Lie algebra $\mathfrak{sl}_2 \mathbb{C}$ has only these two real forms $\mathfrak{sl}_2 \mathbb{R}$ and $\mathfrak{su}_2$?

Question

A Lie algebra is a vector space $\mathfrak{g}$ over some field $F$ together with a binary operation $$\mathfrak{g}\times\mathfrak{g}\to\mathfrak{g}$$ called the Lie bracket satisfying the following axioms: Bilinearity, Alternativity, Jacobi identity, Anticommutativity.

(Correct me if I am wrong)*

• If the Lie algebra over the field $F$ is a complex number, we have a complex Lie algebra.

• If the Lie algebra over the field $F$ is a real number, we have a real Lie algebra.

Given a complex Lie algebra $\mathfrak g$, a real Lie algebra $\mathfrak{g}_0$ is said to be a real form of $\mathfrak g$ if the complexification $$\mathfrak{g}_0 \otimes_{\mathbb{R}} \mathbb{C} \simeq \mathfrak{g}$$ is isomorphic to $\mathfrak{g}$.

A real form need not be unique; for example, $$\mathfrak{sl}_2 \mathbb{C}$$ has two real forms $$\mathfrak{sl}_2 \mathbb{R}$$ and $$\mathfrak{su}_2.$$

**My question is that:

1. how do we show $\mathfrak{sl}_2 \mathbb{C}$ has these and only these two real forms $\mathfrak{sl}_2 \mathbb{R}$ and $\mathfrak{su}_2$? **

2. Can you also explain the operation $\otimes_{\mathbb{R}}$?

Answer 1

Q2: Complexification $(-) \otimes_{\mathbb{R}} \mathbb{C}$ can be thought of very explicitly as follows: the complexified Lie algebra has exactly the same structure constants as the original Lie algebra, but you're now allowed to take complex combinations of the basis elements, not just real ones. Said another way, $\mathfrak{g}_{\mathbb{C}}$ as a vector space has a decomposition $\mathfrak{g} \oplus i \mathfrak{g}$ into real and imaginary parts, and the Lie bracket extends the Lie bracket of $\mathfrak{g}$ linearly.

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Q1A: Abstractly this computation can be carried out using Galois cohomology; we get that the set of real forms of $\mathfrak{sl}_2(\mathbb{C})$ is parameterized by

$$H^1(\text{Gal}(\mathbb{C}/\mathbb{R}), \text{Aut}(\mathfrak{sl}_2(\mathbb{C})) \cong H^1(\mathbb{Z}_2, PGL_2(\mathbb{C}))$$

(it's not entirely obvious that the automorphism group is $PGL_2(\mathbb{C})$ but it follows from the fact that the Dynkin diagram of $\mathfrak{sl}_2(\mathbb{C})$ is $A_1 = \bullet$, so there are no "diagram automorphisms"), where $\mathbb{Z}_2$ acts by complex conjugation on $PGL_2(\mathbb{C})$ in the obvious way. This Galois cohomology group also classifies real forms of $M_2(\mathbb{C})$ (because it also has automorphism group $PGL_2(\mathbb{C})$) and a nontrivial such real form (other than $M_2(\mathbb{R})$) must, by standard facts about central simple algebras, be a $4$-dimensional real division algebra, hence from the Frobenius theorem must be the quaternions $\mathbb{H}$. (Over other fields we get more general quaternion algebras.)

It follows that $H^1(\mathbb{Z}_2, PGL_2(\mathbb{C}))$ has two elements, so there are two real forms, and since $\mathfrak{sl}_2(\mathbb{R})$ and $\mathfrak{su}(2)$ are two real forms (which can be distinguished e.g. by the signature of the Killing form) these are all the real forms. Alternatively, it's actually possible to write down a natural bijection between real forms of $M_2(\mathbb{C})$ and real forms of $\mathfrak{sl}_2(\mathbb{C})$: we define the trace and then pass to the Lie subalgebra of elements of trace zero. Applied to $\mathbb{H}$ this produces the Lie algebra of purely imaginary quaternions $\mathfrak{sl}_1(\mathbb{H}) \cong \mathfrak{sp}(1) \cong \mathfrak{su}(2)$.

In general real forms of simple Lie algebras are classified by Satake diagrams, which are Dynkin diagrams with some extra structure. I don't know anything about this, though.

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Q1B: But that was very high-tech. It's possible to give a lower-tech and fairly direct answer. Suppose $\mathfrak{g}$ is a $3$-dimensional real Lie algebra complexifying to $\mathfrak{sl}_2(\mathbb{C})$. Then $\mathfrak{g}$ must be simple (since $\mathfrak{sl}_2(\mathbb{C})$ is simple). Consider the adjoint action $\text{ad}_X : \mathfrak{g} \to \mathfrak{g}$ of any nonzero $X \in \mathfrak{g}$. By simplicity we must have $\text{tr}(\text{ad}_X) = 0$; equivalently, the image of $\mathfrak{g}$ in $\mathfrak{gl}(\mathfrak{g})$ must have image in $\mathfrak{sl}(\mathfrak{g})$ (otherwise the trace would give a nontrivial abelian quotient). So the eigenvalues of $\text{ad}_X$ add up to $0$, and since $\text{ad}_X(X) = 0$ one of the eigenvalues is $0$, meaning that the other two eigenvalues are either two nonzero real eigenvalues $r, -r$ or two nonzero purely imaginary eigenvalues $is, -is$. We now split into cases:

Case 1: Some $H \in \mathfrak{g}$ has two nonzero real eigenvalues $r, -r$. By scaling $H$ we can assume WLOG that the eigenvalues are $2, -2$ (this isn't really necessary but it's traditional). Let $X, Y$ be the corresponding eigenvectors, so we have $[H, X] = 2X, [H, Y] = -2Y$. The Jacobi identity gives

$$[H, [X, Y]] = [[H, X], Y] + [X, [H, Y]] = 2 [X, Y] - 2 [X, Y] = 0$$

from which it follows that $[X, Y]$ must be a multiple $cH$ of $H$. By rescaling $X$ we can assume WLOG that $[X, Y] = H$, and now we've written down the traditional presentation

$$[H, X] = 2X, [H, Y] = 2Y, [X, Y] = H$$

of $\mathfrak{sl}_2(\mathbb{R})$ (and in fact of $\mathfrak{sl}_2(F)$ for any field $F$; the $2$s matter in characteristic $2$). So $\mathfrak{g} \cong \mathfrak{sl}_2(\mathbb{R})$ in this case.

Case 2: Every $X \in \mathfrak{g}$ has purely imaginary eigenvalues. By scaling appropriately we can find $X \in \mathfrak{g}$ with eigenvalues $i, -i$. In the complexification $\mathfrak{g}_{\mathbb{C}}$ let $Z + iY, Z - iY$ be the corresponding two eigenvectors (it's a general fact that the eigenvectors must be complex conjugates), where $Z, Y\in \mathfrak{g}$: this gives

$$[X, Z \pm iY] = [X, Z] \pm i [X, Y] = -Y \pm iZ$$

hence $[X, Z] = -Y$ and $[X, Y] = Z$; rearranging the first one gives $[Z, X] = Y$. The Jacobi identity gives

$$[X, [Y, Z]] = [[X, Y], Z] + [Y, [X, Z]] = 0$$

so $[Y, Z]$ must be a scalar multiple $cX$ of $X$. By rescaling both $Y$ and $Z$ by $\sqrt{|c|}$ (we can't just rescale one of them this time) we can assume WLOG that $c = \pm 1$. If $c = -1$ we get $[Y, X] = -Z, [Y, Z] = -X$ so $Y$ has eigenvectors $X \pm Z$ with real eigenvalues $\mp 1$ which contradicts our assumption that every element has imaginary eigenvalues. So $c = 1$, and now we've written down the traditional presentation

$$[X, Y] = Z, [Y, Z] = X, [Z, X] = Y$$

of $\mathfrak{su}(2)$ (more or less).

This argument does not use the full strength of the assumption that $\mathfrak{g}$ is a real form of $\mathfrak{sl}_2(\mathbb{C})$: actually all it uses is that $\mathfrak{g}$ is perfect (has trivial abelianization). So this is a classification of $3$-dimensional perfect real Lie algebras (perfect is equivalent to simple in dimension $3$), which fits into the Bianchi classification of all $3$-dimensional real Lie algebras.

Edit, 9/27/22: A recent question has shown me that there's a small gap in this argument; the possibility that $\text{ad}_X$ is nilpotent for every $X \in \mathfrak{g}$ (so that $\text{ad}_X$ never has distinct eigenvalues) needs to be ruled out. This can be done following the argument here from Fulton and Harris although maybe simpler arguments are possible.

Answer 2

Just to add to Qiaochu Yuan's great answer what happens over a more general field $k$ (for safety, let's assume $char(k)=0$, although I think we only really need $char(k) \neq 2$ in what follows):

The case distinction in Q1B is now whether some element in $\mathfrak g$ has two nonzero eigenvalues in $k^*$ or not. In case there is such an element, everything goes exactly like in Case 1 and we get the split form $\mathfrak{sl}_2(k)$.

In Case 2, we now see that all elements of $\mathfrak g$ have a pair of eigenvalues which are of the form $\pm c_1\sqrt a$ for some $c_1 \in k^*$, $a \in k^* \setminus k^{*2}$; by scaling we take one such $X$ whose eigenvalues are $\pm \sqrt a$. Following the exact analogue of Qiaochu's procedure, again we find $Y,Z$ with $[X,Z]=aY$ and $[X,Y]=Z$ as well as $[Y,Z] = cX$ for some $c \in k$. Now if we try the second rescaling procedure, we run into an extra problem since we have to scale $Y,Z$ simultaneously, so we can re-scale that $c$ only by squares, i.e. $c$ "is" any representative of the coset $c \cdot k^{*2} \in k^*/k^{*2}$. This is the moment where over $\mathbb R$ we have just an old-fashioned positive/negative distinction, but over other fields it can get quite intricate. We do, however, still get that $-c$ cannot be a square by an analogous argument as in Qiaochu's answer. Actually, let's call this non-square $b:=-c$. Over $\mathbb R$, we were done there because we can scale to $b=-1$ and get what many people have checked to be a presentation of $\mathfrak{su}_2$. Now however, we need more restriction on $b$. Namely, if we write out the adjoint of a general element

$$xX + yY + zZ \in \mathfrak g \qquad(x,y,z \in k)$$

as a $3\times 3$-matrix, we see its characteristic polynomial is

$$\chi_{ad (xX + yY + zZ)}(T)= T(T^2-ax^2-by^2+abz^2)$$

and we see that we need the quadratic form $aX_1^2+bX_2^2-abX_3^2$ to not represent a square in $k$. Actually, one knows either through general theory of anisotropic Lie algebras or of quadratic forms (or the proof of theorem 4.20 in Keith Conrad's Quaternion Notes) that this is equivalent to either of the following:

1. The equation $Y_0^2-aY_1^2-bY_2^2=0$ has no solution $(0,0,0) \neq (y_0,y_1,y_2) \in k^3$.
2. $b$ is not a norm of the extension $k(\sqrt a)\vert k$.
3. $a$ is not a norm of the extension $k(\sqrt b)\vert k$.

We get a contradiction to the assumption of this case as soon as these are not satisfied. So they are satisfied, and our Lie algebra $\mathfrak g$ is completely described by that basis $X,Y,Z$ and

$$[X,Y]=Z, \quad [Y,Z]=-bX, \quad [Z,X]= -aY$$

which makes it exactly the "totally imaginary part" of the quaternion algebra $(a,b)_k$ with, in K. Conrad's notation (Def. 3.3), $X:=u/2, Y=v/2, Z=uv/2$.

So each $k$-form of $\mathfrak{sl}_2$ which is not $\simeq \mathfrak{sl}_2(k)$ is of that form for some pair $(a,b)$ as above, and conversely, for each pair $(a,b) \in k \times k$ satisfying above conditions 1-3 we get such an "anisotropic form" of $\mathfrak{sl}_2$. If one digs deeper, one finds that two such Lie algebras given by pairs $(a,b)$ and $(a',b')$ respectively, are isomorphic as Lie algebras if and only if the corresponding quaternion (division) algebras are isomorphic $(a,b)_k \simeq (a', b')_k$, if and only if the quadratic forms in condition 1 with the respective pairs $(a,b)$ and $(a',b')$ are congruent up to scalar multiples (called "similar" or "cogredient" in some sources); equivalently, we need an actual congruence of the quadratic forms in four variables $Y_0^2-aY_1^2-bY_2^2+abY_4^2$ and $Y_0^2-a'Y_1^2-b'Y_2^2+a'b'Y_4^2$, which over a general field might be not totally trivial to decide. For $k=\mathbb R$, as Qiaochu's answer shows, there is actually only one, coming from the Hamilton quaternions. Also over $p$-adic fields, there is only one. But e.g. over $k=\mathbb Q$, there are infinitely many non-isomorphic quaternion division algebras (cf. corollary 5.5), hence infinitely many non-isomorphic Lie algebras $\mathfrak g$ for which $\mathbb C \otimes_{\mathbb Q} \mathfrak g \simeq \mathfrak{sl}_2(\mathbb C)$. (This last case was essentially asked and answered here.)

For more approaches, see also https://mathoverflow.net/q/78481/27465.

Answer 3

The elementary proof of Qiaochu is indeed very nice, but since this site also addresses beginners, let me point out a minor inaccuracy, that might be confusing: In the first case, in order to make sure that r is not 0, you have to choose the element X to be semisimple. (Clearly, if you choose a nilpotent element X then the spectrum of ad(X) will be just 0 with multiplicity 3). Thus you need to know that every simple Lie algebra contains a semisimple element. This is an easy consequence of the Jordan-Chevalley decomposition, but it should be mentioned. Accordingly, in the second step, you only know that every semisimple element has non-real spectrum - of course, nilpotent elements still exist! In the c=-1 case one then has to argue not with Y (which might not be semisimple), but rather its semisimple part in the abstract Jordan-Chevalley decomposition of Y. This does not affect the argument at all, nor the one by Torsten, but it does affect the comment after the proof, that we only use that the Lie algebra is perfect. This is only true if we already know that a 3-dimensional perfect real Lie algebra is closed under Jordan-Chevalley decomposition and hence contains semisimple elemens.