Krystal Guo

algebraic graph theorist, assistant professor of discrete math at University of Amsterdam and researcher at QuSoft.

Homepage: https://krystalguo.com
Google Scholar: https://scholar.google.com/citations?user=9AIoBl8AAAAJ&;
Twitter: https://twitter.com/guo_krystal

Last Notes

2026-05-17 14:10:13 UTC

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Hmm, yes, the drawings from these eigenvalues are just the usual drawings using the roots of unity.
Here's a much larger (but maybe more natural?) construction.
The truncation of the line graph of the cube is also a polyhedral graph and it has cubic factors in its characteristic polynomial. It's a cubic planar graph on 48 vertices, with 26 faces; 12 of degree 4, 8 of degree 6 and 6 of degree 8.
The graph6_string: 'op_HOC??G@?@?G??_@O?@?C???G?_??AC??I???G?G????`?@????C?_??@npub1ag6…4qtc???@??@?@???OG???GO_???c?@????????G??A?@?????@??C????????C_???G?O?????@npub1ag6…4qtc??C???C????CG_???_??A?????c?????O?@????@?G?????A@?G????A@'
Here's an embedding using the eigenvectors of one of the cubic roots (they happen to have multiplicity 3).
https://media.mathstodon.xyz/media_attachments/files/116/590/307/158/699/924/original/3182471db88f91d3.mp4

2026-05-17 12:48:29 UTC

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I'm talking about eigenvalues/eigenvectors of the adjacency matrix, though for regular graphs, the adjacency and Laplacian matrices are interchangeable.
Yes, the 1-skeleton of the 7-prism (2 faces of length 7 and 7 faces of length 4) has eigenvalues that are roots of a cubic. Its minimal polynomial is:
(x - 3)(x - 1)(x^3 - 2x^2 - x + 1)^2 (x^3 + 4x^2 + 3x - 1)^2.
The graph is the Cartesian product of the 7-cycle with an edge. The 9-prism also has eigenvalues that are roots of a cubic.
In general, the n-prism "lifts" the eigenvalues of the n-cycle, whose characteristic polynomial (of the adj matrix) by given the Chebyshev polynomial of the 1st kind. This is a good way to make graphs with interesting Galois automorphisms, because the eigenvalues are then (functions of) cosines of rational multiples of pi.

2026-05-17 11:05:07 UTC

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Cool! Is there something special about \sqrt{5}? There are other polytopes (whose 1-skeleton) have irrational eigenvalues with multiplicity 3.

2026-05-16 20:50:01 UTC

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Sorry, one of the messages didn't send. So, the squished-looking ones are because I computed the eigenvalues using SageMath's default, which tries very hard to output an integer eigenvector, which is why the entries are nice. But they're not orthogonal or anything. Now, I found the eigenvalues with linalg from scipy. Now, the eigenvectors for sqrt(5) produce the icosahedron and those for -sqrt(5) produce the great icosahedron. Each of these eigenspaces happens to have dimension 3, so they produce coordinates in 3D.
The map swapping sqrt(5) and -sqrt(5) is the (only) non-trivial Galois automorphism of the splitting field of the eigenvalues, so this map takes an eigenvector for sqrt(5) to an eigenvector of -sqrt(5), because the adjacency matrix has rational (in fact, integer) entries, which are invariant under all Galois automorphisms.
https://media.mathstodon.xyz/media_attachments/files/116/586/218/666/052/675/original/942c14afe8f8ad0d.mp4

2026-05-16 18:38:48 UTC

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it only let me attach one video at a time from my phone.
https://media.mathstodon.xyz/media_attachments/files/116/585/705/931/018/743/original/3c0ab5d5a1d53d3c.mp4

2026-05-16 18:37:53 UTC

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ah yes, they do look squished to me. The reason is that I’m computing the eigenvalues in SageMath, which is trying very hard to find integer eigenvectors, so there are more 0s than 1s.
Now I think about it, your map of swapping sqrt(5) to - sqrt(5) is just the single nontrivial Galois automorphism of the splitting field of the eigenvalues. This means if I apply it to an eigenvector of -sqrt(5), I should get an eigenvector of sqrt(5).
So now, I’ve computed the eigenvectors in scipy linalg (which gives orthonormal eigenvectors). The eigenvectors of -sqrt(5) give the great icosahedron and of sqrt(5) give the icosahedron. New video.
https://media.mathstodon.xyz/media_attachments/files/116/585/697/022/664/220/original/25873dfcab2ea4ba.mp4

2026-05-16 16:45:14 UTC

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How did I produce these coordinates? The answer is eigenmagicks! The eigenvalues of the icosahedral graph are 5, -1, sqrt(5) and -sqrt(5). The multiplicity of the irrational eigenvalues are each 3. So I found an eigenbasis for the -sqrt(5) eigenspace, which is 3 vectors. Then I just used the v-th entry of these 3 vectors to obtain (x,y,z) coordinates for vertex v.
The least eigenvalue of a graph appears to be a rigid graph invariant, particularly when its absolute value is small. Graphs with least eigenvalue at least -2 or at least -3, admit geometric realizations in lattices (excepting finitely many sporadic graphs).

2026-05-16 16:40:12 UTC

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I think you're correct. As a graph theorist, though I maybe don't see the different between the icosahedron and the great icosahedron 😂
As a graph in SageMath, the icosahedron has 12 vertices and the neighbours are as follows:
0 [1, 5, 7, 8, 11]
1 [0, 2, 5, 6, 8]
2 [1, 3, 6, 8, 9]
3 [2, 4, 6, 9, 10]
4 [3, 5, 6, 10, 11]
5 [0, 1, 4, 6, 11]
6 [1, 2, 3, 4, 5]
7 [0, 8, 9, 10, 11]
8 [0, 1, 2, 7, 9]
9 [2, 3, 7, 8, 10]
10 [3, 4, 7, 9, 11]
11 [0, 4, 5, 7, 10]
Vertex positions version 1:
0: (1.0, 0.0, 0.0),
1: (0.0, 1.0, 0.0),
2: (0.0, 0.0, 1.0),
3: (-1.0, 0.0, 0.0),
4: (0.618034, 1.0, 0.618034),
5: (-1.0, -0.618034, 0.618034),
6: (0.618034, -0.618034, -1.0),
7: (-0.618034, 0.618034, 1.0),
8: (-0.618034, -1.0, -0.618034),
9: (1.0, 0.618034, -0.618034),
10: (0.0, -1.0, 0.0),
11: (0.0, 0.0, -1.0)
This produces a drawing of the great icosahedron. I scaled so that they are on the unit sphere and I've also plotted the unit sphere (in light blue), see attached video 1.
Vertex positions version 2:
0: [1, 0, 0],
1: [0, 1, 0],
2: [0, 0, 1],
3: [-1, 0, 0],
4: [-1.618034, 1, -1.618034],
5: [-1, 1.618034, -1.618034],
6: [-1.618034, 1.618034, -1],
7: [1.618034, -1.618034, 1],
8: [1.618034, -1, 1.618034],
9: [1, -1.618034, 1.618034],
10: [0, -1, 0],
11: [0, 0, -1]
This produces a drawing of the icosahedron, again scaled to be on the unit sphere (light blue), see attached video 2.
https://media.mathstodon.xyz/media_attachments/files/116/585/232/072/958/256/original/1f2695c74ce62b8c.mp4
https://media.mathstodon.xyz/media_attachments/files/116/585/232/055/658/335/original/8fd156b203dfd1ed.mp4