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Is is possible to add a skewed grid in a plot? Consider the following image; enter image description here

The grey axes are the normal axes ($\mathbf{e}_1$ and $\mathbf{e}_2$), the black axes are transformations $\mathbf{e}'_1=2\mathbf{e}_1 - \mathbf{e}_2$ and $\mathbf{e}'_2=-\mathbf{e}_1 + \mathbf{e}_2$ and the blue arrow is $(1,1)$ in the $\mathbf{e}$-base and $(2,3)$ in the $\mathbf{e}'$-base. I would like to insert a (red) grid that is “aligned” with the $\mathbf{e}'$-base, i.e. skewed so that one clearly sees that the blue arrow has coordinates $(2,3)$ in the transformed coordinate system. TIA.

MWE:

o = {0, 0};
e1 = {1, 0};
e2 = {0, 1};
e1p = 2 e1 - e2;
e2p = -e1 + e2;
v = {1, 1};
a0 = Graphics[{Blue, Arrow[{o, v}]}, GridLines -> Automatic, PlotRange -> {{-1.5, 4.5}, {-3, 2}}, Frame -> True];
a1 = Graphics[{Black, Opacity[.25], Arrow[{o, e1}], Arrow[{o, e2}]}];
a2 = Graphics[{Black, Arrow[{o, e1p}], Arrow[{o, e2p}]}];
k1 = Graphics[{Red, Arrow[{o, 2 e1p}]}];
k2 = Graphics[{{Red, Arrow[{2 e1p, 2 e1p + 3 e2p}]}, {Red, Arrow[{2 e1p, 2 e1p + 1 e2p}]}, {Red, Arrow[{2 e1p, 2 e1p + 2 e2p}]}}];
Show[a0, k1, k2, a1, a2]

The code might not be the best, improvements are most welcome!

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4 Answers 4

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With ColorFunction -> (White &)you can have a white background:

o = {0, 0};
e1 = {1, 0};
e2 = {0, 1};
e1p = 2 e1 - e2;
e2p = -e1 + e2;
v = {1, 1};
a0 = Graphics[{Blue, Arrow[{o, v}]}, GridLines -> Automatic, 
   PlotRange -> {{-1.5, 4.5}, {-3, 2}}, Frame -> True];
a1 = Graphics[{Black, Opacity[.25], Arrow[{o, e1}], Arrow[{o, e2}]}];
a2 = Graphics[{Black, Arrow[{o, e1p}], Arrow[{o, e2p}]}];
k1 = Graphics[{Red, Arrow[{o, 2 e1p}]}];
k2 = Graphics[{{Red, Arrow[{2 e1p, 2 e1p + 3 e2p}]}, {Red, 
     Arrow[{2 e1p, 2 e1p + 1 e2p}]}, {Red, 
     Arrow[{2 e1p, 2 e1p + 2 e2p}]}}];
pl = ContourPlot[.2, {x, -2, 4}, {y, -3, 2}, 
   MeshFunctions -> {(e1p.{-#2, #1}) &, (e2p.{-#2, #1}) &}, 
   Mesh -> {Range[-5, 5], Range[-5, 5]}, ColorFunction -> (White &)];
Show[pl, a0, k1, k2, a1, a2]

enter image description here

With ContourPlot[0.2,...,ColorFunction->Hue you can have any color you want by changing 0.2:

Using MeshStyleyou can specify the display of mesh lines. E.g.

pl = ContourPlot[.2, {x, -2, 4}, {y, -3, 2}, 
   MeshFunctions -> {(e1p.{-#2, #1}) &, (e2p.{-#2, #1}) &}, 
   Mesh -> {Range[-5, 5], Range[-5, 5]}, ColorFunction -> (White &), 
   MeshStyle -> {{Magenta, Opacity[0.5]}, {Green, Opacity[0.5]}}];
Show[pl, a0, k1, k2, a1, a2]

enter image description here

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Another way is shifting the infiniteline InfiniteLine[{0, 0}, e1p] and InfiniteLine[{0, 0}, e2p]

o = {0, 0};
e1 = {1, 0};
e2 = {0, 1};
e1p = 2 e1 - e2;
e2p = -e1 + e2;
v = {1, 1};
a0 = Graphics[{Blue, Arrow[{o, v}]}, 
   PlotRange -> {{-1.5, 4.5}, {-3, 2}}, Frame -> True];
a1 = Graphics[{Black, Opacity[.25], Arrow[{o, e1}], Arrow[{o, e2}]}];
a2 = Graphics[{Black, Arrow[{o, e1p}], Arrow[{o, e2p}]}];
k1 = Graphics[{Red, Arrow[{o, 2 e1p}]}];
k2 = Graphics[{{Red, Arrow[{2 e1p, 2 e1p + 3 e2p}]}, {Red, 
     Arrow[{2 e1p, 2 e1p + 1 e2p}]}, {Red, 
     Arrow[{2 e1p, 2 e1p + 2 e2p}]}}];
skew = Graphics[{Opacity[.5], 
    Table[{{Green, InfiniteLine[j*e2p, e1p]}, {Cyan, 
       InfiniteLine[j*e1p, e2p]}}, {j, -5, 5}]}];
Show[skew, a0, k1, k2, a1, a2, PlotRange -> {{-2, 4}, {-3, 2}}]

enter image description here

ParametricPlot also work.

skew2 = ParametricPlot[x*e1p + y*e2p, {x, -8, 8}, {y, -8, 8}, 
  Mesh -> {Range[-8, 8]}, PlotStyle -> White, BoundaryStyle -> None, 
  Frame -> False, Axes -> False, PlotRange -> {{-2, 4}, {-3, 2}}, 
  MeshStyle -> {Cyan, Green}]
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We can use e1p and e2p to define two lists of transformation functions and use them to translate InfiniteLine[{o, e2p}] and InfiniteLine[{o, e2p}] to get the grid lines with desired slants.

trs1 = # e1p & /@ Range[-7, 7];
trs2 = # e2p & /@ Range[-7, 8];

Graphics[{{ 
   Sequence @@ CurrentValue[{StyleDefinitions, "GraphicsGridLines"}],
   Translate[InfiniteLine[{o, e2p}], #] & /@ trs1,
   Translate[InfiniteLine[{o, e1p}], #] & /@ trs2},
  Black, Opacity[.25], Arrow[{o, e1}], Arrow[{o, e2}],
  Opacity[1], Arrow[{o, e1p}], Arrow[{o, e2p}],
  Blue, Arrow[{o, v}],
  Red, Arrow[{e1p, 2 e1p}],
  Arrowheads[{0, Automatic, Automatic, Automatic}], 
  Arrow[{2 e1p, 2 e1p + 3 e2p}]},
 PlotRange -> {{-2, 5}, {-3, 2}}, Frame -> True, ImageSize -> Large]

enter image description here

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  • $\begingroup$ A fine solution (as always!). The code is bit difficult for me to understand, but that is a fault on my side(!). Would it be possible to change the ticks/tick labels also so they agree with the skewed lines? $\endgroup$
    – mf67
    Mar 11, 2021 at 17:20
  • 1
    $\begingroup$ @mf67, please see the updated version. If you want additional ticks on the left frame, try using FrameTicks -> {{Range[-3, 5, .5], Automatic},{ Automatic, Automatic}}. $\endgroup$
    – kglr
    Mar 12, 2021 at 20:21
  • $\begingroup$ Not sure I understand, but should not the 3 on the x-axis be 0 and 1 on the y-axis be 0 in the skewed system's tick labels? $\endgroup$
    – mf67
    Mar 13, 2021 at 0:22
  • $\begingroup$ @mf67, right,now i see what you mean. That's good new question. $\endgroup$
    – kglr
    Mar 13, 2021 at 0:31
  • 1
    $\begingroup$ Obviously there is a bug in Translate + InfiniteLine, because Translate[InfiniteLine[{o, e2p}], trs1] should wok too. $\endgroup$
    – Alexey Popkov
    Mar 13, 2021 at 0:51
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Of course, we can do this with graphics primitives. However, the lazy way is to use ContourPlot, only for the sake of the Mesh. Here is your example:

a1 = Graphics[{Black, Opacity[.25], Arrow[{o, e1}], Arrow[{o, e2}]}];
a2 = Graphics[{Black, Arrow[{o, e1p}], Arrow[{o, e2p}]}];
k1 = Graphics[{Red, Arrow[{o, 2 e1p}]}];
k2 = Graphics[{{Red, Arrow[{2 e1p, 2 e1p + 3 e2p}]}, {Red, 
     Arrow[{2 e1p, 2 e1p + 1 e2p}]}, {Red, 
     Arrow[{2 e1p, 2 e1p + 2 e2p}]}}];
pl = ContourPlot[1, {x, -2, 4}, {y, -3, 2}, 
   MeshFunctions -> {(e1p.{-#2, #1}) &, (e2p.{-#2, #1}) &}, 
   Mesh -> {Range[-5, 5], Range[-5, 5]}];
Show[{pl, a0, k1, k2, a1, a2}, Axes -> True]

enter image description here

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1
  • 1
    $\begingroup$ Is there a way to control the yellow "paper" color and the color of the mesh? $\endgroup$
    – mf67
    Mar 9, 2021 at 21:11

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