What is the motivation for complexifying a Lie algebra?

In quantum mechanical angular momentum the commutation relations

$$[J_x,J_y]=iJ_z, \quad [J_y,J_z] = iJ_x,\quad [J_z,J_x] = iJ_y$$

become, on complexifying (arbitrarily defining $J_{\pm} = J_x \pm iJ_y$)

$$[J_+,J_-] = 2J_z,\quad [J_z,J_\pm] = \pm 2J_z.$$

and then everything magically works in quantum mechanics. This complexification is done for the Lorentz group also, as well as in the conformal algebra.

There should be a unified reason for doing this in all cases explaining why it works, & further some way to predict the answers once you do this (without even doing it), though I was told by a famous physicist there is no motivation :(


Short answer: complexifications facilitate representation theory.

In physics, we typically want to find representations of a Lie algebra $\mathfrak g$, and often times determining the representations of its complexification $\mathfrak g_\mathbb C$ is easier. Moreover, we have the following theorem (see ref 1. Proposition 4.6) which tells us that determining the representations of the complexification allows us to determine the representations of the original algebra.

Theorem. Let $\mathfrak g$ be a real Lie algebra, and let $g_\mathbb C$ be its complexification. Every finite-dimensional complex representation $\pi$ of $\mathfrak g$ has a unique extension to a complex-linear representation $\pi_\mathbb C$ of $\mathfrak g_\mathbb C$ \begin{align} \pi_\mathbb C(X+iY) = \pi(X) + i\pi(Y) \end{align} for all $X,Y\in\mathfrak g$. Furthermore, $\pi_\mathbb C$ is irreducible as a representation of $\mathfrak g_\mathbb C$ if and only if $\pi$ it is irreducible as a representation of $\mathfrak g$.

Example. Angular momentum in QM

In the case of the angular momentum in quantum mechanics, what physics books are doing mathematically is attempting to find representations of $\mathfrak {su}(2)$ acting on the Hilbert space of a given physical system. The complexification of $\mathfrak{su}(2)$ is $\mathfrak{sl}(2,\mathbb C)$, and $\mathfrak{sl}(2,\mathbb C)$ has a nice basis $J_\pm, J_z$ which has no counterpart in $\mathfrak{su}(2)$ and which makes determining representations much easier. The structure relations in the $J_\pm, J_z$ basis allow one to use "raising" and "lowering" operators.

Example. Lorentz algebra

In relativistic quantum field theory, we look for representations of $\mathfrak{so}(1,3)$. It turns out, quite happily, that when we complexify this algebra, it splits into a direct sum of complexified angular momentum algebras: \begin{align} \mathfrak{so}(1,3)_\mathbb C \cong \mathfrak{sl}(2,\mathbb C)\oplus \mathfrak{sl}(2,\mathbb C), \end{align} and since we already know the representation theory of the complexified angular momentum algebra so well, this makes studying the representations of the Lorentz algebra easy.


  1. Hall, Lie Groups, Lie Algebras, and Representations
  • 1
    $\begingroup$ I came here to ask the same question and I'm still somewhat unsatisfied: Namely, not all complex representations of $so(1,3)$ (or, equivalently, of its complexification $sl(2, \mathbb{C}) \oplus sl(2, \mathbb{C})$) give rise to a corresponding real representation (in the sense that they are the unique extension of a real representation). So I'm wondering why these representations are not being ruled out right from the start but, instead, are considered to be physically possible (at least in principle). $\endgroup$
    – balu
    Feb 20 '18 at 20:08
  • $\begingroup$ …Contrast this with the following approach: Instead of complexifying $so(1, 3)$, consider the isomorphism $so(1, 3) = sl(2, \mathbb{C})$ (as real Lie algebras) and then study the complex $sl(2, \mathbb{C})$ and its representations (all of which descent to real representations of $sl(2, \mathbb{C}) = so(1, 3)$). I assume the reason we're not doing this in reality is that not every real representation of $so(1, 3)$ can be obtained that way, so we thereby might be leaving out physically sensible representations. In contrast, when complexifying we're seemingly doing the opposite: Overcounting. $\endgroup$
    – balu
    Feb 20 '18 at 20:22
  • 1
    $\begingroup$ Alternative reference: Brian C. Hall, arxiv.org/abs/math-ph/0005032 Prop. 5.5. $\endgroup$
    – Qmechanic
    Oct 15 '21 at 11:40

From a mathematical perspective, to develop Lie algebra representation theory most efficiently, we need the field $\mathbb{F}$ of the Lie algebra to be algebraically closed. See e.g. Ref. 1, where this assumption is used already in the beginning of Chapter II.

The situation for Lie algebras is similar to when we in linear algebra try to diagonalize, say, a real normal matrix. Such a matrix is always diagonalizable in an orthonormal set of eigenvectors, but the eigenvectors and eigenvalues could be complex. Even for physical systems which are manifestly real in nature, such complex eigenvectors and complex eigenvalues are often useful concepts.

In more detail, for an $n$-dimensional Lie algebra $\frak{g}$, we would like something similar to a Chevaller-basis to exists. This means (among other things) that it should be possible to pick a Cartan subalgebra (CSA) $\frak{h}$ with generators $H_i$, $i=1,\ldots, r$; where $r$ is the rank of $\frak{g}$; and supplemented with basis elements $E_a$, $a=1, \ldots n-r$, $$ {\frak g}~=~{\rm span}_{\mathbb{F}} \left( \{ H_i | i=1,\ldots, r\} \cup \{ E_a | a=1,\ldots, n- r\}\right) ,$$ with the property that the Lie bracket $[E_a,H_i]$ is proportional to $E_a$. The $E_a$ play the role of raising and lowering operators, or equivalently, creation and annihilation operators.

All finite-dimensional semisimple complex Lie algebras has a Chevaller-basis.

Example: The Lie algebra $sl(2,\mathbb{C})$: Think of $H_i$ as $J_3$, and $E_a$ as $J_{\pm}$.

From a physical perspective weights in the facts that e.g.

  1. quantum theory uses complex Hilbert spaces, cf. this Phys.SE post and links therein;

  2. the complex Lie group $SL(2,\mathbb{C})$ happens to be the (double cover of the) restricted Lorentz group $SO^{+}(3,1)$, cf. e.g. this Phys.SE post;

  3. one may speculate that it is easier to construct physically sensible theories based on the category of (complex) analytic functions rather than, say, the category of real smooth functions.


  1. J.E. Humphreys, Intro to Lie Algebras and Representation Theory, Graduate texts in Math 9, Springer Verlag.
  • 3
    $\begingroup$ @bolbteppa You are a tough person to please. The classification of complex semi-simple Lie algebras (due to Cartan and Killing) is a lot easier than the real version. Looking for (inequivalent) real forms of the complex Lie algebra leads to the classification of real Lie algebras. (see Josh's answer). $\endgroup$
    – suresh
    Oct 15 '14 at 2:21

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