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A tangent space $T_{m,M}$ is defined as the set of all linear derivations at a point $m$ on a manifold $M$. Linear derivations are operators that satisfy the Leibniz rule, i.e. $f,g \in F_{m,M}, O(f,g) = f*O(g)+g*O(f)$ where $F_{m,M}$ is the space of smooth functions defined on a subset of $M$ defined on a subset including $m$. Operators are defined to map from $F_{m,M} \to \mathbb{R}$

The coordinate basis $\{t_i\}$ of tangent vectors are defined as $t_i(f) = \frac{\partial f}{\partial x^i}$ where the $x^i$ are coordinate functions belonging to some chart of $M$.

I read that the coordinate basis apparently spans the space of all linear derivations at m. It's obvious that derivatives belong to the space, but how can you show that the partial derivatives span the space of linear derivations?

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Derivations by definition are linear operators, and as you pointed out, they are maps from $F_{m,M}$ to the reals.

Since you are in the neighborhood of a point $m$ labeled by local coordinates $x_i$, each basis vector in the tangent plane corresponds to a local coordinate. Each of which make a tangent with the surface at that point (by definition), therefore the tangent plane (space) consists of all linear combinations of the natural basis in $T_{m,M}$.

Does that answer your question?

I mean that to show they span the linear space is to claim they form a complete basis. If I only have N coordinates, I need at most N basis vectors. These vectors need also to be linearly independent, but that's trivially true because they are associated with independent coordinates. It's sort of a ''by definition'' proof.

Credit for answering this goes to a friend that I recently asked about this post, I have asked if he wants to be named.

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This is half of the content of Proposition 3.2 in Lee's Introduction to Smooth Manifolds. In summary, and using the notation of the question here:

Hint The given definition of derivation at local, so by choosing smooth coordinates of the given manifold $M$ centered at the given point $m \in M$, we need only prove the claim for $M = \Bbb R^n$ and $m = {\bf 0}$.

For a derivation $X$ on $M$ at ${\bf 0}$, set $v^i := X(x^i)$, where $x^i$ is the usual $i$th coordinate function. By applying Taylor's Theorem (with remainder) to expand an arbitrary smooth function $f$ on a neighborhood of ${\bf 0}$, show that $X = v^i t_i$, which in particular implies $X \in \operatorname{span}\{t_i\}$.

Fix a smooth, real-valued function $f$ on a neighborhood of ${\bf 0}$. By Taylor's Formula, there are smooth functions $g_i$ on a neighborhood of $m$ such that $g_i({\bf 0}) = 0$ and $$f(x) = f(m) + \sum_{i = 1}^n t_i(f) x^i + \sum_{i = 1}^n g_i(x) x^i .$$ Each summand of the last term is a product of functions that vanish at ${\bf 0}$, so applying $X$ to $f$ gives $$X(f) = X\left(\sum_{i = 1}^n t_i(f) x^i\right) = \sum_{i = 1}^n t_i(f) X(x^i) = \sum_{i = 1}^n v^i t_i(f) .$$

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  • $\begingroup$ Could you elaborate on what g_i(x) is? It's not very clear to me. Also, I can't really follow the last line. It's not clear to me how X distributes the way you show in the second equality, and it's not clear why the other terms of f(x) in the expansion are irrelevant. Perhaps I just need to get around to reading that book for the context, but at least for posterity, can you please clarify further? $\endgroup$ – Mr. HelloBye Nov 12 '18 at 22:40
  • $\begingroup$ Here the $g_i$ are just smooth functions whose existence is guaranteed by Taylor's Theorem. Forst the constant term, since $x \mapsto f(m)$ is constant, we have $X(x \mapsto f(m)) = 0$. For the "quadratic" term, $\sum_i g_i(x) x^i$, by linearity and the definition of derivation, we get $$X\left(\sum_i g_i(x) x^i\right) = \sum_i X(g_i(x) x^i) = \sum_i [X(g_i(x)) (0) + (0) X(x^i)] = 0$$. $\endgroup$ – Travis Nov 12 '18 at 23:23
  • $\begingroup$ Do you mean that the $g_i(x)$ are the Peano form of the remainder? $\endgroup$ – Mr. HelloBye Nov 13 '18 at 1:47

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