From Wikipedia:

The Lie algebra of the Lie group of invertible upper triangular matrices is the set of all upper triangular matrices, not necessarily invertible, and is a solvable Lie algebra.

I don't understand why the invertibility condition on the matrices in the Lie group disappears for the Lie algebra.

To solve for the Lie algebra of a Lie group, I usually differentiate (at the identity) the "conditions" that define the group, and these become constraints on the elements in the Lie algebra. For instance, for $\mathfrak{so}(n)$, differentiating the condition $\det A = 1$ on the identity leads to the constraint $\mathrm{tr}\, A = 0$; similarly, the constraint of skew-symmetry is obtained from differentiating the other condition that matrices in $SO(n)$ must satisfy, namely $A^TA=I$.

Is a different method being employed for the invertible upper triangular matrices? I sense that there is something really simple going on here, but I'm drawing a blank.


Here the "the conditions that define" the Lie group are $a_{ij}=0, i<j$ differentiate that you obtain again $a_{ij}=0$ with no other constraints.

Generally, you can't have an invertibility constraint in a Lie algebra since it is a vector space.

  • $\begingroup$ So you're saying that the condition $\det A \neq 0$ is ignored here when computing the Lie algebra? $\endgroup$ – Anthony Apr 4 '16 at 13:28
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    $\begingroup$ Yes, it is ignored here as it is ignored when you calculate the Lie algebra of $SO(n)$ by setting $A^TA=I$ and differentiate it. You have just to consider the equation which defines the group as a closed subgroup of $Gl(n,R)$. $\endgroup$ – Tsemo Aristide Apr 4 '16 at 13:32
  • $\begingroup$ Ah, of course! This makes total sense. Thank you. $\endgroup$ – Anthony Apr 4 '16 at 13:45

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