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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 " \, $.

Any suggestions? Thanks in advance!

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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.

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  • $\begingroup$ 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? $\endgroup$ – userr777 Oct 22 '18 at 6:38
  • $\begingroup$ Yes, you obtain something in $\mathfrak g$. And what I explained is why that something is an upper triangular matrix. $\endgroup$ – José Carlos Santos Oct 22 '18 at 7:28
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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.

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