Let $G$ be the Lie group of all upper triangular real matrices (over $\mathbb{R}$) with positive diagonal elements. Denote $\mathfrak{g}$ its Lie algebra.
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Do we have surjectivity of $\exp : \mathfrak{g}\to G$ ? and what about injectivity?
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Is $\mathfrak{g}$ equal to the algebra of all upper triangular matrices?
Best Answer
The exponential map $\exp\colon \mathfrak{n}^+\rightarrow N$ from the nilpotent Lie algebra of all stricly upper-triangular matrices to the nilpotent group of all upper uni-triangular matrices is given by polynomial maps with inverse given by matrix logarithm. Hence $\exp$ is bijective in this case. This is no longer true for upper-triangular matrices. The question when $\exp$ is injective is easier than the question when $\exp$ is surjective. For injectivity we have very nice criteria, given for example at this MSE-question:
If $G$ is a real Lie group with Lie algebra $\mathfrak{g}$, then the following are equivalent:
Here $\mathfrak{e}$ is the 3-dimensional Lie algebra with basis $(H,X,Y)$ and bracket $[H,X]=Y$, $[H,Y]=-X$, $[X,Y]=0$. It is isomorphic to the Lie algebra of the group of isometries of the plane. Its central extension $\tilde{\mathfrak{e}}$ is defined as the 4-dimensional Lie algebra defined by adding a central generator $Z$ and the additional nonzero bracket $[X,Y]=Z$.
Concerning surjectivity the situation is more complicated, see On surjectivity of exponential map for Lie groups, or Exponential map is surjective for compact connected Lie group.
So, for 1. Yes, $\exp$ is surjective, because every upper-triangular real matrix with positive diagonal entries is invertible and is the square of an invertible matrix, see here.
For 2.) Yes.