# The Lie Algebra of Invertible Upper Triangular Matrices

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.

• So you're saying that the condition $\det A \neq 0$ is ignored here when computing the Lie algebra? Apr 4, 2016 at 13:28
• 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 $Gl(n,R)$. Apr 4, 2016 at 13:32
• Ah, of course! This makes total sense. Thank you. Apr 4, 2016 at 13:45
• So this Lie Algebra is just the $\{0\}$ vector space for $0 \in Gl(n,R)$? Apr 25, 2020 at 0:51
• Hello, I just had the same question - but I realised I had been misreading - we want invertible matrices, not det M = 1 matrices! Perhaps this might help others. Of course, the first does not lead to constraints on the Lie algebra as suggested by Tsemo Aristide, but the latter leads to the usual tr A = 0 on the Lie algebra. Tsemo Aristide made a typo when he says SO(n), he means O(n).
– dla
Feb 10, 2022 at 13:23

I think that the infinitesimal point of view (i.e. synthetic differential geometry/smooth infinitesimal analysis) makes it quite straightforward to find Lie algebras of matrix groups.

Let $$K$$ be a commutative ring, and let $$\varepsilon$$ be a nilsquare infinitesimal, i.e. $$\varepsilon^2 =0$$. Essentially we're just using the familiar algebraic geometry approach to derivatives via dual numbers.

Let's first find the Lie algebra $$\mathfrak{gl}_n(K)$$ of the general linear group $$\operatorname{GL}_n(K)$$. Because this is a matrix group then $$X \in \mathfrak{gl}_n(K)$$ if and only if $$I + \varepsilon X \in \operatorname{GL}_n(K).$$ In other words, we want $$\varepsilon \mapsto I + \varepsilon X$$ to be an infinitesimal curve through the identity of the Lie group (this is not just a hand-wavey explanation, but a truly rigorous approach in SDG).

We easily see that for any square matrix $$X$$, we have $$(I + \varepsilon X) (I-\varepsilon X)= I - \varepsilon^2 X^2 = I + 0X = I.$$ Hence $$\mathfrak{gl}_n(K)$$ is just the lie algebra of all square matrices.

Now let's tackle the group $$\operatorname{RT}_n(K)$$ of $$n \times n$$ right triangular matrices with entries in $$K$$. Using the same technique, we want to find all matrices $$R \in \mathfrak{gl}_n(K)$$ such that $$I + \varepsilon R \in \operatorname{RT}_n(K)$$. Well, obviously we need $$R$$ to be right triangular so that this expression is right triangular. But that's all we need: we already know that $$I + \varepsilon R$$ is invertible by our previouse argument.

The same argument will easily show that the strictly right triangular matrices are the Lie algebra of the group of right unitriangular matrices.