# Understanding the definition of Lie derivative

I'm reading Warner. "Foundations of Differentiable Manifolds and Lie Groups." In p. 69, it gives the definition of the Lie derivative as follows:

2.24 Definition (summerized) Fix a smooth vector field $$X$$ on a differentiable manifold $$M$$. Let $$Y$$ be another smooth vector field on $$M$$. The Lie derivative of $$Y$$ with respect to $$X$$ at $$m$$, which is denoted by $$(L_XY)_m$$, is $$(L_XY)_m = \lim_{t\to 0} \frac{dX_{-t}(Y_{X_t(m)}) - Y_m} {t}$$.

Here, $$X_t(m) = \gamma_m(t)$$, where $$\gamma_m$$ is an integral curve of $$X$$ such that $$0$$ is in the domain of $$\gamma_m$$ and $$\gamma_m(0) = m$$. Also, $$Y_m = Y(m)$$.

I see that the function $$t\mapsto\frac{dX_{-t}(Y_{X_t(m)}) - Y_m} {t}$$ is a function from the subset of $$\mathbb R$$ into $$M_m$$, the tangent space to $$M$$ at $$m$$. But to define a limit of such function, there should be a topology on $$M_m$$, right? But I haven't learned any inner product, norm, metric, or topology given on $$M_m$$. So what is the topology on $$M_m$$?

• Is $M_m$ the tangent space to $M$ at the point $m$? So a finite-dimensional vector space? – Lord Shark the Unknown Feb 17 at 10:50
• @Lord Yes, $M_m$ is the tangent space to $M$ at $m$, so $\operatorname{dim}(M_m)$ = $\operatorname{dim}(M)$ so that it is finite. – zxcv Feb 17 at 10:56
• You know that on a finite-dimensional real vector space, there is only one sensible topology.... – Lord Shark the Unknown Feb 17 at 11:03