Why is $\mathfrak{sl}(n)$ the algebra of traceless matrices?

Question

I'm studying Lie algebras as purely algebraic objects, without much of a background in the differential geometry surrounding Lie groups.

The definition of $\mathfrak{sl}(n)$ has been given to me as the algebra of traceless $n\times n$ matrices with multiplication $$[A,B]=AB - BA$$ With my limited knowledge of differential geometry, I roughly understand that $\mathfrak{sl}(n)$ is defined as the tangent space of the Lie group $SL(n)$ at the identity, but I don't see how this leads to the definition of $\mathfrak{sl}(n)$ as the traceless $n\times n $ matrices.

I was hoping that someone would be able to roughly explain (or point me in the direction of a not-too-differential-geometry-technical article):

1) What is the reason $\mathfrak{sl}(n)$ is taken to be the algebra of traceless $n \times n$ matrices?

2) What is the connection between $\mathfrak{sl}(n)$ and $SL(n)$?

3) What is the motivation for this definition?

Answer 1

For a Lie group $G$, there is an exponential map $\operatorname{exp}:g\to G$ (where $g$ is the corresponding Lie algebra) such that $\operatorname{exp}\{tA\}$ is just the one-parameter subgroup of $G$ with tangent vector $A$ at $e$. For matrix groups, this is just the usual matrix exponential $e^{At}$. Now note that $\operatorname{det}e^A=e^{\operatorname{tr}A}$.

Answer 2

There is a purely geometrical explanation. First, let $n=2$, and consider $\mathrm{SL}(2,\mathbb{R})$.

Let $X : [0,1) \to \mathrm{SL}(2,\mathbb{R})$ be a smooth path, with $X(0) =E$, where $E$ is the identity.

$$X(t) = \left[\begin{array}{cc} a(t) & b(t) \\ c(t) & d(t) \end{array}\right]$$ where $\det[X(t)] = (ad-bc)(t) = 1$ for all $t$. The derivative $X'(t)$, as $t$ tends to zero, gives a tangent vector to $\mathrm{SL}(2,\mathbb{R})$ at $E$. We have $$X'(t) = \left[\begin{array}{cc} a'(t) & b'(t) \\ c'(t) & d'(t) \end{array}\right]$$

Since $ad-bc \equiv 1$ we can differentiate to give $$a'd+ad'-b'c-bc'\equiv 0$$ Since $X(0)=E$ we know that $a(0)=1$, $b(0)=0$, $c(0)=0$ and $d(0)=1$. Hence: $$\begin{eqnarray*} a'd+ad'-b'c-bc' &\equiv& 0 \\ \\ a'(0)\,d(0)+a(0)\,d'(0)-b'(0)\,c(0)-b(0)\,c'(0) &=& 0 \\ \\ a'(0)+d'(0) &=& 0 \end{eqnarray*}$$ This tells us that $X'(0)$ must be a traceless matrix.

General Dimension

Jacobi's Formula tells us that

$$\det(X)' = \mathrm{tr}\! \left[ \mathrm{adj}(X)X' \right] $$

Since $\det(X) \equiv 1$ we have $\det(X)'\equiv 0$. When $t=0$, $X=E$ and so $\mathrm{adj}(X) = E$. Hence $$\mathrm{tr}\!\left[X'(0)\right]=0$$

Answer 3

Fix $A\in \mathfrak{sl}(n)$. Then, $\exp(tA)=e^{tA}\in SL(n,\mathbb{R})$. Thus, $\det(e^{tA})=1$ for every $t\in\mathbb{R}$. Also, let $J$ be the Jordan Canonical Form of $A$, with $A=PJP^{-1}$. This way,differentiating both sides: $$0=\left.\frac{d}{dt}\right|_{t=0}\det(e^{tA})=\left.\frac{d}{dt}\right|_{t=0}\det(Pe^{tJ}P^{-1})=\left.\frac{d}{dt}\right|_{t=0}\det(e^{tJ})=(I).$$ Note that $Spec(tJ)=Spec(tA)$ and that $Spec(e^{tA})=e^{tSpec(A)}$. Moreover, the determinant of a triangular matrix is the product of its diagonal entries, which in the case of the Jordan Form they are exactly the matrix' eigenvalues. Thus, $$\det(e^{tJ})=\prod_{\lambda\in\operatorname{Spec(A)}}e^{tA}.$$ This way, $$(I)=\left.\frac{d}{dt}\right|_{t=0}\prod_{\lambda\in\operatorname{Spec(A)}}e^{tA}=\left.\frac{d}{dt}\right|_{t=0}e^{\sum_{\lambda\in Spec(A)}t\lambda}=\left.\frac{d}{dt}\right|_{t=0}e^{t\cdot tr(A)}=tr(A).$$

Therefore $\mathfrak{sl}(n)\subset\{A\in\mathfrak{gl}(n):tr(A)=0\}$. Note that, a trace is a rank 1 linear function, both these spaces are $n^2-1$ dimensional. Thus, they coincide.