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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
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2 Answers

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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Heisenberg groups can be viewed as functors from commutative rings to groups. For example, you can start with the upper triangular matrix construction over integers (without the factor of $\frac12$ you included), and pass to other commutative rings by base change. If you're not concerned with coordinate independence, then all (finite dimensional) Heisenberg groups over all fields arise this way.

For your second question, if you want to construct a Heisenberg group in a coordinate free way, you take any projective $R$-module with a symplectic form, and define the central extension of $M$ by $R$ by taking the 2-cocycle to be the form you chose. Since bilinearity implies the cocycle condition, this construction is a functor on symplectic modules over a commutative ring, i.e., module symplectomorphisms are taken to group isomorphisms. This is essentially the same as your first construction, but with $\omega$ instead of $\frac12\omega$. This construction can be generalized to the setting of schemes and similar objects by using locally free sheaves, and one may also twist the center $R$ to be a line bundle - this appears, e.g., in the beginning of Lysenko's paper on metaplectic bundles.

If $2$ is invertible in $R$, then one may consider the cocycle $\frac12\omega$ instead of $\omega$, and there is a straightforward isomorphism between the central extensions as groups, given by dilating the center. If your reason for wanting the $\frac12$ normalization is so you can have compatibility with exponentiation, then that reason disappears when working in small characteristic, where exponentiation doesn't make sense.

To answer your first question in a little more generality, we need to consider a projective $R$-module $M$ equipped with a symplectic form $\omega$ and a polarization $M \cong U \oplus U^\ast$, where $U^\ast = \operatorname{Hom}_R(U,R)$ and both $U$ and $U^\ast$ are isotropic. Given these data, you may construct a group of $3 \times 3$ upper-triangular matrices $\left( \begin{smallmatrix} 1 & u & r \\ 0 & 1 & \phi \\ 0 & 0 & 1 \end{smallmatrix} \right)$, where $u$ ranges over elements of $U$, $\phi$ ranges over elements of $U^\ast$, and $r$ ranges over elements of $R$. The multiplication between elements of $U$ and $U^\ast$ is given by the duality pairing: $u\phi = \phi(u)$. It is straightforward to check that this gives you the same 2-cocycle as the symplectic form. In particular, the central extension is independent of the choice of polarization.

When $R$ is a field, polarizations always exist for symplectic vector spaces. This can be proved by showing that you can always extend an isotropic space and its dual by adding basis vectors, if the spaces are not Lagrangian.

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The appropriate group cohomology cocycles in characteristic $2$ are not the same as symplectic forms, though. The space is bilinear forms modulo forms of type $f(a,b)=g(a+b)-g(a)-g(b)$. This maps onto the space of symplectic forms by $s(a,b)=f(a,b)-f(b,a)$, but the kernel, symmetric bilinear forms modulo forms of type $f(a,b)=g(a+b)-g(a)-g(b)$, has nontrivial $2$-part, since the inversion formula $g(a)=f(a,a)/2$ is only defined in $\mathbb Z[1/2]$. –  Will Sawin Jun 15 '12 at 18:33
The antisymmetrizing map does not quite land in symplectic forms, due to the nondegeneracy condition. However, I see that I was mistaken in thinking I could avoid the asymmetric lift of $\omega$. –  S. Carnahan Jun 16 '12 at 2:33
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