# Is the set of all Heisenberg Matrices Connected?

The set of all $3\times 3$ matrices of the form $$\begin{pmatrix} 1 & a & b\\ 0 & 1 & c\\ 0 & 0 & 1\\ \end{pmatrix}$$ forms a group surely, are they connected?

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Sure. Each is path connected to the identity, by a linear segment, replace $a,b,c$ by $at,bt,ct$ with $0 \leq t \leq 1.$ – Will Jagy Sep 4 '12 at 3:33
@WillJagy +1. Why don't you post that as an answer? – Alex Becker Sep 4 '12 at 3:41
@AlexBecker done. Good thinking. – Will Jagy Sep 4 '12 at 3:52

Sure. Each is path connected to the identity, by a linear segment, replace $a,b,c$ by $at,bt,ct$ with $0 \leq t \leq 1.$

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The obvious bijection from $\mathbf R^3$ (or is that $\mathbf C^3$? you didn't say) to this set of matrices, sending $(a,b,c)$ to the matrix you wrote, is a homeomorphism (continuous with continuous inverse). Therefore your question is equivalent to whether $\mathbf R^3$ (or $\mathbf C^3$) is connected, and this is true. The group structure plays no role at all for this.

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Recall that

$$\mathfrak{h} = \left\{ \left(\begin{array}{ccc} 0 & a & b \\ 0 & 0 & c \\ 0 & 0& 0 \end{array}\right) : a,b,c \in \Bbb{R} \right\}$$

is the Lie algebra of the Heisenberg group $H$. Now take any $h \in \mathfrak{h}$ and upon calculating $\exp h$ you will see easily why $\exp : \mathfrak{h} \to H$ is surjective. Since $\mathfrak{h}$ is a linear space it follows that $H$ is the continuous image of a path connected space and hence is path connected.

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what is the lie algebra of a group dear sir? – miosaki Sep 4 '12 at 4:12
@KutukKatuk The lie algebra of $H$ is defined to be the set of all $3 \times 3$ matrices $A$ such that $\exp(tA) \in H$ for all real numbers $t$. It is a linear space. – user38268 Sep 4 '12 at 7:14

Let $\mathcal{H}\subset\mathbb{R}^{3\times 3}$ be the set of matrices of the form given in the question.

The map $i: \mathbb{R}^3 \to \mathcal{H}$ defined by $$i((a,b,c)) = \begin{pmatrix} 1 & a & b\\ 0 & 1 & c\\ 0 & 0 & 1\\ \end{pmatrix}$$ is a continuous, affine bijection between $\mathbb{R}^3$ and $\mathcal{H}$. Since $\mathbb{R}^3$ is connected, it follows that $\mathcal{H} = i(\mathbb{R}^3)$ is connected (in fact it is affine). (The same is true if $\mathbb{R}$ is replaced by $\mathbb{C}$, of course.)

In fact, it is straightforward to see that if $A = i((a,b,c))$, and $A' = i((a',b',c'))$, then $t \mapsto i((a+t(a'-a), b+t(b'-b), c+t(c'-c))$, $t \in [0,1]$ is a path between $A$ and $A'$. Hence $\mathcal{H}$ is path connected.

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