomas Finley

16:19

@Jakobian thanks! I get it.

Semiclassical

16:35

@Obliv in quantum physics it's more often the spatial FT, since that's what defines the relation between position space and momentum space

in circuits you don't have space so you just see transforms w/r/t to time, but often that's the Laplace transform b/c you want to handle signals that start at some time 0

in electrodynamics you see both

the only annoying point is conventions. in physics, if you do both space and time FTs you get $$\begin{equation}
f(\mathbf{x},t) \propto\int_\mathbb{R}\int_{\mathbb{R}^3} e^{i(\mathbf k \cdot \mathbf x - \omega t)} \widehat f(\mathbf k,\omega)\ \mathrm d^3k\,\mathrm d\omega
\end{equation}$$

(technically this is the inverse FT---I'm grabbing from another source b/c i'm lazy)

main thing to notice is that the space variables transform with one sign in the exponent while the time variable transforms with the opposite sign

which makes sense on physics grounds, b/c said exponent (the phase) has level curves of the form $\mathbf{k}\cdot \mathbf{x}-\omega t=$ const

eh, that explanation isn't as helpful as i want it to be. mostly just focus on the different conventions

you also see physicists often getting lazy and writing $f(\mathbf{k},\omega)$ as tho $f$ is the same function

usually i don't mind that kind of lazy physics convention, but for FTs i really do prefer to keep them separate

honestly the spatial FT is easier for me to make sense of in most contexts, if only b/c the temporal FT brings up the question of what a negative frequency means

Thomas Finley

17:13

0

Q: Let $R$ be a ring and $P$ be an ideal containing an ideal $I$ of $R.$ Show that $P$ is a maximal ideal of $R$ iff $P/I$ is a maximal ideal of $R/I.$

Let $R$ be a ring and $P$ be an ideal containing an ideal $I$ of $R.$ Show that $P$ is a maximal ideal of $R$ iff $P/I$ is a maximal ideal of $R/I.$ My attempt so far is as follows: Let $P/I$ be a maximal ideal of $R/I$ then, let $U$ be a ideal of $R$ such that $P\subseteq U\subseteq R.$ This mea...

abstract-algebra ring-theory maximal-and-prime-ideals

Need some help with this, please.

Jakobian

17:33

@ThomasFinley The map $J\mapsto J/I$ from the ideals of $R$ containing $I$ to ideals of $R/I$ is an order-preserving bijection

this proves the claim

Thomas Finley

@Jakobian Do you mean this: Since, $\phi:R\to R/I$ such that \phi(x)=x+I$ is an onto homomorphism so, there is a 1-1 correspondance between the set of ideals of $R$ containing $I$ to the set of ideals of $R/I.$ ---I know this fact. But, I don't get how you use this to prove the claim?

Jakobian

@ThomasFinley I don't know what you wrote, your LaTeX is broke

em, broken I should say

Thomas Finley

@Jakobian Sorry, I am fixing it.

Do you mean this: Since, $\phi:R\to R/I$ such that $\phi(x)=x+I$ is an onto homomorphism so, there is a 1-1 correspondance between the set of ideals of $R$ containing $I$ to the set of ideals of $R/I.$ Furthermore this correspondance can be achieved by associating each ideal $W'$ of $R/I$ with an ideal $W=\{x\in R:\phi(x)\in W'\}$ of $R.$---I know this fact. But, I don't get how you use this to prove the claim?

Jakobian

17:50

@ThomasFinley sounds about right

Thomas Finley

Since, $P$ is a maximal ideal of $R$ containing $I$ so, it gets associated to the ideal $W'=\{\phi(x):x\in P\}=\{x+I:x\in P\}=P/I.$

Jakobian

The correspondence is not only bijective, its also order-preserving

this is important, its order-preserving

that means, that if you take an ideals $J_1, J_2$ containing $I$ such that $J_1\subseteq J_2$, then $J_1/I \subseteq J_2/I$

The order on the set of all ideals, is the order by inclusion

For an ideal to be maximal, it means to be maximal among all proper ideals with respect to this order

order-preserving maps preserve maximality of elements

Thomas Finley

@Jakobian Ok, so generally speaking what's the notion of "order preserving"?

@Jakobian Things are starting to make sense slightly

Jakobian

If $S_1, S_2$ are two posets, then $f:S_1\to S_2$ is called order preserving if $x\leq y$, $x, y\in S_1$ implies that $f(x)\leq f(y)$

so those are basically, non-decreasing maps

They preserve the relation "$\leq$" that is being talked about

this is similar to the concept of, say, homomorphism, but for posets

those are basically homomorphisms for posets

Thomas Finley

Oh, the thing is I haven't studied anything about posets till now. Is there any other way to prove the original claim in the OP?

If there's not, then I think I must move on

and maybe say, proceed with the other problems

Jakobian

17:58

I'm giving you the definition of order-preserving etc. just because you've asked me to, this is not that important for the problem

The important thing is that the correspondence between ideals this and that is order-preserving, in the sense I've mentioned above

even without all the talk about posets

you can show this yourself, without the need to know these definitions

Thomas Finley

@Jakobian My mistake, then what is important to 'know' to solve the problem is that- besides there being a 1-1 correspondance between the set of ideals of $R$ containing $I$ to the set of ideals of $R/I,$ we also have that if $J_1$ is an ideal in $R$ and $J_2$ is an ideal in $R/I$ such that $J_1$ corresponds to $J_2$ then, if $J_1$ is maximal in $R$ we have $J_2$ maximal in $R/I$ as well. --I think I got you correctly.

Jakobian

Yes

And this can be shown from the fact that this correspondence is order-preserving, that is if two ideals $A, B$ are included in each other, say $A\subseteq B$, then $A/I\subseteq B/I$

where $A/I = \{x+I : x\in A\}$ (this is very common notation)

Semiclassical

exam rambling incoming

Thomas Finley

I think the general case is: If $\phi:R\to R'$ with kernel $I$ is an onto homomorphism then there is a 1-1 correspondence between the set of ideals of $R$ containing $I$ to the set of ideals of $R'.$ Furthermore this correspondance can be achieved by associating each ideal $W'$ of $R/I$ with an ideal $W=\{x\in R:\phi(x)\in W'\}$ of $R.$ Now, if $J_1$ is an ideal in $R$ and $J_2$ is an ideal in $R'$ such that $J_1$ corresponds to $J_2$ then, if $J_1$ is

maximal in $R$ we have $J_2$ maximal in $R'$ as well.

Am I correct?

Jakobian

@ThomasFinley If $\phi:R\to R'$ is surjective homomorphism, then $R'\cong R/I$ for $I = \text{ker}(\phi)$, so this is not actually much more general

(honestly, not at all more general)

Semiclassical

18:08

problem asked, in effect, to solve for x given $y=1/(1-x)-1/(1+x)$; they could assume $x$ was small to get simpler algebra

Jakobian

But yes, if $J$ is maximal in $R$, and $\phi$ is a surjective homomorphism, then $\phi(J)$ is also maximal

the way you write the correspondence between the ideals is slightly confusing, and not that important

Thomas Finley

@Jakobian How can we prove this then?

Jakobian

prove what

Thomas Finley

Never have I came accross this theorem. I mean the order preserving part... But more or less it seems intuitive enough

Semiclassical

most students in effect misinterpreted the problem as $y=1/(1-x)-1\approx x$

and a small handful tried to solve as $y/2\approx 1/(1-x)-1\approx x$

Jakobian

18:12

@ThomasFinley What you're writing here is inverse of the correspondence, but the easier way to write this is using images $\phi(I)$ for an ideal $I$ contained in the kernel of $\phi$

Semiclassical

funny thing is: in the small $x$ limit, the original equation becomes $y\approx 1+x-(1-x)=2x$

so even tho that last approach is strictly-speaking wrong, it's correct in the small $x$ limit

Thomas Finley

@Jakobian yeah, and I just realized that. But for now, should I take the result "Now, if $J_1$ is an ideal in $R$ and $J_2$ is an ideal in $R'$..." for granted and a mere consequence of intuition and consider the problem done or will you advocate for proving it rigorously?

I may sound strange, but I am in need of sone suggestions regarding this.

(I realized my previous generalisation had an error. The correct thing is:

If $\phi:R\to R'$ with kernel $I$ is an onto homomorphism then there is a 1-1 correspondence between the set of ideals of $R$ containing $I$ to the set of ideals of $R'.$ Furthermore this correspondance can be achieved by associating each ideal $W'$ of $R'$ with an ideal $W=\{x\in R:\phi(x)\in W'\}$ of $R.$ Now, if $J_1$ is an ideal in $R$ and $J_2$ is an ideal in $R'$ such that $J_1$ corresponds to $J_2$ then, if $J_1$ is maximal in $R$ we have $J_2$ maximal in $R'$ as well.)

Semiclassical

another funny / silly aspect of this. simplifying the correct equation further yields $x=y(1-x^2)$. if one substitutes $x=yz$ and $t=-y^2$, then this becomes $z=1+tz^2$. which is mostly notable in that solving for $z$ and expanding in powers of $t$ gives $$z=1+t+2t^2+5t^3+14t^4+\cdots$$

which is exactly the ordinary generating function for the Catalan numbers!

blah, should be $x=(y/2)(1-x^2)$, $x=(y/2)z$, $t=-(y/2)^2$

so the simplified answer is indeed $x=y/2$, but the way in which the exact answer differs from this approximation is precisely encoded by the Catalan numbers

Thorgott

18:45

@Koro they probably want to say that forms generalize the cross product in R^3

robjohn

20:06

@Pedro Is this a random walk where the steps are $+1$ and $-1$ with equal probability?

Ted Shifrin

20:22

hi @robjohn and rehi @Pedro

robjohn

@TedShifrin Hey, there. It's supposed to start raining soon. Perhaps 3" this week.

Ted Shifrin

We are unlikely to get much ….

Pedro

@robjohn Yes.

Of course $\mathbb P(\tau \geqslant k) = \mathbb P(\tau_a\geqslant k, \tau_{-b}\geqslant k)$, so one might want to compute a joint distribution. It is known that $\mathbb E(t^{\tau_m}) = f_1(t)^{|m|}$ where $f_1$ is the pmf of $\tau_1$, and this is basically computed with Catalan numbers: if I remember correctly, $\mathbb P(\tau_1 = 2j+1) = \frac{1}{2^{2j+1}} C_j$.

Indeed, if you look at a coin toss sequence that first hits $1$ at $2j+1$, what you're looking at is at a Catalan path from $0$ to $2j$ among all possible $2^{2j+1}$ paths, plus the condition of going up in the last moment, which immediately gives the answer.

leslie townes

20:48

okay, why are the catalan numbers appearing in two distinct ways on the same screen of chat

Pedro

@leslietownes cambridge.org/core/books/catalan-numbers/…

Jakobian

@leslietownes Perhaps Pedro is a student of Semiclassical

Pedro

@robjohn A useful fact you may already know: if $M_n$ is the maximum of the $S_k$ up to time $n$, then $M_n\geqslant a$ iff $\tau_a \leqslant n$, where $a > 0$. So alternatively one needs to think about the maximum and minimum processes of the walk.

Jakobian

@leslietownes perhaps you have an idea why is this version of implicit function theorem stated in such an awkward way?

It touches upon Banach spaces, which I consider you an expert at

leslie townes

no idea. what is the source?

Jakobian

20:55

Dieudonne, Smith, Eilenberg - Treatise on Analysis 1

leslie townes

oh, other weirdos too, haha. i was gonna say, we should ask ted.

Jakobian

I've asked Ted, but it didn't help me to understand this

leslie townes

i have long since given up attempting to infer authorial intent from unusual phrasing, but you'd expect dieudonne to give some idea of what that is for

maybe it is for nothing and we are spinning in an indifferent, unforgiving, godless cosmos

i wonder if it's clearer in the original french. this strikes me as something that could conceivably have its sense destroyed by a translator who is perfectly fluent but not trained in math

Jakobian

My impression was that, they're stating that because they are claiming there is no situation so that there is some smaller open set on which two or more solutions exist, just not extendable further

other versions of this theorem seem to show that there exists a neighbourhood of $(x_0, y_0)$ instead, on which $u$ is the unique function that happens to be continuously differentiable

To be honest, I think both are slightly different statements

leslie townes

your impression makes sense, it just feels like a competent author would have included something that makes whatever point they are trying to emphasize more clearly, and dieudonne is generally a competent author, so...

if anyone is visited by dieudonne's ghost this holiday season, ask about this

Jakobian

21:10

Perhaps there is also an issue of going into infinite-dimensional Banach spaces, but I can't be sure without carefully analyzing the proof

leslie townes

perhaps, but again i don't think a competent author would put just this very delicately constructed house of cards into the hypotheses with no explanation of why it looks that way

this is a rare moment where we'll ask dieudonne to "dieu better"

Jakobian

I don't think Dieudonne cares. He's writing/doing calculus on Banach spaces

leslie townes

he can't help that he's french

Jakobian

I mean I had to dig into why Dieudonne assumed his functions are continuous too. Turned out that if $f$ is "differentiable" in the sense that there exists some linear map, not necessarily continuous, which satisfied the definition of derivative, then its continuity is equivalent to continuity at that point

Which is an important assumption to prove chain rule

Nowhere did I see Dieudonne emphasizing this

leslie townes

idk, read a french book, find yourself asking french questions

Jakobian

21:14

I definitely think this is a valuable book, but it takes a careful reader that can truly appreciate the contents, and not get confused while doing so

A one that's "mathematically mature" enough. I think I wasn't as mature to be able to appreciate the contents of it in the past, but I am now

Sha Vuklia

Anyone here familiar with $C^*$ algebras?

Jakobian

Leslie is

Sha Vuklia

ok cool, well let me try: given a $C^*$ algebra $A$, and $a\in A_+$ (i.e., $a^*=a$ and $\sigma(a)\subset [0,\infty))$, one can show that there exists a unique $b \in A_+$ such that $a=b^2$. Assume now that for some $c\in A$, we have $ac=ca$. I want to show that $bc=cb$. One way of doing it would be that I know furthermore that $b\in C^*(a)$, where $C^*(a)$ is the $C^*$-subalgebra generated by $a$.

I am thinking maybe $C^*(a)$ is the closures of polynomials in $a$, in which case we get commutativitity. However, I'm not entirely sure about this

leslie townes

huh, do you have the continuous functional calculus? i can't remember if it's any easier to do sqrt( ) than it is any continuous function.

Koro

Let $g:U\to R$ be continuous $U\subset R^n$. Define $\phi:U\to R^n$ as $\phi(u_1,u_2,...,u_n)= (u_1,u_2,...,u_n, g(u_1,...,u_n))$. I want to show that the volume of $\phi$ is $\int_U \sqrt{(1+\sum_{i=1}^n (\frac{\partial g}{\partial u_i})^2}$.

leslie townes

21:29

but regardless of whether you have it or not, yes, b is indeed a norm limit of polynomials in a, which is one way of showing that second part.

Thorgott

you mean the graph of $\phi$, I assume, and this only makes sense once the right definition of "volume" is provided

Ted Shifrin

No, Koro has it right.

Koro

@Thorgott $\phi:U\to R^{n+1}$ there. It was a typo.

Thorgott

as far as I'm concerned, maps don't have volumes, but I guess that is a point of contention

Koro

@Thorgott So I assign volume to a map known as parametrised n surface in R^{n+k}.

Jakobian

21:32

I think they talk about volume because $\phi$ is an embedding

Koro

This is done as follows:

Jakobian

So its like identifying curve with its image

Koro

(Parametrised n-surface): The map $\phi: U\to R^{n+k}$, where $U\subset R^n$ is open, is said to be (,) if derivative map $D\phi(p)$ is full rank for every p in U.

Ted Shifrin

Haven’t you done volume/surface area with the Gram determinant?

Sha Vuklia

@leslietownes I've only started looking at that theory today, so I'm not comfortable with it

Koro

21:37

Define $E_i= D\phi_p(e_i)$, where $e_i=$ unit vector in R^n with ith entry 1 and 0 otherwise. Define $V(\phi)= \int_U \sqrt{det(E_i. E_j)}$. This $V(\phi)$ is called the volume of $\phi$.

@TedShifrin I probably know it by different name?

Did you mean : $\sqrt{det(E_i. E_j)}$?

Sha Vuklia

@leslietownes I'm hesitating between showing that or just finding the right theorems from continuous functional calculus

I'm kinda hoping to find the right result for a one-line proof

leslie townes

okay. i think that no matter what spectrum(A) is, you can fairly explicitly construct a sequence of real polynomials converging to sqrt(x) on spectrum(A). and if you have such a sequence p_n(x), then the sequence p_n(x^2) will converge uniformly to the identity function on the spectrum of any positive square root of A.

Koro

Re-linking my question should it get lost:

leslie townes

so if you have the spectral mapping theorem for polynomials (which is elementary) and that sequence of polynomials for the square root, that should be enough to do the exercise in a kind of 'bare handed' way.

Koro

10 mins ago, by Koro

Let $g:U\to R$ be continuous $U\subset R^n$. Define $\phi:U\to R^n$ as $\phi(u_1,u_2,...,u_n)= (u_1,u_2,...,u_n, g(u_1,...,u_n))$. I want to show that the volume of $\phi$ is $\int_U \sqrt{(1+\sum_{i=1}^n (\frac{\partial g}{\partial u_i})^2}$.

Sha Vuklia

21:40

okay thanks, I will have a better look at the spectral mapping theorem. I'll let you know if I'm still stuck with this one

Koro

So I used the definition above and computed $E_i= [e_i, \frac{\partial g}{\partial u_i}]$. So the det(E_i. E_j)= det( a symmetric matrix with entries $1+(\frac{\partial g}{\partial u_i})^2$ at the diagonal and entries $E_i. E_j$ for $i\ne j$)

I don't see how it simplifies to the desired one.

Ted Shifrin

@Koro Yes. Then you’re done.

Work it out for $n=2$.

Koro

yeah, I'm trying to see it for n=2.

it works for n=2. :)

the off diagonal terms cancel out after expanding

probably not so easy to observe this for n>2

probably some induction type of proof.

Ted Shifrin

Do $n=3$. Then induction?

Or Grsm-schmidt.

Sha Vuklia

@leslietownes maybe they ask for sqrt because we don't necessarily assume that our $C^*$ algebra in unital

however, I'm a bit too behind on the theory, I should first have a better look at some basic results, because right now I'm not able to reason with it

Koro

21:55

@TedShifrin yeah that should work.

leslie townes

@ShaVuklia the idea is that [however constructed] "b" is the image of "sqrt" under "the functional calculus for a" [meaning: the unique star-isomorphism from C(spec(a)) into the Cstar algebra that sends 1 to the identity operator and the identity function to a]. more concretely b = lim p_n(a) where p_n is any sequence of polynomials chosen as above. and if c is any positive square root then c = sqrt(c^2) = sqrt(a) [perhaps more concretely c = lim p_n(c^2) = lim p_n(a) = b]

where the first equality ["functional calculus respects composition"] comes out of a spectral mapping theorem.

if you are doing non unital algerbas you can work in a unitization or use the [eventual, maybe not easy to see] fact that sqrt( ) vanishes at 0 so you do not need the identity as part of your functional calculus [so those real polynomials can have zero constant term]

Sha Vuklia

okok, that's all making a lot of sense. I think I like the unitization route right now, and I will just use that on faith right now, and do some verifications later this week :)

then I feel confident enough now to invoke the spectral mapping theorem

so thank you!

leslie townes

i don't think it's amenable to being a "one line" proof unless you have something like a general statement of the functional calculus for things that might not be in a unital algebra (which is not hard but is sometimes not how textbooks first state it).

Koro

What does it mean to say that a map $f:S^2\to S^2$ is orientation preserving?

For a manifold M, take a point p in M and look at charts (patches) $\phi: U\to V$ and $\psi: U'\to V'$ such that $p\in V\cap V'$. We say that M has positive orientation if $\phi\circ \psi^{-1}$ has positive derivative.

Thorgott

you want to specify orientations on the manifolds

Koro

22:14

I think I got it.

Suppose that f(p)= -p. Let $N: S^2\to S^2$ be a Gauss map of the surface $S^2$.

Can I say that $N(f(p))= -N(p)$?

it seems true intuitively. Normal at a point can be outward (away from the centre) or towards the centre.

yup true. We usually take $N(p)= p \|p\|^{-1}$

Koro

22:53

For a), I do the following: I can take $\phi(t, z)= (\cos t, \sin t, z)$ where $0<t<2\pi$ and $|z|<1$ for a local parametrisation of a point not on x-axis. For other points, I define a \phi similarly with some adjustments to t.

The definition that I use for a map f:S--> \tilde S to be orientation preserving is the following: If for positively oriented bases v_1, ...,v_n\in S_p, the vectors dv_1,..., dv_n are positively oriented.

that is, if N is the smooth unit normal vector field on S, then det(v_1,..., v_n, N(p))>0 implies det(df(v_1),...,df(v_n), N_2(f(p))>0.

in b), I want to find such N_2.

Putting $N_2=N$, I get this last determinant as $x\sqrt{1-z^2}$

which vanishes on yz plane.

that's why $N_2=N$ doesn't work.

But x=0 would mean that I'm mapping to north/south pole.

and that's not allowed.

so x can't be 0.

nope. x=0 would mean |y|=1. So $\phi(0,y,z)= (0, y\sqrt{1-z^2},z)$. Since z is never 1, the second term is non zero. So \phi never maps to north/south pole.

So N_2= N doesn't work.

Thorgott

23:26

$S_2$ is an open subset of a sphere

you should be able to write down a unit normal vector field on a sphere

and then at worse you have to multiply it by $-1$

Ted Shifrin

@Koro Re your volume of the graph question. Here's the right way to do the linear algebra. We have $E_i = (e_i,a_i)$, $i=1,\dots,n$. Let $E_{n+1} = (-a_1,-a_2,\dots,1)$. Then $E_i\cdot E_{n+1} = 0$ for all $i$. Consider the matrix $A$ whose rows are $E_1,\dots,E_{n+1}$. With simple row operations, you can see that $\det A = 1+\|a\|^2$. Look at $AA^\top$. It has a very nice block form, and your determinant is the determinant of the big block.

You were thinking about that new problem before, weren't you? This is the equi-areal projection from the sphere (less poles) to cylinder.

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