I'm reading this paper by Fernandez and de Leon. I was having trouble regarding the computation of a basis of left-invariant 1-forms of the Lie group $G$ described below.

Let $G$ be a matrix Lie group of dimension 6 whose elements are of the form

$$A=\begin{bmatrix} e^t & 0 & xe^t & 0 & 0 & y_1 \\ 0 & e^{-t} & 0 & xe^{-t} & 0 & y_2 \\ 0 & 0 & e^t & 0 & 0 & z_1 \\ 0 & 0 & 0 & e^{-t} & 0 & z_2 \\ 0 & 0 & 0 & 0 & 1 & t \\ 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix}$$ where $t,x,y_1,y_2,z_1,z_2\in \mathbb{R}$. A global system of coordinates of $G$ are then given by $$t(A)=t, x(A)=x, y_i(A)=y_i, z_i(A)=z_i, 1\leq i\leq 2.$$ Then they say that a standard computation reveals that the following 1-forms on $G$ constitute a basis for the space of left-invariant 1-form: $$\alpha=dt, \beta=dx, \gamma_1=e^{-t}dy_1-xe^{-t}dz_1, \gamma_2=e^tdy_2-xe^tdz_2, \delta_1=e^{-t}dz_1, \delta_2=e^tdz_2.$$

Because I do not know any standard computation to get the basis for the space of left-invariant 1-forms on a Lie group, I have to ask how do they calculate? Is there really a standard procedure to obtain such a basis, or is it really just by trial methods?

Anyone who knows this please help me... I really wanted to learn this. Thanks in advance

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    $\begingroup$ I believe there is a typo. The fourth form should be $\gamma_2 = e^tdy_2 -xe^tdz_{{\bf 2}}$. $\endgroup$ Jun 1 at 17:08

2 Answers 2


One option for computing a left-invariant coframe on $G$ is to compute the Maurer–Cartan form---that is, the left-invariant $\mathfrak{g}$-valued $1$-form $\omega$ on $G$ characterized via the identification $T_e G \cong \mathfrak{g}$ by $\omega_e = \operatorname{id}_\mathfrak{g}$---and then read off the independent components.

Explicitly, $$\omega := A^{-1} dA$$ and in our case the nonzero entries of the first row of $\omega$ are $$\omega_{11} = dt, \qquad \omega_{13} = dx, \qquad \omega_{16} = e^{-t} (dy_1 - x \,dz_1),$$ which in your labeling are $\alpha, \beta, \gamma_1$, respectively.

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    $\begingroup$ This is definitely the better method. $\endgroup$ Jun 1 at 17:14
  • $\begingroup$ @Travis Wilse Can you explain why this method always works? and by reading off the independent component do you mean choose any 6 independent entries of the matrix $A^{-1}dA$? $\endgroup$
    – Uncool
    Jun 8 at 6:12
  • $\begingroup$ @Uncool Given any element $\eta \in \mathfrak{g}^*$---in particular, any element that projects an element of $\mathfrak{g}$ to an entry of any matrix representation thereof---$\eta \circ \omega$ is a left-invariant 1-form on $G$. And yes: More precisely, given a basis $(\eta^k)$ of $\mathfrak{g}^*$, $\eta^k \circ \omega$ is a left-invariant coframe of $G$. $\endgroup$ Jun 9 at 0:46

This is not as simple as Travis Willse's approach. It's a bit more hands on, which means it takes a bit more work, but it may give some insight into why the left-invariant one-forms appear as they do.

Let $M \in G$ be a point with coordinates $(a, b, c_1, c_2, d_1, d_2)$. Then there is a diffeomorphism $f_M : G \to G$ given by $A \mapsto MA$. A one form $\eta$ on $G$ is left-invariant if $f_M^*\eta = \eta$ for all $M \in G$. Performing the matrix multiplication $MA$, one see that in the given coordinates, we have

$$f_M(t, x, y_1, y_2, z_1, z_2) = (t + a, x + b, y_1e^a + z_1be^a + c_1, y_2e^{-a}+z_2be^{-a} + c_2, z_1e^{a} + d_1, z_2e^{-a} + d_2).$$


\begin{align*} f_M^*dt &= d(t+a) = dt\\ f_M^*dx &= d(x+b) = dx\\ f_M^*dy_1 &= d(y_1e^a + z_1be^a + c_1) = e^ady_1 + be^adz_1\\ f_M^*dy_2 &= d(y_2e^{-a}+z_2be^{-a} + c_2) = e^{-a}dy_2 + be^{-a}dz_2\\ f_M^*dz_1 &= d(z_1e^{a} + d_1) = e^adz_1\\ f_M^*dz_2 &= d(z_2e^{-a} + d_2) = e^{-a}dz_2. \end{align*}

So we see that $\alpha := dt$ and $\beta := dx$ are left-invariant.

The last two equations have a right hand side which is very similar to the original form, so let's focus on those next. The fact that the right hand side is a rescaling of the original form indicates that $dz_i$ are not left-invariant, but we can make them left-invariant by multiplication with an appropriate function. As the coefficient of the right hand side only depends on $a$, we're looking for functions $g_i(t)$ such that $g_i(t)dz_i$ left-invariant. Note that $f_M^*(g_2(t)dz_2) = g_2(t+a)e^{-a}dz_2$. To obtain a left-invariant form, we need $g_2(t+a)e^{-a} = g_2(t)$, and hence $g_2(t+a) = g_2(t)e^a$, so we must have $g_2(t) = e^t$. Likewise, $g_1(t)dz_2$ is left-invariant if and only if $g_1(t+a) = g_1(t)e^{-a}$, so $g_1(t) = e^{-t}$. Therefore $\delta_1 := e^{-t}dz_1$ and $\delta_2 := e^t dz_2$ are left-invariant.

Finally, let's consider the middle two equations. By the considerations in the previous paragraph, we can rescale by $e^{\mp t}$ to deal with the common $e^{\pm a}$ factor:

\begin{align*} f_M^*(e^{-t}dy_1) &= e^{-t}dy_1 + be^{-t}dz_1 = e^{-t}dy_1 + b\delta_1\\ f_M^*(e^tdy_2) &= e^tdy_2 + be^tdz_2 = e^tdy_2 + b\delta_2. \end{align*}

The only way that $b\delta_i$ can arise from the pullback of a form is if that form involves a term $h_i(x)\delta_i$ (compare with the discussion in the previous paragraph). As

$$f_M^*(e^{-t}dy_1 + h_1(x)\delta_1) = e^{-t}dy_1 + b\delta_1 + h_1(x+b)\delta_1 = e^{-t}dy_1 + (h_1(x+b) + b)\delta_1,$$

the form $e^{-t}dy_1 + h_1(x)\delta_1$ is left-invariant if and only if $h_1(x+b) = h_1(x) - b$, so we must have $h_1(x) = -x$. Likewise, $e^tdy_2 + h_2(x)\delta_2$ is left-invariant if and only if $h_2(x+b) = h_2(x) - b$ so $h_2(x) = -x$. Therefore $\gamma_1 := e^{-t}dy_1 - x\delta_1 = e^{-t}dy_1 -xe^{-t}dz_1$ and $\gamma_2 := e^tdy_2 - x\delta_2 = e^tdy_2 - xe^tdz_1$ are left-invariant.

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    $\begingroup$ This answer gives some great insight into how left-invariant $1$-forms behave in coordinates. I recommend to op computing using this method at least once in your life, to get a stronger sense of what's going on mechanically. $\endgroup$ Jun 2 at 0:12
  • $\begingroup$ This is nice... I hope I can perform my computation using this method. $\endgroup$
    – Uncool
    Jun 2 at 6:36

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