---- Let $M$ be a [[smooth manifold]] and $TM$ its [[tangent bundle]]. > [!definition] Definition. ([[vector field]]) > A **vector field on $M$** is a [[section]] of the natural projection $\pi:TM \to M$, i.e., a [[continuous]] map $X:M \to TM$, usually written $p \mapsto X_{p}$, with the property that $X_{p} \in T_{p}M$ for all $p \in M$. > A **rough vector field** on $M$ is a [[vector field]] on $M$ is map $X:M \to TM$ that satisfies the above property except it might not be [[continuous]]. > With $\big( U, (x^{i}) \big)$ a smooth [[coordinate chart]] for $M$, we write the value of $X$ at any point $p \in U$ in terms of the [[tangent space at a point of a smooth manifold|coordinate basis]] $X_{p}=X^{i}(p) \ \frac{ \partial }{ \partial x^{i} } |_{p}.$ One has that $X$ is a **smooth vector field** iff $X^{i}$ is smooth for all $i$. > > > The collection of vector fields on $M$ is variously denoted $\mathscr{V}(M)$, $\mathscr{X}(M)$, $\Gamma(TM)$, $\Omega^{0}_{M}(TM)$. It carries two structures: > - That of an (infinite-dimensional) $\mathbb{R}$-[[Lie algebra]] (see below); > - That of a $C^{\infty}(M)$-[[module]], [[the space of differential 1-forms is dual to that of vector fields|which is]] [[natural transformation|naturally]] [[dual vector space|dual]] to the space $\Omega^{1}(M)$ of [[differential form|1-forms]]. Note that the two are not *a priori* [[isomorphism|isomoprhic]] in the absence of additional structure (e.g. a [[metric tensor|metric]]) on $M$. > > Note that these two structures are not compatible, e.g. the Lie bracket of vector fields on $M$ is not $C^{\infty}(M)$-[[bilinear map|bilinear]]. > > [!equivalence] Vector fields as derivations of $C^{\infty}(M)$. > Recall that a tangent vector $X_{p} \in T_{p}M$ at may be [[tangent vector to a smooth manifold|interpreted]] as a [[derivation]] at $p$, $X_{p}:C^{\infty}(M) \to \mathbb{R}$. > Explicitly, if $X_{p}$ looks in coordinates like $X_{p}=X^{i}_{p} \frac{ \partial }{ \partial x^{i} } |_{p}$ for $X_{p}^{i} \in \mathbb{R}$, then one has $X_{p}(f)=(D_{X_{p}}f) |_{p}, \text{ where } (D_{X_{p}} f )|_{p}:= X_{p}^{i} \frac{ \partial f }{ \partial x^{i} } (p).$ Letting $p$ smoothly range over $M$ defines a smooth [[vector field]] $X:M \to TM$. Correspondingly, the [[vector field]] $X$ defines a [[derivation]] $C^{\infty}(M) \to C^{\infty}(M)$, via[^1] $X(f):= (D_{X_{(\cdot)}} f )|_{(\cdot)}.$ > In coordinate-free language, $Xf$ is the [[the space of differential 1-forms is dual to that of vector fields|evaluation]] $Xf:=(df)(X)$. ^equivalence > [!definition] Definition. (Lie bracket of vector fields) > We can then make sense of a **Lie bracket of vector fields** $[X,Y]$ by considering $X,Y$ as derivations $C^{\infty}(M)\to C^{\infty}(M)$ and taking their [[commutator]] $X \circ Y - Y \circ X$. Concretely, this means $[X,Y](f)=X \big( Y(f) \big)-Y\big( X(f) \big) : M \to \mathbb{R}.$ > for all $f \in C^{\infty}(M)$.[^3] This formula is of little usefulness for computations, however, because it requires one to compute terms involving second derivatives of $f:M \to \mathbb{R}$ which always cancel out. The below coordinate formula for the [[Lie algebra|Lie bracket]] already accounts for such cancellations: > > $[X,Y]= \left( X^{i}\frac{ \partial Y^{j} }{ \partial x^{i} } - Y^{i} \frac{ \partial X^{j} }{ \partial x^{i} } \right) \frac{ \partial }{ \partial x^{j} } $ > or more concisely:[^4] $[X,Y]=(XY^{j}-YX^{j}) \frac{ \partial }{ \partial x^{j} }. $ > > [!basicproperties] Properties of the Lie bracket of vector fields. > - It is, in fact, a [[Lie algebra|Lie bracket]]. > - For $f,g \in C^{\infty}(M)$, $[fX, gY]=fg[X,Y] + (fXg)Y - (gYf)X.$ > - (Naturality) If $X_{1}$ is $(F:M \to N)$-related to $Y_{1}$ and $X_{2}$ to $Y_{2}$, then $[X_{1}, X_{2}]$ is $F$-related to $[Y_{1}, Y_{2}]$. > - (Pushforward) As a special case of the above: if $F$ is a [[diffeomorphism]], $Y_{i}=F_{*}X_{i}=(dF)X_{i}$, then $F_{*}[X_{1}, X_{2}]=[F_{*}X_{1}, F_{*}X_{2}]$ ^properties > [!basicexample] > Because the standard coordinate vector fields on $\mathbb{R}^{n}$ [[equality of mixed partials for C2 functions|satisfy]] $[\partial_{i}, \partial_{j}]=0,$ the coordinate vector fields in a chart $(U, \varphi)$ on $M$ satisfy $[\partial_{i}, \partial_{j}]=0$. ^basic-example - [ ] there is some more stuff going on here with Lie algebra and the like, need to do all that (later: [[Lie algebra of a Lie group|left-invariant vector field]] covers that?) ---- #### [^1]: The Leibniz rule $X(fg)=X(f)g + fX(g) \text{ for all }f,g \in C^{\infty}(M)$ is satisfied because it is satisfied at each $p \in M$ (because $X_{p}$ is a derivation at $p$). [^2]: The 'interior product' extends this duality to arbitrary differential forms, but this is not covered in our course. [^3]: Crucially, $X \circ Y$ and $Y \circ X$ themselves need not be [[vector field|vector fields]] (they need not be [[derivation|derivations]], cf. Lee pg. 186). The key fact is that their [[commutator]] *is* a [[derivation]] (Lee has the easy computation). [^4]: Recall that $XY^{j}$ is to be interpreted as the [[vector field]] $X$ acting on $Y^{j} \in C^{\infty}(U)$. ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` Given a map $\hat{\theta}:\mathscr{V}(M) \to C^{\infty}(M)$, define a differential 1-form $\begin{align} \theta:M & \to T^{*}M \\ p & \mapsto (X_{p} \mapsto \hat{\theta}(X)(p)) \end{align}$