How do I make sense of terms $X^j\partial_j(Y^i)$ in the Lie bracket of vector fields?

Question

I am trying to understand the definition of the Lie bracket of vector fields. According to two presumably independent sources (Amari/Nagaoka 2000 and wikipedia), the Lie bracket of two vector fields $X$ and $Y$ is

$$ \begin{array}{rcl} [X,Y]&=&\sum_{i=1}^{n}\left(X(Y^{i})-Y(X^{i})\right)\partial_{i}\\ &=&\sum_{i=1}^{n}\sum_{j=1}^{n}\left(X^{j}\partial_{j}(Y^{i})-Y^{j}\partial_{j}(X^{i})\right)\partial_{i} \end{array} $$

where $\partial_{i}$ is the natural basis $\frac{\partial}{\partial\xi^{i}}$ for the tangent space at point $p$, and $X^{i}$ as well as $Y^{i}$ are the corresponding coordinates, i.e.

$$ X=\sum_{i=1}^{n}X^{i}\partial_{i} $$

$$ Y=\sum_{i=1}^{n}Y^{i}\partial_{i} $$

Here is what I do not understand. There is probably a simple answer for this. $[X,Y]$ is a vector field, so $\left(X(Y^{i})-Y(X^{i})\right)$ are the coordinates corresponding to a tangent space at point $p$ of the manifold. These are supposed to be real numbers. $X$ and $Y$ are vectors in the tangent space at point $p$; $X^{i}$ and $Y^{i}$ are real numbers. How am I supposed to read an expression of the form $X(Y^{i})$, a vector times a real number? The corresponding problem in the expansion is $X^{j}\partial_{j}(Y^{i})$ -- I am not sure how to read this expression. $X^{j}\partial_{j}$ makes perfect sense, it's a vector in the tangent space. But how do you multiply it by a real number? Here is the wikipedia link:

Lie bracket of vector fields

Answer 1

For a vector field $X$ and a function $f$, it is custom to let $X(f)$ denote the derivative of $f$ in the direction $X$.

Answer 2

this expression is consequence of direct computation: Given a local coordinates say $\{v_{i}\}|_{i=1}^{n}$. Then the vector field $X,Y$ can be locally expressed w.r.t coordinates basis : $X=X^{i}\frac{\partial}{\partial v^{i}},Y=Y^{i}\frac{\partial}{\partial v^{i}}$. Then, by definition,$[X,Y]$ is a vector field and can be expressed wrt local coordinates: $$[X,Y](f)=XY(f)-YX(f)=(X^{i}\frac{\partial}{\partial v^{i}}) (Y^{j}\frac{\partial f}{\partial v^{j}})-(Y^{i}\frac{\partial }{\partial v^{i}}) (X^{j}\frac{\partial f}{\partial v^{j}})=X^{i}(Y^{j}\frac{\partial }{\partial v^{i}} \frac{\partial f}{\partial v^{j}}+\frac{\partial Y^{j}}{\partial v^{i}}\frac{\partial f}{\partial v^{j}}) -Y^{i}(X^{j}\frac{\partial }{\partial v^{i}} \frac{\partial f}{\partial v^{j}}+\frac{\partial X^{j}}{\partial v^{i}}\frac{\partial f}{\partial v^{j}})=(X^{i}\frac{\partial Y^{j}}{\partial v^{i}}-Y^{i}\frac{\partial X^{j}}{\partial v^{i}})\frac{\partial }{\partial v^{j}}(f)$$ Therefore, we can represent $[X,Y]=(X^{i}\frac{\partial Y^{j}}{\partial v^{i}}-Y^{i}\frac{\partial X^{j}}{\partial v^{i}})\frac{\partial }{\partial v^{j}}$

Answer 3

Sometimes it is helpful to see vector fields as operators acting on functions . The action is that of a partial derivative.

With a vector field $V = V^i \partial_i$ and a function $f$:

$V(f) = V^i \frac{\partial f}{\partial_i}$

The components $V^i$ of the vector field are functions as well, so they, too, can be operated upon by another field:

$X(Y) = \left(X^j \frac{\partial Y^i}{\partial_j}\right)\partial_i$

The result is a vector field as well with components $X(Y)^i = X^j \frac{\partial Y^i}{\partial_j}$. Being a vector field, $X(Y)$ could again act on yet another field.

Take, for example, the coordinate basis fields for polar coordinates:

$X = \partial_r = x/r \partial_x + y / r \partial_y\\ Y = \partial_\phi = -y\partial_x + x \partial_y$

The components of the fields are function of $x$ and $y$ and this is what $X$ acting on $Y$ yields:

$X(Y) = \left(\frac{x}{r}\frac{\partial(-y)}{\partial_x} + \frac{y}{r}\frac{\partial(-y)}{\partial_y}\right)\partial_x + \left(\frac{x}{r}\frac{\partial x}{\partial_x} + \frac{y}{r}\frac{\partial x}{\partial_y}\right)\partial_y = -\frac{y}{r}\partial_x + \frac{x}{r}\partial_y = \frac{1}{r}\partial_\phi$

The vector field $X(Y)$ answers the question: How does the field $Y$ change in direction of $X$?

For example, an azimuthal basis vector of polar coordinates at point $P$ changes in the direction of the local radial basis vector at $P$: Its direction is unchanged but its length changes: $X(Y) = \partial_r(\partial_\phi) = 1/r\partial_\phi$. This becomes especially clear for points $P$ on the $x$ axis: $\partial_r$ points to $+x$ direction, $\partial_\phi$ points to $+y$ direction. The vectors of $Y$ get longer farther away from the origin but don't change direction. This is just what $X(Y)$ tells you. Note that $X(Y)$ tells you what gets added to $Y$.

Hope that helps!

Answer 4

There is an isomorphism between vector fields $X\in\Gamma(TM)$ defined as smooth sections, i.e. $M\to TM$, and derivations $\mathrm{Der}(C^\infty(M))$. We can write this isomorphism as $X\mapsto \nabla_X$, where $\nabla_X:C^\infty(M)\to C^\infty(M)$ is such that $$(f(p), \nabla_X f(p)) = \mathrm df\circ X(p),$$ where here $f:M\to\mathbb R$ and thus $\mathrm df:TM\to T\mathbb R\simeq \mathbb R\times\mathbb R$.

Now, when we write the expression for $X$ in coordinates, $X=X^i \partial_i$, we can interpret this expression in two different ways:

1. as a product of functions $X^i:p\mapsto X^i(p)$ and vector fields $\partial_i$, defined so that $$\partial_i|_p\equiv \partial_i(p)\in \{p\}\times T_p M.$$
2. as a product of maps $X^i$ and derivations $\partial_i\in\mathrm{Der}(C^\infty(M))$ such that $$\partial_i f: p \mapsto \partial_i f(p)\in\mathbb R.$$

The second meaning is what people mean when writing things such as $X(f)$: $$X(f) \equiv \eta_2\circ\mathrm df\circ X \equiv X^i \partial_i f: M\ni p\mapsto X^i(p)\partial_i f(p) \in\mathbb R,$$ where $\eta_2:T\mathbb R\to\mathbb R$ maps pairs to their second elements: $\eta_2((p,v))\equiv v$.

Things such as $Y(X(f))$ are defined similarly: $$Y(X(f)) = Y^i \partial_i (X^j \partial_jf) = Y^i (\partial_i X^j)\partial_j f+Y^i X^j \partial_i\partial_j f.$$