Is the complexification of $\mathfrak{sl}(n, \mathbb{C})$ itself?

Question

More precisely if I complexify the algebra, is the following isomorphism true? $$\mathfrak{sl}(n, \mathbb{C})_\mathbb{C} \cong \mathfrak{sl}(n, \mathbb{C})$$ Given that the complex dimension is already $n^2-1$, and each element of the matrix can already be complex, do we gain anything by complexifying here?

Answer 1

The complexification of $\mathfrak{sl}_n(\mathbb{C})$ is $\mathfrak{sl}_n(\mathbb{C}) \times \mathfrak{sl}_n(\mathbb{C})$.

This follows from the observation that $\mathfrak{sl}_n(\mathbb{C})$ is itself the complexification of $\mathfrak{sl}_n(\mathbb{R})$, so its complexification can be computed as

$$\left( \mathfrak{sl}_n(\mathbb{R}) \otimes_{\mathbb{R}} \mathbb{C} \right) \otimes_{\mathbb{R}} \mathbb{C} \cong \mathfrak{sl}_n(\mathbb{R}) \otimes_{\mathbb{R}} \left( \mathbb{C} \otimes_{\mathbb{R}} \mathbb{C} \right)$$

and then computing that

$$\mathbb{C} \otimes_{\mathbb{R}} \mathbb{C} \cong \mathbb{C}[x]/(x^2 + 1) \cong \mathbb{C}[x]/((x + i)(x - i)) \cong \mathbb{C} \times \mathbb{C}.$$

So we get two copies of the complexification of $\mathfrak{sl}_n(\mathbb{R})$.

This generalizes to an arbitrary complex Lie algebra $\mathfrak{g}$ but it's a bit more subtle. The general statement is that the complexification $\mathfrak{g} \otimes_{\mathbb{R}} \mathbb{C}$ of the underlying real Lie algebra of $\mathfrak{g}$ is $\mathfrak{g} \times \overline{\mathfrak{g}}$ where $\overline{\mathfrak{g}}$ is the conjugate - the same real Lie algebra as $\mathfrak{g}$ but with the conjugate complex multiplication. This follows from repeating the above calculation but using $\mathfrak{g} \otimes_{\mathbb{C}} (\mathbb{C} \otimes_{\mathbb{R}} \mathbb{C})$ instead and being more careful about complex structures.

If $\mathfrak{g}$ is already the complexification of a real Lie algebra (equivalently, has a real form) then it's isomorphic to its conjugate. In particular every semisimple (even reductive) complex Lie algebra has this property. Probably there are solvable complex Lie algebras that don't but I don't know any off the top of my head.

This also isn't specific to Lie algebras and also applies to commutative algebras, associative algebras, vector bundles, group representations, all sorts of stuff.

Given that the complex dimension is already $n^2-1$, and each element of the matrix can already be complex, do we gain anything by complexifying here?

When we complexify we ignore the existing complex structure and freely add a new one. Concretely you can think about this as follows: replace each complex number $a + bi$ in a matrix entry of an element of $\mathfrak{sl}_n(\mathbb{C})$ with the $2 \times 2$ real matrix $\left[ \begin{array}{cc} a & b \\ -b & a \end{array} \right]$. This is how you can think about $\mathfrak{sl}_n(\mathbb{C})$ as a real Lie algebra (in fact as a real Lie subalgebra of $\mathfrak{sl}_{2n}(\mathbb{R})$). When we complexify a second time we allow each of these real numbers to also be complex.

Answer 2

I would just like to add to the very good existing answers one basic subtlety which, I think, needs to be pointed out more strongly and explicitly.

Namely, note that if $\mathfrak g$ is a Lie algebra over $\mathbb C$, then of course we have

$$\mathbb C \otimes_{\mathbb C} \mathfrak g \simeq \mathfrak g.$$

But when we say we "complexify" such a Lie algebra $\mathfrak g$ (which already is a complex Lie algebra), we are actually not doing that, at all. Rather, we are doing two things: We first apply scalar restriction, and then scalar extension to it. The shorthand notation for that is

$$\mathbb C \otimes_{\color{red}{\mathbb R}} \mathfrak g \qquad \qquad (1)$$

and what is hidden in there is that the right hand side is no longer the complex Lie algebra $\mathfrak g$, but rather "$\mathfrak g$ viewed as a real Lie algebra". I have found it enlightening (although, admittedly, tedious) to keep track of that with explicit notation, e.g. $\mathrm{Res}_{\mathbb C\vert \mathbb R} ( \mathfrak g)$ (for "restriction"). So if $\mathfrak g$ is an $n$-dimensional complex Lie algebra, then $\mathrm{Res}_{\mathbb C\vert \mathbb R} (\mathfrak g)$ is a $\color{red}{2}n$-dimensional real Lie algebra. And we should properly denote $(1)$ as

$$\mathbb C \otimes_{\color{red}{\mathbb R}} \color{red}{\mathrm{Res}_{\mathbb C\vert \mathbb R}} ( \mathfrak g)$$

Of course this looks horribly cumbersome, but when looked at it this way, it will hopefully become clearer why this does not just give back $\mathfrak g$. Also, it is the restriction step which is responsible for doubling the dimension.

I wrote much more about extension and restriction in my answer here. It is a bit verbose, but it also points out that in general, we cannot even expect that restriction followed by extension (or, for that matter, the other way around, which can give something very different again) would just give several copies of the original Lie algebra, like it does here. In a way, we are still in a good case.

Answer 3

For any real finite-dimensional vector space $V$ with dimension $n$, $\dim V_{\Bbb C}$ is $2n$ over $\Bbb R$ (and $n$ over $\Bbb C$). So, that isomorphism is not true: $\mathfrak{sl}(n,\Bbb C)_{\Bbb C}\not\simeq\mathfrak{sl}(n,\Bbb C)$, since$$\dim\mathfrak{sl}(n,\Bbb C)_{\Bbb C}=2\dim\mathfrak{sl}(n,\Bbb C)\neq\dim\mathfrak{sl}(n,\Bbb C).$$