# Sandbox for drafts of long, complex posts [closed]

This sandbox is intended for saving drafts of long, complex posts, especially posts whose composition takes a long time. It serves to localize to one thread the front-page "bumps" caused by edits to drafts of such posts, so that they may be easily ignored. Also, it helps to guard against losing longly-composed posts due to system crashes.

When you are happy with your draft here, you may simply copy the code and paste it to the desired location.

## Proper Use of the Sandbox

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The sandbox has been closed to prevent the creation of new answers. There are more than enough existing answers for users to edit over, and this will greatly reduce the frequency at which we request that the answers be disassociated from specific users.

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• take note of the URL of the answer provided by clicking the share button, or
• save a copy of your draft locally (or even "in the cloud").
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is sufficient. Periodically users may go through and free up answer slots that have not been edited in, say, over one month. But you can aid in the smooth running of this sandbox by clearing away your drafts when you are finished with them.

6. Do not "claim" multiple answers concurrently. Since this post is closed, the answers are a limited resource. If you really must compose several long, complex posts at the same time, you can still use a single answer, separating the different drafts using Markup: horizontal rules (---) and/or headings (# Header 1 #) are natural choices.

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## closed as off-topic by Arthur Fischer♦Sep 18 '14 at 10:37

• This question does not appear to be about Mathematics Stack Exchange or the software that powers the Stack Exchange network within the scope defined in the help center.
If this question can be reworded to fit the rules in the help center, please edit the question.

I have added a [sandbox] tag to allow people ignore it more easily (via software support of ignoring tags), and since it seems that we have two sandboxes now, a tag may seem a bit more in place here. –  Asaf Karagila Jul 18 '12 at 8:35
(+1) For thinking outside the (sand)box. –  cardinal Jul 18 '12 at 19:40
At the suggestion of the moderators, I have gone and changed the associated owners of all the answers here to the Community user. This way, the original owners will not receive excess pings for each time another user uses the draft space for their work. Enjoy! –  Grace Note Oct 5 '12 at 14:45
To prevent crashes I've found the "Bookmarks to disable/enable MathJax", provided in here, pretty useful. –  leo Dec 17 '12 at 18:03
This "sandbox" is being closed to prevent the creation of new answers. To start a draft, simply edit one of the existing free answers. –  Arthur Fischer Sep 18 '14 at 10:37

This answer is free to be used

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$$p \in \mathscr{A}\mathscr{v}\mathscr{a}\mathscr{i}\mathscr{l}\mathscr{a}\mathscr{b}\mathscr{l}\mathscr{e}$$

$$p \in \{\rm{this\ answer}\}$$

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This answer is free for anyone to use

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This answer is free for anyone to use

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This answer is free to for anyone to use.

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This answer is free for anyone to use

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This answer is free to for anyone to use

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(Still has a flaw!)

It holds the relations: $$\mathrm{ad}_\varepsilon(A)=\delta\tau_{+\varepsilon}[A]e^{i\varepsilon H}=e^{i\varepsilon H}\delta\tau_{-\varepsilon}[A]$$

They are derivations: $$\mathrm{ad}_\varepsilon(AB)=\mathrm{ad}_\varepsilon(A)B+A\mathrm{ad}_\varepsilon(B)$$

And they vanish on: $$\mathrm{ad}_\varepsilon(e^{itH})=i[\delta H_\varepsilon,e^{itH}]=0$$

By iteration one gets: $$\mathrm{ad}_\varepsilon^N(A)=\delta\tau_{+\varepsilon}^N[A]e^{Ni\varepsilon H}=e^{Ni\varepsilon H}\delta\tau_{-\varepsilon}^N[A]$$

Also they commute: $$\tau_\varepsilon,\mathrm{id}\in\mathcal{B}(\mathcal{B}(\mathcal{H})):\quad\tau_\varepsilon\circ\mathrm{id}=\mathrm{id}\circ\tau_\varepsilon$$

And they preserve: $$\tau:\mathbb{R}\to\mathcal{B}(\mathcal{B}(\mathcal{H})):\quad\tau^{\varepsilon+\varepsilon'}=\tau^\varepsilon\circ\tau^{\varepsilon'}$$

By Newton's formula: $$\delta\tau_\varepsilon^N=\frac{1}{\varepsilon^N}\sum_{n=0}^N\binom{N}{n}(-1)^{N-n}\tau^{n\varepsilon}$$

So one derives at: $$\mathrm{ad}_\varepsilon^N(A)=\frac{1}{\varepsilon^N}\sum_{n=0}^N\binom{N}{n}(-1)^{N-n}\tau^{n\varepsilon}[A]e^{iN\varepsilon H}$$

Concluding formula.

Taylor Expansion

Regard an expansion: $$F_\varepsilon\in\mathcal{C}^N(\mathbb{R},E):\quad F_\varepsilon=P^\varepsilon_K+R^\varepsilon_K$$

For Taylor polynomial:* $$\frac{1}{\varepsilon^N}\sum_{n=0}^N\binom{N}{n}(-1)^{N-n}P^\varepsilon_K(n\varepsilon)\stackrel{K=N-1}{=}0$$

Suppose one has: $$\|F_\varepsilon^{(N)}(n\varepsilon s)\|^{\varepsilon\neq0}_{s\in[0,1]}<\infty:\quad F_\varepsilon^{(N)}(n\varepsilon s)\stackrel{\varepsilon\to0}{\to} F_0^{(N)}(0)$$

For Taylor remainder:* $$\lim_{\varepsilon\to0}\frac{1}{\varepsilon^N}\sum_{n=0}^N\binom{N}{n}(-1)^{N-n}R_K(n\varepsilon)\stackrel{K=N-1}{=}F_0^{(N)}(0)$$

Concluding expansion.

By the previous thread: $$\left.\frac{\mathrm{d}^N}{\mathrm{d}t^N}\right|_{t=n\varepsilon s}\tau^t[A]e^{iN\varepsilon H}\varphi=\tau^{n\varepsilon s}[\mathrm{ad}^N(A)]e^{iN\varepsilon H}\varphi$$

They admit a dominant: $$\|\tau^{n\varepsilon s}[\mathrm{ad}^N(A)]e^{iN\varepsilon H}\varphi\|\leq\|\mathrm{ad}^N(A)\|\cdot\|\varphi\|$$

And converge pointwise: $$\tau^{n\varepsilon s}[\mathrm{ad}^N(A)]e^{iN\varepsilon H}\varphi\stackrel{\varepsilon\to0}{\to}\mathrm{ad}^N(A)\varphi$$

So the above gives: $$\mathrm{ad}^N(A)\varphi=\lim_{\varepsilon\to0}\mathrm{ad}_\varepsilon^N(A)\varphi=:\mathrm{ad}_0^N(A)\varphi$$

The dominant bounds: $$\|\mathrm{ad}_\varepsilon(A)\|_{\varepsilon\neq0}\leq\|\mathrm{ad}_0^N(A)\|=\|\mathrm{ad}^N(A)\|<\infty$$

Regard the core: $$\mathcal{D}^M:=\bigcap_{m=0}^M\mathcal{D}(H^m):\quad\overline{(H^m)_{\mathcal{D}^M}}=H$$

And regular functions: $$\eta(\varphi,\psi):=\langle\tau[A]\varphi,\psi\rangle\in\mathcal{C}^M(\mathbb{R},\mathbb{C})$$

By induction one gets: $$\eta^{(M)}_0(\varphi,\psi)=i^M\sum_{m=0}^M\binom{M}{m}(-1)^{M-m}\langle AH^m\varphi,H^{M-m}\psi\rangle$$

Note that it holds: $$\eta^{(m)}_{n\varepsilon s}(\varphi,\psi)=\eta^{(m)}_0(e^{-in\varepsilon sH}\varphi,e^{-in\varepsilon sH}\psi)$$

They admit a dominant: $$|\langle\tau^{n\varepsilon s}[A]e^{iN\varepsilon H}H^m\varphi,H^{M-m}\psi\rangle|\leq\|A\|\cdot\|H^m\varphi\|\cdot\|H^{M-m}\psi\|$$

And converge pointwise: $$\langle\tau^{n\varepsilon s}[A]e^{iN\varepsilon H}H^m\varphi,H^{M-m}\psi\rangle\stackrel{\varepsilon\to0}{\to}\langle AH^m\varphi,H^{M-m}\psi\rangle$$

So the above gives: $$\eta_0^{(N)}(\varphi,\psi)=\lim_{\varepsilon\to0}\langle\mathrm{ad}_\varepsilon^N(A)\varphi,\psi\rangle=:\langle\mathrm{ad}_0^N\varphi,\psi\rangle$$

That gives the bound: $$|\eta^{(N)}_\theta(\varphi,\psi)|=\lim_{\varepsilon\to0}|\langle\mathrm{ad}_\varepsilon^N(A)e^{-i\theta H}\varphi,e^{-i\theta H}\psi\rangle|\leq\|\mathrm{ad}_\varepsilon^N(A)\|_{\varepsilon\neq0}\|\varphi\|\cdot\|\psi\|$$

Set another expansion: $$\eta_\theta(\varphi,\psi)=\sum_{l=0}^{L=N-1}\frac{1}{l!}\eta^{(l)}_0(\varphi,\psi)\theta^l+\frac{N}{N!}\theta^{N}\int_0^1(1-s)^{(N-1)}\eta^{(N)}_{\theta s}(\varphi,\psi)\mathrm{d}s$$

Note the trivial bound: $$|\eta_\theta(\varphi,\psi)|=|\langle\tau^\theta[A]\varphi,\psi\rangle|\leq\|A\|\cdot\|\varphi\|\cdot\|\psi\|$$

That implies bounds:** $$|\eta^{(l)}_0(\varphi,\psi)|\leq\|\eta^{(l)}_0\|\cdot\|\varphi\|\cdot\|\psi\|$$

Especially one has: $$|\langle iA\varphi,H\psi\rangle-\langle iAH\varphi,\psi\rangle|=|\eta^{(1)}_0(\varphi,\psi)|\leq\|\eta^{(1)}_0\|\cdot\|\varphi\|\cdot\|\psi\|$$

**Here is a flaw: Summands not positive!

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# Normal Operators

(Second problem still unsolved.)

Problem

Given a Hilbert space $\mathcal{H}$.

Consider normal operators: $$N:=\int\lambda\mathrm{d}E(\lambda)\quad N:=\int\lambda\mathrm{d}E(\lambda)$$

Suppose they commute: $$E'E=EE'\iff N'N=NN'$$

They give rise to: $$E_0:=E\otimes E':=EE'$$

They represent as: $$N^{(\prime)}=\iint\pi^{(\prime)}(\lambda,\lambda')\mathrm{d}E_0(\lambda,\lambda')$$

Then is it possible: $$N'=\eta(N)\lor N=\eta'(N')$$

Denote for shorthand: $$\mathcal{F}(N^{(\prime)}):=\{\eta(N^{(\prime)}):\eta\in\mathcal{B}(\mathbb{C})\}=:\mathcal{F}(E^{(\prime)})$$

For functional calculus: $$\mathcal{F}(E)\cup\mathcal{F}(E')\subseteq\mathcal{F}(E_0)$$

Then does there exist: $$N_0^*N_0=N_0N_0^*:\quad\mathcal{F}(N)\cup\mathcal{F}(N')\subseteq\mathcal{F}(N_0)$$

(Positive or negative results?)

Attempt

Given the Hilbert space $\mathbb{C}^4$.

Consider normal operators: $$N:=\begin{pmatrix}1&0\\0&-1\end{pmatrix}\oplus\begin{pmatrix}0&0\\0&0\end{pmatrix}\quad N':=\begin{pmatrix}0&0\\0&0\end{pmatrix}\oplus\begin{pmatrix}1&0\\0&-1\end{pmatrix}$$

Then they commute: $$N'N=\begin{pmatrix}0&0\\0&0\end{pmatrix}\oplus\begin{pmatrix}0&0\\0&0\end{pmatrix}=NN'$$

But neither gives the other: $$\eta(N)=\begin{pmatrix}\eta(1)&0\\0&\eta(-1)\end{pmatrix}\oplus\begin{pmatrix}\eta(0)&0\\0&\eta(0)\end{pmatrix}\neq\begin{pmatrix}0&0\\0&0\end{pmatrix}\oplus\begin{pmatrix}1&0\\0&-1\end{pmatrix}=N'$$ $$\eta'(N')=\begin{pmatrix}\eta'(0)&0\\0&\eta'(0)\end{pmatrix}\oplus\begin{pmatrix}\eta'(1)&0\\0&\eta'(-1)\end{pmatrix}\neq\begin{pmatrix}1&0\\0&-1\end{pmatrix}\oplus\begin{pmatrix}0&0\\0&0\end{pmatrix}=N'$$

Concluding counterexample.

Does the second assertion fail too?

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This answer is free to for anyone to use.

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This answer is free to for anyone to use.

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# Polar Decomposition

Problem

Given Hilbert spaces $\mathcal{H}$ and $\mathcal{K}$.

Consider a closed operator: $$A:\mathcal{D}(A)\subseteq\mathcal{H}\to\mathcal{K}:\quad A=A^{**}$$

And its decompositions: $$A=J|A|:\quad(J)^*(J)=1_{\overline{\mathcal{R}|A|}}$$ $$A^*=J_*|A^*|:\quad ({J_*})^*(J_*)=1_{\overline{\mathcal{R}|A^*|}}$$

Note that one has:* $$\overline{\mathcal{R}|A|}=\overline{\mathcal{R}A^*}\quad\overline{\mathcal{R}|A^*|}=\overline{\mathcal{R}A}$$

Then they are adjoints: $$|A|J^*=A^*=J_*|A^*|\implies J^*=J_*$$

How can I check this?

Attempt (I)

Given Hilbert spaces $\mathcal{H}$ and $\mathcal{K}$.

Consider partial isometries: $$J:\mathcal{H}\to\mathcal{K}:\quad JJ^*J=J$$

Then one obtains: $$J^*J=P\implies|J|=J^*J\implies J=J|J|$$ $$JJ^*=Q\implies|J^*|=JJ^*\implies J^*=J^*|J^*|$$

So the result is positive.

Attempt (II)

Given a Hilbert space $\mathcal{H}$.

Consider selfadjoints: $$H:\mathcal{D}H\subseteq\mathcal{H}\to\mathcal{H}:\quad H=H^*$$

Then one obtains: $$H=H^*=J_*|H^*|=J_*|H|\implies J_*=J$$

So the result is positive.

Attempt (III)

Given the Hilbert space $\mathbb{C}^2$.

Consider the nilpotent: $$N:\mathbb{C}^2\to\mathbb{C}^2:\quad N:=\begin{pmatrix}0&1\\0&0\end{pmatrix}$$

Then one obtains: $$N^*N=\begin{pmatrix}0&0\\0&1\end{pmatrix}\implies|N|=\begin{pmatrix}0&0\\0&1\end{pmatrix}\implies N=\begin{pmatrix}0&1\\0&0\end{pmatrix}|N|$$ $$NN^*=\begin{pmatrix}1&0\\0&0\end{pmatrix}\implies|N^*|=\begin{pmatrix}1&0\\0&0\end{pmatrix}\implies N^*=\begin{pmatrix}0&0\\1&0\end{pmatrix}|N^*|$$

So the result is positive.

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This answer is free for anyone to use

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The set $\frac 1n \mathbb Z = \left \{\frac xn :x \in \mathbb Z \right \}$ supports the equivalence relation "modulo $m$", where $2 \le m \in \mathbb Z.$ We will represent the resulting quotient group as $\frac 1n \mathbb Z_{mn}.$ For example

$\frac 16 \mathbb Z_{18} = \left\{ \overline{\frac 06}, \, \overline{\frac 16}, \, \overline{\frac 26}, \, \dots, \, \overline{\frac {17}{6}}\ \right\} = \left\{ \overline{0}, \, \overline{\frac 16}, \, \overline{\frac 13}, \, \overline{\frac 12}, \, \overline{\frac 23}, \, \overline{\frac 56}, \, \overline{1}, \, \overline{\frac 76}, \, \overline{\frac 43}, \, \overline{\frac 32}, \, \overline{\frac 53}, \, \overline{\frac {11}6}, \, \overline{2}, \, \overline{\frac {13}6}, \, \overline{\frac 73}, \, \overline{\frac 52}, \, \overline{\frac 83}, \, \overline{\frac {17}6} \right\}$

where the corresponding equivalence relation is "modulo $3$" and the denominators must be divisors of $6$.

From this point on, when we write $\frac 1N \mathbb Z_M,$ it will be assumed that $N|M.$

Define $\mathbb M_N = \left\{ \mathbf n = \begin{bmatrix} n_1\\ n_2\\ n_3\\ \vdots\\ n_N\\ \end{bmatrix} \in \mathbb Z^{N \times 1} : \text{ for all } 1\le i\le N, \, n_i \ge 1 \right\}.$

For $\mathbf u, \mathbf v \in \mathbb M_N$, we will express $\displaystyle \mathcal G = \prod_{i=1}^N \frac{1}{u_i} \mathbb Z_{v_i}$ as $\mathcal G = \frac{1}{\mathbf u} \mathbb Z_{\mathbf v}$

and $\mathbf u, \mathbf v \in \mathbb Z^{A \times B}$ define $\mathbf u \equiv \mathbf v \pmod {\mathbf a}$ if and only if for all $1\le i\le N$ and for all $1 \le j \le B,$ $\mathbf u_{ij} \equiv \mathbf v_{ij} \pmod{\mathbf a_i}.$ In other words, $\mathbf u \equiv \mathbf v \pmod {\mathbf a}$ if and only if the corresponding columns of $\mathbf u$ and $\mathbf v$ are all congruent modulo $\mathbf a.$

For all $\mathbf x \in \mathbb Z^{N \times 1},$ define $\mathbf D_{\mathbf x} = \text{diag}(\mathbf x).$ Then

For all $\mathbf a \in \mathbb M_A$ and $\mathbf u, \mathbf v \in \mathbb Z^{A \times B}, \mathbf u \equiv \mathbf v \pmod{\mathbf a}$ if and only if there exists $\hat{\mathbf u} \in \mathbf Z^{A \times B}$ such that $\mathbf v = \mathbf u + \mathbf D_{\mathbf a} \hat{\mathbf u}$

We will be concerned with the finite abelian group $\displaystyle \mathcal G = \sum_{i=1}^A \mathbb Z_{a_i}$ for some $\mathbf a \in \mathbb M_A.$ We will associate $\mathcal G$ with the set $\mathbb Z^{A \times 1} / \mathbf a$ and we will associate $\mathcal G^B$ with the set $\mathbb Z^{A \times B} / \mathbf a$

It's going to turn out that $\mathbb Z^{A \times B} / \mathbf a$ is not "dense" enough. We are going to need finite abelian groups with fractions in them. My goal is to use no more than I absolutely have to.

Arithmetic in these groups is most easily explained by pointing out that the mapping $f:\frac 1n \mathbb Z_{mn} \to \mathbb Z_{mn}$ defined by $f(\xi) = n\xi$ is an additive-group isomorphism with $f^{-1}(y) = \overline{\frac {y}{n}}$.

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Let $k\mathbf{Top}$ denote the category of all maps between $k$-spaces (here a map shall always mean a continuous funtion), where a $k$-space is a space $X$ with the final topology for all test-maps into it, and a test-map is a map $t:K\to X$ with $K$ compact Hausdorff.
There is a coreflector $k:\mathbf{Top}\to k\mathbf{Top}$ sending each space $X$ to its $k$-ification, adding all sets with open preimage under every test-map to the topology. The unit of this adjunction is the identity on a $k$-space, that's why $k$ is called a right-adjoint-left-inverse.

We know that $k\mathbf{Top}$ is Cartesian closed. This means if $Y$ is a $k$-space, we have a bijection $$k\mathbf{Top}(X\times_k Y, Z) \cong k\mathbf{Top}(X,k(Z^Y))$$ where $X\times_k Y$ is the product in $k\mathbf{Top}$, obtained by applying $k$ to the product in $\mathbf{Top}$, and $Z^Y$ is endowed with the test-open topology (generated by the sets $W(t,U)=\{f:Y\to Z\mid \text{Im}(ft)\subseteq U\}$ for test-maps $t$ and open $U$ in $Z$).

One can show that if $X$ is a $k$-space, then $X\times I$ is a $k$-space, so there is no need to apply $k$ to the usual product (In fact, this works for $I$ replaced by any locally compact Hausdorff space, but let's stick to $I$). Now my question is: Is also $Z^I$ a $k$-space?

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# Wave Operators

Given Hilbert spaces $\mathcal{H}_0$ and $\mathcal{H}$.

Consider Hamiltonians: $$H_\#:\mathcal{D}H_\#\to\mathcal{H}_\#:\quad H_\#=H_\#^*$$

Denote their evolutions: $$U_\#(t)^*=U_\#(-t)=U_\#(t)^{-1}$$

Regard operators: $$J_0\in\mathcal{B}(\mathcal{H}_0,\mathcal{H})\quad J\in\mathcal{B}(\mathcal{H},\mathcal{H}_0)$$

Assume the limits: $$\Omega_0^\pm\{J_0\}\varphi_0:=\lim_{t\to\infty}U(t)^*J_0U_0(t)\varphi_0\quad(\varphi_0\in\mathcal{D}H_0)$$

Asymptotic inverses: $$JJ_0\cong_01_0:\iff\lim_{t\to\pm\infty}\{JJ_0-1_0\}U_0(t)=0$$ $$J_0J\cong1:\iff\lim_{t\to\pm\infty}\{J_0J-1\}U(t)=0$$

Then for the ranges: $$\mathcal{R}\Omega_0^\pm\{J_0\}=\mathcal{H}\iff JJ_0\cong_01_0\quad J_0J\cong1$$

Then the limits exist: $$\Omega^\pm\{J\}\varphi:=\lim_{t\to\pm\infty}U_0(t)^*JU(t)\varphi\quad(\varphi\in\mathcal{H})$$

Especially one obtains: $$\Omega\{J\}\Omega_0\{J_0\}=1_0\quad\Omega_0\{J_0\}\Omega\{J\}=1$$

How can I prove this?

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Given real $λ$ and $ε > 0$ and real $n \to \infty$:

$\frac{1}{n}\sum_{k \ge 0} f(\frac{k}{n}) \left( (1-\frac{λ}{n})^k-e^{-λ\frac{k}{n}} \right)$

$\ = \frac{1}{n}\sum_{k \ge 0} f(\frac{k}{n}) \left( e^{k\ln(1-\frac{λ}{n})} -e^{-λ\frac{k}{n}} \right)$

$\ \in \frac{1}{n}\sum_{k \ge 0} f(\frac{k}{n}) \left( e^{k(-\frac{λ}{n}+[\frac{ε}{n}])} -e^{-λ\frac{k}{n}} \right)$

$\ = \frac{1}{n}\sum_{k \ge 0} f(\frac{k}{n}) e^{-λ\frac{k}{n}} ( e^{[\frac{εk}{n}]} - 1 )$

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