For each pair of the following Lie algebras over $\Bbb R$, decide whether or not they are isomorphic:
- the Lie algebra $\Bbb R^3$, where the Lie bracket is given by the vector product;
- the upper triangular $2 \times 2$ matrices over $\Bbb R$;
- the strictly upper triangular $3 \times 3$ matrices over $\Bbb R$;
- $L = \{x \in \mathfrak{gl}(3, \Bbb R) : x^t = −x)\}$.
So far I'm pretty sure the upper triangular and strictly upper triangular matrices are not isomorphic, because the Lie algebra of strictly upper triangular matrices is nilpotent and the Lie algebra of upper triangular matrices are not.
Also, for $x^t$ in (4), does this just mean $x$ to any number $t$, or does $t$ have a specific meaning in an abstract algebra context?
If it is any number $t$, then I have $x=0$ and $x=-1$ as valid x values for that set, where $0=-0$, and $-1$ raised to any even integer $= 1$.
Working on this further, I believe that (2) is not isomorphic to any other, because it is of dimension $2$, where all others are of dimension $3$.
Best Answer
You are correct that (3) is nilpotent but (2) is not, and hence they are not isomorphic.
When you calculated that the Lie algebra $\mathfrak{n}_3$ in (3) is nilpotent ($\mathfrak{n}_3$ is called the Heisenberg algebra), you computed that its derived Lie algebra $[\mathfrak{n}_3, \mathfrak{n}_3]$ is $$[\mathfrak{n}_3, \mathfrak{n}_3] = \left\{\pmatrix{\cdot&\cdot&b\\&\cdot&\cdot\\&&\cdot} : b \in \Bbb R \right\} \cong \Bbb R .$$
This suggests computing the derived Lie algebra $$[\mathfrak{g}, \mathfrak{g}]$$ for the Lie algebras in (1) and (4) and comparing them to those of (2) and (3), which recovers Lord Shark the Unknown's hint in the comments.
It is not true, by the way, that the Lie algebra $\mathfrak{t}_2$ in (2) has dimension $2$; it has dimension $3$, since it has basis $$\left\{\pmatrix{a&\cdot\\&\cdot}, \pmatrix{\cdot&b\\&\cdot}, \pmatrix{\cdot&\cdot\\&c}\right\} .$$ Putting your intuition more precisely: The inclusion map $\mathfrak{t}_2 \hookrightarrow \mathfrak{gl}(2, \Bbb R)$ is a $2$-dimensional representation of $\mathfrak{t}_2$, whereas the inclusion map $\mathfrak{n}_3 \hookrightarrow \mathfrak{gl}(3, \Bbb R)$ is a $3$-dimensional representation.
Finally, it was mentioned in the comments, but for a matrix $x \in \mathfrak{gl}(n, \Bbb R)$, $x^t$ just denotes transpose of $x$.
(In the last two comments, we used the (canonical) identification of $\mathfrak{gl}(n, \Bbb R)$ with the space $M(n, \Bbb R)$ of $n \times n$ matrices, equipped with the usual matrix commutator as Lie bracket.)