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I'm studying for a qualifying exam and just reviewed the following theorem:

Let $G$ be a Lie group with Lie algebra $\mathfrak{g}$. Suppose $\mathfrak{h}\leq \mathfrak{g}$ is a subalgebra. Then there exists a unique connected Lie subgroup $H \leq G$ such that $T_1H=\mathfrak{h}$.

I'm now trying to apply this theorem to do the following exercise:

Let $H=\left\{\left( \begin{array}{ccc} 1 & x & z \\ 0 & 1 & y \\ 0 & 0 & 1 \end{array}\right) \Bigg\vert x,y,z \in \mathbb{R} \right\}$ and let $ Z =\left\{\left( \begin{array}{ccc} 1 & 0 & z \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right) \Bigg\vert z \in \mathbb{R} \right\}$ so that $Z$ is a Lie subgroup of the nilpotent Lie group $H$. ($Z$ is the center of $H$.) Find all connected $2$-dimensional Lie subgroups of $H$ that contain $Z$. Describe them as both submanifolds of $H$ and as groups.

The first thing I did was find their corresponding Lie algebras: $$\mathfrak{h}=T_1H=\left\{\left( \begin{array}{ccc} 0 & x & z \\ 0 & 0 & y \\ 0 & 0 & 0 \end{array}\right) \Bigg\vert x,y,z \in \mathbb{R} \right\} ~\text{ and }~\mathfrak{z}=T_1Z=\left\{\left( \begin{array}{ccc} 0 & 0 & z \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right) \Bigg\vert z \in \mathbb{R} \right\}$$

Next we need to find all the $2$-dimensional subalgebras of $\mathfrak{h}$ containing $\mathfrak{z}$. Here's where I lose confidence in my solution; I think that for any $a,b \in \mathbb{R}$ the following are all the $2$-dim. subalgebras (containing $\mathfrak{z}$):

  1. $\operatorname{span}\left\{ \left( \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right), ~\left( \begin{array}{ccc} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right) \right\}$

  2. $\operatorname{span}\left\{ \left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{array}\right), ~\left( \begin{array}{ccc} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right) \right\}$

  3. $\operatorname{span}\left\{ \left( \begin{array}{ccc} 0 & a & 0 \\ 0 & 0 & b \\ 0 & 0 & 0 \end{array}\right), ~\left( \begin{array}{ccc} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right) \right\}$

These are all subalgebras because the bracket of basis elements is the zero matrix, so by bi-linearity of the bracket they are closed under bracketing.

In the proof of the theorem the corresponding Lie subgroup to a Lie algebra is the connected leaf of the resulting foliation which contains $1 \in G$. This seems hard to find so instead I think we can just "find" the Lie subgroups which will have the corresponding Lie algebras:

  1. $\left\{ \left( \begin{array}{ccc} 1 & x & z \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right) \Big\vert x,z \in \mathbb{R} \right\}$
  2. $\left\{ \left( \begin{array}{ccc} 1 & 0 & z \\ 0 & 1 & y \\ 0 & 0 & 1 \end{array}\right) \Big\vert y,z \in \mathbb{R} \right\}$
  3. $\left\{ \left( \begin{array}{ccc} 1 & ta & z \\ 0 & 1 & tb \\ 0 & 0 & 1 \end{array}\right) \Big\vert t,z \in \mathbb{R} \right\}$

Each of these sets are groups with multiplication. The first two are diffeomorphic to $\mathbb{R}^2$, but I'm not sure how to see what the third Lie group is as a manifold.

Any corrections, comments or hints would be greatly appreciated!

EDIT: Following the comment of Sunghyuk, $$\operatorname{exp} \left( \begin{array}{ccc} 0 & a & 0 \\ 0 & 0 & b \\ 0 & 0 & 0 \end{array}\right) = \left( \begin{array}{ccc} 1 & a & \frac{ab}{2} \\ 0 & 1 & b \\ 0 & 0 & 1 \end{array}\right)$$ and we could instead write the corresponding Lie subgroup instead as $\left\{ \left( \begin{array}{ccc} 1 & ta & z+\frac{t^2ab}{2} \\ 0 & 1 & tb \\ 0 & 0 & 1 \end{array}\right) \Big\vert t,z \in \mathbb{R} \right\}$ which is the same Lie subgroup as 3. but perhaps gives us more intuition on how to view it as a manifold?

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  • $\begingroup$ There is a minor typo: $Z$ is a Lie "subgroup", not a subalgebra. Just wanted to point it out. $\endgroup$ – Henry May 3 '17 at 15:16
  • $\begingroup$ Have you tried taking the exponential map? $\endgroup$ – Henry May 3 '17 at 15:20
  • $\begingroup$ Am I missing something? Your description of the third Lie group (call it $G_3$) yields a very simple diffeomorphism $\mathbb{R}^2\to G_3$. What are asking? $\endgroup$ – Amitai Yuval May 3 '17 at 21:15
  • $\begingroup$ @AmitaiYuval I don't see what the map is between $G_3$ and $\mathbb{R}^2$ is. $\endgroup$ – ThePiper May 4 '17 at 2:43
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Let $G$ denote group number 3. Define $\varphi:\mathbb{R}^2\to G$ by$$(t,z)\mapsto\left(\begin{array}{ccc}1&ta&z\\0&1&tb\\0&0&1\end{array}\right).$$ Assuming $(a,b)\neq(0,0)$, the map $\varphi$ is a diffeomorphism.

Edit: If you want a diffeomorphism which is also a group isomorphism, you just need to use the exponential map, which you already described yourself. The diffeomorphism will be$$(t,z)\mapsto\left(\begin{array}{ccc}1&ta&z+\frac{t^2ab}{2}\\0&1&tb\\0&0&1\end{array}\right).$$A straightforward computation verifies that this is indeed a group isomorphism.

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  • $\begingroup$ This isn't a Lie group homomorphism though, shouldn't we be able to see that they are isomorphic as Lie groups? $\endgroup$ – ThePiper May 4 '17 at 15:11
  • $\begingroup$ @ThePiper I added something to my answer. $\endgroup$ – Amitai Yuval May 4 '17 at 19:57

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