Levi-Civita symbol: 3D matrix
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.
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
add a comment |
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.
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
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
10 hours ago
@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
10 hours ago
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
10 hours ago
add a comment |
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.
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
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.
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
matrices 3d tikz-matrix
asked 10 hours ago
SebastianoSebastiano
10.5k42061
10.5k42061
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
10 hours ago
@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
10 hours ago
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
10 hours ago
add a comment |
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
10 hours ago
@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
10 hours ago
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
10 hours ago
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
10 hours ago
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
10 hours ago
@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
10 hours ago
@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
10 hours ago
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
10 hours ago
Well, I was thinking about how to solve this and I read that word ;-) lol I knew I had to write something :D
– JouleV
10 hours ago
add a comment |
2 Answers
2
active
oldest
votes
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}
Please see point (1) in my comment to the answer by @marmot.
– barbara beeton
9 hours ago
@barbarabeeton Is it better now?
– Ignasi
9 hours ago
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
8 hours ago
add a comment |
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}
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. ;-)
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
9 hours ago
@barbarabeeton Thanks! (1) can be addressed in a very simple way: loadingmathtools
and usingbegin{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
9 hours ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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}
Please see point (1) in my comment to the answer by @marmot.
– barbara beeton
9 hours ago
@barbarabeeton Is it better now?
– Ignasi
9 hours ago
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
8 hours ago
add a comment |
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}
Please see point (1) in my comment to the answer by @marmot.
– barbara beeton
9 hours ago
@barbarabeeton Is it better now?
– Ignasi
9 hours ago
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
8 hours ago
add a comment |
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}
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}
edited 7 hours ago
answered 9 hours ago
IgnasiIgnasi
94.1k4170313
94.1k4170313
Please see point (1) in my comment to the answer by @marmot.
– barbara beeton
9 hours ago
@barbarabeeton Is it better now?
– Ignasi
9 hours ago
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
8 hours ago
add a comment |
Please see point (1) in my comment to the answer by @marmot.
– barbara beeton
9 hours ago
@barbarabeeton Is it better now?
– Ignasi
9 hours ago
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
8 hours ago
Please see point (1) in my comment to the answer by @marmot.
– barbara beeton
9 hours ago
Please see point (1) in my comment to the answer by @marmot.
– barbara beeton
9 hours ago
@barbarabeeton Is it better now?
– Ignasi
9 hours ago
@barbarabeeton Is it better now?
– Ignasi
9 hours ago
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
8 hours ago
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
8 hours ago
add a comment |
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}
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. ;-)
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
9 hours ago
@barbarabeeton Thanks! (1) can be addressed in a very simple way: loadingmathtools
and usingbegin{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
9 hours ago
add a comment |
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}
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. ;-)
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
9 hours ago
@barbarabeeton Thanks! (1) can be addressed in a very simple way: loadingmathtools
and usingbegin{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
9 hours ago
add a comment |
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}
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. ;-)
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}
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. ;-)
edited 9 hours ago
answered 9 hours ago
marmotmarmot
106k4127242
106k4127242
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
9 hours ago
@barbarabeeton Thanks! (1) can be addressed in a very simple way: loadingmathtools
and usingbegin{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
9 hours ago
add a comment |
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
9 hours ago
@barbarabeeton Thanks! (1) can be addressed in a very simple way: loadingmathtools
and usingbegin{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
9 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
9 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
9 hours ago
@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
9 hours ago
@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
9 hours ago
add a comment |
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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
10 hours ago
@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
10 hours ago
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
10 hours ago