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

  1. Do not post a new answer! We wish all the answers on this page to be owned by the Community user (so that only a non-sentient bot is informed of edits to these answers). Posting a new answer will make you the owner, meaning that you will be notified whenever another user makes an edit to that answer.

    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.

  2. Do not delete answers! Deleting seems like a reasonable option, but there are no "hard deletions" on Stack Exchange, and users with sufficient privileges will still see your supposedly deleted postings. Deleted answers will be undeleted and cleared for the use of others.

  3. Do look for an answer which indicates that it is free and then edit it to your heart's content. If none appears available, take over the one that has been left unchanged the longest (which will appear at the bottom of the page if you order answers by "activity").

  4. Do not expect your draft to remain untouched for days. There are no guarantees that your draft will be the latest revision if you return days later. While users will try not to step over others' toes, it may happen that an unfinished draft is edited out. Your draft will, however, still exist as a revision of the answer it was made in. If your drafting is expected to take place over a longer period of time, either

    • 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").
  5. Do clear your draft when you are finished. This includes removing all $\LaTeX$ from your answers. Replacing all code with a simple statement like

    This answer is free for anyone to use

    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.

  7. Do not create new such sandboxes. The point of having a unique such sandbox is that it minimizes the noise on the front page when the sandbox is edited. If there were multiple sandboxes they will frequently occupy numerous front page slots, pushing other topics off the front page, and increasing noise.


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

17 Answers 17

Mourre Adjoint


(Still has a flaw!)

Adjoint Formula

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.

Adjoint Variation

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

Concluding adjoint variation.

Mourre Adjoint

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

Concluding Mourre adjoint.

*See the thread: Binomial

**Here is a flaw: Summands not positive!


This answer is free foranyone to use


This answer is free to be used


Normal Operators


(Second problem still unsolved.)


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?)


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?




This answer is free to for anyone to use


Polar Decomposition


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.

*See the thread: Ranges


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?


This answer is free for anyone to use




The remaining case is the one where $x_B(0) \neq 0$ and $\tilde y_B(0) > 0$. Since $\tilde y'_B(t) < -1$ when $\tilde y_B(t) > 0$, the trajectory reaches the line $\tilde y = 0$ in finite time. Let $T > 0$ be the earliest such time.

Assume for the sake of contradiction that $(x_B(T), \tilde y_B(T)) = (0,0)$. We define sequences $(t_n)_{n \geq 0}$ of times and $(u_n)_{n \geq 0}$ of $\tilde y$-coordinates (plus, for convenience, $x_n = x_B(t_n)$ and $\tilde y_n = \tilde y_B(t_n)$) as follows. The initial conditions will be $t_0 = 0$ and $u_0 = \tilde y_B(0)$. Then, suppose $t_n < T$ and $u_n > 0$ are such that $(x_n, \tilde y_n)$ and $(x_0, u_n)$ are at the same polar angle. Choose $t_{n+1} \in (t_n, T)$ such that \begin{equation} \tag{*} \label{ratio} \frac{\tilde y_n}{x_n} = \frac{F_2(x_{n+1}, \tilde y_{n+1})}{F_1(x_{n+1}, \tilde y_{n+1})} \equiv \frac{\tilde y_{n+1} + \sqrt{x_{n+1}^2 + \tilde y_{n+1}^2}}{x_{n+1}}; \end{equation} such a time exists due to Cauchy's mean value theorem. Since $t_n < T \implies \tilde y_n > 0$, the above ratio has the same sign as $x_n$ and thus $x_0$. It also has the same sign as $x_{n+1}$ (see item 2 of the earlier list). Thus, there exists a $u_{n+1} > 0$ such that $(x_0, u_{n+1})$ has the same polar angle as $(x_{n+1}, \tilde y_{n+1})$.

However, observe that by \eqref{ratio} and item 2, $$ \frac{u_n}{x_0} = \frac{u_{n+1} + \sqrt{x_0^2 + u_{n+1}^2}}{x_0}, $$ which implies $u_{n+1} < u_n - x_0$. This means that $u_n < u_0 - n x_0$ for all $n \geq 0$, contradicting the fact that the $u_n$ are all positive.

Thus even in this last case, $B$ is eventually side-by-side with $A$, after which $B$ is doomed to fall further and further behind $A$ forever.



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?


\newpage \section{Central extensions} \subsection*{General terminologies:} An injective group homomorphism is called a monomorphism, and a surjective group homomorphism is called an epimorphism. These are denoted by the mapping symbol $\hookrightarrow$ and $\twoheadrightarrow$ respectively. This definition holds in the categories $\text{FinGrp},\text{AbGrp},\text{Grp}$ \subsection*{Definition of an exact sequence:} A sequence of homomorphisms, is called exact if we have:

$$1\overset{\phi_1}{\to} N \overset{\phi_2}{\to} G \overset{\phi_3}{\to} H \overset{\phi_4}{\to} 1$$ where we have the following:

$$\text{Im}(\phi_1) =\ker(\phi_2)$$ $$\text{Im}(\phi_2) =\ker(\phi_3)$$ $$\text{Im}(\phi_3) =\ker(\phi_4)$$

\subsection*{Quick Example of a short exact sequence}

$$0\hookrightarrow \Bbb Z \overset{\times n}{\hookrightarrow} \mathbb Z \twoheadrightarrow \mathbb Z/n \twoheadrightarrow 0$$

$$\text{Im}(\phi_1) = 0 = \ker(\phi_2)$$ $$\text{Im}(\phi_2) = n\mathbb Z = \ker(\phi_3)$$ $$\text{Im}(\phi_3) = \mathbb{Z}/n = \ker(\phi_4)$$

\subsection*{Definition of an extension:} An extension of a group $G$ by a group $A$ is given by an exact sequence of group homomorphisms:

$$1\hookrightarrow A \hookrightarrow E \twoheadrightarrow G \twoheadrightarrow 1$$

This extension is called central if $A$ is abelian and the image of the monomorphism from $A\hookrightarrow E$ is in the center of $E$. \newpage \subsection*{Example of a central extension: Discrete Heisenberg group}

The Heisenburg group is also a lie group. It is not only a group but with topology inherited from $\Bbb R^3$ it becomes a smooth manifold with applications in Fourier analysis, quantum mechanics and other subjects. We will consider the discrete Heisenburg group, for simplicity. $$$$ Why do we care about central extensions? Finding group extensions is hard. Split extensions are the easiest to classify, these are semi-direct products, and general extensions are not much better than impossible to classify. This Heisenburg group is an example of one that doesn't split.


$$H_3(\mathbb{Z})= \begin{bmatrix}1&x&z\\0&1&y\\0&0&1 \end{bmatrix}|a,b,c\in \mathbb Z$$

[don't need to say]: We have the relations $z=xyx^{-1}y^{-1},xz=zx,yz=zy$

[don't need to say]: It has generators $$x=\begin{bmatrix}1&1&0\\0&1&0\\0&0&1\end{bmatrix},y=\begin{bmatrix}1&0&0\\0&1&1\\0&0&1\end{bmatrix},z=\begin{bmatrix}1&0&1\\0&1&0\\0&0&1\end{bmatrix}$$

The center of this is: $\begin{bmatrix}1&0&z\\0&1&0\\0&0&1\end{bmatrix}$

and we have a central extension from the short exact sequence: $$I\hookrightarrow N = Z(H_3(\Bbb Z)) \hookrightarrow H_3{\Bbb Z} \twoheadrightarrow \Bbb H_3(\Bbb Z)/N \twoheadrightarrow I$$

Since $N$ is the center, this is a central extension as long as we can show that the sequence is exact:

What we are working with in disguise: $$1\hookrightarrow \Bbb Z \hookrightarrow H_3{\Bbb Z} \twoheadrightarrow \Bbb Z^2 \twoheadrightarrow 1$$

$$\text{Im}(\phi_1)=I=\ker(\phi_2)$$ $$\text{Im}(\phi_2)=N=\ker(\phi_3)$$ $$\text{Im}(\phi_3)=H_3(\mathbb Z)/N=\ker(\phi_4)$$

The explicit maps are:

$$\phi_2: (0,0,z)\mapsto \begin{bmatrix}1&0&z\\0&1&0\\0&0&1\end{bmatrix},\phi_3: \begin{bmatrix}1&x&z\\0&1&y\\0&0&1\end{bmatrix}\mapsto (x,y,0)$$

[don't need to say:] $1=(0,0,0)$ [don't need to say:] $H/N$ is not a subgroup, it is a quotient group


Let $M$ be a monoid. A homomorphism $\varphi:M\to N$ recognizes $L\subset M$, if $L=\varphi^{-1}(\varphi(L))$. A monoid $N$ recognizes $L \subset M$, if there exists a homomorphism $\varphi:M\to N$ that recognizes $L$. A subset of $M$ is recognizable, if it is recognized by a finite monoid. The family $\operatorname{REC}(M)$ of recognizable sets over $M$ is defined as the set of recognizable subsets of $M$.

A similar definition can be found in "J.-E. Pin, Mathematical Foundations of Automata Theory", and already "S. Eilenberg, Automata, Languages, and Machines" defines recognizable sets.

A deterministic $M$-automaton $\mathcal A=(Q,\cdot,q_0,F)$ consists of a set of states $Q$, a transition function $\cdot:Q\times M\to Q$, an initial state $q_0\in Q$, and a set of final states $F\subset Q$. For $q\in Q$ and $u,v\in M$, the transition function has to satisfy $$\begin{array}{rcl}(q\cdot u)\cdot v &=& q \cdot (uv)\\ q\cdot 1 &=& q\end{array}$$ The subset of $M$ accepted by $\mathcal A$ is $$L(\mathcal A) = \{u\in M|q_0\cdot u\in F\}$$

If $Q$ is a finite set, then $\mathcal A$ is said to have finitely many states.

I found this definition in chapter 7 Automatentheorie of the German text book "Volker Diekert, Manfred Kufleitner, Gerhard Rosenberger: Diskrete algebraische Methoden: Arithmetik, Kryptographie, Automaten und Gruppen”. There we also find the following theorem:

Theorem 7.10 Let $L\subset M$. The following statements are equivalent:

  1. $L$ is recognizable, i.e. $L\in \operatorname{REC}(M)$.
  2. $L$ gets accepted by a deterministic $M$-automaton with finitely many states.
  3. The "minimal automaton $\mathcal A_L$" has finitely many states.
  4. The "syntactic monoid $\operatorname{Synt}(L)$" is finite.

I have the impression that this definition of deterministic $M$-automata is unsuitable for characterizing deterministic finite state transducers as deterministic $\Sigma^*\times \Sigma^*$-automata.


A presheaf on a category $J$ is a functor $J^{\text{op}}\to\mathbf{Set}$. The category of presheaves on $J$, denoted $\hat J$, with natural transformations as arrows is an example of functor category, and every category $J$ embeds into $\hat J=\mathbf{Set}^{J^{\text{op}}}$ via the Yoneda embedding $Y:x\mapsto J(-,x)$.

Note that your category $\cal A$ can be identified with $\hat{\Bbb N}$ since $\Bbb N=\Bbb N^{\text{op}}$ and every endomorphism determines a functor from the monoid $\Bbb N$ to $\mathbf{Set}$.

A nice property of a $\hat J$ for small $J$ is that it is Cartesian closed. Given any presheaves $B$ and $C$, if there exists an exponential $C^B$, then by Yoneda's lemma $$ C^B(x)=\text{Nat}(J(-,x),C^B-) = \hat J(Y(x),C^B)\cong \hat J(Y(x)×B,C) $$ and the arrow $C^B(x)\to C^B(y)$ is the function precomposing with $Yf×1_B$. So we only need to check that this presheaf is really the exponential. So we define $ε:C^B×B\Rightarrow C$ as $$ \begin{align} ε_x:\hat J(J(-,x)×B,C)×B(x) &\to C(x) \\ (\sigma,b)\qquad &\mapsto \sigma_x(1_x,b) \end{align} $$ It is easy to check that $ε$ is natural.

Note how similar this is to the construction in your case. Setting $J=\Bbb N, B=α,$ and $C=β$, and noting that $Y(\bullet)$ is the functor sending every $k\in \Bbb N$ to the addition "$...+k$", we get $β^α(\bullet)=\mathcal A(s×α,β)$. Precomposing with $s^k×1_α$ is the same as causing $h$ to be applied to $(n+k,a)$ instead of $(n,a)$. And the transformation $ε_x$ sends a map $h$ and an element $a\in α(\bullet)$ to $h(0,a)$.


This answer is free for anyone to use