4
$\begingroup$

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.

  1. Do we have surjectivity of $\exp : \mathfrak{g}\to G$ ? and what about injectivity?

  2. Is $\mathfrak{g}$ equal to the algebra of all upper triangular matrices?

$\endgroup$
2
$\begingroup$

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:

  1. $\exp$ is injective
  2. $\exp$ is bijective
  3. $\exp$ is a real analytic diffeomorphism
  4. $G$ is solvable, simply connected, and $\mathfrak{g}$ does not admit $\mathfrak{e}$ as subalgebra of a quotient.
  5. $G$ is solvable, simply connected, and $\mathfrak{g}$ does not admit $\mathfrak{e}$ or $\tilde{\mathfrak{e}}$ as subalgebra

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.