= Solution
Multiplying the matrices gives the <Heisenberg group> law
$$
(x,y,z)(a,b,c)=(x+a,y+b,z+c+xb),\qquad
(x,y,z)^{-1}=(-x,-y,-z+xy).
$$
Differentiating at the identity gives all strictly upper-triangular matrices. With $X=E_{12}$, $Y=E_{23}$ and $Z=E_{13}$, the <commutator> gives
$$
\boxed{\mathfrak g=\operatorname{span}\{X,Y,Z\},\qquad [X,Y]=Z,\quad [X,Z]=[Y,Z]=0.}
$$
This is the <Heisenberg Lie algebra>.
The right <Maurer-Cartan form> is
$$
dg\,g^{-1}=X\,dx+Y\,dy+Z\,(dz-y\,dx).
$$
Thus the <right-invariant coframe of the real Heisenberg group> is $\sigma^1=dx$, $\sigma^2=dy$, $\sigma^3=dz-y\,dx$. Under a right translation,
$$
R_{(a,b,c)}(x,y,z)=(x+a,y+b,z+c+xb),
$$
we have $dx'=dx$, $dy'=dy$, and $dz'-y'dx'=dz-y\,dx$. All three <right-invariant differential forms> are preserved.
Every <right-invariant Riemannian metric> is determined by an arbitrary <inner product> at the identity. In this coframe its most general expression is
$$
\boxed{h=H_{ij}\,\sigma^i\sigma^j,\qquad H=H^{\mathsf T}>0,\quad H_{ij}\text{ constant}.}
$$
Equivalently,
$$
\begin{aligned}
h={}&H_{11}dx^2+H_{22}dy^2+H_{33}(dz-y\,dx)^2\\
&+2H_{12}dx\,dy+2H_{13}dx(dz-y\,dx)+2H_{23}dy(dz-y\,dx).
\end{aligned}
$$
Products here are symmetric products. If a pseudo-Riemannian <metric tensor> is intended, replace positive definiteness by nondegeneracy. Since every right translation preserves the coframe, it is an <isometry>. The action is faithful, and $a\mapsto R_{a^{-1}}$ is a group homomorphism, embedding a copy of $G$ into the <isometry group>.
The one-parameter right translations produce the <Killing frame for a right-invariant Heisenberg metric>:
$$
\boxed{K_X=\partial_x,\qquad K_Y=\partial_y+x\partial_z,\qquad K_Z=\partial_z.}
$$
Their flows are respectively $(x+t,y,z)$, $(x,y+t,z+xt)$ and $(x,y,z+t)$. Each preserves $h$, and their <Lie brackets of vector fields> are $[K_X,K_Y]=K_Z$, with the other two zero. This explicitly realizes the <Heisenberg Lie algebra>. These <Killing vector fields> are left-invariant vector fields; the vector fields dual to the right-invariant coframe are instead $\partial_x+y\partial_z,\partial_y,\partial_z$, and should not be substituted for these generators.
For the diagonal case, put
$$
h=\alpha\,dx^2+\beta\,dy^2+\chi(dz-y\,dx)^2,\qquad \alpha,\beta,\chi>0.
$$
The <Kaluza-Klein decomposition along a Killing field> uses $V=h(K_X,K_X)=\alpha+\chi y^2$. Completing the square in $dx$ gives the <Heisenberg metric Kaluza-Klein reduction>:
$$
\boxed{V=\alpha+\chi y^2,\qquad A=-\frac{\chi y}{\alpha+\chi y^2}\,dz,
\qquad \gamma=\beta\,dy^2+\frac{\alpha\chi}{\alpha+\chi y^2}\,dz^2.}
$$
These depend only on the quotient coordinates $(y,z)$ and satisfy $h=V(dx+A)^2+\gamma$. For a diagonal indefinite <metric tensor>, the same expressions hold wherever $V\ne0$; a null <Killing vector field> cannot be treated with this completed-square decomposition.
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