November 2020 - mathfun@mailman.xmission.com

Re: [math-fun] A pretty surface with surprising symmetry
by Dan Asimov 03 Nov '20

03 Nov '20

I think that for compact orientable surfaces M and N, the condition that 𝜒(M) = d 𝜒(N) for some d ∊ Z+ implies that M is a covering space of N. So as Allan was hoping, the converse is true. Of course there are also lots of maps between surfaces that aren't covering maps. If g ≥ h then there is always a continuous map M_g —> M_h. Many of these are "branched covering" maps, which if the surfaces are given appropriate metrics can be realized as holomorphic maps between Riemann surfaces. I.e., they preserve angles, except at finitely many points where the angle is multiplied by a positive integer > 1. These are called "branched coverings". Particularly beautiful maps are based on the so-called Weierstrass P-function, which defines a branched covering S^2 —> T^2. —Dan Allan Wechsler wrote: ----- Oh, Dan, that's perfect. It slots into a place in my brain so nicely that I think I must have known it before and forgotten it. It explains the presence of a 2-to-1 map from the sphere to the projective plane, maps of arbitrary index from the torus to itself, the *absence *of maps between the sphere and pretty much everything else *except* the projective plane. But there must be all sorts of weird maps from, say, the triple donut (characteristic -4) to the double donut (characteristic -2). Though, come to think of it, your theorem doesn't demonstrate that such maps are always possible, only that sometimes they are clearly impossible. -----

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[math-fun] 73 (was Re: A pretty surface with surprising symmetry)
by Keith F. Lynch 03 Nov '20

03 Nov '20

Dan Asimov <dasimov(a)earthlink.net> wrote: > Or consider the number 73. Not particularly conducive to symmetry, > you might think. 73 is a palindrome in Morse Code: --... ...-- It's also a common greeting among hams.

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Re: [math-fun] A pretty surface with surprising symmetry
by Dan Asimov 03 Nov '20

03 Nov '20

If this "d-to-1 map" h : X —> Y is a covering map, then the covering space (the domain) will have d times the number of vertices, edges, and faces of the image. This means 𝜒(X) = d 𝜒(Y). I presume we're talking about compact orientable surfaces. Since for them, 𝜒 = 2 - 2g, this tells us that 2 - 2g_X = d (2 - 2g_Y), or g_X = d g_Y - d + 1 —Dan Jim Propp wrote: ----- What is the relation between the genus of X and the genus of Y when there is a d-to-1 map from X to Y? (Assume that around each y in Y we can find a disk whose preimage consists of d disks.) Do we have genus(X) = d genus(Y) ? -----

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Re: [math-fun] A pretty surface with surprising symmetry
by Dan Asimov 03 Nov '20

03 Nov '20

I think both the genus g and the Euler characteristic bear on what symmetries a surface can have. The (orientable, compact) surface M_g of genus g always has a metric with an isometry of order g: Just arrange the holes in a circle. The surface of genus g = K + 1 always has a metric with an isometry of order K: Arrange one hole in the middle and the other K around it. The surface M_3 of genus 3 (with X(M_g) = -4) has a metric with an isometry of order 7. M_3 cannot be exhibited in 3-space with this 7-fold symmetry. It has other metrics, like one having 192 symmetries, including a subgroup of order 64. It has no metric with both a 7-fold symmetry and a subgroup of order 64, since a metric surface of genus g can have at most 168 * (g-1) symmetries and 7*64 = 448 > 336 would violate that theorem. —Dan Allan Wechsler wrote: ----- My thought is that if the genus is 73, the Euler characteristic is -144. 73 may seem like a number that is not very conducive to fancy symmetry, but -144 seems a lot more promising. I'm guessing that the Euler characteristic is more important to allowable symmetries than the genus is. Other examples can be harvested from the thickened skeleta of the platonic solids treated as surfaces. I expect the genus plus one to be a nicely divisible number. The tetrahedron's skeleton has genus 3; 4 has better divisibility. The cube has genus 5 but 6 is better. The octahedron: 7 versus 8. The dodecahedron: 11 versus 12. The icosahedron: 19 versus 20. In other words, I don't think the symmetries are juggling a set whose size is the genus; I think the relevant size is the genus plus 1. -----

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Re: [math-fun] A pretty surface with surprising symmetry
by Dan Asimov 01 Nov '20

01 Nov '20

Michael Kleber wrote: ----- I have no intuition for where the genus of 73 comes from. Can you offer any way to intuit that, beyond just doing a calculation? Is that somehow 1 + 36 + 36, with a 36 for each 1-skeleton? ----- Not sure how to go beyond a calculation. But that calculation is this: The 1-skeleton of the 24-cell has 24 vertices and 96 edges, so its Euler characteristic is X = V - E = -72. Same for the dual 1-skeleton. Since a (tubular) neighborhood of the 1-skeleton in the 3-sphere S^3 can be deformed by shrinking down to the 1-skeleton itself, that also has Euler characteristic equal to -72. Now the tubular neighborhoods of the 1-skeleton and the dual 1-skeleton can be inflated until they just overlap in their common boundary, which is the surface in question, M_g. Now we can express the Euler characteristic of the 3-sphere S^3 (which, like for every odd-dimensional sphere, is 0) as the sum of the Euler characteristics of each tubular neighborhood, minus the Euler characteristic X(M_g) of their overlap, the surface M_g: 0 = X(S^3) = -72 + -72 - X(M_g) so X(M_g) = -144. Since the Euler characteristic and genus of an orientable (compact) surface M_g are always related by the equation X(M_g) = 2 - 2g it follows that g = 73. Sorry this isn't very intuitive. —Dan ----- I'm always struck by instances in math unexpected symmetry. ... ... Or consider the number 73. Not particularly conducive to symmetry, you might think. But recall the regular 4D polytope called the 24-cell: It's 24 octahedra arranged face-to-face to form the 3-dimensional sphere S^3. In fact it has 24 vertices, 96 edges, 96 2-dimensional faces, and 24 3-dimensional octahedra. This makes it self-dual. So if you draw the vertices and edges of this 24-cell on the 3-sphere S^3, you can also draw the *dual* vertices and edges as well — they will form a disjoint 1-dimensional figure in S^3. Now if you removed both of these "1-skeleta" from S^3, the portion of S^3 that remains is topologically equivalent to the product of a surface M_g of genus g with the real line R. And it turns out that the genus of g M_g is 73. Even nicer, it turns out that the surface of genus 73 that is halfway between the two dual 1-skeleta in S^3 is a *minimal surface*, meaning that if you drew a small simple closed curve C about any of its points, the surface minimizes the area among all surfaces in S^3 whose boundary is the curve C. Even nicer yet, this minimal surface M_73 has an isometry with itself for any isometry of the 24-cell. One of the octahedra can be taken to any of the 24 octahedra, and one of its triangles can go to any of the 8 triangular faces in any of 6 dihedral way, making for 24 x 24 x 2 = 1152 isometries. Nicest of all, the two halves of S^3 created when M_73 is removed can be interchanged, which carries M_73 to itself, giving it another 1152 isometries for a grand total of 2304 altogether. Who said 73 wasn't symmetrical. —Dan -----

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[math-fun] Quaternion eigenvectors, Part Deux
by Henry Baker 01 Nov '20

01 Nov '20

The following exposition is known [Hitzer], but not well known, as Wikipedia & other obvious sources don't cover it. It is well known that every 4D rotation R can be expressed as Q -> FQG, for quaternions F,G which depend upon the rotation R. It is also well known that every 4D rotation can be expressed as the *independent* rotations of 2 planes orthogonal to one another, each with their own separate rotation. Unfortunately, it isn't so well known how to elegantly find the rotation R and the 2 orthogonal subspaces given the two quaternions F,G. Given *unit* quaternions F,G, we can extract the 'scalar' and 'vector' parts of F,G as follows: s(F) = (F+F')/2 ; scalar part of F s(G) = (G+G')/2 ; scalar part of G V(F) = (F-F')/2 ; 'pure' vector part of F V(G) = (G-G')/2 ; 'pure' vector part of G where F',G' are the 'conjugates' of F,G, respectively. We will also need the 'pure', 'unit' vectors f,g: f = V(F)/|V(F)| = V(F)/sqrt(|V(F)|^2) = V(F)/sqrt(-V(F)^2) g = V(G)/|V(G)| = V(G)/sqrt(|V(G)|^2) = V(G)/sqrt(-V(G)^2) (Since |V|^2 = VV' = V(-V) = -V^2 for any *pure* vector V.) We now have s(f)=s(g)=0 and |f|=|g|=1. Now |F|=|G|=1, so we can express F,G as: F = exp(f*a) = cos(a)+f*sin(a), for some real angle a. G = exp(g*b) = cos(b)+g*sin(b), for some real angle b. Here's the tough part: finding the eigenvectors of the linear operator Q->FQG. Consider the quaternions 1+fg and f-g. Some observations: s(f-g) = 0 ; pure vector <=> 0 scalar part s(1+fg) = s(1)+s(fg) = 1-(f.g) ; (f.g) is vector dot product = 1-cos(theta) ; theta is angle between f,g = 2 sin(theta/2)^2 n(f-g) = (f-g)(f-g)' = (f-g)(f'-g') = ff'+gg'-gf'-fg' = 1 + 1 - (gf'+fg') = 2 + fg + gf = 2 + 2 s(fg) = 2 (1 + s(fg)) = 2 (1 - f.g) = 2 (1-cos(theta)) ; theta is angle between f,g = 4 sin(theta/2)^2 So |f-g| = 2 sin(theta/2). n(1+fg) = (1+fg)(1+fg)' = (1+fg)(1+g'f') = 1 + fg + g'f' + fgg'f' = 2 + 2 s(fg) = 2 + 2 (-f.g) = 2 - 2 f.g = 2 (1-f.g) = 2 (1-cos(theta)) = 4 sin(theta/2)^2 So |1+fg| = 2 sin(theta/2). dot(f-g,1+fg) = ((f-g)(1+fg)'+(1+fg)(f-g)')/2 = ((f-g)(1+g'f')+(1+fg)(f'-g'))/2 = ((f-g-gg'f+fg'f')+(f'-g'+fgf'-fgg'))/2 = (f+fg'f'-g-f'+f'-g'+fgf'-f)/2 = (fg'f'-g-g'+fgf')/2 = (f(-g)(-f)-g-(-g)+fg(-f))/2 = (fgf-fgf)/2 = 0 So, 1+fg is perpendicular to f-g in 4D, but both have the same length. Consider the operation of f()g on 1+fg and f-g: f(1+fg)g = (f+ffg)g = fg+ffgg = fg+(-1)(-1) = fg+1 = 1+fg f(f-g)g = (ff-fg)g = ffg-fgg = (-1)g-f(-1) = -g+f = f-g So, 1+fg and f-g are *eigenvectors* for the eigenvalue +1 ! Similarly, consider 1-fq and f+g: s(f+g) = 0 ; pure vector <=> 0 scalar part s(1-fg) = s(1)-s(fg) = 1+(f.g) ; (f.g) is vector dot product = 1+cos(theta) ; theta is angle between f,g = 2 cos(theta/2)^2 n(f+g) = (f+g)(f'+g') = ff'+gg'+fg'+gf' = 1 + 1 - fg - gf = 2 - (fg+gf) = 2 - 2 s(fg) = 2 - 2 (-f.g) = 2 + 2 f.g = 2 (1+f.g) = 2 (1+cos(theta)) = 4 cos(theta/2)^2 So |f+g| = 2 cos(theta/2). n(1-fg) = (1-fg)(1-fg)' = (1-fg)(1-g'f') = 1 - fg - g'f' + fgg'f' = 1 - fg - gf + 1 = 2 - (fg+gf) = 2 - 2 s(fg) = 2 - 2 (-f.g)) = 2 + 2 f.g = 2 + 2 cos(theta) = 2 (1+cos(theta)) = 4 cos(theta/2)^2 So |1-fg| = 2 cos(theta/2). dot(f+g,1-fg) = ((f+g)(1-fg)'+(1-fg)(f+g)')/2 = ((f+g)(1-g'f')+(1-fg)(f'+g'))/2 = ((f+g-fg'f'-gg'f')+(f'+g'-fgf'-fgg'))/2 = (f+g-fg'f'-f'+f'+g'-fgf'-f)/2 = (-fg'f'-fgf'+g+g')/2 = (-fgf+fgf)/2 = 0 So f+g is perpendicular to 1-fg in 4D, and both have the same length. Consider the operation of f()g on f+g and 1-fg: f(f+g)g = ffg+fgg = (-1)g+f(-1) = -g-f = -(f+g) f(1-fg)g = (f-ffg)g = fg-ffgg = fg-(-1)(-1) = fg-1 = -(1-fg) So, f+g and 1-fg are *eigenvectors* for the eigenvalue -1 ! Finally, the two 2D subspaces spanned by (1+fg,f-g) and (f+g,1-fg) respectively are orthogonal to one another: dot(f+g,f-g) = ((f+g)(f-g)'+(f-g)(f+g)')/2 = ((f+g)(f'-g')+(f-g)(f'+g'))/2 = (ff'-gg'+gf'-fg'+ff'-gg'-gf'+fg')/2 = (gf'-fg'-gf'+fg')/2 = 0 dot(1-fg,1+fg) = ((1-fg)(1+fg)'+(1+fg)(1-fg)')/2 = ((1-fg)(1+g'f')+(1+fg)(1-g'f'))/2 = (1-fg+g'f'-fgg'f'+1-g'f'+fg-fgg'f')/2 = (1-fg+g'f'-1+1-g'f'+fg-1)/2 = 0 dot(f+g,1+fg) = ((f+g)(1+fg)'+(1+fg)(f+g)')/2 = ((f+g)(1+g'f')+(1+fg)(f'+g'))/2 = (f+g+fg'f'+gg'f'+f'+g'+fgf'+fgg')/2 = (f+g+fg'f'+f'+f'+g'+fgf'+f)/2 = (f+g+fgf-f-f-g-fgf+f)/2 = 0 dot(f-g,1-fg) = ((f-g)(1-fg)'+(1-fg)(f-g)')/2 = ((f-g)(1-g'f')+(1-fg)(f'-g'))/2 = (f-g-fg'f'+gg'f'+f'-g'-fgf'+fgg')/2 = (f-g-fgf+f'+f'-g'+fgf+f)/2 = (f-g-fgf-f-f+g+fgf+f)/2 = 0 Since our four 4D vectors (1+fg,f-g,f+g,1-fg) are all orthogonal to one another, and the dimension of the quaternions is four, these four vectors span the space of all quaternions. Although our four 4D vectors (1+fg,f-g,f+g,1-fg) are orthogonal, they are NOT normalized, as |1+fg| = |f-g| = 2 sin(theta/2) |1-fg| = |f+g| = 2 cos(theta/2) Nevertheless, it is remarkable and convenient that the 2 vectors spanning the +1 subspace have the *same length*, and the 2 vectors spanning the -1 subspace also have the same length. Now that we have basis vectors for the entire space and the two orthogonal subspaces, we would still love to quickly 'split' a given quaternion into two parts, the part with eigenvalue +1 and the part with eigenvalue -1. Now consider the following *projection* operations: Pp: Q -> (Q+fQg)/2 Pm: Q -> (Q-fQg)/2 A projection operation is idempotent: Pp(Pp(Q)) = Pp((Q+fQg)/2)) = ((Q+fQg)/2 + f(Q+fQg)g/2)/2 = (Q/2 + fQg/2 + fQg/2 + ffQgg/2)/2 = (Q/2 + fQg + (-1)Q(-1)/2)/2 = (Q+fQg)/2 = Pp(Q) Pm(Pm(Q)) = Pm((Q-fQg)/2) = ((Q-fQg)/2 - f(Q-fQg)g/2)/2 = (Q/2 -fQg/2 - fQg/2 + ffQgg/2)/2 = (Q/2 -fQg + Q/2)/2 = (Q-fQg)/2 = Pm(Q) These project operators cancel one another: Pp(Pm(Q)) = Pp((Q-fQg)/2) = ((Q-fQg)/2 + f(Q-fQg)g/2)/2 = (Q/2 - fQg/2 + fQg/2 -ffQgg/2)/2 = (Q/2 - Q/2)/2 = 0 Pm(Pp(Q)) = Pm((Q+fQg)/2) = ((Q+fQg)/2 - f(Q+fQg)g/2)/2 = (Q/2 + fQg/2 - fQg/2 - ffQgg/2)/2 = (Q/2 - Q/2)/2 = 0 Now consider the operation of Pp, Pm on our four basis vectors (1+fg,f-g,f+g,1-fg): Pp(1+fg) = (1+fg + f(1+fg)g)/2 = (1 + fg + fg + ffgg)/2 = (2+2fg)/2 = 1+fg Pp(f-g) = (f-g + f(f-g)g)/2 = (f - g + ffg - fgg)/2 = (f - g - g + f)/2 = (2f - 2g)/2 = f-g Pp(f+g) = (f+g + f(f+g)g)/2 = (f + g + ffg + fgg)/2 = (f + g - g - f)/2 = 0 Pp(1-fg) = (1-fg + f(1-fg)g)/2 = (1 - fg + fg - ffgg)/2 = 0 Pm(1+fg) = (1+fg - f(1+fg)g)/2 = (1 + fg - fg -ffgg)/2 = 0 Pm(f-g) = (f-g - f(f-g)g)/2 = (f - g - ffg + fgg)/2 = (f - g + g - f)/2 = 0 Pm(f+g) = (f+g - f(f+g)g)/2 = (f + g - ffg - fgg)/2 = (f + g + g + f)/2 = f+g Pm(1-fg) = (1-fg - f(1-fg)g)/2 = (1 - fg - fg + ffgg)/2 = (1 - 2fg + 1)/2 = (2-2fg)/2 = 1-fg So we can use Pp, Pm to separate a given quaternion into 2 components: Q = Pp(Q)+Pm(Q). Pp(Q)+Pm(Q) = (Q+fQg)/2 + (Q-fQg)/2 = Q/2 + fQg/2 + Q/2 - fQg/2 = Q This analysis shows the two eigenvalues +-1 operating on their own separate 2D subspaces and components: fQg = f(Pp(Q)+Pm(Q))g = f(Pp(Q))g + f(Pm(Q))g = (+1)Pp(Q) + (-1)Pm(Q) = Pp(Q) - Pm(Q) But wait, there's more! n(Q) = QQ' = n(Pp(Q)+Pm(Q) = n(Pp(Q)) + n(Pm(Q)) = Pp(Q)Pp(Q)' + Pm(Q)Pm(Q)' So the 'norm' n(Q) also splits into the sum of the norms of the 2 projected components. We have so far restricted ourselves to the pure unit vectors f,g. But F,G are full (unit) quaternions complete with scalar parts: F = exp(f*a) = cos(a)+f*sin(a), for some real angle a. G = exp(g*b) = cos(b)+g*sin(b), for some real angle b. Given Q=Pp(Q)+Pm(Q), we can analyze FQG as follows: FQG = F(Pp(Q)+Pm(Q))G = F(Pp(Q))G + F(Pm(Q))G = exp(f*a)Pp(Q)exp(g*b) + exp(f*a)Pm(Q)exp(g*b) = exp(f*(a-b))Pp(Q) + exp(f*(a+b))Pp(Q) = exp(f*a) (exp(-f*b)Pp(Q) + exp(f*b)Pp(Q)) or FQG = Pp(Q)exp(g*(b-a)) + Pm(Q)exp(g*(b+a)) = (Pp(Q)exp(-g*a) + Pm(Q)exp(g*a)) exp(g*b) So we can either express the result of FQG in terms of exp(f*something) or exp(g*something). We can now relax the requirement that |F|=|G|=1. The projection operators still work correctly, but we must insert |F| and |G| where appropriate into our equations, e.g., FQG = |F|exp(f*a)Pp(Q)|G|exp(g*b) + |F|exp(f*a)Pm(Q)|G|exp(g*b) = |F|exp(f*(a-b))Pp(Q)|G|+ |F|exp(f*(a+b))Pp(Q)|G| = |F|exp(f*a) (exp(-f*b)Pp(Q) + exp(f*b)Pp(Q))|G| = F (exp(-f*b)Pp(Q) + exp(f*b)Pp(Q)) |G| Or, FQG = |F|Pp(Q)exp(g*(b-a))|G| + |F|Pm(Q)exp(g*(b+a))|G| = |F| (Pp(Q)exp(-g*a) + Pm(Q)exp(g*a)) exp(g*b)|G| = |F| (Pp(Q)exp(-g*a) + Pm(Q)exp(g*a)) G ---------------------------------- [Hitzer] Google "Hitzer" "quaternion" in scholar.google.com to pull up a number of relevant articles by Eckhard Hitzer, e.g., Eckhard Hitzer, "The orthogonal planes split of quaternions and its relation to quaternion geometry of rotations". 30th International Colloquium on Group Theoretical Methods in Physics (Group30), Journal of Physics: Conference Series 597 (2015) 012042, doi:10.1088/1742-6596/597/1/012042. https://iopscience.iop.org/article/10.1088/1742-6596/597/1/012042/pdf

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Re: [math-fun] A pretty surface with surprising symmetry
by Dan Asimov 01 Nov '20

01 Nov '20

As another math-fun member very kindly pointed out: 24*24*2 = 1152 (not 1052 as I had it). And that makes 24*24*2*2 = 48^2 = 2304 (not 2104 as I had it). Now fixed below. —Dan -------- Quoting Dan Asimov <dasimov(a)earthlink.net>: > I'm always struck by instances in math unexpected symmetry. > For instance, consider ways of coloring the tiles of the > hexagonal tessellation of the plane with n colors so that > no two tiles of the same color share an edge. > > One way to do this involves 7 colors. It could be like this: > > 0 1 2 3 4 5 6 0 1 2 3 4 5 6 0 1 2 > > 5 6 0 1 2 3 4 5 6 0 1 2 3 4 5 6 0 > > 2 3 4 5 6 0 1 2 3 4 5 6 0 1 2 3 4 > > 0 1 2 3 4 5 6 0 1 2 3 4 5 6 0 1 2 > > 4 5 6 0 1 2 3 4 5 6 0 1 2 3 4 5 6 > > 2 3 4 5 6 0 1 2 3 4 5 6 0 1 2 3 4 > > 6 0 1 2 3 4 5 6 0 1 2 3 4 5 6 0 1 > > 4 5 6 0 1 2 3 4 5 6 0 1 2 3 4 5 6 > > 1 2 3 4 5 6 0 1 2 3 4 5 6 0 1 2 3 > > 6 0 1 2 3 4 5 6 0 1 2 3 4 5 6 0 1 > > 3 4 5 6 0 1 2 3 4 5 6 0 1 2 3 4 5 > > 1 2 3 4 5 6 0 1 2 3 4 5 6 0 1 2 3 > > 5 6 0 1 2 3 4 5 6 0 1 2 3 4 5 6 0 > > 3 4 5 6 0 1 2 3 4 5 6 0 1 2 3 4 5 > > (and repeat). So here's a tessellation by 6-sided figures that is colored > with 7 colors, each color forming a subset of the hexagons isometric to > that of any other color. > > Or consider the number 73. Not particularly conducive to symmetry, you might > think. But recall the regular 4D polytope called the 24-cell: It's 24 > octahedra arranged face-to-face to form the 3-dimensional sphere S^3. In fact > it has 24 vertices, 96 edges, 96 2-dimensional faces, and 24 3-dimensional > octahedra. This makes it self-dual. > > So if you draw the vertices and edges of this 24-cell on the 3-sphere S^3, > you can also draw the *dual* vertices and edges as well ? they will form > a disjoint 1-dimensional figure in S^3. Now if you removed both of these > "1-skeleta" from S^3, the portion of S^3 that remains is topologically > equivalent to the product of a surface M_g of genus g with the real line R. > And it turns out that the genus of g M_g is 73. > > Even nicer, it turns out that the surface of genus 73 that is halfway between > the two dual 1-skeleta in S^3 is a *minimal surface*, meaning that > if you drew > a small simple closed curve C about any of its points, the surface minimizes > the area among all surfaces in S^3 whose boundary is the curve C. > > Even nicer yet, this minimal surface M_73 has an isometry with itself for > any isometry of the 24-cell. One of the octahedra can be taken to any of the > 24 octahedra, and one of its triangles can go to any of the 8 > triangular faces > in any of 6 dihedral way, making for 24 x 24 x 2 = 1152 isometries. > > Nicest of all, the two halves of S^3 created when M_73 is removed can be > interchanged, which carries M_73 to itself, giving it another 1152 isometries > for a grand total of 2304 altogether. > > Who said 73 wasn't symmetrical. > > ?Dan -----

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[math-fun] One parameter quintic solution
by Bill Gosper 01 Nov '20

01 Nov '20

This is probably ho-hum, but while trying to simplify a Chowla-surdberg, I found that 0 = a + 80 x + 20 x^3 + x^5 has solutions {((-a - Sqrt[4096 + a^2])^(1/5) E^((2 I \[Pi])/5))/2^( 1/5) + ((-a + Sqrt[4096 + a^2])^(1/5) E^((2 I \[Pi])/5))/2^( 1/5), ((-a + Sqrt[4096 + a^2])^(1/5) E^(-((2 I \[Pi])/5)))/2^( 1/5) + ((-a - Sqrt[4096 + a^2])^(1/5) E^(-((4 I \[Pi])/5)))/2^( 1/5), ((-a - Sqrt[4096 + a^2])^(1/5) E^(-((2 I \[Pi])/5)))/2^( 1/5) + ((-a + Sqrt[4096 + a^2])^(1/5) E^(-((4 I \[Pi])/5)))/2^( 1/5), (-a + Sqrt[4096 + a^2])^(1/5)/2^( 1/5) + ((-a - Sqrt[4096 + a^2])^(1/5) E^((4 I \[Pi])/5))/2^( 1/5), (-a - Sqrt[4096 + a^2])^(1/5)/2^( 1/5) + ((-a + Sqrt[4096 + a^2])^(1/5) E^((4 I \[Pi])/5))/2^(1/5)} Is it from a two-parameter family? C.f. https://en.wikipedia.org/wiki/Quintic_function#Other_solvable_quintics --rwg

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