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I have seen two constructions called the Heisenberg group. If $k$ is a field of characteristic not equal to $2$ and $V$ is a $2n$-dimensional vector space over $k$ with symplectic form $\omega : V \times V \to k$, one of these groups is $V \times k$ with the operation $(v,a) \cdot (w,b) = (v+w,a+b+\frac12 \omega(v,w))$. The other group is the subgroup of $\text{GL}_{n+2}(k)$ consisting of matrices with $1$s along the diagonal and $0$s elsewhere, except for the top row and rightmost column. This construction has the advantage of working over $k$ of arbitrary characteristic, but unfortunately uses coordinates.

I have two questions:

  1. When $k$ does not have characteristic $2$, are these two groups isomorphic?

  2. Is there a coordinate-free construction of the Heisenberg group in characteristic $2$ (preferably in terms of a symplectic vector space over $k$)?

Edit: Will Sawin has answered the first question in the affirmative. The second question still remains, so let me phrase it more precisely: if $k$ is a field of characteristic $2$ and $V$ a symplectic vector space over $k$, can we give a coordinate-free construction of the Heisenberg group in terms of $V$ which is isomorphic to the matrix group described above? (I am not even sure that we can always choose a symplectic basis in characteristic $2$.)

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I am not convinced that the description you're looking for exists in characteristic $2$. In characteristic not equal to $2$, the Heisenberg Lie algebra is $V \otimes k$ with Lie bracket coming from the symplectic form and the Heisenberg group is obtained by exponentiating it, which really does require division by $2$. More generally it seems to me that Lie algebras are definitely the wrong thing to consider in characteristic $2$ (at least the motivating argument I know for writing down the Lie algebra axioms fails in characteristic $2$). – Qiaochu Yuan Jun 14 '12 at 21:45
Sorry, that should be $V \oplus k$. – Qiaochu Yuan Jun 14 '12 at 21:45
@Qiaochu: OK, you convinced me that this is the wrong thing to look at. I found a couple of papers which handle the case of finite fields of characteristic $2$: it seems that the correct thing to do is replace the field with its associated ring of Witt vectors of length $2$. I wonder what the right approach is over other fields of characteristic $2$? – Justin Campbell Jun 14 '12 at 22:33
I thought I gave a negative answer to the second question. Given a Heisenberg group, you get a vector space (the central quotient) with a symplectic form (the commutator). But you have other data, not determined by the symplectic form. In addition to $xyx^{-1}y^{-1}$ mapping you to the center, in characteristic $2$, $x^2$ also does the job. So your vector space has a quadratic form as well. – Will Sawin Jun 16 '12 at 17:54
up vote 4 down vote accepted

Part 1:

To show that this is isomorphic to the other one, fix a symplectic basis for the symplectic vector space. The isomorphism sets the coefficients of the basis elements of $v$ equal to the matrix entires in the top row and rightmost column other than the top-right corner, and $a$ equal to the entry in the top-right corner minus an adjustment which is the product of the top row and rightmost column of the matrix divided by two. For simplicity, look at three-by-three matrices:

$$\begin{pmatrix} 1 & x_1 & z_1 \\ 0 & 1 & y_1 \\ 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & x_2 & z_2 \\ 0 & 1 & y_2 \\ 0 & 0 & 1 \end{pmatrix}=\begin{pmatrix} 1 & x_1+x_2 & z_1+z_2+x_1y_2 \\ 0 & 1 & y_1+y_2 \\ 0 & 0 & 1 \end{pmatrix} $$

Set $a=z+xy/2$, then:


This is the formula for the first definition.

This gives the isomorphism.

Part 2:

One cannot solve this problem, using just a symplectic structure, in characteristic $2$. The problem is, given a symplectic vector space, to canonically form a group, isomorphic to the Heisenberg group, with a surjection onto the vector space. A two-dimensional vector space over $\mathbb F_2$ has a unique symplectic structure, but there are three different surjections from the Heisenberg group $D_4$ to it.

I guess one could solve it by placing some additional structure on the space. I'm not sure what that structure should be.

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If you mean what I think you mean by "symplectic basis," then it seems to me that the group operations (even after removing the $\frac12$) do not agree. I don't know if I can or should write out matrix computations in a comment box, but I think in the $3 \times 3$ case that the matrix product is not compatible with $\omega(q,p)=−1$, where $p,q$ is a symplectic basis for $V$. One can see this at a glance, since no minus signs appear anywhere in the matrix product. – Justin Campbell Jun 14 '12 at 19:22
Also, I don't agree that there is no reason to put the $\frac12$ there: it appears when we exponentiate the Heisenberg Lie algebra. Is the group we obtain after removing the $\frac12$ isomorphic to what we started with? – Justin Campbell Jun 14 '12 at 19:27
Your first objection is correct. My formula for the isomorphism is wrong, and the correct formula looks nicer with a minus sign in place. Removing the $1/2$ is an isomorphism - just multiply all values of $a$ by $2$ while keeping $v$ in place. However that prevents the matrix=coordinate-free isomorphism from working in characteristic two. So I don't know what to do there. – Will Sawin Jun 14 '12 at 19:45
Indicentally, for formulas like yours, there's no need to restrict to $3\times 3$ matrices. Just declare that the $y_i$s are column vectors (elements of some vector space $V$), the $x_i$s are row vectors (elements of $V^*$), their product means their dot product (contraction), and the middle $1$ is the unit matrix. – Theo Johnson-Freyd Jun 15 '12 at 3:06

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