# A group action of the Heisenberg group with special symmetries

Suppose we look at the Heisenberg group $H_{d}$ as a matrix group of upper triangular matrices over the ring $\mathbb{Z}/d\mathbb{Z}$. You can even choose $d$ to be prime if you want. A natural irrep of $H_{d}$ acting on $\mathbb{C}^{d}$ maps the group elements into the "shift" and "phase" operators, plus roots of unity. More specifically, the two natural generators map the orthonormal basis vectors from $j \to j+1\mod d$, and the Fourier transform of that operation, plus overall phases by roots of unity. The question is this:

Can you find a unit vector $v$ such that $|(v,U_g v)| = c$ for all g not in the center of $H_{d}\ ?$ One can solve for the constant: $c=\frac{1}{\sqrt{d+1}}$.

Numerics suggests that these vectors exist in all the dimensions $< 67$, hence they may exist in every dimension, but the form of the vectors contains no (obvious) hint as to how to prove this.

This problem seems extremely truculent and any help is greatly appreciated!

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Wiki'd as open problem, see David Speyer's answer and attached comments, below. –  Scott Morrison Nov 23 '09 at 19:33
Z_d is mildly ambiguous; do you mean Z/dZ? –  Qiaochu Yuan Dec 1 '09 at 17:44

Just to check, this is the same as the conjecture of Zauner, yes?

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Yes; I'm surprised that you've heard of it. From my experience, people don't seem to know about this problem. –  Steve Flammia Oct 27 '09 at 21:46
I've heard about it from Henry Cohn, who thinks a lot about sphere packing. It was also mentioned in the comments to Scott Aaronson's post "The ten most annoying questions in quantum computing." scottaaronson.com/blog/?p=112 I have no ideas how to attack it, though. –  David Speyer Oct 27 '09 at 22:15
I would be interested in knowing what progress has been made on this problem, and what are some reasonable approaches. How would you feel about refocusing the question that way and making this a community wiki, as proposed here tea.mathoverflow.net/discussion/8/… ? –  David Speyer Oct 29 '09 at 13:35
Sure. Should I just edit the original statement, or start a new thread? –  Steve Flammia Oct 29 '09 at 22:58