In Differential Geometry $\frac{\partial}{\partial x}$ is a vector field. Specifically, it is the coordinate induced basis vector field for the total space of the tangent bundle TM. The definition I know is:

let $(M,\mathcal{O},\mathcal{A})$ be a smooth manifold and $(U,x)$ be a chart $$ \left( \frac{\partial f}{\partial x^i} \right)_p = \partial_i \left( f \circ x^{-1} \right) (x(p)) $$ where the smooth function $f: M \to \mathbb{R} $ is a smooth function at $p \in U \subset M$. Now in the domain of the chart $U$ I can decompose a vector field $X = X^i \frac{\partial}{\partial x^i}$ in its components.

Now the discussion here got me to think about the basis vectors. Basically I want to ask wether or not it is important to distinguish between $\frac{\partial}{\partial x^i}$ in the above notation and on $\mathbb{R}^d$ (the undergrad notion of partial deriative). And where this would be relevant.

If I apply a vector field $X$ to another one $Y$ and write its components, I have $$ (XY)^i = X^m \left( \frac{\partial Y^i}{\partial x^m} \right) $$ and suddenly the basis vector field notion turned into the partial derivative of the component functions, which are in a subset of $\mathbb{R}^d$. As far as I know, a vector can act on a function from $M \to \mathbb{R}$ and not on functions in $\mathbb{R}^d$. Where do I have to draw the line, or DO I have to draw a line between these 2 notions? For the time being I just use them dependent on the argument. If the argument is on the manifold I expand the formula above, otherwise I use the undergrad notion.

  • $\begingroup$ Thinking about the domain of the components of a vector field I think I got an idea: Is it true, that the 'partial derivative' is the same, since the coordinates are functions from $M \to \mathbb{R}^d$, and all the difference is that on coordinates we have the multivariable partial diff., where in the vector basis case it is the abstract notion of a vector? Basically, I would have to never bother making a difference? $\endgroup$ – mike Apr 6 '16 at 19:53

Actually it is quite simple: One simply has to observe which kind of map the components of a vector are. A vector field $X$ can be written in compenents w.r.t. the chart $x$ as $$ X = X^i \left( \frac{\partial}{\partial x^i} \right) $$ where the "partial derivative" symbol denotes the basis vectors. Investigating this expression we note, that the components are real numbers for each point on the manifold (or one point if X is a single vector instead of a field), thus $$ X^i : M \to \mathbb{R}$$ and $$\frac{\partial}{\partial x^i} : C^{\infty}(M) \to C^{\infty}(M) .$$

Therefore, the vectorfield is easily identified as element of the total space of the tangent bundle: $$ \frac{\partial}{\partial x^i} \in TM$$ and since the components vary smoothly along the manifold, if and only if the vector field is smooth, we see, that $$ X^i \in C^{\infty}(M).$$

Thus I conclude, that applying the vector to the components is simply applying the vector to a function on the manifold - exactly what they are meant for..

It remains to be answered, if there actually exists a case, where the notions differ, but my initial (and most important for calculations) example turns out to be easily treated.

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