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Sandbox

My First Slide

Markdown+itex2MML Sandbox

Play around below. Your changes will, periodically, be rolled back.

First Page

Some examples

(1)

\mathop{min} w_h p_h + w_r p_r + w_l p_l

(2)

\left\{ \begin{aligned} \nabla \times \vec{\mathbf{B}} - \frac{1}{c}\frac{\partial\vec{\mathbf{E}}}{\partial t} &= \frac{4\pi}{c}\vec{\mathbf{j}} \\ \nabla \cdot \vec{\mathbf{E}} &= 4 \pi \rho \\ \nabla \times \vec{\mathbf{E}}+\frac{1}{c}\frac{\partial\vec{\mathbf{B}}}{\partial t} &= \vec{\mathbf{0}} \\ \nabla \cdot \vec{\mathbf{B}} &= 0 \end{aligned} \right.

Here’s an equation

(3)

{\int_{-\infty}^\infty e^{-a x^2/2} \mathrm{d}x} = \sqrt{\frac{2\pi}{a}}

which we can later refer¹ back to as (3).

Aligned equations:

(4)

\begin{aligned} a+b &= b+a \\ a+(b+c) &= (a+b)+c \end{aligned}

The Dirac equation (boxed):

\boxed{(i\slash{D}+m)\psi = 0}

Here’s the table of Clifford² algebras over $\mathbb{R}$ :

$j$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
$\mathcal{C}\ell_{j}^-$	$\mathbb{R}$	$\mathbb{C}$	$\mathbb{H}$	$\mathbb{H}\oplus\mathbb{H}$	$\mathbb{H}(2)$	$\mathbb{C}(4)$	$\mathbb{R}(8)$	$\mathbb{R}(8)\oplus\mathbb{R}(8)$	$\mathbb{R}(16)$
$\mathcal{C}\ell_{j}^+$	$\mathbb{R}$	$\mathbb{R}\oplus\mathbb{R}$	$\mathbb{R}(2)$	$\mathbb{C}(2)$	$\mathbb{H}(2)$	$\mathbb{H}(2)\oplus\mathbb{H}(2)$	$\mathbb{H}(4)$	$\mathbb{C}(8)$	$\mathbb{R}(16)$

where the generators of $\mathcal{C}\ell_{j}^\pm$ satisfy

\gamma_i\gamma_j +\gamma_j \gamma_i =\pm 2\delta_{i j}

and $\mathcal{C}\ell_{n+8}^\pm = \mathcal{C}\ell_n^\pm \otimes \mathbb{R}(16)$ .

(5)

\lim_{n \to \infty} \sum_{k=1}^n \frac{1}{k^2} = \frac{\pi^2}{6}

(6)

\mathbf{V}_{1} \times \mathbf{V}_{2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\\\ \frac{\partial X}{\partial u} & \frac{\partial Y}{\partial u} & 0 \\ \frac{\partial X}{\partial v} & \frac{\partial Y}{\partial v} & 0 \\ \end{vmatrix}

Second Page

More Examples

(7)

\nabla \times \vec{E} = - \frac {\partial \vec{B}}{\partial t}

(8)

\oint \mathbf{B}\cdot \mathrm{d}\mathbf{l} = \href{https://en.wikipedia.org/wiki/Amp%C3%A8re%27s_circuital_law}{\mu_0 I_\text{enc}}

$H^1(\mathcal{Z}, \mathcal{O}(-k))$ Let $G=(V,E)$ be a graph, with $w:V\to [0,1]$ a weight function.

(9)

\{Q_i, Q_j\} = \delta_{ij}\mathcal{H}.

Third Page

Theorems

Definition

Let $H$ be a subgroup of a group $G$ . A left coset of $H$ in $G$ is a subset of $G$ that is of the form $x H$ , where $x \in G$ and $x H = \{ x h : h \in H \}$ .

Similarly a right coset of $H$ in $G$ is a subset of $G$ that is of the form $H x$ , where $H x = \{ h x : h \in H \}$ .

Lemma

Let $H$ be a subgroup of a group $G$ , and let $x$ and $y$ be elements of $G$ . Suppose that $x H \cap y H$ is non-empty. Then $x H = y H$ .

Proof

Let $z$ be some element of $x H \cap y H$ . Then $z = x a$ for some $a \in H$ , and $z = y b$ for some $b \in H$ . If $h$ is any element of $H$ then $a h \in H$ and $a^{-1}h \in H$ , since $H$ is a subgroup of $G$ . But $z h = x(a h)$ and $xh = z(a^{-1}h)$ for all $h \in H$ . Therefore $z H \subset x H$ and $x H \subset z H$ , and thus $x H = z H$ . Similarly $y H = z H$ , and thus $x H = y H$ , as required.

Lemma

Let $H$ be a finite subgroup of a group $G$ . Then each left coset of $H$ in $G$ has the same number of elements as $H$ .

Proof

Let $H = \{ h_1, h_2,\ldots, h_m\}$ , where $h_1, h_2,\ldots, h_m$ are distinct, and let $x$ be an element of $G$ . Then the left coset $x H$ consists of the elements $x h_j$ for $j = 1,2,\ldots,m$ . Suppose that $j$ and $k$ are integers between $1$ and $m$ for which $x h_j = x h_k$ . Then $h_j = x^{-1} (x h_j) = x^{-1} (x h_k) = h_k$ , and thus $j = k$ , since $h_1, h_2,\ldots, h_m$ are distinct. It follows that the elements $x h_1, x h_2,\ldots, x h_m$ are distinct. We conclude that the subgroup $H$ and the left coset $x H$ both have $m$ elements, as required.

Theorem

(Lagrange’s Theorem). Let $G$ be a finite group, and let $H$ be a subgroup of $G$ . Then the order of $H$ divides the order of $G$ .

Proof

Each element $x$ of $G$ belongs to at least one left coset of $H$ in $G$ (namely the coset $x H$ ), and no element can belong to two distinct left cosets of $H$ in $G$ (see Lemma 1). Therefore every element of $G$ belongs to exactly one left coset of $H$ . Moreover each left coset of $H$ contains $|H|$ elements (Lemma 2). Therefore $|G| = n |H|$ , where $n$ is the number of left cosets of $H$ in $G$ . The result follows.

Corollary

Let $x$ be an element of a finite group $G$ . Then the order of $x$ divides the order of $G$ .

Theorem

Let $f : \Delta \longrightarrow \Delta,$ where $\Delta=\{z\in\mathbb{C}: \vert z \vert \lt 1\}$ , be analytic with $a \in \Delta$ . Then

\left\vert\frac{f(z)-f(a)}{1-\overline{f(a)}f(z)}\right\vert\le \left\vert\frac{z-a}{1-\overline{a}z}\right\vert

for all $\vert z \vert \le 1$ and

\frac{\vert f'(a)\vert}{1-\vert f(a)\vert^2}\le \frac{1}{1-\vert a \vert^2}.

Furthermore, equality holds iff $f$ realizes a conformal mapping of $\Delta$ onto itself.

Proof

Let $w=\frac{z-a}{1-\overline{a}z}$ and put $\phi(w)=\frac{f(z)-f(a)}{1-\overline{f(a)}f(z)}$ . Define for $\abs{b}\lt 1$ $C_b(z)=\frac{z-b}{1-\overline{b}z}.$ All conformal maps from $\Delta$ to itself, sending $b$ to $0$ , are of the form $C_b(z)e^{i\gamma}$ for $\gamma\in[0,2\pi].$ In this notation, $\phi(w)=C_{f(a)}\circ f \circ C_a^{-1}(w),$ where $C_a^{-1}$ is the inverse of $C_a$ as a function. Note that $C_a(z)$ is conformal, so it has an inverse. It is clear that $\phi(0)=C_{f(a)}\circ f \circ C_a^{-1}(0)=C_{f(a)}(f(a))=0$ . Since $C_a^{-1}: \Delta \longrightarrow \Delta$ and $f : \Delta \longrightarrow \Delta$ and $C_{f(a)}: \Delta \longrightarrow \Delta,$ then $\vert\phi(w)\vert\lt 1$ for $\vert w\vert \lt 1 .$ Applying Schwarz’s lemma, we obtain $\vert\phi(w)\vert\le \vert w \vert$ for $\vert w \vert \le 1$ . Furthermore, if equality holds, then $f(z)=e^{i\gamma'} z$ for $\gamma'\in [0,2\pi]$ . Therefore,

(10)

\left\vert\frac{f(z)-f(a)}{1-\overline{f(a)}f(z)}\right\vert\le \left\vert\frac{z-a}{1-\overline{a}z}\right\vert

for all $\vert z \vert\le 1.$ Rearranging, we obtain

\left\vert\frac{f(z)-f(a)}{z-a}\right\vert\le\left\vert\frac{1-\overline{f(a)}f(z)}{1-\overline{a}z}\right\vert.

If we take the limit as $z$ tends to $a$ , we obtain

\left\vert f'(a)\right\vert \le \left\vert\frac{1-\vert f(a)\vert ^2}{1-\vert a \vert^2}\right\vert=\frac{1-\vert f(a)\vert^2}{1-\vert a \vert^2},

\frac{\vert f'(a)\vert}{1-\vert f(a)\vert^2}\le \frac{1}{1-\vert a \vert^2}.

As said above, if equality holds in (10), then Schwarz’s lemma tells us that $\phi(w)=e^{i\gamma'}w$ . Thus, $\phi(w)=C_{f(a)}\circ f \circ C_a^{-1}(w)=e^{i\gamma'}w,$ so $f(z)=C_{f(a)}^{-1}(e^{i\gamma'}C_a(z))$ . Since $e^{i\gamma'}C_a(z)$ is conformal, $C_{f(a)}^{-1},$ the inverse function of $C_{f(a)}$ , is conformal, and a composition of conformal maps is conformal, then $f$ is a conformal map of $\Delta$ onto itself. Conversely, if $f$ is a conformal map of $\Delta$ onto itself, then $\phi(w)=C_{f(a)}\circ f \circ C_a^{-1}(w)=e^{i\gamma}C_b(w),$ since a composition of conformal maps is conformal and because all conformal maps from $\Delta$ onto itself are of the form $e^{i\gamma}C_b(w).$ We also know that $\phi(0)=0,$ so $b=0$ . Therefore,

\phi(w)=e^{i\gamma}C_0(w)=e^{i\gamma}w \Leftrightarrow \vert\phi(w)\vert=\vert w \vert \Leftrightarrow \left\vert\frac{f(z)-f(a)}{1-\overline{f(a)}f(z)}\right\vert=\left\vert\frac{z-a}{1-\overline{a}z}\right\vert

for all $\vert z \vert\le 1$ . In sum, equality holds in (10) iff $f$ is a conformal map from $\Delta$ to itself.

Remark

Someone needs to code \abs, i.e. \left\vert # \right\vert. This is terribly annoying. Also, when using $$, why must one enter a line before and after.

SVG graphics, created in SVG-Edit:

E_\Sigma

A_0

$K^0\overline{K}^0$ Mixing

Yet More examples

SVG:

Animated SVG

Complicated commutative diagrams (equations in SVG)

SVG in equations.

In $SU(3)$ , $\begin{svg} <svg xmlns="http://www.w3.org/2000/svg" width="30" height="16" viewBox="0 0 30 16"> <desc>Rank-2 Symmetric Tensor Representation</desc> <g transform="translate(5,5)" fill="#FCC" stroke="#000" stroke-width="2"> <rect width="10" height="10"/> <rect width="10" height="10" x="10"/> </g> </svg> \end{svg}\includegraphics[width=2em]{young1} \otimes \begin{svg} <svg xmlns="http://www.w3.org/2000/svg" width="20" height="16" viewBox="0 0 20 16"> <desc>Fundamental Representation</desc> <g transform="translate(5,5)" fill="#FCC" stroke="#000" stroke-width="2"> <rect width="10" height="10"/> </g> </svg> \end{svg}\includegraphics[width=1em]{young2} = \begin{svg} <svg xmlns="http://www.w3.org/2000/svg" width="30" height="26" viewBox="0 0 30 26"> <desc>Adjoint Representation</desc> <g transform="translate(5,5)" fill="#FCC" stroke="#000" stroke-width="2"> <rect width="10" height="10"/> <rect width="10" height="10" x="10"/> <rect width="10" height="10" y="10"/> </g> </svg> \end{svg}\includegraphics[width=2em]{young3} \oplus \begin{svg} <svg xmlns="http://www.w3.org/2000/svg" width="40" height="16" viewBox="0 0 40 16"> <desc>Rank-3 Symmetric Tensor Representation</desc> <g transform="translate(5,5)" fill="#FCC" stroke="#000" stroke-width="2"> <rect width="10" height="10"/> <rect width="10" height="10" x="10"/> <rect width="10" height="10" x="20"/> </g> </svg> \end{svg}\includegraphics[width=3em]{young4}$ .

Cases:

r_{a+1} = \begin{cases} 0 & \text{with prob.}\quad \exp(-\theta r_a) \\ \max \lbrace \delta r_a, z \rbrace & \text{with prob.}\quad 1 - \exp(-\theta r_a) \end{cases}

Stretchy Brackets:

q_a(z) = \sigma_a^{-1} \exp{\left[ -\frac{\gamma + z}{\sigma_a} \right]}

Linearity of Quadrature Rules

\sum_{i = 1}^N {\left( {\alpha f(x_i ) + \beta g(x_i )} \right)w_i } = \alpha \sum_{i = 1}^N {f(x_i )w_i } + \beta \sum_{i = 1}^N {g(x_i )w_i }

Linearity of Integrals

{\int_a^b {\left( {\alpha f(x)\, + \beta g(x)} \right)dx = } \alpha \int_a^b {f(x)\,dx} + \beta \int_a^b {g(x)\,dx} }

Can we talk about $x_i^2$ inline? What about $\int_a^b x^2\,dx$ ? Inline fractions $\frac{x-x_2}{x_1-x_2}$ ?
Big fractions

p_3 (x) = \left( {\frac{1}{2}} \right)\frac{{\left( {x - \frac{1}{2}} \right)\left( {x - \frac{3}{4}} \right)\left( {x - 1} \right)}}{{\left( {\frac{1}{4} - \frac{1}{2}} \right)\left( {\frac{1}{4} - \frac{3}{4}} \right)\left( {\frac{1}{4} - 1} \right)}} + \left( {\frac{1}{2}} \right)\frac{{\left( {x - \frac{1}{2}} \right)\left( {x - \frac{3}{4}} \right)\left( {x - 1} \right)}}{{\left( {\frac{1}{4} - \frac{1}{2}} \right)\left( {\frac{1}{4} - \frac{3}{4}} \right)\left( {\frac{1}{4} - 1} \right)}} + \left( {\frac{1}{2}} \right)\frac{{\left( {x - \frac{1}{2}} \right)\left( {x - \frac{3}{4}} \right)\left( {x - 1} \right)}}{{\left( {\frac{1}{4} - \frac{1}{2}} \right)\left( {\frac{1}{4} - \frac{3}{4}} \right)\left( {\frac{1}{4} - 1} \right)}} + \left( {\frac{1}{2}} \right)\frac{{\left( {x - \frac{1}{2}} \right)\left( {x - \frac{3}{4}} \right)\left( {x - 1} \right)}}{{\left( {\frac{1}{4} - \frac{1}{2}} \right)\left( {\frac{1}{4} - \frac{3}{4}} \right)\left( {\frac{1}{4} - 1} \right)}}

Diagram

\begin{matrix} P_1(Y) &\to& P_1(X) \\ \downarrow &\Downarrow\mathrlap{\sim}& \downarrow \\ T' &\to& T \end{matrix}

\mathcal{} versus \mathscr{}
$\begin{gathered} \mathcal{ABCDEFGHIJKLMNOPQRSTUVWXYZ}\\ \text{versus}\\ \mathscr{ABCDEFGHIJKLMNOPQRSTUVWXYZ} \end{gathered}$

(11)

v_1

“This is my text”, says Anymouse.

Ruby code example:

    class Person
      attr_reader :name, :age
      def initialize(name, age)
        @name, @age = name, age
      end
      def <=>(person) # Comparison operator for sorting
        @age <=> person.age
      end
      def to_s
        "#@name (#@age)"
      end
    end
 
    group = [
      Person.new("Bob", 33), 
      Person.new("Chris", 16), 
      Person.new("Ash", 23) 
    ]
 
    puts group.sort.reverse

A Python example:

ListOfStrings(
              title = _("A List of some strings"),
              help  = _("A List of strings"),
              orientation = "horizontal"
              )

Tikz Pictures

A new Tikz feature to play around with. Requires an~~feature?~~ ~~to play around with. Requires an~~ additional install (in addition to Instiki) to render the Tikz code into SVG.

Here’s an example of some Tikz code

\begin{tikzpicture}[decoration={markings,
mark=at position .5 with {\arrow{>}}}]
\usetikzlibrary{arrows,shapes,decorations.markings}
\begin{scope}[scale=2.0]
\node[Bl,scale=.75] (or1) at (8,3) {};
\node[scale=1] at (8.7,2.9) {$D3$ brane};
\node[draw,diamond,fill=yellow,scale=.3] (A1) at (7,0) {}; 
\draw[dashed] (A1) -- (7,-.7);
\node[draw,diamond,fill=yellow,scale=.3] (A2) at (7.5,0) {}; 
\draw[dashed] (A2) -- (7.5,-.7);
\node[draw,diamond,fill=yellow,scale=.3] (A3) at (8,0) {}; 
\draw[dashed] (A3) -- (8,-.7);
\node[draw,diamond,fill=yellow,scale=.3] (A4) at (8.5,0) {}; 
\draw[dashed] (A4) -- (8.5,-.7);
\node[draw,diamond,fill=yellow,scale=.3] (A5) at (9,0) {};
\draw[dashed] (A5) -- (9,-.7);
\node[draw,circle,fill=aqua,scale=.3] (B) at (9.5,0) {};
\draw[dashed] (B) -- (9.5,-.7);
\node[draw,regular polygon,regular polygon sides=5,fill=purple,scale=.3] (C1) at (10,0) {}; 
\draw[dashed] (C1) -- (10,-.7);
\node[draw,regular polygon,regular polygon sides=5,fill=purple,scale=.3] (C2) at (10.5,0) {};
\draw[dashed] (C2) -- (10.5,-.7);
\draw (6.8,-.7) -- (6.8,-.9) to (9.2,-.9) to (9.2,-.7);
\draw (9.8,-.7) -- (9.8,-.9) to (10.7,-.9) to (10.7,-.7);
\draw[->-=.75] (C2) to (10.2,.35);
\draw[->-=.75] (C1) to (10.2,.35);
\node[scale=.6] at (9.9,.35) {$(2,2)$};
\draw[->-=.7] (B) to (9.6,.7);
\draw (10.2,.35) to (9.6,.7);
\node[scale=.6] at (9.35,.9) {$(4,0)$};
\draw[->-=.5] (9.1,.8) to (A5);
\draw (9.6,.7) to (9.1,.8) to (A5);
\draw (9.1,.8) to [out=170,in=280] (8.3,1.45);
\draw[dashed] (8.3,1.45) to (8.1,2.5);
\draw[->-=.5] (8.1,2.5) to (or1);
\node[scale=.75] at (7.7,2.7) {$(3,0)$};
%\draw (11.4,2.4) to [out=180,in=90] (6.2,-.5) to [out=90,in=0] (or1) -- cycle;
\node[scale=.75] at (8,-1.1) {A-type};
\node[scale=.75] at (9.5,-1.1) {B-type};
\node[scale=.75] at (10.25,-1.1) {C-type};
\draw[dashed] (8.7,.6) to [out=180,in=90] (6.2,-.55) to [out=270,in=180] (8.7,-1.6) to [out=0,in=270] (11.2,-.55) to [out=90,in=0] (8.7,.6) -- cycle;
\node[scale=1] at (12,.6) {$E_6$ singularity};
\end{scope}
\end{tikzpicture}

which, when processed, produces

\begin{tikzpicture}[decoration={markings, mark=at position .5 with {\arrow{>}}}] \usetikzlibrary{arrows,shapes,decorations.markings} \begin{scope}[scale=2.0] \nodeBl,scale=.75 at (8,3) {}; \node[scale=1] at (8.7,2.9) { $D3$ brane}; \nodedraw,diamond,fill=yellow,scale=.3 at (7,0) {}; \drawdashed – (7,-.7); \nodedraw,diamond,fill=yellow,scale=.3 at (7.5,0) {}; \drawdashed – (7.5,-.7); \nodedraw,diamond,fill=yellow,scale=.3 at (8,0) {}; \drawdashed – (8,-.7); \nodedraw,diamond,fill=yellow,scale=.3 at (8.5,0) {}; \drawdashed – (8.5,-.7); \nodedraw,diamond,fill=yellow,scale=.3 at (9,0) {}; \drawdashed – (9,-.7); \nodedraw,circle,fill=aqua,scale=.3 at (9.5,0) {}; \drawdashed – (9.5,-.7); \nodedraw,regular polygon,regular polygon sides=5,fill=purple,scale=.3 at (10,0) {}; \drawdashed – (10,-.7); \nodedraw,regular polygon,regular polygon sides=5,fill=purple,scale=.3 at (10.5,0) {}; \drawdashed – (10.5,-.7); \draw (6.8,-.7) – (6.8,-.9) to (9.2,-.9) to (9.2,-.7); \draw (9.8,-.7) – (9.8,-.9) to (10.7,-.9) to (10.7,-.7); \draw->-=.75 to (10.2,.35); \draw->-=.75 to (10.2,.35); \node[scale=.6] at (9.9,.35) { $(2,2)$ }; \draw->-=.7 to (9.6,.7); \draw (10.2,.35) to (9.6,.7); \node[scale=.6] at (9.35,.9) { $(4,0)$ }; \draw->-=.5 to (A5); \draw (9.6,.7) to (9.1,.8) to (A5); \draw (9.1,.8) to out=170,in=280; \drawdashed to (8.1,2.5); \draw->-=.5 to (or1); \node[scale=.75] at (7.7,2.7) { $(3,0)$ }; %\draw (11.4,2.4) to out=180,in=90 to out=90,in=0 – cycle; \node[scale=.75] at (8,-1.1) {A-type}; \node[scale=.75] at (9.5,-1.1) {B-type}; \node[scale=.75] at (10.25,-1.1) {C-type}; \drawdashed to out=180,in=90 to out=270,in=180 to out=0,in=270 to out=90,in=0 – cycle; \node[scale=1] at (12,.6) { $E_6$ singularity}; \end{scope} \end{tikzpicture}

~~And two more:~~

\begin{tikzpicture} \useasboundingbox (-4,-3) rectangle (4,4.75); \drawthick – (0,4) – ( 3.46410,-2) – (-3.46410,-2); \drawthick circle (2); \drawthick – (0,-2); \drawthick – ( 1.73205,1); \drawthick – (-1.73205,1); \filldraw (-1.73205,1) circle (3pt) node[anchor=south east, scale=1.5] { $e_1$ }; \filldraw ( 1.73205,1) circle (3pt) node[anchor=south west, scale=1.5] { $e_2$ }; \filldraw (0,-2) circle (3pt) node[anchor=north, scale=1.5] { $e_3$ }; \filldraw (0,0) circle (3pt) node[anchor=west, xshift=7pt, scale=1.5] { $e_4$ }; \filldraw (3.46410,-2) circle (3pt) node[anchor=north, scale=1.5] { $e_5$ }; \filldraw (-3.46410,-2) circle (3pt) node[anchor=north, scale=1.5] { $e_6$ }; \filldraw (0,4) circle (3pt) node[anchor=south, scale=1.5] { $e_7$ }; \end{tikzpicture}

\begin{tikzpicture}[every tqft/.style={fill=orange,fill opacity=0.25}, tqft/every boundary component/.style={fill=yellow,fill opacity=.5}, tqft/every upper boundary component/.style={draw,thin,blue}, tqft/every lower boundary component/.style={draw,dashed,thin,blue}] \usetikzlibrary{tqft} \begin{scope}[tqft/every incoming boundary component/.style={fill=green, fill opacity= .25}] \pic[tqft/pair of pants,draw,name=a,at={(0,0)}]; \pic[tqft/reverse pair of pants,draw,anchor=incoming boundary 2,name=e,at={(8,0)}]; \end{scope} \begin{scope}[tqft/every outgoing boundary component/.style={draw,thin,blue,fill=yellow}] \pic[tqft/reverse pair of pants,draw,anchor=incoming boundary 1,name=b,at=(a-outgoing boundary 2)]; \end{scope} \begin{scope}[tqft/every incoming boundary component/.style={fill=green, fill opacity= .25}] \pic[tqft/cylinder,draw,anchor=outgoing boundary 1,name=d,at=(b-incoming boundary 2)]; \end{scope} \path (b-incoming boundary 2) ++(1.5,0) node[font=\Huge] {(=)}; \begin{scope}[tqft/every outgoing boundary component/.style={draw,thin,blue,fill=yellow}] \pic[tqft/cylinder,draw,anchor=incoming boundary 1,name=c,at=(a-outgoing boundary 1)]; \pic[tqft/pair of pants,draw,anchor=incoming boundary 1,name=f,at=(e-outgoing boundary 1)]; \end{scope} \end{tikzpicture}

~~\begin{tikzpicture}\definecolor{mycolor}{RGB}{255,51,76}\drawcolor=mycolor circle (4ex);\end{tikzpicture}~~

You can also refer to it as (3). Chacun à son goût!. ↩

For more information, see Wikipedia. ↩

Revision from March 18, 2025 16:13:03 by Yu

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MAN Wiki Sandbox (Rev #2, changes)

Sandbox

My First Slide

Markdown+itex2MML Sandbox

First Page

Some examples

Second Page

More Examples

Third Page

Theorems

Definition

Lemma

Proof

Lemma

Proof

Theorem

Proof

Corollary

Theorem

Proof

Remark

SVG graphics, created in SVG-Edit:

Yet More examples

Tikz Pictures

MAN Wiki
Sandbox (Rev #2, changes)