13.2 The Heisenberg group

13.2 The Heisenberg group

Concept links · terms present in this machine draft; source roles are unverified: Lie algebra · Lie group · Lie bracket

Exponentiating matrices in ${\mathfrak{h}}_3$ gives

$$\exp \left(\begin{array}{ccc}0& x& z\\ 0& 0& y\\ 0& 0& 0\end{array}\right)=\left(\begin{array}{ccc}1& x& z+\frac{1}{2}xy\\ 0& 1& y\\ 0& 0& 1\end{array}\right)$$

so the group with Lie algebra ${\mathfrak{h}}_3$ will be the group of upper triangular 3 by 3 real matrices with 1 on the diagonal, and this group will be the Heisenberg group $H_3$ . For our purposes though, it is better to work in exponential coordinates $(i.e.$ , labeling a group element with the Lie algebra element that exponentiates to it). In these coordinates the exponential map relating the Heisenberg Lie algebra ${\mathfrak{h}}_{2d+1}$ and the Heisenberg Lie group $H_{2d+1}$ is just the identity map, and we will use the same notation

$$\left(\left(\genfrac{}{}{0ex}{}{x}{y}\right),z\right)$$

for both Lie algebra and corresponding Lie group elements.

Matrix exponentials in general satisfy the Baker-Campbell-Hausdorf formula, which says

$$e^A{e}^B=e^{A+B+\frac{1}{2}[A,B]+\frac{1}{12}[A,[A,B]]-\frac{1}{12}[B,[A,B]]+\dots }$$

where the higher terms can all be expressed as repeated commutators. This provides one way of showing that the Lie group structure is determined (for group elements expressible as exponentials) by knowing the Lie bracket. For the full formula and a detailed proof, see chapter 5 of [42]. One can easily check the first few terms in this formula by expanding the exponentials, but the dificulty of the proof is that it is not at all obvious why all the terms can be organized in terms of commutators.

For the case of the Heisenberg Lie algebra, since all multiple commutators vanish, the Baker-Campbell-Hausdorf formula implies for exponentials of elements of ${\mathfrak{h}}_3$

$$e^A{e}^B=e^{A+B+\frac{1}{2}[A,B]}$$

(a proof of this special case of Baker-Campbell-Hausdorf is in section 5.2 of [42]). We can use this to explicitly write the group law in exponential coordinates:

Definition (Heisenberg group, $d = 1 )

$.TheHeisenberggroup$H _ { 3 }$isthespace$\mathbf { R } ^ { 3 } = \mathbf { R } ^ { 2 } \oplus \dot { \mathbf { R } }$ with the group law

$$\left(\left(\genfrac{}{}{0ex}{}{x}{y}\right),z\right)\left(\left(\genfrac{}{}{0ex}{}{x^{\prime }}{y^{\prime }}\right)z^{\prime }\right)=\left(\left(\genfrac{}{}{0ex}{}{x+x^{\prime }}{y+y^{\prime }}\right),z+z^{\prime }+\frac{1}{2}(x{y}^{\prime }-y{x}^{\prime })\right)\text{\hspace{1em}(13.2)}$$

The isomorphism between ${\mathbf{R}}^2\oplus \mathbf{R}$ with this group law and the matrix form of the group is given by

$$\left(\left(\genfrac{}{}{0ex}{}{x}{y}\right),z\right)\leftrightarrow \left(\begin{array}{ccc}1& x& z+\frac{1}{2}xy\\ 0& 1& y\\ 0& 0& 1\end{array}\right)$$

Note that the Lie algebra basis elements $X,Y,Z$ each generate subgroups of $H_3$ isomorphic to R. Elements of the first two of these subgroups generate the full group, and elements of the third subgroup are “central”, meaning they commute with all group elements. Also notice that the non-commutative nature of the Lie algebra (equation 13.1) or group (equation 13.2) depends purely on the factor $x{y}^{\prime }-y{x}^{\prime }$

The generalization of this to higher dimensions is:

Definition (Heisenberg group)

The Heisenberg group $H_{2d+1}$ is the space ${\mathbf{R}}^{2d+1}$ with the group law

$$\left(\left(\genfrac{}{}{0ex}{}{\mathbf{x}}{\mathbf{y}}\right),z\right)\left(\left(\genfrac{}{}{0ex}{}{{\mathbf{x}}^{\prime }}{{\mathbf{y}}^{\prime }}\right),z^{\prime }\right)=\left(\left(\genfrac{}{}{0ex}{}{\mathbf{x}+{\mathbf{x}}^{\prime }}{\mathbf{y}+{\mathbf{y}}^{\prime }}\right),z+z^{\prime }+\frac{1}{2}(\mathbf{x}\cdot {\mathbf{y}}^{\prime }-\mathbf{y}\cdot {\mathbf{x}}^{\prime })\right)$$

where $\mathbf{x},{\mathbf{x}}^{\prime }\mathbf{y},{\mathbf{y}}^{\prime }\in {\mathbf{R}}^d$

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