How are $SU(n)$, $SL(n)$ and $\mathfrak{sl}(n,\mathbb{C})$ related?

Question

In the case of the root system $A_{n-1}$, I want to understand the correspondence between the Lie group and the Lie algebra. Please help me understand the relationship between the Lie groups $SU(n)$ and $SL(n)$ and the Lie algebra $\mathfrak{sl}(n,\mathbb{C})$.

This is what I think I understand. The special unitary group $SU(n)$ is the Lie group of $n\times n$ unitary matrices with determinant $1$. A matrix is unitary if its conjugate transpose is also its inverse: $UU^*=U^*U=I$. The rows form an orthonormal basis of $\mathbb{C}^n$, and so do the columns, and the rows and columns are orthonormal with respect to each other.

There is a correspondence between a Lie group $G$ and its Lie algebra $\mathrm{Lie}(G)$ given by:

$\mathrm{Lie}(G)= \{ X\in M_n(\mathbb{C})|e^{tX}\in G \textrm{ for all }t\in \mathbb{R} \}$

The formula $\mathrm {det}(e^X)=e^{\mathrm {tr} (X)}$ and the condition that the $\mathrm{det}(U)=1$ for all $U\in SU(n)$ means that $\mathrm{tr}(X)=0$ for all $X\in \mathrm{Lie}(SU(n))$. Which is to say, the Lie algebra $\mathrm{Lie}(SU(n))$ is a subset of the special linear algebra $\mathfrak{sl}(n)=\{X\in M_n(\mathbb{C})|\mathrm{tr}(X)=0\}$.

The condition $UU^*=I$ for $U\in SU(n)$ translates into the condition $X-X^*=0$ for $X\in \mathrm{Lie}(SU(n))$. This means that $x_{ij}=\overline{x}_{ji}$, and in particular, $x_{ii}$ is purely imaginary. These are the skew Hermitian matrices.

Combining the two conditions, I think we should have that $\mathrm{Lie}(SU(n))$ is the set of skew Hermitian $n \times n$ matrices in $M_n(\mathbb{C})$ with zero trace. That would be perhaps $\mathfrak{su}(n,\mathbb{C})$.

But the Lie algebra associated with the root system $A_{n-1}$ is given by $\mathfrak{sl}(n,\mathbb{C})$. I suppose that makes sense given from Victor Kac's Lecture 14 that the related Cartan subalgebra $\mathfrak{h}$ is the set of all traceless diagonal matrices.

I see at the Wikipedia article on Lie group-Lie algebra correspondence that for root system $A_n$, the compact Lie group is $SU(n+1)$ and the complexification of the associated Lie algebra is $\mathfrak{sl}(n+1)$. I don't understand the relevance here of complexification but I imagine that may be what I'm not understanding.

The table at the Wikipedia article on Classical group lists, for the root system $A_{n-1}$, with a complex Lie algebra, the group $SL(n,\mathbb{C})$ and the maximal compact subgroup as $SU(n)$. So I suppose an issue here is that the corresponding group would be $SL(n,\mathbb{C})$, but it is not compact, so we need to restrict ourselves to the maximal compact subgroup. Even if that's the case, I still need to understand why, starting from $SU(n)$, the correspondence doesn't take us to $\mathfrak{su}$. Is that not semi-simple? Is that the problem?

Please, what is the connection between $SU(n)$, $SL(n)$, $\mathfrak{sl}(n)$ and $A_{n-1}$?

I would also appreciate an explanation of what this means in the cases $n=2$ and $n=3$.

Thank you!

Answer 1

Very short answer: You have to be very precise about which base field, $\mathbb R$ or $\mathbb C$, you are considering in each case.

Over $\mathbb C$, there is the Lie group $SL_n(\mathbb C)$ and its Lie algebra $\mathfrak{sl}_n(\mathbb C)$, and every Cartan subalgebra of this will have roots which form a system of type $A_{n-1}$. There is extensive literature on this.

Over $\mathbb R$ however, one can e.g. look at the Lie groups $SL_n(\mathbb R)$, which have Lie algebra $\mathfrak{sl}_n(\mathbb R)$, but also at the Lie groups $SU(n)$ and their Lie algebras $\mathfrak{su}_n$ -- note that elements of these are often written as certain matrices with complex entries, but they are not complex Lie groups resp. algebras, but real ones. Notice in particular that $\mathfrak{su}_n$, which indeed can be identified with the traceless skew-hermitian $n\times n$-matrices, is not a vector space over $\mathbb C$, but over $\mathbb R$ (of dimension $n^2-1$).

Now it turns out that the non-isomorphic real Lie groups $SL_n(\mathbb R)$ and $SU_n$ both have complexification (isomorphic to) $SL_n(\mathbb C)$. They are so-called real forms of $SL_n(\mathbb C)$. Likewise but even simpler to see -- on the Lie algebra level, complexification is just done by tensoring with $\mathbb C$ -- both $\mathbb C \otimes_{\mathbb R}\mathfrak{su}_n$ and $\mathbb C \otimes_{\mathbb R}\mathfrak{sl}_n(\mathbb R)$ are isomorphic to $\mathfrak{sl}_n(\mathbb C)$, i.e. both $\mathfrak{su}_n$ and $\mathfrak{sl}_n(\mathbb R)$ are real forms of $\mathfrak{sl}_n(\mathbb C)$.

For $n=2$, $SL_2(\mathbb R)$ and $SU_2$ are (up to isomorphism) the only real forms of $SL_2(\mathbb C)$. For higher $n$ though, and for other classes of Lie groups / algebras, there are usually more real forms. The last example here is a real form of $\mathfrak{sl}_3(\mathbb C)$, called $\mathfrak{su}_{1,2}$, which is neither isomorphic to $\mathfrak{sl}_3(\mathbb R)$ nor to $\mathfrak{su}_3$.

It's quite common in the literature when speaking of root systems, what is meant is actually the root system of the complexification. In that terminology, both $SL_n(\mathbb R)$ and $SU_n$ (or their Lie algebras) have root system $A_{n-1}$. However, there is also the notion of relative or restricted or real or $k$-rational (here for $k=\mathbb R$) root systems; in this case, the relative root system of $SL_n(\mathbb R)$ would still be $A_{n-1}$, whereas the relative root system of $SU_n$ is empty (which is always the case for compact semisimple groups). More on those "relative roots" e.g. here, where I tried to compute all examples of real forms where that restricted root system is of type $BC$ (something that can never happen for complex Lie groups / algebras).

One further thing to note: By a fantastic coincidence (?), for each complex simple Lie algebra, there is up to iso exactly one real form which is compact (e.g. above, $\mathfrak{su}_n$ is the compact real form of $\mathfrak{sl}_n(\mathbb C)$). Also, there is always exactly one so-called "split" real form, whose restricted roots are just the same as the roots of the complexified version (e.g. above $\mathfrak{sl}_n(\mathbb R)$ is the split real form of $\mathfrak{sl}_n(\mathbb C)$). In a way, these two are extreme cases on opposite ends of a spectrum. As noted above, in general there are many more cases "between" them. They are classified by so-called "Satake diagrams", which are like an upgrade of the Dynkin diagrams: the underlying Dynkin diagram of a Satake diagram tells us of what type ($A_n, B_n, C_n, ..., G_2$) the complexification is, and the extra ornaments that make it a Satake diagram (black vs. white nodes, and arrows) encode which real form of that complex type we have. See further references and examples here or here.

Added: It's maybe worthwhile to note that beyond everything mentioned above, the (Lie group / Lie algebra)-correspondence is also not one-to-one, over any ground field. Rather, for one given semisimple Lie algebra there is a lattice of connected groups that "sits" over it, with an adjoint (centreless) one at the bottom and a simply connected one on top. E.g. over $\mathbb C$,

$PSL_2(\mathbb C)$ (adjoint) and $SL_2(\mathbb C)$ (simply connected) share the Lie algebra $\mathfrak{sl}_2(\mathbb C)$;

whereas over $\mathbb R$,

$PSL_2(\mathbb R)$ (adjoint), $SL_2(\mathbb R)$, $Mp_2(\mathbb R)$ (the metaplectic group), ... , $\overline {SL_2(\mathbb R)}$ (the simply connected universal cover of $SL_2(\mathbb R)$), with the "..." being infinitely more in-between, all share the Lie algebra $\mathfrak{sl}_2(\mathbb R)$ (compare last three sentences here);

whereas the compact real one has only two manifestations again:

$PSU_2$ (adjoint, and happens to be $\simeq SO_3(\mathbb R)$) and $SU_2$ (simply connected) share the Lie algebra $\mathfrak{su}_2$.

If one allows even disconnected groups, then there's infinitely many more groups sitting over each Lie algebra, but that's basically stuff like

$SL_2(\mathbb C) \times$ (your favourite finite group) still has Lie algebra $\mathfrak{sl}_2(\mathbb C)$.

Answer 2

The group $SL(n,\mathbb C)$ is a complex Lie group whose Lie algebra is $\mathfrak{sl}(n,\mathbb C)$. The group $SU(n)$ is a compact real Lie group whose Lie algebra is $\mathfrak{su}(n)$, the Lie algebra of all skew-symmetric $n\times n$ complex matrices with null trace. It turns out that its complexification (that is, $\mathfrak{su}(n)\bigotimes\mathbb C$) is isomorphic to $\mathfrak{sl}(n,\mathbb C)$. The same thing occurs with the real Lie group $SL(n,\mathbb R)$: the complexification of its Lie algebra $\mathfrak{sl}(n,\mathbb R)$ is isomorphic to $\mathfrak{sl}(n,\mathbb C)$. What's special about $SU(n)$ is that it is (up to isomorphism) the only compact and connected Lie group such that the complexification of its Lie algebra is isomorphic to $\mathfrak{sl}(n,\mathbb C)$ ($SL(n,\mathbb R)$ is connected, but it is not compact).