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Levi-Civita symbol: 3D matrix

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Levi-Civita symbol: 3D matrix


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20















During the current review of the tensors I have arrived at a page of Wikipedia where you can see the symbol of Levi-Civita in a beautiful three-dimensional matrix.



enter image description here



I hope that nobody will be angry with me if I do not produce any MWE but for me it would be nice to see the construction of a matrix so made and can be made available to other users.










share|improve this question


















  • 2





    Nobody will be angry, don't worry. If I hadn't reached my 40-vote limit, you would have got my +1. Your question is very interesting (and hard).

    – JouleV
    yesterday













  • @JouleV LOL LOL when I thought about what to write, I sincerely thought about you :-). Isn't it that you write to me that I'm 10k and I don't produce even a minimum MWE?

    – Sebastiano
    yesterday






  • 1





    Well, I was thinking about how to solve this and I read that word ;-) lol I knew I had to write something :D

    – JouleV
    yesterday
















20















During the current review of the tensors I have arrived at a page of Wikipedia where you can see the symbol of Levi-Civita in a beautiful three-dimensional matrix.



enter image description here



I hope that nobody will be angry with me if I do not produce any MWE but for me it would be nice to see the construction of a matrix so made and can be made available to other users.










share|improve this question


















  • 2





    Nobody will be angry, don't worry. If I hadn't reached my 40-vote limit, you would have got my +1. Your question is very interesting (and hard).

    – JouleV
    yesterday













  • @JouleV LOL LOL when I thought about what to write, I sincerely thought about you :-). Isn't it that you write to me that I'm 10k and I don't produce even a minimum MWE?

    – Sebastiano
    yesterday






  • 1





    Well, I was thinking about how to solve this and I read that word ;-) lol I knew I had to write something :D

    – JouleV
    yesterday














20












20








20


3






During the current review of the tensors I have arrived at a page of Wikipedia where you can see the symbol of Levi-Civita in a beautiful three-dimensional matrix.



enter image description here



I hope that nobody will be angry with me if I do not produce any MWE but for me it would be nice to see the construction of a matrix so made and can be made available to other users.










share|improve this question














During the current review of the tensors I have arrived at a page of Wikipedia where you can see the symbol of Levi-Civita in a beautiful three-dimensional matrix.



enter image description here



I hope that nobody will be angry with me if I do not produce any MWE but for me it would be nice to see the construction of a matrix so made and can be made available to other users.







matrices 3d tikz-matrix






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked yesterday









SebastianoSebastiano

10.5k42062




10.5k42062








  • 2





    Nobody will be angry, don't worry. If I hadn't reached my 40-vote limit, you would have got my +1. Your question is very interesting (and hard).

    – JouleV
    yesterday













  • @JouleV LOL LOL when I thought about what to write, I sincerely thought about you :-). Isn't it that you write to me that I'm 10k and I don't produce even a minimum MWE?

    – Sebastiano
    yesterday






  • 1





    Well, I was thinking about how to solve this and I read that word ;-) lol I knew I had to write something :D

    – JouleV
    yesterday














  • 2





    Nobody will be angry, don't worry. If I hadn't reached my 40-vote limit, you would have got my +1. Your question is very interesting (and hard).

    – JouleV
    yesterday













  • @JouleV LOL LOL when I thought about what to write, I sincerely thought about you :-). Isn't it that you write to me that I'm 10k and I don't produce even a minimum MWE?

    – Sebastiano
    yesterday






  • 1





    Well, I was thinking about how to solve this and I read that word ;-) lol I knew I had to write something :D

    – JouleV
    yesterday








2




2





Nobody will be angry, don't worry. If I hadn't reached my 40-vote limit, you would have got my +1. Your question is very interesting (and hard).

– JouleV
yesterday







Nobody will be angry, don't worry. If I hadn't reached my 40-vote limit, you would have got my +1. Your question is very interesting (and hard).

– JouleV
yesterday















@JouleV LOL LOL when I thought about what to write, I sincerely thought about you :-). Isn't it that you write to me that I'm 10k and I don't produce even a minimum MWE?

– Sebastiano
yesterday





@JouleV LOL LOL when I thought about what to write, I sincerely thought about you :-). Isn't it that you write to me that I'm 10k and I don't produce even a minimum MWE?

– Sebastiano
yesterday




1




1





Well, I was thinking about how to solve this and I read that word ;-) lol I knew I had to write something :D

– JouleV
yesterday





Well, I was thinking about how to solve this and I read that word ;-) lol I knew I had to write something :D

– JouleV
yesterday










2 Answers
2






active

oldest

votes


















23














Something like that?



documentclass[tikz,border=3.14mm]{standalone}
usepackage{mathtools}
usetikzlibrary{matrix,backgrounds,3d}
usepackage{tikz-3dplot}
%definecolor{mygreen}{RGB}{12,252,12}
begin{document}
tdplotsetmaincoords{75}{20}
begin{tikzpicture}[tdplot_main_coords]
begin{scope}[canvas is xz plane at y=1,transform shape]
node[inner sep=0pt,text=green!70!black,opacity=0.8] (mat1)
{$displaystylebegin{pmatrix*}[r]
0 & 1 & 0 \
-1 & 0 & 0 \
0 & 0 & 0 \
end{pmatrix*}$};
begin{scope}[on background layer]
fill[green!70!black,opacity=0.2] ([xshift=8.5pt]mat1.south west)
coordinate (blb) to[out=140,in=-140,looseness=0.7]
([xshift=8.5pt]mat1.north west) coordinate (tlb) --
([xshift=-8.5pt]mat1.north east) coordinate (trb)
to[out=-40,in=40,looseness=0.7] ([xshift=-8.5pt]mat1.south east)
coordinate (brb)
-- cycle;
end{scope}
end{scope}
%
begin{scope}[canvas is xz plane at y=0,transform shape]
node[inner sep=0pt,text=red,opacity=0.8] (mat2) {$displaystyle
begin{pmatrix*}[r]
0 & 0 & -1 \
0 & 0 & 0 \
1 & 0 & 0 \
end{pmatrix*}$};
begin{scope}[on background layer]
fill[red,opacity=0.2] ([xshift=8.5pt]mat2.south west) to[out=140,in=-140,looseness=0.7]
([xshift=8.5pt]mat2.north west) -- ([xshift=-8.5pt]mat2.north east)
to[out=-40,in=40,looseness=0.7] ([xshift=-8.5pt]mat2.south east) -- cycle;
end{scope}
end{scope}
%
begin{scope}[canvas is xz plane at y=-1,transform shape]
node[inner sep=0pt,text=blue,opacity=0.8] (mat3) {$displaystyle
begin{pmatrix*}[r]
0 & 0 & 0 \
0 & 0 & 1 \
0 & -1 & 0 \
end{pmatrix*}$};
begin{scope}[on background layer]
fill[blue,opacity=0.2]
([xshift=8.5pt]mat3.south west) coordinate (blf)
to[out=140,in=-140,looseness=0.7]
([xshift=8.5pt]mat3.north west) coordinate (tlf)
-- ([xshift=-8.5pt]mat3.north east) coordinate (trf)
to[out=-40,in=40,looseness=0.7] ([xshift=-8.5pt]mat3.south east)
coordinate (brf) -- cycle;
end{scope}
end{scope}
foreach X in {tl,tr,br}
{draw[thin,orange] (X f) -- (X b);}
begin{scope}[on background layer]
draw[thin,orange] (blf) -- (blb);
end{scope}
node[left] at (mat3.west) {$varepsilon_{ijk}=$};
end{tikzpicture}
end{document}


enter image description here



EDIT: Aligned the entries right, big thanks to Barbara Beeton. (I just wonder why no one complained that the Levi-Civita tensor is not a tensor, but a tensor density. ;-)



2nd EDIT: Response to Anush's comment (well-taken! ;-).



documentclass[tikz,border=3.14mm]{standalone}
usepackage{mathtools}
usetikzlibrary{matrix,backgrounds,3d}
usepackage{tikz-3dplot}
begin{document}
tdplotsetmaincoords{75}{20}
begin{tikzpicture}[tdplot_main_coords]
begin{scope}[canvas is xz plane at y=1,transform shape]
node[inner sep=0pt,text=green!70!black,opacity=0.8] (mat1)
{$displaystylebegin{pmatrix*}[r]
0 & hphantom{-}1 & hphantom{-}0 \
-1 & 0 & 0 \
0 & 0 & 0 \
end{pmatrix*}$};
begin{scope}[on background layer]
fill[green!70!black,opacity=0.2] ([xshift=8.5pt]mat1.south west)
coordinate (blb) to[out=140,in=-140,looseness=0.7]
([xshift=8.5pt]mat1.north west) coordinate (tlb) --
([xshift=-8.5pt]mat1.north east) coordinate (trb)
to[out=-40,in=40,looseness=0.7] ([xshift=-8.5pt]mat1.south east)
coordinate (brb)
-- cycle;
end{scope}
end{scope}
%
begin{scope}[canvas is xz plane at y=0,transform shape]
node[inner sep=0pt,text=red,opacity=0.8] (mat2) {$displaystyle
begin{pmatrix*}[r]
hphantom{-}0 & hphantom{-}0 & -1 \
0 & 0 & 0 \
1 & 0 & 0 \
end{pmatrix*}$};
begin{scope}[on background layer]
fill[red,opacity=0.2] ([xshift=8.5pt]mat2.south west) to[out=140,in=-140,looseness=0.7]
([xshift=8.5pt]mat2.north west) -- ([xshift=-8.5pt]mat2.north east)
to[out=-40,in=40,looseness=0.7] ([xshift=-8.5pt]mat2.south east) -- cycle;
end{scope}
end{scope}
%
begin{scope}[canvas is xz plane at y=-1,transform shape]
node[inner sep=0pt,text=blue,opacity=0.8] (mat3) {$displaystyle
begin{pmatrix*}[r]
hphantom{-}0 & 0 & hphantom{-}0 \
0 & 0 & 1 \
0 & -1 & 0 \
end{pmatrix*}$};
begin{scope}[on background layer]
fill[blue,opacity=0.2]
([xshift=8.5pt]mat3.south west) coordinate (blf)
to[out=140,in=-140,looseness=0.7]
([xshift=8.5pt]mat3.north west) coordinate (tlf)
-- ([xshift=-8.5pt]mat3.north east) coordinate (trf)
to[out=-40,in=40,looseness=0.7] ([xshift=-8.5pt]mat3.south east)
coordinate (brf) -- cycle;
end{scope}
end{scope}
foreach X in {tl,tr,br}
{draw[thin,orange] (X f) -- (X b);}
begin{scope}[on background layer]
draw[thin,orange] (blf) -- (blb);
end{scope}
begin{scope}[canvas is xz plane at y=0,transform shape]
node[left] at (mat2.west -| mat3.west) {$varepsilon_{ijk}=$};
end{scope}
end{tikzpicture}
end{document}


enter image description here






share|improve this answer


























  • Two main differences with the image in the question: (1) in the negative entries, the digits are not aligned (and the spacing between columns adjusted to make them visually uniform), and (2) the tops and bottoms of the parentheses are not connected. of these, (1) detracts more from the appearance, although the meaning isn't affected; I happen to velue graceful appearance.

    – barbara beeton
    yesterday











  • @barbarabeeton Thanks! (1) can be addressed in a very simple way: loading mathtools and using begin{pmatrix*}[r]. (2) I do not understand. In Sebastiano's screen shot there are these four lines. Of course, if you'd ask me what they are good for, I'd admit that this is a very good question. ;-) Will revise my answer to address (1), thanks again!

    – marmot
    yesterday











  • The $-1$ in the front bottom middle doesn’t look aligned with the $0$ and $1$ in the othe two arrays. In the OPs example they are all nicely lined up.

    – Anush
    22 hours ago






  • 1





    @Anush Yes, but this a question of what the OP wants. On can definitely convince LaTeX to typeset the matrices in the way you suggest. I was using some standard routine because it produces some standard output the community seems to have agreed upon. One reason why I wrote the solution in this way is that anyone can adjust the matrices without knowing anything about TikZ, and also because this way one can use orthographic projections, which we cannot subject a tikz matrix to (at least not in straightforward way). I guess Wikipedia would have used orthographic projections if they could.

    – marmot
    13 hours ago






  • 1





    @marmot Yes. I have to say your solution is so lovely I am trying to think of an excuse to use it now!

    – Anush
    13 hours ago



















21














More or less:



documentclass[tikz,border=2mm]{standalone} 
usetikzlibrary{positioning, matrix}
usepackage{amsmath}

newcommand{arrayfilling}[2]{
fill[#2!30, opacity=.5] ([shift={(1mm,1mm)}]#1.north west) coordinate(#1auxnw)--([shift={(1mm,1mm)}]#1.north east)coordinate(#1auxne) to[out=-75, in=75] ([shift={(1mm,-1mm)}]#1.south east)coordinate(#1auxse)--([shift={(1mm,-1mm)}]#1.south west)coordinate(#1auxsw) to[out=105, in=-105] cycle;
fill[#2!80!black, opacity=1] (#1auxne) to[out=-75, in=75] (#1auxse) to[out=78, in=-78] cycle;
fill[#2!80!black, opacity=1] (#1auxnw) to[out=-105, in=105] (#1auxsw) to[out=102, in=-102] cycle;
}


begin{document}
begin{tikzpicture}[font=ttfamily,
mymatrix/.style={
matrix of math nodes, inner sep=0pt, color=#1,
column sep=-pgflinewidth, row sep=-pgflinewidth, anchor=south west,
nodes={anchor=center, minimum width=5mm,
minimum height=3mm, outer sep=0pt, inner sep=0pt,
text width=5mm, align=right,
draw=none, font=small},
}
]

matrix (C) [mymatrix=green] at (6mm,5mm)
{0 & 1 & 0 \ -1 & 0 & 0\ 0 & 0 & 0\};
arrayfilling{C}{green}

matrix (B) [mymatrix=red] at (3mm,2.5mm)
{0 & 0 & -1 \ 0 & 0 & 0\ 1 & 0 & 0\};
arrayfilling{B}{red}

matrix (A) [mymatrix=blue] at (0,0)
{0 & 0 & 0 \ 0 & 0 & 1\ 0 & -1 & 0\};
arrayfilling{A}{blue}

foreach i in {auxnw, auxne, auxse, auxsw}
draw[brown, ultra thin] (Ai)--(Ci);

node[below left=-1mm and 5mm of B.west] {$epsilon_{ijk} =$};
end{tikzpicture}
end{document}


enter image description here






share|improve this answer


























  • Please see point (1) in my comment to the answer by @marmot.

    – barbara beeton
    yesterday











  • @barbarabeeton Is it better now?

    – Ignasi
    yesterday











  • Yes, better. one might quibble that there is more space than necessary, compared with the space at the right. I think it's best to pay attention to what is actually present in each column, but that does require more attention.

    – barbara beeton
    yesterday











  • In the same way as the comment that I have added to marmot I can not establish, given that they are two codes at the same level, which is the best. You have been very good. Thank you very much.

    – Sebastiano
    9 hours ago











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









23














Something like that?



documentclass[tikz,border=3.14mm]{standalone}
usepackage{mathtools}
usetikzlibrary{matrix,backgrounds,3d}
usepackage{tikz-3dplot}
%definecolor{mygreen}{RGB}{12,252,12}
begin{document}
tdplotsetmaincoords{75}{20}
begin{tikzpicture}[tdplot_main_coords]
begin{scope}[canvas is xz plane at y=1,transform shape]
node[inner sep=0pt,text=green!70!black,opacity=0.8] (mat1)
{$displaystylebegin{pmatrix*}[r]
0 & 1 & 0 \
-1 & 0 & 0 \
0 & 0 & 0 \
end{pmatrix*}$};
begin{scope}[on background layer]
fill[green!70!black,opacity=0.2] ([xshift=8.5pt]mat1.south west)
coordinate (blb) to[out=140,in=-140,looseness=0.7]
([xshift=8.5pt]mat1.north west) coordinate (tlb) --
([xshift=-8.5pt]mat1.north east) coordinate (trb)
to[out=-40,in=40,looseness=0.7] ([xshift=-8.5pt]mat1.south east)
coordinate (brb)
-- cycle;
end{scope}
end{scope}
%
begin{scope}[canvas is xz plane at y=0,transform shape]
node[inner sep=0pt,text=red,opacity=0.8] (mat2) {$displaystyle
begin{pmatrix*}[r]
0 & 0 & -1 \
0 & 0 & 0 \
1 & 0 & 0 \
end{pmatrix*}$};
begin{scope}[on background layer]
fill[red,opacity=0.2] ([xshift=8.5pt]mat2.south west) to[out=140,in=-140,looseness=0.7]
([xshift=8.5pt]mat2.north west) -- ([xshift=-8.5pt]mat2.north east)
to[out=-40,in=40,looseness=0.7] ([xshift=-8.5pt]mat2.south east) -- cycle;
end{scope}
end{scope}
%
begin{scope}[canvas is xz plane at y=-1,transform shape]
node[inner sep=0pt,text=blue,opacity=0.8] (mat3) {$displaystyle
begin{pmatrix*}[r]
0 & 0 & 0 \
0 & 0 & 1 \
0 & -1 & 0 \
end{pmatrix*}$};
begin{scope}[on background layer]
fill[blue,opacity=0.2]
([xshift=8.5pt]mat3.south west) coordinate (blf)
to[out=140,in=-140,looseness=0.7]
([xshift=8.5pt]mat3.north west) coordinate (tlf)
-- ([xshift=-8.5pt]mat3.north east) coordinate (trf)
to[out=-40,in=40,looseness=0.7] ([xshift=-8.5pt]mat3.south east)
coordinate (brf) -- cycle;
end{scope}
end{scope}
foreach X in {tl,tr,br}
{draw[thin,orange] (X f) -- (X b);}
begin{scope}[on background layer]
draw[thin,orange] (blf) -- (blb);
end{scope}
node[left] at (mat3.west) {$varepsilon_{ijk}=$};
end{tikzpicture}
end{document}


enter image description here



EDIT: Aligned the entries right, big thanks to Barbara Beeton. (I just wonder why no one complained that the Levi-Civita tensor is not a tensor, but a tensor density. ;-)



2nd EDIT: Response to Anush's comment (well-taken! ;-).



documentclass[tikz,border=3.14mm]{standalone}
usepackage{mathtools}
usetikzlibrary{matrix,backgrounds,3d}
usepackage{tikz-3dplot}
begin{document}
tdplotsetmaincoords{75}{20}
begin{tikzpicture}[tdplot_main_coords]
begin{scope}[canvas is xz plane at y=1,transform shape]
node[inner sep=0pt,text=green!70!black,opacity=0.8] (mat1)
{$displaystylebegin{pmatrix*}[r]
0 & hphantom{-}1 & hphantom{-}0 \
-1 & 0 & 0 \
0 & 0 & 0 \
end{pmatrix*}$};
begin{scope}[on background layer]
fill[green!70!black,opacity=0.2] ([xshift=8.5pt]mat1.south west)
coordinate (blb) to[out=140,in=-140,looseness=0.7]
([xshift=8.5pt]mat1.north west) coordinate (tlb) --
([xshift=-8.5pt]mat1.north east) coordinate (trb)
to[out=-40,in=40,looseness=0.7] ([xshift=-8.5pt]mat1.south east)
coordinate (brb)
-- cycle;
end{scope}
end{scope}
%
begin{scope}[canvas is xz plane at y=0,transform shape]
node[inner sep=0pt,text=red,opacity=0.8] (mat2) {$displaystyle
begin{pmatrix*}[r]
hphantom{-}0 & hphantom{-}0 & -1 \
0 & 0 & 0 \
1 & 0 & 0 \
end{pmatrix*}$};
begin{scope}[on background layer]
fill[red,opacity=0.2] ([xshift=8.5pt]mat2.south west) to[out=140,in=-140,looseness=0.7]
([xshift=8.5pt]mat2.north west) -- ([xshift=-8.5pt]mat2.north east)
to[out=-40,in=40,looseness=0.7] ([xshift=-8.5pt]mat2.south east) -- cycle;
end{scope}
end{scope}
%
begin{scope}[canvas is xz plane at y=-1,transform shape]
node[inner sep=0pt,text=blue,opacity=0.8] (mat3) {$displaystyle
begin{pmatrix*}[r]
hphantom{-}0 & 0 & hphantom{-}0 \
0 & 0 & 1 \
0 & -1 & 0 \
end{pmatrix*}$};
begin{scope}[on background layer]
fill[blue,opacity=0.2]
([xshift=8.5pt]mat3.south west) coordinate (blf)
to[out=140,in=-140,looseness=0.7]
([xshift=8.5pt]mat3.north west) coordinate (tlf)
-- ([xshift=-8.5pt]mat3.north east) coordinate (trf)
to[out=-40,in=40,looseness=0.7] ([xshift=-8.5pt]mat3.south east)
coordinate (brf) -- cycle;
end{scope}
end{scope}
foreach X in {tl,tr,br}
{draw[thin,orange] (X f) -- (X b);}
begin{scope}[on background layer]
draw[thin,orange] (blf) -- (blb);
end{scope}
begin{scope}[canvas is xz plane at y=0,transform shape]
node[left] at (mat2.west -| mat3.west) {$varepsilon_{ijk}=$};
end{scope}
end{tikzpicture}
end{document}


enter image description here






share|improve this answer


























  • Two main differences with the image in the question: (1) in the negative entries, the digits are not aligned (and the spacing between columns adjusted to make them visually uniform), and (2) the tops and bottoms of the parentheses are not connected. of these, (1) detracts more from the appearance, although the meaning isn't affected; I happen to velue graceful appearance.

    – barbara beeton
    yesterday











  • @barbarabeeton Thanks! (1) can be addressed in a very simple way: loading mathtools and using begin{pmatrix*}[r]. (2) I do not understand. In Sebastiano's screen shot there are these four lines. Of course, if you'd ask me what they are good for, I'd admit that this is a very good question. ;-) Will revise my answer to address (1), thanks again!

    – marmot
    yesterday











  • The $-1$ in the front bottom middle doesn’t look aligned with the $0$ and $1$ in the othe two arrays. In the OPs example they are all nicely lined up.

    – Anush
    22 hours ago






  • 1





    @Anush Yes, but this a question of what the OP wants. On can definitely convince LaTeX to typeset the matrices in the way you suggest. I was using some standard routine because it produces some standard output the community seems to have agreed upon. One reason why I wrote the solution in this way is that anyone can adjust the matrices without knowing anything about TikZ, and also because this way one can use orthographic projections, which we cannot subject a tikz matrix to (at least not in straightforward way). I guess Wikipedia would have used orthographic projections if they could.

    – marmot
    13 hours ago






  • 1





    @marmot Yes. I have to say your solution is so lovely I am trying to think of an excuse to use it now!

    – Anush
    13 hours ago
















23














Something like that?



documentclass[tikz,border=3.14mm]{standalone}
usepackage{mathtools}
usetikzlibrary{matrix,backgrounds,3d}
usepackage{tikz-3dplot}
%definecolor{mygreen}{RGB}{12,252,12}
begin{document}
tdplotsetmaincoords{75}{20}
begin{tikzpicture}[tdplot_main_coords]
begin{scope}[canvas is xz plane at y=1,transform shape]
node[inner sep=0pt,text=green!70!black,opacity=0.8] (mat1)
{$displaystylebegin{pmatrix*}[r]
0 & 1 & 0 \
-1 & 0 & 0 \
0 & 0 & 0 \
end{pmatrix*}$};
begin{scope}[on background layer]
fill[green!70!black,opacity=0.2] ([xshift=8.5pt]mat1.south west)
coordinate (blb) to[out=140,in=-140,looseness=0.7]
([xshift=8.5pt]mat1.north west) coordinate (tlb) --
([xshift=-8.5pt]mat1.north east) coordinate (trb)
to[out=-40,in=40,looseness=0.7] ([xshift=-8.5pt]mat1.south east)
coordinate (brb)
-- cycle;
end{scope}
end{scope}
%
begin{scope}[canvas is xz plane at y=0,transform shape]
node[inner sep=0pt,text=red,opacity=0.8] (mat2) {$displaystyle
begin{pmatrix*}[r]
0 & 0 & -1 \
0 & 0 & 0 \
1 & 0 & 0 \
end{pmatrix*}$};
begin{scope}[on background layer]
fill[red,opacity=0.2] ([xshift=8.5pt]mat2.south west) to[out=140,in=-140,looseness=0.7]
([xshift=8.5pt]mat2.north west) -- ([xshift=-8.5pt]mat2.north east)
to[out=-40,in=40,looseness=0.7] ([xshift=-8.5pt]mat2.south east) -- cycle;
end{scope}
end{scope}
%
begin{scope}[canvas is xz plane at y=-1,transform shape]
node[inner sep=0pt,text=blue,opacity=0.8] (mat3) {$displaystyle
begin{pmatrix*}[r]
0 & 0 & 0 \
0 & 0 & 1 \
0 & -1 & 0 \
end{pmatrix*}$};
begin{scope}[on background layer]
fill[blue,opacity=0.2]
([xshift=8.5pt]mat3.south west) coordinate (blf)
to[out=140,in=-140,looseness=0.7]
([xshift=8.5pt]mat3.north west) coordinate (tlf)
-- ([xshift=-8.5pt]mat3.north east) coordinate (trf)
to[out=-40,in=40,looseness=0.7] ([xshift=-8.5pt]mat3.south east)
coordinate (brf) -- cycle;
end{scope}
end{scope}
foreach X in {tl,tr,br}
{draw[thin,orange] (X f) -- (X b);}
begin{scope}[on background layer]
draw[thin,orange] (blf) -- (blb);
end{scope}
node[left] at (mat3.west) {$varepsilon_{ijk}=$};
end{tikzpicture}
end{document}


enter image description here



EDIT: Aligned the entries right, big thanks to Barbara Beeton. (I just wonder why no one complained that the Levi-Civita tensor is not a tensor, but a tensor density. ;-)



2nd EDIT: Response to Anush's comment (well-taken! ;-).



documentclass[tikz,border=3.14mm]{standalone}
usepackage{mathtools}
usetikzlibrary{matrix,backgrounds,3d}
usepackage{tikz-3dplot}
begin{document}
tdplotsetmaincoords{75}{20}
begin{tikzpicture}[tdplot_main_coords]
begin{scope}[canvas is xz plane at y=1,transform shape]
node[inner sep=0pt,text=green!70!black,opacity=0.8] (mat1)
{$displaystylebegin{pmatrix*}[r]
0 & hphantom{-}1 & hphantom{-}0 \
-1 & 0 & 0 \
0 & 0 & 0 \
end{pmatrix*}$};
begin{scope}[on background layer]
fill[green!70!black,opacity=0.2] ([xshift=8.5pt]mat1.south west)
coordinate (blb) to[out=140,in=-140,looseness=0.7]
([xshift=8.5pt]mat1.north west) coordinate (tlb) --
([xshift=-8.5pt]mat1.north east) coordinate (trb)
to[out=-40,in=40,looseness=0.7] ([xshift=-8.5pt]mat1.south east)
coordinate (brb)
-- cycle;
end{scope}
end{scope}
%
begin{scope}[canvas is xz plane at y=0,transform shape]
node[inner sep=0pt,text=red,opacity=0.8] (mat2) {$displaystyle
begin{pmatrix*}[r]
hphantom{-}0 & hphantom{-}0 & -1 \
0 & 0 & 0 \
1 & 0 & 0 \
end{pmatrix*}$};
begin{scope}[on background layer]
fill[red,opacity=0.2] ([xshift=8.5pt]mat2.south west) to[out=140,in=-140,looseness=0.7]
([xshift=8.5pt]mat2.north west) -- ([xshift=-8.5pt]mat2.north east)
to[out=-40,in=40,looseness=0.7] ([xshift=-8.5pt]mat2.south east) -- cycle;
end{scope}
end{scope}
%
begin{scope}[canvas is xz plane at y=-1,transform shape]
node[inner sep=0pt,text=blue,opacity=0.8] (mat3) {$displaystyle
begin{pmatrix*}[r]
hphantom{-}0 & 0 & hphantom{-}0 \
0 & 0 & 1 \
0 & -1 & 0 \
end{pmatrix*}$};
begin{scope}[on background layer]
fill[blue,opacity=0.2]
([xshift=8.5pt]mat3.south west) coordinate (blf)
to[out=140,in=-140,looseness=0.7]
([xshift=8.5pt]mat3.north west) coordinate (tlf)
-- ([xshift=-8.5pt]mat3.north east) coordinate (trf)
to[out=-40,in=40,looseness=0.7] ([xshift=-8.5pt]mat3.south east)
coordinate (brf) -- cycle;
end{scope}
end{scope}
foreach X in {tl,tr,br}
{draw[thin,orange] (X f) -- (X b);}
begin{scope}[on background layer]
draw[thin,orange] (blf) -- (blb);
end{scope}
begin{scope}[canvas is xz plane at y=0,transform shape]
node[left] at (mat2.west -| mat3.west) {$varepsilon_{ijk}=$};
end{scope}
end{tikzpicture}
end{document}


enter image description here






share|improve this answer


























  • Two main differences with the image in the question: (1) in the negative entries, the digits are not aligned (and the spacing between columns adjusted to make them visually uniform), and (2) the tops and bottoms of the parentheses are not connected. of these, (1) detracts more from the appearance, although the meaning isn't affected; I happen to velue graceful appearance.

    – barbara beeton
    yesterday











  • @barbarabeeton Thanks! (1) can be addressed in a very simple way: loading mathtools and using begin{pmatrix*}[r]. (2) I do not understand. In Sebastiano's screen shot there are these four lines. Of course, if you'd ask me what they are good for, I'd admit that this is a very good question. ;-) Will revise my answer to address (1), thanks again!

    – marmot
    yesterday











  • The $-1$ in the front bottom middle doesn’t look aligned with the $0$ and $1$ in the othe two arrays. In the OPs example they are all nicely lined up.

    – Anush
    22 hours ago






  • 1





    @Anush Yes, but this a question of what the OP wants. On can definitely convince LaTeX to typeset the matrices in the way you suggest. I was using some standard routine because it produces some standard output the community seems to have agreed upon. One reason why I wrote the solution in this way is that anyone can adjust the matrices without knowing anything about TikZ, and also because this way one can use orthographic projections, which we cannot subject a tikz matrix to (at least not in straightforward way). I guess Wikipedia would have used orthographic projections if they could.

    – marmot
    13 hours ago






  • 1





    @marmot Yes. I have to say your solution is so lovely I am trying to think of an excuse to use it now!

    – Anush
    13 hours ago














23












23








23







Something like that?



documentclass[tikz,border=3.14mm]{standalone}
usepackage{mathtools}
usetikzlibrary{matrix,backgrounds,3d}
usepackage{tikz-3dplot}
%definecolor{mygreen}{RGB}{12,252,12}
begin{document}
tdplotsetmaincoords{75}{20}
begin{tikzpicture}[tdplot_main_coords]
begin{scope}[canvas is xz plane at y=1,transform shape]
node[inner sep=0pt,text=green!70!black,opacity=0.8] (mat1)
{$displaystylebegin{pmatrix*}[r]
0 & 1 & 0 \
-1 & 0 & 0 \
0 & 0 & 0 \
end{pmatrix*}$};
begin{scope}[on background layer]
fill[green!70!black,opacity=0.2] ([xshift=8.5pt]mat1.south west)
coordinate (blb) to[out=140,in=-140,looseness=0.7]
([xshift=8.5pt]mat1.north west) coordinate (tlb) --
([xshift=-8.5pt]mat1.north east) coordinate (trb)
to[out=-40,in=40,looseness=0.7] ([xshift=-8.5pt]mat1.south east)
coordinate (brb)
-- cycle;
end{scope}
end{scope}
%
begin{scope}[canvas is xz plane at y=0,transform shape]
node[inner sep=0pt,text=red,opacity=0.8] (mat2) {$displaystyle
begin{pmatrix*}[r]
0 & 0 & -1 \
0 & 0 & 0 \
1 & 0 & 0 \
end{pmatrix*}$};
begin{scope}[on background layer]
fill[red,opacity=0.2] ([xshift=8.5pt]mat2.south west) to[out=140,in=-140,looseness=0.7]
([xshift=8.5pt]mat2.north west) -- ([xshift=-8.5pt]mat2.north east)
to[out=-40,in=40,looseness=0.7] ([xshift=-8.5pt]mat2.south east) -- cycle;
end{scope}
end{scope}
%
begin{scope}[canvas is xz plane at y=-1,transform shape]
node[inner sep=0pt,text=blue,opacity=0.8] (mat3) {$displaystyle
begin{pmatrix*}[r]
0 & 0 & 0 \
0 & 0 & 1 \
0 & -1 & 0 \
end{pmatrix*}$};
begin{scope}[on background layer]
fill[blue,opacity=0.2]
([xshift=8.5pt]mat3.south west) coordinate (blf)
to[out=140,in=-140,looseness=0.7]
([xshift=8.5pt]mat3.north west) coordinate (tlf)
-- ([xshift=-8.5pt]mat3.north east) coordinate (trf)
to[out=-40,in=40,looseness=0.7] ([xshift=-8.5pt]mat3.south east)
coordinate (brf) -- cycle;
end{scope}
end{scope}
foreach X in {tl,tr,br}
{draw[thin,orange] (X f) -- (X b);}
begin{scope}[on background layer]
draw[thin,orange] (blf) -- (blb);
end{scope}
node[left] at (mat3.west) {$varepsilon_{ijk}=$};
end{tikzpicture}
end{document}


enter image description here



EDIT: Aligned the entries right, big thanks to Barbara Beeton. (I just wonder why no one complained that the Levi-Civita tensor is not a tensor, but a tensor density. ;-)



2nd EDIT: Response to Anush's comment (well-taken! ;-).



documentclass[tikz,border=3.14mm]{standalone}
usepackage{mathtools}
usetikzlibrary{matrix,backgrounds,3d}
usepackage{tikz-3dplot}
begin{document}
tdplotsetmaincoords{75}{20}
begin{tikzpicture}[tdplot_main_coords]
begin{scope}[canvas is xz plane at y=1,transform shape]
node[inner sep=0pt,text=green!70!black,opacity=0.8] (mat1)
{$displaystylebegin{pmatrix*}[r]
0 & hphantom{-}1 & hphantom{-}0 \
-1 & 0 & 0 \
0 & 0 & 0 \
end{pmatrix*}$};
begin{scope}[on background layer]
fill[green!70!black,opacity=0.2] ([xshift=8.5pt]mat1.south west)
coordinate (blb) to[out=140,in=-140,looseness=0.7]
([xshift=8.5pt]mat1.north west) coordinate (tlb) --
([xshift=-8.5pt]mat1.north east) coordinate (trb)
to[out=-40,in=40,looseness=0.7] ([xshift=-8.5pt]mat1.south east)
coordinate (brb)
-- cycle;
end{scope}
end{scope}
%
begin{scope}[canvas is xz plane at y=0,transform shape]
node[inner sep=0pt,text=red,opacity=0.8] (mat2) {$displaystyle
begin{pmatrix*}[r]
hphantom{-}0 & hphantom{-}0 & -1 \
0 & 0 & 0 \
1 & 0 & 0 \
end{pmatrix*}$};
begin{scope}[on background layer]
fill[red,opacity=0.2] ([xshift=8.5pt]mat2.south west) to[out=140,in=-140,looseness=0.7]
([xshift=8.5pt]mat2.north west) -- ([xshift=-8.5pt]mat2.north east)
to[out=-40,in=40,looseness=0.7] ([xshift=-8.5pt]mat2.south east) -- cycle;
end{scope}
end{scope}
%
begin{scope}[canvas is xz plane at y=-1,transform shape]
node[inner sep=0pt,text=blue,opacity=0.8] (mat3) {$displaystyle
begin{pmatrix*}[r]
hphantom{-}0 & 0 & hphantom{-}0 \
0 & 0 & 1 \
0 & -1 & 0 \
end{pmatrix*}$};
begin{scope}[on background layer]
fill[blue,opacity=0.2]
([xshift=8.5pt]mat3.south west) coordinate (blf)
to[out=140,in=-140,looseness=0.7]
([xshift=8.5pt]mat3.north west) coordinate (tlf)
-- ([xshift=-8.5pt]mat3.north east) coordinate (trf)
to[out=-40,in=40,looseness=0.7] ([xshift=-8.5pt]mat3.south east)
coordinate (brf) -- cycle;
end{scope}
end{scope}
foreach X in {tl,tr,br}
{draw[thin,orange] (X f) -- (X b);}
begin{scope}[on background layer]
draw[thin,orange] (blf) -- (blb);
end{scope}
begin{scope}[canvas is xz plane at y=0,transform shape]
node[left] at (mat2.west -| mat3.west) {$varepsilon_{ijk}=$};
end{scope}
end{tikzpicture}
end{document}


enter image description here






share|improve this answer















Something like that?



documentclass[tikz,border=3.14mm]{standalone}
usepackage{mathtools}
usetikzlibrary{matrix,backgrounds,3d}
usepackage{tikz-3dplot}
%definecolor{mygreen}{RGB}{12,252,12}
begin{document}
tdplotsetmaincoords{75}{20}
begin{tikzpicture}[tdplot_main_coords]
begin{scope}[canvas is xz plane at y=1,transform shape]
node[inner sep=0pt,text=green!70!black,opacity=0.8] (mat1)
{$displaystylebegin{pmatrix*}[r]
0 & 1 & 0 \
-1 & 0 & 0 \
0 & 0 & 0 \
end{pmatrix*}$};
begin{scope}[on background layer]
fill[green!70!black,opacity=0.2] ([xshift=8.5pt]mat1.south west)
coordinate (blb) to[out=140,in=-140,looseness=0.7]
([xshift=8.5pt]mat1.north west) coordinate (tlb) --
([xshift=-8.5pt]mat1.north east) coordinate (trb)
to[out=-40,in=40,looseness=0.7] ([xshift=-8.5pt]mat1.south east)
coordinate (brb)
-- cycle;
end{scope}
end{scope}
%
begin{scope}[canvas is xz plane at y=0,transform shape]
node[inner sep=0pt,text=red,opacity=0.8] (mat2) {$displaystyle
begin{pmatrix*}[r]
0 & 0 & -1 \
0 & 0 & 0 \
1 & 0 & 0 \
end{pmatrix*}$};
begin{scope}[on background layer]
fill[red,opacity=0.2] ([xshift=8.5pt]mat2.south west) to[out=140,in=-140,looseness=0.7]
([xshift=8.5pt]mat2.north west) -- ([xshift=-8.5pt]mat2.north east)
to[out=-40,in=40,looseness=0.7] ([xshift=-8.5pt]mat2.south east) -- cycle;
end{scope}
end{scope}
%
begin{scope}[canvas is xz plane at y=-1,transform shape]
node[inner sep=0pt,text=blue,opacity=0.8] (mat3) {$displaystyle
begin{pmatrix*}[r]
0 & 0 & 0 \
0 & 0 & 1 \
0 & -1 & 0 \
end{pmatrix*}$};
begin{scope}[on background layer]
fill[blue,opacity=0.2]
([xshift=8.5pt]mat3.south west) coordinate (blf)
to[out=140,in=-140,looseness=0.7]
([xshift=8.5pt]mat3.north west) coordinate (tlf)
-- ([xshift=-8.5pt]mat3.north east) coordinate (trf)
to[out=-40,in=40,looseness=0.7] ([xshift=-8.5pt]mat3.south east)
coordinate (brf) -- cycle;
end{scope}
end{scope}
foreach X in {tl,tr,br}
{draw[thin,orange] (X f) -- (X b);}
begin{scope}[on background layer]
draw[thin,orange] (blf) -- (blb);
end{scope}
node[left] at (mat3.west) {$varepsilon_{ijk}=$};
end{tikzpicture}
end{document}


enter image description here



EDIT: Aligned the entries right, big thanks to Barbara Beeton. (I just wonder why no one complained that the Levi-Civita tensor is not a tensor, but a tensor density. ;-)



2nd EDIT: Response to Anush's comment (well-taken! ;-).



documentclass[tikz,border=3.14mm]{standalone}
usepackage{mathtools}
usetikzlibrary{matrix,backgrounds,3d}
usepackage{tikz-3dplot}
begin{document}
tdplotsetmaincoords{75}{20}
begin{tikzpicture}[tdplot_main_coords]
begin{scope}[canvas is xz plane at y=1,transform shape]
node[inner sep=0pt,text=green!70!black,opacity=0.8] (mat1)
{$displaystylebegin{pmatrix*}[r]
0 & hphantom{-}1 & hphantom{-}0 \
-1 & 0 & 0 \
0 & 0 & 0 \
end{pmatrix*}$};
begin{scope}[on background layer]
fill[green!70!black,opacity=0.2] ([xshift=8.5pt]mat1.south west)
coordinate (blb) to[out=140,in=-140,looseness=0.7]
([xshift=8.5pt]mat1.north west) coordinate (tlb) --
([xshift=-8.5pt]mat1.north east) coordinate (trb)
to[out=-40,in=40,looseness=0.7] ([xshift=-8.5pt]mat1.south east)
coordinate (brb)
-- cycle;
end{scope}
end{scope}
%
begin{scope}[canvas is xz plane at y=0,transform shape]
node[inner sep=0pt,text=red,opacity=0.8] (mat2) {$displaystyle
begin{pmatrix*}[r]
hphantom{-}0 & hphantom{-}0 & -1 \
0 & 0 & 0 \
1 & 0 & 0 \
end{pmatrix*}$};
begin{scope}[on background layer]
fill[red,opacity=0.2] ([xshift=8.5pt]mat2.south west) to[out=140,in=-140,looseness=0.7]
([xshift=8.5pt]mat2.north west) -- ([xshift=-8.5pt]mat2.north east)
to[out=-40,in=40,looseness=0.7] ([xshift=-8.5pt]mat2.south east) -- cycle;
end{scope}
end{scope}
%
begin{scope}[canvas is xz plane at y=-1,transform shape]
node[inner sep=0pt,text=blue,opacity=0.8] (mat3) {$displaystyle
begin{pmatrix*}[r]
hphantom{-}0 & 0 & hphantom{-}0 \
0 & 0 & 1 \
0 & -1 & 0 \
end{pmatrix*}$};
begin{scope}[on background layer]
fill[blue,opacity=0.2]
([xshift=8.5pt]mat3.south west) coordinate (blf)
to[out=140,in=-140,looseness=0.7]
([xshift=8.5pt]mat3.north west) coordinate (tlf)
-- ([xshift=-8.5pt]mat3.north east) coordinate (trf)
to[out=-40,in=40,looseness=0.7] ([xshift=-8.5pt]mat3.south east)
coordinate (brf) -- cycle;
end{scope}
end{scope}
foreach X in {tl,tr,br}
{draw[thin,orange] (X f) -- (X b);}
begin{scope}[on background layer]
draw[thin,orange] (blf) -- (blb);
end{scope}
begin{scope}[canvas is xz plane at y=0,transform shape]
node[left] at (mat2.west -| mat3.west) {$varepsilon_{ijk}=$};
end{scope}
end{tikzpicture}
end{document}


enter image description here







share|improve this answer














share|improve this answer



share|improve this answer








edited 13 hours ago

























answered yesterday









marmotmarmot

106k5128242




106k5128242













  • Two main differences with the image in the question: (1) in the negative entries, the digits are not aligned (and the spacing between columns adjusted to make them visually uniform), and (2) the tops and bottoms of the parentheses are not connected. of these, (1) detracts more from the appearance, although the meaning isn't affected; I happen to velue graceful appearance.

    – barbara beeton
    yesterday











  • @barbarabeeton Thanks! (1) can be addressed in a very simple way: loading mathtools and using begin{pmatrix*}[r]. (2) I do not understand. In Sebastiano's screen shot there are these four lines. Of course, if you'd ask me what they are good for, I'd admit that this is a very good question. ;-) Will revise my answer to address (1), thanks again!

    – marmot
    yesterday











  • The $-1$ in the front bottom middle doesn’t look aligned with the $0$ and $1$ in the othe two arrays. In the OPs example they are all nicely lined up.

    – Anush
    22 hours ago






  • 1





    @Anush Yes, but this a question of what the OP wants. On can definitely convince LaTeX to typeset the matrices in the way you suggest. I was using some standard routine because it produces some standard output the community seems to have agreed upon. One reason why I wrote the solution in this way is that anyone can adjust the matrices without knowing anything about TikZ, and also because this way one can use orthographic projections, which we cannot subject a tikz matrix to (at least not in straightforward way). I guess Wikipedia would have used orthographic projections if they could.

    – marmot
    13 hours ago






  • 1





    @marmot Yes. I have to say your solution is so lovely I am trying to think of an excuse to use it now!

    – Anush
    13 hours ago



















  • Two main differences with the image in the question: (1) in the negative entries, the digits are not aligned (and the spacing between columns adjusted to make them visually uniform), and (2) the tops and bottoms of the parentheses are not connected. of these, (1) detracts more from the appearance, although the meaning isn't affected; I happen to velue graceful appearance.

    – barbara beeton
    yesterday











  • @barbarabeeton Thanks! (1) can be addressed in a very simple way: loading mathtools and using begin{pmatrix*}[r]. (2) I do not understand. In Sebastiano's screen shot there are these four lines. Of course, if you'd ask me what they are good for, I'd admit that this is a very good question. ;-) Will revise my answer to address (1), thanks again!

    – marmot
    yesterday











  • The $-1$ in the front bottom middle doesn’t look aligned with the $0$ and $1$ in the othe two arrays. In the OPs example they are all nicely lined up.

    – Anush
    22 hours ago






  • 1





    @Anush Yes, but this a question of what the OP wants. On can definitely convince LaTeX to typeset the matrices in the way you suggest. I was using some standard routine because it produces some standard output the community seems to have agreed upon. One reason why I wrote the solution in this way is that anyone can adjust the matrices without knowing anything about TikZ, and also because this way one can use orthographic projections, which we cannot subject a tikz matrix to (at least not in straightforward way). I guess Wikipedia would have used orthographic projections if they could.

    – marmot
    13 hours ago






  • 1





    @marmot Yes. I have to say your solution is so lovely I am trying to think of an excuse to use it now!

    – Anush
    13 hours ago

















Two main differences with the image in the question: (1) in the negative entries, the digits are not aligned (and the spacing between columns adjusted to make them visually uniform), and (2) the tops and bottoms of the parentheses are not connected. of these, (1) detracts more from the appearance, although the meaning isn't affected; I happen to velue graceful appearance.

– barbara beeton
yesterday





Two main differences with the image in the question: (1) in the negative entries, the digits are not aligned (and the spacing between columns adjusted to make them visually uniform), and (2) the tops and bottoms of the parentheses are not connected. of these, (1) detracts more from the appearance, although the meaning isn't affected; I happen to velue graceful appearance.

– barbara beeton
yesterday













@barbarabeeton Thanks! (1) can be addressed in a very simple way: loading mathtools and using begin{pmatrix*}[r]. (2) I do not understand. In Sebastiano's screen shot there are these four lines. Of course, if you'd ask me what they are good for, I'd admit that this is a very good question. ;-) Will revise my answer to address (1), thanks again!

– marmot
yesterday





@barbarabeeton Thanks! (1) can be addressed in a very simple way: loading mathtools and using begin{pmatrix*}[r]. (2) I do not understand. In Sebastiano's screen shot there are these four lines. Of course, if you'd ask me what they are good for, I'd admit that this is a very good question. ;-) Will revise my answer to address (1), thanks again!

– marmot
yesterday













The $-1$ in the front bottom middle doesn’t look aligned with the $0$ and $1$ in the othe two arrays. In the OPs example they are all nicely lined up.

– Anush
22 hours ago





The $-1$ in the front bottom middle doesn’t look aligned with the $0$ and $1$ in the othe two arrays. In the OPs example they are all nicely lined up.

– Anush
22 hours ago




1




1





@Anush Yes, but this a question of what the OP wants. On can definitely convince LaTeX to typeset the matrices in the way you suggest. I was using some standard routine because it produces some standard output the community seems to have agreed upon. One reason why I wrote the solution in this way is that anyone can adjust the matrices without knowing anything about TikZ, and also because this way one can use orthographic projections, which we cannot subject a tikz matrix to (at least not in straightforward way). I guess Wikipedia would have used orthographic projections if they could.

– marmot
13 hours ago





@Anush Yes, but this a question of what the OP wants. On can definitely convince LaTeX to typeset the matrices in the way you suggest. I was using some standard routine because it produces some standard output the community seems to have agreed upon. One reason why I wrote the solution in this way is that anyone can adjust the matrices without knowing anything about TikZ, and also because this way one can use orthographic projections, which we cannot subject a tikz matrix to (at least not in straightforward way). I guess Wikipedia would have used orthographic projections if they could.

– marmot
13 hours ago




1




1





@marmot Yes. I have to say your solution is so lovely I am trying to think of an excuse to use it now!

– Anush
13 hours ago





@marmot Yes. I have to say your solution is so lovely I am trying to think of an excuse to use it now!

– Anush
13 hours ago











21














More or less:



documentclass[tikz,border=2mm]{standalone} 
usetikzlibrary{positioning, matrix}
usepackage{amsmath}

newcommand{arrayfilling}[2]{
fill[#2!30, opacity=.5] ([shift={(1mm,1mm)}]#1.north west) coordinate(#1auxnw)--([shift={(1mm,1mm)}]#1.north east)coordinate(#1auxne) to[out=-75, in=75] ([shift={(1mm,-1mm)}]#1.south east)coordinate(#1auxse)--([shift={(1mm,-1mm)}]#1.south west)coordinate(#1auxsw) to[out=105, in=-105] cycle;
fill[#2!80!black, opacity=1] (#1auxne) to[out=-75, in=75] (#1auxse) to[out=78, in=-78] cycle;
fill[#2!80!black, opacity=1] (#1auxnw) to[out=-105, in=105] (#1auxsw) to[out=102, in=-102] cycle;
}


begin{document}
begin{tikzpicture}[font=ttfamily,
mymatrix/.style={
matrix of math nodes, inner sep=0pt, color=#1,
column sep=-pgflinewidth, row sep=-pgflinewidth, anchor=south west,
nodes={anchor=center, minimum width=5mm,
minimum height=3mm, outer sep=0pt, inner sep=0pt,
text width=5mm, align=right,
draw=none, font=small},
}
]

matrix (C) [mymatrix=green] at (6mm,5mm)
{0 & 1 & 0 \ -1 & 0 & 0\ 0 & 0 & 0\};
arrayfilling{C}{green}

matrix (B) [mymatrix=red] at (3mm,2.5mm)
{0 & 0 & -1 \ 0 & 0 & 0\ 1 & 0 & 0\};
arrayfilling{B}{red}

matrix (A) [mymatrix=blue] at (0,0)
{0 & 0 & 0 \ 0 & 0 & 1\ 0 & -1 & 0\};
arrayfilling{A}{blue}

foreach i in {auxnw, auxne, auxse, auxsw}
draw[brown, ultra thin] (Ai)--(Ci);

node[below left=-1mm and 5mm of B.west] {$epsilon_{ijk} =$};
end{tikzpicture}
end{document}


enter image description here






share|improve this answer


























  • Please see point (1) in my comment to the answer by @marmot.

    – barbara beeton
    yesterday











  • @barbarabeeton Is it better now?

    – Ignasi
    yesterday











  • Yes, better. one might quibble that there is more space than necessary, compared with the space at the right. I think it's best to pay attention to what is actually present in each column, but that does require more attention.

    – barbara beeton
    yesterday











  • In the same way as the comment that I have added to marmot I can not establish, given that they are two codes at the same level, which is the best. You have been very good. Thank you very much.

    – Sebastiano
    9 hours ago
















21














More or less:



documentclass[tikz,border=2mm]{standalone} 
usetikzlibrary{positioning, matrix}
usepackage{amsmath}

newcommand{arrayfilling}[2]{
fill[#2!30, opacity=.5] ([shift={(1mm,1mm)}]#1.north west) coordinate(#1auxnw)--([shift={(1mm,1mm)}]#1.north east)coordinate(#1auxne) to[out=-75, in=75] ([shift={(1mm,-1mm)}]#1.south east)coordinate(#1auxse)--([shift={(1mm,-1mm)}]#1.south west)coordinate(#1auxsw) to[out=105, in=-105] cycle;
fill[#2!80!black, opacity=1] (#1auxne) to[out=-75, in=75] (#1auxse) to[out=78, in=-78] cycle;
fill[#2!80!black, opacity=1] (#1auxnw) to[out=-105, in=105] (#1auxsw) to[out=102, in=-102] cycle;
}


begin{document}
begin{tikzpicture}[font=ttfamily,
mymatrix/.style={
matrix of math nodes, inner sep=0pt, color=#1,
column sep=-pgflinewidth, row sep=-pgflinewidth, anchor=south west,
nodes={anchor=center, minimum width=5mm,
minimum height=3mm, outer sep=0pt, inner sep=0pt,
text width=5mm, align=right,
draw=none, font=small},
}
]

matrix (C) [mymatrix=green] at (6mm,5mm)
{0 & 1 & 0 \ -1 & 0 & 0\ 0 & 0 & 0\};
arrayfilling{C}{green}

matrix (B) [mymatrix=red] at (3mm,2.5mm)
{0 & 0 & -1 \ 0 & 0 & 0\ 1 & 0 & 0\};
arrayfilling{B}{red}

matrix (A) [mymatrix=blue] at (0,0)
{0 & 0 & 0 \ 0 & 0 & 1\ 0 & -1 & 0\};
arrayfilling{A}{blue}

foreach i in {auxnw, auxne, auxse, auxsw}
draw[brown, ultra thin] (Ai)--(Ci);

node[below left=-1mm and 5mm of B.west] {$epsilon_{ijk} =$};
end{tikzpicture}
end{document}


enter image description here






share|improve this answer


























  • Please see point (1) in my comment to the answer by @marmot.

    – barbara beeton
    yesterday











  • @barbarabeeton Is it better now?

    – Ignasi
    yesterday











  • Yes, better. one might quibble that there is more space than necessary, compared with the space at the right. I think it's best to pay attention to what is actually present in each column, but that does require more attention.

    – barbara beeton
    yesterday











  • In the same way as the comment that I have added to marmot I can not establish, given that they are two codes at the same level, which is the best. You have been very good. Thank you very much.

    – Sebastiano
    9 hours ago














21












21








21







More or less:



documentclass[tikz,border=2mm]{standalone} 
usetikzlibrary{positioning, matrix}
usepackage{amsmath}

newcommand{arrayfilling}[2]{
fill[#2!30, opacity=.5] ([shift={(1mm,1mm)}]#1.north west) coordinate(#1auxnw)--([shift={(1mm,1mm)}]#1.north east)coordinate(#1auxne) to[out=-75, in=75] ([shift={(1mm,-1mm)}]#1.south east)coordinate(#1auxse)--([shift={(1mm,-1mm)}]#1.south west)coordinate(#1auxsw) to[out=105, in=-105] cycle;
fill[#2!80!black, opacity=1] (#1auxne) to[out=-75, in=75] (#1auxse) to[out=78, in=-78] cycle;
fill[#2!80!black, opacity=1] (#1auxnw) to[out=-105, in=105] (#1auxsw) to[out=102, in=-102] cycle;
}


begin{document}
begin{tikzpicture}[font=ttfamily,
mymatrix/.style={
matrix of math nodes, inner sep=0pt, color=#1,
column sep=-pgflinewidth, row sep=-pgflinewidth, anchor=south west,
nodes={anchor=center, minimum width=5mm,
minimum height=3mm, outer sep=0pt, inner sep=0pt,
text width=5mm, align=right,
draw=none, font=small},
}
]

matrix (C) [mymatrix=green] at (6mm,5mm)
{0 & 1 & 0 \ -1 & 0 & 0\ 0 & 0 & 0\};
arrayfilling{C}{green}

matrix (B) [mymatrix=red] at (3mm,2.5mm)
{0 & 0 & -1 \ 0 & 0 & 0\ 1 & 0 & 0\};
arrayfilling{B}{red}

matrix (A) [mymatrix=blue] at (0,0)
{0 & 0 & 0 \ 0 & 0 & 1\ 0 & -1 & 0\};
arrayfilling{A}{blue}

foreach i in {auxnw, auxne, auxse, auxsw}
draw[brown, ultra thin] (Ai)--(Ci);

node[below left=-1mm and 5mm of B.west] {$epsilon_{ijk} =$};
end{tikzpicture}
end{document}


enter image description here






share|improve this answer















More or less:



documentclass[tikz,border=2mm]{standalone} 
usetikzlibrary{positioning, matrix}
usepackage{amsmath}

newcommand{arrayfilling}[2]{
fill[#2!30, opacity=.5] ([shift={(1mm,1mm)}]#1.north west) coordinate(#1auxnw)--([shift={(1mm,1mm)}]#1.north east)coordinate(#1auxne) to[out=-75, in=75] ([shift={(1mm,-1mm)}]#1.south east)coordinate(#1auxse)--([shift={(1mm,-1mm)}]#1.south west)coordinate(#1auxsw) to[out=105, in=-105] cycle;
fill[#2!80!black, opacity=1] (#1auxne) to[out=-75, in=75] (#1auxse) to[out=78, in=-78] cycle;
fill[#2!80!black, opacity=1] (#1auxnw) to[out=-105, in=105] (#1auxsw) to[out=102, in=-102] cycle;
}


begin{document}
begin{tikzpicture}[font=ttfamily,
mymatrix/.style={
matrix of math nodes, inner sep=0pt, color=#1,
column sep=-pgflinewidth, row sep=-pgflinewidth, anchor=south west,
nodes={anchor=center, minimum width=5mm,
minimum height=3mm, outer sep=0pt, inner sep=0pt,
text width=5mm, align=right,
draw=none, font=small},
}
]

matrix (C) [mymatrix=green] at (6mm,5mm)
{0 & 1 & 0 \ -1 & 0 & 0\ 0 & 0 & 0\};
arrayfilling{C}{green}

matrix (B) [mymatrix=red] at (3mm,2.5mm)
{0 & 0 & -1 \ 0 & 0 & 0\ 1 & 0 & 0\};
arrayfilling{B}{red}

matrix (A) [mymatrix=blue] at (0,0)
{0 & 0 & 0 \ 0 & 0 & 1\ 0 & -1 & 0\};
arrayfilling{A}{blue}

foreach i in {auxnw, auxne, auxse, auxsw}
draw[brown, ultra thin] (Ai)--(Ci);

node[below left=-1mm and 5mm of B.west] {$epsilon_{ijk} =$};
end{tikzpicture}
end{document}


enter image description here







share|improve this answer














share|improve this answer



share|improve this answer








edited yesterday

























answered yesterday









IgnasiIgnasi

94.2k4172314




94.2k4172314













  • Please see point (1) in my comment to the answer by @marmot.

    – barbara beeton
    yesterday











  • @barbarabeeton Is it better now?

    – Ignasi
    yesterday











  • Yes, better. one might quibble that there is more space than necessary, compared with the space at the right. I think it's best to pay attention to what is actually present in each column, but that does require more attention.

    – barbara beeton
    yesterday











  • In the same way as the comment that I have added to marmot I can not establish, given that they are two codes at the same level, which is the best. You have been very good. Thank you very much.

    – Sebastiano
    9 hours ago



















  • Please see point (1) in my comment to the answer by @marmot.

    – barbara beeton
    yesterday











  • @barbarabeeton Is it better now?

    – Ignasi
    yesterday











  • Yes, better. one might quibble that there is more space than necessary, compared with the space at the right. I think it's best to pay attention to what is actually present in each column, but that does require more attention.

    – barbara beeton
    yesterday











  • In the same way as the comment that I have added to marmot I can not establish, given that they are two codes at the same level, which is the best. You have been very good. Thank you very much.

    – Sebastiano
    9 hours ago

















Please see point (1) in my comment to the answer by @marmot.

– barbara beeton
yesterday





Please see point (1) in my comment to the answer by @marmot.

– barbara beeton
yesterday













@barbarabeeton Is it better now?

– Ignasi
yesterday





@barbarabeeton Is it better now?

– Ignasi
yesterday













Yes, better. one might quibble that there is more space than necessary, compared with the space at the right. I think it's best to pay attention to what is actually present in each column, but that does require more attention.

– barbara beeton
yesterday





Yes, better. one might quibble that there is more space than necessary, compared with the space at the right. I think it's best to pay attention to what is actually present in each column, but that does require more attention.

– barbara beeton
yesterday













In the same way as the comment that I have added to marmot I can not establish, given that they are two codes at the same level, which is the best. You have been very good. Thank you very much.

– Sebastiano
9 hours ago





In the same way as the comment that I have added to marmot I can not establish, given that they are two codes at the same level, which is the best. You have been very good. Thank you very much.

– Sebastiano
9 hours ago


















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