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Let $T^a$ (with $a = 1,2,\ldots,n$) be a set of generators of a Lie group that satisfy the commutation relations: \begin{equation} [T^a,T^b] = i \sum_{c=1}^n f^{abc} T^c \,, \end{equation} where $f^{abc}$ are called the structure constants.

Is there a one-to-one correspondence between the set of structure constants (up to an overall constant factor and the labelling) and the Lie group? If yes, then does somebody know of an overview somewhere on the web that gives you all the Lie groups and its structure constants?

(I am, obviously, a beginner in this field.)

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  • $\begingroup$ Do you mean Lie algebra instead of Lie groups? $\endgroup$
    – user99914
    Nov 17, 2013 at 12:19
  • $\begingroup$ I don't know the difference very well as I am new to this branch of mathematics. Is there no one-to-one correspondence between Lie algebras and Lie groups either? $\endgroup$
    – Tom
    Nov 17, 2013 at 12:52
  • $\begingroup$ I am interested in this: "does somebody know of an overview somewhere on the web that gives you all the Lie groups and its structure constants?" Any good resource? $\endgroup$
    – wonderich
    Dec 19, 2013 at 3:31

1 Answer 1

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I will try to answer to the OP by describing at first the relationship between Lie algebras and structure constants. After it, I will move to the problem of integrating a given Lie algebra to a simply connected Lie group. This should help in understanding the relationship between structure constans and Lie groups.

  • On Lie algebras and structure constants

Let $G$ be a Lie group and $\mathfrak g$ be its Lie algebra. We restrict to the finite dimensional case for simplicity. Let $\{e_i\}$ be a basis of $\mathfrak g$ (a basis of infinitesimal generators for $G$): then

$$[e_i,e_j]=f^k_{ij}e_k,$$

where $f^k_{ij}$ are called the structure constants of the Lie group $G$. We are using Einstein's convention on repeated indices. Note that, choosing any other basis $\{v_i\}$ of $\mathfrak g$, we obtain new structure constants

$$[v_i,v_j]=\tilde{f}^k_{ij}v_k,$$

with $$\tilde{f}^k_{ij}=A_{in}A_{jm}f^s_{nm}A^{-1}_{ks}$$ if $v_i=A_{in}e_n$.

Any Lie algebra $\mathfrak g$ is determined by its structure constants modulo the transformations $f^s_{nm}\mapsto \tilde{f}^s_{nm}$ given above; such transformations determine an equivalence relation. In summary, any two Lie algebras (of the same dimension) with the same structure constants are isomorphic as Lie algebras.

  • On Lie algebras and Lie groups

The Lie algebra $\mathfrak g$ captures the local properties of $G$ as it needs only the connected component of the identity in $G$. To answer to your question, it is then necessay to talk about "local vs. global".

On the local level, one has the fundamental theorems of Lie (I do not discuss them here). On the global level, one needs some work. The theorem below is due to Cartan, and it can be proven using Ado’s theorem, or the Levi-Malcev decomposition. It says that every Lie algebra of finite dimension over $\mathbb k = \mathbb R, \mathbb C$ is isomorphic to the Lie algebra of a Lie group with some non trivial topological properties. This Lie group is unique up to isomorphisms.

Theorem. If $\mathfrak g$ is a Lie algebra of finite dimension over $\mathbb k = \mathbb R, \mathbb C$, there is a connected simply connected Lie group $G$ such that $\mathfrak g = \operatorname{Lie}(G)$. $G$ is determined uniquely up to isomorphism.

One can translate the above theorem into equivalence of the categories from the subcategory of connected simply connected Lie groups and the category of finite dimensional Lie algebras over the base field.

In summary, for any connected simply connected Lie group $G$ one has a unique (up to isomorphisms) finite dimensional Lie algebra $\mathfrak g$ isomorphic to $\operatorname{Lie}(G)$; this Lie algebra is determined by its structure constants as above.

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  • $\begingroup$ Thanks a lot Avitus for clearing that up! $\endgroup$
    – Tom
    Nov 18, 2013 at 20:47

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