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I am very well aware of how you compute Lie derivatives or tensor fields and just vector fields via the Lie bracket, but what confuses me is when there are addition operation involved in say the tensor field or the vector fields for example, how would you compute something like this?

$$\mathcal L_X h=L_X \bigg(\frac{x}{x+1}dx\otimes dx+y^2 dy\otimes dy + \sin z dz\otimes dz\bigg)$$

Where the tensor $h$ is a $2-$tensor defined on some manifold $N =\mathbb{R}\times \mathbb{R}^2: x\in\mathbb{R}$ and $(y,z) \in \mathbb{R}^2.$ I would appreciate a clear example so that I can compute Lie derivatives in a better way.

Michael Albanese
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Aurora Borealis
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1 Answers1

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The Lie derivative is a derivation of the algebra of tensor fields. That means

  • $\mathcal{L}_X(cT) = c\mathcal{L}_XT$ for all $c \in \mathbb{R}$,
  • $\mathcal{L}_X(T_1 + T_2) = \mathcal{L}_X(T_1) + \mathcal{L}_X(T_2)$, and
  • $\mathcal{L}_X(T_1\otimes T_2) = \mathcal{L}_X(T_1)\otimes T_2 + T_1\otimes\mathcal{L}_X(T_2)$.

By the second dot point, we have

\begin{align*} \mathcal L_X h &= \mathcal{L}_X \left(\frac{x}{x+1}\,dx\otimes dx+y^2\, dy\otimes dy + \sin z\, dz\otimes dz\right)\\ &= \mathcal{L}_X \left(\frac{x}{x+1}\,dx\otimes dx\right) +\mathcal{L}_X(y^2\, dy\otimes dy) + \mathcal{L}_X(\sin z\, dz\otimes dz). \end{align*}

Now each of these three terms can be further expanded using the third dot point. I will only do the first one.

$$\mathcal{L}_X\left(\frac{x}{x+1} dx\otimes dx\right) = \mathcal{L}_X\left(\frac{x}{x+1}\right)dx\otimes dx + \frac{x}{x+1}\mathcal{L}_X(dx)\otimes dx + \frac{x}{x+1}dx\otimes\mathcal{L}_X(dx).$$

It follows from Cartan's magic formula that for any differential form $\omega$, we have $\mathcal{L}_X(d\omega) = d(\mathcal{L}_X\omega)$. So now the computation reduces to calculating the Lie derivative of a function $f$ which is given by $\mathcal{L}_X(f) = X(f)$.

If $X = X^x\partial_x + X^y\partial_y + X^z\partial_z$, then

$$\mathcal{L}_X\left(\frac{x}{x+1}\right) = X\left(\frac{x}{x+1}\right) = X^x\partial_x\left(\frac{x}{x+1}\right) = \frac{X^x}{(x+1)^2}$$

and

$$\mathcal{L}_X(dx) = d(\mathcal{L}_X x) = d(X(x)) = d(X^x\partial_xx) = dX^x$$

so

\begin{align*} \mathcal{L}_X\left(\frac{x}{x+1} dx\otimes dx\right) &= \mathcal{L}_X\left(\frac{x}{x+1}\right)dx\otimes dx + \frac{x}{x+1}\mathcal{L}_X(dx)\otimes dx + \frac{x}{x+1}dx\otimes\mathcal{L}_X(dx)\\ &= \frac{X^x}{(x+1)^2}\, dx\otimes dx + \frac{x}{x+1}\, dX^x\otimes dx + \frac{x}{x+1}\, dx\otimes dX^x. \end{align*}

Michael Albanese
  • 99,526
  • Thank you very much, now I have one question. Why is: $$\mathcal{L}_X\left(\frac{x}{x+1} dx\otimes dx\right) = \mathcal{L}_X\left(\frac{x}{x+1}\right)dx\otimes dx + \frac{x}{x+1}\mathcal{L}_X(dx)\otimes dx + \frac{x} {x+1}dx\otimes\mathcal{L}_X(dx).$$ When using the third point there are only two terms on the RHS, but here we have 3 terms? – Aurora Borealis Apr 14 '18 at 12:26
  • I dont see how the term $\frac{x}{x+1}\mathcal L_{X}(dx)\otimes dx $ arises. The remaining two terms seem clear from the third point, but the centre term seems to confuse me. – Aurora Borealis Apr 14 '18 at 12:33
  • There are three terms because functions are $0$-tensors and $\frac{x}{x+1}dx\otimes dx = \frac{x}{x+1}\otimes dx\otimes dx$. – Michael Albanese Apr 14 '18 at 12:45
  • Oo ok I get the algebra after expanding we get three terms, but why are we allowed to write$$ \frac{x}{x+1} dx\otimes dx = \frac{x}{x+1}\otimes dx \otimes dx?$$ – Aurora Borealis Apr 14 '18 at 12:49