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 $2$.)



Heisenberg group in characteristic two
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:


$a\_{12}=z\_1+z\_2+x\_1y\_2-(x\_1+x\_2)(y\_1+y\_2)/2=a\_1+a\_2+(x\_1y\_2-x\_2y\_1)/2$


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.


