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I noticed that the characterizations of the Lie algebras of matrix Lie groups can be obtained by differentiation. For example:

$$O(n) = XX^t = \mathbb{1} \implies \mathfrak{o}(n) = X + X^t = \mathbb{0}$$

$$SO(n) = XX^t = \mathbb{1},\; \text{det}(X) = 1 \implies \mathfrak{so}(n) = X + X^t = \mathbb{0},\; tr(X)=0$$

but it works also for $U(n), SU(n), Sl(n,\mathbb{K})$.

Does this work for a general Lie group?

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Yes I know this, but I was wondering if there exists a theorem stating this in a formal way –  Abramo Jun 4 '12 at 16:27
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I guess you'd have to be more specific about what you want. I admit that this is a hard thing to do, at times! –  Dylan Moreland Jun 4 '12 at 16:30
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I would like to know if there exist a theorem saying: Given a Lie group described by a set of equations, the Lie algebra is described by taking the derivatives of those equation –  Abramo Jun 4 '12 at 17:13
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Maybe you're looking for the idea that the Lie algebra of $G$ is the tangent space at the identity. It's discussed here: en.wikipedia.org/wiki/… –  mt_ Jun 4 '12 at 17:17

2 Answers 2

up vote 6 down vote accepted

The matrix groups can be defined as regular level sets of functions $f: GL(n) \to M$ where $M = \mathbb R$ (e.g. for $SL(n)$) or $M = n\times n$ matrices (e.g. for $O(n)$). In general, if you have a map $f: M \to N$ and $y \in N$ is a regular value then $f^{-1}(y)$ is a submanifold and $T_xf^{-1}(y) = \ker df_x : T_x M \to T_y N$. So in the case of $O(n)$ for example you're differentiating the function $X \mapsto XX^t$ at the identity to see that its Lie algebra (identifiable with $T_e O(n)$ is skew-symmetric matrices.

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I just got to the same solution by myself, just before reading your answer. In any case thanks! –  Abramo Jun 4 '12 at 18:37
    
Great answer Eric! I know this is a bit late of a comment but couldn't we do the same thing for the set of upper triangular matrices $B\in GL(n)$? What would be the map $f$ in this case? Could matrix groups be defined as regular level sets of functions only in the real setting or is this true in the complex setting as well? –  math-visitor Jun 30 '12 at 8:56

A Lie group is also a differentiable manifold; in particular we can define its tangent space at the identity. Intuitively, the tangent space is the set of directions $v$ such that if you start at the identity in $G$ and move infinitesimally in direction $v$, you stay in $G$.

Let's say that $G$ is defined by the vanishing of some differentiable function $f$: so $G = \{ X: f(X)=0\}$. The tangent space consists of matrices $M$ such that $f(I + \epsilon M) =0$ to first order in $\epsilon$: using the Taylor expansion and neglecting higher terms, what we want is that $df_I(M)=0$ where $df_I$ is the total derivative at the identity. Elements of the kernel of the total derivative are exactly those vectors which can be realised as $\gamma'(0)$ for some differentiable $\gamma: (-1,1)\to G$: the usual definition of tangent space is the equivalence classes of such functions $\gamma$ under the relation $\gamma_1\sim \gamma_2$ iff $\gamma_1'(0)=\gamma_2'(0)$.

The Lie algebra associated to $G$ has as its underlying vector space the tangent space at the identity, so it really is obtained by differentiating and setting the total derivative equal to zero.

Looking at your $O(n)$ example, the "$f$" is $f(X)=XX^t-I$. To find the total derivative at the identity, we ought to have $f(I+\epsilon M) = I + \epsilon df_I(M) +$ higher order terms. To order one in epsilon we have $f(I+\epsilon M)= \epsilon (M + M^t)$, so the total derivative at the identity is $M+M^t$. The vanishing of this is exactly the condition you gave for being in the Lie algebra of $O(n)$.

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