# Problem computing Lie algebra of unitriangular matrices

Let $$G$$ the Lie group of the upper unitriangular matrices, i.e. \begin{align} G : = \{ A= (a_{ij})_{ij} \, \, |\, a_{ii} = 1 \, \, \, \forall \, i \, \, \text{and} \, \, a_{ij} = 0 \, \, \forall \, i>j \} \end{align}

Computing the Lie algebra of G.

I think that the result is the set of the strictly upper triangular matrices. I tried also to prove that \begin{align} LG = \{A= (a_{ij})_{ij} \, \, |\, a_{ij} = 0 \, \, \forall \, i\geq j \} \end{align} The proof of $$\, "\supseteq \, "$$ I think's it's okay, but I have some problem to show the inclusion $$\, "\subseteq " \,$$.

Yes, $$\mathfrak g$$ is the space of strictly upper triangular matrices. If $$M$$ is such a matrix, then it is clear that$$(\forall t\in\mathbb{R}):e^{tM}\in G.\tag1$$On the other hand, if $$(1)$$ holds then, differentiating $$e^{tM}$$ and taking $$t=0$$, one gets that $$M$$ is upper triangular. But if one of the entries of the main diagonal of $$M$$ was not $$0$$, the entry at the same position of $$e^{tM}$$ would not be $$1$$. Therefore, $$M$$ us strictly upper triangular.
• Not sure that I have understood how can we conclude that $M$ is upper triangular. If I take the derivative of $e^{tM}$ I obtain something in $LG$, or not? – userr777 Oct 22 '18 at 6:38
• Yes, you obtain something in $\mathfrak g$. And what I explained is why that something is an upper triangular matrix. – José Carlos Santos Oct 22 '18 at 7:28
There is no need to mess with the exponential map if you make the simple observation that $$G$$ is the affine subspace $$I+\{\text{strictly upper-triangular matrices}\}$$ in $$\operatorname{End}(\mathbb{R}^n)$$. Thus the tangent space at identity corresponds is canonically identified with the tangent space to the vector subspace of strictly upper-triangular matrices at $$0$$, which is of course the space of strictly upper triangular matrices.