# 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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## 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

Pascal tetrahedral slice for $(a+b+c)^0$
$\begin{matrix} 1 \end{matrix}$

Pascal tetrahedral slice for $(a+b+c)^1$
$\begin{matrix} &a\\ b&&c \end{matrix}$

Pascal tetrahedral slice for $(a+b+c)^2$
$\begin{matrix} &&a^2\\ &2ab&&2ac\\ b^2&&2bc&&c^2 \end{matrix}$

Pascal tetrahedral slice for $(a+b+c)^3$
$\begin{matrix} &&&a^3\\ &&3a^2b&&3a^2c\\ &3ab^2&&6abc&&3ac^2\\ b^3&&3b^2c&&3bc^2&&c^3 \end{matrix}$

Pascal tetrahedral slice for $(a+b+c)^4$
$\begin{matrix} &&&&a^4\\ &&&4a^3b&&4a^3c\\ &&6a^2b^2&&12a^2bc&&6a^2c^2\\ &4ab^3&&12ab^2c&&12abc^2&&4ac^3\\ b^4&&4b^3c&&6b^2c^2&&4bc^3&&c^4 \end{matrix}$

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available for use. available for use.

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This answer is free to for anyone to use.

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Feel free to edit this answer.

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This answer is free to for anyone to use

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This post is free to use for everyone.

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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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$\frac{\left(\frac{1+\sqrt5}2\right)^{N} -\left(\frac{1-\sqrt5}2\right)^{N}}{\sqrt5} =\frac{\left(\frac{1+\sqrt5}2\right)^{N-1} -\left(\frac{1-\sqrt5}2\right)^{N-1}}{\sqrt5} +\frac{\left(\frac{1+\sqrt5}2\right)^{N-2} -\left(\frac{1-\sqrt5}2\right)^{N-2}}{\sqrt5}$

$\left(\frac{1+\sqrt5}2\right)^{N} -\left(\frac{1-\sqrt5}2\right)^{N} =\left(\frac{1+\sqrt5}2\right)^{N-1} -\left(\frac{1-\sqrt5}2\right)^{N-1} +\left(\frac{1+\sqrt5}2\right)^{N-2} -\left(\frac{1-\sqrt5}2\right)^{N-2}$

$\left(\frac{1+\sqrt5}2\right)^{N} -\left(\frac{1+\sqrt5}2\right)^{N-1} -\left(\frac{1+\sqrt5}2\right)^{N-2} =\left(\frac{1-\sqrt5}2\right)^{N} -\left(\frac{1-\sqrt5}2\right)^{N-1} -\left(\frac{1-\sqrt5}2\right)^{N-2}$

$(1+\sqrt5)^{N}-2(1+\sqrt5)^{N-1}-4(1+\sqrt5)^{N-2} =(1-\sqrt5)^{N}-2(1-\sqrt5)^{N-1}-4(1-\sqrt5)^{N-2}$

$(1+\sqrt5)^{N-2}((1+\sqrt5)^2-2(1+\sqrt5)-4) =(1-\sqrt5)^{N-2}((1-\sqrt5)^2-2(1-\sqrt5)-4)$

$(1+\sqrt5)^{N-2}(6+2\sqrt5-2-2\sqrt5-4) =(1-\sqrt5)^{N-2}(6-2\sqrt5-2+2\sqrt5-4)$

$(1+\sqrt5)^{N-2}·0=(1-\sqrt5)^{N-2}·0$

$0=0$

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Question:

For $k \ge 1$ and $d \ge 2$, let ${}^f_d\Omega^t_{d+k} (d;c_k(d - 1))$ be a face of the tridiagonal Birkhoff polytope $\Omega^t_{d+k}$, as described in this question.

What are the combinatorial types of the facets of ${}^f_d\Omega^t_{d+k} (d;c_k(d - 1))$?

We begin with the facets associated with main diagonal entries (i.e., of the form $a_{j,j} = 0$). First, observe that the potential number of facets of ${}^f_d\Omega^t_{d+k} (d;c_k(d - 1))$ associated with main diagonal entries is $d - 1$ - i.e., corresponding to the unconstrained main diagonal elements from $a_{2,2}$ to $a_{d+k-1,d+k-1}$.

We consider three main cases (with sub-cases):

Case $0$: A facet associated with a main diagonal entry $a_{j,j}$ in the interior of a part of $c_k(d - 1)$ - i.e., neither the first nor last entry in the part. (Note that this implies that the part containing $a_{j,j}$ has length at least three.) Let's count these (potential) facets. From the $d - 1$ unconstrained entries, subtract $2k$ entries, representing the first and last entry in each part of $c_k(d - 1)$. If there are parts of $c_k(d - 1)$ equal to $1$, we have over-subtracted, as the first and last entries in these parts are not distinct. Say there are $k_1$ parts of $c_k(d - 1)$ equal to $1$; we arrive at a count of $d - 1 - 2k + k_1$. The combinatorial type of a Case $0$ facet meets the definition of the class of $\Omega^t_{d+k}$ faces specified here; specifically, it is the face ${}^f_{d-1}\Omega^t_{d+k} (d-1;c_{k+1}(d - 2))$, where $c_{k+1}(d - 2)$ is formed from $c_k(d - 1)$ by replacing the part containing $a_{j,j}$ with two parts - the number of entries in the part preceding $a_{j,j}$, and the number of entries in the part following $a_{j,j}$.

Case $1a$: The facet $a_{2,2} = 0$, where the first part of $c_k(d - 1) \gt 1$.

Case $1b$: The facet $a_{2,2} = 0$, where the first part of $c_k(d - 1) = 1$.

Case $1c$: The facet $a_{d+k-1,d+k-1} = 0$, where the last part of $c_k(d - 1) \gt 1$.

Case $1d$: The facet $a_{d+k-1,d+k-1} = 0$, where the last part of $c_k(d - 1) = 1$.

Case $2a$: The facet $a_{j,j} = 0$, where $2 \lt j \lt d + k - 1$, and $a_{j,j}$ is the first entry of a part of $c_k(d - 1) \gt 1$.

Case $2b$: The facet $a_{j,j} = 0$, where $2 \lt j \lt d + k - 1$, and $a_{j,j}$ is the last entry of a part of $c_k(d - 1) \gt 1$.

Case $2c$: The facet $a_{j,j} = 0$, where $2 \lt j \lt d + k - 1$, and $a_{j,j}$ is the entry representing a part of $c_k(d - 1) = 1$.

Next, we consider facets associated with superdiagonal matrix entries (of the form $a_{j,j+1} = 0$). (As tridiagonal doubly stochastic matrices are symmetric, $a_{j,j+1} = 0$ if and only if the corresponding subdiagonal entry $a_{j+1,j} = 0$ as well.) To review, ${}^f_d\Omega^t_{d+k} (d;c_k(d - 1))$ is the intersection of $k - 1$ facets of $\Omega^t_{d+k}$ with equations: $$a_{n_i,n_i} = 0 for i = 1, ... k - 1$$, where $$n_1 \gt 2; n_{i+1} \gt n_i + 1 for i = 1, ... k - 2; n_{k-1} \lt d + k - 1$$ If $k \gt 1$, we have one or more constrained main diagonal elements $a_{n_i,n_i} = 0$. We will show that $a_{n_{i} - 1,n_i} = 0$ and $a_{n_i,n_{i} + 1} = 0$ are lower-dimensional faces (not facets) of ${}^f_d\Omega^t_{d+k} (d;c_k(d - 1))$. Case A: $a_{n_{i} - 1,n_i} = 0$ (plus we already have $a_{n_i,n_i} = 0$). Because all matrices within $\Omega^t_{d+k}$ are tridiagonal doubly stochastic, we conclude that $a_{n_{i} + 1,n_i} = 1$. Again, because all matrices within $\Omega^t_{d+k}$ are tridiagonal doubly stochastic (with all non-negative entries), we conclude that $a_{n_{i} + 1,n_{i} + 1} = 0$ and $a_{n_{i} + 1,n_{i} + 2} = 0$. However, $a_{n_{i} + 1,n_{i} + 1}$ is an unconstrained main diagonal entry included in Case 1 or 2 above; thus we know that $a_{n_{i} + 1,n_{i} + 1} = 0$ is a facet of ${}^f_d\Omega^t_{d+k} (d;c_k(d - 1))$. Case B: $a_{n_{i} + 1,n_{i} + 2} = 0$ (and $a_{n_i,n_i} = 0$) is similar to Case A and will be omitted.

Case $3a$: The facet $a_{1,2} = 0$, where the first part of $c_k(d - 1) \gt 1$.

Case $3b$: The facet $a_{1,2} = 0$, where the first part of $c_k(d - 1) = 1$.

Case $3c$: The facet $a_{d+k-1,d+k} = 0$, where the last part of $c_k(d - 1) \gt 1$.

Case $3d$: The facet $a_{d+k-1,d+k} = 0$, where the last part of $c_k(d - 1) = 1$.

Case $4a$: The facet $a_{n_{i} + 1,n_{i} + 2} = 0$, where $n_{i} \lt d + k - 2$, and $a_{n_{i} + 1,n_{i} + 1}$ is the first entry of a part of $c_k(d - 1) \gt 2$.

Case $4b$: The facet $a_{n_{i} - 2,n_{i} - 1} = 0$, where $3 \lt n_{i}$, and $a_{n_{i} - 1,n_{i} - 1}$ is the last entry of a part of $c_k(d - 1) \gt 2$.

Case $4c$: The facet $a_{n_{i} + 1,n_{i} + 2} = 0$, where $n_{i+1} = n_i + 3$, so that the entries $a_{n_{i} + 1,n_{i} + 1}$ and $a_{n_{i} + 2,n_{i} + 2}$ form a part of $c_k(d - 1) = 2$.

Case $4d$: The facet $a_{j,j+1} = 0$, where $1 \lt j \lt j + 2 \lt n_1$.

Case $4e$: The facet $a_{j,j+1} = 0$, where $n_{k-1} \lt j \lt j + 2 \lt d + k$.

Case $4f$: The facet $a_{j,j+1} = 0$, where, for $1 \le i \lt k - 1$, $n_{i} + 1 \lt j \lt j + 2 \lt n_{i+1}$.

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$$\begin{array}{c|c|c|} \hfill d & \text{1} & \text{2} & \text{3} & \text{4} & \text{5} & \text{6} & \text{7} & \text{8} & \text{9} & \text{10} & \text{11} & \text{12} & \text{13} & \text{14} & \text{15} \\ n\ mod\ d \\ \hline \text{1 mod d} & 0 & \\ \hline \text{2 mod d} & 0 & 0 \\ \hline \text{3 mod d} & 0 & 1 & 0 \\ \hline \text{4 mod d} & 0 & 0 & 1 & 0 \\ \hline \text{5 mod d} & 0 & 1 & 2 & 1 & 0 \\ \hline \text{6 mod d} & 0 & 0 & 0 & 2 & 1 & 0 \\ \hline \text{7 mod d} & 0 & 1 & 1 & 3 & 2 & 1 & 0 \\ \hline \text{8 mod d} & 0 & 0 & 2 & 0 & 3 & 2 & 1 & 0 \\ \hline \text{9 mod d} & 0 & 1 & 0 & 1 & 4 & 3 & 2 & 1 & 0 \\ \hline \text{10 mod d} & 0 & 0 & 1 & 2 & 0 & 4 & 3 & 2 & 1 & 0 \\ \hline \text{11 mod d} & 0 & 1 & 2 & 3 & 1 & 5 & 4 & 3 & 2 & 1 & 0 \\ \hline \text{12 mod d} & 0 & 0 & 0 & 0 & 2 & 0 & 5 & 4 & 3 & 2 & 1 & 0 \\ \hline \text{13 mod d} & 0 & 1 & 1 & 1 & 3 & 1 & 6 & 5 & 4 & 3 & 2 & 1 & 0 \\ \hline \text{14 mod d} & 0 & 0 & 2 & 2 & 4 & 2 & 0 & 6 & 5 & 4 & 3 & 2 & 1 & 0 \\ \hline \text{15 mod d} & 0 & 1 & 0 & 3 & 0 & 3 & 1 & 7 & 6 & 5 & 4 & 3 & 2 & 1 & 0 \\ \hline \text{16 mod d} & 0 & 0 & 1 & 0 & 1 & 4 & 2 & 0 & 7 & 6 & 5 & 4 & 3 & 2 & 1 \\ \hline \end{array}$$

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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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$P=3AC-B^2$
$Q=27A^2D-9ABC+2B^3$
$R=\sqrt{4P^3+R^2}$

$$X=\frac{\sqrt[3]{4(-Q+R)}+\sqrt[3]{4(-Q-R)}-2AB}{6A}$$

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I have been looking at representing isomorphisms between finite abelian groups using linear transformations. My studies indicated that the transformation matricies sometimes needed to consists of rational numbers. Below is a sort of summary of my context.

The set $\frac 1v \mathbb Z = \left \{\frac xv :x \in \mathbb Z \right \}$ is a group with respect to addition and supports the equivalence relation $\pmod m$ where $2 \le m \in \mathbb Z.$ In particular, $\frac xn \equiv \frac yn \pmod u \iff x \equiv y \pmod{uv}.$ For any $\rho \in \frac 1n \mathbb Z,$ the resulting equivalence class is $\bar{\rho} = \{\rho + uz : z \in \mathbb Z \}.$ We will represent the resulting quotient group as $\frac 1n \mathbb Z_{uv}.$ For example

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

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

From this point on, when we write $\frac 1v \mathbb Z_u,$ it will be assumed that $u$ and $v$ have been chosen so that

• $1 \le v \lt u$
• $v|u$ and
• The equivalence relation is $\left(\text{mod } \dfrac uv \right).$

Note that $\left|\frac 1v \mathbb Z_u \right| = u$.

We will be interested in finite abelian groups of the form $\displaystyle \mathcal G = \prod_{i=1}^m \frac{1}{b_i} \mathbb Z_{a_i}.$ We will treat the members of $\mathcal G$ as column vectors $\bar{\mathbf{g}} = \left( \overline{g_1}, \; \overline{g_2}, \; \overline{g_3}, \; \cdots, \; \overline{g_m} \right)^T.$ If we let $\mathbf a = \left( a_1, \; a_2, \; a_3, \; \cdots, \; a_m \right)^T$ and $\mathbf b = \left( b_1, \; b_2, \; b_3, \; \cdots, \; b_m \right)^T$ then we will express this as $\mathcal G = \frac{1}{\mathbf b} \mathbb Z^{m \times 1}_{\mathbf a}.$ The group $\mathcal G^n$ is also a group and it's elements can be represented by $m \times n$ matrices whose columns are members of $\mathcal G$. We will express this as $\mathcal G^n = \frac{1}{\mathbf b} \mathbb Z^{m \times n}_{\mathbf a}.$

For any $\mathbf c = \left( c_1, \; c_2, \; c_3, \; \cdots, \; c_v \right)^T \in \mathbb Z^v$, we define $$\mathbf D_{\mathbf c} = \begin{pmatrix} c_1 & 0 & 0 & \cdots & 0 \\ 0 & c_2 & 0 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & c_v \end{pmatrix}$$

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This post is available for use.

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A general Delta complex $X$ can be described combinatorially as a sequence of sets $X_0,X_1,\dots$ together with maps $d_i:X_n\to X_{n-1}$ for any $i\in[n]=\{0,\dots,n\}$ such that $d_j d_i = d_i d_{j+1}$ whenever $j\ge i$. The elements of $X_n$ are called $n$-simplices of $X$, and the maps $d_i$ are face maps of $X$.
A delta complex $X$ has a realization $|X|$ which is constructed as follows: For any $n$-simplex $\sigma$ take a standard-$n$-simplex $\Delta^n_\sigma$. Now identify its $i$-th subsimplex with the standard-$(n-1)$-simplex $\Delta^{n-1}$ corresponding to the face $d_i x$ using the linear inclusion $\Delta^{n-1} \to \Delta^n$ which preserves the ordering of the vertices. For example, you may have the delta complex $X_0=\{v\} \leftleftarrows X_1=\{e\}$, where the two arrows represent $d_0$ and $d_1$. Then you have a line $\Delta^1$ and a point $\Delta^0$, and both endpoints of the line are glued to that point, so we get a circle.

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• Definition of a non-degenerate bilinear form
• Definition of the orthogonal group $O(n,\Bbb C)$
• Suppose $f\in M_c(\Bbb C)$ is a non-singular set $$L_f=\{x\in M_n(\Bbb C):x^Tf+fx=0\}$$ Show tha $\rm{tr}(x)=0,\forall x\in L_f$
• Give the definition of $e^A$ for $A\in M_n(\Bbb C)$. Write down $e^A$ in the case $A=E_{ij}$ the $n\times n$ matrix with a $1$ in position $i,j$.
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