Real forms and the groups they integrate to

A complex Lie algebra does not specify a real symmetry group. First choose a real form, then a connected group integrating that real algebra, then the representation category. These choices have different consequences: $\mathfrak{su}(2)$ and $\mathfrak{sl}(2,\mathbb R)$ both complexify to $\mathfrak{sl}(2,\mathbb C)$, but their matrix groups have different topology and different unitary representation theories.

All algebras below are finite-dimensional in characteristic zero. Algebra representations are complex-linear on a complex vector space unless stated otherwise; the Lie algebra acting may itself be real. A subscript $0$ on an orthogonal group selects its identity component.

Complexification and realification

A real form of a complex algebra $\mathfrak g$ is a real Lie algebra $\mathfrak g_0$ with an isomorphism

$$\mathfrak g_0\otimes_{\mathbb R}\mathbb C\simeq\mathfrak g.$$

Equivalently, after specifying the inclusion, it is the fixed algebra of a conjugate-linear involutive automorphism $\sigma$ of $\mathfrak g$. Indeed, every $X\in\mathfrak g$ has the unique decomposition

$$X=\frac{X+\sigma X}{2} +i\frac{X-\sigma X}{2i},$$

with both displayed components fixed by $\sigma$. Consequently $\dim_{\mathbb R}\mathfrak g_0=\dim_{\mathbb C}\mathfrak g$. For $\mathfrak{sl}(n,\mathbb C)$, $\sigma(X)=\overline X$ fixes $\mathfrak{sl}(n,\mathbb R)$, while $\sigma(X)=-X^\dagger$ fixes $\mathfrak{su}(n)$. The latter operation preserves the commutator despite reversing matrix multiplication: the minus sign is essential. See Etingof, §§39.2–39.4, especially Theorem 39.6, for real structures and their classification.

Realification instead forgets complex scalar multiplication. It doubles the dimension and is denoted $\mathfrak g_{\mathbb R}$. Its complexification is $\mathfrak g\oplus\overline{\mathfrak g}$. In the matrix example there is an explicit complex-linear isomorphism

$$(\mathfrak{sl}(2,\mathbb C)_{\mathbb R})\otimes_{\mathbb R}\mathbb C \longrightarrow\mathfrak{sl}(2,\mathbb C)\oplus\mathfrak{sl}(2,\mathbb C), \qquad X\otimes z\longmapsto(zX,z\overline X).$$

Here the extension is complex-linear in the tensor factor $z$; it does not assert that $X\mapsto\overline X$ is complex-linear. Its images of $X\otimes1$ and $(iX)\otimes1$ independently generate both summands. Thus the six-real- dimensional Lorentz algebra is not a third real form of a single complex $A_1$.

A matrix comparison in rank one

Put

$$H=\begin{pmatrix}1&0\\0&-1\end{pmatrix},\quad E=\begin{pmatrix}0&1\\0&0\end{pmatrix},\quad F=\begin{pmatrix}0&0\\1&0\end{pmatrix}.$$

The commutators are $[H,E]=2E$, $[H,F]=-2F$, $[E,F]=H$. They define the same complex algebra in both following cases:

Real algebra	Real basis	Real dimension	Complexification
$\mathfrak{su}(2)$	$iH,\ E-F,\ i(E+F)$	3	$A_1$
$\mathfrak{sl}(2,\mathbb R)$	$H,\ E+F,\ E-F$	3	$A_1$
$\mathfrak{sl}(2,\mathbb C)_{\mathbb R}$	$H,E,F,iH,iE,iF$	6	$A_1\oplus A_1$

The Killing form $B(X,Y)=\operatorname{tr}(\operatorname{ad}X\operatorname{ad}Y)$ on either three-dimensional form is $4\operatorname{tr}(XY)$. In the first basis its Gram matrix is $\operatorname{diag}(-8,-8,-8)$; in the second it is $\operatorname{diag}(8,8,-8)$. A real algebra isomorphism preserves this signature, so the two forms cannot be isomorphic over $\mathbb R$. Complexifying removes this obstruction because multiplication of basis vectors by $i$ changes the signs.

For a real semisimple algebra, negative-definite Killing form characterizes the compact form. The rank-one classification has precisely these two real forms; $\mathfrak{su}(1,1)\simeq\mathfrak{sl}(2,\mathbb R)$ is another realization of the second, not a third case. See Etingof, Proposition 39.8, Exercise 40.5, and §40.3(1).

The complex rank of both is one. Their real ranks are zero and one: in a Cartan decomposition $\mathfrak g_0=\mathfrak k\oplus\mathfrak p$, real rank is the dimension of a maximal abelian subspace of $\mathfrak p$. For the split example, $\theta(X)=-X^T$ gives $\mathfrak k=\mathbb R(E-F)$ and $\mathfrak p=\mathbb RH+\mathbb R(E+F)$; a maximal commuting subspace of $\mathfrak p$ is the line $\mathbb RH$. The compact form has $\mathfrak p=0$. This is not the dimension of a maximal compact torus: $SO(2)$ still has a one-dimensional torus.

Global groups and maximal compact subgroups

A real algebra has a unique connected simply connected integrating Lie group, and other connected integrations are quotients by discrete central subgroups. The covering-group lesson derives why this additional choice matters for representations.

Connected matrix group	Real Lie algebra	Maximal compact subgroup	Fundamental group
$SU(2)$	$\mathfrak{su}(2)$	$SU(2)$	$0$
$SO(3)$	$\mathfrak{su}(2)$	$SO(3)$	$\mathbb Z_2$
$SL(2,\mathbb R)$	$\mathfrak{sl}(2,\mathbb R)$	$SO(2)$	$\mathbb Z$
$SL(2,\mathbb C)$ as a real group	$\mathfrak{sl}(2,\mathbb C)_{\mathbb R}$	$SU(2)$	$0$
$SO_0(1,3)$	$\mathfrak{so}(1,3)$	$SO(3)$	$\mathbb Z_2$

For the special linear matrix groups the topology follows directly from polar decomposition. Write $g=UP$ with $P=(g^\dagger g)^{1/2}$ positive Hermitian, $\det P=1$, and $U\in SU(2)$. The logarithm identifies the allowed $P$ with traceless Hermitian matrices, a real three-dimensional vector space. The map $UP\mapsto UP^{1-t}$ retracts $SL(2,\mathbb C)$ onto $SU(2)\simeq S^3$. Over $\mathbb R$, replace the dagger by transpose: the positive symmetric factor has two parameters, and the same argument retracts $SL(2,\mathbb R)$ onto $SO(2)\simeq S^1$. These are manifold decompositions, not direct-product group decompositions.

The maximal-compact assertion for these matrix groups is proved in Conrad–Landesman, Appendix D.2, especially Remark D.2.1. The polar construction and its topology are also given in Etingof, §43.1, Corollary 43.2 and Example 43.4.

The universal cover of $SL(2,\mathbb R)$ is not the matrix group in this table. Lifting its rotation subgroup unwraps the circle to a line. If $z$ denotes the central lift of a rotation by $\pi$, the maps to $SL(2,\mathbb R)$ and $PSL(2,\mathbb R)$ kill $\langle z^2\rangle$ and $\langle z\rangle$ respectively. Do not infer that this covering group contains the same compact circle or has the same fundamental group. See its atlas record.

The Lorentz spin map, explicitly

Use Minkowski signature $(+---)$ and identify a vector with the Hermitian matrix

$$X=x^0I+x^1\sigma_1+x^2\sigma_2+x^3\sigma_3 =\begin{pmatrix}x^0+x^3&x^1-ix^2\\x^1+ix^2&x^0-x^3\end{pmatrix}.$$

Then $\det X=(x^0)^2-(x^1)^2-(x^2)^2-(x^3)^2$. For $A\in SL(2,\mathbb C)$, $X\mapsto AXA^\dagger$ preserves Hermiticity and this determinant. It therefore defines a real linear Lorentz transformation. Connectedness places its image in $SO_0(1,3)$.

If it acts trivially, evaluating at $X=I$ makes $A$ unitary, after which $AXA^{-1}=X$ for all Hermitian $X$ forces $A$ to be scalar; determinant one leaves $A=\pm I$. Thus its kernel is exactly $\{\pm I\}$. The differential has zero kernel and both real Lie algebras have dimension six. The image is an open subgroup of the connected target, hence is the whole target. We obtain

$$SL(2,\mathbb C)/\{\pm I\}\simeq SO_0(1,3).$$

The preceding polar argument makes the source simply connected, so this is the universal double cover. Restricting to $SU(2)$ fixes $x^0$ and gives the usual $SU(2)\to SO(3)$ rotation map. Dividing the polar decomposition by its central sign gives $SO_0(1,3)\simeq SO(3)\times\mathbb R^3$ as manifolds and justifies the final row of the table. The matrix argument proves the map, not just an algebra coincidence; compare the map record and Etingof, §40.1(4).

What happens to representations

The complex highest-weight module $\operatorname{Sym}^n\mathbb C^2$ restricts to either $SU(2)$ or $SL(2,\mathbb R)$, has dimension $n+1$, and lets the central $-I$ act by $(-1)^n$. It descends to the respective quotient by $\{\pm I\}$ exactly for even $n$. Its dimension and central sign do not determine whether its action preserves a positive-definite Hermitian form.

On $SU(2)$ the defining Hermitian form and its symmetric powers are invariant. On $SL(2,\mathbb R)$ the element $\operatorname{diag}(e^t,e^{-t})$ has weights $e^{(n-2k)t}$ for $0\le k\le n$. For $n>0$ some eigenvalues have modulus other than one, so no positive-definite invariant Hermitian form exists. This is a direct obstruction to unitarity of these finite-dimensional modules; it does not exclude infinite-dimensional unitary representations.

Finite-dimensional complex Lorentz modules instead carry two half-integer labels $(j_L,j_R)$ because the complexification has two $A_1$ factors. The spin-cover center acts by $(-1)^{2j_L+2j_R}$; only integral $j_L+j_R$ descends to $SO_0(1,3)$. The vector is $(\tfrac12,\tfrac12)$ and the two Weyl spinors are $(\tfrac12,0)$ and $(0,\tfrac12)$. These describe field indices, not a classification of particle Hilbert spaces. Continue with spacetime representations.

Use the atlas and tools within their category

The curated real-form index supplies 19 semisimple real-form records and selected integrations, rather than all noncompact groups. The highest-weight workbench computes compact-category invariants; its output is not a noncompact unitary representation classifier. Use roots and weights for its label conventions, global-form descent for central kernels, and groups, cosets, and classifying spaces before using a topological quantity in a physics argument.