The first Heisenberg group $H^1$ is a Lie group that can be realized as a group of matrices $\begin{pmatrix}1&x^1&x^3\\ 0&1&x^2\\ 0&0&1\end{pmatrix}$ where $(x^1, x^2, x^3) \in \mathbb{R}^3$ and the group operation is the multiplication of matrices. Then $H^1$ has a natural global chart $\begin{pmatrix}1&x^1&x^3\\ 0&1&x^2\\ 0&0&1\end{pmatrix} \mapsto (x^1,x^2,x^3)$. Let $X$, $Y$, and $Z$ be left-invariant vector fields such that at the identity $e \in H^1$ we have $X_e=(\frac{\partial}{\partial x^1}), Y_e=(\frac{\partial}{\partial x^2})$ and $Z_e=(\frac{\partial}{\partial x^3})$. Let $\langle \cdot, \cdot \rangle$ be the left-invariant Riemannian metric such that $\{X, Y, Z\}$ is an orthonormal global frame with respect to $\langle \cdot, \cdot \rangle$. Determine the components functions $g_{ij} =\langle \partial/\partial x^i, \partial/\partial x^j \rangle$ of the Riemannian metric $\langle \cdot, \cdot \rangle$.
In this problem I don't have a clue how to start. I know that a left-invariant vector field $X$ is completely determined by it's value $X_e$ since $$X_g = \ell_{g*}(X_e)$$ where $\ell_g$ is the left multiplication map. I don't really now how to use this map to find the components of the Riemannian metric. What is it that I need to compute here to get the component functions out of $g$?
At each $x=\begin{pmatrix}1&x^1&x^3\\ 0&1&x^2\\ 0&0&1\end{pmatrix}$, to calculate $g_{ij}(x)$, note that $g(x) = (\ell_{x^{-1}}\ )_* g_e$. \begin{split} g_{ij}(x) &= g_x \left( \frac{\partial }{\partial x^i}\bigg|_x, \frac{\partial }{\partial x^j}\bigg|_x\right) \\ &= (\ell_{x^{-1}})^*g_e \left( \frac{\partial }{\partial x^i}\bigg|_x, \frac{\partial }{\partial x^j}\bigg|_x\right) \\ &= g_e \left( (\ell_{x^{-1}})_*\frac{\partial }{\partial x^i}\bigg|_x, (\ell_{x^{-1}})_*\frac{\partial }{\partial x^j}\bigg|_x\right) \end{split}
Note that $$x^{-1} = \begin{pmatrix}1&-x^1&1+ x^1x^2-x^3\\ 0&1&-x^2\\ 0&0&1\end{pmatrix},\ \ (\ell_{x^{-1}})_*X =x^{-1} X$$ (for the second formula see here) and thus
\begin{align} (\ell_{x^{-1}})_* \frac{\partial }{\partial x^1}\bigg|_x &= \begin{pmatrix}1&-x^1&1+ x^1x^2-x^3\\ 0&1&-x^2\\ 0&0&1\end{pmatrix} \begin{pmatrix}0& 1&0\\ 0&0&0\\ 0&0&0\end{pmatrix}= \begin{pmatrix}0& 1&0\\ 0&0&0\\ 0&0&0\end{pmatrix} = \frac{\partial }{\partial x^1}\bigg|_e\\ (\ell_{x^{-1}})_* \frac{\partial }{\partial x^2}\bigg|_x &= \begin{pmatrix}1&-x^1&1+ x^1x^2-x^3\\ 0&1&-x^2\\ 0&0&1\end{pmatrix} \begin{pmatrix}0& 0&0\\ 0&0&1\\ 0&0&0\end{pmatrix}=\begin{pmatrix}0& 0&-x^1\\ 0&0&1\\ 0&0&0\end{pmatrix}= \frac{\partial }{\partial x^2}\bigg|_e- x^1 \frac{\partial }{\partial x^3}\bigg|_e\\ (\ell_{x^{-1}})_* \frac{\partial }{\partial x^3}\bigg|_x &= \begin{pmatrix}1&-x^1&1+ x^1x^2-x^3\\ 0&1&-x^2\\ 0&0&1\end{pmatrix} \begin{pmatrix}0& 0&1\\ 0&0&0\\ 0&0&0\end{pmatrix}=\begin{pmatrix}0& 0&1\\ 0&0&0\\ 0&0&0\end{pmatrix} = \frac{\partial }{\partial x^3}\bigg|_e \end{align} Since $\{\frac{\partial }{\partial x^1}\bigg|_e, \frac{\partial }{\partial x^2}\bigg|_e, \frac{\partial }{\partial x^3}\bigg|_e\}$ is by defintion orthonormal, we have
$$g(x) = \begin{pmatrix} 1 & 0 & 0 \\ 0 &1+(x^1)^2 & -x^1 \\ 0 & -x^1 & 1\end{pmatrix}.$$