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I have elementary questions about Lie groups and their associated Lie algebras.

Let $G=GL(n,\mathbb{C})$. Then associated to this Lie group is the Lie algebra $M_n(\mathbb{C})$ with the commutator relation $[x,y]=xy-yx$ or we can define its Lie algebra to be $M_n(\mathbb{C})$ with left-invariant vector fields.

  1. When would we want to use one over the other? Are they equivalent? Is the Lie algebra with left-invariant vector fields only defined over $\mathbb{R}$?

  2. Why aren't we working with right-invariant vector fields?

  3. $M_n(\mathbb{C})$ can be thought of as $n^2$-dimensional complex Lie algebra or $2n^2$-dimensional real Lie algebra with complex structure. What is an example of an even dimensional real Lie algebra that cannot have a complex structure? Does there exist such an example using a subset of matrices?

  4. Could there exist other Lie algebras (other than the ones mentioned above) for $GL(n,\mathbb{C})$?

Thank you.

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You could be more explcit about what you mean by "with left-invariant vector fields": vector fields are not part of the language of Lie algebras in the way the Lie bracket is (although some sets of vector fields do have a natural Lie algebra structure). Then if you make explicit how the Lie bracket is defined in terms of left-invariant vector fields, it is merely an exercise to check whether this Lie bracket is the same as the one defined by the commutator (it is). Can you do it? –  Marc van Leeuwen Jul 9 '12 at 12:03
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1 Answer

up vote 9 down vote accepted

Let me answer each of your questions:

(1) The first characterization of the Lie algebra of $G$ is convenient because it is concrete, i.e., it is very useful in practice. For example, one would use the first characterization of the Lie algebra of $G$ when one explictly wishes to compute the differentials of Lie algebra homomorphisms. However, the second characterization of the Lie algebra of $G$ is convenient because it is abstract, i.e., it is very useful in theory.

The two characterizations are equivalent. The proof of this equivalence is a standard result in Lie theory. The Lie algebra of $G$ is a complex Lie algebra (because, for example, $G$ is a complex Lie group).

(2) We could certainly work with right-invariant vector fields and the theory would remain the same. The reason for the preference of "left" in place of "right" is the same as the reason for the preference of composing functions from right to left rather than from left to right; however, this is more a matter of tradition than a matter of mathematics.

(3) Let $\mathfrak{u}(n)$ denote the real vector space consisting of all $n\times n$-matrices with complex entries that are skew-hermitian.

Exercise 1: Prove that $\mathfrak{u}(n)$ is a real Lie algebra but that it is not even a complex vector space and thus cannot be a complex Lie algebra.

Exercise 2: Prove that $\mathfrak{u}(n)$ is the Lie algebra of $U(n)$.

Exercise 3: Prove that the dimension of the real Lie algebra $\mathfrak{u}(n)$ is $\frac{n(n+1)}{2}$. In particular, the real Lie algebra $\mathfrak{u}(n)$ is even-dimensional if and only if $n\equiv 0,3 \pmod 4$.

(4) No. A Lie group $G$ has a unique Lie algebra by definition. Edit: Jim Conant (rightly) pointed out below in the comments that one can associate more than one Lie algebra to a Lie group. In addition to the "standard Lie algebra of a Lie group", one can associate the Lie algebra consisting of all smooth vector fields on the Lie group $G$. In fact, this construction is valid even when $G$ is not necessarily a Lie group but only a smooth manifold.

Exercise 4: Prove that the Lie algebra consisting of all smooth vector fields on the smooth manifold $M$ is equivalent to the Lie algebra consisting of all derivations $C^{\infty}(M)\to C^{\infty}(M)$.

Exercise 5: Prove that the Lie algebra of a Lie group $G$ is a subalgebra of the Lie algebra of $G$ consisting of all smooth vector fields on $G$.

I hope this helps! I appreciate that some of my explanations are probably not as complete as is required for someone asking these questions. Please feel free to ask for further explanation if you wish.

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Regarding (4), the OP didn't specify how to associate a Lie algebra to a Lie group, and there are more Lie algebras you can construct from a Lie group. For instance, you can take the infinite-dimensional Lie algebra of all smooth vector fields on the Lie group. –  Grumpy Parsnip Jul 9 '12 at 12:40
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@JimConant Thank you very much; of course, you are correct. I think this slipped my mind when I answered the question. I have added this point to my answer above (with due acknowledgement) and I have also included two relevant exercises. –  Amitesh Datta Jul 9 '12 at 13:54
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Regarding 3, are you claiming that $\mathfrak{u}(n)$ is not naturally equipped with a complex structure or that it cannot be equipped with any complex structure whatsoever? (Clearly it is not a complex subspace of $M_n(\mathbb{C})$ but it doesn't follow that it can't have some more exotic complex structure.) –  Qiaochu Yuan Jul 9 '12 at 14:49
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@QiaochuYuan Thank you for your comment; you are correct, of course, that the language I used was a little unfortunate as the intended meaning is ambiguous. I am claiming that $\mathfrak{u}(n)$ is not a complex subspace of $M_{n}(\mathbb{C})$. Also, I believe that a real Lie algebra $\mathfrak{g}$ possesses the structure of a complex Lie algebra if and only if there exists a map $T:\mathfrak{g}\to\mathfrak{g}$ such that $[T(X),Y]=T([X,Y])$ for all $X,Y\in \mathfrak{g}$ and $T^2=-\text{Id}_{\mathfrak{g}}$. –  Amitesh Datta Jul 10 '12 at 3:18
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@Qiaochu If $\mathfrak{u}(n)$ possesses a complex structure, then $n\equiv 0\pmod 4$. Indeed, if $\mathfrak{c}(n)$ is the Cartan subalgebra of $\mathfrak{u}(n)$ consisting of all diagonal matrices in $\mathfrak{u}(n)$, then $\mathfrak{c}(n)$ is fixed by $T:\mathfrak{u}(n)\to \mathfrak{u}(n)$. In particular, $\mathfrak{c}(n)$ possesses a complex structure. We conclude that $n$ is even. Of course, the dimension of $\mathfrak{u}(n)$ is $\frac{n(n+1)}{2}$ which must also be even if $\mathfrak{u}(n)$ possesses a complex structure. Therefore, $n\equiv 0\pmod 4$. –  Amitesh Datta Jul 12 '12 at 3:38
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