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The left-invariant metric of the Heisenberg group can be given as:

$ds^2 = \frac{1}{4} a^2 \text{dx}^2 y^2-\frac{1}{2} a^2 \text{dx} \text{dy} x y+a^2 \text{dx} \text{dz} y+\frac{1}{4} a^2 \text{dy}^2 x^2-a^2 \text{dy} \text{dz} x+a^2 \text{dz}^2+\text{dx}^2+\text{dy}^2$,

where $a$ is a constant.

I was interested in "visualizing" this geometry, say for $a=1$. My first idea was to embed it into Euclidean space $ds^2 = dx^2 + dy^2 + dz^2$. So, I tried to diagonalize the above metric to obtain:

$ds^2 = dx^2 + \left[\frac{1}{8} \left(a^2 \left(x^2+y^2+4\right)-\sqrt{a^4 \left(x^2+y^2+4\right)^2+8 a^2 \left(x^2+y^2-4\right)+16}+4\right)\right]dy^2 + \left[\frac{1}{8} \left(a^2 \left(x^2+y^2+4\right)+\sqrt{a^4 \left(x^2+y^2+4\right)^2+8 a^2 \left(x^2+y^2-4\right)+16}+4\right)\right]dz^2$,

and then matching this to the Euclidean metric, which gives $x = \pm iy$.

I don't know if this approach is correct, since other sources online seem to suggest that the Heisenberg group embedded in $\mathbb{R}^3$ consists of "spheres".

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