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I am trying to understand the notion (and notation) of the Lie derivative on a general manifold by trying to convert the notation the concrete example of the Lie group O(n).

Let $X,Y$ be smooth vector fields on a smooth manifold $M$, $p \in M$ and the local flow $\psi_t: U \rightarrow M$ of X in a neighborhood $U$ of $p$. Then the $\textit{Lie derivative}$ and thus the Lie bracket is defined as:

$$ [X,Y]_p := \mathcal{L}_X(Y_p) = \lim_{t\rightarrow 0} \frac{(d\psi)_{-t}\; Y_{\psi_t(p)} - Y_p}{t}$$

In words: one uses the pullback of the vector field $Y$ along the flow of $X$ to define this derivative.

Now, for $M=O(n)$, with associated Lie algebra $\mathfrak{o}(n)=\{ X \in M(n,\mathbb{R}) : X = -X^T \}$, the Lie bracket for $A,B \in \mathfrak{o}(n)$ can be written as:

$$[A,B] = \lim_{t\rightarrow 0} \frac{\gamma(t) - B}{t}, $$

where the curve $\gamma$ is given by

$$ \gamma(t) = \exp(tA)B\exp(-tA). $$

Unfortunately, I got stuck in defining the abstract notation appropriately to arrive at the latter expression.

Does someone has an idea how to define $\psi$ and the vector fields in this situation? Or, does someone know a better $\textit{concrete}$ example where one can well understand the Lie derivative?

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Vector fields are equivalent to derivations (en.wikipedia.org/wiki/Derivation_(abstract_algebra)) on the algebra $C^{\infty}(M)$ of smooth functions, and then the Lie derivative is just the commutator of the corresponding derivations. –  Qiaochu Yuan Aug 8 '12 at 21:13

1 Answer 1

The Lie algebra is the tangent space of the Lie group at the identity. The exponential map maps the vector fields ($A$, $B$) in the tangent space to the local flow $\exp(tA)$ generated by them on the Lie group. In this case, the explicit form of the exponential map is written as the generalized power series expansion of the exponential function for matrices.

The unit circle $U(1)$ is a Lie group with the imaginary axis as the Lie algebra, and the exponential map takes the imaginary numbers to the unit circle. The Lie bracket $[ia, ib]$ commutes, and we can see it in $\exp(ita) ib \exp(-ita) - ib = 0$.

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