Heisenberg Lie algebra

The three-dimensional Heisenberg Lie algebra has a basis $a,b,c$ with $[a,b]=c$ and $c$ central. It is a two-step nilpotent Lie algebra.

The real Heisenberg group consists of upper-unitriangular $3\times 3$ real matrices. In coordinates $(q,r,s)$ its multiplication is $(q,r,s)(q^{\prime },r^{\prime },s^{\prime })=(q+q^{\prime },r+r^{\prime },s+s^{\prime }+q{r}^{\prime })$.

For a commutative ring $R$ with an identity element, use coordinatesThese matrices form a group under multiplication, withThese formulas require no division, so they remain valid over rings with zero divisors. The center of a group is $\{A(0,y,0):y\in R\}$: the commutator has only central coordinate $x{z}^{\prime }-x^{\prime }z$, and testing $x^{\prime }=0,z^{\prime }=1$ and $x^{\prime }=1,z^{\prime }=0$ forces $x=z=0$. For finite $R$, its group order is $|R{|}^3$. This includes real, integer and modular versions of the Heisenberg group; the ring $\mathbb{Z}/4\mathbb{Z}$ must not be confused with the finite field of four elements.

In the Heisenberg group over a commutative ring with $R=\mathbb{Z}/4\mathbb{Z}$, the subset $H=\{A(t,y,t):t,y\in R\}$ is an Abelian subgroup of group order 16. Its product has central coordinate $y+y^{\prime }+t{t}^{\prime }$, symmetric in the two factors. Its inverse element is $A(-t,t^2-y,-t)$. The map $A(x,y,z)\mapsto x-z$ is a surjective group homomorphism to the additive group of $R$, with kernel of a group homomorphism $H$. Thus $H$ is a normal subgroup and has four cosets in the 64-element group.

Unlike the odd-prime case, division by two is unavailable. In fact take $u=A(1,0,1)$ and $v=A(2,0,2)$. Then $u^n=A(n,(\genfrac{}{}{0ex}{}{n}{2}),n)$, where the binomial coefficient is computed as an integer before reduction modulo four. Hence $u^4=A(0,2,0)$ and $u$ has element order eight. The element $v$ has element order two and is not a power of $u$: the powers with outer coordinates two have central coordinates one or three. Their generated subgroups intersect trivially, so the 16-element Abelian group is the direct product of groups $C_8\times C_2$.

In coordinates $h(x,y,z)=\left(\begin{array}{ccc}1& x& y\\ 0& 1& z\\ 0& 0& 1\end{array}\right)$, the Heisenberg group law is $h(x,y,z)h(a,b,c)=h(x+a,y+b+xc,z+c)$. Commutation for every $(a,b,c)$ forces $xc=az$, hence $x=z=0$. The center of a group is therefore the $y$-axis. The surjective group homomorphism $h(x,y,z)\mapsto (x,z)$ has exactly this kernel of a group homomorphism, giving the stated quotient group by the first isomorphism theorem.

For the Heisenberg group law $(x,y,z)(a,b,c)=(x+a,y+b,z+c+xb)$, the right Maurer-Cartan form has coefficients $dx,dy,dz-ydx$. They satisfy $d{\sigma }^1=d{\sigma }^2=0$ and $d{\sigma }^3={\sigma }^1\wedge {\sigma }^2$. A general right-invariant Riemannian metric is a constant positive-definite symmetric matrix in this coframe.

Every right-invariant Riemannian metric on the Heisenberg group has Killing vector fields ${\partial }_x$, ${\partial }_y+x{\partial }_z$ and ${\partial }_z$. Their Lie brackets of vector fields reproduce $[X,Y]=Z$ and the central relations. Their flows are right translations. The dual frame of the right-invariant coframe is different and has the opposite Lie-bracket sign.

The modular Heisenberg group consists of upper-unitriangular $3\times 3$ matrices over the finite field ${\mathbb{F}}_p$. In coordinates $M(x,y,z)$, multiplication is $M(x,y,z)M(x\prime ,y\prime ,z\prime )=M(x+x\prime ,y+y\prime +xz\prime ,z+z\prime )$, and its order is $p^3$. Its center of a group is $\{M(0,y,0)\}\cong C_p$; the quotient group by the centre is $C_p\times C_p$, represented by the $x,z$ coordinates. The elementary generators $a=M(1,0,0)$, $b=M(0,0,1)$, $c=M(0,1,0)$ satisfy $a^p=b^p=c^p=e$, $ab{a}^{-1}{b}^{-1}=c$, and $c$ central. For $p=2$, $(ab{)}^2=c$, so the whole group does not have exponent two.

For multiplication $(x,y,z)(a,b,c)=(x+a,y+b,z+c+xb)$, the left-invariant frame isIts Lie brackets of vector fields satisfy $[E_1,E_2]=E_3$ and the other basis brackets vanish.

The horizontal smooth distribution on the real Heisenberg group is spanned by $E_1={\partial }_x$ and $E_2={\partial }_y+x{\partial }_z$. It is not an involutive distribution because their bracket is ${\partial }_z$, but their brackets span the full tangent space. A horizontal curve satisfies $\dot{x}=\alpha$, $\dot{y}=\beta$, and $\dot{z}=x\beta$.

Every point $(P_x,P_y,P_z)$ can be reached from the identity by a smooth curve tangent to the Heisenberg horizontal distribution. Put $a=2(P_z-P_x{P}_y/2+P_x)$ and chooseTheir integrals give $x(1)=P_x$, $y(1)=P_y$, and $z(1)={\int }_0^{1}x(t)\beta (t)dt=P_x{P}_y/2-P_x+a/2=P_z$. The third coordinate measures the effect of the noncommuting horizontal directions; integrability would instead confine horizontal curves to a two-dimensional leaf.

The integer Heisenberg group consists of upper-unitriangular three-by-three integer matrices. In coordinates $(x,y,z)$ its multiplication isIt is the semidirect product ${\mathbb{Z}}^2{⋊}_A\mathbb{Z}$ forand its abelianization is ${\mathbb{Z}}^2$.

Write $M(a,b,c)=\left(\begin{array}{ccc}1& a& b\\ 0& 1& c\\ 0& 0& 1\end{array}\right)$ with integer coordinates. Multiplication in this parametrization is $M(a,b,c)M(A,B,C)=M(a+A,b+B+aC,c+C)$. The subgroup $H=\{M(0,0,t):t\in \mathbb{Z}\}$ has right cosets obtained by multiplying on the left by its elements, so its right cosets are exactly the displayed sets with fixed $(a,b)$. In particular, right cosets are indexed by ${\mathbb{Z}}^2$. Left cosets differ: $M(a,b,c)H$ fixes $a$ and $b-ac$, since right multiplication changes $b$ by $at$ and $c$ by $t$. This also exhibits why $H$ is not a normal subgroup of the Integer Heisenberg group.

In the real Heisenberg group with multiplication $(x,y,z)(x^{\prime },y^{\prime },z^{\prime })=(x+x^{\prime },y+y^{\prime },z+z^{\prime }+x{y}^{\prime })$, the displayed set is a discrete cocompact subgroup. Writing $a=(k,0,0)$, $b=(0,k,0)$, $c=(0,0,k)$ gives $[a,b]=c^k$ and $c$ central. Hence its abelianization is ${\mathbb{Z}}^2\oplus \mathbb{Z}/k$. The right quotient maps to ${\mathbb{R}}^2/k{\mathbb{Z}}^2$ by its first two coordinates; each fibre is the central circle $\mathbb{R}/k\mathbb{Z}$.

Write the Integer Heisenberg group with central generator $c=[a,b]$, so $ab=cba$. Moving $a$ successively past powers of $b$ givesFor positive exponents this follows by two inductions; inverses give the remaining signs. The representation $a=I+E_{12}$, $b=I+E_{23}$, $c=I+E_{13}$ checks the identity and shows $c$ has infinite order.

On $L^2(\mathbb{R})$, translations and multiplication by phases give the Schrödinger representation of the Heisenberg group. The central coordinate acts by an overall phase.

On $\mathbb{C}[x]$, the operators $d/dx$, multiplication by $x$, and the identity satisfy $[d/dx,x]=1$. Sending $a,b,c$ to these operators gives a faithful irreducible infinite-dimensional representation of the Heisenberg Lie algebra.