Structure constants in Lie algebra

[Primroses]

I am studying QFT and got stuck on one question which may be simple. Why is structure constants (in Lie algebra) real and independent on representations? Can you give me a detailed proof?

Thanks!

 

[fzero]

I don't believe that the structure constants are necessarily real for a general Lie algebra. If the generators $X^a$ are Hermitian, then we can define real structure constants ${f^{ab}}_c$ through the definition

$$ [ X^a, X^b ] = i {f^{ab}}_c X^c.$$

In another basis, these structure constants will transform like a (2,1) tensor under the change of basis and may no longer be real.

Also, it is not logically the case that the structure constants are independent of the representation. Rather, a definition of the Lie algebra invariably involves specifying some representation, typically called the defining representation. This defines the structure constants. Then any other representation must satisfy commutation relations consistent with these structure constants. So the logic is that the structure constants are fundamental, while to show that some other collection of matrices is actually a representation of the Lie algebra requires that their commutators conform to the structure constants. It is impossible to discuss a representation (other than the defining one) without knowing the structure constants.

 

[dextercioby]

Hi, fzero is of course right.

In the abstract algebraic approach, the structure constants are in the definition of an abstract Lie algebra (associative algebra over a field + Lie product) and of course are not restricted to be real or complex. Since one encounters Lie algebras in connection with Lie groups, once one has specified the group (its manifold real or complex) he has automatically determined the structure constants of the group/algebra.

 

[George Jones]

Lie algebras exist independently of representations. Often, physicists are concerned with real Lie algebras, e.g., $\mathfrak{su}\left(2\right)$.

Suppose we have a real Lie algebra $\mathcal{L}$, and let $\left\{ x_i \right\}$ be a basis for the Lie algebra. Since any element of the Lie algebra is a a real linear combination of the basis elements, each commutator is real linear combination of the basis elements. In particular, the commutators of the basis elements are linear real combinations of the basis elements, i.e.,

$$\left[ x_i , x_j\right] = c^k_{jk} x_k .$$

A representation of the real Lie algebra $\mathcal{L}$ on a complex vector space $V$ is a mapping $\rho$ from $\mathcal{L}$ into the set of linear operators on $V$, i.e., each Lie algebra element is associated (represented) by an operator on $V$. This association is linear and preserves commutators. Consequently, the structure constants are real and independent of representation.

In more detail: Let each Lie algebra basis element $x_i$ be represented by operator $X_i$, i.e., $X_i = \rho \left( x_i \right)$. Then,

$$\begin{align}
\left[ X_i , X_j \right] &= \left[\rho\left(x_i \right), \rho\left(x_j \right)\right]\\
&= \rho\left( \left[ x_i , x_j \right] \right)\\
&= \rho\left( c^k_{jk} x_k \right)\\
&= c^k_{jk} X_k
\end{align}$$

So, the same structure constants for the real Lie algebra $\mathcal{L}$ appear in all representations of $\mathcal{L}$. Fixing the structure constants for a Lie algebra $\mathcal{L}$ fixes the same real structure constants in all representations.

Changing the basis for $\mathcal{L}$ will change the structure constants, i.e., $\left[ x'_i , x'_j\right] = c'^k_{jk} x'_k$, but the structure constants will be the new real $c'^k_{jk}$ in all representations.

It is possible, at the representation level, to consider operators $Y_m$ that are complex linear combinations of the generators $X_i$, and then $\left[ Y_i , Y_j\right] = d^k_{jk} x'_k$ can have complex $d^k_{jk}$, but I would not call these structure constants of the original Lie algebra, as there exist no corresponding elements $y_i$ of the real Lie algebra $\mathcal{L}$ with $\left[ y_i , y_j\right] = d^k_{jk} y_k$.