Problem: How to prove the vector field identity $[v,fw]=(L_vf)w+f[v,w]$?

  • Thread starter Thread starter Chris L T521
  • Start date Start date
Click For Summary
The problem involves proving the vector field identity \([v, fw] = (L_v f)w + f[v, w]\) for smooth vector fields \(v\) and \(w\) on a smooth manifold \(M\), with \(f\) as a smooth function. The solution demonstrates the identity by expanding the left-hand side using the properties of vector fields and the Lie derivative. It shows that the expression simplifies to the right-hand side through careful manipulation of derivatives and applying the definition of the Lie derivative. The proof confirms the validity of the identity, highlighting the relationship between the Lie bracket and the Lie derivative. This identity is significant in differential geometry and the study of vector fields.
Chris L T521
Gold Member
MHB
Messages
913
Reaction score
0
Here's this week's problem!

-----

Problem
: Let $v$ and $w$ be smooth vector fields on a smooth manifold $M$ and let $f$ be a smooth function. Prove that\[[v,fw]=(L_vf)w+f[v,w].\]

-----

 
Physics news on Phys.org
No one answered this week's problem. You can find my solution below.

[sp]For $v,w$ smooth vector fields on $M$ and $f$ a smooth function, we have
\[\begin{aligned}{[v,fw]} &= (v(fw)^i-fwv^i)\frac{\partial}{\partial x^i}\\ &= \left(\sum v_j\frac{\partial}{\partial x^j}(fw)^i-fwv^i\right)\frac{\partial}{\partial x^i}\\ &= \left(\sum v_j\left(\frac{\partial f}{\partial x^j}w^i + f\frac{\partial w^i}{\partial x^j}\right)-fwv^i\right)\frac{\partial}{\partial x^i}\\ &= \left(L_v fw^i +f\sum v_j\frac{\partial w^i}{\partial x^j}-fwv^i\right)\frac{\partial}{\partial x^i}\\ &=L_v f w^i\frac{\partial}{\partial x^i}+(fvw^i-fwv^i)\frac{\partial}{\partial x^i}\\ &= (L_v f)w+f[v,w].\end{aligned}\][/sp]
 

Similar threads

Replies
15
Views
2K
Replies
1
Views
1K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 1 ·
Replies
1
Views
4K
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K