3

I saw this table here on Wikipedia and it states that the Lie algebra of the special linear group $SL_n(\mathbb C)$ is the group of traceless matrices $\mathfrak{sl}_n$.

I know the definition of a Lie algebra: given a matrix group $G \subseteq GL_n(\mathbb K)$ its Lie algebra is defined to be the tangent space at $I$.

From this definition it is immediately clear what dimension the Lie algebra has to have. What is unclear to me is:

Assuming I don't already know the answer, how do I arrive at the insight that this tangent space is traceless matrices?

I skipped through Tapp's book in search of hints and found that the exponential map is a homomorphism from the Lie algebra into the Lie group. I also found a lemma that states that

$$ \det e^A = e^{\mathrm{trace}(A)}$$

That is unfortunately only half-useful: If I already know what the Lie algebra is then this lets me prove that the image of the Lie algebra under the homomorpism defined by the exponential map is indeed in $SL_n$.

So it's kind of not so useful. Surely there must be geometric insight?

learner
  • 1,967

2 Answers2

8

Here's one way to think of things:

For each element of the tangent space at $I \in G$, we can construct a flow, and thus a corresponding one-parameter subgroup $A(t)$.

We can further show that for any Lie group $G \subset GL(n,\Bbb C)$, every one-parameter subgroup has the form $A(t) = e^{tX}$ for some matrix $X = A'(0)$.

Thus, we may conclude that $X \in \mathfrak g \iff e^{tX} \in G$ for all $t \in \Bbb R$, when $G \subset GL(n, \Bbb C)$.

In this particular case, we may state that $$ X \in \mathfrak{sl}_n \iff e^{tX} \in SL_n \quad \forall t \in \Bbb R \iff \det(e^{tX}) = 1 \quad \forall t \in \Bbb R \\ \iff e^{t \cdot \text{tr}(X)} = 1 \quad \forall t \in \Bbb R \iff \text{tr}(X) = 0 $$ Thus, $\mathfrak{sl}_n$ consists precisely of the traceless matrices.

Note in particular the importance of considering all $t \in \Bbb R$. For example, if we take $$ X = \pmatrix{2 \pi i &0 \\0&2 \pi i} $$ we find that $X \notin \mathfrak{sl}_2$, even though $e^X = I \in SL_2$.

Ben Grossmann
  • 225,327
2

Hint. Consider a smooth curve $\gamma:\mathbb R\to SL$ in your group which at O is at the identity element. What is its tangent vector at zero?

As the function $t \mapsto \det\gamma(t)$ is constant, its derivative at 0 is 0. Compute it.

Mariano Suárez-Álvarez
  • 135,076